Our cosmology course is well under way and it's a lot of fun so far! The class direction overall is towards describing the inflationary universe by means of quantum field theory, but this week we're focused on relativity. We're allowed to work on our homework together, however, I'm spending most of my time in the lab this semester, so I'll be posting my homework notes here. If you'd like to grab bits and pieces, make suggestions, or contribute, the whole shooting match will also be archived on github.

Our first homework contains a problem that involves accelerating reference frames. The question is, given the transformation between the lab and the accelerating reference frame, figure out if the line element $ds^2$ is preserved. There are a few interesting aspects to this problem. First, while the transform looks similar to Rindler coordinates, it's not, (as ar as I can tell.) Second, looking into Rindler coordinates a bit, they seem to maintain an invariant $ds^2$, which kind of makes sense, because the calculation is made moving form one instantaneous inertial rest frame to another in succession. This also kind of doesn't make sense because mentions of acceleration fairly scream, non-inertial.

3. We need to find out if $ds^2$, the four-space line element is invariant under a given transformation.

The given transformation is

$t' = \dfrac{1}{g}\left(e^{gz}\right)sinh\left(gt\right)$

$z' = \dfrac{1}{g}\left(e^{gz}\right)cosh\left(gt\right)$

So,

$dt' = \dfrac{1}{g}g\left(e^{gz}\right)sinh\left(gt\right)dz + \dfrac{1}{g}g\left(e^{gz}\right)cosh\left(gt\right)dt$

$dz' = \dfrac{1}{g}g\left(e^{gz}\right)cosh\left(gt\right)dz + \dfrac{1}{g}g\left(e^{gz}\right)sinh\left(gt\right)dt$

$dt^{\prime 2} = e^{2gz}sinh^2\left(gt\right)dz^2 + e^{2gz}cosh^2\left(gt\right)dt^2$

Where the mixed differentials have been omitted because they will cancel with the mixed terms from $dz^{\prime 2}$.

$dz^{\prime 2} = e^{2gz}cosh^2\left(gt\right)dz^2 + e^{2gz}sinh^2\left(gt\right)dt^2$

$ds^{\prime 2} = dt^{\prime 2} - dz^{\prime 2} = e^{2gz}sinh^2\left(gt\right)dz^2 + e^{2gz}cosh^2\left(gt\right)dt^2 - $

$\left(e^{2gz}cosh^2\left(gt\right)dz^2 + e^{2gz}sinh^2\left(gt\right)dt^2\right)$

$= e^{2gz}\left(sinh^2\left(gt\right) - cosh^2\left(gt\right)\right)dz^2 + e^{2gz}\left(cosh^2\left(gt\right) - sinh^2\left(gt\right)\right)dt^2$

$ds^{\prime 2} = e^{2gz}\left(dt^2 - dz^2\right)$

Consequently, $ds^{\prime 2}$ is not invariant in this metric.

Just as a quick check, what if we had factored the $gz$ term back in as a phase to the $sinh$ and $cosh$ functions right away? How would that have changed things? Can the $e^{gz}$ even be factored back in as a phase?

$sinh$ can be written as

$sinh\left(x\right) = \dfrac{e^x +e^{-x}}{2}$

Factoring the $e^{gz}$ back in gives us

$\dfrac{e^{gt + gz} +e^{-gt + gz}}{2}$

So, it looks like life is good. We can't get back to a single $sinh$ or $cosh$ when you take the $gz$ back into the expression.

The metric given for this problem looks like the Davies and Birrell form of the Rindler coordinates. However, if we write down Rindler coordinates from Rindler's "Special Relativity" book, (1960, p. 41), they are:

$t = \dfrac{1}{g}sinh\left(g \tau\right)$

$z = \dfrac{1}{g}cosh\left(g \tau\right) - \dfrac{1}{g}$

While the missing leading exponential term, (since it's not there), will not cause the final result to not be invariant in $ds^2$, other issues may. Playing the same $dt^2 - dz^2$ trick as above, (even though we're not looking at the primed, but the unprimed frame), results in

$ds^2 = dt^2 - dz^2 = cosh^2\left(g\tau\right)d\tau^2 - sinh^2\left(g\tau\right)d\tau^2$

or $ds^2 = d\tau^2$

This seems to indicate that Rindler coordinates maintain an invariance in $ds$ even though they incorporate acceleration and a non-inertial frame. It would also seem to indicate that the metric given in this problem is not in fact the same as Rindler coordinates.

