Potential Energy formula in Special Relativty

In summary: 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.
  • #36
FactChecker said:
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.
 
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  • #37
Sagittarius A-Star said:
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.
 
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  • #38
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.
 
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  • #39
FactChecker said:
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.
 
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  • #40
pervect said:
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?
 
  • #41
FactChecker said:
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:

FactChecker said:
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.)

FactChecker said:
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.

FactChecker said:
(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.
 
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  • #42
pervect said:
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.

pervect said:
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.
 
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  • #43
PeterDonis said:
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?
 
  • #44
FactChecker said:
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.
 
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  • #45
PeterDonis said:
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.
 
  • #46
pervect said:
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.
 
  • #47
pervect said:
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.
 
  • #48
PeterDonis said:
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.
 
  • #49
PAllen said:
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.
 
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  • #50
Ibix said:
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##?
 
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  • #51
Sagittarius A-Star said:
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##?
0.8c as measured by who, and interpreted by who? Assuming that we accept a measurement made by an observer at height ##h## at rest in the accelerating (Rindler) frame, then we can adapt @jartsa's approach.

The proper acceleration experienced by the floor of the lift is ##\alpha##, and that experienced by an observer at rest at height ##h## above it is ##\alpha_h=\alpha/(1+\alpha h/c^2)##. We release an object from rest at a height ##h+H## above the floor. As it passes height ##h##, the observer there measures its kinetic energy to be ##m\alpha_hH##, implying a Lorentz gamma of ##\gamma_h##, where ##\gamma_h-1=\alpha_hH/c^2##.

Now we can write down the potential energy gain from falling that last distance ##h##. It is $$\begin{eqnarray*}
E&=&m\alpha (h+H)-m\alpha_hH\\
&=&\gamma_hm\alpha h
\end{eqnarray*}$$where the results in the previous paragraph were used to eliminate ##H## and ##\alpha_h##.

So you do indeed need a factor of the initial ##\gamma## measured by the higher observer. However, it's worth noting that pseudo-gravitational time dilation is in play here, and the lower observer can reasonably argue that the higher observer's local measurement of the velocity isn't the one he would have made. So some care is needed about exactly how you are defining that initial velocity.
 
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  • #52
FactChecker said:
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?
But I've shown, how this works somewhere in these forums yesterday using the example of a twin in circular motion. It's a simple example where you can do the calculation in both the IRF (restframe of the stay-at-home twin) and the accelerated (local) frame (rest frame of the twin in circular motion). For this you don't need a coverage of the entire Minkowski space. The there used cylindrical Born coordinates are all that's needed to do the calculation in the accelerating frame. You don't even need the complete construction of the local frame (i.e., to provide a tetrad for the accelerated twin since we are only interested for the proper time of the twins).
 
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  • #53
PeterDonis said:
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.
Can you give a reference, where the Born coordinates are extended to the entire Minkowski space? I thought, it's limited to the cylinder, where ##\omega r<c##. Indeed in the standard Born coordinates, the point ##r>1/\omega##, ##\varphi=\text{const}##, ##z=\text{const}## moves on a space-like world line and thus cannot be the world line of "an observer".

For the usual Langevin observers you can start with cylinder coordinates in a global rest frame,
$$\mathrm{d}s^2=\mathrm{d} t^2 -\mathrm{d} r^2 - r^2 \mathrm{d} \varphi^2 - \mathrm{d} z^2.$$
Then the tetrad field of the Langevin observers at any point within the said cylinder can be defined as
$$e_0=\frac{1}{\sqrt{1-\omega^2 r^2}} \partial_t +\frac{\omega}{\sqrt{1-\omega^2 r^2}} \partial_{\varphi},$$
$$e_1=\partial_r,$$
$$e_2=\frac{\omega r}{\sqrt{1-\omega^2 r^2}} \partial_t + \frac{1}{r \sqrt{1-\omega^2 r^2}} \partial_{\varphi},$$
$$e_3=\partial_z,$$
which are obviously only defined for ##\omega r<1##.

