Relativity fails with Magnetic Force

The question is whether this frame-dependent definition of the electric current has anything to do with the actual physics in this context. If the magnetic field is generated by electrons moving in the electric field of ions (as I suggested), then obviously it wouldn't, as the current is given by the relative velocity between the two, which is frame-independent (I mentioned already earlier mechanical friction as a comparable analogy). I at least don't know any examples where a magnetic field would be produced in the presence of electrons (or generally speaking one kind of charge) alone.

It is not irrelevant. The question is whether an orbiting electron would in practice also produce a magnetic field in the absence of the atomic nucleus.

This is basically just casting the cited derivations in a different form. The crucial point of the violation of charge invariance remains (as addressed earlier in detail)

What you derived here is essentially the linear boost of a non-rotating wheel. If you want to derive how, according to the Lorentz transformation, the shape of an initially stationary wheel with radius R looks when it is subject to a rotational + a translational velocity, then you have to do a boost that combines the two. You only did a translational boost, so it is not surprising that your shape is the same as for a non-rotating wheel.
So effectively, you would have to make the velocity v in the gamma factor angle dependent (according to the sum of the two velocities involved).

But the vertical segments should be unaffected by the translational boost, so there is no way that the latter could shift electrons from the vertical into the horizontal sections.

Thomas

 

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The main problem with the variant of electrodynamics you're suggesting is that it can't actually be stated in any simple way. You're relying too much on human judgements about what to call a "wire", "the system", etc. that interacting particles don't know about. At the end of the day, electrodynamics is a theory of the behaviour of moving charges and the electromagnetic field. If you take a classical (ie. non-quantum) point of view, then that means that the dynamical variables you have available to you are the charges, masses, positions, and velocities of the charged particles - ie. \(q_{i} \, , \; m_{i} \, , \; \bar{x}_{i}(t) \, , \; \bar{v}_{i}(t)\). The charge density and flux are just macroscopic summaries of these dynamical variables and are easily defined in terms of them. How would you define the appropriate "current" that should appear in Maxwell's equations, as well as the \(\bar{v}\) in the Lorentz force, purely in terms of these variables characterising N charged particles?

When considering the experimental basis for electrodynamics you should also keep QED in mind. Quantum electrodynamics reduces to classical electrodynamics in the classical realm (at least to within quantum corrections of order \(\hbar\)). The relevant points about QED here are:

• The electron-photon interaction is correctly described by the photon field/electron four-current coupling term which appears in the QED lagrangian - ie. \(\mathcal{L}_{\text{int} \,=\, - \, e \, A_{\mu} \, j^{\mu}\). In the classical limit, the electron four-current multiplied by the electric charge reduces to the (frame-dependent) charge density and flux that appear in Maxwell's equations.
• QED is generally regarded to be one of the most accurate theories in the history of physics.

So what you're suggesting is likely to be a more complicated theory (assuming you actually succeed in formulating it properly, it will violate the superposition principle and suffer from non-locality) which won't play well with Occam's razor, and it will almost certainly conflict with QED.

The reason for this is simply the orders of magnitude involved. If you split the force on a charge into contributions due to the electric and magnetic fields, you'll find that in typical situations the magnitude of the magnetic contribution to the force is of the order of a relativistic correction compared with the electric contribution. This is visible in the calculation I presented in post #3 in this thread, for example. The result is that, under non-relativistic conditions, the electric field completely dominates over the effects of the magnetic field, unless the electric field is extremely weak. So you'll usually only easily observe magnetic effects close to matter that doesn't possess a significant overall net charge.

The point of it is that it's a far more general result applicable to arbitrary charge distributions. In fact it applies just as well to other distributions (eg. mass densities and fluxes).

For finite charge distributions there's no violation. For example, in one spatial dimension the charge distribution associated with a point charge q at rest at the origin is \(\rho(x, \, t) \,=\, q \, \delta(x)\), with \(j = 0\), and applying the transformation I posted just gets you:

\( \begin{align} \rho^{\prime}(x^{\prime}, \, t^{\prime}) \,&=\, q \, \delta( x^{\prime} + v t^{\prime} ) \\ \\ \\ j^{\prime}(x^{\prime}, \, t^{\prime}) \,&=\, - \, q v \, \delta( x^{\prime} + v t^{\prime} ) \end{align} \)​

So every point charge q is mapped onto a point charge q (ie. the same charge) with a different velocity. You've only found a problem where you ignore the fact that the total charge of an infinite wire isn't well defined and try to define a total charge anyway (its undefined because the only way of attributing a total charge is via some sort of limit, and different ways of calculating the limit will get you different results). The bottom line is that Lorentz transformations can't create or destroy charges.

I never assume that the wheel is initially non-rotating. In fact I never even assume we're dealing with a wheel - just that a particle is following a trajectory that satisfies \(x(t)^{2} + y(t)^{2} = R^{2}\) at all times. I explicitly introduced a parameter \(\theta\) which you could manipulate in any way you like to study otherwise arbitrary trajectories constrained on the wheel rim. For example, for uniformly rotating motion, set \(\theta_{\text{p}}(t) \,=\, \omega t \,+\, \theta_{0}\). I didn't need to assume that \(\theta\) was constant in order to obtain the ellipse constraint equation in the boosted frame. After the transformation, the parameter disappears when you square and add both equations, regardless of what you assume about it.

They're affected by the relativity of simultaneity effect. To see this, take a vertical trajectory of the form:

\( \begin{align} x(t) \,&=\, x_{0} \\ \\ \\ y(t) \,&=\, y_{0} \,+\, u t \end{align} \)​

The inverse of a Lorentz boost along the x-axis is:

\( \begin{align} t \,&=\, \gamma \bigl( t^{\prime} \,+\, \frac{v}{c^{2}} x^{\prime} \bigr) \\ \\ \\ x \,&=\, \gamma \bigl( x^{\prime} \,+\, v t^{\prime} \bigr) \\ \\ \\ y \,&=\, y^{\prime} \end{align} \)​

Substituting these into the trajectory yields:

\( \begin{align} \gamma \bigl( x^{\prime} \,+\, v t^{\prime} \bigr) \,&=\, x_{0} \\ \\ \\ y^{\prime} \,&=\, y_{0} \,+\, u \gamma \bigl( t^{\prime} \,+\, \frac{v}{c^{2}} x^{\prime} \bigr) \end{align} \)​

Finally, some simple algebra will get you:

\( \begin{align} x^{\prime}(t^{\prime}) \,&=\, x_{0}^{\prime} \,-\, v t^{\prime} \\ \\ \\ y^{\prime}(t^{\prime}) \,&=\, y_{0}^{\prime} \,+\, u^{\prime} t^{\prime} \end{align} \)​

where:

\( \begin{align} x_{0}^{\prime} \,&=\, \frac{x_{0}}{\gamma} \\ \\ \\ u^{\prime} \,&=\, \frac{u}{\gamma} \\ \\ \\ y_{0}^{\prime} \,&=\, y_{0} \,+\, \frac{u v}{c^2} x_{0} \end{align} \)​

The last identity illustrates the relativity of simultaneity effect: if you have two vertical trajectories as defined above such that \(y_{1}(t) = y_{2}(t)\) at all times (ie. same \(y_{0}\) and \(u\), but different \(x_{0}\)'s), then the transformed trajectories will differ by: \(y_{1}^{\prime}(t^{\prime}) \,-\, y_{2}^{\prime}(t^{\prime}) \,=\, \frac{u v}{c^2} \Delta x_{0}\).

This result is due to the \(- \frac{v}{c^2} x\) term that appears in the Lorentz boost. The y coordinates may not be touched, but the times they're reparameterized in terms of aren't the same and this results in events at different x coordinates ending up out of sync.

Additionally, the reduced vertical velocity \(u^{\prime} \,=\, \frac{u}{\gamma}\) illustrates the time dilation effect, and the reduced initial x coordinate \(x_{0}^{\prime} \,=\, \frac{x_{0}}{\gamma}\) is simply length contraction. Note also that if you take a number of trajectories that all share the same \(x_{0}\), then they'll all share the same \(x^{\prime}\) coordinate at any given time in the boosted frame. In other words, a vertical wire segment would be transformed onto a (moving) vertical wire segment in the boosted frame, unlike what's illustrated on your diagram.

 

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The definition would hardly be more complicated than for N particles in the frame-dependent view: the electric current (and thus the resulting magnetic field) would simply be given by the momentary relative velocities of two collisionally interacting charges (whether this would have to be unlike charges or not would have to determined by experiments). And naturally this would also adhere to the superposition principle, i.e. only if there is some systematic net motion of the charges relatively to each other would there be a net magnetic field.
And the velocity v in the magnetic force term F=q*vxB could for instance simply be the velocity of the charge q to the center of mass of the N particles creating the magnetic field (so for a wire it would approximately be the reference frame in which the wire is at rest).

Yes, and that's exactly the reason why the frame-dependent view is pretty much academic anyway in this context, because in order for magnetic forces to become significant and measurable in practice you need overall charge neutrality i.e. a superposition of two currents of oppositely charged particles. And that makes the net current in practice frame-independent as the velocity difference of two particles is frame-independent (at least for velocities v<<c). I am just going still a step further and suggesting that it is not the difference of two frame-dependent currents that is relevant here, but the current defined by the relative motion of the particles in the course of their actual physical interaction.

Obviously, the overall charge of a single particle (or a finite distribution in general) is unaffected by the Lorentz transformation. That's exactly the reason why an infinite wire is used in these derivations discussed, as this scenario allows more easily to sweep any violation of charge invariance under the carpet.

First of all, let me just mention that your transformation above was actually incorrect: if you use the primed variables for the boosted frame and assume the unprimed variables as given, then your Lorentz transformation should be (using g for gamma) x'=g*(x-vt) ; t'=g*(t-vx/c^2) . You would have to solve this algebraically for x,t and then plug this into the circle equation, which gives a different equation to the one you gave (still a quadratic equation in x',t', but not one in terms of (x'+vt')^2).

But anyway, the more crucial point is that, as I indicated already, the Lorentz transformation goes from the rest frame of a particle to the boosted frame. This means it is actually inconsistent to assume that the coordinate of the particle in the rest frame is time dependent. You would have to choose the rest frame such that everything is time independent (i.e. the wheel is neither translating nor rotating). Consider for instance a square 'wheel' (which is easier to treat here) consisting of 4 rods of length L orientated parallel to the coordinate axes and centered on the origin. The end points of the top horizontal rod then have the coordinates

(1) x1=L/2 ; x2=-L/2 ; i.e. x1-x2 = L.

We now consider the square in a frame where it is rotating. Noting that the rotational velocity is always in direction of the length of the rod, the Lorentz transformation is generally

(2) x'=g*(x-vt)
(3) t'=g*(t-vx/c^2).

where g is the gamma factor associated with the rotational velocity.

Solving (3) for t and inserting into (2) gives after a little algebra

(4) x'=(x-vt')/g ,

i.e. applied to the end points of the rod

(5) x1'=(L/2-vt')/g ; x2'=(-L/2-vt')/g ; i.e. x1'-x2' = L/g .

In other words, in the rotating frame the rod should be length contracted by the gamma factor. This obviously not only applies to all sides of a square, but also if you generalize this to an n-polygon or indeed a circle (where n is infinite).

Now the point is that if you have additionally a translational velocity, the absolute value for the resultant velocity (and thus also the gamma factor) becomes different for the upper and lower rods which this should have different lengths in the corresponding reference frame. So the shape of the rotating object in the boosted frame should not be symmetric anymore in a direction perpendicular to the translational boost.

