# Transformation law of the energy momentum tensor

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• filip97

#### filip97

We have 4-tensor of second rank. For example energy-momentum tensor ##T_μν##

, which is symmetric and traceless. Then

##T_{μν}=x_μx_ν+x_νx_μ##

where ##x_μ##
is 4-vector. Every 4- vector transform under Lorentz transform as (12,12). If we act on ## T_{μν}##

, by representation( with homomorphism) then we have

##D(T_{μν})=(D(x_μ)D(x_ν))+(D(x_ν)D(x_μ))##

,or

Tμν
transform as
##(\cfrac{1}{2},\cfrac{1}{2})⊕(\cfrac{1}{2},\cfrac{1}{2})=(1⊕0)⊕(1⊕0)=1⊕1⊕0⊕0##

But we have trace of two components ##0##
. On wikipedia write that traceless symmetric tensors transform on representation ##(1,1)=2⊕0##

.

Where is error?

Let's start with vectors (which don't transform at all, because they are invariant objects). The contravariant vector components transform with a Lorentz matrix ##{\Lambda^{\mu}}_{\nu}## according to
$$\bar{V}^{\mu}={\Lambda^{\mu}}_{\nu} V^{\nu}.$$
In terms of the usual naming of representations that's the transformation under ##(1/2,1/2)##. The covariant components accordingly transform as
$$\bar{V}_{\mu}={\Lambda_{\mu}}^{\nu} V_{\nu}.$$
Indices are raised and lowered with the Minkowski pseudo-metric components, ##(\eta_{\mu \nu})=(\eta^{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)##. Further a Lorentz matrix obeys
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=\eta_{\rho \sigma}.$$
From this it follows that 2nd-rank tensor components transform as
$$\bar{T}^{\mu \nu} = {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} T^{\rho \sigma},$$
i.e. under ##(1/2,1/2) \otimes (1/2,1/2)## as it must be.

It's clear that you can decompose any 2nd-rank tensor in a symmetric traceless part, an antisymmetric, and the trace part. Start with the symmetric and antisymmstric decomposition:
$$T^{\mu \nu}=\frac{1}{2} (T^{\mu \nu}+T^{\nu \mu}) + \frac{1}{2} (T^{\mu \nu}-T^{\nu \mu})=S^{\mu \nu} + A^{\mu \nu}.$$
The symmetric part can be decomposed in a traceless and a trace part,
$$S^{\mu \nu} = S^{\mu \nu}-\frac{1}{4} \eta_{\rho \sigma} S^{\rho \sigma} \eta^{\mu \nu} +\frac{1}{4} \eta_{\rho \sigma} S^{\rho \sigma} \eta^{\mu \nu}.$$
It's clear that the trace-part components are invariant themselves, i.e., they are scalars.

Dale
Ok, this is partialy cleared.
##(1/2,1/2)⊗(1/2,1/2)=1/2⊗1/2⊗1/2⊗1/2=(1\oplus 0)⊗(1\oplus 0)=##
##=(1⊗1)\oplus(0\otimes 0)=2\oplus 0\oplus0##, and on Wikipedia write that
traceless symmetric second rank tensor transformed as ##(1,1)=2\oplus 0\neq 2\oplus 0\oplus0##. We have traces in both (in first representation we have two times ##0## component, and in second once).

Moderator's note: Moved to relativity forum.

For example energy-momentum tensor ##T_{μν}##

, which is symmetric and traceless.

The stress-energy tensor is symmetric, but it is not always traceless.

Then

$$T_{μν}=x_μx_ν+x_νx_μ$$

where ##x_μ## is 4-vector.

This is not true in general either.

As others have pointed out, the symmetric stress energy tensor is not necessarily trace free. For example, an idealized pressureless fluid has a stress-energy tensor of

$$\begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

which is obviously not trace free.

Like any other symmetric rank 2 tensor, the stress-energy tensor is diagonalizeable, though. See for instance wiki,

In analogy with the theory of symmetric matrices, a (real) symmetric tensor of order 2 can be "diagonalized".

Thus a coordinate transformation can put the tensor in the form.

$$\begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & P_x & 0 & 0 \\ 0 & 0 & P_y & 0 \\ 0 & 0 & 0 & P_z \end{bmatrix}$$

While you could further decompose it into a trace-free part, and the trace multipled by an identity matrix, the physical interpretation is easiest without decomposing it , in my opinion.

