# Potential Energy formula in Special Relativty

TL;DR Summary
What is the formula for potential energy of an object, falling with relativitic velocity in a pseudo-gravitational field in an accelerated frame?
Reference frame is an accelerated frame in SR (uniformly accelerated with "g" in flat spacetime). An object is falling with relativitic velocity of up to 0.8 c in the pseudo-gravitational field in this frame.

From Newton's theory, I know the formula for potential energy in such a scenario:

##W = m * g * h##.

For small velocities, this formula should also be usable in SR in the mentioned scenario. But what is at relativistiv velocities?

Shall I use in the formula for potential energy in hight "h" the rest mass (m₀) or the formerly called "relativistic mass" (m₀ * γ)?

##W = m_0 * g * h## or
##W = m_0 * \gamma * g * h## ?

Reason for the question: In the following paper about the "twin paradox", I don't understand the reason for including a factor γ in the formula for pseudo-gravitational potential, see equation (8), then compare with equation (3) in:
https://arxiv.org/ftp/arxiv/papers/1002/1002.4154.pdf

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2022 Award
Imagine an object released at height ##h## above the floor of a lift that is accelerating with proper acceleration ##\alpha## and is instantaneously at rest in an inertial frame. Continuing to work in that inertial frame, coordinate acceleration of the lift floor is ##a=\alpha/\gamma^3##, giving $$v(t)=\frac{\alpha t}{\sqrt{1+\alpha^2 t^2/c^2}}$$and hence the position of the floor is $$x(t)=\frac {c^2}\alpha\left(\sqrt{1+\alpha^2t^2/c^2}-1\right)$$Setting ##x(t)=h##, solving for ##t##, and substituting that into ##v(t)## gives us a messy ##v## but a ##\gamma## factor of ##1+\alpha h/c^2##. Thus in the frame of the floor, the kinetic energy of a mass ##m## falling a height ##h## is ##(\gamma-1)mc^2=m\alpha h##.

So it looks to me like there's no gamma factor. I'll take a look at the paper now.

• • Abhishek11235, vanhees71, Sagittarius A-Star and 1 other person
2022 Award
I'm not seeing any factors of ##\gamma## in equation 8 of Grøn's paper. ##L_0## is the rest distance to Alpha Proxima (does he mean Proxima Centauri?), which seems in accord with my result above.

I'm not seeing any factors of ##\gamma## in equation 8 of Grøn's paper. ##L_0## is the rest distance to Alpha Proxima (does he mean Proxima Centauri?), which seems in accord with my result above.
The ##\gamma## is implicitely included in ##L_0##, because formula (8) is meant for the traveling twin's frame, in which the distance of the twins is ##L## from equation (3) because of length contraction, and not ##L_0##.

2022 Award
First, I don't think equation (8) is "for" any frame - it's a statement of an invariant, the ratio of the proper times along the worldlines and ##L_0## has the meaning of rest length defined in equation 3. Second, the only time the rocket is ##L_0## away from the Earth is at turnaround when it's momentarily at rest anyway. Third, this is a smoothly accelerating rocket - it doesn't have a frame in the SR "inertial frame" sense, so there isn't a unique ##\gamma## factor attributable to it, so it would make no sense to put one into this formula without an integral of some sort.

Mentor
Reference frame is an accelerated frame in SR

This is not a sufficient specification. There are many possible "accelerated frames". You need to be more specific about which one you are talking about--for example, Rindler coordinates?--before we can answer your question.

In the following paper

The paper doesn't give a sufficient specification of what frame it is using either. I note that it does not appear to be peer-reviewed, which makes me suspect that the author knows other experts would be able to raise issues with it.

The paper also claims that GR is necessary to solve the twin paradox. This is false.

• vanhees71
Mentor
this is a smoothly accelerating rocket

The paper only actually applies equation (8) in the limit of infinitely large acceleration, i.e., instantaneous turnaround. For that case it reduces to equation (10). In other words, the paper is showing that, in the limit of infinitely large acceleration, the aging of the Earth twin in the traveling twin's "rest frame" (which the paper, as I noted in my previous post, does not actually define) does not approach zero, but instead approaches a finite positive limiting value.