But your equations exactly show that the electron is still in the vertical segment in the boosted frame as well: differentiate the coordinates with regard to time and you have

dx'/dt' = v
dy'/dt' = u' ,

i.e. the x-component of the velocity is just the boost velocity. The only velocity associated with the electric current is in the y-coordinate, which means that the electron can not possibly be in the horizontal section.

Thomas

 

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How close do the charges have to pass before they interact in this way? How long is "momentary"? What if two negative charges are passing by a positive one, or M positive charges are passing near N negative ones? You're implicitly introducing a lot of extra parameters in your seemingly simple definition, and there's no evidence that they're actually necessary. This is what I mean by "complicated" - you need all these extra qualifiers to define the "current", each of which takes away from the predictive power of the alternative theory you're advocating, and none of which are necessary in electrodynamics.

If the magnetic field around a wire isn't the magnetic field due to the ions + the magnetic field due to the electrons, superposition is violated.

An obvious problem with this is that the "magnetic field" you're defining this way isn't locally defined, since you need to bring in the velocity of a system that could be a considerable distance away (and, worse, at a retarded time). The "frame-dependent" magnetic field, on the other hand, is completely defined at a given location by its action on test charges at that location, without reference to any distant system, which is the whole point of the operational definition involving test charges. This makes it a much more useful quantity.

Well if you don't want to consider the symmetry properties of electrodynamics or the implications the operational definition of the electromagnetic field has for its transformation properties, then no-one is forcing you to. If you consider the frame-dependence of the electromagnetic field an "academic" issue, why are you tiring to propose an (arguably equally academic) alternative to it? Why are you participating in this thread?

You've yet to demonstrate any violation of charge invariance in need of sweeping under the carpet. The charge/flux transformation I sketched out is locally derived based entirely on the way space and time coordinates transform, and will preserve charge invariance as long as the continuity equation is satisfied. At this point, I don't even see what you're trying to argue. What hasn't been settled?

Do this and you'll obtain the inverse relations I applied. For example:

\( \begin{align} x^{\prime} \,&=\, \gamma \bigl( x \,-\, v t \bigr) \quad \quad &(1a) \\ \\ \\ t^{\prime} \,&=\, \gamma \bigl( t \,-\, \frac{v}{c^2} x \bigr) \quad \quad &(1b) \end{align} \)​

Rearranging (1a) for x gives:

\( x \,=\, \frac{x^{\prime}}{\gamma} \,+\, v t \quad \quad \quad \quad (2) \)​

and substituting into (1b), simplifying, and rearranging just gives:

\( t \,=\, \gamma \bigl( t^{\prime} \,+\, \frac{v}{c^2} x^{\prime} \bigr) \quad \quad \quad \quad (3a) \)​

Similarly it's possible show (eg. by substituting (3a) into either (1a) or (1b)) that:

\( x \,=\, \gamma \bigl( x^{\prime} \,+\, v t^{\prime} \bigr) \quad \quad \quad \quad (3b) \)​

In fact this is a consistency requirement. Relativity would be in serious trouble if inverting a Lorentz boost such as (1) gave anything other than the Lorentz boost (3) (at the very least, anything else would contradict reciprocity).

As another consistency requirement, it's also possible to show that successive Lorentz boosts yield another boost, just like combining two rotations yields another rotation. In general if coordinate systems A and B are related by a Lorentz transformation, and B and C are related by a Lorentz transformation, then the direct transformation between A and C is also a Lorentz transformation. The easiest way to see how this works for boosts is to re-express them in terms of the rapidity parameter.

Mathematically, the set of all Lorentz transformations possesses a well studied and documented group structure.

There's no such restriction on Lorentz transformations. In fact their main defining property (that they leave the speed of light invariant) contradicts this: in relativity, anything moving at the speed of light doesn't have a rest frame.

I'd suggest you consider learning to manipulate Minkowski diagrams as a way of gaining clearer insights about Lorentz transformations. Minkowski diagrams are a way of visualising Lorentz boosts by representing them on space-time diagrams (which are just like the distance vs. time graphs we all had to draw in high school, except for the convention of representing time on the vertical axis).

This is only true at the centre of each rod (though the difference becomes negligible for many-sided regular polygons).

What you're describing isn't a Lorentz transformation at all (which relate the coordinates of inertial frames passing each other at constant velocity, so v should be a constant, fixed parameter). As a coordinate transformation, it doesn't even make much sense. If you take v as variable (depending on y for instance), then (2) describes a coordinate system moving with a shear rate with respect to the "rest" frame, rather than rotating. You seem to be applying four different such "shears" to the four rods. Whatever you're doing, you're not applying the same coordinate transformation consistently to the entire wheel.

This doesn't affect your argument, but the correct expression should be: \(x^{\prime} \,=\, \frac{x}{\gamma} \,-\, v t^{\prime}\)

There's no symmetry principle analogous to relativity or invariance under fixed rotations which you can apply to spinning objects, and different objects will be affected in different ways by continuous rotation. In the case of a rotating wheel, the rim will want to contract, but effective centrifugal forces will pull outward and if the wheel has spokes they will also resist contracting. A rotating wheel could shrink or expand in radius, or (more likely) break apart due to the stresses, or the spokes could punch through the rim, depending on the wheel's composition, structure, and angular velocity.

You're making a lot of unsupported assumptions about what your (not well-defined) "rotating" transformation will yield when combined with a real Lorentz transformation, and the result isn't even of any significance anyway. We're interested in what the rotating wheel looks like in an inertial frame in which it is moving, given what the wheel looks like in an inertial frame in which it is spinning but otherwise at rest (ie. with all the effects rotation has on the wheel's structure already contained in its description).

Only if you assume that the electrons are always in the vertical segment in the rest frame (a vertical segment that extends to infinity?), which isn't what we're discussing. An electron will only travel along a vertical segment up to the corners, then travels along a horizontal segment. The important issue is when this happens. Based on the equations, if electron (A) reaches the top horizontal segment (located at, say, \(y = 0\)), at the same time as electron (B) leaves it in the rest frame, then in the boosted frame (B) leaves the horizontal segment only a time \(\Delta t^{\prime} \,=\, \frac{\gamma v L}{c^{2}}\) after (A) enters it. During this time interval both electrons are in the horizontal segment in the boosted frame. So the average number of electrons in each horizontal segment is frame-dependent due to relativity of simultaneity.

 

All these questions are obviously potentially relevant if you look at individual particles, but the situation would be similar with the usual definition of a 'current', simply because the velocities are not constant due to the particle interactions . What we observe as a 'current' (with either interpretation) is just the average of all particles over a given time. The point is that if you adopt the interaction based view, the equations of electrodynamics would be frame-invariant by design, so there would be no need for any transformation equations for the electromagnetic field (and this would make, as far as I am concerned, electrodynamics much easier).

If you assume that the magnetic field is created by the interaction of the positive and negative charges, then there is no contribution from either species alone. The superposition would come only from the sum of the individual interactions. And to pick up on the paragraph you edited out: yes, the resultant magnetic force on a test charge would then effectively be a three-body force (which makes sense to me, as there are for instance no magnetic monopoles, i.e. the magnetostatic force is undoubtedly a higher order phenomenon than the electrostatic force (which is also indicated by its much smaller magnitude).

According to Maxwell's equations, one can not define a local interaction without reference to a distant system (whether you assume it to be frame-dependent or frame-independent).

I thought I had demonstrated it already: if one assumes that a wire becomes homogeneously charged due to sole fact of viewing it from a different reference frame, then this amounts to a violation of charge invariance as the total charge is different in the two reference frames. If on the other hand one enforces the charge invariance as a separate condition, then all that could result is a higher order multipole field, but the total charge (the monopole field) would be unchanged.

It is quite obvious that the transformation equations for the electromagnetic field must violate charge invariance, because, according to Maxwell's equations, a change in the electric field basically means a change in the charge (for a stationary situation at least), independently of what happens to the magnetic field in the transformation.

What I meant is that the frame-dependent theory is academic because in practice you need both an electron and ion current in order to be able to neutralize the electrostatic field and thus be able to measure the magnetic field. So effectively the net current will always be frame-independent anyway (at least for small velocities).

Of course, being 'academic' is not a decisive argument against a theory, and also, this applies here only to the definition of the 'current' (that determines the magnetic field); the frame-dependence due to the velocity v in the magnetic force would exist independently of this, but here the problem is that this enforces a velocity dependent transformation between the electric and magnetic force, which violates charge invariance (and thus is not acceptable).

Yes, sorry, that was my mistake. I somehow got an additional gamma factor in there. But there is actually something else incorrect in your approach, which is to square (and add) the equations. You can see that this leads to wrong results from the simple constant velocity equation

(1) x = u*t

In the primed frame (moving with velocity v relatively to the unprimed frame), this would then be

(2) g*(x'+vt') = u*g*(t'+vx'/c^2) , with g the gamma factor.

If you solve this for x', you get

(3) x' = (u-v)*t'/ (1-uv/c^2) ,

(which is nothing else than the usual velocity addition equation).

Note that according to (3), x' will change if one replaces u with -u, but obviously this will not happen if you square Eq.(2) and solve the resulting quadratic equation for x', because there won't be any term linear in u any more. So the latter solution obviously would be incorrect.

Alternatively to Eq.(3) you can of course do two separate boosts, one with velocity u (to get from the frame where the particle is at rest to the unprimed frame) and one with velocity v (to get from the unprimed to the primed frame).

The point is that for a rotating wheel, u is different for all points on the wheel, so you have to evaluate (3) for each point (and each corresponding value of u) separately . There can not be a single Lorentz transformation of a rotating wheel.
Einstein himself actually implies this in his 1905 paper, §4 when he says "A rigid body which, measured in a state of rest, has the form of a sphere, therefore has in a state of motion--viewed from the stationary system--the form of an ellipsoid".. As a rotating wheel is not in a state of rest, the conclusion does therefore not apply in this case (as follows from the consideration above).

How should a wheel break apart in one reference frame but not the other?

Your equations of motion x(t)=x0 and y(t)=y0+ut imply that the particle is always in the vertical segment in the rest frame. And the result for the Lorentz transformed coordinates ( x'(t')=x0'-vt' and y'(t')=y0'+u't') shows that it is also always in the vertical section in the boosted frame (because in the x-dimension the particle has no velocity apart from the boost velocity).

Thomas

 

Mathematically, the current density associated with a single classical (ie. non-quantum) point charge q following the trajectory \(\bar{x}(t)\) is just:

\( \bar{j}(\bar{r} ,\, t ) \,=\, q \, \frac{\text{d}\bar{x}}{\text{d}t} \, \delta \bigl( \bar{r} \,-\, \bar{x}(t) \bigr) \)​

The current is usually treated as a course average (for convenience, because experimentally we can't track the exact positions of large numbers of electrons anyway, and because nowadays we know that electrons aren't point particles in the classical sense), but Maxwell's equations don't actually require this. There's no ambiguity on how current should be interpreted in electrodynamics - the local averages are just concepts borrowed from fluid mechanics.

Electrodynamics is already relativistically invariant by construction anyway. In Newtonian physics, an equation is automatically invariant under rotations by construction if it is a vector equation built out of vectors, scalars, and vector and scalar products, even though the components of vectors are dependent on the orientation of the coordinate axes. Minkowski's approach is just a generalization of this which makes it trivially easy to put together relativistic theories. There's no need to try to "force" physical quantities to be invariant any more than Newtonian physics needed to define rotationally invariant vectors.

