B Potential Energy formula in Special Relativty

[Sagittarius A-Star]

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

 

[Ibix]

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.

 

[Ibix]

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.

 

[Sagittarius A-Star]

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 $L_0$.

 

[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. 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.

 

[PeterDonis]

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.

 

[PeterDonis]

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.

 

[Sagittarius A-Star]

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:
https://www.hilarispublisher.com/op...the-twinsin-the-paradox-2090-0902-1000218.pdf
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.

 

[Sagittarius A-Star]

I note that it does not appear to be peer-reviewed

Does this appear to be peer-reviewed?
https://www.hilarispublisher.com/op...the-twinsin-the-paradox-2090-0902-1000218.pdf
It has almost the same content, see for example equation (22) in there.

 

[pervect]

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.

 

[Sagittarius A-Star]

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

 

[Ibix]

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.

 

[Ibix]

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.

 

[Ibix]

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.

 

[PeterDonis]

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 [27] 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.

 

[PeterDonis]

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.

 

[Sagittarius A-Star]

Also note that reference [27] 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 [27] seems to be freely available and also downloadable as PDF:
https://www.researchgate.net/profile/Oyvind_Gron2/publication/230923551_The_twin_paradox_in_the_theory_of_relativity/links/0a85e53bf245d59b08000000/The-twin-paradox-in-the-theory-of-relativity.pdf

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

 

[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.

 

[PeterDonis]

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 $1.1 / 0.8 = 1.375$.

 

[PeterDonis]

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.

 

[Sagittarius A-Star]

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?

https://www.researchgate.net/profile/Oyvind_Gron2/publication/230923551_The_twin_paradox_in_the_theory_of_relativity/links/0a85e53bf245d59b08000000/The-twin-paradox-in-the-theory-of-relativity.pdf

 

[PeterDonis]

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.

 

[Sagittarius A-Star]

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 $-v$?

 

[PeterDonis]

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.

 

[Sagittarius A-Star]

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:
https://www.researchgate.net/profile/Oyvind_Gron2/publication/230923551_The_twin_paradox_in_the_theory_of_relativity/links/0a85e53bf245d59b08000000/The-twin-paradox-in-the-theory-of-relativity.pdf

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):
https://www.hilarispublisher.com/op...the-twinsin-the-paradox-2090-0902-1000218.pdf

 

[PeterDonis]

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.

 

[pervect]

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.

 

[Ibix]

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.

 

[Sagittarius A-Star]

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.

 

[FactChecker]

Here, reference [27] seems to be freely available and also downloadable as PDF:
https://www.researchgate.net/profile/Oyvind_Gron2/publication/230923551_The_twin_paradox_in_the_theory_of_relativity/links/0a85e53bf245d59b08000000/The-twin-paradox-in-the-theory-of-relativity.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.

 

[Ibix]

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.

 

[vanhees71]

Indeed, it's much simpler to calculate the proper times of both twins using the coordinates of the IRF.

 

[FactChecker]

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.

 

[PeterDonis]

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.

 

[vanhees71]

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.

 

[Sagittarius A-Star]

So maybe Einstein knew what he was talking about when he used pseudo-gravitational potential to explain the twin paradox.

Yes, definitely! His "twin paradox" explanation of 1918 is a valid approach to explain the asymmetry of the scenario, if you skip there his statements related to "Mach's principle". But it is not the only possible explanation alternative.

For Einstein, understanding of non-inertial reference frames was very important. Reason: It was his agenda to extend the principle of relativity from inertial frames to general frames, and to develop via the principle of equivalence "bottom-up" his GR.

 

[PeterDonis]

For Einstein, understanding of non-inertial reference frames was very important. Reason: It was his agenda to extend the principle of relativity from inertial frames to general frames, and to develop via the principle of equivalence "bottom-up" his GR.

It's worth noting that this agenda of Einstein's also created confusion and argument about whether non-inertial frames in flat spacetime were part of SR or GR. The modern answer is that they are part of SR, but there are many papers in the literature which say otherwise.

 

[vanhees71]

I think that confusion was the cause for Einstein's "detour". He had the right idea already in 1912 concerning general covariance and got confused for some time due to the hole argument. Only finally in 1915 he came back to general covariance, and he was almost scooped by Hilbert getting the final field equations arounc the same time via the action principle.

 

[pervect]

So maybe Einstein knew what he was talking about when he used pseudo-gravitational potential to explain the twin paradox.

