# Relativity fails with Magnetic Force

Discussion in 'Physics & Math' started by martillo, May 24, 2009.

1. ### przyksquishyValued Senior Member

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3,171
I did, as far as I could tell, exactly what you did on your website. If you did something different, you should be more explicit. And I didn't "merely" give a counter-example. I followed that with an explanation of why the substitution you performed was wrong.

Do that, and then if you differentiate both sides by x' you'll just obtain:
$\frac{\alpha}{c} \,=\, \frac{\alpha}{c}$​

The reason you can't treat the differentials like algebraic quantities is because they're differentials with respect to different parameters. If I've understood your website correctly then, for instance, you try to use the identity x[sub]2[/sub] = v t[sub]2[/sub] to write something like this:
$\frac{\partial \tau_{2}}{\partial x_{2}} \,=\, \frac{1}{v} \, \frac{\partial \tau_{2}}{\partial t_{2}}$​
You assume that this is the same $\frac{\partial \tau_{2}}{\partial t_{2}}$ that appears on the left of equation 10 (I have no idea what you did for the right hand side):
$\frac{1}{2} \biggl[ a_{2} v \, \frac{\partial \tau_{2}}{\partial x_{2}} \,+\, a_{2} \, \frac{\partial \tau_{2}}{\partial t_{2}} \biggr]$​
from which you conclude that the left hand side of equation 10 is just:
$a_{2} \, \frac{\partial \tau_{2}}{\partial t_{2}}$​

Your non-standard notation aside (you really mean things like $\frac{\partial \tau}{\partial x}(x_{2},\, t_{2})$ if you want to explicit where you're evaluating the differentials - the very notation you use betrays you don't really understand what you're doing), partial derivatives don't work that way. The "real" partial derivative with respect to t[sub]2[/sub] means:
$\frac{\partial \tau_{2}}{\partial t_{2}} \,=\, \lim_{h \rightarrow 0} \, \frac{1}{h} \, \Bigl[ \tau(t_{2} + h,\, x_{2}) \,-\, \tau(t_{2},\, x_{2}) \Bigr]$​
However the one you obtain by substitution really means:
$\frac{\partial \tau_{2}}{\partial t_{2}} \,=\, \lim_{h \rightarrow 0} \, \frac{1}{h} \, \Bigl[ \tau(t_{2},\, v t_{2} + vh ) \,-\, \tau(t_{2},\, v t_{2}) \Bigr]$​
The key point is that these derivatives are taken with respect to different parameters of the function $\tau$. It is an error to pretend these two objects are the same. You've just fooled yourself into thinking they are by giving them the same name. If you used a more correct notation for your partial derivatives, and if you understood the difference between differentiating with respect to a parameter and evaluating the differential at a point, you'd have avoided this pitfall.

This is meaningless. The "equation" explicitly evaluates $\tau$ along those paths, and for $\tau \,=\, \bigl( t \,-\, \frac{v}{c^{2}}x \bigr)$, you just obtain:
$\frac{\alpha x^{\prime}}{c} \,=\, \frac{\alpha x^{\prime}}{c}$​
The left hand side of the constraint equation is equal to the right hand side of the constraint equation (and for any value of $\alpha$). They're the same. That's what it means for the function $\tau$ to satisfy the constraint, pretty much by definition. That's really all there is to it. There is nothing in the world you can say that will change this fact. How many different ways do you need this explained to you? You may as well try to prove, using some convoluted argument, that the equality 2 + 2 = 4 somehow isn't satisfied.

If you don't reverse the direction of the light signal, then you're contradicting the argument Einstein used to establish his synchronicity constraint in the first place. First your idea of determining $\alpha$ by differentiating the synchronicity constraint, and now you want to plug nonsense into it? I don't know why you keep this pretense up. It's clear you haven't understood anything in Einstein's paper at all.

Last edited: Apr 24, 2010

3. ### przyksquishyValued Senior Member

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3,171
So how does that prevent a current forming similar to the one that would form in the loop? Or something even more complicated, now that there are no walls and the fluid can do basically anything? You're just assuming this doesn't happen.

In fact your objection here is absolutely ridiculous. You treat the submerged object and fluid as if they form a rigid object. When I give you an example of where the object and fluid actually are likely to behave like a rigid object, you don't get the same result as on your website, and you complain using exactly the same retort I'm using against your own derivation in the first place: fluids in general aren't rigid objects.

