The Canonical energy momentum tensor is given by
$$T_{\mu\nu} = \frac{\partial {\cal L}}{\partial (\partial^\mu \phi_s)} \partial_\nu \phi_s – g_{\mu\nu} {\cal L}. $$
A priori, there is no reason to believe that the EM tensor above is symmetric. To symmetrize it we do the following trick.
To any EM tensor we can add the following term without changing its divergence and the conserved charges:
$${\tilde T}_{\mu\nu} = T_{\mu\nu} + \partial^\beta \chi_{\beta\mu\nu}, $$
where $\chi_{\beta\mu\nu} = – \chi_{\mu\beta\nu}$. The antisymmetry of $\chi$ in its $\mu\beta$ indices implies that ${\tilde T}_{\mu\nu}$ is conserved. Also, all the conserved charges stay the same.
Now even though $T_{\mu\nu}$ is not a symmetric tensor, it is possible to choose $\chi_{\beta\mu\nu}$ in such a way so as to make ${\tilde T}_{\mu\nu}$ symmetric. It can be shown that choosing
$$\chi_{\lambda\mu\nu} = – \frac{i}{2}\left[ \frac{\delta {\cal L}}{\delta (\partial^\mu \phi_r) } (I_{\nu\lambda})_{rs} \phi_s + \frac{\delta {\cal L}}{\delta (\partial^\lambda \phi_r) } (I_{\mu\nu})_{rs} \phi_s + \frac{\delta {\cal L} }{\delta (\partial^\nu \phi_r) } (I_{\mu\lambda})_{rs} \phi_s \right]$$
makes the new EM tensor symmetric. Here $(I_{\mu\nu})_{rs}$ is the representation of the Lorentz Algebra under which the fields $\phi_s$ transform.
Here's my question – Is it possible to obtain the symmetric EM tensor directly from variational principles by adding a total derivative term to the Lagrangian. In other words, by shifting ${\cal L} \to {\cal L} + \partial_\mu X^\mu$, and choosing $X^\mu$ appropriately, can we exactly get the shift in the EM tensor required, in order to make the canonical EM tensor symmetric?
What I've done so far – It is possible to show that under a shift in the Lagrangian by a total derivative, one shifts the EM tensor by $T_{\mu\nu} \to T_{\mu\nu} + \partial^\lambda \chi_{\lambda\mu\nu}$ where
$$\chi_{\lambda\mu\nu} = \frac{1}{2} \frac{\delta X_\lambda}{\delta (\partial^\mu \phi_r)} \partial_\nu \phi_r – \frac{1}{2} \frac{ \delta X_\mu }{\delta (\partial_\lambda \phi_r)} \partial_\nu \phi_r + X_\mu g_{\lambda\nu} – X_\lambda g_{\mu\nu} \,. $$
What I wish to do next – I now have a differential equation that I wish to solve:
\begin{align}
&\frac{1}{2} \frac{\delta X_\lambda}{\delta (\partial^\mu \phi_r)} \partial_\nu \phi_r – \frac{1}{2} \frac{ \delta X_\mu }{\delta (\partial_\lambda \phi_r)} \partial_\nu \phi_r + X_\mu g_{\lambda\nu} – X_\lambda g_{\mu\nu} \\
&~~~~~~= – \frac{i}{2}\left[ \frac{\delta {\cal L}}{\delta (\partial^\mu \phi_r) } (I_{\nu\lambda})_{rs} \phi_s + \frac{\delta {\cal L}}{\delta (\partial^\lambda \phi_r) } (I_{\mu\nu})_{rs} \phi_s + \frac{\delta {\cal L} }{\delta (\partial^\nu \phi_r) } (I_{\mu\lambda})_{rs} \phi_s \right] \,.
\end{align}
Any ideas on how to solve this?
Best Answer
OP's question (v7) asks:
No, that project is doomed already for E&M with the Maxwell Lagrangian density
$$ {\cal L}_0~:=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\tag{1} $$
with
$$ F_{\mu\nu}~=~A_{\nu,\mu}-A_{\mu,\nu}, \qquad \frac{\partial{\cal L}_0}{\partial A_{\mu,\nu}}~\stackrel{(1)}{=}~ F^{\mu\nu}.\tag{2}$$
The vacuum EL equations read
$$ 0~\approx~F^{\mu\nu}{}_{,\nu}~=~ d^{\mu}(A^{\nu}_{,\nu})-d_{\nu}d^{\nu}A^{\mu}\tag{3} $$
In E&M, the canonical SEM tensor is$^1$
$$\begin{align} \Theta^{\mu}{}_{\nu}~:=~&\delta^{\mu}_{\nu}{\cal L}_0+\left(-\frac{\partial{\cal L}_0}{\partial A_{\alpha,\mu}}+ d_{\beta}\frac{\partial{\cal L}_0}{\partial A_{\alpha,\mu\beta}}\right) A_{\alpha,\nu} -\frac{\partial{\cal L}_0}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\nu\beta}\cr ~\stackrel{(1)}{=}~&\delta^{\mu}_{\nu}{\cal L}_0+F^{\mu\alpha}A_{\alpha,\nu}~,\end{align}\tag{4}$$
while the symmetric SEM tensor is
$$ T^{\mu}{}_{\nu}~=~\delta^{\mu}_{\nu}{\cal L}_0+F^{\mu\alpha}F_{\nu\alpha}.\tag{5}$$
So the difference is$^2$
