Calculating Surface Area of Schwarzschild Black Hole w/Weyl Coordinates

[user1139]

TL;DR

Unable to find surface area of Black Hole using Weyl coordinates

Recently, I was tasked to find the surface area of the Schwarzschild Black Hole. I have managed to do so using spherical and prolate spheroidal coordinates. However, my lecturer insists on only using Weyl canonical coordinates to directly calculate the surface area.

The apparent problem arises from the term g_zzg_φφ due to the presence of the factor ρ². Upon setting ρ=0, the integrand becomes zero for −M<z<M thus invalidating the surface area integral. I was unsuccessful in my attempts to get rid of the ρ² factor.

Can someone advise me on how to proceed, if it is at all possible?

 

[PeterDonis]

Weyl canonical coordinates

Do you mean these?

https://en.wikipedia.org/wiki/Weyl_metrics#Schwarzschild_solution

 

[user1139]

Yup I'm referring to that

 

[PeterDonis]

Yup I'm referring to that

Ok, then it would be helpful if you would post explicitly your calculations so far. I am still unclear from your OP about exactly what you are doing.

 

[user1139]

Ok, then it would be helpful if you would post explicitly your calculations so far. I am still unclear from your OP about exactly what you are doing.

Let the surface area of the Schwarzschild Black Hole be $A$

$A= 2\pi\int \sqrt{g_{zz}g_{\psi\psi}} \, dz$

Here,

$g_{zz}=(\frac{\sqrt{\rho^2+(z-M)^2}-(z-M)}{\sqrt{\rho^2+(z+M)^2}-(z+M)})(\frac{[\rho^2+(\sqrt{\rho^2+(z-M)^2}-(z-M))(\sqrt{\rho^2+(z+M)^2}-(z+M))]^2}{[\rho^2+(\sqrt{\rho^2+(z-M)^2}-(z-M))^2][\rho^2+(\sqrt{\rho^2+(z+M)^2}-(z+M))^2]})$

and

$g_{\psi\psi}=\rho^2(\frac{\sqrt{\rho^2+(z-M)^2}-(z-M)}{\sqrt{\rho^2+(z+M)^2}-(z+M)})$

The surface of the black hole is defined for $\rho=0$ and $-M\leq z \leq M$

For $-M<z<M$, substituting $\rho=0$ causes the integrand to become zero while for $-M\leq z \leq M$, substituting $\rho=0$ causes the integrand to become "blow up" at the ends

Hence, I am not really sure how to find the surface area of the black hole using only Weyl coordinates, if it is at all possible.

 

[PeterDonis]

Let the surface area of the Schwarzschild Black Hole be $A$

$A= 2\pi\int \sqrt{g_{zz}g_{\psi\psi}} \, dz$

Where are you getting this formula from?

 

[user1139]

Where are you getting this formula from?

The formula when using spherical coordinates is

$\int\int \sqrt{g_{\theta\theta}g_{\phi\phi}} \, d\phi d\theta$

In Weyl cooridnates, only $z$ and $\phi$ vary when finding the surface area. Hence, I switched the metric components and the differential accordingly.

 

[PeterDonis]

In Weyl cooridnates, only $z$ and $\phi$ vary when finding the surface area.

Ok, you wrote $\psi$ instead of $\phi$ before, so that confused me.

Try rewriting the formula for $\rho$ in terms of $L$ and $M$, as defined on the Wikipedia page. You should find that there is a cancellation in the formula for $g_{\phi \phi}$ in terms of $L$ and $M$ that means $g_{\phi \phi}$ is not equal to $0$ at the horizon, even though $\rho = 0$ there. (Note that you can't just conclude that $g_{\phi \phi} = 0$ because $\rho = 0$ at the horizon, since the denominator of $g_{\phi \phi}$ also goes to zero at the horizon, so you have $0 / 0$, which is undefined unless you can do some algebra to make a cancellation that gives a finite answer.)

 

[user1139]

Ok, you wrote $\psi$ instead of $\phi$ before, so that confused me.

Try rewriting the formula for $\rho$ in terms of $L$ and $M$, as defined on the Wikipedia page. You should find that there is a cancellation in the formula for $g_{\phi \phi}$ in terms of $L$ and $M$ that means $g_{\phi \phi}$ is not equal to $0$ at the horizon, even though $\rho = 0$ there. (Note that you can't just conclude that $g_{\phi \phi} = 0$ because $\rho = 0$ at the horizon, since the denominator of $g_{\phi \phi}$ also goes to zero at the horizon, so you have $0 / 0$, which is undefined unless you can do some algebra to make a cancellation that gives a finite answer.)

I am unsure on how to rewrite the formula for $\rho$ in terms of $L$ and $M$, is it possible for you to show me partial workings?

 

[PeterDonis]

I am unsure on how to rewrite the formula for $\rho$ in terms of $L$ and $M$

The Wikipedia page gives $\rho$ in terms of $r$ and $M$, and $r$ in terms of $L$ and $M$. Just substitute.

 

[user1139]

The Wikipedia page gives $\rho$ in terms of $r$ and $M$, and $r$ in terms of $L$ and $M$. Just substitute.

How would you deal with $\sin\theta$?

 

[PeterDonis]

How would you deal with $\sin\theta$?

Equation (15) on the Wikipedia page also gives $z$ in terms of $\theta$, which can easily be inverted to give $\theta$ in terms of $z$ (since $r = 2M$ is constant on the horizon).

 

[user1139]

Equation (15) on the Wikipedia page also gives $z$ in terms of $\theta$, which can easily be inverted to give $\theta$ in terms of $z$ (since $r = 2M$ is constant on the horizon).

If you are going to substitute $r=2M$ eventually, why not do the same with the equation for $\rho$?

 

[PeterDonis]

If you are going to substitute $r=2M$ eventually, why not do the same with the equation for $\rho$?

You can do that, as long as you don't conclude that you can just plug $\rho = 0$ into the equation for $g_{\phi \phi}$. As has already been shown, that doesn't work.

 

[PeterDonis]

If you are going to substitute $r=2M$ eventually, why not do the same with the equation for $\rho$?

Perhaps this answer might be helpful: doing this is what got you into trouble in the first place, since it leads you to set $\rho = 0$ everywhere. So maybe it might be worth trying not doing it and see where that gets you.

 

[PeterDonis]

Equation (15) on the Wikipedia page also gives $z$ in terms of $\theta$, which can easily be inverted to give $\theta$ in terms of $z$ (since $r = 2M$ is constant on the horizon).

If the $r = 2M$ substitution in the equation relating $z$ and $\theta$ and the equation for $\rho$ is really a stumbling block for you, try substituting $r = L + M$ instead (since that is another one of the equations in (15)).