# Potential Energy formula in Special Relativty

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

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
Gold Member
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
2020 Award
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.

kent davidge
vanhees71
Gold Member
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)##.

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Sagittarius A-Star
pervect
Staff Emeritus
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
Staff Emeritus
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 travelling 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 travelling 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.