Was Einstein lucky when not considering twin paradox as paradox?

[stevendaryl]

The stipulations are the ones that the reader is supposed to know

I would say that this discussion is all about what it means for a coordinate system to be "valid", and that a lot of misconceptions about SR result from not being clear about what that means.

 

[stevendaryl]

Don't you also have to assume transformation laws for the objects of the equations of physics? Or are you bundling that into what you mean by 'equations of physics'?

I realize that I'm on a little shaky grounds here. I'm not positive that what I said was wrong, but I'm not positive that what I said was correct, either.

The question, which I don't know the answer to, is: What goes wrong if you make the incorrect assumption about the transformation properties of scalars, vectors, tensor, etc.?

It might be worth while to work out an example.

Suppose you have an equation that is correct in one coordinate system, for example, the geodesic equation:

$m \frac{d^2 x^\mu}{ds^2} = F^\mu$

Let me introduce $U^\mu = \frac{dx^\mu}{ds}$. Then it's

$m \frac{d U^\mu}{ds} = F^\mu$

The difficulty is that although I've written $F^\mu$ as a vector, it may actually be a combination of a vector force together with connection coefficients.

So, change coordinates to $x'^\alpha$ and define $L^\mu_\alpha = \partial_\alpha x^\mu$. Then

$U^\mu = L^\mu_\alpha U^\alpha$ and we get an equation for $U^\alpha$:

$m (L^\mu_\alpha \frac{dU^\alpha}{ds} + (\partial_\beta L^\mu_\alpha) U^\alpha U^\beta) = F^\mu$

Multiplying by the inverse transformation matrix gives:
$m (\frac{dU^\alpha}{ds} + (L^{-1})^\alpha_\mu (\partial_\beta L^\mu_\gamma) U^\gamma U^\beta) = (L^{-1})^\alpha_\mu F^\mu$

So, we can define the "effective 4-force" in the new coordinates to be:

$F'^\alpha = (L^{-1})^\alpha_\mu F^\mu - m((L^{-1})^\alpha_\mu (\partial_\beta L^\mu_\gamma) U^\gamma U^\beta)$

So in the new coordinate system, we have the equation of motion:

$m \frac{dU^\alpha}{ds} = F'^\alpha$

So now I'm not sure--where is it necessary to know which things are vectors, tensors or scalars? I think the problem comes in with the physical meaning of $F'^\alpha$. Even if $F^\mu$ had some simple interpretation, in terms of scalar or vector fields, the new "force" $F'^\mu$ would have a very complicated definition. But the equations of motion would still work, wouldn't they, even though you've misidentified a connection coefficient term as a 4-vector?

 

[PAllen]

I realize that I'm on a little shaky grounds here. I'm not positive that what I said was wrong, but I'm not positive that what I said was correct, either.

The question, which I don't know the answer to, is: What goes wrong if you make the incorrect assumption about the transformation properties of scalars, vectors, tensor, etc.?

It might be worth while to work out an example.

Suppose you have an equation that is correct in one coordinate system, for example, the geodesic equation:

$m \frac{d^2 x^\mu}{ds^2} = F^\mu$

Let me introduce $U^\mu = \frac{dx^\mu}{ds}$. Then it's

$m \frac{d U^\mu}{ds} = F^\mu$

The difficulty is that although I've written $F^\mu$ as a vector, it may actually be a combination of a vector force together with connection coefficients.

So, change coordinates to $x'^\alpha$ and define $L^\mu_\alpha = \partial_\alpha x^\mu$. Then

$U^\mu = L^\mu_\alpha U^\alpha$ and we get an equation for $U^\alpha$:

$m (L^\mu_\alpha \frac{dU^\alpha}{ds} + (\partial_\beta L^\mu_\alpha) U^\alpha U^\beta) = F^\mu$

Multiplying by the inverse transformation matrix gives:
$m (\frac{dU^\alpha}{ds} + (L^{-1})^\alpha_\mu (\partial_\beta L^\mu_\gamma) U^\gamma U^\beta) = (L^{-1})^\alpha_\mu F^\mu$

So, we can define the "effective 4-force" in the new coordinates to be:

$F'^\alpha = (L^{-1})^\alpha_\mu F^\mu - m((L^{-1})^\alpha_\mu (\partial_\beta L^\mu_\gamma) U^\gamma U^\beta)$

So in the new coordinate system, we have the equation of motion:

$m \frac{dU^\alpha}{ds} = F'^\alpha$

So now I'm not sure--where is it necessary to know which things are vectors, tensors or scalars? I think the problem comes in with the physical meaning of $F'^\alpha$. Even if $F^\mu$ had some simple interpretation, in terms of scalar or vector fields, the new "force" $F'^\mu$ would have a very complicated definition. But the equations of motion would still work, wouldn't they, even though you've misidentified a connection coefficient term as a 4-vector?