The electromagnetic field already transforms in other ways anyway. As vector quantities, the components of the electric and magnetic field vectors depend on how you orient the coordinate axes. The Doppler effect implies that the Fourier modes of the electric and magnetic fields are variant under Lorentz boosts. A more mundane example follows from basic dimensional analysis: the electric field has SI units of kg·m·s[sup]-3[/sup]·A[sup]-1[/sup], and the value of the electric field will transform if, for example, you switch from metres to kilometres (x' = 1000 x). In physics, coordinate-system dependent quantities are the norm, not the exception, and it is a remarkable property when a theory turns out to possess global symmetries such as Lorentz invariance despite this. I don't see why this should make a theory any more complicated: it simply means that you have a choice among many different-but-equivalent ways of describing the same physical system.

Also I'm sceptical that what you're suggesting would actually result in an invariant theory. The problem is that your quantities, based on velocity differences, aren't exactly Lorentz invariants. They're exact Galilean invariants. Plugging Galilean invariant quantities into an otherwise Lorentz invariant theory will most likely result in something that's invariant under neither set of transformations.

We already have an interaction model for electrodynamics. Quantum electrodynamics has been around in its modern form since about the 1950's and is incorporated in the Standard Model of particle physics. In addition to fully reproducing electrodynamics in the classical regime, it's the theory used to predict Feynman rules for QED processes such as electron-positron annihilation, Compton scattering, and Bremsstrahlung. Because we have a successful underlying quantum theory of electrodynamics, proposing modifications to the classical theory would be a bit like proposing modifications to Kepler's laws and ignoring the issue of whether these modifications were compatible with the underlying theory of gravitation. While all the other arguments I've been employing are primarily cosmetic, this one isn't. Classical electrodynamics isn't a standalone theory and you can't simply modify it in isolation and blindly assume you'll end up with something consistent with the rest of physics.

Also, how would you explain the dipolar magnetic fields around individual electrons due to their spin?

How so? Maxwell's equations are completely local. For instance, according to Gauss' law, the divergence of the electric field \(\bar{\nabla} \cdot \bar{E}\) at a point \(\bar{x}_{0}\) is proportional to the charge density \(\rho(\bar{x}_{0})\) at that exact same point \(\bar{x}_{0}\), and so on with the other three equations. Similarly, the Lorentz force on a moving charge is only dependent on the electric and magnetic fields at that point (or flipped around for test charges, this is a completely local definition of the electromagnetic field). The only non-local interactions are the ones you're advocating. Also note that non-local, instantaneous interactions contradict relativity (more specifically, information of any kind propagating faster than light can lead to causality violations in relativity - this is the reason the theory has to impose c as a universal speed limit).

I've already explained why this isn't the case: you need to account for the wire's end points. In order to satisfy charge conservation, you need to account for the fact that, for example, charge might drain from one end of the wire and collect at the other. Let's say that the wire extends from \(x = 0\) to \(x = L\) in its rest frame, with a current \(j = j_{0}\) flowing within the wire. The charge draining and collecting at the ends could be modelled with two Dirac delta's, so we could model the wire as:

\( \begin{align} \rho(x , \, t) \,&=\, j_{0} t \, \bigl[ \delta(x \,-\, L) \,-\, \delta(x) \bigr] & \\ \\ \\ j(x , \, t) \,&=\, j_{0} \quad \quad &\text{for } 0 \,\leq\, x \,\leq\, L \end{align} \)​

The total charge is given by integrating over a region containing the wire:

\( Q \,=\, \int \text{d}x \, \rho(x , \, t) \,=\, j_{0} t \,-\, j_{0} t \,=\, 0 \)​

Plugging an inverse Lorentz boost into the first equation yields this:

\( \rho(x^{\prime}, \, t^{\prime}) \,=\, j_{0} \, ( t^{\prime} + \frac{v}{c^{2}} x^{\prime} ) \, \Bigl[ \, \delta \bigl( x^{\prime} + v t^{\prime} - \frac{L}{\gamma} \bigr) \,-\, \delta \bigl( x^{\prime} + v t^{\prime} \bigr) \, \Bigr] \quad \quad \quad \quad \text{for } - v t^{\prime} \,\leq\, x^{\prime} \,\leq\, \frac{L}{\gamma} - v t^{\prime} \)​

(if you're checking this, then this uses \(\delta( \alpha x) \,\equiv\, \frac{\delta(x)}{\alpha}\)). Since \(f(x) \delta(x \,-\, \alpha) \equiv f(\alpha) \delta(x \,-\, \alpha) \), we can also equivalently express this as:

\( \rho(x^{\prime}, \, t^{\prime}) \,=\, j_{0} \, \Bigl( \frac{t^{\prime}}{\gamma^{2}} \,-\, \frac{v}{c^{2}} \, \frac{L}{\gamma} \Bigr) \, \delta \bigl( x^{\prime} + v t^{\prime} - \frac{L}{\gamma} \bigr) \,-\, j_{0} \, \frac{t^{\prime}}{\gamma^{2}} \, \delta \bigl( x^{\prime} + v t^{\prime} \bigr) \)​

Finally we just evaluate the transformed charge density:

\( \rho^{\prime} \,=\, \gamma ( \rho \,-\, \frac{v}{c^{2}} j) \)​

with \(j \,=\, j_{0}\), obtaining:

\( \rho^{\prime}(x^{\prime}, \, t^{\prime}) \,=\, j_{0} \, \Bigl( \frac{t^{\prime}}{\gamma} \,+\, \frac{v L}{c^{2}} \Bigr) \, \delta \bigl( x^{\prime} + v t^{\prime} - \frac{L}{\gamma} \bigr) \,-\, j_{0} \, \frac{t^{\prime}}{\gamma} \, \delta \bigl( x^{\prime} + v t^{\prime} \bigr) \,-\, \gamma \frac{v}{c^{2}} j_{0} \)​

for \(- v t^{\prime} \,\leq\, x^{\prime} \,\leq\, \frac{L}{\gamma} - v t^{\prime}\), with \(\rho \,=\, 0\) elsewhere.

We find out if this violates charge invariance by integrating:

\( Q^{\prime} \,=\, \int \text{d}x^{\prime} \, \rho^{\prime}(x^{\prime}, \, t^{\prime}) \,=\, j_{0} \, \Bigl( \frac{t^{\prime}}{\gamma} \,+\, \frac{v L}{c^{2}} \Bigr) \,-\, j_{0} \, \frac{t^{\prime}}{\gamma} \,-\, \gamma \, \frac{v}{c^{2}} \, j_{0} \, \frac{L}{\gamma} \,=\, 0 \)​

Physically, what's happening is that the endpoints have a net charge in the boosted frame. While in the rest frame they were:

\( \begin{align} q_{0}(t) \,&=\, - j_{0} t \quad \quad &\text{and} \\ \\ \\ q_{L}(t) \,&=\, + j_{0} t \end{align} \)​

so that \(q_{0}(t) \,+\, q_{L}(t) \,=\, 0\), in the boosted frame they're:

\( \begin{align} q^{\prime}_{0}(t^{\prime}) \,&=\, - j_{0} \, \frac{t^{\prime}}{\gamma} \quad \quad &\text{and} \\ \\ \\ q^{\prime}_{L}(t^{\prime}) \,&=\, + j_{0} \, \Bigl( \frac{t^{\prime}}{\gamma} \,+\, \frac{v L}{c^{2}} \Bigr) \end{align} \)​

so that \(q^{\prime}_{0}(t^{\prime}) \,+\, q^{\prime}_{L}(t^{\prime}) \,=\, j_{0} \, \frac{v L}{c^{2}}\), which exactly compensates the charge on the rest of the wire.

In general, charge invariance is fully satisfied as long as the local continuity equation is satisfied.

Why are you applying the constant velocity equation to something that isn't in rectilinear motion? If you transform the wrong trajectory you'll get the wrong result.

This only holds for particles in rectilinear motion. In general only the differential form of this equation:

\( \text{d}x^{\prime} \,=\, \frac{u \,-\, v}{1 \,-\, \frac{uv}{c^{2}}} \, \text{d}t^{\prime} \)​

will hold.

Variable substitution doesn't work this way.

So what do you expect to happen to, for instance, a non-rotating spherical planet with a net atmospheric current "rotating" on the outside of the planet? In a boosted frame, are you claiming that the planet is an ellipse, but that the atmosphere transforms differently and doesn't stick to the surface of the planet? What about a human being? Are you claiming that a human body would be length contracted, but that the flowing blood could be transformed somewhere other than the person's blood vessels? The way you're applying Lorentz transformations obviously leads to contradictions and nonphysical results, and since you yourself know this (given your transformation of the rectangular wire as an example) I really don't see the point of arguing this. If, on the other had, you apply Lorentz transformations the same way as any other variable substitution just like everyone else does (and as depicted on Minkowski diagrams) without making up your own rules about them, you don't get such nonsensical results.

You're denying the antecedent.

It doesn't. I was stating that you can't deduce what happens to a rotating wheel given a non-rotating wheel from symmetry principles alone, so the idea of transforming from a rotating frame to a non-rotating frame doesn't make much sense. If you weren't doing this anyway, then I misunderstood your approach and you can ignore this comment.

Those equations of motion are only valid along a portion of the complete trajectory. The full trajectory in the rest frame might be something more like:

\( \begin{align} \left\lbrace \begin{matrix} x(t) \,&=\, 0 \\ \\ \\ y(t) \,&=\, u t \\ \end{matrix} \right. \quad \quad &\text{for }& 0 \,\leq\, t \,\leq\, \frac{Y}{u} \\ \\ \\ \\ \\ \\ \left\lbrace \begin{matrix} x(t) \,&=\, u t \\ \\ \\ y(t) \,&=\, Y \\ \end{matrix} \right. \quad \quad &\text{for }& \frac{Y}{u} \,\leq\, t \,\leq\, \frac{X \,+\, Y}{u} \\ \\ \\ \\ \\ \\ \left\lbrace \begin{matrix} x(t) \,&=\, X \\ \\ \\ y(t) \,&=\, Y \,-\, u t \\ \end{matrix} \right. \quad \quad &\text{for }& \frac{X + Y}{u} \,\leq\, t \,\leq\, \frac{X \,+\, 2 Y}{u} \\ \\ \\ \\ \\ \\ \left\lbrace \begin{matrix} x(t) \,&=\, X \,-\, u t \\ \\ \\ y(t) \,&=\, 0 \\ \end{matrix} \right. \quad \quad &\text{for }& \frac{X \,+\, 2 Y}{u} \,\leq\, t \,\leq\, \frac{2 X \,+\, 2 Y}{u} \end{align} \)​

This is obviously not a particularly appetizing problem to apply a Lorentz transformation to by brute force, and I never said I was calculating the transformation of the full trajectory. You made a blanket statement along the lines that Lorentz boosts could have no effect on vertical trajectories, which isn't entirely true: in particular, the transformed initial coordinate \(y^{\prime}_{0}\) depends on \(x_{0}\).

Applied to the rectangular wire problem, electrons on the right-hand vertical segment \(x_{0} \,=\, X\) will have their trajectories "delayed" relative to the electrons on the left-hand segment \(x_{0} \,=\, 0\) in the boosted frame. In particular, electrons on the right-hand segment will begin that portion of their trajectory at a delayed time compared with electrons entering from the left-hand segment. Although I haven't explicitly calculated it, this delay is consistent with the electrons spending longer on the top horizontal segment and leaving it at a delayed time, which answers your original question about where the extra electrons come from.

