B Potential Energy formula in Special Relativty

[Ibix]

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.

 

[vanhees71]

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).

 

[vanhees71]

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?

 

[Sagittarius A-Star]

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.

 

[FactChecker]

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.

 

[PAllen]

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.

 

[PeterDonis]

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.

 

[PeterDonis]

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.

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).

 

[PeterDonis]

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.

 

[Sagittarius A-Star]

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.

 

[Ibix]

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?

 

[PeterDonis]

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.

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.

 

[vanhees71]

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 $\omega r' \geq 1$?

 

[PeterDonis]

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.

 

[Ibix]

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.

 

[vanhees71]

Sure, that's of course right.

 

[pervect]

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.

 

[pervect]

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.

 

[PeterDonis]

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.

 

[PeterDonis]

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.

 

[pervect]

In order to do a simple check of the proposed energy formula, I decided to consider a simple case. Suppose we set up an accelerating frame, based on a reference rocket acclelerating at 1 light year/year^2, approximately 9.5 meters/second^2. What velocity do we need to launch a projectile at in order to reach a "height" of 1 light years above our reference rocket?

This is an example of geometric units. We measure all distances in light years, and all times in years. As a consequence, the numerical value of "c" is 1, as is the numerical value of "a".

We predict from the formula that $dx/d\tau$ must equal $\sqrt{3}$ for an object to have the same energy as an object moving with $dx/d\tau=0$ at a height of 1. At a height of 1, $\gamma_g = 1+ax/c^2 = 1+1 = 2$, meaning an object at that height has twice its rest energy.

In order for $dx/d\tau$ to be $\sqrt{3}$, we find $v = dx/dt = \sqrt{3}/2$ , giving $dx/d\tau$ = (dx/dt) \, (dt/d$\tau$) = $\sqrt{3}$.

Here's the plot I did on a Rindler chart. The chart charts the course of the two rockets, our reference rocket starting at d=1, and a second rocket 1 light year "higher" in the accelerated frame. These plots are done in the inertial frame co-moving with the rockets at t=0.

These two rockets are plotted on the graph in black. The projectile launched at a velocity of $\sqrt{3} / 2$ follows a straight line inertial trajectory of $d = \frac{\sqrt{3}}{2}\,t+1$, plotted in green.

The plot confirms the prediction. The green line representing the launched projectile just catches up to the black line of the higher rocket.

The equations are fairly simple if one wants to go into more depth. The reference rocket follows a hyperbola $d^2-t^2=1$. The rocket one light year above the reference rocket follows a hyperbola of $d^2-t^2 = 4$. There are various articles on "hyperbolic motion" that should explain this more - I'm not going to look for any further references unless there are questions. The green line does indeed meet the black line at d=4, t=$2 \,\sqrt{3}$

 

[PeterDonis]

Here is proper time.

Actually, for the observer at $x = 1 / \alpha$ in Rindler coordinates, the observer's proper time is the same as Rindler coordinate time (since this observer has no time dilation in Rindler coordinates). So in the approximation you are using for your energy formula, Rindler coordinate time can be substituted for proper time.

 

[pervect]

Actually, for the observer at x=1/α in Rindler coordinates, the observer's proper time is the same as Rindler coordinate time (since this observer has no time dilation in Rindler coordinates). So in the approximation you are using for your energy formula, Rindler coordinate time can be substituted for proper time.

I don't think this is true.

For instance, a clock near the Rindler horizon will be ticking much differently than the coordinate time there. In general one needs to calculate $d\tau$ from the line element.

The formula for energy was originally derived in Rindler line eleement, but when we modified the energy formula to

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

(which I do prefer) the line element in the accelerated coordinate system (t,x,y,z) becomes:

$$ds^2 = -c^2\,d\tau^2 = -(c^2 + a^2\,x^2) dt^2 + dx^2 + dy^2 + dz^2$$
or with the +--- convention, which may be easier for this case,

$$d\tau^2 = (1+ a^2\,x^2 / c^2)dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$

 

[pervect]

I had a few more thoughts.

First of all I wanted to give an overview of the concepts used to derive the energy I worked on earlier, formula, now that I'm reasonably hapy with said formula. Previously, I had simply suggested that it was similar to methods used to calculate the energy of test masses orbiting in the Schwarzschild geometry of GR without giving more details.

