0
$\begingroup$

I'm trying to understand the transition from the 1st line of the Lagrangian to the second. we substitute for $\eta$ but how is the multiplication happening here? if I multiply the terms into the matrix elements, won't I get a matrix whose elements are the terms to the right of the matrix here?enter image description here

$\endgroup$
  • $\begingroup$ Recall that repeated indices are summed over. The term $\eta^{\mu \nu}\partial_{\mu} \phi \partial_{\nu} \phi$ is a scalar, not a matrix. $\endgroup$ – preferred_anon Aug 20 '18 at 10:01
0
$\begingroup$

Since $\eta^{\mu \nu}$ is non-zero for diagonal elements only, we only need to sum those terms when $\mu = \nu$. Therefore, we have \begin{align} &\qquad \frac{1}{2} \eta^{\mu \nu} \; \partial_{\mu}\phi \; \partial_{\nu} \phi \\ &= \frac{1}{2} \Bigg[ \eta^{00} \partial_{0}\phi \; \partial_{0} \phi \; + \; \eta^{11} \partial_{1}\phi \; \partial_{1} \phi \; + \; \eta^{22} \partial_{2}\phi \; \partial_{2} \phi \; + \; \eta^{33} \partial_{3}\phi \; \partial_{3} \phi \Bigg]\\ &= \frac{1}{2} \Bigg[ (+1)\partial_{t}\phi \; \partial_{t} \phi \; + \; (-1) \partial_{x}\phi \; \partial_{x} \phi \; + \; (-1) \partial_{y}\phi \; \partial_{y} \phi \; + \; (-1) \partial_{z}\phi \; \partial_{z} \phi \Bigg] \\ &= \frac{1}{2} \dot{\phi}^{\, 2} - \frac{1}{2}(\nabla\phi)^2 \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.