I feel this should be quite straight forward as I have learnt some tensor stuff before but I can't quite see how to do this. So the Euler-Lagrange equation for a field is Please Register or Log in to view the hidden image! Using the Lagrangian for the scalar field Please Register or Log in to view the hidden image! in the E-L equation yields the Klein Gordon equation Please Register or Log in to view the hidden image! However I'm having trouble plugging the Lagrangian into the left hand side of the E-L equation. I can see you can pull down the other indice by putting in the metric tensor and I can kind of see then how you get the result from the product rule but only in a handwaving way and I am unsure about how you differentiate with respect to a tensor like that and in particular the manipulation of the indices when you do it.
This is a bit confusing the first time you see it. If it helps \(\partial_{\mu} = \frac{\partial}{\partial x^{\mu}}\) Indices that are on the bottom on the top are on the top when on the bottom. Then \(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} = \frac{\partial \partial_{\mu}\psi \partial^{\mu}\psi}{\partial (\partial_{\mu} \psi)} = \partial^{\mu}\psi\). Another way to think of it is to just replace \(\partial_{\mu} \psi \equiv x_{\mu}\). This may make things a bit more clear Please Register or Log in to view the hidden image!
Cheers. I was always under the impression an index couldn't be repeated more than twice. Also the Lagrangian has a factor of 1/2 out the front so if you got that result wouldn't the Klein-Gordon equation also have a factor of 1/2 in front of one of the terms. Surely you also have to differentiate the second term aswell.
One way of doing this is to write \(\partial_\alpha\varph\partial^\alpha\varph=g^{ \alpha\nu}\partial_\alpha\varph\partial_\nu\varph\), where \(g^{\alpha\nu}\) is the spacetime metric, also written as \(\eta^{\alpha\nu}\) in flat space. I'm writing it with alphas instead of mu's because I'm summing over the indices labelled by alpha, whereas I'm going to take a partial derivative with respect to \(\partial_\mu\varph\) for some fixed index mu. Then the partial derivative of this expression with respect to \(\partial_\mu\varph\) is \(g^{\alpha\nu}\left[\partial_\alpha\varph\delta^\mu_\nu+\partial_\nu \varph\delta^\mu_\alpha\right]\), which follows naturally from the product rule for derivatives. If you now contract indices between the Kronecker deltas and the metric tensor, you easily end up with \(2\partial^\mu\varph\), which is what you need in order to derive the Klein-Gordon equation (remember there's a factor of 1/2 in the Lagrangian to cancel with this factor of 2).
Few weeks back,I also got stuck with it.I did it like this: \(\frac{d}{dt}\ x^i\ x_i\) differentiation this and using metric tensor in appropriate places,I found the expression equal to \(\ 2\ x^i\)
You're wrong about that because \(\frac{d}{dt}(x^{i}x_{i})\) has no free index, but \(2x^{i}\) does. \(\frac{d}{dt}(x^{i}x_{i}) = 2\dot{x}_{i}x^{i}\) and \(2x^{i} = \frac{\partial}{\partial x_{i}}(x^{j}x_{j})\) Two quite different results.
