I am trying to calculate the Noether current, more specifically, the energy density of the Yang-Mills-Higgs Lagrangian. Please refer to the equations in the Harvey lectures on Magnetic Monopoles, Duality, and Supersymmetry. I am trying to get the equation 1.25 from the Lagrangian 1.20. My Lagrangian is as follows: $$\mathcal{L}=-\frac{1}{4}Tr(F_{\mu \nu}F^{\mu \nu})+\frac{1}{2}Tr(D_\mu \Phi D^\mu \Phi)-V(\Phi)$$

As the Lagrangian is invariant under gauge transformation, the Noether current is

$$\Theta = \frac{\partial \mathcal{L}}{\partial(\partial_\mu A_\nu)}\partial_\nu A_\mu + \frac{\partial \mathcal{L}}{\partial(\partial_\mu \Phi)}\partial_\nu \Phi $$

The first derivative of the current can be calculated as follows:

$$\frac{\partial(F_{\alpha\beta}F^{\alpha\beta})}{\partial(\partial_\mu A_\nu)}=2F^{\alpha \beta}\frac{\partial F_{\alpha \beta}}{\partial{(\partial_\mu A_\nu)}}$$ Expanding out $F_{\alpha\beta}$, and using $\frac{\partial(\partial_\alpha A_\beta)}{\partial(\partial_\mu A_\nu)}=\delta^\mu_\alpha\delta^\nu_\beta$, we get $$\frac{\partial(F_{\alpha\beta}F^{\alpha\beta})}{\partial(\partial_\mu A_\nu)}=2F^{\alpha\beta}(\delta^\mu_\alpha\delta^\nu_\beta-\delta^\mu_\beta\delta^\nu_\alpha)=4F^{\mu\nu}$$

Exactly in the same way, the calculation for the second derivative can be done. So I get the following: $$\Theta ^{\mu \nu}=-F^{\mu\nu} \partial_\nu A_\mu + D_\mu \Phi \partial_\nu \Phi$$

From this how do I get the answer written in Harvey's lectures. $$\Theta^{\mu \nu}=F^{\mu \rho}F^{\rho \nu}+D^\mu \Phi D^\nu \Phi$$

I have tried hard but I am not getting the 'extra' terms in the answer to cancel and give me my anser, or I am making some fundamental error. Thanks in advance.

EDIT: As Ron has pointed out, we are expected to calculate the symmetric energy tensor. Please could anyone tell me how to get the symmetric stress-energy tensor directly and from the canonical energy tensor, what is the conceptual difference between the two? If possible suggest a reference.

  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/27048/2451 $\endgroup$
    – Qmechanic
    Jul 13, 2012 at 13:01
  • $\begingroup$ The question in the edit is a new question, and this requires an elaborate answer, you should post it separately. $\endgroup$
    – Ron Maimon
    Jul 13, 2012 at 19:23
  • $\begingroup$ The Wikipedia page on the Belinfante-Rosenfeld stress tensor is useful. Differentiating the action with respect to g is the quickest way to get the correct stress tensor, the difference is explained on Wikipedia, it is due to the angular momentum density in the fields not being properly incorporated in the canonical stress tensor to give the right relation between the angular moemntum and the stress tensor you expect on general grounds from the fact that a rotation on a distant axis is a translation. $\endgroup$
    – Ron Maimon
    Jul 13, 2012 at 20:08
  • $\begingroup$ @Qmechanic: I am sorry, I didn't follow your link, it makes my answer unnecessary. $\endgroup$
    – Ron Maimon
    Jul 14, 2012 at 3:40

1 Answer 1


Your answer is not going to be the same as the answer in the paper, because you are calculating the canonical stress-energy tensor, which is conserved, but which is not symmetric, and which has a complicated relation to the angular momentum tensor. The issue is that there two different conserved quantities, the stress energy tensor, and the angular momentum tensor, and the information in the two partially overlaps for a theory with both rotational/Lorentz and translation invariance.

The standard easy fix is to calculate the stress tensor by differentiating with respect to $g_{\mu\nu}$. This is how you automatically get a symmetric stress tensor with the right properties that it makes the angular momentum tensor by just multiplying by x factors

$$ L_{\mu\nu\alpha} = x_{\alpha} T_{\mu\nu} - x_{\mu}T_{\alpha\nu} $$

In some convention, and assuming you are using a symmetric stress tensor. The derivative of the action with respect to $g_{\mu\nu}$ gives Harvey's result.

I should point out that you missed a term of $-\eta^{\mu\nu} L$ in your stress tensors, both in yours and in Harvey's. This doesn't affect the question.

  • $\begingroup$ Please could you explain why the $−\eta_{\mu \nu}L$ term is there. In my second equation which is the formula for the noethers current, shouldnt this term not be there? This term is because of the change in the Lagrangian due to the symmetry transformation, but as the lagrangian is invariant under the gauge transformation, we should only consider the change in action. Also, please could you give me a reference to read about the 'canonical' and 'symmetric' energy-tensors, what I have done is the only stress-energy tensor I know. $\endgroup$
    – user7757
    Jul 13, 2012 at 10:43
  • $\begingroup$ @ramanujan_dirac: That term is present because of the shift in boundary conditions when you do a translation (in the canonical case). To understand this, derive the standard formula for conservation of energy in 1d Lagrangian: $H = p\dot{q} -L $. It's the same as the reason for -L. Translations are spacetime symmetries, and moving the integration region affects the boundaries. In the GR way, it's because of the metric factor $\sqrt{g}$ in the density. In either case, this is always the right formula. Look up Belinfante-Rosenfeld stress-tensor for this. $\endgroup$
    – Ron Maimon
    Jul 13, 2012 at 15:49

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