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utesfan100
2012-Sep-04, 09:31 PM
This question is motivated by a desire to understand the math required for an ATM theory I am working on. Thus the methods of solution are of more interest to me than any actual calculations.

From Wikipedia I find that the Reissner–Nordström metric (http://en.wikipedia.org/wiki/Reissner-Nordstr%C3%B6m_metric) is expressed as:
c^2d\tau^2=\alpha^2c^2dt^2-\frac{dr^2}{\alpha^2}-r^2d\Omega^2
where
\alpha^2=1-\frac{2GM}{rc^2}+\frac{GkQ^2}{r^2c^4}
d\Omega^2=d\theta^2+\sin^2\theta d\phi^2

I understand that this represents a 4x4 diagonal array, \eta_{\mu\nu}=diag(\alpha^2,\alpha^{-2},r^2,r^2\sin^2\theta).

Does this mean the inverse metric is given by\eta^{\mu\nu}=diag(\alpha^{-2},\alpha^2,r^2,r^2\sin^2\theta)?

I assume this includes the weight of the electric field, which is defined from the potential. This article also gives the electromagnetic potential as A_\mu=(\frac{Q}{r},0,0,0)

Would A^\mu=(\alpha^{-2}\frac{Q}{r},0,0,0)? If so, why did we not define A^\mu=(\frac{Q}{r},0,0,0), or even split the difference with A_\mu=(\alpha\frac{Q}{r},0,0,0) and A^\mu=(\alpha^{-1}\frac{Q}{r},0,0,0)?

utesfan100
2012-Sep-04, 09:50 PM
My end goal of this thread is to understand how to determine the bending of an electron moving at 0.9999c at infinity by a Sun-like object of 2e30 kg and a charge of +100 C with a closest approach of 7.5e8 m, and a similar nearly extremal object with a charge of +4e18 C.

I anticipate that the former will be nearly the bending of light in the Schwarzschild metric, with the second having a correction twice that expected by the classical theory due to charge.

Selfsim
2012-Sep-04, 11:01 PM
This question is motivated by a desire to understand the math required for an ATM theory I am working on. ????

slang
2012-Sep-04, 11:50 PM
????

No, let's NOT try to goad utesfan100 into explaining his ATM theory, or how this relates to it, here. Let's just try to answer his questions, if we can. For a rather large value of "we", not including "me". If that's not what you intended, Selfsim, consider this a warning for others not to go that direction.

Cougar
2012-Sep-05, 12:35 AM
...with a charge of +4e18 C...

Wolfram interprets this as degrees Celsius, which I doubt is your intent.

utesfan100
2012-Sep-05, 01:02 AM
Wolfram interprets this as degrees Celsius, which I doubt is your intent.
In SI, I understand Celsius to be °C, C is Coulomb the unit of electric charge.

utesfan100
2012-Sep-05, 06:01 PM
I had a few sign errors in the OP, and have emphasized my questions.

This question is motivated by a desire to understand the math required for an ATM theory I am working on. Thus the methods of solution are of more interest to me than any actual calculations.

From Wikipedia I find that the Reissner–Nordström metric (http://en.wikipedia.org/wiki/Reissner-Nordstr%C3%B6m_metric) is expressed as:
c^2d\tau^2=\alpha^2c^2dt^2-\frac{dr^2}{\alpha^2}-r^2d\Omega^2
where
\alpha^2=1-\frac{2GM}{rc^2}+\frac{GkQ^2}{r^2c^4}
d\Omega^2=d\theta^2+\sin^2\theta d\phi^2

I understand that this represents a 4x4 diagonal array, \eta_{\mu\nu}=diag(\alpha^2,-\alpha^{-2},-r^2,-r^2\sin^2\theta).

Question 1: Does this mean the inverse metric is given by\eta^{\mu\nu}=diag(\alpha^{-2},-\alpha^2,-r^2,-r^2\sin^2\theta)?

