Dr. Gyula Bene Department for Theoretical Physics, Loránd Eötvös University Pázmány Péter sétány 1/A, 1117 Budapest 6. week
Energy-momentum tensor
Another derivation of energy-momentum tensor is given that yields a symmetric \(T_{ik}\).
Train of thought:
An infinitesimal coordinate transformation is performed.
Since the Lagrangian is a scalar, the action of matter cannot change due
to that coordinate transformation, i.e., \(\delta S=0\).
The formal change of action is derived and set equal to zero.
Matter fields satisfy the equations of motion, hence their changes do not
contribute to the change of action in first order.
A nontrivial change is due to the change of metric. Result: the covariant
four divergence of a symmetric tensor (\(T_{ik}\)) vanishes. Note that this
does not automatically imply a conservation theorem.
\[\begin{align}
S=\frac{1}{c}\int \Lambda\sqrt{-g}\;d\Omega
\end{align}\]
Infinitesimal coordinate transformation:
\[\begin{align}
x'^i=x^i+\xi^i
\end{align}\]
(\(\xi^i\) is small)
\[\begin{align}
dx'^i=dx^i+\frac{\partial \xi^i}{\partial x^l}dx^l=\left(\delta^i_l+\frac{\partial \xi^i}{\partial x^l}\right)dx^l
\end{align}\]
\[\begin{align}
g'^{ik}(x'^l)=g^{mn}(x^l)\left(\delta^i_m+\frac{\partial \xi^i}{\partial x^m}\right)\left(\delta^k_n+\frac{\partial \xi^k}{\partial x^n}\right)\approx g^{ik}(x^l)+g^{im}\frac{\partial \xi^k}{\partial x^m}
+g^{kn}\frac{\partial \xi^i}{\partial x^n}
\end{align}\]
Rearranging the left hand side:
\[\begin{align}
g'^{ik}(x'^l)=g'^{ik}(x^l+\xi^l)=g'^{ik}(x^l)+\frac{\partial g'^{ik}}{\partial x^l}\xi^l\approx g'^{ik}(x^l)+\frac{\partial g^{ik}}{\partial x^l}\xi^l
\end{align}\]
Hence
\[\begin{align}
g'^{ik}(x^l)=g^{ik}(x^l)-\frac{\partial g^{ik}}{\partial x^l}\xi^l+g^{im}\frac{\partial \xi^k}{\partial x^m}
+g^{kn}\frac{\partial \xi^i}{\partial x^n}
\end{align}\]
\[\begin{align}
\xi^{i;k}+\xi^{k;i}&=g^{kn}\frac{\partial \xi^i}{\partial x^n}+g^{km}\Gamma^i_{ml}\xi^l
+g^{im}\frac{\partial \xi^k}{\partial x^m}+g^{im}\Gamma^k_{ml}\xi^l \\
&=g^{im}\frac{\partial \xi^k}{\partial x^m}+g^{kn}\frac{\partial \xi^i}{\partial x^n}
+\left(g^{km}g^{in}+g^{im}g^{kn}\right)\frac{1}{2}\left(\frac{\partial g_{nm}}{\partial x^l}+\frac{\partial g_{nl}}{\partial x^m}
-\frac{\partial g_{ml}}{\partial x^n}\right)\xi^l
\\
&=g^{im}\frac{\partial \xi^k}{\partial x^m}+g^{kn}\frac{\partial \xi^i}{\partial x^n}
+g^{km}g^{in}\frac{\partial g_{nm}}{\partial x^l}\xi^l=g^{im}\frac{\partial \xi^k}{\partial x^m}+g^{kn}\frac{\partial \xi^i}{\partial x^n}
-g^{km}g_{nm}\frac{\partial g^{in}}{\partial x^l}\xi^l \\
&=g^{im}\frac{\partial \xi^k}{\partial x^m}+g^{kn}\frac{\partial \xi^i}{\partial x^n}
-\frac{\partial g^{ik}}{\partial x^l}\xi^l
\end{align}\]
Therefore
\[\begin{align}
g'^{ik}(x^l)=g^{ik}(x^l)+\delta g^{ik}(x^l)\;,\quad\quad \delta g^{ik}=\xi^{i;k}+\xi^{k;i}