(By the way, in case you were wondering, the powers that be here at SciForums added \(\LaTeX\) support a while ago, so you can format equations by typing code surrounded by [noparse]\([/noparse] tags. For example, [noparse][tex]E = m c^{2}\)[/noparse] renders as \(E = m c^{2}\). There's a thread about it [thread=61223]here[/thread] and a tutorial here if you're interested in using it, though it can be a bit time-consuming getting things to format properly.)

 

I am not blindly modifying electrodynamics. On the contrary, it appears to me that the present frame-(observer)-dependent interpretation of certain variables in the laws of electrodynamics has been adopted more or less by accident and without much thought. So there is every reason to try and adopt a more physical interpretation of these equations that uniquely defines all variables independent of the observer.

The question is whether an electron would also have a magnetic moment if it would not be rotating in an electric field (e.g. of the atomic nucleus). Obviously, rotation is also a kind of motion (and would thus produce a magnetic field with my interpretation).

They are only written as 'pseudo-local' in their differential form. From the integral form of Maxwell's equations it is apparent that they are in fact not local (e.g. Gauss's law states that the electric flux through a closed surface is equal the enclosed charged in the corresponding volume; since the volume of the surface itself is zero, the equation is therefore anything but local).

And I said 'pseudo-local' in case of the differential form, because due to the presence of spatial derivatives they can not really be local: one can not define a derivative by just one point alone; a derivative always implies non-locality (be it of an infinitesimally small extension).

Yes, you said this already before (and I agreed with it), but the point is that these treatments purporting to derive the relativistic charging of the wire (i.e. the transformation equations for the electromagnetic field) do not bother about a strict application of the Lorentz transformation like you did it (and I did it on my web page Magnetic Fields and Lorentz Force ), and as a result they violate charge invariance, or lead to other violations of physical (and logical) laws (see also below).

That's no problem. Just consider the regions near the top and the bottom of the wheel (where x is close to zero). In the vicinity of these points, the motion can be approximately described by x=ut, and the equation x' = (u-v)*t'/ (1-uv/c^2) would then be applicable in the boosted frame (which implies that the absolute value of x' changes when u changes sign (i.e. when we consider corresponding points near the top and the bottom of the wheel).

Well, you really shouldn't ask me that. I am just applying the Lorentz transformation here (be it in the sloppy way as in these 'wire charging' derivations, or more strictly like you suggested it) and find that either way one obtains physical impossibilities here (see also below).

As is evident from your equations further above (post #62), the only thing that would happen to the vertical sections is that they would shift up or down (depending on the relative signs of u and v) by an amount uv/c^2*x0. This a) wouldn't bring any charges into (or out of) the horizontal sections, and b) would lead to the physical impossibility of the end points of the horizontal and vertical sections of the current loop becoming disconnected.

This again is a physical impossibility, because the only way to have a higher electron density in the top section and a lower density in the bottom section is (assuming the current is constant in the whole loop) is to speed up/ slow down the electrons accordingly, i.e. it would require a physical force. This would be inconsistent with the fact that in the rest frame there is nor force acting on the electrons (assume for instance a superconducting loop).

But before you get too puzzled about all these inconsistencies that are apparent here: the reason for this is simply that the Lorentz transformation is erroneous in the first place. It has been derived mathematically incorrectly from the postulate of the invariance of c (see my page Mathematical Flaws in Einstein's 'On the Electrodynamics of Moving Bodies').

Thomas

 

Your basis for this view basically amounts to you, for some reason, feeling uncomfortable with frame-dependent quantities. Euclidean vectors are coordinate-system dependent. Are they non-physical? In Minkowski's formulation of electrodynamics, the electromagnetic field F belongs to a class of physical quantities called (rank 2 antisymmetric) tensors. They're geometrical entities whose components are projections onto a particular coordinate system in a similar sense to Euclidean vectors. In fact, believe it or not, in the language of differential forms all four of Maxwell's equations can be reduced to:

\( \begin{align} \text{d} \ast F \,&=\, J \\ \\ \\ \text{d}F \,&=\, 0 \end{align} \)​

If you want a completely frame-independent expression of electrodynamics, this is it.

I was talking about the electron's spin, not its orbital angular momentum. Lone electrons are magnetic dipoles. This dipole moment is detectable in Stern-Gerlach-type experiments and is known to cause electrons to precess in an external magnetic field (Larmor precession).

You can integrate any equation over arbitrary volumes. That doesn't mean that the theory is non-local. The point is that for a non-local theory, the differential form either won't exist or will equate physical quantities evaluated at different places at the same time.

As long as the relations aren't between points a finite distance apart, the theory is local. The derivatives dictate the way causal relations propagate in the theory. In electrodynamics the result is well-known: causality propagates along the light cone. The only "theories" that could satisfy your "point defined" locality are trivial equalities like \(f(x,\, t) \,=\, g(x,\, t)\), with no possibility of causal relations or (predictable) dynamics whatsoever.

I really don't understand this reasoning: you've just agreed that the Lorentz transformation acting on a finite wire preserves charge invariance. We see that there are time-dependent charges at the end points of the wire that the derivations you linked to wouldn't want to worry about, so the obvious solution is just to take an infinitely long wire. Even if the websites didn't use an infinitely long wire, they could still use an arbitrarily long finite one (whose complete transformation I've just calculated) rendering the effect of the endpoint charges negligible compared with the charge along the "body" of the wire, and still reach the same conclusions they did. Where's the problem with charge invariance?

In general you could develop the motion of a particle moving on any part of the wheel around any time \(t_{0}\) as:

\( \bar{x}(t) \,=\, ( t \,-\, t_{0} ) \bar{u}_{0} \,+\, \bar{x}_{0} \)​

where \(\bar{x}_{0}\) and \(\bar{u}_{0}\) are respectively the particle's position and velocity at the time \(t_{0}\), and are treated as constants. If you apply a Lorentz transformation (skipping to the result) you'll get a transformed trajectory of the form:

\( \bar{x}^{\prime}(t^{\prime}) \,=\, ( t^{\prime} \,-\, t^{\prime}_{0} ) \bar{u}^{\prime}_{0} \,+\, \bar{x}^{\prime}_{0}(t^{\prime}) \)​

approximately valid for times around \(t^{\prime}_{0} \,=\, \gamma ( t_{0} \,-\, \frac{v}{c^{2}} x_{0} )\). \(\bar{x}_{0}\), which is a point on the wheel edge being treated as a constant (ie. effectively a fixed point on the edge of a hypothetical non-rotating circle with the same outline as the wheel), will map onto the corresponding point \(\bar{x}^{\prime}_{0}\) on a length-contracted moving ellipse. Specifically its x-component becomes:

\( x^{\prime}_{0}(t^{\prime}) \,=\, \frac{x_{0}}{\gamma} \,-\, v t^{\prime} \)​

The difference between \(x^{\prime}_{0}\) and the ellipse centre at \(x_{\text{c}}^{\prime} = - v t^{\prime}\) is just the regular length-contracted \(\frac{x_{0}}{\gamma}\). So a trajectory developed in the vicinity of the wheel edge maps onto a trajectory developed in the vicinity of the ellipse edge.

This is where you should see something's wrong in your application or interpretation of Lorentz boosts. The Lorentz-transformed coordinates x' and t' are invertible continuous functions of x and t. Have you seriously considered your stance here? You're claiming that a pair of continuous functions can produce discontinuities!

The Lorentz transformation predicts this. The velocities are given by the velocity addition formula, and the "excess" velocity compared with the moving wire is just the boosted velocity minus v:

\( \frac{ \pm \, u \,+\, v }{1 \, \pm \frac{uv}{c^{2}}} \,-\, v \,=\, \frac{1 \,\mp\, \frac{uv}{c^{2}}}{1 \,+\, \frac{uv}{c^{2}}} \)​

The norm of this quantity is sign dependent, with the result that electrons will spend longer on the upper segment than on the lower one.

There has to be a force in the rest frame. The electrons wouldn't turn the corners otherwise.

Please keep in mind that a) the Lorentz transformation did not originate with Einstein and b) there are many different ways of deriving the Lorentz transformation, including more modern methods treating the Lorentz group as a Lie group. In the 1970's one particular derivation was published by a Jean-Marc Lévy-Leblond [sup](ref)[/sup] that doesn't even invoke the invariance of c postulate.

In (a) defining \(\tau\) as a function of \(x^{\prime}\) makes sense as \(x^{\prime} \,=\, \text{cst}\) is the simplest way of tracking the time coordinate moving along with k. In what will ultimately become the Lorentz boost, he's after \(t^{\prime}(x,\, t)\) evaluated along a path \(x \,=\, v t \,+\, c\) - ie. following the moving clock. The partial derivative \(\frac{\part \tau}{\part t}\) with \(x^{\prime}\) constant is essentially a hydrodynamic derivative. Physically, it's the moving clock's time dilation rate.

For (b), you get (4) from (3) by applying first order Taylor developments like:

\( \tau(0,\, t \,+\, \varepsilon) \,=\, \tau(0,\, t) \,+\, \varepsilon \, \frac{\part \tau}{\part t} \,+\, \ldots \)​

With all the derivatives evaluated at \((0,\, t)\). The constant terms end up cancelling, and the higher order terms vanish in the limit \(x^{\prime} \rightarrow 0\), leaving only the first order derivatives.

Also, concerning the "flaws" you descibe on your other pages, take the light-cone definition for example. It (probably also with some qualifiers about Lorentz transformations being linear and so on) defines the Lorentz group as the group of all transformations satisfying:

\( c^{2} t^{\prime 2} \,-\, x^{\prime 2} \,-\, y^{\prime 2} \,-\, z^{\prime 2} \,=\, c^{2} t^{2} \,-\, x^{2} \,-\, y^{2} \,-\, z^{2} \)​

You could choose to restrict your attention to transformations that preserve the time coordinate \(t^{\prime} \,=\, t\), in which case the definition reduces to:

\( x^{\prime 2} \,+\, y^{\prime 2} \,+\, z^{\prime 2} \,=\, x^{2} \,+\, y^{2} \,+\, z^{2} \)​

or transformations which preserve the norms of Euclidean vectors. In other words, the Lorentz group is defined to contain the full rotation group, and consequently a Lorentz invariant theory is automatically invariant under rotations. A general Lorentz transformation is fully specified by six parameters (basically three boost velocities and three angles). Whenever I've said I was applying a "Lorentz boost along the x-axis", this meant that, for convenience, I was choosing to orient both coordinate systems the same way and such that the boost acted along the x-axis.

There's also the ten-parameter Poincaré group, which is basically defined as the Lorentz group with every possible translation added, so a Poincaré-invariant theory, as part of its definition, is automatically invariant under translations. By Noether's theorem, this means (Lagrangian based) Poincaré-invariant theories are relativistic theories with energy, momentum, and angular momentum conservation laws.

 

Minor errata:

Apparently I can't do basic algebra or even dimensional analysis. The correct result should be:

\( \pm \, \frac{u}{\gamma^{2} \bigl( 1 \,\pm\, \frac{uv}{c^{2}}\bigr)} \)​

Also, since I've drawn myself into posting anyway:

in case it's not clear, I'm specifically referring to this page and in particular this paragraph:

The "cross combination" (3a) with (4b) and (3b) with (4a) defines a Lorentz boost combined with either a 180 degree rotation (around eg. the y axis) or a parity transformation (inversion of the spatial axes). Applied as a passive transformation, it relates two coordinate systems moving past each other who's x-axes are pointing in opposite directions. Only degenerate combinations eg. relating both (3a) and (3b) with (4a) are ruled out by the general requirement that a coordinate transformation should be a one-to-one (ie. invertible) remapping of the coordinates.