• Ibix
First, I don't think equation (8) is "for" any frame - it's a statement of an invariant, the ratio of the proper times along the worldlines and ##L_0## has the meaning of rest length defined in equation 3.
Yes. But in another paper the same formula is used, and there they speak of a "graviational potential difference". So it must refer there to the accalerated twin's frame, see equation (22) in:
But there, they use also the non-length-contracted distance, what I don't understand.
Second, the only time the rocket is ##L_0## away from the Earth is at turnaround when it's momentarily at rest anyway.
Not in the limit of infinitely large acceleration, that @PeterDonis pointed out in post #7.
Third, this is a smoothly accelerating rocket - it doesn't have a frame in the SR "inertial frame" sense, so there isn't a unique ##\gamma## factor attributable to it, so it would make no sense to put one into this formula without an integral of some sort.
Except ##\gamma## cancels out because ##\gamma * L = L_0##, which might be constant in the limit of infinitely large acceleration.

Staff Emeritus
Newtonian methods aren't really going to cut it here.

In the Schwarzschild geometry, there is a fairly well known "effective potential" formulation. This is discussed in MTW's "gravitation", or online at https://www.fourmilab.ch/gravitation/orbits/

In the Schwarzschild geometry, it looks like

$$\left( \frac{dr}{d\tau} \right) ^2 + V^2(r) = E^2$$

You can get the value for V^2(r) from the reference, but it's not really relevant to the topic. The basic point is that there is a constant of motion that we call energy, and the proper velocity, ##dr/d\tau##, can be calculated as a function of the r-coordinate and the energy.

If my calculations are correct (and I'm not too confident yet they are), for the Rindler metric in geometric units with the coordinates (t,x) we write the line element as:

$$ds^2 = -\alpha^2 x^2 d\tau^2 + dx^2$$

the corresponding equation of motion is

$$\left( \frac{dx}{d\tau} \right)^2 = \frac{E^2}{\alpha^2 x^2} - 1$$

Note that x=0 in the RIndler metric is the "Rindler horizon". In the region where the metric is well behaved x > 0.

We can put this in the form

$$\alpha^2 x^2 \left( \frac{dr}{d\tau} \right) ^2 + \alpha^2 x^2 = E^2$$

which writes the total energy as the sum of a kinetic term, related to the proper velocity ##dx/d\tau##, and another term based only on position, x.

$$\alpha^2 x^2 \left( \frac{dr}{d\tau} \right) ^2 + \alpha^2 x^2 = E^2$$
which writes the total energy as the sum of a kinetic term, related to the proper velocity , and another term based only on position, x.
I am confused with a units plausibility-check. Reason: On the right side of the equation is an energy², on the left side no mass nor energy, but only space and time values.

2022 Award
On closer inspection, I don't like any of this.

Let's imagine a twin paradox where the traveller is already traveling at ##v## when passing the stay-at-home, and turns around with proper acceleration ##g## before returning at the same speed. The elapsed time for the stay-at-home is the time for the inertial phases (call this ##T## for each) plus the turnaround time. Using the expression for ##v(t)## in #2, the stay-at-home's coordinate time for turnaround is ##2v\gamma/g##, where ##\gamma## is associated with the inertial phase's speed ##v##. Thus the stay-at-home's elapsed time is ##2T+2v\gamma/g##.

The traveller experiences time ##2T/\gamma## during the inertial phases. I didn't write an expression for proper time in #2, but it can be obtained by deriving ##\gamma(t)##, observing that ##d\tau/dt=\gamma## and integrating. The result is ##\tau=(c/a)\mathrm{asinh}(a(t-t_0)/c)##, where ##t_0## is the coordinate time when the ship is momentarily at rest. We can plug in the start and end coordinate times of acceleration (##t_0\pm v\gamma/g##) from above to get a proper time under acceleration of ##(2c/g)\mathrm{asinh}(v\gamma/c)## and a total elapsed time for the traveller of ##2T/\gamma+(2c/g)\mathrm{asinh}(v\gamma/c)##. Note that this is not the same as Gron's equation 7, which appears to be just the coordinate time divided by the ##\gamma## associated with the initial speed ##v##.