I assume this includes the weight of the electric field, which is defined from the potential. This article also gives the electromagnetic potential as A_\mu=(\frac{Q}{r},0,0,0)

Question 2: Would A^\mu=(\alpha^{-2}\frac{Q}{r},0,0,0)? If so, why did we not define A^\mu=(\frac{Q}{r},0,0,0), or even split the difference with A_\mu=(\alpha\frac{Q}{r},0,0,0) and A^\mu=(\alpha^{-1}\frac{Q}{r},0,0,0)?

VonBelmont
2012-Sep-05, 06:19 PM
This question is motivated by a desire to understand the math required for an ATM theory I am working on. Thus the methods of solution are of more interest to me than any actual calculations.

From Wikipedia I find that the Reissner–Nordström metric (http://en.wikipedia.org/wiki/Reissner-Nordstr%C3%B6m_metric) is expressed as:
c^2d\tau^2=\alpha^2c^2dt^2-\frac{dr^2}{\alpha^2}-r^2d\Omega^2
where
\alpha^2=1-\frac{2GM}{rc^2}+\frac{GkQ^2}{r^2c^4}
d\Omega^2=d\theta^2+\sin^2\theta d\phi^2

I understand that this represents a 4x4 diagonal array, \eta_{\mu\nu}=diag(\alpha^2,\alpha^{-2},r^2,r^2\sin^2\theta).

Does this mean the inverse metric is given by\eta^{\mu\nu}=diag(\alpha^{-2},\alpha^2,r^2,r^2\sin^2\theta)?

I assume this includes the weight of the electric field, which is defined from the potential. This article also gives the electromagnetic potential as A_\mu=(\frac{Q}{r},0,0,0)

Would A^\mu=(\alpha^{-2}\frac{Q}{r},0,0,0)? If so, why did we not define A^\mu=(\frac{Q}{r},0,0,0), or even split the difference with A_\mu=(\alpha\frac{Q}{r},0,0,0) and A^\mu=(\alpha^{-1}\frac{Q}{r},0,0,0)?

I thought the term \frac{Q}{r} had a negative term in it since

1 - \frac{2M}{r} = -\frac{Q}{r}

Just thinking out loud. It's been a while since I read anything on this, I am trying to work out your answers.

utesfan100
2012-Sep-05, 08:24 PM
I thought the term \frac{Q}{r} had a negative term in it since

1 - \frac{2M}{r} = -\frac{Q}{r}

Just thinking out loud. It's been a while since I read anything on this, I am trying to work out your answers.
It is my understanding that with mass the sign is negative because the bending of space is attractive, for charge this is positive because the force is repulsive.

This difference may also be due to using a signature of (+,-,-,-) rather than (-,+,+,+). The latter is used in most texts on special relativity, while the former is dominant in literature on electromagnetism. The former is also what is usually used in the metrics of black holes, regardless of charged and rotation. Indeed, my original post implied a signature of (+,+,+,+), which is clearly in error and was fixed in the post #7.

We will get a sign change for the test cases above because the central mass is positively charged, and the negligible body passing by is negatively charged.

VonBelmont
2012-Sep-05, 10:03 PM
It is my understanding that with mass the sign is negative because the bending of space is attractive, for charge this is positive because the force is repulsive.

This difference may also be due to using a signature of (+,-,-,-) rather than (-,+,+,+). The latter is used in most texts on special relativity, while the former is dominant in literature on electromagnetism. The former is also what is usually used in the metrics of black holes, regardless of charged and rotation. Indeed, my original post implied a signature of (+,+,+,+), which is clearly in error and was fixed in the post #7.

We will get a sign change for the test cases above because the central mass is positively charged, and the negligible body passing by is negatively charged.

