# Symmetrizing the Canonical Energy-Momentum Tensor

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?

• Related: physics.stackexchange.com/q/27048/2451 and links therein. – Qmechanic Sep 19 '12 at 14:57
• Symmetrizability is equivalent with Lorentz invariance. Thus you need to assume that in your arguments. – Arnold Neumaier Sep 19 '12 at 17:12
• Can you explain that a bit more? I don't undestand what you're trying to say. Thanks! – Prahar Sep 23 '12 at 17:00
• Without the assumption of Lorentz invariance of the action, there is no symmetric e/m tensor, and the standard recipe fails. Lorentz invariance gives you additional properties that you must exploit in your derivation; otherwise you will not be able to arrive at the conclusion (since it might not hold). - If you reply to a comment, you should mention the name, like in @Prahar, so that the original commenter is informed. I noticed your comment only by chance (and hence very late). – Arnold Neumaier Nov 1 '12 at 17:58
• @Prahar I know it's been a long time ago, but how did you obtain the second-to-last expression? – Drake Marquis May 31 '16 at 20:44

Is it possible to obtain a symmetric stress-energy-momentum (SEM) tensor directly from the canonical SEM tensor by adding a total derivative term to the Lagrangian? In other words, by shifting $$\Delta{\cal L}=d_\mu X^\mu$$, and choosing $$X^\mu$$ appropriately, can we exactly get the shift in the SEM tensor required, in order to make the canonical SEM tensor symmetric?

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

$$F_{\mu\nu}~=~A_{\nu,\mu}-A_{\mu,\nu}, \qquad \frac{\partial{\cal L}_0}{\partial A_{\mu,\nu}}~\stackrel{(1)}{=}~ F^{\mu\nu}.\tag{2}$$

$$0~\approx~F^{\mu\nu}{}_{,\nu}~=~ d^{\mu}(A^{\nu}_{,\nu})-d_{\nu}d^{\nu}A^{\mu}\tag{3}$$

In E&M, the canonical SEM tensor is$$^1$$

$$\Theta^{\mu}{}_{\nu}~:=~\delta^{\mu}_{\nu}{\cal L}_0+\left(-\frac{\partial{\cal L}_0}{\partial A_{\alpha,\mu}}+ d_{\beta}\frac{\partial{\cal L}_0}{\partial A_{\alpha,\mu\beta}}\right) A_{\alpha,\nu} -\frac{\partial{\cal L}_0}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\nu\beta}~\stackrel{(1)}{=}~\delta^{\mu}_{\nu}{\cal L}_0+F^{\mu\alpha}A_{\alpha,\nu}~,\tag{4}$$

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$$^2$$

$$T^{\mu}{}_{\nu} -\Theta^{\mu}{}_{\nu}~\stackrel{(4)+(5)}{=}~ F^{\mu\alpha}A_{\nu,\alpha}~=~ d_{\alpha}(F^{\mu\alpha}A_{\nu}) - \underbrace{F^{\mu\alpha}{}_{,\alpha}}_{~\approx~0}A_{\nu}$$ $$~\stackrel{?}{\approx}~\delta^{\mu}_{\nu}\Delta{\cal L}+\left(-\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu}}+ d_{\beta}\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}\right) A_{\alpha,\nu} -\frac{\partial\Delta{\cal L}}{\partial A_{\alpha,\mu\beta}}A_{\alpha,\nu\beta}\tag{6}$$

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. $$\Box$$

• Hello just curios: As you claim it is not possible, in A.O. Barut - Electrodynamics and classical theory of fields he is doing precisely that, deriving the symmetric EM tensor from the variation principle. Maybe I understand your answer wrong? – Diger Apr 27 '19 at 10:59
• I tried to clarify the answer. It sounds like Barut is talking about the metric/Hilbert SEM tensor, which is not the topic of this post. – Qmechanic Apr 27 '19 at 11:48