Moving on, this statement:

suggests you've misinterpreted what the Lorentz transformation is supposed to be. The idea is that the primed and unprimed coordinate systems (modulo one's preferences about orientations and where to put the origins) are supposed to reflect the way distances and time intervals are measured in each of the two frames. The transformation is universal and independent of what system it is being applied to simply because we don't define or measure the positions and velocities of light signals in a fundamentally different way than any other position or velocity. If we measure that the speed of light is always c regardless of the observer's velocity, the idea is that this implies something about the effect velocity has on our measuring instruments, which in turn implies something about the physics governing the structure of those measuring instruments.

Finally, here:

The problem is that you've only done half the transformation. Because of the relativity of simultaneity effect, comparing x'[sub]L+[/sub] and x'[sub]L-[/sub] for the same t isn't the same as comparing them for the same t'. You've just done the former, while condition (8) does the latter.

On this page the issue you describe, despite what you say, isn't actually of the same nature as your issue with the light cone definition:

You're forgetting that Einstein is looking for a single transformation that both maps (1) onto (2) and maps (1a) onto (2a). So while (1), (2), (1a), and (2a) are specific trajectories (ie. "not the same x"), he's using them to put more general constraints on the Lorentz transformation itself. In (3) and (4), x' and t' are the same implicit functions of the parameters x and t. In obtaining them he's also implicitly assumed that the transformation he's looking for is linear. You can essentially read (3) and (4) as requiring that the Lorentz boost's eigenvectors be (1, -c) and (1, +c).

And I really don't understand how you can conclude that the Lorentz transformation itself is erroneous. When we say that the Lorentz transformation leaves c invariant, we don't mean that Einstein wrote some beautiful derivation of it. We mean that it leaves c invariant as a statement of fact, regardless of what anyone might have said back in 1905. I had a math professor a couple of years ago who's advice was that we shouldn't worry too much about rigour when deriving anything. His reasoning was that non-rigorous derivations get the right result 90% of the time, and it is usually much easier to check in retrospect that the result solves the initial problem than to worry over subtleties at every step of the derivation. With respect to Einstein's derivations, whatever his techniques he got what he'd set out for. It's not difficult to check that the Lorentz boost indeed maps the trajectory \(x \,=\, \pm\, ct\) onto the corresponding \(x^{\prime} \,=\, \pm\, ct^{\prime}\).

 

(I wanted to comment to comment on the LaTex issue already in my last post, but then forgot about it again)


It is not only that the formatting can be time-consuming, but it is also the difference between seeing something like this


[noparse]
The inverse of a Lorentz boost along the x-axis is:

\( \begin{align} t \,&=\, \gamma \bigl( t^{\prime} \,+\, \frac{v}{c^{2}} x^{\prime} \bigr) \\ \\ \\ x \,&=\, \gamma \bigl( x^{\prime} \,+\, v t^{\prime} \bigr) \\ \\ \\ y \,&=\, y^{\prime} \end{align} \)​

Substituting these into the trajectory yields:

\( \begin{align} \gamma \bigl( x^{\prime} \,+\, v t^{\prime} \bigr) \,&=\, x_{0} \\ \\ \\ y^{\prime} \,&=\, y_{0} \,+\, u \gamma \bigl( t^{\prime} \,+\, \frac{v}{c^{2}} x^{\prime} \bigr) \end{align} \)​

[/noparse]


and this


[noparse]
You can see that this leads to wrong results from the simple constant velocity equation

(1) x = u*t

In the primed frame (moving with velocity v relatively to the unprimed frame), this would then be

(2) g*(x'+vt') = u*g*(t'+vx'/c^2) , with g the gamma factor.

If you solve this for x', you get

(3) x' = (u-v)*t'/ (1-uv/c^2) ,
[/noparse]


when you are reading email notifications or archived posts for instance.

LaTex is simply not supported well enough to make the formatting effort worthwhile in my opinion. Even many web servers don't support LaTEx.
I always keep backups of my own posts because sometimes I use these as a basis for posts in other forums or newsgroups, or for my own web pages, and then it is convenient if I can just copy and paste them rather than having to re-format everything because LaTex is not supported.

In case of longer posts/articles I also prefer to have everything (text and formulae) in the same font (apart from special symbols obviously) because I feel it aids the understanding. If the contrast between the text and formulae is too strong, people may tend to focus on either of them and neglect the other.
For my own web pages I only use some HTML formatting, which is much more widely supported, but unfortunately this seems to be disabled on Sciforums.

Thomas

 

It would be generally unphysical (and indeed logically absurd) to assume that the interaction between two particles could depend in any way on the observer (e.g. the latter's state of motion). The interaction must be describable by reference to the interacting particles alone. I mentioned already the friction force, which in case of the drag of objects in gases or fluids for instance is velocity dependent. There the velocity in the force equation is quite obviously the relative velocity of the object to the medium. As such it is independent of the reference frame, and nobody would try here to suggest a theory where the velocity is frame-dependent. The question is, why is the velocity in the magnetic force equation is taken as frame dependent (i.e. as physically undefined)? I don't know any positive reason for this. It just seems to be a case of sloppy formulation to me.

I was talking about the electron spin as well. What I meant was that a point on the spinning electron sphere moves as much with regard to the atomic nucleus due to the spin motion as it is moving due to the orbital motion. Both motions should create a magnetic moment with this interpretation. But there shouldn't be a magnetic moment for a free electron, and indeed the Stern-Gerlach effect is only observed for atoms and ions, not free electrons (see http://msc.phys.rug.nl/quantummechanics/stern.htm , although there they give a different interpretation of this fact).

I am not quite getting your argument. My understanding of 'local' would be that the value of a function at a given point does not in any way depend on what goes on at other points. If that function involves any kind of integral, then it isn't strictly local anymore in this sense.

Yes, so? I would like to remind you that it was you who has been advocating point-locality for Maxwell's equations (I quote "according to Gauss' law, the divergence of the electric field at a point is proportional to the charge density at that exact same point"). The electric field (which determines the physical force on a particle) has indeed nothing whatsoever to do with the charge density at that point, but on the contrary, only with the integral over the charge densities at other points (if the latter would be zero, the electric field would be zero).

On what grounds are you saying that the endpoint charges are negligible? It is exactly them that make the wire overall neutral, i.e. the wire would not have any monopole field, but would behave like an electric quadrupole for sufficiently large distances. I have pointed out why this is not compatible with Maxwell's equations (which predict a magnetic monopole field for a finite line current at large distances).

As you said, we are dealing with a pair of functions (one for the horizontal sections and one for the vertical sections) and there is no reason to assume that they together form one continuous function. The point is simply that you have two different transformations for both sections (you could probably make the whole thing continuous by rounding off the corners of the rectangle, but still the result would look very odd, not to mention the physical problems with this (see below)).

Yes, obviously, there is a force associated with turning the corners, but this adds to zero for the complete loop. There is no net force in the rest frame, but there would have to be one in the boosted frame if the electrons should speed up in the lower horizontal section and slow down in the upper (or vice versa). Actually, I would hesitate to call it a force, as there is nothing that could do the physical work required. I would just call it a violation of the energy and momentum conservation laws.



But more fundamentally, it occurred to me now that the purported differential length contractions for the electrons and ions should actually be impossible for logical reasons alone: if I define the location of an electron in the wire's rest frame by being in the vicinity of a given ion, then I can't see how it could be in the vicinity of a different ion in the boosted frame, because you could objectively define 'vicinity' for instance by assuming that they physically interact somehow. And you can't have an interaction of two particles in one reference frame but not the other.

Thomas

 

(note: a fundamental discussion of the Lorentz transformation itself would probably be worth a separate thread, but I'll be away soon for a couple of weeks, so I don't want to open a new one, but respond to it here)

Yes, I know that the Lorentz transformation has been formulated in various forms before Einstein, but Einstein was the first who attempted an axiomatic derivation of it. Without some kind of axiomatic approach it is not possible to 'derive' the Lorentz in the sense of the word (the reference you gave doesn't derive it either; it derives a number of solutions of which the Lorentz transformation is just one (one of the others being the Galilei transformation), and in order for the Lorentz transformation to apply one would need to postulate further that there is a general limiting velocity for objects, for which there are no grounds (the physical requirement is merely that the speed of light is invariant); so if anything this article makes the Lorentz transformation actually implausible).

It is absolutely incorrect to use x' as an argument of tau. When Einstein says in his paper "We first define tau as a function of x', y, z, and t" then this is obviously completely non-sensical as y,z and t are variables in the stationary frame but x' is a variable in the moving frame i.e. in the frame of the light source and the (co-moving) mirror (it is simply the (fixed) distance between the two).
The arguments of tau have to be x,y,z,t (i.e. x'+vt,y,z,t).

The point is that with the quadratic constraint

x[sup]2[/sup] +y[sup]2[/sup] +z[sup]2[/sup] = c[sup]2[/sup]*t[sup]2[/sup]
x'[sup]2[/sup] +y'[sup]2[/sup] +z'[sup]2[/sup] = c[sup]2[/sup]*t'[sup]2[/sup]

any information about the direction of propagation of the light signal has been lost. Any point on the unprimed sphere could in this way be correlated to any point on the primed sphere. This is obviously non-sensical as it would allow the light signals to travel in arbitrary directions in the two frames and still satisfy the constraint.
The correct constraint is (limiting the problem just to the x-coordinate)

x=ct <=> x'=ct'
x=-ct <=> x'=-ct'

Any squaring of the latter equations obviously produces additional solutions that are not part of the problem (and indeed are physically not acceptable).

Of course, you can adopt any convention regarding the coordinate axes you want, but choosing to have them pointing in the opposite direction would merely mean that the constraint now is (referring to the previous paragraph)

x=ct <=> x'=-ct'
x=-ct <=> x'=ct'

Instead of the coordinate of the light signal having equal signs (with opposite signs not allowed), they now have opposite signs (with equal signs not allowed). However, when you square the equations, you bring in again the disallowed cases.

Yes, I am well aware about what the Lorentz transformation is supposed to do, but the point is that the whole problem arises solely from the special nature of the propagation of light signals. And why should this be a reason to revise the concepts of space and time in general? As an analogy, if you throw a boomerang and find that its path differs very much from that of other objects, would you try to develop a new theory of gravity that is able to explain both? I don't think so. You would rather develop a special theory for the boomerang that works with the present theory of gravity.

It occurred to me already that I should have better given the Lorentz transformation in terms of x',t' rather than x',t. I corrected this now on my web page. It doesn't change anything about the conclusion though because for

x' = x/g -vt'

x' still doesn't change sign when x changes sign, thus violating the original constraints for the propagation of light signals.

The issue is essentially of the same nature, both for Einsteins 'algebraic' derivation and the 'light sphere' derivation. In both cases the signs for the directions of propagation of the light signals are mixed up (for the 'light sphere' derivation it is due to squaring the original constraints, in the 'algebraic' derivation by adding up mutually exclusive equations).