\end{align}\]
Similarly (since
\(g'^{ik}g'_{kl}=\delta^i_l\)
)
\[\begin{align}
g'_{ik}(x^l)=g_{ik}(x^l)+\delta g_{ik}(x^l)\;,\quad\quad \delta g_{ik}=-\xi_{i;k}-\xi_{k;i}
\end{align}\]
\[\begin{align}
\delta S&=\frac{1}{c}\int \left\{\frac{\partial \left(\sqrt{-g}\Lambda\right)}{\partial g^{ik}}\delta g^{ik}
+\frac{\partial \left(\sqrt{-g}\Lambda\right)}{\partial \left(\frac{\partial g^{ik}}{\partial x^l}\right)}\delta \left(\frac{\partial g^{ik}}{\partial x^l}
\right)\right\}d\Omega \\
&=\frac{1}{c}\int \left\{\frac{\partial \left(\sqrt{-g}\Lambda\right)}{\partial g^{ik}}
-\frac{\partial }{\partial x^l}\left(\frac{\partial \left(\sqrt{-g}\Lambda\right)}{\partial \left(\frac{\partial g^{ik}}{\partial x^l}\right)}\right)\right\}\delta g^{ik}\;d\Omega
\end{align}\]
Let
\[\begin{align}
T_{ik}=\frac{2}{\sqrt{-g}}\left\{\frac{\partial \left(\sqrt{-g}\Lambda\right)}{\partial g^{ik}}
-\frac{\partial }{\partial x^l}\left(\frac{\partial \left(\sqrt{-g}\Lambda\right)}{\partial \left(\frac{\partial g^{ik}}{\partial x^l}\right)}\right)\right\}
\end{align}\]
(Symmetric!)
Thus we have
\[\begin{align}
\delta S=\frac{1}{2c}\int T_{ik}\delta g^{ik}\;\sqrt{-g}\;d\Omega
\end{align}\]
Inserting the previous expression of \(\delta g^{ik}\):
\[\begin{align}
\delta S=\frac{1}{2c}\int T_{ik}\left(\xi^{i;k}+\xi^{k;i}\right)\;\sqrt{-g}\;d\Omega
\end{align}\]
or
\[\begin{align}
\delta S&=\frac{1}{c}\int T_{ik}\xi^{i;k}\;\sqrt{-g}\;d\Omega=\frac{1}{c}\int T^k_{i}\xi^{i}_{;k}\;\sqrt{-g}\;d\Omega \\
&=\frac{1}{c}\int \left(T^k_{i}\xi^{i}\right)_{;k}\;\sqrt{-g}\;d\Omega
-\frac{1}{c}\int \xi^{i}\;T^k_{i;k}\;\sqrt{-g}\;d\Omega
\end{align}\]
In the first term we make use of the identity about the four divergence of
vectors. We get
\[\begin{align}
\int \left(T^k_{i}\xi^{i}\right)_{;k}\;\sqrt{-g}\;d\Omega=\int \frac{\partial }{\partial x^k}\left(\sqrt{-g}\;T^k_{i}\xi^{i}\right)\;d\Omega
\end{align}\]
Applying Gauss's theorem, we get zero.
The rest:
\[\begin{align}
\delta S=-\frac{1}{c}\int \xi^{i}\;T^k_{i;k}\;\sqrt{-g}\;d\Omega
\end{align}\]
Since \(\xi^{i}\) is arbitrary, we have
\[\begin{align}
T^k_{i;k}=0\;.
\end{align}\]
In flat spacetime this is identical with the continuity equation of
the energy-momentum tensor. Examples: Electromagnetic field:
Lagrangian(without \(\sqrt{-g}\)):
\[\begin{align}
\Lambda=-\frac{\epsilon_0}{4}F_{ik}F^{ik}\end{align}\]
Energy-momentum tensor:
\[\begin{align}
T_{ik}=\epsilon_0\left(-F_{il}F_k^l+\frac{1}{4}F_{lm}F^{lm}g_{ik}\right)\end{align}\]
Macroscopic matter (dust, gas, fluid etc.)
\[\begin{align}
T_{ik}=(p+\epsilon)u_iu_k-p\;g_{ik}\end{align}\]
Here \(p\) stands for pressure, \(\epsilon\) for energy density (energy per unit volume),
\(u^i\) for the four-velocity of matter.
bene@arpad.elte.hu
