Relativity fails with Magnetic Force

Since when are the language and conventions of physicists determined by you? Only you are having difficulty with the Lorentz force, so here I am, telling you specially because only you need it, that the velocity in the Lorentz force in Maxwell's theory is frame-dependent and was never intended to be an invariant quantity. Anyone with a good background in physics and a decent understanding of electrodynamics will tell you the same.

Note by the way that in addition to all the problems with your alternative interpretation of electrodynamics you keep ignoring, you've completely ignored the issue of internal consistency. You have no right to assume that you can just piggyback on to the equations of a successful theory like electrodynamics, "reinterpret" them in order to get properties you like, and that you'll end up with a consistent theory. In fact it is easy to see that in your case interpreting Maxwell's equations in such a way as to otain invariant electric and magnetic fields (your motivation for interpreting the Lorentz force velocity as invariant) results in a self-contradictory theory.

Take the simple case of a single point-like charge moving through a vacuum for instance. The electric field around this charge will be time-dependent since it moves along with the charge. If you're measuring the electric field, you'll find it increases as the charge approaches you, reaches a maximum, and drops again as the charge passes you by. So we have \(\frac{\partial \bar{E}}{\partial t} \neq 0\). Also according to you, there should be no magnetic field around a moving charge (since it's invariant and there's no magnetic field around the charge in its rest frame). So \(\bar{B} = 0\) and therefore \(\bar{\nabla} \times \bar{B} = 0\).

However, one of Maxwell's equations (specifically Maxwell's correction to Ampere's law) states we should have:

\(\bar{\nabla} \times \bar{B} \,=\, \frac{1}{c^{2}} \, \frac{\partial \bar{E}}{\partial t}\)​

everywhere in the vacuum around the moving charge, so your theory contradicts itself.

Similarly, requiring \(\bar{E} = 0\) around a moving bar magnet would lead to a contradiction with Faraday's law of induction:

\(\bar{\nabla} \times \bar{E} \,=\, -\, \frac{\partial \bar{B}}{\partial t}\)​

Maxwell's equations require frame-dependent electric and magnetic fields for consistency.

So to summarise your situation here, you are proposing a more complicated, poorly defined, and internally inconsistent theory in place of electrodynamics with no evidence in your favour and which is inconsistent with the rest of physics. Yet you still insist on discussing this as if you had a viable contender to electrodynamics?

Yes it does. Classical electrodynamics is not regarded as a fundamental theory and it certainly doesn't exist in isolation. It is nowadays seen as a macroscopic approximation to quantum electrodynamics.

Kepler's laws predated Newtonian gravitation. Does that mean it's OK to blindly postulate a "reinterpretation" of Kepler's laws that's inconsistent with the underlying theory of Newtonian gravitation?

It is. It's just you who isn't familiar with electrodynamics. You can't deny this: two or three pages ago I found out you didn't even know about the energy/momentum of the electromagnetic field (their densities, the Poynting vector, and so on), and you've still got a page up on your website somewhere stating the fields must be evaluated instantaneously (ie. not at a retarded time) in order to avoid violations of conservation laws, as if energy conservation in electrodynamics wasn't already a well-established topic. You've also got Coulomb and Biot-Savart's laws incorrectly listed as two of Maxwell's equations.

It is formulated in exactly the same way every other theory with a frame-dependent velocity is. Again, you can't claim an ambiguity in a definition only you are having difficulty with. The appropriate thing to do in such a situation is to find and ask someone with a better understanding of the theory than you.

This seems to be a recurring issue with you. A few years ago it took several posters three or four pages of thread just to explain you'd misunderstood a basic conditional probability concept in a thread originally about a simple, completely standard, and hardly controversial highschool-level homework exercise. It took over three pages because you kept trying to "reinterpret" the problem in a way that would produce the incorrect answer you'd found, despite the repeated explanations by people actually familiar with the relevant probability theory. You're doing exactly the same here with electrodynamics and your personal interpretation of the invariance of c problem.

A velocity referred to some coordinate system is a velocity with respect to any physical object at rest in that coordinate system. The "abstraction" is only in the realisation that it doesn't really matter what those physical "marker" objects happen to be.

I haven't said or implied anything about the Stern-Gerlach effect for over three months now. Why are you putting words in my mouth?

If you aren't implying there must be an alternative explanation then where's the issue?

What "misunderstanding"? It's again you who is unfamiliar with concepts most physicists would be expected to be familiar with. Locality just means you don't have a quantity or field at position \(\bar{x}\) coupled in an irreducible manner to some quantity at position \(\bar{y}\) some finite distance away. The integral form of Maxwell's equations can be reduced to the differential form which is purely local.

Note that while non-locality in a field theory is undesirable in itself, the real issue is that your interpretation of the Lorentz force results in a non-local operational definition of the magnetic field. Why should the magnetic field here be defined partially in terms of what a source perhaps kilometres or even lightyears away that has already emitted the magnetic field is doing? There's just no good reason to define the magnetic field this way, and we simply don't.

I'm not saying anything about those derivations. In post 81 you stated:

This is an outright denial that the finite wire I applied the Lorentz transformation to has a charge density of \(- \frac{\gamma j v}{c^{2}}\) everywhere except at its endpoints in the boosted frame - and with no overall violation of charge invariance. You've been denying or avoiding this ever since.

Your treatment is itself thoroughly unconvincing hand-waving and doesn't do Minkowski's formulation of electrodynamics any justice.

I've already proved quite rigorously that your transformation of the finite wire (which you don't even adequately describe on your site) was incorrect. This is a much more serious error than your application of electro- and magnetostatics to a non-static source. The wire that should appear on your page is described in post 66 and illustrated in a diagram in post 79. It a) doesn't support your conclusions (it's not an electric quadrupole) and b) could be substituted into the derivations you link to (in place of their infinite wires) without altering their conclusions, just by arranging for the endpoints to be sufficiently far away from the test electron.

In other words, you're applying an inverse Galilean transformation to a moving relativistic wire in order to apply energy conservation to it in... which frame? Do I really need to explain what you're doing wrong here?

 

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The energy and momentum of massive particles diverge to infinity as they approach the speed of light, so they'll never be able to exceed c. There's well over half a century of experimental high energy physics confirming this behaviour.

The trajectory of each moving buoy is its coordinates as measured by an array of buoys at rest. Say that at time t the moving buoy with the coordinate x' painted on it (the "x' buoy") is passing the rest buoy of coordinate given by:

\(x \,=\, vt \,+\, b(x^{\prime})\)​

where b = b(x') is an invertible function telling us exactly which trajectory the x' buoy is following. Each x' buoy is equipped with a clock who's display can be described by some function T = T(x'; t), giving the time displayed on it at time t. Then the coodinate transformation, which is completely determined by these two functions b and T, is:

\( \begin{align} x^{\prime}(x,\,t) \,&=\, b^{-1}(x \,-\, vt) \\ t^{\prime}(x,\,t) \,&=\, T \bigl(b^{-1}(x \,-\, vt) ;\, t \bigr) \end{align} \)​

What I've described here is exactly what a coordinate transformation is intended to be. If a light pulse is measured in both frames using each frame's array of buoys, then the light pulse's coordinates transform according to the above transformation.

I'm only considering one detector and light pulse viewed in two reference frames. If you use different coordinate transformations for the light signal and detector, you find that the light signal gets absorbed somewhere other than where the detector is located in a moving reference frame.

For a similar example, imagine you have two mirrors at rest located at x[sub]1[/sub] = +L and x[sub]2[/sub] = -L, with a light signal bouncing periodically between them. Suppose we view this from the perspective of someone moving at velocity v with respect to the mirrors. According to the Galilean transformation, the mirrors move along the trajectories x[sub]1[/sub]' = -vt + L and x[sub]2[/sub]' = -vt - L, yet according to you the the light signal's trajectory is invariant and it would still bounce periodically between +L and -L. In the moving frame the light gets left behind by the moving mirrors and just reflects off thin air.

If you don't apply the same coordinate transformation consistently to everything, you predict nonsense. I've been telling you this for months now.

You're just repeating your old false analogy here. In fact I'm not even clear on how you think the Lorentz transformation applies to this setup at all: the Lorentz transformation is supposed to relate the coordinates of a given event in two different reference frames. Here you're just considering multiple light signals in one reference frame. Just what are you applying the Lorentz transformaton to?

You've derived this constraint based on your false analogy. The Lorentz transformation simply doesn't flip the sign of the x' coordinate when you flip the sign of x. That's OK, since it doesn't leave t' fixed. It only needs to flip the ratio \(\frac{x^{\prime}}{t^{\prime}}\) when \(\frac{x}{t}\) is flipped. Anything more and you are imposing constraints on more than just invariance of the speed of light.