The basic idea behind the calculation is Noether's theorem. Noether's theorem says, roughly speaking, that a conserved energy is associated with a time-translation symmetry. Fun historical fact - it was concerns over the concept of energy in General Relativity raised by Hilbert, Felix Klein, and others that led to the development of the theorem. It's quite an interesting story, also interesting is the sexism that Noether had to deal with in her everyday work, such as not being able to lecture or be paid for her work by the university.

The time-translation symmetry in this case is the time-translation symmetry of the metric. So writing down the metric (or line element) allows us to calculate the energy simply from the time-translation symmetry associated with it. Well, perhaps the details get a bit complex, but the idea is simple.

A notable feature of the energy relation is that the energy - including the rest energy - goes to zero at the Rindler horizion, where (1+ax/c^2) goes to zero. And below that, it goes negative. What does this mean, physically? My argument is that it means that the concept of an accelerated frame is breaking down at the Rindler horizon. Certainly there are problems about thinking of an object "at rest" at said horizon, it's a null surface. Light could have a constant Rindler coordinate, but it cannot meaningfully be said to be "at rest".

Furthermore, I will go so far as to suggest that since the expression for energy basically only depends on the existence of the time translation symmetry, any sort of coordinates that describe an accelerated frame that explicitly have this symmetry are going to experience similar problems. It has been suggested that there may be a way around this. I suppose I can't absolutely rule out the possibility, but I won't be convinced that there is a way around the issue, which I have called a size constraints on the size of an acclerated frame, until I see a well-behaved line element associated with such a frame. So far, we haven't seen any such thing.

 

[PeterDonis]

a clock near the Rindler horizon will be ticking much differently than the coordinate time there

I meant the observer at $x = 1 / \alpha$ in your original formulation, where the Rindler horizon is at $x = 0$. In the revised formulation I gave, which you have now adopted, this observer is at $x = 0$, and the Rindler horizon is at $x = - c^2 / a$.

 

[Sagittarius A-Star]

$$d\tau^2 = (1+ a^2\,x^2 / c^2)dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$

In the first term of the right side of the equation I see an issue with units plausibility check (1+ ...). When I compare to the book "Lecture Notes on the General Theory of Relativity" from Gron, equation (4.47), also a square is missing there.

Proposal:
$$d\tau^2 = (1+ a\,x / c^2)^2dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$
From that, he derives a time dilation formula.

 

[kent davidge]

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.

You say this, but the mentioned coordinates (Born) were shown to be singular at $r = 0$. So let me ask: when you physicists say that a coordinate system covers all of spacetime, is it fine to neglect a single pole like the one in this case?

 

[vanhees71]

Well, the coordinate singularity of Born coordinates at $r=0$ are really not an issue. You can simply reintroduce "Cartesian coordinates" via $x=r \cos \varphi$, $y=r \sin \varphi$, and you have a singularity free chart everywhere. From a mathematical point of view it's a chart over all spacetime.

If you want to describe observers by worldlines given by three of the coordinates held fixed and the fourth to be used as the time coordinate, you are restricted to the cylinder $\omega r<1$. Then $t$ is the time coordinate of course. At $\omega r=1$ you have a light-like curve, and for $\omega r>1$ the coordinate lines are all spacelike.

 

[Ibix]

You say this, but the mentioned coordinates (Born) were shown to be singular at $r = 0$.

Fair point. Examples without this issue are available - simply define $x'=x-x_0\sin(t/t_0)$ and otherwise keep Einstein coordinates. Or you could use radar coordinates based on a non-inertial but not eternally accelerating worldline, if you prefer a less contrived example.

 

[vanhees71]

In the first term of the right side of the equation I see an issue with units plausibility check (1+ ...). When I compare to the book "Lecture Notes on the General Theory of Relativity" from Gron, equation (4.47), also a square is missing there.

Proposal:
$$d\tau^2 = (1+ a\,x / c^2)^2dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$
From that, he derives a time dilation formula.

