Time elapsed along an accelerated world line

1. Apr 3, 2015

PWiz

Since an accelerated frame is at rest in some inertial frame (MCRF) only at at a particular instance, I clearly have to use integration to calculate the total proper time elapsed for the accelerated frame. If I take an infinitesimally small spacetime interval along an arbitrary timelike path, then $dτ^2=dt^2-dx^2-dy^2-dz^2=(1-\frac {dx^2} {dt^2} -\frac {dy^2} {dt^2} - \frac {dz^2} {dt^2})$ $dt^2$.
Now my main question is how to parametrize this is terms of another variable $θ$. If I apply the chain rule on the equation above, then
$dτ = \sqrt{\frac {dt^2-dx^2-dy^2-dz^2}{dθ^2}}$ $dθ$ (note: I'm not using polar coordinates here). I'm a little unsure here: can I write the numerator as $-η_{μν} x_μ x_ν$?
If yes, then does the equation become $$τ = \int \sqrt{-η_{μν} \frac{dx_μ dx_ν}{dθ^2}} dθ$$ ?
EDIT: I'd accidentally written the numerator in the expression as $$x_μ x_ν$$ and forgotten to add the $d$ part.

Last edited: Apr 3, 2015
2. Apr 3, 2015

Orodruin

Staff Emeritus
No, you are changing coordinate differentials for coordinates. If you put the differentials instead it will be true, but you can simply apply the chain rule a few times instead.

3. Apr 3, 2015

PWiz

Aaargh, I missed the $d$, sorry I meant $$τ = \int \sqrt{-η_{μν}\frac{dx_μ dx_ν}{dθ^2}} dθ$$. So this is correct?

4. Apr 3, 2015

Orodruin

Staff Emeritus
Essentially, you would typically not write the denominator as $d\theta^2$ but instead use derivatives.

5. Apr 3, 2015

Orodruin

Staff Emeritus
Oh, and the indices on your coordinate differentials should be contravariant.

6. Apr 3, 2015

PWiz

Alright thanks. So the correct way to write it is $$τ = \int \sqrt{-η_{μν} \frac{dx^μ}{dθ} \frac{dx^ν}{dθ}} dθ$$?

7. Apr 3, 2015

Orodruin

Staff Emeritus
Yes, assuming your metric convention is -+++.

8. Apr 3, 2015

PWiz

Yes, I'm using the Minkowski signature. I actually wanted to write down the matrix for complete clarity, but I'm not able to do it with Latex (I tried reading the guide as well). Can you please post it so that I can take a look at the code?

9. Apr 3, 2015

DrGreg

You mean
$$τ = \int \sqrt{ - \begin{bmatrix}\dot t & \dot x & \dot y & \dot z\end{bmatrix} \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix}\dot t \\ \dot x \\ \dot y \\ \dot z\end{bmatrix} } \, d\theta$$where
$$\dot t = \frac{dt}{d\theta}$$etc? (In units where c = 1).

10. Apr 3, 2015

PWiz

Yes! Thanks :)