In fact if you consider the Lorentz transformation's predictions for the transformation of light signals a little more carefully, you should see that it is just (correctly) predicting the Doppler effect.

Evaluating \(\tau(x^{\prime} \,+\, vt,\, t)\) doesn't produce nonsense. If you want to keep track of the variable substitution explicitly, then you can formally define:

\(\psi \,:\, \mathbb{R}^{2} \,\rightarrow\, \mathbb{R}^{2} \,:\, (x^{\prime},\, t) \,\rightarrow\, (x^{\prime}+vt,\, t)\)​

If you like, what Einstein effectively does is determine the function \(\tau \,\circ\, \psi\) (up to a multiplicative constant), which is a function of x' and t. Since \(\psi\) is invertible, you get \(\tau\) back by composing with \(\psi^{-1}\) (ie. substituting back x' = x - vt):

\(\tau \,=\, (\tau \,\circ\, \psi) \,\circ\, \psi^{-1}\)​

or:

\(\tau(x,\, t) \,=\, \tau \circ \psi (x-vt,\, t)\)​

Fine then. If you have an equality f(x') = g(x') and derive both sides with respect to x' you get another equality:

\( \frac{\text{d}f}{\text{d}x^{\prime}}(x^{\prime}) \,=\, \frac{\text{d}g}{\text{d}x^{\prime}}(x^{\prime}) \)​

Happy? Differentiating a constraint will never produce a new constraint, regardless of what you happen to be calling the variable you're differentiating with respect to.

It isn't supposed to. Read Einstein's paper. He only later determines \(\gamma\) by imposing reciprocity (starting at "For this purpose we introduce a third system of co-ordinates K'"). He requires that a transformation of parameter v followed by a transformation of parameter -v produces the identity transformation. Only the Lorentz transformation satisfies this in addition to invariance of c.

Einstein mastered Riemannian geometry in his lifetime. I'm pretty sure you haven't.

Then you've misunderstood the invariance of c condition. The constraints x=ct <=> x'=ct' and x=ct <=> x'=ct' can be rearranged to:

\( \begin{gather} \frac{x}{t} \,=\, c \; \Leftrightarrow \; \frac{x^{\prime}}{t^{\prime}} \,=\, c \\ \frac{x}{t} \,=\, -c \; \Leftrightarrow \; \frac{x^{\prime}}{t^{\prime}} \,=\, -c \end{gather} \)​

It is a constraint only on the ratio of the spatial and temporal coordinates. Requiring x' = x and t' = t would be better described as an "invariance of the trajectory of light" condition. There is no justification for imposing this.

Circular reasoning. Your claim of a problem is based on your assumption that they're only defined on the light cone.

No they don't. f = 0 and g = 0 are just the trajectories of two different light signals. Note that Einstein also explicitly states:

which is perfectly true. If anything, imposing these conditions everywhere (which does permit Einstein to add and subtract them) would be a stronger constraint than merely imposing these conditions on the light cone (though in practice it's no stronger than necessary - we need invariance of c for all trajectories and not just those that happen to pass through the origin). You can check that there's really no foul play here: the Lorentz transformation really does satisfy these constraints for all x and t, specifically for:

\(\lambda \,=\, \frac{1}{\mu} \,=\, \frac{\sqrt{1 + \frac{v}{c}}}{\sqrt{1 - \frac{v}{c}}}\)​

Intuitively, these parameters measure stretching/compression along the light cone. You might recognise them as none other than the relativistic Doppler shift factors, which we know for a fact are velocity-dependent and generally different from 1.

No, but experimentation is what physics is based on. If you're indifferent to experimental results that confirm relativity, and you insist that nonrelativistic physics must be correct (truth?) despite the complete absence of evidence for it under relativistic conditions, you're not doing physics.

If someone claimed they knew a trick for winning the lottery and then went ahead and won it ten times in a row, would you put that down to coincidence?

This "point" is so frequently abused that it's made its way as #16 on the Baez crackpot index. Creationists in particular are notorious for trying to denounce Evolution as "just a theory".

Einstein's constraints are the ones experimental results require. Yours are not. Why do you think Relativity was ever accepted as a theory in the first place? Do you really think we all can't read what constraints Einstein imposes?

Informed questioning is a good thing. Uninformed "questioning" is pointless and annoying (think "backseat driving"). Unlike you, I've learned a lot of the basics of experimental high energy physics from people actually involved in the CMS experiment at the LHC and the former DELHI experiment at LEP, as well as IceCube (not all high energy physics is done in accelerators). I've seen and manipulated the kind of data that accelerators collect. My opinions are informed ones based on knowledge of the methods of this branch of experimental physics. On the other hand, your clearly uninformed opinions are based on nothing more than the fact accelerator experiments consistently produce results you don't want to have to deal with.

I'd say it's rather you who's forgetting that the primary purpose of accelerators is to measure interaction cross-sections. Take LEP (an electron-positron collider) for instance, which was the accelerator I had in mind when replying earlier. At LEP, the beam intensities were low enough such that no collisions occured in the majority of bunch crossings. If you're under the impression that the calorimeters were detecting some total energy deposit due to hundreds, thousands, or more simultaneous collisions and that I'm dividing this energy by some theoretical "number of collisions" to get the "energy per collision", your impression is wrong. Collisions at LEP were single, isolated events. I think there were at most a few collisions at LEP every second (out of around 45,000 bunch crossings per second), with the frequency of multiple simultaneous collisions dropping off rapidly following a Poisson distribution. To justify the 91 GeV collision energy figure, one of the simplest events at LEP consisted of the detection of an electron and a positron each emitted back-to-back at 45 GeV from the collision point.

I note you also ignored the beam dumps and that there are known examples of damage that the beams can cause. There are other indicators: for instance if charged particles were less energetic they would be much more sensitive to energy loss due to ionization deposits and would spiral much more quickly when passing through matter in a magnetic field than they do (the rate of change of curvature due to energy loss is actually measured and used to reconstruct parameters such as a particle's mass in at least some experiments). If the centre-of-mass energy at LEP were only 0.5 MeV, the secondary particles emitted from collisions most likely couldn't even punch their way through the tracker system, much less leave localised deposits of dozens of GeV in the calorimeters. Also, sometimes the interaction causes the beam particles simply to deflect off one another and they themselves end up in the calorimeters at a small emission angle from the beam, so their energies can be measured directly. In proton colliders, it's actually the individual quarks that take part in interactions and the remaining fragments of the original protons sometimes show up in the detectors at low angles in the form of hadronic jets.

The point is, there are many independent indicators of the energies of accelerator beams available to experimenters. Your idea that accelerators could have been circulating 0.25 MeV electrons and 450 MeV protons for the last fifty years without anyone noticing is just plain ignorant.

By the way, you know that you don't need to go to SLAC or CERN to see relativistic electrons in action, right? You can find them at kinetic energies of up to around 300 keV in transmission electron microscopes, and electrons of at least several MeV are used in radiotherapy. We've literally got relativistic electrons out there treating cancers.

So now you're resorting to conspiracy theories? This is really getting ridiculous.

 

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\(\tau\) is a function of the two independent parameters x and t.

Which, unless you think 2 = 3 = 4, is an incorrect result. The "substituting" you perform on the partial derivatives is not a valid mathematical operation. Open any textbook on introductory analysis and you'll find nothing that justifies it.

Whatever you think you're doing, you're doing it incorrectly. It should be evident from post 96 that your method could be applied to produce a contradiction using any function, so you've found nothing unique to the Lorentz transformation. If this isn't evident to you, let's do it explicitly. Suppose I rename my path:

\( \begin{align} x_{1}(t) \,&=\, t \\ y_{1}(t) \,&=\, 2t \end{align} \)​

and define a new one:

\( \begin{align} x_{2}(t) \,&=\, 3t \\ y_{2}(t) \,&=\, 0 \end{align} \)​

Then, for f(x, y) = x + y we have:

\( f\bigl(x_{1}(t),\, y_{1}(t)\bigr) \,=\, f\bigl(x_{2}(t),\, y_{2}(t)\bigr) \,=\, 3t \quad \forall t \in \mathbb{R} \)​

If I tell someone that I'm thinking of a linear function (ie. f(0, 0) = 0 and its partial derivatives are constant functions) that satisfies \(f\bigl(x_{1}(t),\, y_{1}(t)\bigr) \,=\, f\bigl(x_{2}(t),\, y_{2}(t)\bigr)\) for the paths defined above, they can derive both sides by t:

\( \frac{\partial f}{\partial x} \,+\, 2\, \frac{\partial f}{\partial y} \,=\, 3\, \frac{\partial f}{\partial x} \quad \text{(*)} \)​

which simplifies to \(\frac{\partial f}{\partial x} \,=\, \frac{\partial f}{\partial y}\), and they can correctly deduce the function I'm thinking of up to an overall multiplicative constant.

