How to construct stress-energy tensor for a system?

In summary: be a long circulating cylinder of light" (p.1308) where "system" refers to the solenoidal laser beam.
  • #1
ngkamsengpeter
195
0
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.
 
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  • #2
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}##.
 
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  • #3
WannabeNewton said:
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}##.
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.
 
  • #4
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##.
 
  • #5
WannabeNewton said:
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.
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?
 
  • #6
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.
 
  • #7
ngkamsengpeter said:
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?

##\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.
 
  • #8
WannabeNewton said:
##\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.
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
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 (page 28 onwards)

Do you own any texts on general relativity? Most of them go into detail on fluid flow as well.
 
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  • #11
WannabeNewton said:
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 (page 28 onwards)

Do you own any texts on general relativity? Most of them go into detail on fluid flow as well.
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.
 
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  • #12
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).
 
  • #13
WannabeNewton said:
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).
Thanks. From the paper, it mentioned exterior and interior solution, what is the meaning and difference between these two solutions?
 
  • #14
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
 
  • #15
WannabeNewton said:
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
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?
 
  • #16
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.
 

1. What is a stress-energy tensor?

A stress-energy tensor is a mathematical object used in physics to describe the distribution and flow of energy and momentum in a system. It is a rank-2 tensor that contains information about the density of energy and momentum in each direction at a given point in space and time.

2. How is a stress-energy tensor constructed?

A stress-energy tensor is constructed by calculating the energy and momentum densities at each point in space and time for a given system. This involves integrating over the energy and momentum of all the particles in the system and taking into account the effects of pressure and stress.

3. What is the importance of the stress-energy tensor?

The stress-energy tensor is important in physics because it is a fundamental tool for understanding how energy and momentum are distributed and how they interact with each other in a system. It is a key component in many theories, including general relativity and quantum field theory.

4. Can the stress-energy tensor be used to make predictions?

Yes, the stress-energy tensor can be used to make predictions about the behavior of a system. By solving the equations that describe the evolution of the tensor, we can make predictions about the energy and momentum flow in a system over time.

5. How is the stress-energy tensor related to the conservation laws of energy and momentum?

The stress-energy tensor is closely related to the conservation laws of energy and momentum. In fact, the components of the tensor are directly linked to these laws, with the energy component representing the energy density and the momentum components representing the flow of momentum in each direction. By examining the evolution of the stress-energy tensor, we can see how these conservation laws are manifested in a system.

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