As mentioned above already, the constraint given by the invariance of c demands that in the case of light signals x' changes sign when x changes sign (in other words, two signals sent into opposite directions from the origin must have the same distance from the latter at any time in each reference frame). This is evidently not fulfilled by the Lorentz transformation x' = x/g -vt'(unless v=0).
The Lorentz transformation fulfills the constraints x=ct <=> x'=ct' and x=-ct <=> x'=-ct' separately, but the symmetry of the situation demands that they are being fulfilled together, which is not the case (of course, algebraically one can not have x=ct and x=-ct at the same time; that's why one correctly should use two different variables for the two directions here, i.e. the constraints should be written as x1=ct <=> x1'=ct' and x2=-ct <=> x2'=-ct').

Thomas

P.S.: As mentioned above already, I'll be away soon for a couple of weeks and might not be able to reply to any posts until after my return.

 

Well if it's inconvenient for you, I can limit my use of TeX, but this might become a bit inconvenient if I want to post matrix equations for instance (Lorentz transformations are linear and their matrix representations can be simpler to work with in some cases). In any case I don't use any intensive math in my next two posts.

It was disabled a while ago after someone abused it in threads causing redirects:
[THREAD]61055[/THREAD]

 

Nobody's saying the interaction depends on the observer. All different observers can do is describe the same physical system in different ways. Different reference frames are just different protocols for describing events - they're not alternate realities.

You're comparing an effective theory of an interaction between two surfaces with a completely different theory about a single object in an external field.

Because the velocity of a particle is frame-dependent, and it is the only velocity available at the interaction point.

You don't think an object's velocity relative to an observer is a measurable quantity?

As your own efforts show, it is not possible to "interpret" the velocity, magnetic field, or current as invariant quantities without reinterpreting electrodynamics as a fundamentally different theory that just happens to share the same equations. And even there you've had to be somewhat agnostic about some of the details, for instance in claiming that the magnetic field is produced by an interaction of some unspecified nature, and vague statements about this only occurring when the overall charge in some vicinity is neutral. That's a lot of unspecified parameters that simply don't exist in electrodynamics, and there's no evidence for any new kind of interaction at atomic distances. Particle accelerator experiments nowadays are sensitive to distance scales smaller than the characteristic dimensions of the proton and yet there's no evidence for anything wanting in the electroweak sector of the Standard Model.

Also if you try to reinterpret the quantities in electrodynamics you're still left with a strange situation: if you keep the same equations then they'll still possess the same symmetries. This means that if you've got a mathematical solution to the equations given by fields E, B, rho, and j, then you can apply the Lorentz transformation to all the quantities and obtain an E', B', rho', and j' that automatically constitute another solution to the same equations. How would you interpret this solution?

Why "shouldn't" there be? Also how would you go about calculating the Zeeman effect, for instance? Remember that these theories about electron spin that you're dismissing as "wrong" are accepted because they yield correct quantitative predictions.

Well if you're referring to the electron dispersal then it's something I should have thought of before posting: there's a Lorentz force on the moving electrons.

This isn't really a useful concept since the whole point of physical theories is to make predictions - ie. relate a system's state to some set of initial conditions, which could implicitly define functions relating the system's state at different points. The point of a locally defined theory is that you should be able to apply it on a local level without reference to what's going on a large distance away. The local expressions of electrodynamics allow you to work locally in an "external" (ie. background) electromagnetic field. In what you're proposing, its not difficult to imagine situations in which an observer would need psychic knowledge of what's going on outside a room for instance just to be able to measure the electromagnetic field, whereas in electrodynamics the electromagnetic field is operationally completely locally defined by the Lorentz force independently of whatever might be producing it.

So, if there's an ambiguity, the way you choose to interpret "that exact same point" should take precedence over the example I gave? I was contrasting with you stating that the Lorentz force on a particle at one point could depend explicitly on a system a large distance away.

I was referring to the derivations you posted about that are located here:
http://galileo.phys.virginia.edu/classes/252/rel_el_mag.html
http://physics.weber.edu/schroeder/mrr/MRRtalk.html
You've been claiming that these sites use infinitely long wires only because it allows them to violate charge invariance. This isn't the case: they could use a sufficiently long finite one. If the length of the wire is much greater than the test charge's distance from the wire, then the effects of the endpoints can be rendered negligible compared with the "body" of the wire. The asymptotic behaviour of the electric field only dominates at distances much larger than the characteristic dimensions of the wire.

The endpoints aren't equally charged, so the wire I described would if anything behave like an electric dipole at large distances in all frames.

I defined the functions in such a way that the trajectory would be continuous by specifying their domain of validity (set t = Y/u for the first two sets for instance). But the point I was making was that the Lorentz boost itself is continuous: it's a bog-standard linear transformation. It's Jacobian matrix is defined, constant, and non-singular everywhere and Lorentz transformations are trivially equal to their first order Taylor developments. Graphically Lorentz boosts can be represented on Minkowski diagrams:

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Here the red curve is some particle's worldline (graph of its position over time). To "apply" a Lorentz boost, you simply read off the new coordinates on the diagram. There's no way this procedure could ever lead to the kinds of discontinuities or event omissions you've been describing. If you think you've got an example of an inconsistency in relativity, try depicting and interpreting the situation on one of these diagrams.

Incidentally while I'm drawing Minkowski diagrams there's a point that can be depicted quite clearly using them. Here's how Lorentz boosts act on spatial vectors:

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As you can see, spatial vectors aren't an invariant concept simply because the spatial planes t = cst and t' = cst aren't parallel. So if you want a theory that's relativistically covariant by construction, requiring vector quantities like the electric and magnetic fields and the current to remain invariant is actually inconsistent with Lorentz transformations, and simply the wrong way to go about it geometrically.

There's nothing going on here that is actually new to either the wire or relativity: just think of a system of satellites in circular orbit around a large planet for instance. Even in non-relativistic physics, the individual satellites don't have constant speed or kinetic energy, except in the centre of mass frame. That doesn't mean energy isn't conserved.

So? This isn't actually violated by relativity. If an electron passes in the vicinity of a particular ion in one frame, expressing the situation in different coordinates doesn't alter that. It's just a question of when events occur and in what order they occur. Here's a Minkowski diagram depicting two electrons (green) passing two ions (red) for example:

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We can always ask the mods to split the thread if necessary.

They're both derivations in the loose physicist's sense of the word: you've got a general principle you want to work on, but you apply other information whenever it's practical. For example, one of Einstein's postulates reads something like "the laws of physics are the same in all inertial reference frames", so in practice he can apply information about existing theories in order to rule out "unphysical" transformations. For example, notice that dilations x' = ax and t' = at for some constant 'a' would trivially preserve the speed of light (as well as every other speed), but they're obviously not a symmetry of nature. Einstein implicitly eliminates this possibility when he imposes reciprocity; the Lévy-Leblond derivation does this when he requires the Lorentz boosts to form a one-parameter group. The Lévy-Leblond derivation uses the fact that Galilean invariance has been definitively ruled out experimentally (even if he doesn't explicitly say this), leaving only Lorentz transformations.

I don't think I'd call either derivation "axiomatic" in the mathematical sense of the word.

There is no known way to transmit any information of any kind faster than the speed of light in a vacuum.

Given that Galilean invariance has been experimentally ruled out, you basically have a choice between Lorentz covariance and no covariance of any kind. The fact that c is a limiting velocity is nothing new: in relativity information propagating faster than light turns out to violate causality.

You can define tau in terms of any parameters you like. x and x' are well defined in terms of one another so you can switch between both parameterisations of the function at will. The fact that he's using x' and the Lorentz transformation is defined in terms of x just means he's going to have to substitute x back in at some later stage of the derivation.

If you want a more "meaningful" interpretation of the variable x', just rename it 'b' and write the relation as x = vt + b. You can think of tau(b, t) as a series of clocks following trajectories x = vt + b, where b is a parameter uniquely identifying a specific clock: tau(b, t) is the time shown on the clock at the time t following the trajectory x = vt + b. The idea is that these clocks mark out time in the moving frame, so once he's got tau he's got the Lorentz transformation for t' as t'(x, t) = tau(x - vt, t).

The light cone definition isn't meant to specify a specific Lorentz transformation, and the coordinates in the definition aren't those of a specific trajectory. It is a criterion: if you have a coordinate transformation that satisfies the definition for all coordinates, then you can put that transformation into your bag of Lorentz transformations.

I've never seen this "light sphere" definition as you've posted it before. The correct light-cone definition defines the Lorentz group as the set of all linear isometries that preserve the quadratic form c² t² - x² - y² - z² for all t, x, y, and z.

Alternatively, the definition I'm most familiar with defines the Lorentz group as the group of all 4x4 matrices L such that L[sup]T[/sup] n L = n, where L[sup]T[/sup] is the transpose of L and the diagonal matrix n = diag(1, -1, -1, -1) is the Minkowski metric. By investigating its Lie group, it is not difficult to show that the only solutions to the latter definition are the various combinations of boosts, rotations, parity transformations, and time reversals. It's the definition that you're most likely to see in textbooks covering Minkowski geometry and it is under these transformations that "relativistic" theories are invariant by construction, so if hypothetically it was discovered that these various definitions of the Lorentz group were inequivalent, L[sup]T[/sup] n L = n would be the definitive one.

As mentioned earlier, these constraints are satisfied by the dilation x' = a x and t' = a t for a constant a, which isn't a Lorentz transformation. Both the mathematical definitions I posted above exclude it. Also note that light pulses moving in opposite directions in one frame in general don't in all frames. For example if you fire two light pulses in opposite directions along the y axis, they'll both gain an equal x velocity component under a boost along the x axis.

I'd want theories of atmospheric physics and aerodynamics in an external gravitational field that could explain both. Turning a blind eye when experimental results contradict your theories is a very bad way to learn anything in physics.

Also relativity isn't just based on the observed invariance of the speed of light. At least one of the "direct" predictions of relativity - time dilation - has been observed in clocks (Hafele-Keating, GPS) and the lifetimes of relativistic particles (eg. muons). Even this is just the icing of the cake - at heart, relativity is a theory that imposes symmetries on other theories, so the real test of relativity is how consistent relativistic theories are with reality and compared with their non-relativistic counterparts (where they exist). For instance it was already known by the 1930's that the relativistic model of the Hydrogen atom yielded better predictions for the atom's energy spectrum than the corresponding non-relativistic model did, and it is well known that non-relativistic Newtonian mechanics simply fails at high speed - in particular relativistic particles obey the relativistic relations for energy and momentum, while the non-relativistic definitions of energy and momentum lose all their physical significance under these conditions.

Historically, some of the best evidence comes from relativistic quantum mechanics. As you know, Lorentz transformations mix the space and time coordinates between different frames, and it turns out that imposing Lorentz invariance on quantum mechanics has some surprising qualitative implications. The most basic is that, in quantum field theories, relativity severely limits the classes of particles we can have (usually to scalars, spinors, and vector bosons in interacting theories), and (in conjuction with a few other requirements such as renormalisability and the existence of an energy ground state) limits us to a handful of allowable interaction terms. And yet these are enough to model all known non-gravitational interactions that have been observed in high energy particle experiments.

