Question about Wave equation for light in nonflat spaces

  • Thread starter dand5
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Please correct me if I am wrong.

Solutions to the linear wave equation:

[itex]
\large\frac{\partial \Psi}{\partial t} = \frac{1}{c^{2}}\frac{\partial \Psi}{\partial x^{i}}
[/itex]

are sinusoidal waves of constant wavelength, i.e. they describe light
traveling in a flat space. But when light enters a region that is curved,
it is blue shifted, meaning that the above equation doesn't model
the light wave correctly. So my question is, if the wavelength of the light
becomes variable in a curved spacetime, is there a different more elaborate wave equation that models this? and if so, can it be arrived at from the metric, say for instance for a space with a Schwartzchild geometry?

Thank you for any replies.

P.S. I am not really sure what the relationship or even if one exists between Minkowski's metric and the wave equation is, but they seem to
have a very similar form, that is what prompted the question.
 

pervect

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dand5 said:
Please correct me if I am wrong.

Solutions to the linear wave equation:

[itex]
\large\frac{\partial \Psi}{\partial t} = \frac{1}{c^{2}}\frac{\partial \Psi}{\partial x^{i}}
[/itex]

are sinusoidal waves of constant wavelength, i.e. they describe light
traveling in a flat space. But when light enters a region that is curved,
it is blue shifted, meaning that the above equation doesn't model
the light wave correctly. So my question is, if the wavelength of the light
becomes variable in a curved spacetime, is there a different more elaborate wave equation that models this? and if so, can it be arrived at from the metric, say for instance for a space with a Schwartzchild geometry?

Thank you for any replies.

P.S. I am not really sure what the relationship or even if one exists between Minkowski's metric and the wave equation is, but they seem to
have a very similar form, that is what prompted the question.

Maxwell's equations in curved space-time look like this - sorry if the tensor notation doesn't look very familiar, you may have to read up a bit (maybe even a lot) to understand the math.

[tex]
\nabla_b F^{ab} = 4 \pi J^a
[/tex]
[tex]
\nabla_c F_{ab} + \nabla_a F_{bc} + \nabla_b F_{ca} = 0
[/tex]

Ordinary derivatives [itex]\frac{\partial}{\partial u}[/itex] are replaced with the covariant derivative [itex]\nabla_u[/itex]

See for instance

http://mathworld.wolfram.com/CovariantDerivative.html

note that [itex]\nabla_b F^{ab}[/itex] is equivalent to [itex]F^{ab}{}_{;b}[/itex] in "semicolon notation" used on this webpage.

You may not be familiar with the Faraday tensor F. This is commonly used in more advanced treatments of electromagnetism, and is described a bit in

http://scienceworld.wolfram.com/physics/ElectromagneticFieldTensor.html

You can see that all the electric and magnetic fields, E and B, are included in one tensor entity, F.

It may give some insight to know that the force on a charge can be computed from the Faraday tensor and the 4-velocity of a charge q

force = [itex]q F^a{}_b u^b[/itex]

Also, F can be computed from the 4-vector version of the magnetic vector potential, A

[tex]F_{uv} = \nabla_u A_v - \nabla_v A_u[/tex]

When F is computed from the vector potential A in this manner, it automatically satisfies the second equation (the one with three terms which sum to zero), leaving only the first equation to be solved for A in terms of J.

[itex]J^a[/itex] is the 4-vector version of the familiar current density, J

Raising and lowering indicies is a tensor thing, and is done with the metric. [itex]g_{ab}[/itex] lowers tensor indicies, i.e. [itex]a_u = g_{uv} a^v[/itex], and [itex]g^{ab}[/itex] raises tensor indices in the same manner.

If you really want to understand this in detail, the first thing to do would be to use an orthonormal coordinate system, replace all the covariant derivatives with ordinary derivatives, use a flat-space Minkowski metric (I gather you are familiar with what that looks like), and write out all the tensor equations with all their components to verify that these equations are indeed equivalent to Maxwell's equations.