The Born coordinates are another coordinate transformation with non-orthogonal coordinate lines but not a way to extend the chart further than the cylinder ##\omega r<1##. So how are you extend the coordinates outside of the cylinder?
 
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  • #54
PeterDonis said:
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.
For infinite acceleration, formula (2.6) 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 delivers the correct result of ##6.4 years##, if you set in it
##h = 4LY##, the uncontracted distance. Gron did double-check this with the "gap" of
##(10 - 2 * 1.8) years## from calculations while the inertial motions.

For finite acceleration, formula (2.5) 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 delivers an almost correct result of ##6.4 years + \Delta t##, if you set in it also ##h = 4LY##, the uncontracted distance. That is, what I don't understand.

Let's define the proper turnaround time as ##\Delta t := 1s##. Then the "stationary" twin ages while the complete travel by ##10 years + \Delta t * \gamma(average) < 10 years + (5/3)s##, because ##\gamma## is smaller as 5/3 while the turnaround.

Let's now re-write the left part of formula (2.5) 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 (##d \tau## instead of ##\Delta \tau##, and ##\Delta t := 1s##):
$$d\tau = 1s + \frac{gh} {c^2} dt$$
Would that provide the correct result, if we set ##h := l(t)##, with ##l(t)## as the contracted distance as function of time and then integrate the formula over ##t##?

At the begin and at the end of the turnaround, ##l(t)## would be 4LY * 3/5, in the middle of the turnaround it would be 4LY + "something". This "something" would be calculated from coordinate-acceleration, twice integrated over time.
 
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  • #55
vanhees71 said:
But I've shown, how this works somewhere in these forums yesterday using the example of a twin in circular motion. It's a simple example where you can do the calculation in both the IRF (restframe of the stay-at-home twin) and the accelerated (local) frame (rest frame of the twin in circular motion). For this you don't need a coverage of the entire Minkowski space. The there used cylindrical Born coordinates are all that's needed to do the calculation in the accelerating frame. You don't even need the complete construction of the local frame (i.e., to provide a tetrad for the accelerated twin since we are only interested for the proper time of the twins).
Yes. That was exactly what I was looking for (once I realized the situation). I definitely took notice of your earlier post. Thanks. In my comment that you referenced, I was just explaining what I had been looking for.
 
  • #56
PeterDonis said:
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.
Actually, MTW use generalized Fermi—Normal coordinates for all cases ( not Fermi-Walker), including the case of rotation. I also specifically discussed radar coordinates which are what Dolby & Gull use, and noted that they can only cover the intersection of the total causal future and total causal past of a world line. In many cases this allows coverage of all spacetime when Fermi-Normal fails, but any situation with either past eternal or future eternal acceleration bounded from below, they do not cover all of spacetime.

In the simple case of eternal uniform acceleration, Rindler, Fermi-Normal, and Radar coordinates have the same coverage and the same simultaneity surfaces. However, radar measures distance along slice differently. Rindler is just Fermi-Normal with the origin translated from a given world line to the horizon.
 
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  • #57
vanhees71 said:
Can you give a reference, where the Born coordinates are extended to the entire Minkowski space?

We had this discussion in another thread. Born coordinates are perfectly valid over all of Minkowski space. The tetrad field that describes the Langevin observers is only valid for sufficiently small radial coordinates (such that a curve with constant spatial coordinates is timelike--such a curve becomes null at a particular value of ##r##, and spacelike outside that value). A coordinate chart is not required to have one timelike and three spacelike coordinates, so the chart remains valid even for values of ##r## where the tetrad field no longer is.
 
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  • #58
Sagittarius A-Star said:
Then the "stationary" twin ages while the complete travel by ##10 years + \Delta t * \gamma(average) < 10 years + (5/3)s##, because ##\gamma## is smaller as 5/3 while the turnaround.

