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I'm new here. Given the Lagrangian,

$ L=\frac{1}{2}∂_{μ}\Phi (∂^{μ}\Phi)^{*} − \frac{λ}{4}(\Phi\Phi^{*} - 1)^{2}$

and its energy-momentum tensor

$T^{\mu}_{\nu}=\frac{\partial L}{\partial(\partial^{\mu}\Phi^{*})}\partial_{\nu} \Phi -δ^{μ}_{ν}L $

For the element $T^{0}_{0}$ I get

$T^{0}_{0}=\frac{1}{2}\dot{\Phi}^{2}+\frac{1}{2}(\nabla{\Phi})^{2}+\frac{λ}{4}(\Phi\Phi^{*} - 1)^{2}$

Which seems to be incorrect, could anyone help me find the correct answer? Thank you for your time!!

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    $\begingroup$ The correct formula for the energy momentum tensor is $T^\mu{}_\nu = \frac{ \partial L }{ \partial ( \partial^\mu \Phi ) } \partial_\nu \Phi + \frac{ \partial L }{ \partial ( \partial^\mu \Phi^* ) } \partial_\nu \Phi^* - \delta^\mu_\nu L$. Also, for canonically normalized fields, your Lagrangian should be $L = \partial_\mu \Phi ( \partial^\mu \Phi )^* - \frac{\lambda}{4} ( \Phi \Phi^* - 1 )^2$. $\endgroup$
    – Prahar
    Commented Mar 5, 2018 at 23:21
  • $\begingroup$ Thank you very much @Prahar !! I'm new to field theory so I have still some questions, I get the expression $\frac{\partial(\partial_{μ} \Phi(\partial^{μ} \Phi)^{*})}{\partial(\partial^{μ}\Phi)}\partial_{ν} \Phi$ what does this differentiation equal to? $\endgroup$ Commented Mar 5, 2018 at 23:34
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    $\begingroup$ It is $(\partial_\mu \Phi)^* \partial_\nu \Phi$. $\endgroup$
    – Prahar
    Commented Mar 6, 2018 at 0:02
  • $\begingroup$ Yes. There is! I made a mistake in my earlier formula. The correct index structure is $T_{\mu\nu} = \frac{ \partial L }{ \partial ( \partial^\mu \Phi )} \partial_\nu \Phi + \frac{ \partial L }{ \partial ( \partial^\nu \Phi^* )} \partial_\mu \Phi^* - \eta_{\mu\nu} L$. To obtain $T^\mu_\nu$ you can now just raise the $\mu$ index. $\endgroup$
    – Prahar
    Commented Mar 6, 2018 at 0:43
  • $\begingroup$ First of all, what you have written is not true. What is true is $\frac{ \partial ( \partial_\mu \Phi (\partial^\mu \Phi)^* )}{ \partial (\partial^\mu \Phi ) } \partial_\nu \Phi = \partial_\mu \Phi^* \partial_\nu \Phi$. Next, since $\eta^{\mu\nu}$ does not depend on the field $\Phi$ there is no problem with perform any contraction with it. $\endgroup$
    – Prahar
    Commented Mar 6, 2018 at 3:01

2 Answers 2

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You aren't applying the energy-momentum tensor in the right way. You should sum over both fields. Check again your formula!

Edit: You have to consider the complex field like an other field.

Edit2: The correct formula is given by $$T^\mu_{\nu} = \frac{\partial L}{\partial (\partial_\mu \phi_a)}\partial_\nu \phi_a - \delta_\nu^\mu L $$ where you have to sum over all fiels $\phi_a = \{\Phi, \Phi^*\}$.

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  • $\begingroup$ Would you mind writing down the right expression? I'm having trouble understanding what you are saying. $\endgroup$ Commented Mar 5, 2018 at 23:12
  • $\begingroup$ Sorry to bother you but could you write the correct expression for the energy-momentum tensor regarding 2 fields then? $\endgroup$ Commented Mar 5, 2018 at 23:18
  • $\begingroup$ Ok done, I edited my post. I hope it is clear now. $\endgroup$
    – booroor
    Commented Mar 5, 2018 at 23:30
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The correct canonical normalized Lagrangian should be $$ L = ( \partial^\mu \Phi)^*( \partial_\mu \Phi ) - \frac{\lambda}{4} ( \Phi^* \Phi - 1 )^4 \,. $$ The canonical energy momentum tensor is $$ T^\mu{}_\nu = \frac{\partial L }{ \partial( \partial_\mu \Phi ) } \partial_\nu \Phi + \frac{\partial L }{ \partial( (\partial_\mu \Phi)^* ) } (\partial_\nu \Phi)^* - \delta^\mu_\nu L . $$ Applying this to the Lagrangian at hand, we find $$ T^\mu{}_\nu = (\partial^\mu \Phi)^* \partial_\nu \Phi + \partial^\mu \Phi ( \partial_\nu \Phi )^* - \delta^\mu_\nu \left[ ( \partial^\alpha \Phi)^*( \partial_\alpha \Phi ) - \frac{\lambda}{4} ( \Phi^* \Phi - 1 )^4 \right] $$

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  • $\begingroup$ I can't begin to thank you enough!!! Have a great day!! $\endgroup$ Commented Mar 6, 2018 at 3:27

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