Proving (partial mu phi)^2 = dot phi^2 - (nabla phi)^2 in Index Manipulations

[spaghetti3451]

Homework Statement

Prove the following: $(\partial_{\mu} \phi)^{2} = \dot{\phi}^{2} - (\nabla \phi)^{2}$.

Homework Equations

The Attempt at a Solution

$(\partial_{\mu} \phi)^{2}$
$= (\partial_{\mu} \phi)(\partial_{\mu} \phi)$
$= (\partial_{0} \phi)(\partial_{0} \phi) + (\partial_{1} \phi)(\partial_{1} \phi) + (\partial_{2} \phi)(\partial_{2} \phi) + (\partial_{3} \phi)(\partial_{3} \phi)$

Am I on the right track?

 

[DEvens]

Nearly there. But you seem to be missing a minus sign. And the definitions of ()^2 has a small issue.

Look in your course material or text or whatever you are using. There should be something called a metric. For the purposes of this question it looks like the metric should be a diagonal square matrix, zeroes off axis, and (1, -1, -1, -1) on the diagonal.

 

[spaghetti3451]

Is this correct?

$(\partial_{\mu} \phi)^{2}$
$= (\partial_{\mu}) (\phi \partial^{\mu} \phi)$
$= (\partial_{0} \phi) (\partial^{0} \phi) + (\partial_{1} \phi) (\partial^{1} \phi) + (\partial_{2} \phi) (\partial^{2} \phi) + (\partial_{3} \phi) (\partial^{3} \phi)$

 

[DEvens]

Is this correct?

$(\partial_{\mu} \phi)^{2}$
$= (\partial_{\mu}) (\phi \partial^{\mu} \phi)$
$= (\partial_{0} \phi) (\partial^{0} \phi) + (\partial_{1} \phi) (\partial^{1} \phi) + (\partial_{2} \phi) (\partial^{2} \phi) + (\partial_{3} \phi) (\partial^{3} \phi)$

Not quite. $\partial_{\mu} \phi$ is a vector. The index is "down." So to get the "square" of such a vector you need the inner product.

$g^{\mu \nu} \partial_{\mu} \phi \partial_{\nu} \phi$

 

[nrqed]

Is this correct?

$(\partial_{\mu} \phi)^{2}$
$= (\partial_{\mu}) (\phi \partial^{\mu} \phi)$
$= (\partial_{0} \phi) (\partial^{0} \phi) + (\partial_{1} \phi) (\partial^{1} \phi) + (\partial_{2} \phi) (\partial^{2} \phi) + (\partial_{3} \phi) (\partial^{3} \phi)$

Yes, this is absolutely correct (I disagree with Devens on that point) . But it is true that it may also be written as g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi although here we are not doing general relativity so we would usually write it as \eta^{\mu \nu} \partial_\mu \phi \partial_\nu \phi. But you don't need that notation if you know what \partial^0, \partial_0, \partial^i and \partial_i mean.

 

[DEvens]

Is this correct?

$(\partial_{\mu} \phi)^{2}$
$= (\partial_{\mu}) (\phi \partial^{\mu} \phi)$ [snips]

I see that I forgot to say what was actually wrong. My bad. It's this line. This is taking the derivative $(\partial_{\mu})$ of $(\phi \partial^{\mu} \phi)$. That is, this is not the same as the previous line. It should be $(\partial_{\mu} \phi) ( \partial^{\mu} \phi)$.

 

[nrqed]

I see that I forgot to say what was actually wrong. My bad. It's this line. This is taking the derivative $(\partial_{\mu})$ of $(\phi \partial^{\mu} \phi)$. That is, this is not the same as the previous line. It should be $(\partial_{\mu} \phi) ( \partial^{\mu} \phi)$.

Sorry, I had misunderstood your point. I thought you were referring to the last line, I had actually ignored the second line, assuming that it was a typo. I agree completely with you.Regards,Patrick