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How to construct stress-energy tensor for a system?

  1. Dec 3, 2013 #1
    Given a particular system, how would one construct the stress-energy tensor? I was reading Mallett's paper and the stress-energy given for an infinitely long circulating cylinder of light is of the form [itex]T_{\mu\nu}=\epsilon \eta_\mu \eta_\nu[/itex] where [itex]\eta_\mu=(\eta_0,0,\eta_2,0)[/itex] and ε is the energy density of laser light with [itex]x^\mu=(t,\rho,\phi,z)[/itex]. Since the laser beam move in helical path, I would think that there is an energy flux in z direction, so T_03 should not be zero but it is zero given by Mallett. Can anyone explain to me how he come out with that stress-energy tensor?
    Thanks.
     
  2. jcsd
  3. Dec 3, 2013 #2

    WannabeNewton

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    You should link the paper if you're going to refer to it. Anyways, that's the stress energy tensor for null dust with wave 4-vector field ##\eta^{\mu} = \eta^t \partial_t + \eta^{\phi} \partial_{\phi}## relative to an appropriate global inertial frame. There is clearly no momentum density in the ##z## direction because the null dust is circulating in the ##\phi## direction. The worldlines of the constituents of the null dust are helices which is not the same thing as the spatial flow of the null dust in the global inertial frame that ##\eta^{\mu}## is written relative to. Each constituent of the null dust is described by a helical worldline in space-time but in space (relative to the original choice of global inertial frame) it simply circulates in the ##\phi## direction, thus giving a cylindrical null dust of circular flow as codified by ##\eta^{\mu}##.
     
    Last edited: Dec 3, 2013
  4. Dec 3, 2013 #3
    The paper link:http://www.phys.uconn.edu/~mallett/Mallett2003.pdf
    I am not quite understand. Is that mean that for constant time, the null dust is circulating in ##\phi## direction and it become helix as time move forward? How does the null dust link to the circulating cylinder of light actually? From the paper, I would think that the laser circulate upwards in z-direction, that's why I thought there is an energy flux in z-direction.
    Can you give some references why the dull dust wave 4-vector field is of the form ##\eta^{\mu} = \eta^t \partial_t + \eta^{\phi} \partial_{\phi}## ?
    Thanks.
     
  5. Dec 3, 2013 #4

    WannabeNewton

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    Thanks for the link. I'll try to look up references on the general formalism of null dust fields but in this specific case it should be more or less self-evident.

    Let me start by asking: why do you think that a given light ray in the aforementioned null dust field circulates "upwards in the ##z## direction"? The paper says that the system consists of axially circulating radiation (p.1308). Just imagine a light ray that's moving in a circle, with such a circulating light ray for each value of ##z## so that we get an infinitely long circulating cylinder of light.

    It seems you are confusing the system in fig. 1 on p.1308 with the system that the author is working with when solving Einstein's equation. Note the author states that "A generalized model of the system just described would be a long circulating cylinder of light" (p.1308) where "system" refers to the solenoidal laser beam.

    Are you familiar with matter dust fields? The energy-momentum tensor used here for the circulating null dust with wave 4-vector field ##\eta^{\mu}## is entirely analogous to that of a circulating matter dust with 4-velocity field ##\xi^{\mu} = \xi^t \partial_t + \xi^{\phi}\partial_{\phi}## and rest mass density ##\rho##.
     
  6. Dec 3, 2013 #5
    Yes. From fig.1 it seems that the laser beam is moving in upwards in z-direction. That's make me confuse. Is that mean the author model the solenoidal laser beam as an infinitely number of circulating cylinder of light stack together?
    I am not familiar with matter dust fields also. I am new in this field. Is that [itex]\eta^t,\eta^\phi[/itex] some sort of constant?
     
  7. Dec 3, 2013 #6

    WannabeNewton

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    The author isn't trying to model the solenoidal laser beam per say but rather a more generalized model consisting of exactly what you described: axially circulating radiation stacked up along the ##z## axis, all relative to the original global inertial frame.
     
  8. Dec 3, 2013 #7

    WannabeNewton

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    ##\eta^t## is the time component of the wave 4-vector field and ##\eta^{\phi}## is the axial component of the wave 4-vector field; ##\eta^t## is the angular frequency as measured by an observer at rest in this frame for example.
     
  9. Dec 3, 2013 #8
    Thanks for you clarification. The fig.1 really make me confuse. [itex]\eta^t[/itex] is the frequency, so [itex]\eta^\phi[/itex] should be the wave vector k in [itex]\phi[/itex] direction right?
     
  10. Dec 3, 2013 #9

    WannabeNewton

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  11. Dec 3, 2013 #10

    WannabeNewton

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    By the way, I have two or three set of notes I can link you to that go into some detail on fluid flow in general relativity but I don't know what your background is so I apologize ahead of time if it doesn't fit your bill.

    http://www.physics.uoguelph.ca/poisson/research/agr.pdf (entirety of chapter 2)
    http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf (section 2.8)
    http://preposterousuniverse.com/grnotes/grnotes-one.pdf [Broken] (page 28 onwards)

    Do you own any texts on general relativity? Most of them go into detail on fluid flow as well.
     
    Last edited by a moderator: May 6, 2017
  12. Dec 4, 2013 #11
    Thanks for your references. I have the carroll books and will go through your references.
    The stress-energy tensor given in the paper is for infinitely long circulating cylinder of light. How about a single circulating cylinder of light or a finite length circulating cylinder of light? Is the stress-energy tensor still the same?
    Thanks.
     
    Last edited by a moderator: May 6, 2017
  13. Dec 4, 2013 #12

    WannabeNewton

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    Sure. The stress-energy tensor is local; just restrict everything to ##z_0 < z < z_1##. If you have Carroll then make sure to read through appendix F (geodesic congruences).
     
  14. Dec 30, 2013 #13
    Thanks. From the paper, it mentioned exterior and interior solution, what is the meaning and difference between these two solutions?
     
  15. Dec 30, 2013 #14

    WannabeNewton

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    The paper is simply alluding to the fact that the metric tensor inside of the rotating cylinder differs from the metric tensor outside of the rotating cylinder (except for the boundary of the cylinder where the two metric tensors must join together smoothly). These metric tensors are solutions to Einstein's equation with the axially circulating null dust field acting as a source; the inside metric tensor is the interior solution and the outside metric tensor is the exterior solution.

    It might be instructive for you to solve for the interior and exterior solutions for a rotating cylinder represented by an axially circulating time-like dust field wherein the fluid elements are composed of matter particles; to make the calculations easy you could work in the Newtonian approximation by assuming a Newtonian fluid. Then you could analyze the physical difference(s) between the interior and exterior solutions so obtained. One of the more interesting effects will be the precession of freely falling gyroscopes in the interior and exterior (the precession in the interior will be in an opposite sense to the precession in the exterior).

    Here are some other standard examples of exact solutions to Einstein's equation in which the energy-momentum source involves some kind of circulating dust field:

    http://en.wikipedia.org/wiki/Van_Stockum_dust
    http://en.wikipedia.org/wiki/Gödel_metric
     
  16. Dec 30, 2013 #15
    What confuse me is what is meant by inside and outside of the circulating cylinder of light. Consider the laser beam with beam radius d and move in a circle with radius r0, the exterior solution mean the solution outside of d or r0?
     
  17. Dec 30, 2013 #16

    WannabeNewton

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    A cylinder only has one radius. The interior solution is within this radius and the exterior solution is outside of this radius, that's all there is to it.
     
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