Derive Escape Velocity GR: Source for Schwarzschild Metric

[snoopies622]

TL;DR Summary

Wondering how to find the escape velocity formula consistent with relativity.

I was surprised to read that the formula for escape velocity — at least for a spherical mass like the Earth — is the same in relativity as it is in classical physics:
$$v_e = (2GM/r)^{1/2}$$
I'm wondering if someone can give me a good source for deriving this. (I assume one takes a radial geodesic starting at rest from infinity using the Schwarzschild metric and finding its velocity at radius r?) Thanks.

 

[PeterDonis]

I'm wondering if someone can give me a good source for deriving this.

Many GR textbooks give this as a homework problem.

(I assume one takes a radial geodesic starting at rest from infinity using the Schwarzschild metric and finding its velocity at radius r?)

Yes.

 

[snoopies622]

Thanks Peter, actually I was hoping for a link to something on the internet since my college physics library is still semi-closed due to the Covid-19.

 

[Ibix]

Carroll's lecture notes, around equations 7.43 to 7.48 for the geodesic equation. Read back a bit for definitions of $E$, $L$, $\epsilon$.

 

[snoopies622]

Found it, thanks!

https://preposterousuniverse.com/wp-content/uploads/grnotes-seven.pdf

 

[snoopies622]

Epilogue:

I had forgotten about this, but using a couple equations found here

https://en.wikipedia.org/wiki/Schwarzschild_geodesics

and substituting, I get that the escape velocity isn't exactly the same as the Newtonian result (see OP) but differs by a factor $(1 - r_{s}/r)$.

All I did was find E by setting dr/dT = 0, h (angular momentum) =0 and r=infinity in the given

$$\frac {dr}{dT} = E^2/m^2 c^2 - (1 - r_{s}/r)(c^2 + h^2 / r^2)$$

and then let $v^2 = (dr/dT)^2 / (dt/dT)^2$

with $(1 - r_{s}/r) = (dt/dT) (E/mc^2)$

 

[PeterDonis]

using a couple equations found here

https://en.wikipedia.org/wiki/Schwarzschild_geodesics

and substituting, I get that the escape velocity isn't exactly the same as the Newtonian result (see OP) but differs by a factor $(1 - r_{s}/r)$.

This is not correct. The formula in your OP is correct.

All I did was find E by setting dr/dT = 0, h (angular momentum) =0 and r=infinity in the given

$$\frac {dr}{dT} = E^2/m^2 c^2 - (1 - r_{s}/r)(c^2 + h^2 / r^2)$$

and then let $v^2 = (dr/dT)^2 / (dt/dT)^2$

with $(1 - r_{s}/r) = (dt/dT) (E/mc^2)$

That doesn't give you escape velocity. Escape velocity is just $dr / dT$. What you are calculating here, $dr / dt$, is just a coordinate velocity and has no direct physical meaning.

 

[snoopies622]

Thanks Peter, I'm relieved to hear this since i just noticed that my dr/dt formula gives an escape velocity of zero at the surface of a Schwarzschild black hole!