If we apply naive reasoning to what "during" the inertial phases means for the traveller's estimate of the stay-at-home's age, then this accounts for ##2T/\gamma^2##. Thus the time experienced by the stay-at-home in what the traveller would call "during the acceleration phase must be" ##2T(1-\gamma^{-2})+2v\gamma/g=2Tv^2/c^2+2v\gamma/g##. This does not look like a nice gravitational potential term to me, although I see that in the limit ##g\rightarrow\infty## it does reduce to ##2L_0v/c^2##.

Note that the actual ratio of elapsed times "during" the acceleration phase is$$\begin{eqnarray*} \frac{\tau_s}{\tau_t}&=&\frac{2Tv^2/c^2+2v\gamma/g}{(2c/g)\mathrm{asinh}(v\gamma/c)}\\ &=&\frac{Tv^2g/c^2+\gamma v}{c\,\mathrm{asinh}(v\gamma/c)}\end{eqnarray*}$$which tends to infinity as ##g## tends to infinity. I think this is because, as @pervect notes, Rindler coordinates aren't well behaved at the Rindler horizon.

TLDR: I think Gron's equation 7 isn't the proper time of the traveller during acceleration, so I don't see the ratio of it to the elapsed time for the stay-at-home as physically significant.

• vanhees71
2022 Award
I am confused with a units plausibility-check. Reason: On the right side of the equation is an energy², on the left side no mass nor energy, but only space and time values.
@pervect is using natural units where ##c=1## - hence ##dr/d\tau## is dimensionless, ##\alpha## has dimensions of length-1, and ##x## has dimensions of length. Also, ##E## is total energy per unit mass, so is dimensionless in this system. The units are therefore consistent.

• Sagittarius A-Star
2022 Award
The paper only actually applies equation (8) in the limit of infinitely large acceleration, i.e., instantaneous turnaround. For that case it reduces to equation (10). In other words, the paper is showing that, in the limit of infinitely large acceleration, the aging of the Earth twin in the traveling twin's "rest frame" (which the paper, as I noted in my previous post, does not actually define) does not approach zero, but instead approaches a finite positive limiting value.
So he's trying to consider the pseudo-gravitational potential difference between something at the Rindler horizon and something above it? Does that makes sense? By analogy with Schwarzschild, no (although I'm not sure the analogy is exact), and pervect's comment excluding the horizon from consideration would certainly suggest not.

Mentor
Does this appear to be peer-reviewed?

Yes.

It has almost the same content, see for example equation (22) in there.

Some of the content appears similar, but the approach is very different, and frankly, I find this paper's general approach even more questionable than that of the Gron paper.

However, my issues with both papers are not relevant to the key point for this thread, which is the one I made in post #7. Both papers derive that result--it's equation (26) in the Guerra/Abreu paper--and both derivations look correct to me (and do not depend on the other stuff in the papers that I find questionable). The only missing piece, as I mentioned before, is that the papers do not explicitly say what "non-inertial frame" they are using; but it seems to me that they are probably (implicitly) using Rindler coordinates to describe the non-inertial "turnaround" portion of the traveling twin's trip. At any rate using those coordinates gives the result they give.

Also note that reference  in this paper is a paper by Gron which was published in a peer-reviewed journal, and which appears to be similar in content (though the title is different and the year of publication is 2006) to the arxiv paper by Gron.

• vanhees71
Mentor
So he's trying to consider the pseudo-gravitational potential difference between something at the Rindler horizon and something above it?