A general coordinate transformation could even make the stress-energy tensor diag(-1,1,1,1) at a point, but basis vectors wouldn't be of unit length in this case, which complicates the physical interpretation.

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vanhees71
A general coordinate transformation could even make the stress-energy tensor diag(-1,1,1,1) at a point

I don't think this is true, at least not for the covariant form ##T_{\mu \nu}##. A general coordinate transformation can make the metric tensor ##g_{\mu \nu}## diag(-1, 1, 1, 1) at a point, but the metric tensor is not the same as the stress-energy tensor.

For the mixed form ##T^\mu{}_\nu##, yes, you can, by picking weird units for the coordinates, make the tensor diag(-1, 1, 1, 1) at a point (at least, you can if the pressures are nonzero to begin with), because raising an index changes the sign of the 0-0 component.

I don't think this is true, at least not for the covariant form ##T_{\mu \nu}##. A general coordinate transformation can make the metric tensor ##g_{\mu \nu}## diag(-1, 1, 1, 1) at a point, but the metric tensor is not the same as the stress-energy tensor.

For the mixed form ##T^\mu{}_\nu##, yes, you can, by picking weird units for the coordinates, make the tensor diag(-1, 1, 1, 1) at a point (at least, you can if the pressures are nonzero to begin with), because raising an index changes the sign of the 0-0 component.

Hmmm - well, there's the obvious counterexample of diag(0,0,0,0) for empty space, now that I think about it. And even diga(##\rho##, 0, 0.0) which I already gave as an example, fails. So - yeah, ignore the bit about diag(-1,1,1,1) but keep the part about being able to diagonalize a rank 2 symmetric tensor.

What I should have said was that you can make non-zero terms in the diagonalization have unit magnitude by an appropriate coordinate transformations if you allow non-unit length basis vectors.

Another remark: The energy-momentum tensor of any relativistic field theory is traceless iff the theory is invariant under scaling transformations. Of course that's the case for the free electromagnetic field.

One should note that this is a very "fragile" symmetry. Already quantizing the free electromagnetic field breaks it, which phenomenon is dubbed "trace anomaly". Its generalization to non-Abelian gauge theories, particularly QCD, gives rise to almost all of the mass of the matter surrounding us, but that's another story.

JD_PM, dextercioby and weirdoguy
$$(\cfrac{1}{2},\cfrac{1}{2})⊕(\cfrac{1}{2},\cfrac{1}{2})=(1⊕0)⊕(1⊕0)=1⊕1⊕0⊕0$$
What does this mean? And where did get it from? I am surprised by the fact that nobody asked you these questions.
Your tensor product has the following UNIQUE Clebsch-Gordan decomposition $$\left( \frac{1}{2} , \frac{1}{2}\right) \otimes \left( \frac{1}{2} , \frac{1}{2}\right) = \left( 1 , 1 \right) \oplus \big\{ \left( 0 , 1 \right) \oplus \left( 1 , 0 \right)\big\} \oplus \left( 0 , 0 \right) .$$ In tensor language (as Vanhees71 explained to you) that corresponds to decomposing a general tensor $T_{\mu\nu}$ into irreducible tensors (i.e., tensors that don’t mix with each other under Lorentz transformations): $$T_{\mu\nu} = \left( \frac{1}{2} \left( T_{\mu\nu} + T_{\nu\mu} \right) - \frac{1}{4}\eta_{\mu\nu}T \right) + \frac{1}{2} \left( T_{\mu\nu} - T_{\nu\mu}\right) + \frac{1}{4}\eta_{\mu\nu}T ,$$ where $T = \eta^{\mu\nu}T_{\mu\nu} \in (0,0)$ is the scalar trace, $\left( \frac{1}{2} \left( T_{\mu\nu} + T_{\nu\mu} \right) - \frac{1}{4}\eta_{\mu\nu}T \right) \in (1 , 1)$ is the symmetric traceless part and $\frac{1}{2} \left( T_{\mu\nu} - T_{\nu\mu}\right) \in \big\{ (0 , 1) \oplus (1 , 0)\big\}$ is the antisymmetric part.

As for the energy-momentum tensor, in general it is neither symmetric nor traceless. In a Poincare symmetric theory, the energy-momentum tensor can be made symmetric. And, in order to make it symmetric and traceless, the theory must be invariant under a larger space-time symmetry group called the Conformal group.

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JD_PM, dextercioby, vanhees71 and 1 other person