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Stress-energy tensor (in SR)

  1. Jan 16, 2014 #1
    According to Wikipedia,
    This definition doesn't sit well with me. Flux is defined as the rate that something passes through an infinitesimal surface, divided by the infinitesimal area of that surface. For example, the current flux (or current density), when dotted with a unit vector, gives the rate that charge passes through an infinitesimal surface with that unit vector, divided by the infinitesimal area.

    But "rate" is not Lorentz invariant, because it involves time which is relative. So the definition above seems to be non-Lorentz invariant (or covariant or whatever). Shouldn't this be a problem?
  2. jcsd
  3. Jan 16, 2014 #2


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    Not it isn't a problem. ##T^{\mu\nu}## is itself Lorentz-covariant and that's all that matters. There's no need for the individual components to be Lorentz-invariant. Does it bother you that the components of the 4-momentum ##p^{\mu}## aren't Lorentz-invariant even though ##p^{\mu}## is itself Lorentz-covariant?
  4. Jan 16, 2014 #3
    The components of [itex]T^{\alpha \beta}[/itex] don't need to be invariant, as you point out. The vector [itex](T^{0 \beta},T^{1 \beta},T^{2 \beta},T^{3 \beta})[/itex], on the other hand, should be Lorentz covariant. But the above definition ensures that it isn't, because this "vector" equals a 4-vector (the amount of 4-momentum passing through a given surface) divided by dt, which isn't a scalar.
  5. Jan 16, 2014 #4


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  6. Jan 16, 2014 #5
    Agreed. I dispute the claim that the stress-energy tensor, at least in the definition above, is constructed exclusively out of scalars and vectors.
  7. Jan 16, 2014 #6


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    Not quite. What will be Lorentz covariant is the contraction of the stress-energy tensor with some specific 4-vector. If we call the 4-vector (actually a covector in this case, since you wrote the SET with upper indexes) ##u_{\beta}##, then what will be Lorentz covariant is the 4-vector ##t^{\alpha}## given by

    t^{\alpha} = T^{\alpha \beta} u_{\beta}

    If you Lorentz transform ##t^{\alpha}##, it transforms like a 4-vector. But if you write out the detailed expression for ##t^{\alpha}##, the factors ##T^{\alpha \beta}## in that expression won't transform individually like a 4-vector; only their products with the appropriate (transformed) components of ##u_{\beta}## will.

    In fact, generalizing this observation makes clear what the "components" of a tensor in a given frame actually are; they are contractions of the tensor with the appropriate basis vectors of that frame. So if I pick a particular frame, the component ##T^{12}##, say, in that frame, really means the contraction ##T^{\alpha \beta} \left( \hat{e}_1 \right)_{\alpha} \left( \hat{e}_2 \right)_{\beta}##, where ##\hat{e}_1## and ##\hat{e}_2## are the frame's basis vectors in the "1" and "2" directions. This contraction does indeed describe the flow of "1" momentum across a surface of constant "2" coordinate *in that frame*; and since it's a contraction, it's a Lorentz scalar, so I could compute it, in principle, in any frame, as long as I have expressions, in the frame I'm computing in, for the ##\hat{e}_1## and ##\hat{e}_1## basis vectors of the frame I'm contracting with the tensor.

    Changing frames means changing basis vectors, which means if I now want to compute ##T'^{12}## in a new, primed frame, I need to compute a *different* contraction, ##T^{\alpha \beta} \left( \hat{e'}_1 \right)_{\alpha} \left( \hat{e'}_2 \right)_{\beta}##, using the different basis vectors of the new, primed frame. This will also be a Lorentz scalar, but a *different* Lorentz scalar, because I contracted the tensor with different vectors.
    Last edited: Jan 16, 2014
  8. Jan 16, 2014 #7


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    I think you misunderstood me. If we have an observer O with 4-velocity ##u^{\mu}## then ##j^{\mu} := T^{\mu}{}{}_{\nu}u^{\nu}## will be Lorentz-covariant because it's built from the contracted indices of a 4-tensor and a 4-vector. This physically corresponds to the energy-momentum density of the matter field represented by ##T^{\mu\nu}## as measured by O.
  9. Jan 16, 2014 #8
    Yes, of course you're correct. That was a silly mistake on my part.

