Derivation of the acceleration in the Eddington-Finkelstein Metric

[happyparticle]

Homework Statement

Derivation of the acceleration in the Eddington-Finkelstein Metric

Relevant Equations

$a^r = \frac{a u^r e}{\sqrt{e^2 + g_{tt}}}$

Hi,

I'm trying to derive the equation (14) $a^r = \frac{a u^r e}{\sqrt{e^2 + g_{tt}}}$ from this article No Way Back: Maximizing survival time below the Schwarzschild event horizon and my algebra is really messy, so I'm wondering if I made some mistakes.

The authors say: "With the above definition of the conserved quantity related to the Killing vector, as well as the 4-velocityand 4-acceleration normalization and orthogonality, a little algebra reveals that for an acceleration of magnitude a"

My equations are:
\begin{equation}
e = g_{tt} u^t + g_{tr} u^r
\end{equation}

\begin{equation}
g_{tt} (u^t)^2 + 2g_{tr}u^t u^r + g_{rr} (u^r)^2 = -1
\end{equation}

\begin{equation}
g_{tt} a^t u^t + g_{tr}a^t u^r + g_{rt} a^r u^t + g_{rr} a^r u^r = 0
\end{equation}

\begin{equation}
g_{tt} (a^t)^2 + 2g_{tr} a^t a^r + g_{rr} (a^r)^2 = a^2
\end{equation}

Then, using equation (3) and (1) I isolated $a^t$:
\begin{equation}
a^t = \frac{-g_{rt} a^r u^t - g_{rr}a^r u^r}{e}
\end{equation}

Afterwards, I plugged (5) in (4). However the thing is starting to get pretty messy and I can't get the same expression as the author and I'm wondering if this is the right way to do it.

 

[ergospherical]

You can work in the orthogonal frame $(L, M)$ introduced in Section 3.3, with $L$ being a time-like unit vector and $M$ being a space-like unit vector. At $\tau = 0$, these have been set to be $L^{\mu}(0) = u^{\mu}$ and $M^{\mu}(0) = a^{\mu}/a$. [WLOG, you can take $\tau = 0$ at any instant along the worldline...]

In the 2d $t$-$r$ subspace, you can express the space-like unit vector as $M \propto \xi + \lambda u$, with $\lambda$ being a constant. To fix $\lambda$, contract $$u \cdot M \propto u \cdot \xi + \lambda u \cdot u = e - \lambda$$At $\tau = 0$, then $u \cdot M \big{|}_{\tau = 0} = L \cdot M \big{|}_{\tau = 0} = 0$, therefore $\lambda = e$. So: $$M \propto \xi + e u$$To enforce that $M$ is a unit vector, you must set:$$M = \frac{\xi + eu}{\sqrt{\xi \cdot \xi + 2 e \xi \cdot u + e^2 u \cdot u}} =\frac{\xi + eu}{\sqrt{\xi \cdot \xi + e^2}}$$Now you are almost done -- can you finish off by finding an expression for $a^{\mu} = a M^{\mu}$? Then, you can just read off $a^{t}$ and $a^{r}$.

 

[TSny]

@ergospherical 's method is really clever. I certainly wouldn't have thought of it.

$$a^t = \frac{-g_{rt} a^r u^t - g_{rr}a^r u^r}{e}$$

I think this is correct. You can get from this to equation (15) in the paper in a few steps. Equation (15) reads $$a^t = [1+u^t e]\frac{a^r}{eu^r}.$$ Your equation can be written as $$a^t =\left[-g_{rt} u^ru^t - g_{rr} (u^r)^2\right]\frac{a^r}{ e u^r}.$$ See if you can get the expressions in the brackets in these two equations to agree by using your equation (2).

 

[happyparticle]

@TSny So I guess the "little algebra" is not that little to get the equation (14).

@ergospherical I'm not sure to fully understand. I was trying to get the expression algebraically as the author did.

 

[ergospherical]

Sorry — you can ignore post #2 if it’s not helpful (it’s just another way to get the same result).

Your work is already on a good track — see @TSny’s comments.

 

[TSny]

@TSny So I guess the "little algebra" is not that little to get the equation (14).

Yes. My approach was similar to yours and it took me about a page of writing to get (15) and another page to get (14).

 

[happyparticle]

@ergospherical Thank you anyway!
@TSny How did you start to get (15)?

 

[TSny]

@TSny How did you start to get (15)?

I believe I did the same as you. Combine your equations (1) and (3) to get (5), as in your first post. Then get (14) (15) as outlined in post #3.

 

[happyparticle]

I think (14) is $a^r$. From your post #3 I found $a^t$.

 

[TSny]

I think (14) is $a^r$. From your post #3 I found $a^t$.

I meant to type (15) in post #8 instead of (14). Thanks for catching this.

 

[happyparticle]

Thank you so much!

 

[ergospherical]

...by finding an expression for $a^{\mu} = a M^{\mu}$?

Now that you solved it -- it is worth to note that:
$$a^{\mu} = \frac{a}{\sqrt{\xi \cdot \xi + e^2}} (\xi^{\mu} + eu^{\mu})$$
In this expression, $\xi \cdot \xi = g_{\mu \nu} \xi^{\mu} \xi^{\nu} = g_{tt} \xi^t \xi^t = g_{tt}$
You may then read off:

$a^{t} = \frac{a(1 + eu^{t})}{\sqrt{g_{tt} + e^2}}$

$a^{r} = \frac{aeu^{r}}{\sqrt{g_{tt}+ e^2}}$