While Einstein undoubtedly "knew what he was doing", in the sense that he got the correct answers and blazed the path we all follow, that doesn't mean that we haven't found better paths to the destination than the one he pioneered. So it's not necessarily a wise idea to try and retrace his footsteps exactly - we really have found better ways in the century following his formulation of the theory.

To be more specific, accelerated frames are not necessary to understand the Twin paradox, and indeed the notion of the existence of an accelerated frame that covers all of space-time turns out to be not totally correct. Accelerated frames do exist, but they only cover a limited region of space-time. This is discucssed in MTW's text, Gravitation, "constraints on the size of the frame of an accelerated observer". (That's from memory, the exact language may be slightly different, in spite of my use of quote marks. I could look it up, if people want the exact wording, or explain further.)

The difficulties that occur with the accelerated frame approach don't involve the turn-around point, which is the one most people focus on. Rather they happen during the outbound journey, before the turnaround. The mathematical relationship between the coordinates of the inertial and accelerating observers is no longer a 1:1 correspondence, which is generally regarded as a requirement for a "frame" of reference to exist. On a space-time diagram, when we draw the lines representing a time coordinate of t=0 in the instantaneous inertial frame of the accelerated observer at the start of his journey, and the line representing time coordinate t=constant at some subsequent proper time later in the journey, the two lines cross, indicating that point, where the lines cross, has two diferent time coordinates in the accelerated frame. In at least one case, there turns out to be an infinite number of time coordinates for the same point. This mathematical ill-behavior quickly turns into apparent physical ill-behavior if one does not realize what is going on.

For a spaceship accelerating at 1 Earth gravity, the difficulties start to arise after about a year of continuous proper acceleration.

 

[FactChecker]

To be more specific, accelerated frames are not necessary to understand the Twin paradox, and indeed the notion of the existence of an accelerated frame that covers all of space-time turns out to be not totally correct. Accelerated frames do exist, but they only cover a limited region of space-time.

I'm not sure what this is saying. I am aware of two approaches that allow the traveling twin to calculate that the stay-at-home twin is older. Both appear to me to use accelerated frames. One hypothesizes a pseudo-gravitational potential. I think that the other stays within SR and calculates the consequence of the accelerated turn-around. (I only have a crude idea of what this calculation would be like.) Are you saying that both of these are limited acceleration-frame approaches and that there is a third approach?

 

[PeterDonis]

I am aware of two approaches that allow the traveling twin to calculate that the stay-at-home twin is older. Both appear to me to use accelerated frames.

Only one of the approaches you mention requires a non-inertial frame:

One hypothesizes a pseudo-gravitational potential.

This one, because a pseudo-gravitational potential is only present in a non-inertial frame. (And it's not "hypothesized", it's always present in a non-inertial frame.)

I think that the other stays within SR and calculates the consequence of the accelerated turn-around.

The pseudo-gravitational potential approach also stays within SR; as has already been pointed out in the other thread we are having on the twin paradox, SR = flat spacetime.

Calculating the actual spacetime length of the trajectories of the two twins can be done in any frame; you could do it in the same non-inertial frame as you used for the pseudo-gravitational approach if you wanted to. But that would be pointless since the calculation is much easier in the inertial rest frame of the stay-at-home twin, so that's how everybody does it.

(I only have a crude idea of what this calculation would be like.)

It's just geometry. In the idealized case of an instantaneous turnaround, you have a triangle in spacetime, and the lengths of each of its three sides are easily calculated using the spacetime interval formula (which is just the spacetime analogue of the Pythagorean theorem, with a minus sign instead of a plus sign before the square of the "space: term). In the more realistic case of a turnaround with finite acceleration, you have to do an integral to calculate the length of that piece of the traveling twin's trajectory.

Have you looked at the Usenet Physics FAQ article on the twin paradox?

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

If not, I strongly suggest doing so.

 

[PeterDonis]

Accelerated frames do exist, but they only cover a limited region of space-time.

This is not quite correct. There are ways of constructing non-inertial frames that do cover all of spacetime. But it is true that the most "natural" ones, the ones people tend to come up with on a first try, will only cover a limited region of spacetime.

This is discucssed in MTW's text, Gravitation, "constraints on the size of the frame of an accelerated observer".

