Verifying the Komar Integral for the Kerr-Newman Solution

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In summary, the Komar integral J(V) is a mathematical expression given by J(V)=\frac{1}{16 \pi G} \oint_{\partial V} dS_{\mu \nu} D^\mu m^\nu, where V is the volume of a spacelike hypersurface \Sigma with boundary \partial V and m=\frac{\partial}{\partial \phi} is the Killing vector field. This Komar integral is associated with the Kerr-Newman solution and can be used to verify that J=Ma. In order to compute J(V), the Christoffel symbols need to be expressed in terms of dS_{\mu \nu} and it may be possible to show a relationship between J(V
  • #1
latentcorpse
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Given the Komar integral

[itex]J(V)=\frac{1}{16 \pi G} \oint_{\partial V} dS_{\mu \nu} D^\mu m^\nu[/itex]
where [itex]V[/itex] is the volume of the spacelike hypersurface [itex]\Sigma[/itex] with boundary [itex]\partial V[/itex] and [itex]m=\frac{\partial}{\partial \phi}[/itex] is the Killing vector field This particula Komar integral is associated with, I am asked to verify that [itex]J=Ma[/itex] for the Kerr-Newman solution with parameter [itex]a[/itex].

I have not really got any idea what to do here since the only definition in my notes of [itex]a[/itex] is [itex]a=\frac{J}{M}[/itex] and so I ended up going round in circles.
I was wondering if I am supposed to extract something from the formula for [itex]J(V)[/itex] that we can write as the mass or the ADM mass or something?

Thanks.
 
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  • #2
You'll want to express [tex]J(V)[/tex] and [tex]E(V)[/tex] in terms of Christoffel symbols. I haven't tried to compute the Christoffel symbols, but since many of them vanish, it's possible that only [tex]dS_{\phi t}[/tex] appears in both expressions.
 
  • #3
fzero said:
You'll want to express [tex]J(V)[/tex] and [tex]E(V)[/tex] in terms of Christoffel symbols. I haven't tried to compute the Christoffel symbols, but since many of them vanish, it's possible that only [tex]dS_{\phi t}[/tex] appears in both expressions.

Does this look as though it is correct:

[itex]D^\mu m^\nu = g^{\mu \rho} D_\rho m^\nu = g^{\mu \rho} ( \partial_\rho m^\nu + \Gamma^\nu{}_{\rho \alpha} m^\alpha ) = g^{\mu \rho} \Gamma^\nu{}_{\rho \phi} m^\phi= g^{\mu \rho} \Gamma^\nu{}_{\rho \alpha}[/itex] where I killed [itex]\partial_\rho m^\nu[/itex] since [itex]m^\nu[/itex] is always a constant (either 0 or 1 depending on whether [itex]\nu=\phi[/itex] or not) and then in the last line [itex]m^\phi=1[/itex]

So we'd have

[itex]J(V)=\frac{1}{16 \pi G} \oint_{\partial V} dS_{\mu \nu} g^{\mu \rho} \Gamma^\nu{}_{\rho \phi}[/itex]

Should I hit the [itex]dS_{\mu \nu}[/itex] with the [itex]g^{\mu \rho}[/itex] or is it better to just leave it in the form that it's already in?

The formula for [itex]E(V)[/itex] is

[itex]E(V)=-\frac{1}{8 \pi G} \oint_{\partial V} dS_{\mu \nu} D^\mu k^\nu[/itex]

I'll not type that out yet until I check I got the first one right though!
 
  • #4
latentcorpse said:
[itex]J(V)=\frac{1}{16 \pi G} \oint_{\partial V} dS_{\mu \nu} g^{\mu \rho} \Gamma^\nu{}_{\rho \phi}[/itex]

That's right.

Should I hit the [itex]dS_{\mu \nu}[/itex] with the [itex]g^{\mu \rho}[/itex] or is it better to just leave it in the form that it's already in?

I'd leave it the way it is until you get further along. Ideally, you'd be able to show relations between [tex] \Gamma^\nu{}_{\rho \phi} [/tex] and [tex]\Gamma^\nu{}_{\rho t}[/tex], but it might be that there are additional terms that vanish due to properties of the surface.
 
  • #5
fzero said:
That's right.



I'd leave it the way it is until you get further along. Ideally, you'd be able to show relations between [tex] \Gamma^\nu{}_{\rho \phi} [/tex] and [tex]\Gamma^\nu{}_{\rho t}[/tex], but it might be that there are additional terms that vanish due to properties of the surface.

By a similar procedure then, I find that

[itex]E(V)=-\frac{1}{8 \pi G} \oint_{\partial V} dS_{\mu \nu} g^{\mu \rho} \Gamma^{\nu}{}_{\rho t}[/itex]

So I guess I need to use a relationship between E and V to go ahead? There isn't any such relationship mentioned in this section of my notes - should I just use the formula for the energy of a system with ang mom J i.e. [itex]E=\frac{J^2}{2M}[/itex]?
 
  • #6
I think that you want to show a relationship between the integrands. You'll need to compute the Christoffel symbols to some extent and understand the surface volume element to see what terms contribute.
 
  • #7
fzero said:
I think that you want to show a relationship between the integrands. You'll need to compute the Christoffel symbols to some extent and understand the surface volume element to see what terms contribute.

What is the relationship between E and J? Did I get that right?