If this person had severe misconceptions about partial derivatives, they might try to "substitute" \(\partial y_{2} \,=\, \partial x_{2}\) into (*), yielding:

\( 2\, \frac{\partial f}{\partial x} \,=\, 3\, \frac{\partial f}{\partial x} \)​

which is a contradiction unless f is a constant function. Since I was thinking of f(x, y) = x + y all along, this person has effectively just proved 2 = 3. What should they then conclude? That the expression x + y is somehow "fundamentally flawed", whatever that means?

Well make up your mind. Does nothing satisfy the constraint or do a whole bunch of functions satisfy the constraint? You can't have it both ways.

 

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Hint: draw a free body diagram for a submerged object. Compare it with a free body diagram for one of the two weights on a set of scales.

Since when do fluids behave anything like a set of scales (a rigid object)?

 

You don't have any evidence for that claim. You are just giving your own interpretation of the historical facts (which differs from my interpretation). Fact is that the equations of electrodynamics as such (without further constraints) are ambiguous regarding the definition of certain variables.

My theory doesn't contradict itself. I have never claimed that the induction laws should be part of it. In my view the latter are not fundamental physical laws but just macroscopic effects (otherwise, the appearance of the time derivative causes for instance logical problems with causality) (electromagnetic waves excepted).
The point is that the existence of the Lorentz force produced by a single charge upon another charge would still have to be proven experimentally.

It is not a 'blind' re-interpretation if you exploit ambiguities in the fundamental definitions. If Kepler's laws had been ambiguously defined in some respect, then Newton's laws might well have been developed on the basis of an incorrect interpretation. In this case it should not only be OK but the duty of any serious physicist to consider the consequences of an alternative interpretation.

The only ones who never have problems with anything are those who never reflect logically on anything.
I am trying to point ways out of the problems, you just claim the latter don't exist.

That is bold claim which is physically unjustified (and as far as I am concerned indeed incorrect).

Are you denying everything you said more than 3 months ago as a matter of principle?
In post #68 you said '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)'.
The point is that the Stern-Gerlach effect is not detectable for lone electrons, contrary to what you are saying.

The (local) derivatives of a function can obviously not be defined if the function is not defined, but the function has to be global not local. So the problem simply can not be purely local.

Because the magnetic field is non-locally defined. The earth's magnetic field in space is for instance determined by processes in the earth's interior (and to a small degree in the atmosphere). The local contribution is virtually zero.

Pardon me? But I used exactly your argument in the first place that there can not be an overall charging of the wire.

Again, you should substantiate your claims through a consistent mathematical treatment of the problem (I have countered all of your points in the course of this discussion repeatedly, and I am not going to repeat myself again).

Energy conservation applies in any frame.


Thomas

 

You are again evading the point: in the Levy-Leblond 'derivation' the Lorentz transformation is only an acceptable solution of the existence of a general limiting velocity c is postulated. That applies to massive particles in general, not only charged ones in accelerators (the behaviour of the latter could well be explained by a velocity dependent magnetic force (as I think I mentioned already earlier).
In any case, it would be a circular argument to try to theoretically justify the existence of the limiting velocity by the relativistic mass increase if you haven't even derived the Lorentz transformation yet.

You still seem to be missing the point: it is not the transformation of the coordinates of the two rows of buoys we are interested in (which is always a Galilei transformation (Einstein's x'=x-vt in his 1905 paper; i.e. b=1 in your formula), but the transformation between the coordinates of a light signal in both of those two reference frames. The latter would only would transform the same way if the speed of light would depend in the same on the reference frame as the speed of a buoy. This is not the case. Indeed, the speed of the light pulse does not depend on the reference frame at all, i.e. the coordinate transformation for the light pulse must be the identity transformation.

Again you are missing the point: there is no such thing as a unique physical entity 'light pulse'. It is just so that a certain atomic/electromagnetic process in the light source causes after a given time an atomic/electromagnetic process in the detector. What goes on in a different detector is unrelated to this and constitutes a completely different event. So there is nothing wrong with the two detector events happening in different places.

x'+vt indeed wouldn't produce nonsense as this is a coordinate in the rest frame (x'+vt=x), but Einstein uses x' as the first coordinate, which is not a coordinate in the rest frame.

If you would differentiate correctly, then yes, you wouldn't get any new constraints. The point is that an incorrect differentiation is instrumental for Einstein's 'derivation' (see again my page http://www.physicsmyths.org.uk/lorentz3.htm ).

I have mastered whatever I had to master, and this was without any great problems (in contrast to Einstein , according to his own words ( http://thinkexist.com/quotation/do_not_worry_about_your_problems_with_mathematics/15457.html )).

No, it is isn't circular reasoning I am afraid. If you claim that the relationship is valid for all x,t, but it leads in fact to algebraic inconsistencies on the light cone (i.e. does not hold there), then one can logically conclude that your claim is false.

This is a hypothetical question. Any lottery/casino would ban you already way before you won the jackpot for the 10th time in a row. So much for your 'power' here.

I was arguing against the crackpot idea that a scientific theory can have any 'power' in the sense of the word. Only an omniscient and omnipotent being could have that. A theory may be as well confirmed experimentally as you want, from a logical point of view there is still the possibility that it is inappropriate or wrong, and that reality may be better approximated by a different theory. Even if there is little evidence for that at the moment, sometimes it helps to 'think big' to make progress in science. People like John Baez are, with all respect, 'small thinkers' who prefer to go along with the mainstream and try to discourage any dissident views by means of some silly 'crackpot rules' rather than engaging in an objective scientific discussion.

With regard to Einstein's 1905 paper, I have been asking myself the question as well why it was ever accepted. My guess is that Einstein's convoluted way of arguing and writing worked for him rather against him, as the referee(s) did not realize that the constraint used by Einstein is not the constraint implied by experiments.

The judgment about what is 'informed' and 'uninformed' depends wholly on your viewpoint. If there are unrecognized theoretical errors in the interpretation of the data, then even the so called 'experts' are not really 'informed' anymore.

As I said, the point is that the cross section depends strongly on the energy. The latter is assumed as known, but if this is incorrect, the measurement of the cross section will be wrong as well.

My idea was that the electro- and magnetostatic interaction force is velocity dependent (as numerically given by the 'relativistic' gamma factor, see my page http://www.physicsmyths.org.uk/dynamics.htm). This means that the energy gained by a charged particle in an electric field will be less than given by the usual potential energy difference. Reversely, this energy will be given off again over the same potential energy difference, so energy detectors will not notice at all that the energy was less than given by the potential energy difference in the analyzing field.

Conspiracy theory? So you deny that GPS is essentially a military project? You deny that this raises issues of conflict of interests as far as scientific transparency is concerned? You deny that for instance until some years ago, the GPS system was in practice sabotaged by the military itself in order to make it less accurate for other users?


Thomas

 

I mentioned already last time that the substitution of the differentials results means that the sum of the derivatives of the two variables is replaced by twice the derivative of one of the variables. In order to make the results consistent, you have therefore to substitute both variables with the other and average the two results.
If you take for instance the two paths (I am merely choosing this in order to avoid problems with the zero in your above example)

x1(t)=2t
y1(t)=2t

x2(t)=t
y2(t)=3t

then you find that the substitution works out correctly, if you do it as I said.

For the relationship discussed on my page http://www.physicsmyths.org.uk/lorentz3.htm (where the function values are not being evaluated as the function is not known) it doesn't actually make any difference at all: if you do the substitution for both variables, add the results, and then substitute the one of the variables back, you end up again with my Eq.(11).

Note by the way that you can do the substitution already for the initial function before you take the derivatives. In the above example, you could substitute the second variable 'y2' by a variable '3*x2' and then apply the chain rule to this. You'll end up with the same result as after substituting the differential dy2 by 3*dx2 later on. So it is not at all a problem of partial differentials here. The substitution follows just from algebraic identities.

Thomas

 

Always have and always will be. Archimedes used the principle for his famous 'weighing' of the fake gold crown. The point is that the displaced fluid always has the same volume as the submerged object. If Archimedes could have made sure that the volumes of the real and fake gold crown are equal, he could indeed have used a normal scale for this.

