stevendaryl
Staff Emeritus
To illustrate how confused it is to think that GR is needed to use coordinates in which an accelerated observer is at rest, take a look at Rindler coordinates. It's a coordinate transformation from

$(x,t) \Rightarrow (X,T)$

where $T = tanh^{-1}(\frac{ct}{x})$ and $X = \sqrt{x^2 - c^2 t^2}$

In terms of the coordinates $(X,T)$, you find that
1. Clocks at "rest" (that is, $X$ is constant) run faster the higher up they sit (larger values of $X$)
2. Light rays bend downwards (toward negative values of $X$).
3. An object dropped from "rest" will accelerate downward (decreasing $X$)
These facts don't require GR, they are derivable from SR alone.

There is no need for a "principle of equivalence" to allow us to use these coordinates, any more than there is a need for a principle of equivalence to use Newtonian physics in polar coordinates.

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Can you provide a reference for this "GR solution" that agrees with the SR comoving frames as an example?

I've seen it done in lots of places, over the years. And I've never seen its solution NOT agree with the co-moving inertial frames solution. I've never seen it agree with the Dolby&Gull solution. It always says that, anytime the traveler's rocket is off, the traveler concludes that the home twin is aging more slowly, as given by the standard time-dilation result. I.e., the GR solution always says that the traveler's perspective is the same as an inertial observer whenever the rocket is off. And that ALL of the fast aging of the home twin, according to the traveler, occurs only while the rocket is firing. That is different than the D&G solution. If there are multiple "equally good" SR solutions for the traveler's perspective, why aren't there multiple "equally good" GR solutions for the traveler's perspective?

I don't remember all the places I've seen that GR solution given, but I do remember that it is quite common. I think the Wikipedia page on the twin paradox does it (and it may give some specific references). And I think that the link given by JessM to John Baez's webpage on the twin paradox gives it (perhaps also with references). I think I might have also seen it in MTW's "Gravitation" book, and in Born's "Einstein's Theory of Relativity" book.

Dale
Mentor
2020 Award
I've seen it done in lots of places, over the years.
I need an example so that I can understand what you are referring to. I simply don't know what you mean by a "GR solution". If it has been done in lots of places then I am sure that I have read one, but I don't recognize it as what you are calling "a GR solution". Every solution that I am aware of is what I would call "an SR solution" since the spacetime is flat.

I'm not saying that you are wrong, I am just saying that I don't know what you are referring to. I suspect that we are talking about the same things and just using different words. Also, are you talking about a quantitative solution, or simply a hand-waving qualitative explanation?

A.T.
These facts don't require GR,
Why is it so important whether some explanation is called a "SR-explanation" or a "GR-explanation"? Isn't SR a special case of GR, and therefore any SR-explanation also a GR-explanation?

As I said, GR in the limit of vanishing spacetime curvature simply IS SR.
Doesn’t this make the equivalence principle part of SR, since it only applies when spacetime curvature is negligible?

stevendaryl
Staff Emeritus
Why is it so important whether some explanation is called a "SR-explanation" or a "GR-explanation"? Isn't SR a special case of GR, and therefore any SR-explanation also a GR-explanation?

Sure. Every SR explanation is also a GR explanation. But not the other way around. So some problems require GR, and others don't. My point is that no problem involving flat spacetime requires GR.

Doesn’t this make the equivalence principle part of SR, since it only applies when spacetime curvature is negligible?

If spacetime is perfectly flat, then you don't need the equivalence principle, because everything can be done using SR alone. What the equivalence principle allows is to solve problems in curved spacetime by breaking spacetime into small regions and then approximating those regions by flat sections of spacetime.

To me, the equivalence principle is really the same, as far as physical content, to the two claims below:
1. Spacetime is curved.
2. "Gravitational forces" are actually connection coefficients due to the choice of a noninertial coordinate system. (If spacetime is curved, then there are no inertial coordinate systems except in the limit of small regions of spacetime).
It's a guess about the nature of gravity. You don't need the equivalence principle to get "inertial forces" (connection coefficients) for a noninertial coordinate system; that's derivable from the equations of motion in inertial coordinates.

harrylin said: "As explained in the summary which was posted twice here: the challenge that was thrown at his feet, was how to produce the same prediction about the twins as in SR, while using coordinates in which the "traveler" is claimed to be in rest all the time. That should be possible according to Einstein's GR postulate, as he there also acknowledged. [..] Following your inclination, I'm unable to discern Maxwell's theory of electrodynamics from Einstein's."

