# Finding Lagrangian density for particle with spin 1 and mass $\mu$

I am trying to solve my first ever QFT problem, which reads as follow:

From the wave equation for a particle with spin 1 and mass $$\mu$$: $$\left[g_{\mu\nu}(\Box+\mu^2)-\frac{\partial}{\partial x^\mu}\frac{\partial}{\partial x^\nu} \right]\phi^\nu(x)=0$$ it is deduced that $$\frac{\partial \phi^\nu}{\partial x^\nu}=0$$ From this, derive the corresponding Lagrangian density: $$L=-\frac{1}{2}\frac{\partial \phi^\nu}{\partial x^\mu}\frac{\partial \phi_\nu}{\partial x_\mu}+\frac{1}{2}\mu^2\phi_\nu\phi^\nu+\frac{1}{2}\left(\frac{\partial \phi^\nu}{\partial x^\nu}\right)^2$$

I am a bit lost as to how to solve it, but I decided to attempt the following:

First I substitute the D'alambertian into the wave equation:

$$\left[g_{\mu\nu}(\frac{\partial^2}{\partial t^2}-\nabla^2+\mu^2)-\frac{\partial}{\partial x^\mu}\frac{\partial}{\partial x^\nu} \right]\phi^\nu(x)=0$$

and now distribute the product: $$g_{\mu\nu}(\frac{\partial^2\phi^\nu(x)}{\partial t^2}-\nabla^2\phi^\nu(x)+\mu^2\phi^\nu(x))-\frac{\partial}{\partial x^\mu}\frac{\partial}{\partial x^\nu}\phi^\nu(x)=0$$

Now I believe that I can contract the time derivative with the spatial ones, since $$\partial^\mu\partial_\nu=\frac{\partial^2}{\partial t^2}-\nabla^2$$

$$g_{\mu\nu}(\partial^\mu\partial_\nu\phi^\nu(x)+\mu^2\phi^\nu(x))-\frac{\partial}{\partial x^\mu}\frac{\partial}{\partial x^\nu}\phi^\nu(x)=0$$

At this point I can definitively see some resemblance between what I have and what I am supposed to have, but I don't really know how to proceed. Is what I have done so far right? How can I continue? If not, how should I appraoch this problem?

• Is this from a textbook? – Qmechanic Sep 11 '19 at 17:23
• No, it is a problem that my teacher gave to my class. I tried looking around for it but came out empty handed, as it is supposed to be a very basic example that most books don't bother to see – Nick Heumann Sep 11 '19 at 17:25
• What Qmechanic is saying is that this problem is not written well. You can’t uniquely “derive” a Lagrangian density from an equation of motion. What you can do is guess a Lagrangian density that gives the desired equation of motion. – knzhou Sep 11 '19 at 17:39
• As a wild guess, what your teacher wants is for you to compute the Euler-Lagrange equations of that Lagrangian and verify you get the first equation. And in the long run what you want to do is get away from this course and learn from a good book instead. – knzhou Sep 11 '19 at 17:40
• oh how odd, because the problem states exactly what I wrote here. I guess he must mean what you said, I'll double check tomorrow then to make sure I got it right – Nick Heumann Sep 11 '19 at 17:41