I will try to obtain the result using another way. It is well known that the Lagrangian density determined up to divergence of some four-vector $\mathcal{L}(x)\to \mathcal{L}(x)+\partial_\mu\psi^\mu(x)$ Let's understand what contribution the second term gives in the energy-momentum tensor. $$\hat T^\nu _\mu=\partial_\rho\Bigr(\frac{\delta {\cal\psi^\rho }}{\delta (\partial^\mu \phi_s)} \partial^\nu \phi_s - g_\mu^\nu {\cal \psi^\rho}\Bigr)=\partial_\rho \chi^{\rho\nu}_ \mu$$ $\psi^\rho$ is arbitrary four-vector, contains in $\phi_s$ and $\partial^\rho\phi_r$. Set that $\psi^\rho=f(\phi^2)\phi_r\partial^\rho\phi_r$.(If I require that the Lagrangian dependence only $\phi_r$ and first derivative of it. It will be general form) We obtain the following result $$\chi^{\rho\nu}_ \mu =g_\mu^\rho {\cal \psi^\nu}-g_\mu^\nu {\cal \psi^\rho}$$ where $g^\mu_\rho=\delta^\mu_\rho$ is a Kronecker symbol. Thus we obtain that energy momentum tensor defined up to such term $T^{\mu\nu}\to T^{\mu\nu}+\partial_\rho\chi^{\rho\mu\nu}$ where $\chi^{\rho\mu\nu}=-\chi^{\mu\rho\nu}$. This fact is a consequence of Lagrangian feature(The Lagrangian density determined up to divergence of some four-vector $\mathcal{L}(x)\to \mathcal{L}(x)+\partial_\mu\psi^\mu(x)$).

Edit

Using previous formula it is easy to obtain that $$\chi^{\mu\rho\nu} =g^{\mu\rho} {\cal \psi^\nu}-g^{\mu\nu} {\cal \psi^\rho}$$ After contraction with $g_{\mu\rho}$ we obtan that $$\psi^\nu=\frac{1}{D-1}\chi^{\mu\rho\nu}g_{\mu\rho}$$ where $D$-is dimensionality of space.

In spite of the Lagrangian contain second derivatives, all of it is true.Because it Lagrangian differ only for full derivative. If you interesting in this question, you should write General relativity. Because action of General relativity which contain Riemann curvature tensor(which contain second derivatives).

• Instead of $g^{\nu}_\mu$ and similar symbols, did you mean to write $\delta^\nu_\mu$ and similar symbols (here, $\delta$ is the Kronecker delta)? – Danu Dec 30 '14 at 17:55
• @Peter If $\psi^\rho$ contains $\partial^\rho \phi_r$, then the Lagrangian contains upto two derivatives of the fields (since it contains $\partial_\mu \psi^\mu$). One then needs to completely modify the standard definitions of the stress tensor since they all assume dependence upto 1 derivative only. I have considered these issues and the results are in my question already. – Prahar Dec 30 '14 at 23:16
• @Peter - Abovementioned issues aside, I agree with your answer, but it does not answer my question. I know that the stress-tensor is determined upto additive terms of the form $\partial_\rho \chi^{\rho\mu\nu}$ and I also know that there always exists a choice of $\chi^{\rho\mu\nu}$ to make it symmetric. My question is - Is there a choice of $\psi^\rho$ such that the corresponding "canonical stress-tensor" is symmetric. – Prahar Dec 30 '14 at 23:20
• @Peter - Let me pose the question in a different (possibly more general) way. I'll use the notations in your answer. Given a $\psi^\rho$, one can always find the corresponding $\chi^{\rho\mu\nu}$ (I have done this already in the question). The question is - Can the reverse process be achieved, i.e. Given any $\chi^{\rho\mu\nu}$, is it true that it can be derived from some $\psi^\rho$. If true, can you explicitly construct such a $\psi^\rho$. – Prahar Dec 30 '14 at 23:23
• I edit a little bit. I tried to answer in our questions. – Peter Jan 6 '15 at 10:35

Yes this is possible. For a suitable choice of the lagrangian the Noether energy momentum tensor is symmetric.

• Note that OP asks about the canonical SEM tensor rather than the metric/Hilbert SEM tensor. – Qmechanic May 23 '19 at 14:30
• My answer addresses the OP's question. – my2cts May 24 '19 at 22:45