In the limit of infinite acceleration, yes, the traveling twin would be at the Rindler horizon. However, it's only a limit, and the limit is mathematically valid. Physically, of course, infinite acceleration is impossible; in any real case, the traveling twin would, of course, have some finite acceleration during the turnaround and his Rindler horizon would be some finite distance below him. The purpose of the limit is simply to make the point that I made in post #7: that the "elapsed time during the turnaround" of the stay-at-home twin does not go to zero in the limit of infinite acceleration and zero elapsed time for the turnaround of the traveling twin; it approaches a finite positive value. So idealizing the turnaround as taking a negligibly short period of time, while it does remove the need to calculate the elapsed turnaround time for the traveling twin (which will indeed be negligibly short), does not allow us to ignore the elapsed time during the turnaround for the stay-at-home twin.

Also note that reference  in this paper is a paper by Gron which was published in a peer-reviewed journal, and which appears to be similar in content (though the title is different and the year of publication is 2006) to the arxiv paper by Gron.
Here, reference  seems to be freely available and also downloadable as PDF:

Equation (2.2) shows the line element of the uniformly accelerated reference frame.

• FactChecker
jartsa
Imagine an object released at height ##h## above the floor of a lift that is accelerating with proper acceleration ##\alpha## and is instantaneously at rest in an inertial frame. Continuing to work in that inertial frame, coordinate acceleration of the lift floor is ##a=\alpha/\gamma^3##, giving $$v(t)=\frac{\alpha t}{\sqrt{1+\alpha^2 t^2/c^2}}$$and hence the position of the floor is $$x(t)=\frac {c^2}\alpha\left(\sqrt{1+\alpha^2t^2/c^2}-1\right)$$Setting ##x(t)=h##, solving for ##t##, and substituting that into ##v(t)## gives us a messy ##v## but a ##\gamma## factor of ##1+\alpha h/c^2##. Thus in the frame of the floor, the kinetic energy of a mass ##m## falling a height ##h## is ##(\gamma-1)mc^2=m\alpha h##.

So it looks to me like there's no gamma factor. I'll take a look at the paper now.

Well I think the object should not be released, but launched at high downwards speed at height ##h## above the floor. Maybe then there would be a gamma factor in the potential energy.

I mean if we are interested about the potential energy of a fast moving object then the object should be moving fast.

Of course the launching at high speed could be done by lifting the object to high altitude and then releasing. When lifted the object would gain potential energy ##m\alpha_1 L## according to person at height h, ##m\alpha_2 L## according to person at floor. L is the lifting distance.

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Mentor
I think Gron's equation 7 isn't the proper time of the traveller during acceleration

It's not exactly that, but it's a plausible approximation. The exact formula if we assume constant proper acceleration is

$$a \tau = 2 \tanh^{-1} v = 2 \omega$$

where ##a## is the proper acceleration, ##\tau## is the proper time of the traveling twin during turnaround, ##v## is the speed of the traveling twin relative to the stay-at-home twin during the inertial legs, and ##\omega## is the rapidity corresponding to that speed. (I'm using units in which ##c = 1##.)

In other words, the correct relativistic intuition is that the change in rapidity is equal to acceleration times proper time, whereas the non-relativistic intuition is that the change in speed is equal to acceleration times time. The latter is where Gron's equation 7 comes from.

How big is the error involved in Gron's approximation? For small ##v##, ##\tanh^{-1} v \approx v##, which is why the non-relativistic approximation works. For ##v = 0.8##, we have ##\tanh^{-1} v = 1.1##, so Gron's formula is underestimating the proper time for that ##v## by a factor ##1.1 / 0.8 = 1.375##.

Mentor
Equation (2.2) shows the line element of the uniformly accelerated reference frame.

That is the Rindler line element for the case where ##x = 0## corresponds to the traveling twin (instead of the Rindler horizon). So Gron does appear to be using Rindler coordinates, but in a slightly different way from the way that has been discussed previously in this thread.