Then, say for the Christoffel symbol in E we get

[itex]\Gamma^\nu{}_{\rho t}=\frac{1}{2} g^{\nu \alpha} ( g_{\rho \alpha , t} + g_{\alpha t, \rho} - g_{\rho t,\alpha })[/itex]
I reckon I can get rid of the first term since no components of the Kerr-Newman metric have t dependence. So that would leave
[itex]\Gamma^\nu{}_{\rho t}=\frac{1}{2} g^{\nu \alpha} ( g_{\alpha t, \rho}-g_\rho t, \alpha})[/itex]
 
  • #8
latentcorpse said:
What is the relationship between E and J? Did I get that right?

I doubt that [tex]E=J^2/(2M)[/tex] holds for a charged black hole. In any case, I don't think you need such a relationship to solve this problem.


Then, say for the Christoffel symbol in E we get

[itex]\Gamma^\nu{}_{\rho t}=\frac{1}{2} g^{\nu \alpha} ( g_{\rho \alpha , t} + g_{\alpha t, \rho} - g_{\rho t,\alpha })[/itex]
I reckon I can get rid of the first term since no components of the Kerr-Newman metric have t dependence. So that would leave
[itex]\Gamma^\nu{}_{\rho t}=\frac{1}{2} g^{\nu \alpha} ( g_{\alpha t, \rho}-g_\rho t, \alpha})[/itex]

As I've been saying, you'll also want to find out what conditions are put on [tex]dS_{\mu\nu}[/tex] from the spacelike hypersurface condition.
 
  • #9
fzero said:
I doubt that [tex]E=J^2/(2M)[/tex] holds for a charged black hole. In any case, I don't think you need such a relationship to solve this problem.




As I've been saying, you'll also want to find out what conditions are put on [tex]dS_{\mu\nu}[/tex] from the spacelike hypersurface condition.

If you check out section 5.3.1 in here:
http://arxiv.org/PS_cache/gr-qc/pdf/9707/9707012v1.pdf
it says just under eqn (5.43) that if V is on a t=const spacelike hypersurface then [itex]dS_\mu m^\mu=0[/itex] for the J integral and that for the E integral (see below eqn (5.41)) that we should pick V as a 2-sphere at spatial infinity.

Is this of any use?
 
  • #10
latentcorpse said:
If you check out section 5.3.1 in here:
http://arxiv.org/PS_cache/gr-qc/pdf/9707/9707012v1.pdf
it says just under eqn (5.43) that if V is on a t=const spacelike hypersurface then [itex]dS_\mu m^\mu=0[/itex] for the J integral and that for the E integral (see below eqn (5.41)) that we should pick V as a 2-sphere at spatial infinity.

Is this of any use?

Yes, but that leads to expressions involving the stress tensor. Maybe it works in the weak source approximation, but I'm not sure that it's valid.
 
  • #11
fzero said:
Yes, but that leads to expressions involving the stress tensor. Maybe it works in the weak source approximation, but I'm not sure that it's valid.

So what would you suggest for finding these conditions then?
 
  • #12
latentcorpse said:
So what would you suggest for finding these conditions then?

As I wrote:

fzero said:
You'll need to compute the Christoffel symbols to some extent and understand the surface volume element to see what terms contribute.
 
  • #13
fzero said:
As I wrote:

Surely I've already done the Christoffel symbols as far as they can be taken though?
I don't know what to say about [itex]dS_{\mu \nu}[/itex]?
 
  • #14
latentcorpse said:
Surely I've already done the Christoffel symbols as far as they can be taken though?

It's possible that there's some simplifications if you actually compute them from knowledge of the metric.

I don't know what to say about [itex]dS_{\mu \nu}[/itex]?

You'll want to use properties of the hypersurface, as in your post #9.

The sorts of simplifications you're looking for are analogous to the way that you'd get back to equ. (5.28) from (5.31) in Townsend's notes.
 

1. What is the Komar integral for the Kerr-Newman solution?

The Komar integral is a mathematical tool that helps in defining the total energy and angular momentum of a rotating and charged black hole, as described by the Kerr-Newman solution. It is based on the concept of the Killing vector, which represents the symmetry of spacetime.

2. Why is it important to verify the Komar integral for the Kerr-Newman solution?

The verification of the Komar integral is crucial because it serves as a consistency check for the solution. It ensures that the solution is physically meaningful and satisfies the equations of general relativity. It also provides a way to calculate the total energy and angular momentum of a black hole, which are important quantities in studying the dynamics of black holes.

3. How is the Komar integral calculated for the Kerr-Newman solution?

The Komar integral is calculated by integrating the energy-momentum tensor and the Killing vector over a closed surface surrounding the black hole. For the Kerr-Newman solution, this integral takes into account the mass, angular momentum, and electric charge of the black hole.

4. Can the Komar integral be used to calculate the energy and angular momentum of other black hole solutions?

Yes, the Komar integral can be used to calculate the energy and angular momentum of any stationary, axisymmetric black hole solution. This includes the Kerr solution, which describes a rotating black hole without charge, and the Reissner-Nordström solution, which describes a non-rotating black hole with charge.

5. Are there any limitations to the Komar integral for the Kerr-Newman solution?

One limitation of the Komar integral is that it only provides a global measure of energy and angular momentum, rather than a local one. This means that it cannot be used to calculate the energy and momentum at a specific point in the spacetime. Additionally, the Komar integral requires a well-defined Killing vector, which may not exist in all black hole solutions.

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