No, I just explained how I'm reaching that conclusion via the Schrodinger equation. It is you who is arbitrarily trying to impose that the wavefunction should be discontinuous. If it was true, in quantum mechanics, that a potential barrier could completely stop a particle then this result should arise naturally when we solve the Schrodinger equation. It shouldn't have to be put in by hand.

False analogy. You're still applying classical arguments to non-classical systems.

I was just using "barrier" as a synonym for potential wall. Its thickness is just its spatial extent.

5. ### tsmidRegistered Senior Member

Messages:
368
I think I told you already earlier that this is incorrect in this case because we don't have a genuine function of two independent variables here. The two variables on both sides of the equation are algebraically connected by the equations x1=x'+v*t1 , t1=x'*a1 ; x2=v*t2 , t2=x'*a2 (where a1=1/(c-v) and a2=1/(c-v) +1/(c+v) ). So it is not possible to vary one variable and hold the other constant (as it would be required for the definition of the partial derivative (as you indicate yourself with your equation)). This is the whole point why the derivation on my page Mathematical Flaws in Einstein's 'On the Electrodynamics of Moving Bodies' (Eqs.(7)-(11)) is appropriate and correct here. As for the explicit substitution of the variables used there: this was only required because I inserted the actual values for a1 and a2 in order to get an equation (Eq.(10)) that can be compared to the one Einstein obtained. I then merely re-substituted the variables in order to get Eq.(11), which shows the condition under which it is consistent at all. However, I could have derived Eq.(11) straightaway using the chain rule when differentiating the 'master equation':

if you have

(1) 1/2 * tau2 = tau1

where

(2) tau1 = tau(x1,t1)
(3) tau2 = tau(x2,t2)

with

(4) x1=(1/a1+v)*t1 , t1=x'*a1

(5) x2=v*t2 , t2=x'*a2 ,

then, if you do differentiate (1) with regard to x', you can immediately write this according to the chain rule as

(6) 1/2 * ( dtau2/dx2 *dx2/dt2 *dt2/dx' + dtau2/dt2 *dt2/dx') = dtau1/dx1 *dx1/dt1 *dt1/dx' + dtau1/dt1 *dt1/dx' ,

or

(7) 1/2 * (dtau2/dt2 *a2 + dtau2/dt2 *a2) = dtau1/dt1 *a1 + dtau1/dt1 *a1)

i.e.

(8) a2 *dtau2/dt2 = 2*a1* dtau1/dt1 (Eq.(11) on my web page)

This equation can obviously only hold true if a2=2*a1 .

Eq.(8) is generally valid, so it can also be applied to a function of the form corresponding to the Lorentz transformation tau = alpha (t-v*x/c^2), which yields

(9) a2 *alpha = 2*a1 *alpha .

This again only holds true if a2=2*a1 . So in general the derivative of tau is non-existent, and therefore tau as well. The agreement that you appeared to get for the specific case of the Lorentz transformation was only due the fact that you applied the total derivative incorrectly by pretending the variables x and t are independent.

None of the equations that Einstein defines ( http://www.fourmilab.ch/etexts/einstein/specrel/www/ ) depends on the light signal being reflected back to the origin. It might as well just carry on in the original direction (by the same distance). The only difference is that the second leg would now also take a time r_AB/(c-v) rather than r_AB/(c+v), which however would result in an equation inconsistent with his original one if one follows through his procedure (namely it would result in dtau/dx'=0).

Thomas

Last edited: May 14, 2010

7. ### tsmidRegistered Senior Member

Messages:
368
On the contrary, if you use one Schrödinger equation (with one continuous wave function) to describe two or more unconnected systems of particles, then this is obviously inappropriate and wrong.

You conveniently omitted the rest of my argument, where I said: "And trying nevertheless to define one is merely a logical contradiction in terms (and just calling it 'quantum mechanical' doesn't change anything about it; the energy conservation law would still be violated for instance)".

In other words, what you are doing is to basically enforce arbitrarily a single continuous wave function for the whole universe, and then 'concluding' from this that particles can miraculously traverse potential differences larger than their kinetic energy.

So how is this 'potential wall' then physically realized in practice (e.g. for the example you mentioned, the scanning tunneling microscope)?.

