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Stress-energy tensor

  1. Oct 22, 2009 #1

    How would go about arguing that the Stress-Energy tensor is actually a tensor based on how it must be linear in both it's arguments?
    I'm thinking it requires one 1-form to select the component of 4-momentum (e.g. [tex] \vec{E}=<\tilda{dt} ,\vec{P}> ) [/tex] and also one 1-form to define the surface (e.g [tex] \tilda{dt} [/tex] defining surfaces of constant t, so giving us densities etc).

    I know that [tex] T^{\alpha \beta}=T(\tilda{dx^{\alpha}}, \tilda{dx^{\beta}}) [/tex]. Not sure how one would argue that it therefore must be linear in these arguments?
  2. jcsd
  3. Oct 22, 2009 #2


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    The intuitive approach I take is that the stress-energy tensor just represents the amount of energy, and momentum, per unit volume.

    If you double the volume, you double the amount of energy and momentum contained (assuming a small volume and that the distribution is smooth when the volume is small enough) which is why it's linear with respect to the vector or one-form that represents the volume.

    We already know that the energy-momentum 4-vector is a vector and is appropriately additive.

    The tricky part is why we represent a volume with a vector or one-form. In the language of differential forms, dx^dy^dz , where ^ is the "wedge product" represents a volume element - but this three form has a dual, which is a vector (or one form).

    You can think of it as representing a volume element by a vector (or one-form, but I think of it as a vector) that points in the time direction perpendicular to the volume, and whose length represents the size of the volume.
  4. Oct 22, 2009 #3
    That's an interesting way of looking at it pervect. I see it quite differently (again from the Schutz book mainly), seeing one-forms as definining constant surfaces, e.g. dx (twiddle) defines surfaces of constant x (basically the same idea as in Vector calc whereby the vector gradient defines surfaces of constant phi, say). With this notion you can then also think of another one form selecting which component of the 4-momentum you want to consider via the relation, e.g. [tex] \vec{E}=<\tilda{dt} ,\vec{P}> ) [/tex] , the one form dt, selects the energy comp.

    So feeding both one forms into T, say for e.g. dt, dx.....to get the [tex] T^{tx} [/tex] component, tells us we want to look at energy flux through constant x sufaces.

    What I don't understand is how linearity is implied by these physical considerations, since what does feeding T, say 2dt mean? does that really mean twice the volume?
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