No. Don't wave your hands. If you're going to do the math, you need to do it correctly. The correct way to handle finite acceleration is to do an integral. Section 3 of the paper derives the integral that needs to be done (the integral of equation 3.5 in the paper) from a Lagrangian, but it can also be done from the Rindler metric.

Sagittarius A-Star said:
At the begin and at the end of the turnaround, ##l(t)## would be 4LY * 3/5, in the middle of the turnaround it would be 4LY + "something". This "something" would be calculated from coordinate-acceleration, twice integrated over time.

No, it's much simpler than that. Because of time symmetry, the middle of the turnaround is the instant where the traveling twin is momentarily at rest with respect to the stay-at-home twin. At that instant, the distance to the stay-at-home twin in the traveling twin's rest frame (assuming we are using Rindler coordinates for that frame, as Gron's paper does) is 4 LY, the same as the distance in the inertial frame that is momentarily comoving with the traveling twin (since that is just the stay-at-home twin's rest frame).
 
  • #59
PAllen said:
MTW use generalized Ferm—Normal coordinates for all cases ( not Fermi-Walker)

Yes, you're right, I should have said Fermi normal coordinates. I was confusing them with Fermi-Walker transport.
 
  • #60
PeterDonis said:
the middle of the turnaround is the instant where the traveling twin is momentarily at rest with respect to the stay-at-home twin. At that instant, the distance to the stay-at-home twin in the traveling twin's rest frame (assuming we are using Rindler coordinates for that frame, as Gron's paper does) is 4 LY
But that can be only valid in the case of infinite acceleration. I am speaking here about finite acceleration (formula 2.5). When the (de-)acceleration starts after having traveled ##4LY/\gamma##, then additional distance is needed for doing the de-acceleration.
 
  • #61
vanhees71 said:
Can you give a reference, where the Born coordinates are extended to the entire Minkowski space?
Wikipedia gives the Born coordinates in terms of cylindrical polars on Minkowski spacetime as$$\begin{eqnarray*}
t'&=&t\\
r'&=&r\\
\phi'&=&\phi-\omega t\\
z'&=&z
\end{eqnarray*}$$The associated coordinate basis is fine at any ##r'##, and is clearly not a tetrad field at any ##r'##. Do you mean something different by Born coordinates?
 
  • #62
Sagittarius A-Star said:
that can be only valid in the case of infinite acceleration

No. It is valid for finite acceleration as well. The finite acceleration case is just as time symmetric as the infinite acceleration case.

Sagittarius A-Star said:
But that can be only valid in the case of infinite acceleration. I am speaking here about finite acceleration (formula 2.5). When the (de-)acceleration starts after having traveled ##4LY/\gamma##, then additional distance is needed for doing the de-acceleration.

You are wrong. Do the math and see. Note that the Gron paper's calculation using the Lagrangian gives the answer I am giving: see the paragraph right after equation 3.7.
 
  • #63
Yes, sorry. I was in the obviously wrong opinion that we discuss accelerated observers with their reference frames, defined by tetrads with the time-like one the tangent-unit vector along his time-like worldline, but now obviously you want to discuss again something else with the Born coordinates (which have only the usual coordinate singularity at ##r=r'=0##).

So what do you want to describe with Born coordinates for ##\omega r' \geq 1##?
 
  • #64
vanhees71 said:
what do you want to describe with Born coordinates

Spacetime. I was giving an example of a non-inertial coordinate chart that covers all of spacetime. I am not claiming that it gives a non-inertial tetrad field that covers all of spacetime.
 
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  • #65
vanhees71 said:
So what do you want to describe with Born coordinates for ##\omega r' \geq 1##?
I thought it was just a counter example to the claim that non-inertial coordinates only cover part of Minkowski space. So nothing particularly practical.
 
  • #66
Sure, that's of course right.
 
  • #67
PAllen said:
I'm curious what specific coordinates you are referring to.

I'm also curious. If Peter digs up the Dolby & Gull paper, and it has a line element, I'd like to see it.