In addition, in the context of quantum field theories, relativity implies the spin-statistics theorem and the CPT theorem. The former predicts the observation that integer spin particles (bosons) should occupy symmetric states and obey Bose-Einstein statistics, while half-integer spin particles (fermions) should occupy antisymmetric states (Pauli exclusion) and obey Fermi-Dirac statistics. The latter predicts invariance under combined charge conjugation, parity, and time reversal. CPT notably implies that particles and antiparticles should have the same masses and mean lifetimes, and the fact that CPT symmetry seems to be satisfied isn't trivial given that C, P, CP, and T symmetry have all been observed to be individually violated. Relativity also implies (or at least very strongly hints at) the existence of antiparticles. Dirac's interpretation of the "conjugate" particles to electrons appearing in his attempt at a relativistic quantum theory as an "anti-electron" notably predated the experimental discovery of the positron by about a year. While we don't have much use for Dirac's theory nowadays, antiparticles similarly appear naturally in more modern relativistic theories.

Basically, wherever we test relativity we get results consistent with relativity despite all the constraints it puts on physical theories. If you want to argue that relativity is wrong, which really just boils down to arguing that physical theories "can't" possess a certain symmetry despite the enormous evidence that they do (this brief listing of evidence I've given isn't exhaustive - it's just what I happen to know off the top of my head), you've got quite a few coincidences to explain even if you ignore quantitative results.

Now you're doing essentially the same thing in reverse: you're flipping the sign of x at the same t' instead of the same t.

To throw some numbers at the problem, let's say v/c is approximately 0.866 such that the Lorentz factor is 2, and work in time units of ct for simplicity. Suppose you emit two photons at (ct, x) = (0, 0) in opposite directions along your x axis and both photons explode (or some other event occurs to them) simultaneously at time ct = 1. Then the two photon explosions take place at the coordinates (ct, x) = (1, 1) and (ct, x) = (1, -1). If you apply a Lorentz boost to these two points, you find that in a moving observer's frame the two explosions occur respectively at (ct', x') = (0.286, 0.286) and (ct', x') = (3.73, -3.73) - ie. both photons explode at different times different distances from the origin (relativity of simultaneity), but their velocities were +/- c in both frames.

If you like, here are a few intermediate steps you could add to Einstein's algebraic derivation, or simply another way of approaching the problem. The idea is to write the problem in "light-cone coordinates", say f = ct + x and g = ct - x for all x and t. These are depicted on the following diagram:

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In these coordinates, all light pulses follow trajectories that either take the form f = constant or g = constant, such as the one depicted on the above diagram. Obviously the most general coordinate transformation that's going to preserve the speed of light is simply to rescale the f and g coordinates: f' = A f, and g' = B g for constants A and B. What happens when you rewrite this transformation in terms of x, ct, x', and ct'?

No problem - I had to finish off an exam session this week anyway.

 

No, I don't mind if you keep on using LaTeX. I can still log on to Sciforums and get the full post from there.
I just wanted to say that I personally prefer to write my posts as platform-independent as possible in case I want to use them elsewhere.

Thomas

 

Note: If I have not replied to any of your points explicitly, then because I feel it is implicitly answered by my other replies (this is my general policy when posting).

A vector is only frame-dependent if you are implicitly making that assumption.
Consider for instance the Coulomb law in the frequently written form F=q1*q2/r^2 . I am sure you are not claiming here that the radius vector r has to be understood as a frame-dependent quantity, because it would mean the Coulomb force between two particles would depend on how you choose the origin of your coordinate system. Obviously, the vector r has here to be understood as the relative vector between the two particles, which is frame-independent. Strictly speaking, one should thus use r=r1-r2 here.
Likewise, it would be similarly non-sensical to assume the the velocity v in the Lorentz-force F=q*vxB is frame-dependent. Since we are here as well dealing with a law describing the force of a certain system on a particle, v should also be a corresponding velocity difference, i.e. be independent of the reference frame.

If the Stern-Gerlach effect can not be detected for free electrons due to the interference of other effects, then this still leaves the possibility that in this case it does not exist at all, simply because a free electron may have no magnetic moment. As far as I am aware, all the evidence for the magnetic moment comes from electrons bound in or interacting with atoms or ions.

Nothing would be going on locally in electrodynamics if there wouldn't be something going on elsewhere.

These derivations have not taken charge invariance into account at all. If they had, they would have come to the same result like I in my own derivation of the relativistic wire charging, namely that there can not be an overall charging of the wire.

Satellites have no constant speed because they are accelerated by the gravitational force. There is no such physical force that could accelerate the electrons in a current loop.

That claim is not consistent with your previous one where you stated stated that, in case of the current loop, an electron located at the corner in the rest frame would be somewhere else in the segment in the boosted frame (which would imply that the electron is in the vicinity of different ions in both frames).

And the reason for this seems to be the point I mentioned earlier, namely that it is incorrect to have a particle moving in the rest frame. In case of the electron, you would have to choose the rest frame such that the electron is actually at rest, and then do the boost from there. This obviously means that you have two different boost velocities for the ions and electrons, and thus the locations won't coincide anymore after the boosts.
So the Minkowski diagram is actually incorrect unless the world lines are vertical.

Thomas

 

Well, x'=x and t'=t (a=1) would obviously satisfy reciprocity, and that is exactly what the invariance of c implies through the constraints x=ct <=> x'=ct' and x=-ct <=> x'=-ct' (where x and x' are the coordinates of a light signal only, not coordinates of material bodies (which are described by the Galilei transformation)).

That is a conclusion that follows from the Lorentz transformation, so it would be a circular argument trying to justify the Lorentz transformation by this.

It's bad enough that he doesn't explicitly say this, because it shows that his 'derivation' isn't actually a derivation. In fact, the way his (and many other) 'derivations' of the Lorentz transformation works is to assume a set of constraints so that the Lorentz transformation must emerge as one solution. However, these constraints have nothing to do with the constraints imposed by nature in the form of the invariance of c. They are much weaker than the latter and thus in fact invalidate it in general. I have illustrated this point on my page discussing Einstein's algebraic derivation of the Lorentz transformation by the following simple mathematical analogy (I have fictitiously attributed this to Einstein to make the analogy clearer):


Consider the equation

(I) y(x)=ax + b

with the constraints

(C1) y(1) = 1
(C2) y(-1) = -1

The task is to determine the coefficients 'a' and 'b' by applying the constraints to (I). Now since (C1) results in 1=a+b and (C2) in 1=a-b, it is obvious that this requires b=0. But since b=0 is not what Einstein likes, he decides to modify the constraints such that (I) is valid for all b, i.e. he changes (C1) and (C2) to

(C1') y(1) = a +b
(C2') y(-1) = -a +b .

"Fine" Einstein says, "now I have a system of equations that is consistent for all 'b' (and 'a' at that)", but unfortunately it has nothing to do with the problem anymore. The task was not to find a set of constraints that are consistent with (1) irrespective of the value of the coefficients, but to apply the constraints (C1) and (C2) to (I) and thus to find the coefficients.

There is a further choice: you can have two transformation: the usual Galilei transformation for the coordinates of material objects, and the identity transformation (x'=x) for the coordinates of light signal. Indeed, why would you assume that you could cover both with just one transformation given the fundamentally different nature of light compared to material objects?

x' (or b if you want) is not related to clocks. It is the (fixed) coordinate of the mirror in the moving frame, so it can not be an independent argument of tau (because the arguments of tau should be rest frame coordinates).
If you use x' (rather than x'+vt) as the first argument, and then apply the chain rule correctly to Einstein's 'master equation' (as done on my page discussing Einstein's 1905 derivation of the Lorentz transformation), you will find that then you won't even get a correct result for v=0 (as far as I recall, there is then an additional factor 2 in the equation; it took me quite a while to figure out where this factor comes from).

As indicated above, you can obtain the Lorentz transformation only by making weaker assumptions than the constraints for the propagation of light demand (x=ct <=> x'=ct' and x=-ct <=> x'=-ct'); and the quadratic form is just one way to have a weaker constraint. But it means that the Lorentz transformation is actually in general not consistent with the original constraint.

If you have line of light detectors with synchronized clocks at the same distances from the origin going into opposite directions, and send out a light signal from the origin into both directions, then, according to the principle of the invariance of c, you should find that the time corresponding detectors (at the same distance) show for the arrival of the light signal is identical, irrespective of whether the light source was stationary or moving when the light signal was emitted. In other words, in any case, the positions of the two light signals must be symmetrical to the origin. Or if you want to formulate this in terms of two reference frames, if you flip the coordinate in one reference frame, it must also flip in the other.

The point I made on my page regarding Einstein's algebraic derivation of the Lorentz transformation is that f and g can not be simultaneously defined for all x and t. f only applies to negative x, g only to positive. Otherwise you get an algebraic inconsistency on the light cone (because you can not have simultaneously x=ct and x=-ct).

The point I was trying to make is that if you observe that one kind of objects (photons) move fundamentally differently to all other objects, it is likely that in fact they are different and thus a different velocity transformation applies in the first place.
One has to remember that our usual concept of 'velocity' (i.e. the velocity dependence of the coordinate transformation) is derived from the behaviour of material objects only, so why would you assume that an immaterial object (photons) should have a velocity dependent transformation as well (when in fact experiments point to the contrary)?

First of all, if a theory is conceptually or otherwise logically inconsistent, a comparison of its predictions to experiments is rather academic and meaningless. The Ptolemaic (geocentric) system of the universe was supported by observations for 2000 years, but nonetheless it turned out to be a wrong theory in the end. So even if your point of view is wrong, the data might still seem to largely confirm it.

Of course, the question is how to explain the experiments that seem to confirm Relativity alternatively. But without having inside knowledge of how these experiments were in detail performed and evaluated, this is difficult if not impossible to decide conclusively. So without further experiments in this respect, this will be to a large degree speculation. But it could for instance be that the time dilation observed for moving clocks and muons is due to the fact that they move in an magnetic field. The potential energy change due to the Lorentz force could simply change the rate of the atomic clock by the observed amount. And the lifetime of muons might be somehow affected by the Lorentz force as well. On my page A Newtonian Relativistic Electrodynamics I have also shown that by making the electro- and magnetostatic forces suitably velocity dependent, one could exactly replicate the observed dynamics of charged particles that is for instance observed in accelerators.

Also, as far as I am aware, by no means all experimental data show clear evidence for the existence of time dilation both in Special and General Relativity, and those that are usually cited in support of this effect often have substantial systematic and/or statistical uncertainties (I can't give any references for this at the moment; this was just the general conclusion to which I came when I looked into Relativity experiments in more detail a few years back).

But again, these experimental aspects do not really touch on the theoretical issues we have been discussing.

Thomas

 

The transformation properties of the components of all quantities, under a given transformation, should be completely specified by their geometry. For instance, under a Lorentz boost the unit x vector, expressed in terms of a boosted frame's basis vectors (in units where c = 1), is:

\( \bar{1}_{x} \,=\, \gamma \bigl( \bar{1}_{x^{\prime}} \,-\, \beta \bar{1}_{t^{\prime}} \bigr) \)​

so a purely spatial vector of the form \(\bar{v} \,=\, v_{x} \bar{1}_{x} \,+\, v_{y} \bar{1}_{y} \,+\, v_{z} \bar{1}_{z}\) will in general gain a t-component in a boosted frame. I illustrated this point graphically in my previous post: no vector quantity can be relativistically invariant.

