Noether current for the Yang-Mills-Higgs Lagrangian - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-17T09:00:01Z https://physics.stackexchange.com/feeds/question/31941 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/31941 7 Noether current for the Yang-Mills-Higgs Lagrangian user7757 https://physics.stackexchange.com/users/0 2012-07-13T06:42:35Z 2016-07-07T20:52:07Z <p>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 <a href="http://arxiv.org/abs/hep-th/9603086/" rel="nofollow">Harvey lectures on Magnetic Monopoles, Duality, and Supersymmetry</a>. I am trying to get the equation 1.25 from the Lagrangian 1.20. My Lagrangian is as follows: $$\mathcal{L}=-\frac{1}{4}Tr(F_{\mu \nu}F^{\mu \nu})+\frac{1}{2}Tr(D_\mu \Phi D^\mu \Phi)-V(\Phi)$$</p> <p>As the Lagrangian is invariant under gauge transformation, the Noether current is </p> <p>$$\Theta = \frac{\partial \mathcal{L}}{\partial(\partial_\mu A_\nu)}\partial_\nu A_\mu + \frac{\partial \mathcal{L}}{\partial(\partial_\mu \Phi)}\partial_\nu \Phi$$</p> <p>The first derivative of the current can be calculated as follows:</p> <p>$$\frac{\partial(F_{\alpha\beta}F^{\alpha\beta})}{\partial(\partial_\mu A_\nu)}=2F^{\alpha \beta}\frac{\partial F_{\alpha \beta}}{\partial{(\partial_\mu A_\nu)}}$$ 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 $$\frac{\partial(F_{\alpha\beta}F^{\alpha\beta})}{\partial(\partial_\mu A_\nu)}=2F^{\alpha\beta}(\delta^\mu_\alpha\delta^\nu_\beta-\delta^\mu_\beta\delta^\nu_\alpha)=4F^{\mu\nu}$$</p> <p>Exactly in the same way, the calculation for the second derivative can be done. So I get the following: $$\Theta ^{\mu \nu}=-F^{\mu\nu} \partial_\nu A_\mu + D_\mu \Phi \partial_\nu \Phi$$</p> <p>From this how do I get the answer written in Harvey's lectures. $$\Theta^{\mu \nu}=F^{\mu \rho}F^{\rho \nu}+D^\mu \Phi D^\nu \Phi$$</p> <p>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.</p> <p><strong>EDIT</strong>: 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. </p> https://physics.stackexchange.com/questions/31941/-/31952#31952 4 Answer by Ron Maimon for Noether current for the Yang-Mills-Higgs Lagrangian Ron Maimon https://physics.stackexchange.com/users/4864 2012-07-13T09:09:44Z 2012-07-13T09:09:44Z <p>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.</p> <p>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</p> <p>$$L_{\mu\nu\alpha} = x_{\alpha} T_{\mu\nu} - x_{\mu}T_{\alpha\nu}$$</p> <p>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.</p> <p>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.</p>