What if you misinterpret your initial equation as a collection of 4 scalar equations and transform it? Then you get the same force at any point as you did in the original coordinates. I guess you could still try to argue that this 'works' with a force definition that says all forces must be measured by instruments at rest in in the starting coordinates. Do you really want to argue that?

 

[stevendaryl]

What if you misinterpret your initial equation as a collection of 4 scalar equations and transform it? Then you get the same force at any point as you did in the original coordinates. I guess you could still try to argue that this 'works' with a force definition that says all forces must be measured by instruments at rest in in the starting coordinates. Do you really want to argue that?

That's sort of what I was getting at: You can do a coordinate change without understanding the nature of the terms involved to get equations in the new coordinate system, but it's the physical interpretation of terms in the new coordinate system that are obscure (or extremely convoluted).

 

[stevendaryl]

What if you misinterpret your initial equation as a collection of 4 scalar equations and transform it? Then you get the same force at any point as you did in the original coordinates. I guess you could still try to argue that this 'works' with a force definition that says all forces must be measured by instruments at rest in in the starting coordinates. Do you really want to argue that?

That particular mixup doesn't make sense to me. I'm assuming that what IS known is the coordinates in the two systems. If you don't know that $x^\mu$ is a coordinate, then I don't know what it would mean to say that you know what the coordinate transformation is.

 

[PAllen]

That particular mixup doesn't make sense to me. I'm assuming that what IS known is the coordinates in the two systems. If you don't know that $x^\mu$ is a coordinate, then I don't know what it would mean to say that you know what the coordinate transformation is.

Sure it does. Consider Fx is a function of all 4 coordinates, so is Fy, so is Fz. Now, you just assume you substitute that new coordinate definitions, treating Fx, Fy, Fz as scalar functions. If you consider them as vector something else; as a covector, something else; as a density, something else.

 

[stevendaryl]

Sure it does. Consider Fx is a function of all 4 coordinates, so is Fy, so is Fz. Now, you just assume you substitute that new coordinate definitions, treating Fx, Fy, Fz as scalar functions. If you consider them as vector something else; as a covector, something else; as a density, something else.

My derivation didn't make any assumptions about the nature of $F^\mu$.

Once again, if $U^\mu$ satisfies

$m \frac{dU^\mu}{ds} = F^\mu$

and

$U^\mu = L^\mu_\alpha U^\alpha$

then

$m (L^\mu_\alpha \frac{dU^\alpha}{ds} + (\partial_\beta L^\mu_\alpha) U^\alpha U^\beta) = F^\mu$

is an equation of motion for $U^\alpha$. You don't have to know anything about how $F^\mu$ transforms. Whether it's 4 scalars, or a 4-vector, or what not doesn't seem to matter.

 

[PAllen]

My derivation didn't make any assumptions about the nature of $F^\mu$.

Once again, if $U^\mu$ satisfies

$m \frac{dU^\mu}{ds} = F^\mu$

and

$U^\mu = L^\mu_\alpha U^\alpha$

then

$m (L^\mu_\alpha \frac{dU^\alpha}{ds} + (\partial_\beta L^\mu_\alpha) U^\alpha U^\beta) = F^\mu$

is an equation of motion for $U^\alpha$. You don't have to know anything about how $F^\mu$ transforms. Whether it's 4 scalars, or a 4-vector, or what not doesn't seem to matter.

But I'm considering F as a field, no expression in terms of U. For example, something like Maxwell's equations. You have these functions of coordinates. To transform the equation to other coordinates, at all, you have to make some assumptions.

 

[harrylin]

I would say that this discussion is all about what it means for a coordinate system to be "valid", and that a lot of misconceptions about SR result from not being clear about what that means.