My point is that both the challenge and the response are based on the misconception that GR is any more (or less) capable of using coordinates in which the traveling twin is at rest than SR.
[..]
Maxwell's theory was already invariant under Lorentz transformations. Einstein's contribution was to develop an analogous theory of mechanics. Before Einstein, we had Maxwell's equations, which were invariant in form under Lorentz transformations, and Newton's laws of motion, which were invariant in form under Galilean transformations. Einstein united the theories by modifying Newton's theory to get one that was invariant under Lorentz transformations, as well. He didn't need to modify Maxwell's equations.
[..]
If you know the equations of motion in one coordinate system, then you know the equations of motion in every coordinate system. You don't need an additional postulate that they are invariant under such and such a transformation, it's just a fact of the equations.[..]
Maxwell's theory assumed the use of Newton's transformations. I think that this is a good example to explain the difference between a theory and the equations of a theory, but apparently it is necessary to elaborate:

- the equations are the same
- the same coordinate transformations are mathematically possible with both theories

Nevertheless they were not the same theory because their physical assumptions differed on an essential point.
For MMX they even made contrary predictions!

Thus I clarified:
" the difference is not in Maxwell's equations; the difference is primarily in the postulates about the coordinate systems in which the equations are claimed to be valid."
"harrylin said: The equivalence principle of 1916 says that acceleration can be modeled as gravity"

That's a muddled way of thinking about it. [..]
[..] What you need the equivalence principle for is to predict that clocks and light beams work in a similar way on the surface of a massive planet. That's the physical content of the equivalence principle, that light curves near a massive planet, and that clocks run at different speeds at different altitudes. [..]
I certainly agree with that; however my dislike for GR "vintage 1916" (as the FAQ calls it) doesn't bring me to deny the facts about it. It appears that you even deny the meaning of the term "GR"! The new theory of gravitation was a fantastic spin-off of Einstein's theory of the general relativity of motion. He put it as follows:

"From these discussions we see, that the working out of the general relativity theory must, at the same time, lead to a theory of gravitation; for we can "create" a gravitational field by a simple variation of the co-ordinate system." (emphasis mine)

However, in the light of this discussion it strikes me that his phrasing here lacks precision, so that it is bound to be misunderstood if one does not already understand what he means. No doubt, if the issue had been brought up, he would have agreed that a mere coordinate transformation will not create a gravitational field. The creation of a gravitational field only occurs in his theory when one reinterprets the accelerating system as a non-accelerating system - one that is "in rest".

A similar situation occurs in SR when one Lorentz transforms from a "stationary" system S to a "moving" system S'. The transformation itself does not yet impose on an observer who is co-moving with S' to pretend not to be moving but to be in rest; however, the observer may do that if he/she so desires. This adaptation of interpretation to the used coordinate system is implied in most discussions.

There is no additional physics involved in interpreting "inertial forces" as gravity. It's just what you name terms in the equations of motion.
A sudden inertial force on a test particle doesn't create a gravitational wave, while - I think - a suddenly "induced gravitational field" certainly must do so. That is the central point of my criticism of Einstein's 1918 paper (which I did not yet present), and with that, of 1916 GR.

I consider that paragraph to be misleading, if it encourages people to think that you need GR to be able to reason as if the accelerated observer is at rest. That is purely SR + a coordinate transformation, and GR adds nothing. Well, what it adds is that you may need the inspiration of GR to get you thinking about the use of noninertial coordinate systems.
That does not merely encourage people to think that, it's what SR as well as 1916GR assert. The claim that the accelerated twin is not changing velocity so that the traveler can be considered to be all the time in rest, leads in SR to the twin paradox! We and others discussed this earlier in this same thread and you even brought it up in your reply. The following is according to SR not true in an SR "rest system":
"light beams appear to curve downward, and [..] clocks run at different speeds in different locations of the rocket."
To illustrate how confused it is to think that GR is needed to use coordinates in which an accelerated observer is at rest, take a look at Rindler coordinates. [..]
Once more: Einstein did not pretend to need GR in order to use coordinates in which an accelerated observer is at rest - quite the contrary, he used such SR coordinates as input for GR!
Perhaps the phrasing "at rest in an accelerated coordinate system" causes confusion. While one certainly can sit down and "rest" in an accelerating rocket, that does not change the physical interpretation of an accelerating system. It may be better to speak of "co-moving" with an accelerated coordinate system.