I think Rindler coordinates can be introduced most easily by defining it via a congruence of time-like world lines, considering points in constant proper acceleration along the $x$ axis of a Minkowski coordinate system $(t,x,y,z)$ with the world-line element
$$\mathrm{d}s^2=\eta_{\mu \nu} \mathrm{d}x^{\mu} \mathrm{d}^{\nu}=\mathrm{d} t^2-\mathrm{d} \vec{x}^2,$$
using natural units with $c=1$. In the following we only need to consider the $(t,x)$ plane, because in the change to Rindler coordinate nothing happens with $y$ and $z$. So we can set the corresponding new coordinates simply $y=\eta$ and $z=\zeta$.

Then the congruence of timelike world lines in the planes parallel to the $tx$ plane is defined by the EoM. of observers starting with 0 velocity at $x=\xi$. The solution in terms of a convenient affine parameter $\lambda$ is
$$t=\xi \sinh(\lambda \alpha), \quad x=\xi \cosh(\lambda \alpha).$$
That it's indeed an affine parameter is easily seen from
$$\dot{t}^2-\dot{x}^2=(\xi \alpha)^2,$$
and indeed $\lambda$ is the proper time of the observer on the worldline starting at $\xi=1/\alpha$.
The line element in the Rindler coordinate $(\lambda,\xi,\eta,\zeta)$ is
$$\mathrm{d} s^2 =\xi^2 \alpha^2 \mathrm{d} \lambda^2-\mathrm{d} \vec{x}^{\prime 2}$$
with $\vec{x}'=(\xi,\eta,\zeta)$.
The alternative form mentioned in #76 is also correct. It just defines $\xi$ differently with the initial point of the congruence of uniformly accelerated observers at $x=0$:
$$t=(1/\alpha +\tilde{\xi}) \sinh(\alpha \lambda), \quad x=(1/\alpha+\tilde{\xi}) \cosh(\alpha \lambda)-1/\alpha$$
leading to the mentioned line element
$$\mathrm{d} s^2=(1+\alpha \tilde{\xi})^2 \mathrm{d} \lambda^2-\mathrm{d} \tilde{x}^2$$
where $\tilde{x}^2=(\tilde{\xi},\eta,\zeta)$.

 

[pervect]

In the first term of the right side of the equation I see an issue with units plausibility check (1+ ...). When I compare to the book "Lecture Notes on the General Theory of Relativity" from Gron, equation (4.47), also a square is missing there.

Proposal:
$$d\tau^2 = (1+ a\,x / c^2)^2dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$
From that, he derives a time dilation formula.

Good catch. I'm pretty sure your psoposal is correct - I'll doulbecheck in the morning, if I remember.

 

[pervect]

In the first term of the right side of the equation I see an issue with units plausibility check (1+ ...). When I compare to the book "Lecture Notes on the General Theory of Relativity" from Gron, equation (4.47), also a square is missing there.

Proposal:
$$d\tau^2 = (1+ a\,x / c^2)^2dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$
From that, he derives a time dilation formula.

I looked it up - MTW uses a similar line element, but in geometric units. And you're right. There's a simpler argument for why you're right though that doesn't need to be looked up.

Basically, the metric coefficient is $-\gamma_g^2 \, dt^2$. This follows directly from the concept of gravitational time dilation . $\gamma_g$ used to be $\alpha x$ in the (geometrized) Rindler coordinates, and the point at which the factor $\gamma_g$ was unity was at $x=1/\alpha$. With the new coordinates, $\gamma_g = 1 + ax/c^2$ and $\gamma_g$ is unity at x=0.

 

[Sagittarius A-Star]

$$d\tau^2 = (1+ a\,x / c^2)^2dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$

With the new coordinates, $\gamma_g = 1 + ax/c^2$ and $\gamma_g$ is unity at x=0.

Yes. Gron derives in his book the following equation (4.50):
$$d\tau = dt\sqrt {(1+ \frac{g\,x} {c^2})^2 - \frac{v^2}{c^2}}$$
This expresses the combined effect of the gravitational and the kinematic time dilation.

• While the traveling twin travels inertially, then $g=0$ and the left term under the square root is "1". That gives the normal time-dilation formula for inertial frames.
• While traveling accelerated, then $g\neq 0$ and the right term under the square root can be neglected for the influence on age difference, if the turnaround-time is defined to be arbitrarily small compared to the inertial travel times.