# Physical meaning of the space-space components $T^{ij}$ of the stress-energy tensor $T^{\mu\nu}$

The time-time component $$T^{00}$$ and the time-space components $$T^{0i}$$ of the energy-momentum tensor $$T^{\mu\nu}$$ are respectively called the Hamiltonian (energy) density and the momentum density. Integrated over all space, $$\int \mathrm d^3x\,T^{00}\equiv E$$ and $$\int \mathrm d^3x\,T^{0i}\equiv P^i$$ respectively represent the energy and components of the physical momentum carried by a field. In field theories, the quantities of interest are mainly $$T^{00}$$ and $$T^{0i}$$ components of the $$T^{\mu\nu}$$ tensor. This is because they naturally arise as the conserved charges associated with time translation and space translation symmetries respectively. Therefore their meaning is kind of obvious.

Is there a similar interpretation of the space-space components $$T^{ij}$$? In cosmology, the diagonal space-space components, $$T^{ii}$$, are called pressures $$p$$. Since $$T^{\mu\nu}$$ is a tensor under Lorentz transformation, $$T^{ij}$$ must transform like a tensor under rotation $$T^{ij}=O^i_{~m}O^j_{~n}T^{mn}.$$ That much is clear. The next target is to understand that the stress tensor $$\sigma^{ij}$$ introduced for a fluid or for elastic deformation is somehow related to $$T^{ij}$$ of a field.

This terminology is used even even during inflation which is described as a classical theory which suggest that this 'pressure' terminology make sense for fields (e.g., the inflaton field)?

• en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor
– user4552
Dec 29, 2019 at 15:22
• Note that the use of Latin indices to mean spatial indices is old-fashioned. Today we normally use Latin indices to mean abstract indices.
– user4552
Dec 29, 2019 at 15:23
• I don't know how true that is, I still see many papers referring to the spatial part of any quantity represented with latin indexes. I do find that the convention of Landau & Lifschitz use the inverted convention of latin for general and greek for spatial (as noted in their classical field theory book). Dec 29, 2019 at 15:28
• @BenCrowell Not that this is very relevant for the question, but that has absolutely not been my experience. Dec 29, 2019 at 15:29
• @Triatticus: Are these papers in GR, or are they papers in things like particle physics or something? I notice that your user profile says you do particle physics. The use of Latin indices to mean spatial indices is fundamentally unsuited to the structure of GR, since we don't normally even expect to have one coordinate that is timelike. E.g., you can have two null coordinates and two spacelike coordinates. If you look at modern books such as Carroll, I think you will find that they do use Latin for abstract indices. (Carroll first uses them for spatial indices, but then switches later.)
– user4552
Dec 29, 2019 at 19:14

All this can be understood by the means of the conservation law

$$\partial_{\mu}T^{\mu\nu}=0$$

for every $$\nu$$. Splitting the $$\mu$$ and $$\nu$$ indices into space-like and time-like directions result to the following equations:

$$\partial_{0}T^{00}+\partial_{i}T^{i0}=0,\qquad \partial_{0}T^{0i}+\partial_{j}T^{ji} = 0$$

Since $$T^{00}$$ is the energy density, it follows from the equation above that $$T^{i0}$$ is the flow of the energy density. In order to see that more clearly just integrate the equation in a space-like volume at fixed $$x^{0}$$:

$$\int_{\Sigma}\left(\partial_{0}T^{00}+\partial_{i}T^{i0}\right)=\partial_{0}\int_{\Sigma}T^{00}+\int_{\partial\Sigma}T^{i0}n_{i}(\partial\Sigma)=\partial_{0}E(\Sigma)+\int_{\partial\Sigma}T^{i0}n_{i}(\partial\Sigma)=0$$

which is saying that the variation of the energy inside the space-like volume $$\Sigma$$ is equal to the flux of the vector field $$T^{i0}$$ in the boundary.

Now, looking at the second equation we have something very similar. The equation in integral form is given by:

$$\partial_{0}P^{i}(\Sigma)+\int_{\partial\Sigma}T^{ji}n_{j}(\partial\Sigma)=0$$

where $$T^{0i}$$ is the density for $$P^{i}$$. The variation of the quantity $$P^{i}(\Sigma)$$ is equal to the flux of the vector field $$T^{ij}$$ (i is fixed here) in the boundary of $$\Sigma$$.

It is important to realize that $$E(\Sigma)$$ and $$P^{i}(\Sigma)$$ rotate into each other under boost transformation, which implies that $$P^{i}(\Sigma)$$ is the momentum of the region $$\Sigma$$ and $$T^{0i}$$ is the momentum density. By the definition of pressure (force per area) we conclude that $$T^{ji}$$ are pressures. In an area element with normal vector $$x_{j}$$, $$T^{ij}$$ is the the pressure at the $$x_{i}$$ direction. This can be translated to the following:

An area element $$dA$$ with normal vector $$n_{j}$$ will feel a force $$f^{i}$$ equal to $$f^{i}=n_{j}T^{ji}dA$$

• I did not understand the last statement of first paragraph. Also you did not explain why $T^{ii}$ is pressure. Dec 29, 2019 at 16:53
• If you write out the conservation law you will see. To absorb or reflect a flow along i of the i-component of momentum takes pressure. Dec 29, 2019 at 17:07
• @mithusengupta123 For each component of $T^{\mu\nu}$, the first index tells whether you're talking about energy or momentum; the second index tells whether you are talking about a density or a flow. If the first index is $0$, it indicates that you're talking about energy and if the first index is $1$ (for example), it indicates that you're talking about the $x$- component of momentum. For the second index, the value $0$ indicates a density and a value $1,2$ or $3$ indicates flow.
– SRS
Jan 1, 2020 at 17:12
• Therefore, $T^{00}$ are $T^{ii}$ are energy density and momentum densities while $T^{12}$ is the flow of $x$-component of momentum in $y$- direction.
– SRS
Jan 1, 2020 at 17:12
• @my2cts Second equation is $\frac{\partial p_1}{\partial t}=-\partial_1T^{11}-\partial_2T^{21}-\partial_3T^{31}$. A decrease in the momentum in the x direction causes a flow of momentum in the x,y and z directions? Jan 1, 2020 at 19:58