That is the Rindler line element for the case where ##x = 0## corresponds to the traveling twin (instead of the Rindler horizon). So Gron does appear to be using Rindler coordinates, but in a slightly different way from the way that has been discussed previously in this thread.
In equation (2.4) he uses ##x = h## in the accelerated frame. The numbers-example in equation (2.6) shows, that he sets ##h = 4 LY##.

Why does he use the non-length-contracted distance in the traveling twin's frame?

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Mentor
Why does he use the non-length-contracted distance in the traveling twin's frame?

In equation (2.4) he is preparing to treat the idealized case of an instantaneous turnaround; in that idealized case, the distance is the non-length-contracted distance, because in that idealized case, the turnaround, in the Rindler coordinates he is using, takes place at the instant at which the traveling twin is momentarily at rest relative to the stay-at-home twin.

In equation (2.4) he is preparing to treat the idealized case of an instantaneous turnaround; in that idealized case, the distance is the non-length-contracted distance, because in that idealized case, the turnaround, in the Rindler coordinates he is using, takes place at the instant at which the traveling twin is momentarily at rest relative to the stay-at-home twin.
Isn't "turnaround" and "momentarily at rest" a contradiction? Even if the ##\Delta t## becomes arbitrarily small, hasn't it to go from velocity ##+ v## until ##-v##?

Mentor
Isn't "turnaround" and "momentarily at rest" a contradiction?

No.

Even if the ##\Delta t## becomes arbitrarily small, hasn't it to go from velocity ##+ v## until ##-v##?

Yes, and at some point in between we will have ##v = 0##. And in the idealized case of an instantaneous turnaround, that is the correct value of ##v## to use in obtaining the distance between the twins that is used in calculating the stay-at-home twin's elapsed time during the turnaround. The reason, briefly, is that ##v = 0## is the point of time symmetry.

The reason, briefly, is that ##v = 0## is the point of time symmetry.

I'll have a closer look and try to understand this. Also, in the Abstract it says, that this involves an assumption, that is not obvious:
Two ways of making this prediction
are presented. The first one is formally very simple. However, it involves an
assumption that is not obvious when the traveling twin stipulates the distance
of his brother. The second method makes use of Lagrangian dynamics in a
uniformly accelerated reference frame. Then no such assumption is necessary.
Source:

In the other paper, this statement is critizied, without giving a detailed correction:
He acknowledges the difficulty of the former, as “it involves an assumption that is not obvious when the traveling twin stipulates the distance of his brother.” Although this is incorrect, it exposes the difficulties with the speech surrounding the standard interpretation of special relativity. A deeper analysis of this question will be given separately in a future publication.
Source (on page 7 of 10):

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Mentor
in the Abstract it says, that this involves an assumption, that is not obvious

In the other paper, this statement is critizied, without giving a detailed correction

As I've said, I actually think there are issues with both papers, so I would not recommend relying on everything they say.

The simplest way to see the "time symmetry" statement I made is to observe that, for any finite acceleration, the spacelike hypersurface that includes the ##v = 0## event (the event at which the traveling twin is momentarily at rest relative to the stay-at-home twin) is also the spacelike hypersurface which is a surface of constant coordinate time in both the Rindler coordinates used to describe the traveling twin's non-inertial "rest frame" during the turnaround, and the stay-at-home twin's inertial rest frame. And everything in the entire scenario is time symmetric about that hypersurface.

In the limit of infinite acceleration, the "turnaround" is now a single point, which lies in the hypersurface just described. So the question is, what value of ##v## should we assign to that point in the limit? The answer is that it should be the value of ##v## that was assigned to the event on the traveling twin's worldline that was in that hypersurface all during the limiting process, i.e., ##v = 0##. That value of ##v## is the one that preserves the time symmetry of the entire scenario--for finite acceleration, ##v = 0## is exactly halfway between the speeds at the start and end of the turnaround. So that property should also be true in the limit of infinite acceleration.

Staff Emeritus
I am confused with a units plausibility-check. Reason: On the right side of the equation is an energy², on the left side no mass nor energy, but only space and time values.

