Taking the non-relativistic limit of a Lagrangian

This question has to do with question 1 on this sheet. I'm trying to self-learn QFT, please be patient with me!

I have problem to get the exact form of the Lagrangian in the non-relativistic limit as stated in the question.

Firstly, I see that a sign in the relativistic Lagrangian is missing there. It should be $$L=(\partial_\mu \phi^*)(\partial^\mu \phi)-m^2|\phi|^2$$

I shall take the signature of the Minkowski metric to be $(+---)$.

Then $$L=(\partial_t \phi^*)(\partial_t \phi)-\nabla\phi^*\cdot \nabla \phi -m^2|\phi|^2$$

Now since $\phi=\exp(-imt)\tilde{\phi}$, we have $$\partial_t \phi=\exp(-imt)(-im\tilde{\phi}+\partial_t \tilde{\phi})$$ giving $$(\partial_t \phi^*)(\partial_t \phi)=m^2|\phi|^2+im[(\partial_t \tilde{\phi})\tilde{\phi}^*-(\partial_t \tilde{\phi}^*)\tilde{\phi}]+\textrm{terms in second order in }\partial_t \textrm{which can be ignored in non-relativistic limit}$$

Noting that $$\nabla\phi^*\cdot \nabla \phi=\nabla\tilde{\phi}^*\cdot \nabla \tilde{\phi}$$ So $$L=im[(\partial_t \tilde{\phi})\tilde{\phi}^*-(\partial_t \tilde{\phi}^*)\tilde{\phi}]-\nabla\tilde{\phi}^*\cdot \nabla \tilde{\phi}$$

To get the form stated in the question, I would need $$(\partial_t \tilde{\phi})\tilde{\phi}^*-(\partial_t \tilde{\phi}^*)\tilde{\phi}=2(\partial_t \tilde{\phi})\tilde{\phi}^*$$

But how? Or have I made a mistake?

• Two Lagrangians differ by the total derivative with respect to time of some function describe the same physical system. – achille hui Aug 27 '13 at 10:26
• @achillehui: ah, thank you! – eek Aug 27 '13 at 11:33