Describing the Schwarzschild metric in Energy

[TrappedLight]

It is known that black holes give up radiation. For very large black holes, this will take a very long time. But nevertheless, they do, including smaller black holes.

 

I am not aware though of any metric which describes this. You may actually rewrite the metric in terms of a rest energy for a non-rotating black hole

 

[math](1 - 2\frac{Gm}{E} \frac{M}{r_s} c^2 dt^{2}) - \frac{dt}{1-2\frac{Gm}{E} \frac{M}{r_s}} - r^2 d\Omega^2[/math]

You won't recognize this exact expression, but you might be aware it is another way to write the metric. You won't recognize it because I derived it. But what is interesting is that we can discuss the energy of the system in the metric. Curiously it cannot be a function of time because of the timelike killing vector, so to describe a change in energy

[math](1 - 2\frac{Gm}{\Delta E} \frac{M}{r_s} c^2 dt^{2})[/math]

it would need to be a function of some other property of the metric. What other parameters might it depend on?

[ajb]

You may actually rewrite the metric in terms of a rest energy for a non-rotating black hole

Can you carefully tell me what is rest energy of a non-rotating black hole?

[TrappedLight]

Can you carefully tell me what is rest energy of a non-rotating black hole?

 

Why, is there something wrong with the statement?

[ajb]

Why, is there something wrong with the statement?

Probabily not. Do you just mean Mc^{2}? The M in the Schwarzschild solution is identified with the mass of the object. As the Schwarzschild solution is static and asymptotically flat all the standard notions of mass/energy agree. I am just wondering what you have really done here.

[TrappedLight]

Sorry I will get back to this soon ajb.

[TrappedLight]

I am just wondering what you have really done here.

 

Sorry I took a while to reply, I haven't been around.

 

Anyway, yeah, I have been wondering what can be done with it. I did notice one thing and that is if the metric describes a system in terms of the energy and is also charged [math]^{\dagger}[/math] and it is in the presence of a gravitational field, this implies that it will radiate due to the equivalence principle. There will be energy lost over a period of time as it radiates away energy, Perhaps an interesting case to consider would be a spinless, charged black hole which is radiating due to being in a gravitational acceleration, then the energy will change in the metric. This approach was inspired by some gravitational papers written by Motz.

 

First of all, we would recall the relativistic Larmor equation

[math]P = \frac{2}{3} \frac{e^2}{M^2c^3} (\frac{dP}{ds})^2[/math]

Mass in relativity is calculated by noticing that you calculate a mass in motion through

[math]m = \frac{m_0}{(1 - \frac{v^2}{c^2})^{\frac{1}{2}}}[/math]

Where [math]m_0[/math] is the ''rest mass.'' The mass may also be seen to be related to the density as

[math]M = \int r^2 \rho dr[/math]

The metric can be written in a generic form involving the metric charge

[math]ds^2 = (1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2})c^2 dt^2[/math]

which is equivalent to the Reissner-Nordstrom metric expression, which involves the same argument for the Schwarzschild metric with the added difference of a metric charge [math]Q[/math].

It would be the expression used to describe spinless but charged black hole particles.

If a charged particle is in a gravitational field, this must imply inertial acceleration. It means if the metric has a description of energy, then it must be emitting radiation since it self-implies it through the presence of a gravitational field.

The radiation emitted is therefore

[math]P = \frac{2}{3} \frac{e^2}{(\frac{m^2}{1 - \beta^2})c^3} \frac{1}{(1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2})} (\frac{dP}{dt})^2[/math]

[math]= \frac{2}{3} \frac{e^2}{c^3} \frac{a_{g}^{2}}{(1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2})^2}[/math]

*Note we are considering a particle of radius [math]R[/math], where [math]r = R[/math]

where [math]a_g[/math] is the gravitational acceleration and [math]\Delta E[/math] characterizes the loss of energy as a system is either blue shifted or red shifted to some observer. This is obviously the same effect known as synchroton radiation. Effects of synchroton radiation has been presumed for black hole jet streams, but it has never been applied quite like this for a particle-scale model, but we have been able to do it by creating this thought experiment. The squared metric term in the denominator appears because there is one such factor in [math]ds^2[/math]. The presence of the square of the metric also ensures that the power emitted by a charge in a gravitational field can increase beyond limit, implying a relationship to the luminosity of such a system. (Which has implications of detecting black holes).

I went on to see if the metric

[math](1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2})c^2 dt^2[/math]

has ever been written in the history of physics. I did find one article which used the expression, after a google search on charged, spinless black hole particles, in the ''exterior geometry'' section:

http://arxiv-web3.library.cornell.edu/pdf/1303.1274.pdf

 

The equation might have diverse applications.

You can calculate the denominator as the gravitational redshift directly, by rewriting the metric as a ratio related to the wavelength

[math]\frac{\lambda}{\lambda_0} = \frac{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}[/math]

This is obtained from

[math]z = \frac{1}{(\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2})}} - 1 = \frac{\lambda - \lambda_0}{\lambda}[/math]

also the change in energy is a natural consequence of the change in wavelength.

The metric then can be rewritten in terms of the gravitational shift in the power equation which measures the radiation emitted

[math]P = \frac{2}{3} \frac{e^2}{(\frac{m^2}{1 - \beta^2})c^3} \frac{1}{\frac{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}{\sqrt{1 - 2\frac{Gm}{\Delta E} \frac{M}{r} + \frac{GQ^2}{c^4 R^2}}}} (\frac{dP}{dt})^2[/math]

[math] = \frac{2}{3} \frac{e^2}{c^3} \frac{a_{g}^{2}}{(\frac{\lambda}{\lambda_0})} [/math]

The metric in the denominator now describes the change in the wavelength of the emitted radiation!

Edit* I should also note that the gravitational acceleration can be justly placed in there, since it is a consequence of the equivalence principle and we are working in these first hand principles since we have a charged particle placed in a gravitational field.

Edited October 4, 2013 by TrappedLight