It might be the signature. I will think about this more. I always get my signs mixed up... though in the case I presented, I believe strongly that is true --- be weary of wiki however, it is not always the best source for accurate measurements.

utesfan100
2012-Sep-06, 07:51 PM
Question 1 is now answered in the negative. I neglected to invert the angular components.

http://faculty.gg.uwyo.edu/dueker/tensor%20curvilinear%20relativity/tensor%20analysis%20intro.pdf

The inverse metric is given by \eta^{\mu\nu}=diag(\alpha^{-2},-\alpha^2,-r^{-2},-r^{-2}csc^2\theta).

Question 2 from post #7 remains open.

utesfan100
2012-Sep-06, 09:40 PM
Reading on curved space times, I think I have identified where my most significant difficulties in the maths arise.

Given the metric and potential of the OP, I wish to derive the equivalent of Maxwell's Equations for a charged black hole. The first step is to generate the Faraday tensor.

In SR we would use F_{\alpha\beta}=A_{\beta,\alpha}-A_{\alpha,\beta}, however due to the curvature of the metric the terms on the right hand side are not guaranteed to be tensors. Thus we must use F_{\alpha\beta}=A_{\beta;\alpha}-A_{\alpha;\beta}.

This expands to F_{\alpha\beta}=A_{\beta,\alpha}-\Gamma^{\gamma}_{\beta\alpha}A_{\gamma}-A_{\alpha,\beta}+\Gamma^{\gamma}_{\alpha\beta}A_{\ gamma}. The symmetry of the lower terms in the Cristoffel symbol shows that the formulation of SR remains valid, despite the curvature of space-time.

F_{\alpha\beta}=\left|\left|\begin{array}{cccc}0&\frac{Q}{r^2}&0&0\\ -\frac{Q}{r^2}&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{array}\right|\right|

Faraday's law and Gauss's law for magnetism are expressed in SR using F_{[\alpha\beta,\gamma]}=0. The same antisymmetry arguments used for the Faraday tensor show this is equivelent to F_{[\alpha\beta;\gamma]}=0, because the Cristoffel symbols cancel.

At first these 64 equations look intimidating. The 16 cases where two or three of the indices are identical produce identities that are 0 regardless of the geometry involved. The 24 cases with only spatial indices give us permutations of 3 distinct equations reducing to Gauss's law for magnetism.
\nabla\cdot\vec{B}=0, which is somewhat trivial here because the magnetic field is 0.

The 24 cases with a timelike index also gives us permutations of 3 distinct equations reducing to Faraday's law.
\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}, This reduces to 0, again because the magnetic field is 0.

Apparently the remaining laws will require studying tensor densities.

Reality Check
2012-Sep-07, 01:02 AM
Given the metric and potential of the OP, I wish to derive the equivalent of Maxwell's Equations for a charged black hole.
I am not sure what you mean. Surely the "Maxwell's Equations for a charged black hole" are Maxwell's equations (http://en.wikipedia.org/wiki/Maxwell's_equations) since they are consistant with GR.

Reality Check
2012-Sep-07, 01:14 AM
Question 2 from post #7 remains open.
Assuming that you have the correct contravariant ("inverse") metric then the answer to quesiton 2 is:
No you do not have correct electrical potential because is is not the contravariant metric multiplied by the covariant electrical potential.
They define the covariant electrical potential because it is simpler then the contravariant form.
You cannot "split the difference" as you put it.

utesfan100
2012-Sep-07, 02:46 AM
I am not sure what you mean. Surely the "Maxwell's Equations for a charged black hole" are Maxwell's equations (http://en.wikipedia.org/wiki/Maxwell's_equations) since they are consistant with GR.
I mean Maxwell's Equations in the context of the Reissner–Nordström metric. If specific parameters are eventually required, those given in post #2 should suffice.

utesfan100
2012-Sep-07, 02:53 AM
Assuming that you have the correct contravariant ("inverse") metric then the answer to quesiton 2 is:
No you do not have correct electrical potential because is is not the contravariant metric multiplied by the covariant electrical potential.
When is a contravariant form of a tensor (say the electric potential) not the contravariant metric multiplied by the covariant form of the same tensor?

