# Potential Energy formula in Special Relativty

## 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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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.

• • Abhishek11235, vanhees71, Sagittarius A-Star and 1 other person
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.

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 travelling twin's frame, in which the distance of the twins is ##L## from equation (3) because of lenght 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
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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
PeterDonis
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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.

pervect
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.

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

Let's imagine a twin paradox where the traveller is already travelling 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
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.

• Sagittarius A-Star
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
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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
PeterDonis
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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
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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PeterDonis
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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
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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 travelling twin's frame?

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PeterDonis
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Why does he use the non-length-contracted distance in the travelling 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##?

PeterDonis
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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 travelling 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.
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