MTW is talking about a specific method of constructing a non-inertial frame, which for constant acceleration is just Rindler coordinates. AFAIK they do not discuss other methods of constructing non-inertial frames centered on an accelerated worldline that do cover all of spacetime. I suspect that is because (a) those methods were not very well explored in 1973, when MTW was written, and (b) those methods do not have some properties that MTW considered desirable and so were not considered.

 

[FactChecker]

Calculating the actual spacetime length of the trajectories of the two twins can be done in any frame; you could do it in the same non-inertial frame as you used for the pseudo-gravitational approach if you wanted to. But that would be pointless since the calculation is much easier in the inertial rest frame of the stay-at-home twin, so that's how everybody does it.

I think I agree with all that, but I would like to clarify something. I have always agreed that the calculation in the inertial rest frame of the stay-at-home twin was simple. As you say, it's essentially just the Pythagorean theorem. The part that I had trouble with (and the part that I think makes it seem like a paradox) was getting the same result using the traveling-twin-centered frame. That is what I consider "solving the twin paradox". Would you say that is an accelerated frame?

 

[PeterDonis]

Would you say that is an accelerated frame?

Any frame in which the traveling twin is at rest for the entire trip must be a non-inertial frame, since the traveling twin is non-inertial during the turnaround. But it's still perfectly possible to get the correct answer in such a frame.

 

[pervect]

This is not quite correct. There are ways of constructing non-inertial frames that do cover all of spacetime. But it is true that the most "natural" ones, the ones people tend to come up with on a first try, will only cover a limited region of spacetime.
MTW is talking about a specific method of constructing a non-inertial frame, which for constant acceleration is just Rindler coordinates. AFAIK they do not discuss other methods of constructing non-inertial frames centered on an accelerated worldline that do cover all of spacetime. I suspect that is because (a) those methods were not very well explored in 1973, when MTW was written, and (b) those methods do not have some properties that MTW considered desirable and so were not considered.

Yes, I'd agree that the specific constraints apply to a specific method of constructing an "accelerated frame". MTW makes this point in a different way, they talk about the ambiguity of the concept of creating an "accelerated frame" before their detailed presentation of the specific method they chose to explain at length.

The way I would describe their approach to creating the accelerated frame at a lay level is that one considers the instantaneous inertial reference frames at various instants of time (proper time) for the accelerated observer. One then uses the space coordinates in the instantaneous inertial frame, and the proper time of the acclelerated observer for the time coordinate. This implies that one is using some particular notion of simultaneity, namely the notion of simultaneity is determined by the specific choice of inertial reference frame at any proper time τ of the accelerated observer. Unfortunately, this approach yields a limited size for the accelerated frame.

It's not hard to illustrate the difficulty with a space-time diagram, as long as one can draw a space-time diagram, and appreciate the significance of "lines of simultaneity" drawn on said diagram. However, I've found that resistance on the part of readers to the idea that simlutaneity is relative tends to prevent them from being able to accomplis either tasks or to follow explanations that show how to do them. It's only necessary to understand that simultaneity is a convention, and that the specific convention can be illustrated by drawing a line of events which meet said convention on a space-time diagram. My guess is that it's the first point (accepting that simutaneity is a convention) that is the difficult one.

Any approach to creating an "accelerated frame" is going to have to deal with the issue that if a spaceship accelerates at a constant (proper) acceleration of 1g, observers on said ship will never actually see any signal emitted from the origin point (which I'll call Earth) emitted later than approximately 1 year after take off. The general formula is that if the acceleraton is g, the last event one sees will occur at time c/g.

So in our hypothetical spaceship, if the spaceship leaves in 2020, their radio recievers (or, if they have telescopes, telescopes) will never receive a signal (or observe events through their telescope) emitted after 2021, no matter how long they continue to accelerate. Once they cut their engines, and/or turn around, they'll start to receive signals emitted after 2021.

It turns out to be very problematical to assign coordinates to events that are outside one's light cone. I believe we've had disccusions in the past. I can no longer ennumerate the various approaches, but I do recall that none of them was really satisfying to me.

Because of these difficuties, I really can't recommend the idea of using accelerated frames in an introduction to special relativity as "explain" the twin paradox. It turns out to be quite complicated to do correctly - and it's not necessary. There are easier ways, that also have the advantage of not inviting problems further down the line.

The approach I favor is to start with the twin paradox as one's founding assumption. Assume that if you have two clocks, if they follow different paths through space-time, they will not necessarily agree when the reunite. Follow this up with the notion that in the flat space-time of special relativity, one can say that the clock that has the maximum reading is the one that undergoes inertial motion.