Thomas

Last edited: May 13, 2010
8. ### przyksquishyValued Senior Member

Messages:
3,171
Well then you're wrong right there from the outset. The Lorentz transformation is a function of two independent variables (actually four in general). It's supposed to be a mapping of the coordinates x and t to the coordinates x'(x, t) and t'(x, t). If you substitute in a particular path, then yes,
$\tau_{1}(x^{\prime}) \,=\, \tau \bigl( x_{1}(x^{\prime}),\, t_{1}(x^{\prime}) \bigr)$​
and
$\tau_{2}(x^{\prime}) \,=\, \tau \bigl( x_{2}(x^{\prime}),\, t_{2}(x^{\prime}) \bigr)$​
are two single-valued functions of x'. But $\tau(x,\,t)$ itself isn't and it's the differentials of $\tau$ with respect to its two independent parameters that appear after you apply the chain rule. Which leads to this:
The whole point of the chain rule is that you separate out how much of the derivative of $\tau_{1}(x^{\prime})$ is due to the partial derivatives of $\tau$ itself and how much is due to the paths. Your statement about the chain rule is a misconception and it is easy to prove that it leads to errors. Just take this function:
$\tau(x,\,t) \,=\, t \,-\, \frac{v}{c^{2}} x$​
(I've explicitly set $\alpha = 1$ - you can take any other value you like if you want) evaluated along this path:
\begin{align} x_{2}(x^{\prime}) \,&=\, v \, a_{2} \, x^{\prime} \\ t_{2}(x^{\prime}) \,&=\, a_{2} \, x^{\prime} \end{align}​
(borrowing your own definition of a[sub]2[/sub]) as an example. If you define
$\tau_{2}(x^{\prime}) \,=\, \tau \bigl( x_{2}(x^{\prime}),\, t_{2}(x^{\prime}) \bigr)$​
and substitute in the definition of $\tau$, you find that $\tau_{2}(x^{\prime}) \,=\, \frac{2 x^{\prime}}{c}$, and so:
$\frac{\text{d}\tau_{2}}{\text{d}x^{\prime}} \,=\, \frac{2}{c}$​
If you apply (your variant of) the chain rule first and then substitute in $\tau$, then according to you:
$\frac{\text{d}\tau_{2}}{\text{d}x^{\prime}} \,=\, 2 a_{2} \, \frac{\partial \tau_{2}}{\partial t_{2}} \,=\, \frac{2}{c-v} \,+\, \frac{2}{c+v}$​
So the way you're applying the chain rule clearly does not produce the correct value for the derivative of $\tau_{2}$.

The derivation on your page effectively "proves" that:
$\frac{\alpha}{c - v} \,+\, \frac{\alpha}{c+v} \,=\, \frac{2 \alpha}{c-v}$​
$\frac{\alpha x^{\prime}}{c} \,=\, \frac{\alpha x^{\prime}}{c}$​
(an identity). Do you not understand that this result is wrong regardless of how you derived it?

The sychronicity constraint is supposed to be a statement deciding when two moving clocks are synchronised. The argument used to derive it makes no sense if you put in a third clock. Incidentally, if you do put in a third clock and require something like $\frac{1}{2} \bigl( \tau_{1} \,+\, \tau_{3} \bigr) \,=\, \tau_{2}$, then you'll just end up with a trivial equation that's true for any linear transformation. It doesn't actually constrain anything.

Er, you mean your procedure. You've already shown you don't understand Einstein's.

Last edited: May 15, 2010
9. ### przyksquishyValued Senior Member

Messages:
3,171
Why? Because you say so? As usual, that's the only argument you've been able to present so far.

No I didn't. I dismissed your argument on the basis that you were reasoning classically about a quantum system. And this:
is classical reasoning about a quantum system. Yes, in classical mechanics, a particle of total energy E can't be located somewhere where V > E. But quantum mechanics doesn't hold that statement to be true in general (it doesn't even hold that statement to be particularly meaningful), so there's no contradiction.

Not me and not arbitrarily. Quantum mechanics imposes this. A discontinuous wave-function can't ever be a solution to the Schroedinger equation. I've already proved this. Why do you keep pretending I'm "arbitrarily" claiming a statement I've actually proved?

Why should this be "miraculous"? Because classical mechanics says it can't happen?

It's context-dependent in general. Quantum tunneling is a generic model that appears in a variety of different contexts (Alphanumeric gave some examples), a lot like the harmonic oscillator does. In scanning tunneling microscopy, the "potential wall" is simply vacuum that bound electrons travel through from a conducting tip to the surface of a material.