Our first homework contains a problem that involves accelerating reference frames. The question is, given the transformation between the lab and the accelerating reference frame, figure out if the line element $ds^2$ is preserved. There are a few interesting aspects to this problem. First, while the transform looks similar to Rindler coordinates, it's not, (as ar as I can tell.) Second, looking into Rindler coordinates a bit, they seem to maintain an invariant $ds^2$, which kind of makes sense, because the calculation is made moving form one instantaneous inertial rest frame to another in succession. This also kind of doesn't make sense because mentions of acceleration fairly scream, non-inertial.

3. We need to find out if $ds^2$, the four-space line element is invariant under a given transformation.

The given transformation is

$t' = \dfrac{1}{g}\left(e^{gz}\right)sinh\left(gt\right)$

$z' = \dfrac{1}{g}\left(e^{gz}\right)cosh\left(gt\right)$

So,

$dt' = \dfrac{1}{g}g\left(e^{gz}\right)sinh\left(gt\right)dz + \dfrac{1}{g}g\left(e^{gz}\right)cosh\left(gt\right)dt$

$dz' = \dfrac{1}{g}g\left(e^{gz}\right)cosh\left(gt\right)dz + \dfrac{1}{g}g\left(e^{gz}\right)sinh\left(gt\right)dt$

$dt^{\prime 2} = e^{2gz}sinh^2\left(gt\right)dz^2 + e^{2gz}cosh^2\left(gt\right)dt^2$

Where the mixed differentials have been omitted because they will cancel with the mixed terms from $dz^{\prime 2}$.

$dz^{\prime 2} = e^{2gz}cosh^2\left(gt\right)dz^2 + e^{2gz}sinh^2\left(gt\right)dt^2$

$ds^{\prime 2} = dt^{\prime 2} - dz^{\prime 2} = e^{2gz}sinh^2\left(gt\right)dz^2 + e^{2gz}cosh^2\left(gt\right)dt^2 - $

$\left(e^{2gz}cosh^2\left(gt\right)dz^2 + e^{2gz}sinh^2\left(gt\right)dt^2\right)$

$= e^{2gz}\left(sinh^2\left(gt\right) - cosh^2\left(gt\right)\right)dz^2 + e^{2gz}\left(cosh^2\left(gt\right) - sinh^2\left(gt\right)\right)dt^2$

$ds^{\prime 2} = e^{2gz}\left(dt^2 - dz^2\right)$

Consequently, $ds^{\prime 2}$ is not invariant in this metric.

Just as a quick check, what if we had factored the $gz$ term back in as a phase to the $sinh$ and $cosh$ functions right away? How would that have changed things? Can the $e^{gz}$ even be factored back in as a phase?

$sinh$ can be written as

$sinh\left(x\right) = \dfrac{e^x +e^{-x}}{2}$

Factoring the $e^{gz}$ back in gives us

$\dfrac{e^{gt + gz} +e^{-gt + gz}}{2}$

So, it looks like life is good. We can't get back to a single $sinh$ or $cosh$ when you take the $gz$ back into the expression.

**Questions**The metric given for this problem looks like the Davies and Birrell form of the Rindler coordinates. However, if we write down Rindler coordinates from Rindler's "Special Relativity" book, (1960, p. 41), they are:

$t = \dfrac{1}{g}sinh\left(g \tau\right)$

$z = \dfrac{1}{g}cosh\left(g \tau\right) - \dfrac{1}{g}$

While the missing leading exponential term, (since it's not there), will not cause the final result to not be invariant in $ds^2$, other issues may. Playing the same $dt^2 - dz^2$ trick as above, (even though we're not looking at the primed, but the unprimed frame), results in

$ds^2 = dt^2 - dz^2 = cosh^2\left(g\tau\right)d\tau^2 - sinh^2\left(g\tau\right)d\tau^2$

or $ds^2 = d\tau^2$

This seems to indicate that Rindler coordinates maintain an invariance in $ds$ even though they incorporate acceleration and a non-inertial frame. It would also seem to indicate that the metric given in this problem is not in fact the same as Rindler coordinates.

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