As I've already stated, this implies that if you want equations that are relativitically invariant by construction, you'll need to adopt four-vectors. In fact there's at least one nice example of how this works even in non-relativistic physics: suppose you set \(\bar{j} \,=\, \rho \bar{1}_{t^{\prime}}\) in some particular rest frame. It's not difficult to see that the Galilean relation for unit time vectors is eg. \(\bar{1}_{t^{\prime}} \,=\, \bar{1}_{t} \,+\, v \bar{1}_{x}\), so the same vector can be written as \(\bar{j} \,=\, \rho \bar{1}_{t} \,+\, \rho v \bar{1}_{x}\). This geometric definition automatically gives charge densities and currents the correct transformation law under Galilean transformations (you automatically get eg. \(j_{x} \,=\, \rho v_{x}\)). This also determines how any other transformation, including Lorentz transformations, will act on associated densities and currents. In fact, in the case of a current density in a neutral wire you can even take my previous diagram as a simple graphical representation of the relativistic charging of a wire, as the current vector gains a component in the time direction.

There's no need to "understand" anything. Any correct statement of Coulomb's law will define r as the distance between the charges. In the Lorentz force, the velocity is just the velocity of the particle in whatever frame is being considered. This isn't open to interpretation: if you interpret the velocity in the Lorentz force as anything other than the frame-dependent velocity of the charge, you are describing a theory that isn't the theory of electrodynamics developed by Maxwell and others.

Also a frame-dependent velocity isn't necessarily a problem in a theory. For instance, in Newtonian physics, the kinetic energy and momentum of a massive particle are both explicit functions of the particle's velocity. Both quantities are frame-dependent (E even varies quadratically with velocity) and play a fundamental role in dynamics (so a system's Hamiltonian is frame-dependent), yet this does not result in a violation of Galilean invariance.

Or not. Unless you have an alternative that can reproduce the predictions that current explanations provide, you don't really have a case. (Then there's also Pauli exclusion: the only way two electrons can occupy the same spatial state eg. in an atom is if they possess an additional internal degree of freedom - namely their spin.)

So? How does this address the fact that your operational definition of the magnetic field isn't locally defined? I'm not arguing against the fact that events in one place can affect events somewhere else (at a later time).

There's no issue to take into account:

• Infinite wires don't have a well defined total charge, so there can't be any issue with charge invariance.
• Finite wires charge without violating charge invariance (I proved this in a previous post using the Lorentz transformation), so those derivations could use a long but finite wire without violating charge invariance.

The authors only use infinitely long wires for simplicity. Your accusation that it somehow allows them to "get away" with violating charge invariance is completely unfounded, since in relativity there's no issue with charge invariance to begin with.

I already explained that your transformation of the wire was incorrect. I don't know if you followed the calculation I gave, but what I showed using the Lorentz transformation was that a finite wire would transform like this:

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The result of your own derivation only makes sense if you're considering a beam of electrons flying through space, entering a wire of ions, passing through it, and flying out the other end. If you are, your argument based on the asymptotic behaviour of the fields (which is already quite dubious to begin with) fails. If you aren't, you can't apply the length contraction formula to the electron beam. Either way you haven't demonstrated anything.

So how do they turn the corners then? Why aren't electrons constantly leaking out of all the wires around me? Reality would seem to disagree with you here since real wires apparently do manage to produce the forces needed to guide electrons around, even if you move the wires.

Also, what exactly is the point you're trying to make? You originally introduced this rectangular wire in order argue a point about the asymptotic behaviour of the electromagnetic field in different frames. Are you now saying that your own argument is based on a nonphysical setup and should be dismissed?

My earlier statement explained this:

which is perfectly consistent with what I'm stating here. Suppose electron A passes ion A as it is turning the left corner at space-time coordinates \((ct=0,\, x=0)\) (call this "Event A") at the same time as electron B passes ion B located at the right-hand corner ("Event B") at \((ct=0,\, x=L)\). Then, according to the Lorentz boost, the coordinates in a boosted frame might be \((ct^{\prime}=0,\, x^{\prime}=0)\) for event A and \((ct^{\prime}=\gamma\beta L,\, x^{\prime}=\gamma L)\) for event B. This alone says nothing about where electron A will be at \(ct^{\prime}=\gamma\beta L\) or where electron B is at \(ct^{\prime}=0\), hence there's no contradiction.

This is precicely why you shouldn't apply multiple mutually inconsistent transformations between the same two frames. It's a brilliant way of guaranteeing you'll get nonsensical results. That's why nobody has ever suggested such an interpretation or theory.

I'm applying Minkowski diagrams and Lorentz transformations the same way everyone else does. This is why many of your arguments against relativity are so unconvincing: everyone would clearly see relativity is inconsistent - if only everyone would adopt your inconsistent interpretation of it.

 

My point was that other values of a also satisfy these constraints. For instance, x' = ct' maps onto ax = act, which is the same trajectory as x = ct. Yet these aren't Lorentz transformations.

I wasn't trying to justify the Lorentz transformation on the basis of a limiting velocity. I meant that this particular consequence of relativity is consistent with what we know about reality. The existence of a limiting velocity (in the sense of a maximum velocity at which information can propagate) is never needed as a postulate in any derivation of the Lorentz transformation that I know of.

In the mathematical sense, no. The main point of his derivation is that there are extremely severe restrictions on the kind of relativity theories we can have. You can have either Galilean or Lorentzian relativity, but that's it. No-one will ever discover another possible relativity theory which satisfies his general constraints.

In fact it shouldn't be surprising that such a generic derivation (based on general constraints like homogeneity and causality) should include Galilean invariance as a solution. After all, Galilean invariance is a perfectly reasonable symmetry for a physical system to obey, and it acted as a staple of physics for over 250 years. Only experimental evidence could rule it out.

The only constraint nature imposes on the speed of light is \(\frac{\Delta x}{\Delta t} = \pm c \Leftrightarrow \frac{\Delta x^{\prime}}{\Delta t^{\prime}} = \pm c\). I don't know how you managed to read the \(x = \pm c t \Leftrightarrow x^{\prime} = \pm c t^{\prime}\) constraints as implying the identity transformation. I certainly didn't. I read them as requiring:

\( \begin{align} x^{\prime}(ct,\, t) \,&=\, c t^{\prime}(ct,\, t) \\ x^{\prime}(-ct,\, t) \,&=\, - c t^{\prime}(-ct,\, t) \end{align} \)​

for all t, which is the invariance of c constraint and is exactly the problem that the Lorentz boost solves.

Since the problem is to find the relationship between two coordinate systems in motion with respect to one another (given an invariant c), the relationship between coordinates generally isn't going to be identical. For instance, the spatial origin x' = 0 of the "moving" frame must move along the trajectory x = vt and so generally isn't going to coincide with the origin of the "rest" frame at x = 0. You also see this in your own setup:

Suppose each detector in each line also has its fixed x or x' coordinate painted on it. You can easily see that the only way a light signal at some point can have the same x coordinate in both frames is if it is detected by two detectors which both have the same x coordinate painted on them. This implies that each detector in the "rest" frame must pass the moving detector with the same coordinate when both detect the light pulse. In general this is only possible if the lines of detectors are stationary with respect to one another, which isn't the problem Einstein is trying to solve. So it's only your identity constraint that contradicts the problem at the outset.

Because the coordinate transformation isn't a property of photons or material objects. It is the unique relation between the coordinate systems defined by two observers. The only way you can have two transformations is if one of the observers either deliberately or unknowingly defines different coordinate systems for measuring light in than everything else. But we don't do this. When we say "light travels 300,000,000 metres every second" we're applying the same definitions of the metre and the second as we do for everything else. In applying contradictory relations between the coordinates of inertial frames, you're either making a glaring mathematical error or you're entertaining particularly bizarre notions of space and time which would make even general relativity look tame.

Are you sure you've correctly understood the derivation? Statements on your site like this:

suggest you haven't. Einstein's "master equation" as you call it is a mathematical statement of the definition of synchronicity he describes in §1 of his paper. At this stage of the derivation, Einstein has not yet imposed the invariance of c condition (at least not beyond assuming he can use light signals to synchronise clocks).

Where are you getting these rules from? Mathematically, defining \(\tau\) as a function of x' also implicitly defines \(\tau\) as a function of x. Of course the final result has to be expressed in terms of x, but in all the intermediate stages of the derivation Einstein is free to use whatever parameterisation of the function he wishes. For some reason you seem to think Einstein is not entitled to apply basic mathematics in his derivations.

By the way, if you absolutely insist on exclusively using rest frame coordinates, then Einstein's synchronicity condition becomes:

\( \frac{1}{2} \Bigl[ \tau\bigl(vt,\, t\bigr) \,+\, \tau\bigl( v(t \,+\, t_{2}),\, t\,+\,t_{2}\bigr) \Bigr] \,=\, \tau\bigl( v(t\,+\,t_{1}) \,+\, x^{\prime},\, t\,+\,t_{1}\bigr) \)​

The difference with what's posted in your subsequent derivation is that, on the right hand side, \(\tau\) is evaluated at the x coordinate v(t + t[sub]1[/sub]) + x' (the location of clock B at time t + t[sub]1[/sub], since its trajectory is x = vt + x') rather than just v(t + t[sub]1[/sub]). If you have doubts about consistency, it's not difficult to check that the Lorentz boost does indeed satisfy this constraint.

I don't see what you mean by this. The relation between (t, x) and (f, g) is bijective and well-defined for all t and x.

True, but this is potentially true of any theory. There's also an important distinction between a theory that can explain a set of observations it has been fitted to explain (ie. curve fitting), and a theory that makes genuine predictions. Historically, whenever the accuracy of astronomical observations improved, the Ptolemaic model had to be modified to account for new observations and nowadays "adding epicycles" has become a synonym for bad science (specifically, trying to save a sick theory by adding parameters and complexity to it).

On the other hand, special relativity predicted time dilation, the relativistic energy/momentum relations, and (combined with quantum mechanics) even the existence of antiparticles before each was discovered experimentally, and relativity has survived a century of observations (most notably in high energy physics) without modification. By any reasonable standard, relativity is a resounding success as a theory. The worst that's likely to happen to it now is for it to go the way of Newton, ie. to turn out to be a good approximate symmetry of physics within a limited experimental domain.

Relativity isn't just confirmed by one or two experiments. The whole of the Standard Model is Lorentz invariant, and it is routinely tested under conditions which should easily distinguish relativity from eg. Galilean invariance. Lorentz invariance is basically considered an experimental fact at least up to about TeV energy scales, or equivalently down to distance scales about [sup]1[/sup]/[sub]1000[/sub][sup]th[/sup] the diameter of the proton. Relativity is basically a fact of life for particle physicists.

Also if you want to replace relativity you've also got to do quite a lot better than just reproduce relativity's correct results. Your alternative should be simpler (in the sense of Occam's razor) and correctly anticipate results that haven't been observed yet that relativity fails to anticipate. If your alternative needs more parameters, lacks theoretical justification, and can only explain results it has been fitted to explain, you don't impress anyone.

This is particularly unlikely for muons. While clocks are macroscopic objects that are indeed largely held together by electric forces, muon decay is a weak interaction process and has little or nothing to do with electrodynamics. And although I don't know specific examples off the top of my head, I'd be surprised if muons were the only particles for which time dilation was observed - I'd expect it's routinely observed in some by-products of collision experiments for instance.

Does this:

mean you think CERN, FermiLab, DESY, and SLAC technicians and engineers don't know the power of their own beams?

For Hafaele and Keating maybe (I also had the impression that the uncertainties were quite large), but I understand that inaccuracies would quickly build up in the GPS system if relativistic effects weren't accounted for. But this is why I mentioned high energy physics as one of the best tests of relativity. In particular, quantum electrodynamics is relativistically invariant and is generally acknowledged as having produced some of the most accurately tested predictions in the history of physics. Do you really feel up to the task of suggesting a viable alternative to modern relativistic quantum field theory?