I should have mentioned (and didn't) that I'm using geometric units where c=1. Also, G, the gravitational constant, is unity, though this is important only for the Schwarzschild metric part of the analysis.

The text this is based on, MTW's "Gravitation", uses the same units, as does the fourmilab website I quoted, which is based on said text. I've also set the value of the test mass to unity, as well i.e. So my E is the energy of a unit mass. The forurmilab website denotes this by ##\tilde{E}## the energy / unit mass, rather than E. I omited the tilde (which isn't rendering well in latex anyway). However, I'll put it back in for this post for consistency with the source.

##\alpha## has units of 1/ distance , so ##\alpha^2 x^2## is dimensionless. ##\tilde{E}## is also dimensionless, because \tilde{E} has the dimensions of energy / mass, i.e. units of c^2, but c is dimensionless.

The key equation is that in the Schwarzschild metric

$$\frac{dt}{d\tau} = \dot{t} = \frac{\tilde{E}}{| g_{00} | } =\frac{ \tilde{E}} {1 - 2M/r}$$

We want ##dt/d\tau## to be positive, and ##g_{00}## is negative, so we take the absolute value of ##g_{00}##.

Note that 1-2M/r in the geometric units is equivalent to 1 - 2 GM/r c^2 in standard units.

In the Rindler metric

$$\frac{dt}{d\tau} = \dot{t} = \frac{\tilde{E}}{ | g_{00} | } = \frac{\tilde{E}} { \alpha^2 x^2}$$

The other relationship uses is the normalization condition for a timelike 4-vector in the -+++ signature metric (with geometric units)

$$g_{00} \dot{t}^2 + g_{11} \dot{x}^2 = -1 \quad g_{00} = -\alpha^2 x^2 \quad g_{11} = 1$$

We simply substitute the expression for ##\dot{t}## into the second equation to get the equation for ##\dot{x}##.

One should be able to confirm that the expression for ##\dot{t}## satisfies the geodesic equation for the Rindler metric, given in wiki, https://en.wikipedia.org/wiki/Rindler_coordinates#Geodesics, namely
$$\ddot{t} + \frac{2}{x} \dot{x} \dot{t} = 0 \quad \ddot{x} + x \dot{t}^2 = 0$$

Putting the missing factors of c back in, this yields

$$\frac{1}{c} \dot{x} = \frac{1}{c^2} \frac{\tilde{E}}{\alpha^2 x^2} - 1$$

which keeps everything dimensionless. Or if you prefer

$$c \, \dot{x} = \frac{\tilde{E}}{\alpha^2 x^2} - c^2$$

which gives everything units of Energy/unit mass, i.e. velocity^2.

• dextercioby and Sagittarius A-Star
2022 Award
Well I think the object should not be released, but launched at high downwards speed at height ##h## above the floor.
Why would "how hard can I throw something at the floor" have anything to do with gravitational potential?
Of course the launching at high speed could be done by lifting the object to high altitude and then releasing. When lifted the object would gain potential energy ##m\alpha_1 L## according to person at height h, ##m\alpha_2 L## according to person at floor. L is the lifting distance.
My calculation was exact; the functional form does not change if you increase ##h##. You are correct that observers at different heights measure different gravitational potential differences, but this is an effect of gravitational redshift. It doesn't have an effect on gravitational potential.

• vanhees71
Or if you prefer
$$c \, \dot{x} = \frac{\tilde{E}}{\alpha^2 x^2} - c^2$$
which gives everything units of Energy/unit mass, i.e. velocity^2.
Thank you! I can read this much easier, as I am not used to read the equations with ##c = 1##. Although I think, that they describe physics in a much more natural way by avoiding artifacts of the unit system.

Homework Helper
Gold Member
Here, reference  seems to be freely available and also downloadable as PDF:

Equation (2.2) shows the line element of the uniformly accelerated reference frame.
So maybe Einstein knew what he was talking about when he used pseudo-gravitational potential to explain the twin paradox.