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stevendaryl
Staff Emeritus
Nevertheless they were not the same theory because their physical assumptions differed on an essential point.
For MMX they even made contrary predictions!

This is sort of a complicated subject, but the way I understand it is this:
• Maxwell's equations are invariant under Lorentz Transformations.
• Newton's equations are invariant under Galilean Transformations.
• Therefore, there is only one frame in which they both take on their usual forms.
What that means is that either Newton's equations or Maxwell's equations have to be modified if you change frames. I think that everyone assumed that it was Maxwell's equations that had to be modified; that they were only valid as is in the frame in which the "ether" was at rest.

But is it really true that the prediction of a nonnull result from the Michaelson-Moreley experiment was a prediction from the theory of electromagnetism? I don't think it was. I think it was a prediction from Newtonian physics. Newtonian physics says that if in one frame, you have an object that travels at speed $c$, then in another frame moving at speed $v$ relative to the first, that object travels at some speed between $c-v$ and $c+v$. The prediction of a nonnull result follows from Newtonian mechanics, not Maxwell's equations.

In any case, could you give a reaction to the following argument:

Assume that the usual equations of Special Relativity holds when written in terms of inertial coordinates $(x,y,t)$. Introduce new coordinates $(X,Y,T)$ related to the first two via:
• $X = \sqrt{x^2 - c^2 t^2}$
• $T = tanh^{-1}(\frac{ct}{x})$
• $x = X cosh(T)$
• $t = \frac{X}{c} sinh(T)$
• $Y = y$
Then when described in terms of these new coordinates,
• When an object is dropped from "rest", it accelerates downwards (towards smaller values of $X$)
• Clocks at "rest" that are "higher up" (at larger values of $X$) run faster.
• Light signals that are initially emitted in the $Y$ direction bend down (towards smaller values of $X$)
So the "general relativistic" effects of "gravitational time dilation" and "bending of light rays" and "objects falling under gravitational fields" seem to me to be purely effects of SR in noninertial coordinates. Not only do you not need the principle of equivalence to treat an accelerated observer as if he were at rest, the principle of equivalence plays no role, whatsoever. (Except maybe it allows you to use the word "gravity" when describing the above effects.)

So my questions are: (1) In what sense is the above discussion of Rindler coordinates not an "SR" argument? (2) How would the treatment be any different in a "GR" derivation?

[..] is it really true that the prediction of a nonnull result from the Michaelson-Moreley experiment was a prediction from the theory of electromagnetism? I don't think it was. I think it was a prediction from Newtonian physics. [..]The prediction of a nonnull result follows from Newtonian mechanics, not Maxwell's equations.
I referred to Maxwell's theory, in contrast with Maxwell's equations (in case you did not know it, Maxwell proposed MMX). And that was exactly my point: emphasizing the important difference between a theory of physics, and equations that are used as part of the theory.
Equations without the theory are like a piece of complicated equipment without the manual. :)

In any case, could you give a reaction to the following argument:
I will later, if it appears to still be useful (although I already provided a link to my reply, before you asked!). For could you please first react to :
- the remainder of my last post? (you apparently stopped reading at 1/3 of my post, while the next 2/3 gave a more elaborate explanation of the same)
- post #84? (it may well be that Phoebelasa explained it in a way that is clearer than the way I explained it; but apparently you overlooked it)

stevendaryl
Staff Emeritus
I referred to Maxwell's theory, in contrast with Maxwell's equations (in case you did not know it, Maxwell proposed MMX). And that was exactly my point: emphasizing the important difference between a theory of physics, and equations that are used as part of the theory.

Right, but as I said, the prediction of a nonnull result has almost nothing to do with Maxwell's theory. It follows from the fact that:
1. Light has speed c in at least one frame.
2. According to Newton's laws, velocity transforms as $c \Rightarrow c\pm v$
So a nonnull result for MMX was mostly a prediction about Newtonian physics, it seems to me.

Right, but as I said, the prediction of a nonnull result has almost nothing to do with Maxwell's theory. It follows from the fact that:
1. Light has speed c in at least one frame.
2. According to Newton's laws, velocity transforms as $c \Rightarrow c\pm v$
So a nonnull result for MMX was mostly a prediction about Newtonian physics, it seems to me.
Newton's theory would have predicted a null result, assuming his corpuscular light hypothesis.
Maxwell's theory predicted a positive result, because he assumed the validity of the Galilean transformations.

stevendaryl
Staff Emeritus
Here's another take on describing what was going on in the MMX:

If you know the equations of physics in one coordinate system $x^\alpha$ and you know how to transform from that coordinate system to a second $x'^\mu$, then you know the equations of physics in the second coordinate system. There is no additional physical assumption required to be able to use the new coordinate system. There is no empirical test as to whether the new coordinate system is "valid" or not.

