- #1
Spinny
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Hi, I need some help understanding basic tensor algebra, especially differentiation. The subject I'm studying is quantum field theory, so I'll use examples from there.
First let's start with a real scalar field. This has a Lagrangian density given by
\mathcal{L} = \frac{1}{2}\partial_{\mu}\phi \partial^{\mu}\phi - \frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4
where \lambda is just a (coupling) constant. We must then have that the Euler-Lagrange equation
\frac{\partial \mathcal{L}}{\partial \phi}-\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)} = 0
coincides with the dynamic equation
(\square + m^2)\phi = -\frac{\lambda}{6}\phi^3
The first part of the Euler-Lagrange equation is rather easy, differentiate with respect to \phi, and this gives
\frac{\partial \mathcal{L}}{\partial \phi} = -m^2\phi - \frac{\lambda}{3!}\phi^3
Then, the second part, which I get to be
\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)} = \frac{1}{2}\partial_{\mu}\partial^{\mu}\phi
Combining these gives
\left( \frac{1}{2}\partial_{\mu}\partial^{\mu}+m^2 \right) \phi = -\frac{\lambda}{6}\phi^3
This is an example from the textbook (Elementary Particles and Their Interactions by Quand Ho-Kim and Pham Xuan Yem), although the calculation has not been carried out explicitly. In this book the operator \square is also defined as \square = \partial^{\mu}\partial_{\mu}, which is not quite what I got. Or is it the same after all? If not, what did I do wrong in the calculation? Or is it perhaps a typo in the book? It would be nice if someone could enlighten me on this. I have more examples (with a vector field), but I'll post that after I've, hopefully, gotten some respons to this problem.
First let's start with a real scalar field. This has a Lagrangian density given by
\mathcal{L} = \frac{1}{2}\partial_{\mu}\phi \partial^{\mu}\phi - \frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4
where \lambda is just a (coupling) constant. We must then have that the Euler-Lagrange equation
\frac{\partial \mathcal{L}}{\partial \phi}-\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)} = 0
coincides with the dynamic equation
(\square + m^2)\phi = -\frac{\lambda}{6}\phi^3
The first part of the Euler-Lagrange equation is rather easy, differentiate with respect to \phi, and this gives
\frac{\partial \mathcal{L}}{\partial \phi} = -m^2\phi - \frac{\lambda}{3!}\phi^3
Then, the second part, which I get to be
\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)} = \frac{1}{2}\partial_{\mu}\partial^{\mu}\phi
Combining these gives
\left( \frac{1}{2}\partial_{\mu}\partial^{\mu}+m^2 \right) \phi = -\frac{\lambda}{6}\phi^3
This is an example from the textbook (Elementary Particles and Their Interactions by Quand Ho-Kim and Pham Xuan Yem), although the calculation has not been carried out explicitly. In this book the operator \square is also defined as \square = \partial^{\mu}\partial_{\mu}, which is not quite what I got. Or is it the same after all? If not, what did I do wrong in the calculation? Or is it perhaps a typo in the book? It would be nice if someone could enlighten me on this. I have more examples (with a vector field), but I'll post that after I've, hopefully, gotten some respons to this problem.