A textbook reference is MTW's "Gravitation", pg 391.
 
Last edited:
28
0
Thanks for the reply. I think I see where this is going but Actually, it is my
electrodynamics that needs a real brush-up, my tensor math is ok, I think.

So, I was wondering, can the stress-energy tensor then be derived
from that force law, something like

force / unit volume for a collection of those charges = divergence of the T

This doesn't seem to be right, since that force depends on the 4-velocity
of the charged particles?

Thanks for any additional replies.
 
Last edited:

pervect

Staff Emeritus
Science Advisor
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The electromagnetic contribution to the stress energy tensor is given by:

[tex]
4 \pi T_{ab} = F_{ac} F_b{}^c-\frac{1}{4} g_{ab}F_{de} F^{de}
[/tex]

Here [itex]T_{ab}[/itex] is the stress-energy tensor, and F has already been described. (Note: the index terms run from southwest to northeast in the second occurence of F, this may be hard to see in the latex).

In flat space-time [itex]g_{ab}[/itex] is replaced by [itex]\eta_{ab}[/itex] (i.e. a flat Minkowski metric).

So far there isn't any problem with the simple prescription of replacing ordinary derivatives with covariant derivatives to go from flat to curved space-time. Note that this is basically the same process you have to go through to work in arbitrary coordinates, like spherical or cylindrical coordinates.

When you have to differentiate quantities more than once, though, be warned that this simple prescription may not always work. This tends to show up when you try to find the equation for the vector potential A in a suitable gauge, for instance - terms called "curvature terms" get added. This happens because in general curved space-times the derivative operators don't commute, (the order of differeniation matters), and the simple prescription is ill-defined and fails.

An more-or-less intuitive description of the stress energy tensor of the electromagnetic field in flat or nearly flat space-time when the metric is Minkowskian or almost so goes something like this.

In appropriate geometric units, and singling out the subscript 0 as representing time and the subscripts 1-3 representing space, we can say:

The total energy density, [itex]T_{00}[/itex] is proportional to E^2+B^2.

The T_{0i} terms are proportional to the Poynting vectors [itex]E \times B[/itex]

The other terms in the stress-energy tensor, which are space-space terms, represent stresses. Specifically, they represent a tension along the field lines, and a pressure in perpendicular to the field lines, both of which are proportional to E^2 + B^2

If by some chance you're wondering where energy conservation fits into the picture, that's another longish post, so I won't say much unless you ask. You might start by looking at the sci.physics.faq

Is energy conserved in General Relativity
 

Stingray

Science Advisor
671
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dand5 said:
So, I was wondering, can the stress-energy tensor then be derived
from that force law, something like

force / unit volume for a collection of those charges = divergence of the T

This doesn't seem to be right, since that force depends on the 4-velocity
of the charged particles?
GR requires that the divergence of the (full) stress-energy tensor vanishes. Splitting up this requirement into the electromagnetic and matter stress-energies, and using Maxwell's equations gives the following conservation law for that matter portion of the stress-energy tensor:
[tex]
\nabla_{b} T^{ab} = F^{ab} J_{b}
[/tex]

The standard Lorentz force law (that pervect wrote out) can be derived from this using the standard approximations. And yes, it is supposed to depend on the particle's 4-velocity. The reason is that the force should be dependent on the electric field in the particle's reference frame, and this is [tex]F^{ab} u_{b}[/tex].
 

Stingray

Science Advisor
671
1
Also, to answer the original question a little more directly, the wave equation for the vector potential in free space in flat spacetime is (in Lorenz gauge and Minkowski coordinates)
[tex]
\partial_{b} \partial^{b} A^{a} = \left( \nabla^{2} -\partial_{t}^{2} \right) A^{a} =0 .
[/tex]

In curved spacetime, this goes over to
[tex]
\nabla_{b} \nabla^{b} A^{a} = R^{a}_{b} A^{b} .
[/tex]
 

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