Thomas

 

You have no evidence for yours.

This is easily remedied: next time, read the text near the equations. Any good formulation of electrodynamics will explicitly state something along the lines of "where \(\bar{v}\) is the velocity of the charge". Unless you simply don't know what the velocity of an object is, there's no ambiguity in the formulation of the Lorentz force. Now of course the theory defined this way may be correct and it may not be. You can always try to argue that Maxwell or whoever didn't consider other potential formulations, or you can try to formulate your own alternative theory and argue that it's consistent with the experimental evidence. That's a seperate issue. But this ridiculous idea that the Lorentz force was ambiguously defined really needs to go.

Then what are the equations of your theory? Do you even have anything that could be considered a real alternative theory to electrodynamics? So far all you seem to have is a collection of ad-hoc claims scattered around your website.

Differential equations don't violate causality - they're needed in any theory that's going to predict the evolution of a dynamical system. If the appearance of \(t + \delta t\) in the derivatives is really bothering you, then note that you can just as well define the time derivative of a function by:

\( \frac{\text{d}f}{\text{d}t}(t_{0}) \,=\, \lim_{\delta t \rightarrow 0} \frac{f(t_{0}) \,-\,f(t_{0} - \delta t)}{\delta t} \)​

For a differentiable function, this is completely equivalent to the more common conventional definition.

Really? Why?

Your alternative to this is your vague idea that magnetic fields are produced by some unspecified "interaction" that only takes place in regions where the "average" charge density is zero. You have no evidence for that claim. This is an issue since your interaction would need to take place at inter-atomic distances, which is already explored territory.

Don't be silly. Kepler's been dead for nearly 400 years and Newtonian gravitation is a theory that stands just fine on its own. Unless you think Kepler was some sort of prophet with mystical powers, there's no a priori reason anything he said should be relevant to any physics today.

On the contrary. If you dug up some statement Kepler made - without all the evidence for gravitation he never lived to see - that seems to suggest another interpretation of Kepler's laws, and you wanted to insist that this interpretation should be taken seriously, the burden of proof would be on you to show a) not only that a corresponding theory of gravitation exists, but b) that it is at least as successful at explaining modern observations as Newtonian gravitation is and c) does so without any added complexity.

I don't think you're a particularly good judge of what constitutes logical thinking. And if you're under the impression you're the only person "thinking", well I think xkcd makes the point better than I do. I went through a phase like that in highschool.

The "problems" you're having are all too easily attributable to your own lack of understanding of electrodynamics and relativity. The blunt answer to your problems is "go learn relativity".

As far as I know, the laws of physics are not affected by whether I choose to mark out my coordinate system with expensive buoys or bits of chewing gum (although this may admittedly affect the precision and quality of experiments I try to conduct).

No. In post #68 I certainly attempted to imply that the Stern-Gerlach effect was observed for electrons. But you replied and corrected my error, and accordingly I have not since based any arguments on the Stern-Gerlach effect. So you are certainly putting words in my mouth when you say I am attempting to imply Stern-Gerlach is observable for lone electrons. The fact that you tried to refute something I haven't been saying instead of addressing what I actually have said in the last three months makes me suspect you're not even bothering to read anything I say carefully. This would explain how you can keep insisting there are "problems" when I solve your problems right in front of you.

The derivative of a function at a point x[sub]0[/sub] is completely defined by f in any arbitrarily small open region that contains x[sub]0[/sub]. That's perfectly sufficient to consider a theory "local".

You've missed the point. I was talking about the Lorentz force as an operational definition of the magnetic field. If I invent a detector that measures the magnetic field by measuring its action on a set of test charges contained in the detector, then it will use the velocity of those charges in its rest frame and give you a measure of the magnetic field in that frame. If we used your definition instead, we'd always have to input the velocity of the "source" into the detector so it could calculate the velocity difference between its electrons and the source. This is already less practical, and it gets worse if you're measuring a magnetic field produced by several sources all moving at different velocities.

Then where's the problem? I'm not claiming the wire gets an overall net charge. I showed that the wire will gain a charge density along most of its body. This will produce an electric field near the body of the wire which is approximately given by Ampere's law as long as the end-points are sufficiently far away. That's all the derivations you linked to require.

I've already told you that Minkowski's formulation of electrodynamics proves quite generally and rigorously that electrodynamics is Lorentz invariant. If you're genuinely interested in understanding this and not just hopelessly prejudiced against relativity, you should be able to find a treatment in any reasonably complete text on electrodynamics. For example it's done in Feynman's Lectures (volume II, chapter 25) and in Jackson's "Classical Electrodynamics" (sections 11.6 and 11.9), and I'm sure there are others. So I know for a fact that the only way you can ever apparently produce an example system that contradicts relativity is if you either misapply electrodynamics or you misapply relativity. On your web page, you do both as well as cutting a lot of corners. That really settles the issue as far as I'm concerned.

How can you possibly justify misapplying relativity?

It does. However what you're probably forgetting (I say "probably" because it is difficult to say for sure where you're going wrong when you don't actually substantiate your claims or calculate anything) is that, in the context of relativistic mechanics, the coordinate system you're using is not an inertial coordinate system - ie. the laws of physics still apply but they don't take their canonical form when expressed in those coordinates. In particular the energy/momentum of an electron won't be given by the same formula as it is in an inertial coordinate system.

 

I could just as well counter that the Galilean transformation requires the postulated absence of a general limiting velocity, and that trying to theoretically justify this based on Newtonian physics is circular since Newtonian physics presupposes Galilean invariance.

The point is, either way you've got two general solutions to Levy-Leblond's conditions, and you need extra, experimentally obtained information in order to pick one over the other.

This assumption is completely unfounded. The placement of the buoys depends on the symmetry properties of the physical laws governing the structure of the buoys and who or whatever put them in place. This is not something you can make a priori assumptions about.

Einstein does not, in his 1905 paper, assume that the laws of physics take their canonical form when expressed in terms of x'. It's just a convenient parameter defined by x'=x-vt, to which Einstein attributes no real physical significance.

For example if they both transform according to the Lorentz velocity addition formula...

Non-sequitur. I've already pointed out that you require a false analogy between invariance of c with respect to the source and invariance of c with respect to the observer in order to conclude this.

The point is that your idea leads to contradictions when you look at single events. What's your response to the example I gave with the two mirrors? Do you really think that's a sensible prediction?

Note that it's even worse: you'll find that when you apply the identity transformation to light and the Galilean transformation to everything else, your predictions depend on where you choose to place the origin of your coordinate system.

As I said, whatever physical interpretation you attribute to x' does not affect the mathematics. From a mathematical standpoint, if you have an expression for \(\tau\) in terms of x and t, then you can substitute x = x' + vt and get an expression for \(\tau\) in terms of x' and t. The partial derivatives of these expressions with respect to x and x' are actually the same:

\( \Bigl( \frac{\partial \tau}{\partial x^{\prime}} \Bigr)_{t} \,=\, \Bigl( \frac{\partial \tau}{\partial x} \Bigr)_{t} \)​

where the quantity outside the parentheses denotes the quantity being kept constant. It's actually the time derivatives that are different:

\( \Bigl( \frac{\partial \tau}{\partial t} \Bigr)_{x^{\prime}} \,=\, \Bigl( \frac{\partial \tau}{\partial t} \Bigr)_{x} \,+\, v \, \Bigl( \frac{\partial \tau}{\partial x} \Bigr)_{t} \)​

which is just the familiar expression for the hydrodynamic derivative.

Your continued insistence on attacking Einstein's derivation here is completely incomprehensible. You yourself are admitting that the correct solution to Einstein's synchronicity constraint is any multiple of the Lorentz transformation, which is exactly what Einstein finds. How can you keep claiming Einstein's derivation is flawed and yours is correct when Einstein gets the correct result and you just got a meaningless contradiction?

You think Taylor expansions are "convoluted", you have some really bizarre ideas about partial derivatives, and I've never seen anyone so completely bedevilled by something as elementary as variable substitution.

Context...?

But since the constraints don't lead to any "algebraic inconsistency" (whatever that's supposed to mean), there's no problem. You keep claiming that the constraints produce "an inconsistency" but you've never been able to explicit or demonstrate one.

I take it the lottery/casino managers don't think it's a coincidence then?

The "just a theory" argument isn't on the crackpot index because Baez believes he's omniscient. It is because everyone already knows that any theory could always be refuted by new evidence. The problem with the "just a theory" argument is that the person making it is typically trying to imply that the debatée or the scientific community in general has accepted the theory in question religiously and unquestioningly - and this is usually not the case.