But here's where physical assumptions come into play: Suppose you have two frames, $F$ and $F'$. You set up corresponding coordinate systems $x^\alpha$ and $x'^\mu$ using some physical convention for measuring distances and times. So you use standard clocks and metersticks at rest in $F$ to define the coordinate system $x^\alpha$, and you use standard clocks and metersticks at rest in $F'$ to define the coordinate system $x'^\mu$. Then you don't know what is the mathematical relationship between the primed and unprimed coordinate systems, without physical assumptions about the nature of clocks and metersticks. So you actually don't know what the equations look like in the new coordinate system until you perform empirical tests.

stevendaryl
Staff Emeritus
Newton's theory would have predicted a null result, assuming his corpuscular light hypothesis.
Maxwell's theory predicted a positive result, because he assumed the validity of the Galilean transformations.

Nobody is talking about Newton's theory of light--I'm talking about his laws of motion.

Anyway, I have to vigorously protest the phrase "the validity of the Galilean transformations". That is a meaningless phrase without some additional stipulations. There is no such thing as a valid or invalid coordinate transformation. You can use whatever coordinates are convenient; any are as "valid" as any other.

The real issue is the specific conventions for setting up coordinates in a frame. If you use physical objects, such as clocks, rods, light signals, or whatever, to measure distance and times, and you use those distances and times as the basis for a coordinate system, then it's an empirical question how such a coordinate system depends on the frame of rest of those clocks and rods. So it's not a matter of the Galilean transformations being "valid", but a matter of whether they correctly describe the relationship between two operationally defined coordinate systems.

PAllen
If you know the equations of physics in one coordinate system $x^\alpha$ and you know how to transform from that coordinate system to a second $x'^\mu$, then you know the equations of physics in the second coordinate system. There is no additional physical assumption required to be able to use the new coordinate system. There is no empirical test as to whether the new coordinate system is "valid" or not.
.
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 have to vigorously protest the phrase "the validity of the Galilean transformations". That is a meaningless phrase without some additional stipulations. [..].
The stipulations are the ones that the reader is supposed to know; else we can't even write "SR" in a discussion, because it is meaningless without a clarification of what it is an abbreviation.

[..] Einstein did not even consider the twin paradox as problematic at all [..] ?
I will now expand on my earlier comments.

Although the twin example was not an issue for him in connection with what he named the "special" theory of relativity, it became an issue for him in connection with his "general" theory. He discussed that in his 1918 paper https://en.wikisource.org/wiki/Dialog_about_Objections_against_the_Theory_of_Relativity , and PhoebeLasa gave a summary of that section as follows:
I think there was a specific reason Einstein used GR to resolve the twin paradox. He wanted to construct an analogous scenario (via the equivalence principle) in which the "rocket-twin" could say that he was absolutely stationary and unaccelerated during the whole time that the twins were separated. When he fired his rocket engine, he was doing it strictly to counteract the spatially-uniform gravitational field that is somehow momentarily switched on, so that the rocket-twin would remain stationary and unaccelerated. That momentarily switched-on gravitational field causes the "home twin" (the twin who has no rocket) to accelerate, reverse course, and move toward the "traveler". The resulting conclusion using this GR scenario is that the rocket-twin will say that the "home-twin" suddenly gets much older while that gravitational field is switched on.

The exact same result (regarding the rocket-twin's conclusion about the home-twin suddenly getting much older during the turnaround) is obtained without recourse to GR (and without any gravitational fields), purely from SR, using a non-inertial reference frame for the rocket-twin which is formed by piecing together multiple inertial frames that are each momentarily co-moving with the rocket-twin at different instants of his life. The rocket-twin is always at the spatial origin of his non-inertial reference frame, but he never contends that he doesn't accelerate. He knows that he accelerates, and reverses course, when he turns on his rocket. And he knows that it is the home-twin who is unaccelerated for the whole trip.

There is a difference between being "always absolutely at rest" (Einstein's GR scenario for the rocket-twin) versus "being always at the spatial origin of your own personal reference frame, but accelerating at will using your rocket engine" (the SR scenario). But what the rocket-twin says about the home-twin suddenly getting much older during the turnaround is exactly the same for both scenarios (even though it's a different twin doing the turnaround in the two cases).