They define the covariant electrical potential because it is simpler then the contravariant form.
You cannot "split the difference" as you put it.
What makes the covariant form of the electric potential simpler than the contravariant form?

Is it possible that the splitting the difference as I describe could be accomplished by a change in gauge?

Reality Check
2012-Sep-07, 03:55 AM
I mean Maxwell's Equations in the context of the Reissner–Nordström metric. If specific parameters are eventually required, those given in post #2 should suffice.
Those are Maxwell's Equations. They do not change in the context of any metric. Post#2 is irrelevant.

ETA: The Reissner–Nordström metric is actually a static solution of the EFE when the energy-momentum tensor is that of an electromagnetic field in free space - the Einstein–Maxwell equations (http://en.wikipedia.org/wiki/Einstein-Maxwell_equations#Einstein.E2.80.93Maxwell_equatio ns).

Reality Check
2012-Sep-07, 04:07 AM
When is a contravariant form of a tensor (say the electric potential) not the contravariant metric multiplied by the covariant form of the same tensor?

Never.
And in hindsight, I see the diag() in the contravariant metric which does make your covariant electrical potential correct. So:


Would A^\mu=(\alpha^{-2}\frac{Q}{r},0,0,0)?

Yes

And:


If so, why did we not define A^\mu=(\frac{Q}{r},0,0,0),

Because that is not A^\mu!



What makes the covariant form of the electric potential simpler than the contravariant form?

The actual form (no alpha squared term).



Is it possible that the splitting the difference as I describe could be accomplished by a change in gauge?
I doubt it. But why would you want to make things more complex?

utesfan100
2012-Sep-07, 04:51 AM
I doubt it. But why would you want to make things more complex?
I don't, but in my ignorance I did the equivalent of splitting the difference in my work on my ATM theory, the math of which is close enough to the above mainstream math to warrant avoiding going into any more detail.

I suppose my lack of understanding may revolve around the question why is the electric potential more like a one-form than a vector? Would I be wrong to suspect that this is because we geometrically view potentials as layers of equal surfaces?

utesfan100
2012-Sep-07, 05:26 AM
Those are Maxwell's Equations. They do not change in the context of any metric. Post#2 is irrelevant.

ETA: The Reissner–Nordström metric is actually a static solution of the EFE when the energy-momentum tensor is that of an electromagnetic field in free space - the Einstein–Maxwell equations (http://en.wikipedia.org/wiki/Einstein-Maxwell_equations#Einstein.E2.80.93Maxwell_equatio ns).
If I can't use these equations to determine the bending of the path of the electron in post #2, then I don't know how they work.

Once I understand how they work, the calculations will look like a tedious drive to the finish line, and I will shift my attention back to my notes.

VonBelmont
2012-Sep-07, 08:02 AM
Reading on curved space times, I think I have identified where my most significant difficulties in the maths arise.

Given the metric and potential of the OP, I wish to derive the equivalent of Maxwell's Equations for a charged black hole. The first step is to generate the Faraday tensor.

In SR we would use F_{\alpha\beta}=A_{\beta,\alpha}-A_{\alpha,\beta}, however due to the curvature of the metric the terms on the right hand side are not guaranteed to be tensors. Thus we must use F_{\alpha\beta}=A_{\beta;\alpha}-A_{\alpha;\beta}.

This expands to F_{\alpha\beta}=A_{\beta,\alpha}-\Gamma^{\gamma}_{\beta\alpha}A_{\gamma}-A_{\alpha,\beta}+\Gamma^{\gamma}_{\alpha\beta}A_{\ gamma}. The symmetry of the lower terms in the Cristoffel symbol shows that the formulation of SR remains valid, despite the curvature of space-time.