$$\begin{align} T^{\mu}{}_{\nu} -\Theta^{\mu}{}_{\nu}~\stackrel{(4)+(5)}{=}&~ F^{\mu\alpha}A_{\nu,\alpha}~=~ d_{\alpha}(F^{\mu\alpha}A_{\nu}) - \underbrace{F^{\mu\alpha}{}_{,\alpha}}_{~\approx~0}A_{\nu} \cr ~\stackrel{?}{\approx}~&\delta^{\mu}_{\nu}\Delta{\cal L}+\left(-\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu}}+ d_{\beta}\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}\right) A_{\alpha,\nu} -\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\nu\beta}\end{align}\tag{6} $$
for some total derivative term $\Delta{\cal L}=d_\mu X^\mu$, where $X^\mu$ depends on $A$ and $\partial A$. The question mark (?) in eq. (6) is OP's question. Note that the continuum equation is unaltered on-shell
$$ d_{\mu}T^{\mu}{}_{\nu} ~\approx~ d_{\mu}\Theta^{\mu}{}_{\nu}~\approx~0. \tag{7} $$
For dimensional reasons $X^\mu$ must be on the form$^3$
$$ X^{\mu} ~=~ a A^{\mu} A^{\nu}_{,\nu} + b A^{\nu} A^{\mu}_{,\nu} + c A^{\nu} A_{\nu}^{,\mu}\tag{8} $$
for some constants $a,b,c$. Then
$$\begin{align} \Delta{\cal L}~&~=~d_\mu X^\mu ~\stackrel{(8)+(10)}{=}~\Delta{\cal L}_1+\Delta{\cal L}_2,\tag{9} \cr \Delta{\cal L}_1~&:=~a (A^{\mu}_{,\mu})^2 + b A^{\nu}_{,\mu} A^{\mu}_{,\nu} + c A^{\nu}_{,\mu} A_{\nu}^{,\mu},\tag{10} \cr \Delta{\cal L}_2~&:=~ (a+b) A^{\mu} A^{\nu}_{,\nu\mu} + c A^{\mu}A_{\mu,\nu}^{,\nu}~\stackrel{(3)}{\approx}~(a+b+c) A^{\mu} A^{\nu}_{,\nu\mu}.\tag{11} \end{align} $$
Consider the last term on the right-hand side of eq. (6):
$$\begin{align}\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\nu\beta} &~=~\frac{\partial\Delta{\cal L}_2}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\nu\beta}\cr &~=~\frac{a+b}{2} \left(A^{\alpha} A^{\mu}_{,\alpha\nu}+A^{\mu} A^{\alpha}_{,\alpha\nu}\right) +c A^{\alpha} A^{,\mu}_{\alpha,\nu} \tag{12}\end{align}$$
Apart from the diagonal term $\delta^{\mu}_{\nu}\Delta{\cal L}_2$, the terms in eq. (12) are the only appearances of 2nd-derivatives on the right-hand side of eq. (6). We conclude that
$$ \Delta{\cal L}_2~=~0\qquad\Leftrightarrow\qquad a+b~=~0\quad\wedge\quad c~=~0.\tag{13}$$
Similar arguments shows that eq. (6) is not possible$^4$. $\Box$
--
$^1$ In eq. (4) we have indicated the canonical SEM tensor for a Lagrangian density with up to 2nd-order derivatives. Some references, e.g. Weinberg QFT, have the opposite notational conventions for $T\leftrightarrow\Theta$. Here we are using the $(-,+,+,\ldots,+)$ Minkowski sign convention.
$^2$ In formula (6) we have neglected terms in $\Delta{\cal L}$ that depends on $\partial^3A$, $\partial^4A$, $\partial^5A$, $\ldots$, etc. Such terms are excluded for various reasons.
$^3$ In retrospect, this answer completely shares the premise/ideology/program/conclusion of this Phys.SE post.
$^4$ Interestingly, if we just take the trace of eq. (6), we get
$$\begin{align} A^{\nu}_{,\mu} A^{\mu}_{,\nu} - A^{\nu}_{,\mu} A_{\nu}^{,\mu} &~=~F^{\mu\alpha}A_{\mu,\alpha}\cr &~\stackrel{?}{\approx}~n \Delta{\cal L}+\left(-\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu}}+ d_{\beta}\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}\right) A_{\alpha,\mu} -\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\mu\beta}\cr &~\stackrel{(9)}{=}~(n-2) \Delta{\cal L}_1+(n-1) \Delta{\cal L}_2+ A_{\alpha,\mu} d_{\beta}\frac{\partial\Delta{\cal L}_2}{\partial A_{\alpha,\mu\beta}}\cr &~\stackrel{(11)}{=}~(n-2) \Delta{\cal L}_1+(n-1) \Delta{\cal L}_2+ \frac{a+b}{2}\left((A^{\mu}_{,\mu})^2+A^{\nu}_{,\mu} A^{\mu}_{,\nu} \right) +c A^{\nu}_{,\mu} A_{\nu}^{,\mu} ,\tag{14}\end{align} $$
which leads to the linear eq. system
$$ \begin{align} 0&~=~a+b+c, \tag{15}\cr -1&~=~(n-1)c\qquad\qquad\qquad\Rightarrow\qquad c~=~-\frac{1}{n-1},\tag{16}\cr 0&~=~(n-2)a +\frac{a+b}{2}\qquad\Rightarrow\qquad a~=~-\frac{1}{2(n-1)(n-2)},\tag{17}\cr 1&~=~(n-2)b +\frac{a+b}{2}\qquad\Rightarrow\qquad b~=~\frac{2n-3}{2(n-1)(n-2)},\tag{18}\end{align} $$ which remarkably has a unique & consistent solution. So it is not enough to just take the trace of eq. (6). However together with eq. (13), we conclude that there is no solution. $\Box$