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I am a bit at loss about how to proceed to find the stress-energy tensor given some distribution of matter. The Wikipedia page gives some examples, and some (inequivalent) definitions for it:

  • Using the Einstein-Hilbert action we get the Hilbert stress-energy tensor $$T^{\mu\nu} = \frac{2}{\sqrt{-g}}\frac{\delta (\mathcal{L}_{\mathrm{matter}} \sqrt{-g}) }{\delta g_{\mu\nu}} = 2 \frac{\delta \mathcal{L}_\mathrm{matter}}{\delta g_{\mu\nu}} + g^{\mu\nu} \mathcal{L}_\mathrm{matter}$$ With this formula, my problem reduces to: what is the matter Lagrangian $\mathcal{L}_\mathrm{matter}$ given a configuration of matter?
  • The canonical stress-energy tensor is the Noether current associated to translations in spacetime. This works very well in Minkowski spacetime, but our manifold could be something different. In that case, what would a "translation in spacetime" be?
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    $\begingroup$ i'm not sure what your question here is. maybe rephrase? $\endgroup$ – luksen Nov 6 '13 at 17:17
  • $\begingroup$ @luksen What I am asking is: given some distribution of matter in space (or spacetime), for example a sphere of uniform mass density and radius $R$, centered at the origin, how do I find the stress-energy tensor for that distribution of matter? $\endgroup$ – Daniel Robert-Nicoud Nov 6 '13 at 17:26
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    $\begingroup$ a matter distribution only will not be enough. the stress energy tensor measures total energy-momentum fluxes. e.g. pure electromagnetic radiation, which is massless (photons), will have non-zero $T_{\mu\nu}$. this is why you need the complete lagrangian, so that it tells you what interactions are possible. From this you can calculate the stress energy tensor via variation wrt the metric as you stated above $\endgroup$ – luksen Nov 6 '13 at 17:49
  • $\begingroup$ In addition to missing out on the electromagnetic part, there's another reason the matter distribution isn't enough. Suppose you give me the matter distribution for a spring. What is the stress energy? Well is the spring at its equilibrium length? This will affect the answer. Even if I knew the equilibrium length, I couldn't find the stress because I don't know the spring constant. In general, to get the stress you need material-dependent properties. These properties are encoded in the lagrangian. You will have different lagrangians for different materials so there isn't a unique answer. $\endgroup$ – Brian Moths Nov 6 '13 at 18:06
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    $\begingroup$ @luksen Ok, so a matter distribution is not enough to get the SE tensor. But what data would be enough? Could you give me an example of derivation for the SE tensor in some special case, just to have an idea of how one can proceed? $\endgroup$ – Daniel Robert-Nicoud Nov 6 '13 at 23:27
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To answer your first question: Particles and fields are separate. Particles are the irreducible excitations of fields. You can only get particles after quantizing fields.

However, you might often see people using particles without quantization of fields (in classical mechanics and GTR). You must understand that these are approximate models obtained by assuming the energy density of fields is concentrated in point like particles.

At the heart of non-quantized physics, we have continuous material fields for photons, electrons, quarks etc. These fields are (most generally tensor fields) of the form $${{\psi}_{(i)}}^{s...u}_{e....g}(x,y,z,t)$$ $(i)$ denotes the type of field (like photon, higgs boson etc.)These include scalars, vectors, co-vectors, spinors, etc. The Lagrangian density $L$ is usually function of the components of these various fields and the metric tensor. One needs to rely on observation and other considerations (like gauge symmetries) to construct a covariant (value always fixed at an event) Lagrangian. So the 'configuration' you speak of depends on all these factors.

Your second question is actually far more interesting. There are 2 SE tensors commonly used used.They differ by the divergence of an antisymmetric tensor. This paper: http://authors.library.caltech.edu/19366/1/GoMa1992.pdf discusses this in detail.

The first is the canonical SE tensor derived as a conserved current using Noether's theorem from the space-time translational invariance of the Lagrangian.

The second type of tensor is derived from considerations of diffeomorphic invariance of the action. It is called Belinfante - Rosenfeld SE tensor. A diffeomorphism is a very sophisticated and generalized notion of a translation. Let vector field $X$ be a generator of general diffeomorphism ${\phi}^*$ and volume of integration is $F$. $X$ vanishes outside $F$. So we have $${\int}_{F}L{\eta}-{\phi}^*(L{\eta}) = 0$$ Thus $${\int}_{F}D_{X}(L{\eta})=0$$ where ${\eta}$ is the volume form (I have suppressed the factor of 1/4!)

Expanding the RHS, we get:

$${\int}_{F}D_{X}(L{\eta}) = {\int}_{F}[(\frac{{\partial}L}{{\partial}{{{\psi}_{(i)}}^{s...u}_{e....g}}}-(\frac{{\partial}L}{{\partial}{{{\psi}_{(i)}}^{s...u}_{e....g;c}}})_{;c})D_{X}{{{\psi}_{(i)}}^{s...u}_{e....g}} + \frac{1}{2}T^{ab}D_{X}g_{ab}]{\eta}=0 $$

As you can see the first term is the Euler-Equation which is equal to zero for every field component, so each term in the first part of the integral vanishes.

Now a basic result that can be directly inferred from the definition of a diffeomorphism is $$D_{X}g_{ab}= X_{a;b} + X_{b;a}$$

Substituting this in the above formula $${\int}_{F}D_{X}(L{\eta}) = {\int}_{F}((T^{ab}X_a)_{;b}-(T^{ab}_{;b})X_a){\eta}=0 $$

The first term can be transformed into a surface integral on the boundary of $F$ and vanishes as $X$ vanishes on the boundary of $F$. This leaves us with $${\int}_{F}D_{X}(L{\eta}) = {\int}_{F}-(T^{ab}_{;b})X_a{\eta}=0 $$

Now the above must always be true for any arbitrary $X$, this is only possible if $(T^{ab}_{;b})=0$ .

This tensor is always symmetric and gauge invariant, so it is far more useful in GTR than the canonical tensor. Refer to the paper linked above to know more details about the subtle differences between the two.

References: Chapter 3, 'Large Scale Structure of Space-Time' by Hawking and Ellis

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  • $\begingroup$ Is the "continuous material field" strictly a scalar field? $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jan 8 '14 at 7:16
  • $\begingroup$ @TorstenHĕrculĕCärlemän no, as I have said, it could be a tensor field too. Take for example the components of the electromagnetic stress-tensor $F_{ab}$, it is a 2-form. But ultimately the Lagrangian density will be a scalar function on the spacetime manifold as all the tensor components will be contracted with other tensors. For example the Lagrangian density of the electromagnetic field in vaccuum is $F_{ab}F^{ab}$ and hence it is invariant under an co-ordinate changes. You can also have scalar fields (klein-gordon) and spinorial fields (dirac field) too. The equations remain the same. $\endgroup$ – dj_mummy Jan 8 '14 at 11:10
  • $\begingroup$ But,physically looking at the situation, isn't it paradoxical to assign a tensor field to matter related fields? $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jan 8 '14 at 12:16
  • $\begingroup$ @TorstenHĕrculĕCärlemän Why do you think so? $\endgroup$ – dj_mummy Jan 8 '14 at 13:34
  • $\begingroup$ Pardon me for my limited knowledge, but I thought matter was rudimentarily a scalar quantity and the stress energy tensor had energy/mass as one of the matrix elements and the rest as either stress or momentum terms. I'm confused right here. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jan 8 '14 at 13:54

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