F_{\alpha\beta}=\left|\left|\begin{array}{cccc}0&\frac{Q}{r^2}&0&0\\ -\frac{Q}{r^2}&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{array}\right|\right|



Looks ok, but I could be wrong. I haven't written it out for Christoffel symbols before. But judging out it has panned out, it looks ok.

utesfan100
2012-Sep-07, 08:53 PM
Looks ok, but I could be wrong. I haven't written it out for Christoffel symbols before. But judging out it has panned out, it looks ok.
Yeah, the statement

Faraday's law and Gauss's law for magnetism are expressed in SR using F_{[\alpha\beta,\gamma]}=0. The same antisymmetry arguments used for the Faraday tensor show this is equivalent to F_{[\alpha\beta;\gamma]}=0, because the Cristoffel symbols cancel.
does involve a bunch of Christoffel symbols as written.

This can be organized a bit as follows:
6F_{[\alpha\beta;\gamma]}=(F_{\alpha\beta;\gamma}-F_{\beta\alpha;\gamma})+(F_{\beta\gamma;\alpha}-F_{\gamma\beta;\alpha})+(F_{\gamma\alpha;\beta}-F_{\alpha\gamma;\beta})

We now examine one of these three identical pieces.
F_{\alpha\beta;\gamma}-F_{\beta\alpha;\gamma}=F_{\alpha\beta,\gamma}-\Gamma^{\delta}_{\alpha\gamma}F_{\delta\beta}-\Gamma^{\delta}_{\beta\gamma}F_{\alpha\delta}-F_{\beta\alpha,\gamma}+\Gamma^{\delta}_{\beta\gamm a}F_{\delta\alpha}+\Gamma^{\delta}_{\alpha\gamma}F _{\beta\delta}
=F_{\alpha\beta,\gamma}+\Gamma^{\delta}_{\alpha\ga mma}F_{\beta\delta}-\Gamma^{\delta}_{\beta\gamma}F_{\alpha\delta}-F_{\beta\alpha,\gamma}-\Gamma^{\delta}_{\beta\gamma}F_{\alpha\delta}+\Gam ma^{\delta}_{\alpha\gamma}F_{\beta\delta}
=F_{\alpha\beta,\gamma}-F_{\beta\alpha,\gamma}+2\Gamma^{\delta}_{\alpha\ga mma}F_{\beta\delta}-2\Gamma^{\delta}_{\beta\gamma}F_{\alpha\delta}

Returning to the original expression:
6F_{[\alpha\beta;\gamma]}=(F_{\alpha\beta,\gamma}-F_{\beta\alpha,\gamma}+2\Gamma^{\delta}_{\alpha\ga mma}F_{\beta\delta}-2\Gamma^{\delta}_{\beta\gamma}F_{\alpha\delta})+(F _{\beta\gamma,\alpha}-F_{\gamma\beta,\alpha}+2\Gamma^{\delta}_{\beta\alp ha}F_{\gamma\delta}-2\Gamma^{\delta}_{\gamma\alpha}F_{\beta\delta})+(F _{\gamma\alpha,\beta}-F_{\alpha\gamma,\beta}+2\Gamma^{\delta}_{\gamma\be ta}F_{\alpha\delta}-2\Gamma^{\delta}_{\alpha\beta}F_{\gamma\delta})
=6F_{[\alpha\beta,\gamma]}+2\Gamma^{\delta}_{\alpha\gamma}F_{\beta\delta}-2\Gamma^{\delta}_{\gamma\alpha}F_{\beta\delta}+2\G amma^{\delta}_{\beta\alpha}F_{\gamma\delta}-2\Gamma^{\delta}_{\alpha\beta}F_{\gamma\delta}+2\G amma^{\delta}_{\gamma\beta}F_{\alpha\delta}-2\Gamma^{\delta}_{\beta\gamma}F_{\alpha\delta}
Showing, by the symmetry of the lower Christoffel Symbol indices,
6F_{[\alpha\beta;\gamma]}=6F_{[\alpha\beta,\gamma]}