I don't accept relativity as unquestioned "truth". Why should I? I didn't invent the theory and it isn't promising me everlasting life, so I have no emotional investment in it. What I told you earlier was that relativity had been confirmed in accelerator experiments at least up to about the TeV energy scale, and that this is an impressive accomplishment for a theory that was formulated long before accelerators were available. Lorentz violations may be discovered beyond that - at the LHC for instance. Only time will tell. As far as the attitude of the scientific community is to this, judge for yourself. See here what kinds of hits a Google search for "Lorentz violation" produces. But the point is that even if relativity turns out only to be an approximation to the truth, it can never be less useful than it has already shown itself to be. It will still continue to be used in sub-TeV particle physics as an approximation simply because the experience of over half a century of particle physicists is that it's useful in that experimental domain - just like we didn't throw out Newtonian and classical physics away when relativity and quantum mechanics came along.

It's because Einstein's constraints are consistent with experiments. Invariance of the speed of light means, obviously, that the speed of light must be invariant. Speed is the norm of velocity, which in turn is defined as the time derivative of position. So "invariance of the speed of light" means that if \(\frac{\text{d}x}{\text{d}t} \,=\, \pm c\), then \(\frac{\text{d}x^{\prime}}{\text{d}t^{\prime}} \,=\, \pm c\). This is what experiments show. And the Lorentz transformation predicts:

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

Which is consistent with this. It's really that simple. Your insistence that relativity contradicts invariance of c is beginning to make me think you must just not know what the words "speed" and "velocity" mean. And as indicated earlier, the identity transformation contradicts the Doppler effect.

Attacking the Lorentz transformation based on Einstein's papers is also pointless. He gave his own derivation of them, but they'd already been discovered by the likes of Lorentz and Poincaré and others. It had already been demonstrated that electrodynamics was Lorentz invariant. The invariance of c could already be explained by the hypothesis of Lorentz-Fitzgerald contraction of the experimental apparatus, which is incorporated into the Lorentz transformation and could be explained if matter was primarily electromagnetic in nature. Einstein didn't discover anything new here and his paper was very much in line with the ideas of other physicists living at the time. Einstein's original contribution was the idea that maybe all of physics - and not just electrodynamics - was Lorentz invariant.

Incidentally, as to Einstein's original formulation of relativity being "convoluted", you might be surprised to learn that most physicists might to some extent agree with you. For instance I mostly agree with the points made in this short article. But it also shows that attacking relativity based on Einstein's original papers is pointless. Something you don't seem to have picked up on is that the mathematicians got their hands on the theory in 1907 and modern formulations of relativity don't particularly resemble Einstein's original formulation.

"Uninformed" means you obviously don't know the first thing about accelerator or particle physics. Your superficial comments regarding "rescaling the cross-section" are so unrelated to the way energy is actually measured that I suspect you're guessing how you think it's measured rather than basing your opinions on any actual knowledge. If you're not just guessing, you really should review your sources.

The cross section is measurable directly. There's nothing really subtle in the way cross sections are measured in accelerators. You basically just count up the number of interactions you see and divide by the time-integrated luminosity of the beams. The "luminosity" in question here is just a measure of the particle density of the beams - i.e. it's not an energy. The measurement of the cross-section is independent of the energy, and vice-versa.

Then for all intents and purposes we may as well take the relativistic energy expression as the definition of the energy. It's the quantity that will appear in an energy conservation expression, while the classical energy expression is of no practical value here. Note that relativistic energy/momentum conservation is also seen in strong and weak interactions. That's not some small "niche" experimental domain. It's three of the four fundamental interactions found in nature.

By the way, since you apparently don't think energy conservation is important, why have you been trying to use it as an argument against the relativistic velocity transformation of the electrons in the wire loop?

Well given the publicly available figure of 38 microseconds/day, you can't accuse the US military of merely withholding information. So what are you accusing them, as well as all the physicists and engineers who were consulted on the project, of?

 

I'm not following. Can you present this as a worked example? (Ideally, can you prove your technique is mathematically valid in general?)

And yet, the Lorentz transformation satisfies the original constraint on your page but it doesn't satisfy equation 11. So equation 11 isn't equivalent to the synchronicity constraint and can't have been correctly derived from it.

 

Archimedes just used the fact that water was a (very nearly) incompressible fluid. But that just defines the difference between a liquid and a gas. You're still a long way from a solid, rigid object such as a set of scales.

I should point out at the outset here, to avoid any risk of misunderstanding, that I am not going to try to argue that the acceleration of a submerged object is given by \(a = \bigl( \frac{\rho V}{m} - 1 \bigr) g\). This is for the simple reason that I don't know what the acceleration of a submerged object is. I wonder if anyone does. I'm too ignorant of fluid mechanics to be able to tell you. However to my knowledge there's no really good reason I know of to believe it will be given by \(a = \bigl( \frac{\rho V}{m} - 1 \bigr) g\). Newton's second law tells us that the acceleration of the submerged object will always be proportional to the net external force acting on it, but there's no real reason to believe the external force on a moving submerged object will be given by Archimedes' principle. You actually hint on why on your site:

ie. the derivation of the buoyancy force requires the assumption that the pressure gradient in the fluid exactly compensates the fluid's weight density, and Archimedes' principle is only necessarily valid in that case. Once an object starts accelerating it will move the fluid around it and we're no longer in hydrostatic equilibrium any more.

So far I haven't really contradicted anything you've said. But your subsequent analysis is wrong. In fact, it is a typical example of where you substitute a real problem for an unrelated analogy. I hope you're not seriously under the impression that when a submerged object of volume V starts to accelerate, there's a corresponding volume V of fluid that's constrained to accelerate in an exactly opposite fashion like on a set of scales. Here's a counter-example:

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​

Here I've depicted a submerged object in a closed looped tube, such that no fluid can flow past the object. The system is constrained in such a way that the submerged object can't move without the fluid in the entire tube moving. If you hold the object in place you will have to counteract the usual buoyant force, but if you release it all the fluid in the tube will be set in motion and the object's acceleration will be much less than that predicted on your site. This is obviously an extreme example, but it demonstrates that the "volume of displaced fluid" and the "volume of fluid set in motion" are two different things.

In general, the fluid is not constrained and you can't reduce the problem to a single degree of freedom. The only thing fluid incompressibility (ie. "volume conservation") implies is that the fluid must have a divergence-free velocity vector field. The correct way to approach the problem would involve solving the Navier-Stokes equation (or at least the Euler equation) with appropriate boundary conditions on the surface of the submerged object. If you've ever had the pleasure of manipulating the Navier-Stokes equation, you'll know it's quite a different animal to deal with than a constrained problem such as a set of scales. Highschool students can solve that sort of problem. By contrast, there's a Clay Mathematics Institute Millennium Problem related to the Navier-Stokes equation. Your site effectively trivialises what's actually quite a hard problem.

While I'm at it, here are a couple more 'easy' examples where you're simply factually incorrect regarding some theoretical issue on your site. The first is one of your comments about quantum tunneling. Specifically this one:

Your other two comments about quantum tunneling are just opinions and I'm not going to try to address them or the issue of experimental evidence in general (I'm really not qualified to do this). But I've picked out statement b) because it indicates a fundamental misunderstanding of quantum mechanics. As far as the formulation of quantum mechanics is concerned, it is a postulate that the state of a system will always evolve according to the Schrödinger equation:

\(\hat{H} | \Psi \rangle \,=\, i \hbar \partial_{t} | \Psi \rangle\)​

The Hamiltonian operator is self-adjoint, so it's diagonalisable and all its eigenvalues are real. That's the essential content of the so-called "time-independent" Schrödinger equation. There's at least one family of mutually orthogonal states \(| \varphi_{n} \rangle\) that span the state space such that:

\(\hat{H} | \varphi_{n} \rangle \,=\, E_{n} | \varphi_{n} \rangle\)​

It's "time-independent" in the sense that if you plug these states into the time-dependent Schrödinger equation, you find they evolove trivially. A system initially in the state \(| \varphi_{n} \rangle\) just evolves as:

\( | \varphi_{n}(t) \rangle \,=\, e^{- i\omega_{n} t} | \varphi_{n} \rangle \)​

where \(\omega_{n}\) is given by the de Broglie relation \(E_{n} = \hbar \omega_{n}\). For a single such state this doesn't change anything, since global phases aren't observable quantities in quantum mechanics. But phase differences are observable. For example, a system that starts out, say, in the state:

\(| \Psi \rangle \,=\, \frac{1}{\sqrt{2}}\Bigl[ |\varphi_{1}\rangle \,+\, |\varphi_{2}\rangle \Bigr]\)​

evolves as:

\(| \Psi(t) \rangle \,=\, \frac{1}{\sqrt{2}}\Bigl[ e^{- i\omega_{1} t} |\varphi_{1}\rangle \,+\, e^{- i\omega_{2} t} |\varphi_{2}\rangle \Bigr]\)​

In this case, if you project onto another basis you'll see oscillations. For example if this is a one particle, one dimensional system, the probability density of finding the particle at location x is given by:

\(P(x;\, t) = | \Psi(x;\, t) |^{2}\)​

(where in general I use the notation \(\chi(x) \equiv \langle x | \chi \rangle\)). Calculation yields:

\( P(x;\, t) \,=\, \frac{1}{2} \Bigl[ | \varphi_{1}(x) |^{2} \,+\, \varphi_{2}(x) |^{2} \Bigr] \,+\, \Re[\varphi_{1}^{\ast}(x)\varphi_{2}(x)] \, \cos\bigl( (\omega_{2}-\omega_{1}) t \bigr) \,+\, \Im[\varphi_{1}^{\ast}(x)\varphi_{2}(x)] \, \sin\bigl( (\omega_{2}-\omega_{1}) t \bigr) \)​

Which clearly oscillates in time.

But it doesn't end there. As I said, the family of energy eigenstates spans the entire state space (ie. it's a basis). You can express any state as a linear combination of energy eigenstates. So if your system starts out in the state \(| \Psi(0) \rangle\), then its time evolution can in general be expressed as:

\( | \Psi(t) \rangle \,=\, \sum_{n} \, e^{-i \omega_{n} t} \, \langle \varphi_{n} | \Psi(0) \rangle \, | \varphi_{n} \rangle \)​

In other words, if you solve the time-independent Schrödinger equation for a problem and get all the \(| \varphi_{n} \rangle\)'s and \(\omega_{n}\)'s, you can also write the general solution to the time-dependent Schrödinger equation. There is no artificial distinction between "time-independent" behaviour and "time-dependent" behaviour in quantum mechanics as you seem to think. The latter are linear combinations of the former, so stationary states most certainly have something to say about the evolution of a system.

Another 'easy' one is your page on energy conservation. Your idea that energy conservation only applies to classical mechanics is simply ill informed and most of your argument is an "affirming the consequent"-type logical fallacy. You show that conservative potentials and action/reaction imply energy/momentum conservation within the context of Newtonian physics, but it is a logical fallacy to conclude energy conservation in a more general theoretical context requires these. In particular this statement:

is just factually incorrect and you need to be pretty ignorant of Noether's theorem and contemporary physics in general to say something like this. I suppose the best way to show you this is just to illustrate the point of Noether's theorem. Suppose we have a system characterised by a single degree of freedom q and we have a Lagrangian that's an explicit function of q and its time derivative:

\(L = L(q,\,\dot{q})\)​

(where \(\dot{q} = \frac{\text{d}q}{\text{d}t}\)). The action over a time period for a particular path \(q(t)\) is given by:

\(S \,=\, \int_{t_{0}}^{t_{1}}\text{d}t \, L \bigl( q(t),\, \dot{q}(t) \bigr)\)​

According to the action principle, for fixed boundary conditions the system will evolve in such a way as to extremise the action. You obtain the equations of motion by varying the action with respect to the path with fixed boundary conditions (ie. \(\delta q(t_{0}) = \delta q(t_{1}) = 0\)) and requiring that the variation be zero:

\( \delta S \,=\, \int_{t_{0}}^{t_{1}} \text{d}t \, \biggl\{ \frac{\partial L}{\partial q} \, \delta q \,+\, \frac{\partial L}{\partial \dot{q}} \, \delta \dot{q} \biggr\} \)​

Integration by parts on the second term in the integral turns this into:

\( \delta S \,=\, \biggl[ \frac{\partial L}{\partial \dot{q}} \, \delta \dot{q} \biggr]_{t_{0}}^{t_{1}} \,+\, \int_{t_{0}}^{t_{1}} \text{d}t \, \biggl\{ \frac{\partial L}{\partial q} \,-\, \frac{\text{d}}{\text{d}t} \Bigl( \frac{\partial L}{\partial \dot{q}} \Bigr) \biggr\} \, \delta q \)​

The boundary term disappears due to the fixed boundary conditions, so if we require \(\delta S = 0\) for arbitrary t[sub]0[/sub] and t[sub]1[/sub] we get the familiar Euler-Lagrange equation:

\( \frac{\partial L}{\partial q} \,-\, \frac{\text{d}}{\text{d}t} \Bigl( \frac{\partial L}{\partial \dot{q}} \Bigr) \,=\, 0 \)​

Now here's the point. Suppose that the Lagrangian just happens to be invariant under translations of q, so:

\(L(q+\delta q,\, \dot{q}) = L(q,\, \dot{q})\)​

This implies:

\(\frac{\partial L}{\partial q} \,=\, 0\)​

And the Euler-Lagrange equations imply:

\( \frac{\text{d}}{\text{d}t} \Bigl( \frac{\partial L}{\partial \dot{q}} \Bigr) \,=\, 0 \)​

So in such a case the conjugate momentum \(p = \frac{\partial L}{\partial \dot{q}}\) is conserved.

Now, if the Lagrangian doesn't depend explicitly on time (ie. the laws of physics are time-invariant), then the Lagrangian can only vary in time through variations in \(q\) and \(\dot{q}\):

\( \frac{\text{d}L}{\text{d}t} \,=\, \frac{\partial L}{\partial q} \, \dot{q} \,+\, \frac{\partial L}{\partial \dot{q}} \, \frac{\text{d}\dot{q}}{\text{d}t} \)​

or (using the product rule and applying the Euler-Lagrange equation):

\( \frac{\text{d}L}{\text{d}t} \,=\, \frac{\text{d}}{\text{d}t} \Bigl( \frac{\partial L}{\partial \dot{q}} \, \dot{q} \Bigr) \)​

Which implies that the energy:

\( E \,=\, \dot{q} \, \frac{\partial L}{\partial \dot{q}} \,-\, L \)​

is conserved.

As you should be able to see, I never needed to assume that the Lagrangian took the form "kinetic energy minus potential energy", so this is a completely general result. In the special case where:

\(L \,=\, \frac{1}{2}m \dot{q}^{2} \,-\, V(q)\)​

the Euler-Lagrange equation reduces to Newton's second law:

\(-\, \frac{\partial V}{\partial q} \,-\, m\ddot{q} \,=\, 0\)​

and the energy:

\( E \,=\, \frac{1}{2} m \dot{q}^{2} \,+\, V(q) \)​

is conserved. If V = 0 (no external potential), then the Lagrangian is invariant under spatial translations and the momentum:

\(p \,=\, m \dot{q}\)​

is also conserved.

Here, I've shown that invariance of the Lagrangian under spatial translation implies momentum conservation, and that time-invariance implies energy conservation. It's also possible to show that rotational invariance implies angular momentum conservation. In general, Noether's theorem proves the existence of a conserved quantity or current associated with every symmetry in the Lagrangian, and easily generalises to N particle systems and even to field theory. It's an extremely powerful result. In modern theories, energy is pretty much defined as the conserved Noether current associated with time-invariance. The point of this is that we have a general definition of energy that will adapt to any new theory that comes along.

This isn't all just academic. Open any textbook on field theory and you'll see Lagrangians all over the place. In fact, specifying a Lagrangian is generally the preferred way of proposing a theory, for what I hope are now obvious reasons. Since it ties in with the topic of this thread, here's an actual example. You can derive Maxwell's equations from the Lagrangian density for classical electrodynamics for "known" charge/current distributions:

\( \mathcal{L} \,=\, \frac{1}{2} \varepsilon_{0} \, \bigl( E^{2} \,-\, c^{2} B^{2} \bigr) \,-\, \phi \rho \,+\, \bar{A} \cdot \bar{j} \)​

by varying the action with respect to the scalar and vector potentials. If you add a kinematical part for a charged particle, you can also derive the Lorentz force.

 

Say the boosted observer is moving left, The end points of the boosted observer are co-located with A and B (A' and B') when the lights are pulsed as is the boosted obvserver and the point ME on the other frame {call the boosted observer also the 'midpoint boosted observer'}.