I thought that it was sufficiently clarified in this thread. And I don't recall ever having seen misconceptions about SR because people did not understand what it means for a coordinate system to be "valid" for the laws of physics.

In this context, it's basically the consequence of applying the principle of relativity (or of relative motion) as formulated by Einstein and earlier by Poincare.
- https://en.wikisource.org/wiki/Science_and_Hypothesis/Chapter_7
For good understanding: it was also expressed as the "impossibility to detect absolute motion", because the same laws of physics are observed in systems that are moving relative to each other.

According to the theory of Special Relativity, this principle is true for the inertial reference systems of classical mechanics, and SR's laws of physics are expressed with respect to a system of that class; similarly, its transformation equations are specified for systems of that class. Consequently the use of a reference system K' in arbitrary motion is "at your own risk". The twin "paradox" in SR illustrates nicely that non-inertial reference systems are not valid systems for application of the Lorentz transformations with SR.

However, according to 1916 GR, reference systems in any state of motion must be valid in that sense, with the physics of that theory - as Einstein defended in 1918:
"It is certainly correct that from the point of view of the general theory of relativity we can just as well use coordinate system K' as coordinate system K."
I discussed that issue in post #140 in this thread.

PS I came across a strange remark in Moller's textbook: he states that according to the special theory of relativity 'the special principle of relativity is valid for all physical laws', and adds the footnote: 'With the exception of the laws of gravitation [..]'.
I do think that gravitation works the same according to SR independent of the month of the year - the Earth's gravitational field does not act differently in March as in September!

 

[stevendaryl]

I thought that it was sufficiently clarified in this thread. And I don't recall ever having seen misconceptions about SR because people did not understand what it means for a coordinate system to be "valid" for the laws of physics.

I would say that's exactly what's going on when people claim that the twin paradox is not resolved by SR, or when they claim that it requires GR to use a coordinate system in which an accelerating observer is at rest. So I would say that this very thread is an example.

 

[stevendaryl]

For good understanding: it was also expressed as the "impossibility to detect absolute motion", because the same laws of physics are observed in systems that are moving relative to each other.

Everyone observes the same laws of physics, no matter what their state of motion. Everyone sees the same universe, after all. It's funny that you say that there is no need to clarify what "valid" means, when it sure seems to me that the concept is muddled.

The issue is in terms of the mathematical form of the equations of motion expressing those laws. For a given law, there may be a set of coordinate systems for which that law takes an exceptionally simple form. If by "valid coordinate system" you mean "a coordinate system in which the laws look simplest", then there might be a limited number of valid coordinate systems. But to me, that's a bizarre criterion. If you do Newton's mechanics using polar coordinates, the form of the equations of motion are changed. There are is an additional term in the equations, sometimes called "centrifugal force":

Instead of $m \frac{d^2 r}{dt^2} = F^r$, you get $m (\frac{d^2 r}{dt^2} - m r (\frac{d\theta}{dt})^2) = F^r$

Does that mean that polar coordinates are not "valid"?

The twin "paradox" in SR illustrates nicely that non-inertial reference systems are not valid systems for application of the Lorentz transformations with SR.

I think that is not a very clear way to put it. The Lorentz transformation is a coordinate transformation connecting two systems of coordinates. If $(x,t)$ is an inertial coordinate system, and $(x',t')$ is a noninertial coordinate system, then they are not related by a Lorentz transformation, in the same way that rectangular coordinates are not related to polar coordinates through a Galilean transformation. But that doesn't say anything about the "validity" of noninertial coordinates.

The Lorentz transformations are mathematics. The physical content comes in when you operationally define two coordinate systems (for example, you define how they would be set up using standard clocks and measuring rods in various states of motion). Then the claim that two operationally defined coordinate systems are related by a Lorentz transformation is an empirical claim.

However, according to 1916 GR, reference systems in any state of motion must be valid in that sense, with the physics of that theory - as Einstein defended in 1918:
"It is certainly correct that from the point of view of the general theory of relativity we can just as well use coordinate system K' as coordinate system K."
I discussed that issue in post #140 in this thread.

Yes, that's the quote that I'm saying is very much misleading. Yes, GR allows you to use coordinate system K', but so does SR. Using system K' in SR is no more problematic than using polar coordinates in Newton's mechanics.

 

[Dale]

This is just going around in circles. Time to move on.