Lorenz condition from equations of motion for massive real vector field

Question

I want to prove that the following real vector field $\phi^{\alpha}$ satisfies the Lorenz condition:

$$ \partial_{\alpha} \phi^{\alpha} = 0 $$

for the following Lagrangian density:

$$ L = - \frac{1}{2} \partial_{\alpha} \phi_{\beta} \partial^{\alpha} \phi^{\beta} + \frac{1}{2} \partial_{\beta} \phi^{\beta} \partial_{\alpha} \phi^{\alpha} + \frac{\mu^2}{2} \phi_{\alpha} \phi^{\alpha}$$

I started by trying to find the equations of motion and upon arriving at them I would take a derivative $ \partial_{\alpha}$ of the whole thing and all terms except the one with $\mu$ as a constant would somehow vanish, subsequently yielding the desired result.

I obtained the equation:

$$ [ ( \square + \mu^2) g_{\alpha \rho} - \partial_{\mu} \partial_{\alpha} ] \phi^{\alpha} =0$$

now having $\rho$ and $\mu$ as free indicies can I just write $\rho$ for both of them?

Answer 1

You need to be very careful with index placement in these kinds of equations. $\partial_a \phi_a = 0$ is not the Lorentz condition. The Lorentz condition is $\partial_a \phi^a$. One $a$ needs to be upstairs to indicate that you are summing over the index $a$ (According to the Einstein Summation convention).

You make a similar mistake in the definition of the Lagrangian. The Lagrangian is a Lorentz scalar, which means that it can't have any free indices. So all terms on the Right-Hand side also need to be Lorentz scalars.

The first term is fine. You have one $\alpha$ and one $\beta$ up and downstairs, which means that they are summed over. But the term

$$\partial_\alpha \phi_\alpha \partial^\beta \phi^\beta$$ is not a Lorentz scalar, because the indices are not summed over. You probably mean something like $$\partial_\alpha \phi^\alpha \partial_\beta \phi^\beta$$

Now, let's look at your last equation

$$- \partial_{\mu} ( \partial^{\mu} \phi_{\rho} ) + \partial_{\mu} ( \partial_{\alpha} \phi^{\alpha} ) + \mu^2 \phi_{\rho} = 0$$

In the first term, the $\mu$ index is summed over and the $\rho$-Index is not. So this object has one free index, which makes it a 1-Tensor, i.e. a vector. In the second term, the $\alpha$ is summed over and the free index is a $\mu$. So now, you are trying to add two Lorentz vectors with different indices. You basically have an object like $$v_\rho + w_\mu$$ which is NOT a tensor! The rule is that you can only add objects with the same free indices, otherwise, it's like trying to add a number to a vector or a vector to a matrix. You also can't have the same index twice, unless it's a dummy index, i.e. an index that appears both up- and downstairs and is summed over.

I recommend practicing with some simpler tensor manipulations and trying this exercise again once you're comfortable with the formalism. With practice, you will develop an eye for index placement, and that will allow you to spot these kinds of mistakes very quickly.