The only glitch that I notice in the otherwise excellent summary above, is the following:
Einstein did not really "want to construct" a scenario in which the "rocket-twin" could say that he was absolutely stationary and unaccelerated during the whole time; instead it was asserted by critics that this must be possible according to Einstein's theory.
Also - and this is essential - he did not "use GR to resolve the twin paradox".
Instead he presented the twin scenario as one of the "Objections against the Theory of Relativity", that is, vintage 1916 GR. And this was his playful defence against accusations that 1916 GR is self-contradictory.

That criticism targeted the General postulate of relativity, according to which "The laws of physics must be so constituted that they should remain valid for any system of co-ordinates moving in any manner." - https://en.wikisource.org/wiki/The_...ain_the_extension_of_the_relativity-postulate.

Einstein accepted the challenge with the comment that the critic's 'assertion is of course indisputable'. And near the end of his defence, Einstein states that the theory 'means for a man who maintains consistency of thought a great satisfaction to see that the concept of absolute motion, to which kinematically no meaning can be attributed, does not have to enter physics'. No precise references are given at al, but almost certainly the objection was triggered by Langevin's 1911 paper which gives the first full "twin" example (using SR) to argue that a change of velocity is qualitatively "absolute". And this is just what GR was meant to make "relative".

In an earlier post I imprecisely (sorry!) stated that Moller and Builder criticized Einstein's 1918 paper. Moller's criticism is positive and in his 1952 textbook "The theory of relativity" (which I now again have at hand) he provides the calculation details that are missing in Einstein's paper. Professor Moller thus taught his students GR "vintage 1916". He surely understood* the issue, as he there explains the general principle of relativity and refers to the succession of key papers such as Einstein 1905, Langevin 1911, Einstein 1918.

Notwithstanding that great defence, I think that Builder's 1957 objection turns it into wood wreck - although he apparently never saw Einstein's paper and misunderstood the reason for the calculation; apparently he based his argument on his readings of Tolman and Moller. Historically, the "twin paradox" discussion is a continuous succession of misunderstandings. Thus, in his paper "The resolution of the clock paradox" (G. Builder, Aust. J. Phys. 10, 246–262, 1957), Builder argues that GR can add nothing to the solution that SR already provides. And he argues that 'any application of the principle of equivalence [..] to such cases would be quite trivial", simply because the calculated fields predict effects from acceleration that were used to calculate those fields in the first place. But he next adds a different objection, in disagreement with his earlier triviality argument(!), and this one I consider pertinent:

The [accelerated] reference system Sm does not correspond to any physical system that is realizable even in principle. This conclusion is not affected by the introduction of the concept of the equivalent gravitational ﬁeld. On the contrary, nothing could demonstrate more clearly the artificiality of the reference system Sm, than the statement that its physical equivalent is a gravitational ﬁeld which is everywhere zero until the instant tm=T', has the potential gxm(1+gxm/2c2) from tm=T' to tm=T’+tau’2, and becomes zero everywhere again at tm=T’ +tau'2.
The concept of such a ﬁeld is completely incompatible with the limiting value c for all velocities measured in inertial reference systems; [..] so that the speciﬁed ﬁeld would have to be created simultaneously at all points in S' and be destroyed simultaneously at all points in So.
Thus the principle of equivalence [..] only accentuates the artiﬁciality of the description of our hypothetical experiment in terms of the coordinates of the accelerated reference system Sm.

Indeed, according to GR any "induced gravitational field" must propagate at the speed of light. On top of that, what Builder overlooked or didn't bother to mention: an infinite speed of induction is also not allowed in a gravitational field according to GR, and that still does not suffice to match SR's predicted Doppler effect, as also the speed of starlight is finite - all the stars Doppler-shift instantly at turnaround.

As far as I know, none of the involved authors (Einstein, Tolman, Moller, ...) addressed that self-contradiction.

*Note: Moller makes the statement that SR "only allows treatment of the physical phenomena in frames of reference in uniform motion". Perhaps he means that SR's laws of physics are only valid in those reference frames, but it is easy to see how it can bring unaware readers to the misunderstanding that SR cannot describe observations from accelerating rockets!

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stevendaryl
Staff Emeritus
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
Staff Emeritus
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
Staff Emeritus
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
Staff Emeritus
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
Staff Emeritus
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.

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!

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stevendaryl
Staff Emeritus