• Sagittarius A-Star
2022 Award
So maybe Einstein knew what he was talking about when he used pseudo-gravitational potential to explain the twin paradox.
I don't doubt it. I'm just very doubtful that it's a simple explanation.

• FactChecker
Gold Member
2022 Award
Indeed, it's much simpler to calculate the proper times of both twins using the coordinates of the IRF.

Homework Helper
Gold Member
I don't doubt it. I'm just very doubtful that it's a simple explanation.
I agree. I realize now that I have confused a few aspects of the Twins Paradox. Here is a summary of what I think has been said here. I hope that I do not butcher some people's inputs because there are a great many details that I am not qualified to understand or explain.
1) The correct answer to the Twins Paradox can be calculated using only SR and the IRF of the non-traveling twin.
2) Within SR, there is no real symmetry in the twins' situations because the traveling twin can detect that he does not remain in an IRF. So one can not use his non-inertial reference frame and SR to calculate the correct answer.
3) In order to calculate the correct answer using the traveling twin's non-inertial reference frame, GR is required. Two approaches for that are to use pseudo-gravitational potential or to use relativistic Lagrangian dynamics. These approaches are taken in the reference given by @Sagittarius A-Star in Post #17. Both approaches give the same answer as the one calculated with SR using the IRF of the non-traveling twin.

I hope that this is a good representation of the situation. Thanks to all for clarifying it for me.

Mentor
1) The correct answer to the Twins Paradox can be calculated using only SR and the IRF of the non-traveling twin.

Yes.

2) Within SR, there is no real symmetry in the twins' situations because the traveling twin can detect that he does not remain in an IRF.

Yes. Or, to put it in frame-independent terms, the traveling twin experiences nonzero proper acceleration during his trip (when he turns around), which breaks the symmetry.

one can not use his non-inertial reference frame and SR to calculate the correct answer.

No. Non-inertial reference frames are perfectly acceptable in SR, as long as spacetime is flat. The Gron paper is incorrect when it claims otherwise. This is one of the issues I have with that paper.

3) In order to calculate the correct answer using the traveling twin's non-inertial reference frame, GR is required.

No. See above.

What is required is to specify which non-inertial reference frame you are using for the traveling twin. There are an infinite number of possible non-inertial reference frames in which the traveling twin is at rest. You need to pick one.

Two approaches for that are to use pseudo-gravitational potential or to use relativistic Lagrangian dynamics. These approaches are taken in the reference given by @Sagittarius A-Star in Post #17.

There aren't actually two approaches in that reference. There is a single choice of non-inertial reference frame, and usage of pseudo-gravitational potential in that frame. The two calculations given are, first, of the idealized limiting case of an instantaneous turnaround with infinite acceleration, and second, the more realistic case of a turnaround with finite acceleration. The difference is that in the first case, there is no change in height of the stay-at-home twin during the turnaround, which simplifies the calculation; the second case (using Lagrangian dynamics) is a more complicated calculation that takes into account the change in height of the stay-at-home twin during a turnaround that takes a finite amount of time according to the traveling twin. In other words, it's the same approach, just at two different levels of idealization/realism.

give the same answer as the one calculated with SR using the IRF of the non-traveling twin

Yes.

• FactChecker
Gold Member
2022 Award
I agree. I realize now that I have confused a few aspects of the Twins Paradox. Here is a summary of what I think has been said here. I hope that I do not butcher some people's inputs because there are a great many details that I am not qualified to understand or explain.

1) The correct answer to the Twins Paradox can be calculated using only SR and the IRF of the non-traveling twin.

2) Within SR, there is no real symmetry in the twins' situations because the traveling twin can detect that he does not remain in an IRF. So one can not use his non-inertial reference frame and SR to calculate the correct answer.

3) In order to calculate the correct answer using the traveling twin's non-inertial reference frame, GR is required. Two approaches for that are to use pseudo-gravitational potential or to use relativistic Lagrangian dynamics. These approaches are taken in the reference given by @Sagittarius A-Star in Post #17. Both approaches give the same answer as the one calculated with SR using the IRF of the non-traveling twin.