The boosted observer presumes his motion is stationary and that all motion of the frame where the lights meet at ME simultaneously is moving. Initially you are correct, partially so when the boosted observer sees the B light first, and perhaps we can understand this observers' conclusion that the lights from A and B were turned on at different times. Then the A' and B' observers show the boosted mid-point observer that the A' and B' light reception times are identical. Even with time dilation and no synchronization of clocks in both frames the mid-point boosted obvserver must now conclude that with the observed simultaneous arrival of the lights at A and B and that the lights were emitted simultaneously in the ME frame must make the boosted stop and reconsider and reconclude that the boosted frame is in motion.

If the lights entered the boosted frame simultaneously and the light arrived at ME simultaneously then the 'boosted frame' must be the [previously]accelerated frame that is in motion.

This does effect the equivalence of inertial frame postulate somewhat in that motion of an inertial frame has been detected in reference to the frame containing ME.
From the first edition of "The Handbook of Astronautical Engineering", Chapter 11 regarding relativity theory extended to rocketry, defines the postulates of relativity. "It is impossible to measure or detect unaccelerated translatory motion in deep space, which is where your experiment was conducted" and paraphrasing of course, that the velocity of light will always be measured as c in all inertial frames.

So, Mantillo is correct when he states that the observer motion and position on an exended frame of reference are used, re relatively theory, to determine different physical events in his own frame as well as frames moving relative to his own.

The boosted observer when analyzing the simultaneous arrival of the lights at ME and the non-simultaneous arrival of the lights at his point on the boosted frame must be supicious. The boosted observer ought to say that if the lights were emitted simultaneously and that the ME frame was moving left then the lights could not have arrived at ME at the same time. This means that either the ME was in motion to the left, but if the boosted observer was co-located at the midpoint of A and B (when the lights were emitted), that is at ME, that the simultaneous arrival of the lights at ME frame contradicts the boosted observers' claim that the ME frame is in motion anhd that the boosted frame is stationary-b under these condition the boosted obwserver must have seen the arrival of thye lights at his position simultaneously.

Under the parametric limits imposed on the experiment by the simultaneous emission from A and B proves the motion of the boosted frame.


This is not to imply the entry of some demon 'preferred frame of reference' when the boosted observer is corrected, but it does say much regarding the equivalence of inertial frames in their capability of measuring any relative motion photon-frame of other than c, the speed of light @ ~ 3x10^8km/sec.

Good work mantillo.

 

Agree , 100%. See my reply to BentheMan above.

 

Very interesting thread, indeed.

Question: To call electron spin as magnetic dipoles is ambiguous (to my thinking up to here). The electron motion moves up or down in the magnetic field as determined by the 'spin state', which for an electron can vary, i.e. + OR -. An electron in a polarized +S statev can adopt a +T state when transitioning through a Stern-Gerlach T segment, and upon exiting an unobstructed T segment, the spin will reform into a +S state. As polarization sets a particlular spin state (+ or -) the reformation of the +S state require previous +S state memory-continuity during the transition through the SG segment. requires a force that reorients the monopole from a +S state orientation which is parallel to the S type SG segment z-axis. The field forces a reorientation of matter to condytruct the +T \state Exiting the T segment and reverting to the +S statethe Therefore, to describe spin states a magnetic dipoles presumes dipoles with different physical properties vis a vis qm and classical models.

Are you considering a dipole somewhat analogous to a magnetized rod with rod ends of opposite polarity?

As above regarding przyk's statement re spin, spin is only observed in the presence of an inhomogeneous magnetic field (which could be constructed by the collision of charges on particles moving wrt to each other) which you state correctly, or so I assume, there should not be a magnetic moment for a free electron. The lack of observed spin for electrons in SG experiments has been described by some as being technically difficult to measure up/down motions imposed due to the relatively rapid speed of electrons through the SG segment.That is the direction of motion change is slight proportional to the stime spent in the mag field) . Bullets fired with muzzle velocities of 500 meters/ sec drop more from gravity in a 500 meter distance than bullets firedc with muzzle velocity of 2500 meters/sec. Using slower electrons with increased field strength should cure the measurement problem (sorry the reference to this is years ago memory). In any event electron spin has not been discarded from unambiguous experimental results.

Finally, "spin motion" is only observed in atoms moving in inhomogeneous mag fields and such have no observable motion. Classically, placing two rectangular magnets together will result in [+-][+-] configuration, where if placed in a [+-][-+] arrangement the magnets repel each other. Spin as a magnetic phenomenon does not behave like moving wires (with internal electrons) in magnetic fields. Spin mechanics is an incomplete science and spin behaves as if the particle has some internal system that exposes either a + or as - magnetic monopole. If the + monopole is exposed, the - monopole remains hidden and vice versus. This seems like the +- monopole characteristic of electrons is such that a + monopole evidenced by + motion while the - monopole is evidenced with - motion etc. and that contact with the magnetic pposite magnetic polarity' seen as - motion. The selection of oa + state means the - state is relegated to a state of unobservability, or as non-local. All of this suggests to me is that spin states are dynamic and seen in a time history, as, . . . +-+-+-+-+- . . . and that contact with the field freezes the state into + or - depending on which state was observable [conversely, which state was non-local] when coming into effective contact with the magnetic field.

+-+-+-+-+-+-+ observed

-+-+-+-+-+-+- non-local

 

I am just claiming the possibility of something in the absence of evidence against it (namely that the velocity in the Lorentz force has not been concisely defined). So the lack of your evidence is my evidence.

You still don't seem to understand that it is not important what is said here, but what is possibly not said here. If it is stated "where v is the velocity of the charge" then this could as well mean that v refers to some preferred reference frame (as somehow physically defined by the system in question), but that this constraint has simply been omitted here (for whatever reason). Claiming that the definition of v as given must be a priori complete is nothing more than wishful thinking.

In case this hasn't become clear yet: it is not my primary aim to formulate alternative theories, but to point out inconsistencies in existing ones (this is why my site is called physicsmyths and not alternativephysics or whatever). I merely try give, where possible, alternative concepts which avoid these inconsistencies and thus may point in the right direction (in my opinion anyway). In this sense, I consider it only as a basis for a fundamental critical discussion, not a fully fledged theory.

This still means that the derivative at t0 depends on the function values at neighboring point t0-dt, which means that the such defined physical quantities can not strictly be local (which in my opinion can only be possible for macroscopic physical quantities but not fundamental ones).

You effectively stated that, as a matter of principle, one can not reinterpret a given theory because it might invalidate a derived theory based on the original one. I am saying that this is an invalid argument. The derived theory can not retrospectively justify the original one as it would implicitly contain any errors or misinterpretations of the latter.

The beginning of your answer really says it all: "I don't think" (may be a Freudian slip?).
But seriously, you shouldn't try to turn this into an exchange of personal attacks. I merely pointed out that you may be mistaken with your view that having problems with something necessarily implies not having dealt with the subject matter deeply enough, but that, on the contrary, the problems could be due to its inherent inconsistencies. Whether it is or not can only be decided by scientific arguments, not further ad-hominem attacks.

Learning does not necessarily mean understanding. You could train a parrot to recite the basic postulates of Relativity. I have merely tried to progress beyond that and go to the roots of the subject matter.

It depends on what you understand under 'local'. Strictly speaking a local event should be defined on a mathematical point. Any finite interval around this point makes the problem thus non-local. So the derivative could not be defined locally in the first place.

Yes, that's the complication which in my view actually exists in reality (and which for for instance could explain Galactic Rotation Curves.

The derivations I linked to assume a homogeneously charged wire without end-point charging of opposite polarity. If they had assumed the latter and treated the problem exactly and consistently, they would have noticed themselves that their approach is flawed (as indicated on my page http://www.physicsmyths.org.uk/lorentzforce.htm ).

Thomas

 

Agreed, but then you can hardly call Levy-Leblonds paper a 'derivation of the Lorentz transformation' (as its title suggests). And it doesn't even give any reference to an experimental proof a general limiting velocity.
But most importantly, one should note that the paper deliberately avoids bringing in the invariance of c (which is the only relevant experimental criterion in this context) as a constraint here. If this had been done, it would have turned out that the Lorentz transformation is not an acceptable solutions as it indeed violates the invariance of c (as pointed out by me earlier (referring for instance to my page http://www.physicsmyths.org.uk/lorentz.htm )).

I beg to disagree. The definition of the coordinate system used to describe physical events must be made a priori. Otherwise there would be a logical self-reference, i.e. the coordinate grid would depend on the very physics it is supposed to describe (in mathematical terms you could say that the problems would not have any independent variables anymore).

He should have better attributed some significance to his equation x'=x-vt, then he would have noticed that it is actually inconsistent with the significance he attributes to his (mathematically incorrectly derived) Lorentz transformation equations.