As other posters have already indirectly pointed out, the simple statement that the inertial observer is the one that has the maximum clock reading is only true in the flat space-time of SR. In the curved space-time of GR, it needs to be modified slightly. The modifications aren't really major, but it confuses the presentation enough that I have resisted giving them. It's rather similar to the way that "the straight line is the shortest distance" between two points needs to be modified for a curved spatial geometery, however.

 

[PeterDonis]

It turns out to be very problematical to assign coordinates to events that are outside one's light cone.

More precisely, that are outside one's horizon--i.e., outside the boundary of the region of spacetime that one can ever receive light signals from. For an observer with constant proper acceleration, that's the Rindler horizon.

 

[PeterDonis]

The modifications aren't really major

More precisely, they aren't for regions of spacetime sufficiently small that there are not multiple timelike geodesics connecting the same pair of events. (In more technical language, for regions of spacetime sufficiently small that they contain no conjugate points.) If you relax that restriction, there are more complexities to deal with.

 

[PAllen]

MTW is talking about a specific method of constructing a non-inertial frame, which for constant acceleration is just Rindler coordinates. AFAIK they do not discuss other methods of constructing non-inertial frames centered on an accelerated worldline that do cover all of spacetime. I suspect that is because (a) those methods were not very well explored in 1973, when MTW was written, and (b) those methods do not have some properties that MTW considered desirable and so were not considered.

I'm curious what specific coordinates you are referring to. The two most common methods of constructing coordinates with an arbitrary world line as the spatial origin cannot cover all of spacetime, in general. MTW consider generalized Fermi-Normal coordinates, which, as you mention, have many nice properties near the world line (metric is Minkowski all along the world line, and the connection values on the world line give its proper acceleration). The other common alternative is radar coordinates, which converge to Fermi-Normal near the axial world line (thus maintaining the same properties) but differ further away and can cover all of spacetime in more cases. However, radar coordinates are still limited to covering the intersection of the causal past of the origin world lie and its causal future. While in many cases where Fermi-Normal are limited, radar can cover all of spacetime, this is not true, in general. In particular, for an eternally uniformly accelerated world line, Fermi-Normal and radar coordinates have exactly the same coverage - one quadrant of spacetime.

A trivial way to cover all spacetime in SR would be to couple an arbitrary origin world line to spatial slices of some arbitrary inertial frame, with position measured on each slice from the origin world line. However, these coordinates would not come close to matching local rulers and clocks for the origin world line. The metric would have off diagonal components, and for an eternally accelerating observer, the diagonal components would be asymptotically zero. While I would be willing to call these general coordinates based on a world line origin, I would not be inclined to call them a non-inertial frame.

Of course, this is all terminology, not physics. It gets to "what is a frame, in general" which is something experts on physicsforums have never agreed.

 

[PeterDonis]

I'm curious what specific coordinates you are referring to.

MTW, IIRC, describe Rindler coordinates for the case of constant acceleration, and Fermi-Walker coordinates for the more general case of varying acceleration. Rindler coordinates are actually equivalent to Fermi-Walker coordinates for constant acceleration (assuming zero rotation). These coordinates are of course limited to only part of spacetime. Radar coordinates are also so limited.

An example of non-inertial coordinates that can cover all of Minkowski spacetime, not just a limited region, would be Born coordinates, in which Langevin observers (observers moving in a circle of constant radius, at constant angular velocity, in Minkowski spacetime) are at rest. These coordinates of course must adopt a simultaneity convention that is not that of any of the momentarily comoving inertial frames of the observers (it is in fact the convention of the inertial frame in which the center of the circular orbits is at rest). They also have non-intuitive properties for values of the radial coordinate greater than or equal to a particular value.

I believe there is a paper by Dolby & Gull in which another scheme for constructing non-inertial coordinates that cover all of spacetime, centered on a particular worldline, is described (not their radar coordinates, a different scheme). But I can't find it right now.

 

[Sagittarius A-Star]

Imagine an object released at height $h$ above the floor ...
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.

My question was not about releasing an object at height $h$, but about an object, that has at height $h$ already the velocity $0.8 c$. So it has at height $h$ a potential energy plus already a kinetical energy.

The question in this case is: Does the potential energy part of it need to include a factor $\gamma$?