I hope that this is a good representation of the situation. Thanks to all for clarifying it for me.

No, GR is only required if you deal with curved spacetime and have real gravitational fields (i.e., if you cannot neglect the real gravitational fields due to energy-momentum-stress distributions of matter). You can discuss the twin paradox in SR without ever needing to enter the advanced issue of using a non-flat pseudo-Riemannian manifold as needed to describe real gravitational fields.

You can also calculate both proper times using the non-inertial reference frame of the traveling twin. It's usually a bit more complicated, but you must get the same results, because what we calculate is independent of any choice of frames or coordinates, and only such quantities make physical sense.

It's as in usual 3D Euclidean space: If you calculate the Euclidean length of a curve using Cartesian coordinates or some other "curvilinear coordinates" like spherical coordinates, doesn't change this length which is a coordinate and frame-independent geometrical quantity.

Let's do the twin paradox for Alice being at rest at ##x=a## (with ##(t,x,y,z)## the usual Galilean coordinates of the IRF) in an IRF and Bob going with constant angular velocity on a circle of radius ##a## in the ##xy##.

Calculation 1: In the IRF (the preferred choice, because it's most simple there).

In the IRF coordinates the world lines read
$$x_A^{\mu} = (t,a,0,0), \quad x_B^{\mu}=(t,a \cos(\omega t),a \sin(\omega t).$$
Note that I defined the angular velocity as measured with the coordinate time ##t##, which at the same time is A's proper time. Using ##t## as the parameter of the world lines you get for the twins' proper times for one full round of B, i.e., when A and B meet again the first time
$$\tau_A=\int_{0}^{2 \pi/\omega} \mathrm{d} t \sqrt{\dot{x}_A(t) \cdot \dot{x}_A(t)}=\frac{2 \pi}{\omega}$$
and
$$\tau_B=\int_0^{2 \pi/\omega} \mathrm{d} t \sqrt{\dot{x}_B(t) \cdot \dot{x}_B(t)}=\frac{2 \pi}{\omega} \sqrt{1-\omega^2 a^2},$$
i.e., Bob's clock reads a smaller time than Alice's, i.e., Bob stays younger.

Calculation 2: In the restframe of Bob

Under restframe of Bob here I understand coordinates such that Bob's worldline is described by setting the spatial coordinates constant. The most simple choice of coordinates are of course cylinder coordinates, given by the transformation
$$(t,x,y,z)=(t',R' \cos(\varphi+\omega t'),R' \sin(\varphi'+\omega t'),z').$$
Then indeed Bob's worldline is given by ##R_B'=a=\text{const}##, ##\varphi_B'=0##, ##z_B'=0.##

Alice's worldline is obviously given in these coordinates by ##R_A'=a##, ##\varphi_A'=-\omega t'##, ##z_A'=0##.

The line element in the rotating frame reads
$$\mathrm{d} s^2=\eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\mu}=\mathrm{d} t^{\prime 2} -R^{\prime 2} (\mathrm{d} \varphi'+\mathrm{d} t')^2-\mathrm{d} z^{\prime 2}. \qquad (1)$$
Seen from the new frame, A and B meet the first time again after ##t'=0## when ##\varphi_A'=-\omega' t'=-2 \pi##, i.e., ##t'=2 \pi/\omega##. So we have from (1) for Alice
$$\tau_A=\int_0^{2 \pi/\omega} \mathrm{d} t'=\frac{2 \pi}{\omega}$$
and for Bob
$$\tau_B=\int_0^{2 \pi/\omega} \mathrm{d} t' \sqrt{1-\omega^2 a^2}=\frac{2 \pi}{\omega} \sqrt{1-\omega^2 a^2},$$
which is in accordance with the much simpler calculation in the Galilean coordinates of Alice's rest frame, which is an IRF.

As the calculation shows from standard multivariable calculus: The length of the time-like worldlines, i.e., the proper times of the twins, by construction CANNOT depend on the choice of coordinates/frames.

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