The only reason that makes you think it is a false analogy is that it is not consistent with the Lorentz transformation. This is not a valid argument if you haven't even derived the Lorentz transformation yet.

The point I am making is that it is not allowed to ask what the trajectory of the light signal is other than in the reference frame of the detector/mirror, because the light signal, not being a material object, is only defined by the interaction with the detector. A different reference frame has to be defined by a separate system of detectors/mirrors, but the interaction of the light with these would constitute different physical events.

No, the point is that you are apparently not noticing that the usual concept of 'velocity' or 'speed' (as given by you) is logically inconsistent with being invariant with regard to moving reference frames. The only thing that makes it 'consistent' is the (physically unjustifiable) re-scaling of the original space and time measuring units in a suitable way (which is essentially cheating). As I said already, any velocity-dependent transformation is inconsistent with the invariance of c. The only consistent transformation is the identity transformation. This may seem unfamiliar, but light just is different from material objects in this respect (and others), and it is not appropriate to try to describe it in terms of concepts valid for particles.

An idea exactly based on Einstein's tendency to insufficiently define variables or change the definitions to something else later on.

As long as you have reversible (symmetric) processes (like charged particles accelerated/decelerated in electric field), it is obviously rather academic for the energy conservation what numerical value you give the energy.

I wasn't implying that energy conservation is not important here. On the contrary, the point I was making is that if the interaction force (potential) is velocity dependent, you will calculate an incorrect potential energy (and thus also an incorrect kinetic energy) when neglecting this fact.

The US military have nothing to do with the figure of 38 microseconds/day. That is a theoretical value predicted by Relativity. My point was that military projects should in my opinion not be considered the most trustworthy ones when it comes to the experimental confirmation of such fundamental theoretical effects (unless there is sufficient transparency, i.e. researchers have access to the original raw data, data analysis algorithms etc).

Thomas

 

I thought it would be obvious enough: just do the same like you did in your post #103 i,e. define a function f(x,y)=x+y but with the paths

x1(t)=2t
y1(t)=2t

x2(t)=t
y2(t)=3t

take the total differential of f(x,y), but now substitute not only the dy's by the dx's (as given by the path equations) but also reversely the dx's by the dy's. If you add these two results and divide by 2, you get correctly 4 for df/dt.

Thomas

 

In your example, the displaced volume is indeed the whole fluid but this has nothing to do with buoyancy any more. A buoyant object can freely move in a medium and that is exactly why the displaced volume is equal to the volume of the object itself. This means that the displace volume must, analogously to a scale, move down when the object moves up and vice versa (like I indicated on my page http://www.physicsmyths.org.uk/buoyancy.htm ).
Your example is more like s solid structure. A freely rotating wheel with a weight attached to its rim would behave very similarly.

So where is then the time-dependent development e.g. in http://en.wikipedia.org/wiki/Quantum_tunnelling ? There is none. The tunneling is fully 'explained' in terms of the time-independent Schrödinger equation. My point is that this is logically flawed, as 'tunneling' means you have the particle initially on one side of the barrier and after some time on the other. So some kind of temporal development is necessary to even qualitatively describe the process.

Obviously, as long as you consider the problem only formally mathematically and leave L physically undefined, you don't have to assume that it is equal to T-V.
The point is that both the Hamilton and Lagrange formalisms have been developed in Classical Mechanics, where they are exactly equivalent to the Newtonian formalism (see e.g. http://www.mathpages.com/home/kmath523/kmath523.htm ). The extension of these principles of Classical Mechanics to other (non-classical) physical phenomena is in the first place a purely speculative postulate. If they seem to 'work' there then this has to be considered as a mere coincidence.
So the view that energy and momentum are more fundamental physical quantities than forces is wrong or at least unfounded. Energy and momentum are merely quantities derived from the path and time integrals over the force field respectively and are in principle not required in classical physics at all (one can happily integrate any equation of motion without ever using the concepts of energy and momentum).

Thomas

 

"Possible" isn't the same thing as "plausible". And you're not just claiming it's "possible". In your same post you link to another one of your pages where you state, and I quote:

I don't think it's obvious.

I'd have thought it was quite clear what it means: the v is the velocity of the charge in whatever coordinate system(s) the equations of the theory of electrodynamics hold. You show that you can use the velocity in any inertial coordinate system by investigating the mathematical symmetry properties of electrodynamics. It turns out that you can use the velocity of the charge in any of a family of coordinate systems related by Lorentz transformations.

And how do you expect to do this if you don't understand the theories in the first place? It's perfectly normal to have trouble reconciling different aspects of a theory you don't have a complete understanding of - it's one of the difficulties involved in learning physics.

So are you retracting your claim about the induction equations violating causality?

The "derived" theory is one of the most accurately tested theories in the history of physics. You should not be proposing modifications to electrodynamics unless you're prepared to demonstrate that you can reproduce all of QED's correct predictions with those modifications. My opinion is that you'll never be able to incorporate all your ad-hoc claims into a consistent classical theory, let alone a quantum one as successful as QED. I don't see this opinion being particularly well challenged.

No, it's a standard idiom. In English, "I don't think X" means "I am not of the opinion that X is true". Given the point I was making, your reply is ironic.

I don't hold this view reflexively. I'm seeing you stumble over some really silly issues that you really should be capable of figuring out yourself.

Correct. Proving the internal consistency of the special theory of relativity amounts to a few simple exercises in group theory (showing that the inverse of a Lorentz transformation is another Lorentz transformation, etc.). Proving the Lorentz covariance of electrodynamics is most easily done by showing that the equations of electrodynamics can be expressed in covariant form. The only remaining issue is the consistency of these theories with experiments, and whether a conceptually simpler (in the sense of Occam's razor) collection of theories could explain the same experimental results. You haven't given an experiment that contradicts relativity and you don't have an alternative, simpler theory. That really doesn't leave much to discuss.

Who says I'm just reciting relativity? I've learned, understood, and to some extent even rederived bits of the theory as well as developing my own set of insights about it. In fact, here's a little anecdote: I first heard that "time slows" and "fast moving objects get heavier" as a child from my father. I'd also heard that it was "impossible to go faster than light" (an idea I didn't like - I was always trying to think of ways of sneaking past this limit). One day (I think I was about thirteen years old) I got the idea that it would be really neat if all these funny relativistic effects were happening in such a way that the speed of light would always seem the same for a moving observer. That's how I first heard about the invariance of c. I guessed it. Please keep that in mind before insinuating I'm just repeating theory or that I got misled by Einstein's "convoluted way of arguing".

You could define locality that way. I've already explained that this isn't a terribly useful definition though: if a theory adheres to this definition of locality it can't allow what's going on at one point to influence neighbouring points. In a sense, space-time would become disjoint - you'd have a universe of points all evolving in isolation, with no interactions allowed.

The point is that there is no minimum finite interval necessary to define the derivative at a point. This is an important distinction. You see it in differential geometry for instance. On a curved geometry (such as in general relativity), the parallel transport of a vector is path-dependent, so the difference between two vectors defined at different points a finite distance apart isn't well defined. However, there's no problem defining differential operators on a curved manifold. In fact, they're precisely what allows you to define relationships between "neighbouring" points on a manifold, or how influences will propagate through space and time in physical theories.

I think it's a bit premature for you to start proclaiming your personal interpretation can explain anything when you so far haven't been able to produce a consistent theory.

Which is perfectly reasonable. An infinitely long wire by definition does not have end points.

They did treat the problem consistently. The infinitely long homogenously charged wire is exactly what you get if you apply the Lorentz transformation to an infinitely long uncharged wire with a homogenous current passing through it. And I've already resolved your objection about charge violation: an infinitely long wire does not have a well-defined total charge. At the level of individual electrons and ions, the total charge you find can be anything from negative to positive infinity depending on the order in which you count up the electrons and the ions. The relevant point is that the Lorentz transformation does not create any new charges. Exactly the same electrons and ions exist in both frames.

It's you who decided there was a need for an explicit calculation for a finitely long wire, and it's you who doesn't get it correct. The transformed wire and wire loop described on your page are not what the Lorentz transformation predicts. You then erroneously apply the methods of electrostatics to your already incorrectly transformed wire even though the system is not static in the boosted frame (or even in the rest frame for that matter). You try to justify this by claiming that this is all fixed by assuming the charges are "locally replaced at the rate they are lost", but you demonstrate no such thing (I've already addressed this a long time ago). I'm really getting tired of pointing these errors out just to see you link to your website as if I hadn't already refuted the arguments there.