If we require that the constant-time part of the metric be spatially flat, and that the metric coefficients be independent of time, we are led to consider line elements of the form

$$ds^2 = -f(x,y,z)*dt^2 + dx^2 + dy^2 + dz^2$$

Either of these restrictions could be relaxed, of course. But both are nice to have.

Then, if we insist that the acceleration only have components in the x-direction, i.e. we insist that ##\Gamma^y{}_{tt} = \Gamma^z{}_{tt} = 0##, we are led to consider line elements of the form

$$ds^2 = -f(x)*dt^2 + dx^2 + dy^2 + dz^2$$

The solutions of this which have zero Riemann curvature, so that they represent the flat space-time of special realtivity, are limited. I am getting only the following possibility:

$$ds^2 = -(c1*x + c2)^2 dt^2 + dx^2 + dy^2 + dz^2$$

as the solution for

$$R_{txtx} = -\frac{1}{4 \, f(x)} \left( -2 \, f''(x)f(x) + f'(x)^2 \right) = 0$$

c1=0 corresponds to the Minkowskii metric, and has ##\Gamma^x{}_{tt} = 0, so it's not an accelerated frame of reference.

c2=0 corresponds to the Rindler metric.

Having both c1 and c2 non=zero doesn't change much. At some value of x, (c1*x+c2) will be zero, leading to the same coordinate singularity that we have with Rindler coordinates.
 
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  • #68
Sagittarius A-Star said:
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##?

One of the key differences of accelerated frames in SR as compared to Newtonian physics is the presence of pseudo-gravitational time dilation. This is part of your question, I think, but you haven't addressed the issue even conceptually.

If we use Rindler coordinates, we can write the peseudogravitational time dilation factor as being proportional to the height above the Rindler horizon, where one might be tempted to think that "time stops". Saying that time stops oversimplified, of course, time doesn't actually stop, we're just labelling our coordinates differently.

For an object accelerating at 1g, i.e. 10 m/s^2, the Rindler horizon is roughly one light year below the object.

Denoting the gravitational time dilation as ##\gamma_g##, we can write:

##\gamma_g = \alpha x##

When x = 1/##\alpha##, the time dilation factor is 1, there is no time dilation. At x=0, we're at the Rindler horizon.

The expression I am getting for a conserved relativistic quantity E that represents energy in this case is:

$$E^2 = \gamma_g^2 \left( \left( m \,c \frac{dx}{d\tau} \right)^2 +\left( m\,c^2 \right)^2\right)$$

or

$$E = \gamma_g \,m\,c^2 \sqrt{ \left( \frac { \frac{dx}{d\tau}} {c} \right)^2 +1}$$

Here ##\tau## is proper time. This is done by methods similar to how we define a conserved energy in a Schwarzschild metric. Unfortunately, it's my own calculation, not something from a textbook.

I don't see anyway to motivate this by a Newtonian discussion. However, we can take the non-relativistic limit. If we chose a height where ##\gamma_g \approx 1##, and let ##v \approx \frac{dx}{d\tau}##, the expression reduces to

$$E \approx m\,c^2 + \frac{1}{2} m\,v^2$$

Note that the numerical value of our conserved energy depend on our reference height in the accelelerated frame.
 
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  • #69
pervect said:
If we require that the constant-time part of the metric be spatially flat, and that the metric coefficients be independent of time

These are very restrictive assumptions, as your analysis of their implications shows.
 
  • #70
pervect said:
Note that the numerical value of our conserved energy depend on our reference height in the accelelerated frame.

Better yet, if you translate the coordinates so ##x = 0## corresponds to proper acceleration ##\alpha## (in "natural" units), then you have

$$
\gamma_g = 1 + \frac{a x}{c^2}
$$

where ##a## is the proper acceleration in conventional units (i.e., ##\alpha = a / c^2##), and your conserved energy for ##|x| \ll 1 / \alpha## becomes

$$
E \approx m c^2 + \frac{1}{2} m v^2 + m a x
$$

which of course is easily recognizable as analogous to the Newtonian formula including gravitational potential energy as a function of height.
 
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