Euler-Lagrange Field Theory Question

[jameson2]

Homework Statement

Given the the Lagrangian density $$L= \frac{1}{2}\partial_\lambda\phi\partial^\lambda\phi + \frac{1}{3}\sigma\phi^3$$
(a)Work out the equation of motion.

(b)Calculate from L the stress tensor: $$T^{\mu\nu}=\frac{\partial L}{\partial(\partial_\mu\phi)}\partial^\nu\phi - g^{\mu\nu}L$$ where g is diagonal with matrix entries (1,-1,-1,-1).

(c)Find the 4-divergence of the stress tensor, $$\partial_\mu T^{\mu\nu}$$

(d)Show that the stress tensor is conserved by demonstrating its 4-divergence is zero when the scalar field obeys its equation of motion i.e.$$\partial_\mu T^{\mu\nu} =0$$

Homework Equations

Euler Lagrange Equation of Motion

The Attempt at a Solution

(a) I think that I have this right : $$\partial_\mu(\partial^\mu\phi)-\sigma \phi^2=0$$

(b)I have that the first term in the stress tensor is $$\frac{\partial L}{\partial(\partial_\mu\phi)}\partial^\nu\phi=\partial^\mu\phi\partial^\nu\phi$$ but I don't know how to treat the second part, i.e. $$g^{\mu\nu}L=g^{\mu\nu}(\frac{1}{2}\partial_\lambda\phi\partial^\lambda\phi + \frac{1}{3}\sigma\phi^3)=?$$
I just need to know how to treat the metric tensor g.

Obviously I haven't got to (c) or (d) yet, as I need the answer to (b).

 

[fzero]

Those two terms make up the stress tensor. There isn't anything in particular that you can do to simplify it.

 

[jameson2]

So my stress tensor is:
$$T^{\mu\nu}=\partial^\mu\phi\partial^\nu\phi-g^{\nu\mu}L$$.

I think I'm close to getting the last parts:
$$\partial_\mu T^{\mu\nu} = \partial_\mu(\partial^\mu\phi\partial^\nu\phi)-g^{\mu\nu}\partial_\mu(\frac{1}{2}\partial_\lambda\phi\partial^\lambda\phi+\frac{1}{3}\sigma\phi^3) = \partial^\mu\phi\partial_\mu(\partial^\nu\phi) + \partial^\nu\phi\partial_\mu(\partial^\mu\phi) - g^{\mu\nu}(\frac{1}{2}\partial^\lambda\phi\partial_\mu(\partial_\lambda\phi)+\frac{1}{2}\partial_\lambda\phi\partial_\mu(\partial^\lambda\phi) + 0)$$

and then if it obeys the equation of motion;

$$\partial_\mu T^{\mu\nu} = 2\sigma \phi^2\partial^\nu\phi - g^{\mu\nu}[\frac{1}{2}\delta^\mu_\lambda\sigma\phi^2\partial_\lambda\phi + \frac{1}{2}\delta^\mu_\lambda\sigma\phi^2\partial_\lambda\phi]$$

but this doesn't seem to equal zero, as required in the last part...

$$\partial_\mu T^{\mu\nu} = 2\sigma \phi^2\partial^\nu\phi -\sigma \phi^2\partial^\nu\phi =\sigma \phi^2\partial^\nu\phi$$

It seems as if I only have something slightly wrong, which is annoying.

 

[jameson2]

Actually, it works out if I do is this way. When I'm applying the partial derivative to the second term of the stress tensor, is this how you do it? $$\partial_\mu (g^{\mu\nu}L) = \partial_\mu(g^{\mu\nu}) L + g^{\mu\nu} \partial_\mu L$$ I'm not sure if this is ok though. I use in the next line $$\partial_\mu(g^{\mu\nu})= \partial^\nu$$ which I'm not sure of.

 

[fzero]

The metric is constant. so

$$\partial_\mu (g^{\mu\nu}L) = \partial_\mu(g^{\mu\nu}) L + g^{\mu\nu} \partial_\mu L = \partial^\nu L$$

I think you also left out a term in $$\partial_\mu T^{\mu\nu}$$ that is of the form $$\partial^\nu (\phi^3)$$ (it's what should be in place of +0 in the 2nd equation of post #3) that should cancel the term you had left over.

 

[jameson2]

I was thinking that since $$\phi$$ is a scalar field, the derivative is just zero. I guess that's wrong then.

 

[fzero]

I was thinking that since $$\phi$$ is a scalar field, the derivative is just zero. I guess that's wrong then.

Being a scalar in this context means that the field $$\phi(x^\mu)$$ is invariant under Lorentz transformations. It doesn't mean that it is constant on spacetime.

 

[jameson2]

Got it, thank you very much for your help.