Stress Tensor in Classical Field Theory

[Max Renn]

Hi,

I have a problem in classical field theory.

I have a Lagrangian density $$\mathcal{L}=\frac{1}{2}\partial_\lambda \phi \partial^\lambda \phi + \frac{1}{3}\sigma\phi^3$$. Upon solving the Euler-Lagrange equation for this density, I get an equation of motion for my scalar field $$\phi (x)$$, where $$x = x^\mu$$ is a space-time coordinate. I figured this is $$\Box \phi - \sigma \phi^2 = 0$$. Now, the problem begins.

I have to calculate the following stress tensor:

$$T^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\partial^\nu \phi - g^{\mu \nu}\mathcal{L}$$.

The metric tensor $$g^{\mu \nu}$$ is the "energy-momentum" type, $$\mathrm{diag}(1, -1, -1, -1)$$. Then I have to find its 4-divergence $$\partial_\mu T^{\mu \nu}$$ and show that it's conserved when $$\phi(x)$$ obeys its equation of motion, i.e. that $$\partial_\mu T^{\mu \nu} = 0$$.

Now, if I didn't know any better, I'd say that

$$T^{\mu \nu} = \frac{1}{2}\partial^\mu \phi \partial^\nu \phi + \frac{1}{3} g^{\mu \nu} \sigma \phi^3$$.

I have some serious doubts, however. If this is correct, I have another problem in that I can't seem to find a zero 4-divergence.

I'm quite new to this sort of thing and I have a feeling it's just a lack of practice with tensors and indices.

 

[fzero]

You have a sign wrong and you didn't quite use that formula correctly: $$T^{\mu \nu} =\partial^\mu \phi \partial^\nu \phi - g^{\mu \nu}\left( \frac{1}{2}\partial^\lambda \phi \partial_\lambda \phi +\frac{1}{3} \sigma \phi^3\right)$$

 

[Max Renn]

Oh! Is that the final form of the stress tensor? I wanted to clear up some things with the metric tensor, using some dubious manipulation of indices. That's ok (although the sign mistake is embarrassing). Thank you. This is my solution:

$$\partial_\mu T^{\mu \nu} = \partial_\mu ( \partial^\mu \phi \partial^\nu \phi ) - \partial_\mu g^{\mu \nu}\left( \frac{1}{2}\partial_\lambda \phi \partial^\lambda \phi +\frac{1}{3} \sigma \phi^3\right)$$

$$= (\partial_\mu \partial^\mu \phi) \partial^\nu \phi + (\partial_\mu \partial^\nu \phi) \partial^\mu \phi - g^{\mu \nu} \left[ \frac{1}{2} (\partial_\mu \partial_\lambda \phi) \partial^\lambda \phi + \frac{1}{2} \partial_\lambda \phi (\partial_\mu \partial^\lambda \phi) + \frac{1}{3} \sigma \partial_\mu (\phi^3) \right].$$

Since $$\Box \phi - \sigma \phi^2 = 0$$,

$$\partial_\mu T^{\mu \nu} = 2 \sigma \phi^2 \partial^\mu \phi - g^{\mu \nu} \left[ \frac{1}{2} \sigma \phi^2 \delta_{\mu \lambda} \partial^\lambda \phi + \frac{1}{2} \sigma \phi^2 \delta^\lambda_\mu \partial_\lambda \phi + \frac{1}{3} \sigma \partial_\mu (\phi^3) \right]$$

$$= 2 \sigma \phi^2 \partial^\mu \phi - g^{\mu \nu} \left[ \sigma \phi^2 \partial_\mu \phi + \frac{1}{3} \sigma \partial_\mu (\phi^3) \right]$$

$$= 2 \sigma \phi^2 \partial^\mu \phi - \sigma \phi^2 \partial^\nu \phi - \frac{1}{3} \sigma \partial^\nu (\phi^3)$$

$$= 2 \sigma \phi^2 \partial^\mu \phi - 2 \sigma \phi^2 \partial^\nu \phi$$

$$=0$$

I have a couple of questions, though. Specifically, I'm uneasy about the step where I say $$\sigma \phi^2 (\partial^\mu \phi + \partial^\nu \phi) = 2 \sigma \phi^2 \partial^\mu \phi$$, where $$\partial^\nu (\phi (x^\mu)^3 = 3 \partial^\nu \phi$$ and, on the last line, where $$\nu$$ goes to $$\mu$$ or vice versa to give zero.

 

[fzero]

Oh! Is that the final form of the stress tensor? I wanted to clear up some things with the metric tensor, using some dubious manipulation of indices. That's ok (although the sign mistake is embarrassing). Thank you. This is my solution:

$$\partial_\mu T^{\mu \nu} = \partial_\mu ( \partial^\mu \phi \partial^\nu \phi ) - \partial_\mu g^{\mu \nu}\left( \frac{1}{2}\partial_\lambda \phi \partial^\lambda \phi +\frac{1}{3} \sigma \phi^3\right)$$

$$= (\partial_\mu \partial^\mu \phi) \partial^\nu \phi + (\partial_\mu \partial^\nu \phi) \partial^\mu \phi - g^{\mu \nu} \left[ \frac{1}{2} (\partial_\mu \partial_\lambda \phi) \partial^\lambda \phi + \frac{1}{2} \partial_\lambda \phi (\partial_\mu \partial^\lambda \phi) + \frac{1}{3} \sigma \partial_\mu (\phi^3) \right]. ~~(*)$$

Since $$\Box \phi - \sigma \phi^2 = 0$$,

$$\partial_\mu T^{\mu \nu} = 2 \sigma \phi^2 \partial^\mu \phi - g^{\mu \nu} \left[ \frac{1}{2} \sigma \phi^2 \delta_{\mu \lambda} \partial^\lambda \phi + \frac{1}{2} \sigma \phi^2 \delta^\lambda_\mu \partial_\lambda \phi + \frac{1}{3} \sigma \partial_\mu (\phi^3) \right]$$

You've got several mistakes here. The free index is $$\nu$$, so you should use the metric to raise the index in the second part of (*):

$$(\partial_\mu \partial^\mu \phi) \partial^\nu \phi + (\partial_\mu \partial^\nu \phi) \partial^\mu \phi - \left[ \frac{1}{2} (\partial^\nu \partial_\lambda \phi) \partial^\lambda \phi + \frac{1}{2} \partial_\lambda \phi (\partial^\nu \partial^\lambda \phi) + \frac{1}{3} \sigma \partial^\nu (\phi^3) \right]. ~~(**)$$

You also seemed to try to claim that $$\partial_\mu \partial_\lambda \phi=\delta_{\mu\lambda} \phi$$, which is completely unjustified.

Instead, you should note that the 2nd, 3rd and 4th terms of (**) are all of the same type, and adding up their coefficients gives zero. Then you want to show that the sum of the 1st and 5th terms is proportional to the equation of motion.