    What you're saying makes much more sense now.

    Basically, my question reduces to this: [itex]j^{\mu} := T^{\mu}{}{}_{\nu}u^{\nu}[/itex] has to be a 4-vector, because that's how tensors work. But we're also defining [itex]j^{\mu}[/itex] as "the flux of the [itex]\alpha[/itex]th component of the momentum vector across a surface" (equating the two [itex]j^{\mu}[/itex]s allows us to find [itex]T^{\mu}_{\nu}[/itex]), and I'm surprised this second definition makes a 4-vector as well, as it must if our definition of the stress energy tensor is consistent.
    Last edited: Jan 16, 2014
  10. Jan 16, 2014 #9


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    Are you familiar with the 4-current density ##j^{\mu}## from electromagnetism? (Sorry about using the same symbol but it's standard notation)

    Do you understand why this is a valid 4-vector? If so, the case with the energy-momentum tensor is entirely analogous.
  11. Jan 16, 2014 #10
    Actually, while I am familiar with the 4-current density I don't understand why it's a valid 4-vector.

    The 4-current density ##j_\mu## (I prefer treating it as a dual vector) is defined such that if ##n^\mu## is unit vector, then ##j_\mu n^\mu = \textrm{rate that charge passes through an infinitesimal surface with unit vector }n^\mu \textrm{ divided by }\delta A##. It isn't intuitively obvious that the right hand side of this equation is a scalar.
  12. Jan 16, 2014 #11


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    This looks like a good spot to post a few of the diagrams I've been working on:

    I am drawing 2-d space-time diagrams for a 2-d ideal gas, made up of collisionless particles. You can think of the diagram as one line per particle, or a scaled diagram as one line per n particles. A real gas would have velocities in random directions, but one can "group" the flows of the same velocity into a sub-diagram, compute the contribution of the sub-diagram to the total flow, then add all the sub-flows together to get the total flow.

    (Sorry if this isn't clear - it invites a few new diagrams, but I'm not sure when I'll have time to get around to them).

    This ideal gas has a 2-d stress energy tensor, with 4 components:

    ##T^{00}## the flow of energy in the t-direction, ##T^{01}##, the flow of energy in the x-direction, ## T^{10}##, the flow of momentum in the t-directoin, and ## T^{11}##, the flow of momentum in the x direciton.

    One can add collisions into gas without much difficulty if they happen at a point, but getting into that would be a needless distraction at this point.

    Each particle can be assumed to carry some momentum and energy along its worldine.

    Let's start with the flow of "stuff" in the t direction:

    As you'll see the "rate" of flow in the t direction is the amount of stuff normalized to a unit length, not a unit time. So the flows in the t-direction are spatial densities - they aren't rates. To address an objection previously raised.

    "Stuff" is deliberately vague. It could be the flow of worldlines themselves (which makes up the number-flux 4-vector)

    For more on the number-flux 4-vector see http://web.mit.edu/edbert/GR/gr2b.pdf

    The flow of "stuff" can also be the flow of energy and momentum, which is what we need for the stress-energy tensor. Each particle in the diagram is assumed to have a certain momentum and energy, to get the flow of energy you multiply the flow of worldlines by the energy per woldline, to get the flow of momentum y ou multiply the flow of worldlines by the momentum per worldline.


    Next up - flows in the "x" direction.


    When you do a Lorentz transform, the relevant quantities are the flow of "stuff" in the t' direction, and the x' direction. These directions are not the same as the t and x direction. So you have to transform the flows. Additionally, if you are computing flows of energy and momentum for the stress-energy tensor, you need to transform the energy and momentum relativistically, as well as the flows.

    I had another diagram that illustrated how to transform the flows, but it appears to have fallen victim to a power outage. Basically, though, if you know the flows in the t and x directions, a linear combination of the two will give you the flows in the t' and x' directions, as long as your flows are conserved flows.

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