Variation of Lagrange Density under field transformation

[Dixanadu]

Homework Statement

Hey guys!

So I have a Lagrangian with two coupled fields like so:

$\mathcal{L} = \frac{1}{2}(\partial_{\mu}\phi_{1})(\partial^{\mu}\phi_{1}) +\frac{1}{2}(\partial_{\mu}\phi_{2})(\partial^{\mu}\phi_{2})-\frac{m_{1}^{2}}{2}(\phi_{1}\phi_{1})-\frac{m_{2}^{2}}{2}(\phi_{2}\phi_{2})-g(\phi_{1}\phi_{2})^2$

where g is a coupling constant.

So I have to find the variation of this lagrangian under the transformation

$\delta\phi_{1}=\epsilon\phi_{2}, \delta\phi_{2}=-\epsilon\phi_{1}$

Homework Equations

The Attempt at a Solution

I don't know what to do - do I just plug these into the Lagrangian? and if I do, how do I compute the new $\partial_{\mu}\phi$?

Thanks a lot guys!

 

[ShayanJ]

You should replace $\phi_i$ with $\phi_i+\delta \phi_i$ in the Lagrangian.

 

[Dixanadu]

Okay so i plugged them in but I'm still a bit stuck.

For example consider this term:
$\frac{1}{2}(\partial_{\mu}(\phi_{1}+\epsilon\phi_{2}))$

How do I compute this? (yes I am a noob with this stuff) Would it be equal to $\frac{1}{2}(\partial_{\mu}\phi_{1}+\epsilon\partial_{\mu}\phi_{2})$?

Also do I omit terms of order $\epsilon^{2}$?

Thanks...

 

[ShayanJ]

Would it be equal to 12(∂μϕ1+ϵ∂μϕ2)\frac{1}{2}(\partial_{\mu}\phi_{1}+\epsilon\partial_{\mu}\phi_{2})?

Yes.

Also do I omit terms of order ϵ2\epsilon^{2}?

Yes.

 

[Dixanadu]

So i went through it all and here's where I am:

$\begin{split}\mathcal{L}=\frac{1}{2}[(\partial_{\mu}\phi_{1})(\partial^{\mu}\phi_{1})] +\frac{1}{2}[(\partial_{\mu}\phi_{2})(\partial^{\mu}\phi_{2})] \\ -\frac{m_{1}^{2}}{2}(\phi_{1}^{2}+2\epsilon\phi_{1}\phi_{2}) -\frac{m_{2}^{2}}{2}(\phi_{2}^{2}-2\epsilon\phi_{1}\phi_{2}) -g(\phi_{1}\phi_{2}-\epsilon\phi_{1}^{2}+\epsilon\phi_{2}^{2}) \end{split}$

Is this looking okay...?

Thanks so much...

 

[ShayanJ]

I get the following:
$\mathcal{L}'=\frac{1}{2} \partial_{\mu} \phi_1 \partial^{\mu} \phi_1+\frac{1}{2} \partial_{\mu} \phi_2 \partial^{\mu} \phi_2-\frac{1}{2}m_1^2(\phi_1+2\epsilon \phi_2)\phi_1-\frac{1}{2}m_2^2(\phi_2-2\epsilon \phi_1)\phi_2-g[\phi_1\phi_2-2\epsilon(\phi_1^2-\phi_2^2)]\phi_1\phi_2$