# Equations of motion and Hamiltonian density of a massive vector field

1. Apr 26, 2016

### spaghetti3451

1. The problem statement, all variables and given/known data

The Lagrangian density for a massive vector field $C_{\mu}$ is given by $\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m^{2}C_{\mu}C^{\mu}$ where $F_{\mu\nu}=\partial_{\mu}C_{\nu}-\partial_{\nu}C_{\mu}$.

Derive the equations of motion and show that when $m \neq 0$ they imply $\partial_{\mu}C^{\mu}=0$.

Further show that $C_{0}$ can be eliminated completely in terms of the other fields by $\partial_{i}\partial^{i}C_{0}+m^{2}C_{0}=\partial^{i}\dot{C}_{i}$.

Construct the canonical momenta $\Pi_{i}$ conjugate to $C_{i}, i=1,2,3$ and show that the canonical momentum conjugate to $C_{0}$ is vanishing.

Construct the Hamiltonian density $\mathcal{H}$ in terms of $C_{0},C_{i}$ and $\Pi_{i}$.

(Note: Do not be concerned that the canonical momentum for $C_0$ is vanishing. $C_0$ is non-dynamical - it is determined entirely in terms of the other fields using $\partial_{i}\partial^{i}C_{0}+m^{2}C_{0}=\partial^{i}\dot{C}_{i}$.)

2. Relevant equations

3. The attempt at a solution

Given the Lagrangian density $\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m^{2}C_{\mu}C^{\mu}$, where $F_{\mu\nu}=\partial_{\mu}C_{\nu}-\partial_{\nu}C_{\mu}$ and $C_{\mu}$ is a massive vector field,

$\frac{\partial \mathcal{L}}{\partial (\partial_{\rho}C_{\sigma})}=\frac{\partial}{\partial (\partial_{\rho}C_{\sigma})}\Big(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\Big)$

$\implies \frac{\partial \mathcal{L}}{\partial (\partial_{\rho}C_{\sigma})}=-\frac{1}{4}\frac{\partial}{\partial (\partial_{\rho}C_{\sigma})}[(\partial_{\mu}C_{\nu}-\partial_{\nu}C_{\mu})(\partial^{\mu}C^{\nu}-\partial^{\nu}C^{\mu})]$

$\implies \frac{\partial \mathcal{L}}{\partial (\partial_{\rho}C_{\sigma})}=-\frac{1}{4}[({\eta^{\rho}}_{\mu}{\eta^{\sigma}}_{\nu}-{\eta^{\rho}}_{\nu}{\eta^{\sigma}}_{\mu})(\partial^{\mu}C^{\nu}-\partial^{\nu}C^{\mu})+(\partial_{\mu}C_{\nu}-\partial_{\nu}C_{\mu})(\eta^{\rho\mu}\eta^{\sigma\nu}-\eta^{\rho\nu}\eta^{\sigma\mu})]$

$\implies \frac{\partial \mathcal{L}}{\partial (\partial_{\rho}C_{\sigma})}=-\frac{1}{4}[\partial^{\rho}C^{\sigma}-\partial^{\sigma}C^{\rho}-\partial^{\sigma}C^{\rho}+\partial^{\rho}C^{\sigma}+\partial^{\rho}C^{\sigma}-\partial^{\sigma}C^{\rho}-\partial^{\sigma}C^{\rho}+\partial^{\rho}C^{\sigma}]$

$=-(\partial^{\rho}C^{\sigma}-\partial^{\sigma}C^{\rho})$

$=-F^{\rho\sigma}$

and

$\frac{\partial \mathcal{L}}{\partial C_{\rho}} = \frac{\partial}{\partial C_{\rho}}(\frac{1}{2}m^{2}C_{\mu}C^{\mu})$

$\implies \frac{\partial \mathcal{L}}{\partial C_{\rho}} = \frac{\partial}{\partial C_{\rho}}(\frac{1}{2}m^{2}\eta^{\mu\nu}C_{\mu}C_{\nu})$

$\implies \frac{\partial \mathcal{L}}{\partial C_{\rho}} = \frac{1}{2}m^{2}\eta^{\mu\nu}\frac{\partial}{\partial C_{\rho}}(C_{\mu}C_{\nu})$

$\implies \frac{\partial \mathcal{L}}{\partial C_{\rho}} = \frac{1}{2}m^{2}\eta^{\mu\nu}({\eta^{\rho}}_{\mu}C_{\nu}+{C_{\mu}\eta^{\rho}}_{\nu})$

$\implies \frac{\partial \mathcal{L}}{\partial C_{\rho}} = \frac{1}{2}m^{2}({\eta^{\rho}}_{\mu}\eta^{\mu\nu}C_{\nu}+{\eta^{\nu\mu}C_{\mu}\eta^{\rho}}_{\nu})$

$\implies \frac{\partial \mathcal{L}}{\partial C_{\rho}} = \frac{1}{2}m^{2}({\eta^{\rho}}_{\mu}C^{\mu}+{\eta^{\rho}}_{\nu}C^{\nu})$

$\implies \frac{\partial \mathcal{L}}{\partial C_{\rho}} = \frac{1}{2}m^{2}(C^{\rho}+C^{\rho})$

$\implies \frac{\partial \mathcal{L}}{\partial C_{\rho}} = m^{2}C^{\rho}$

so that

$\frac{\partial \mathcal{L}}{\partial C_{\nu}}-\partial_{\mu}\Big(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}C_{\nu})}\Big)=0 \implies m^{2}C^{\nu}+\partial_{\mu}F^{\mu\nu}=0$.

So, the equations of motion are $m^{2}C^{\nu}+\partial_{\mu}F^{\mu\nu}=0$.

Am I correct so far?

2. Apr 28, 2016

### CAF123

Yes, it's all fine.

3. Apr 28, 2016

### spaghetti3451

Thanks!

Next, I need to show that, when $m\neq 0$, the equations of motion $m^{2}C^{\nu}+\partial_{\mu}F^{\mu\nu}=0$ imply $\partial_{\mu}C^{\mu}=0$:

$m^{2}C^{\nu}+\partial_{\mu}F^{\mu\nu}=0$

$\implies m^{2}C^{\nu}+\partial_{\mu}(\partial^{\mu}C^{\nu}-\partial^{\nu}C^{\mu})=0$

$\implies m^{2}C^{\nu}+\partial_{\mu}\partial^{\mu}C^{\nu}-\partial_{\mu}\partial^{\nu}C^{\mu}=0$

$\implies m^{2}C^{\nu}+\partial_{\mu}\partial^{\mu}C^{\nu}-\partial^{\nu}\partial_{\mu}C^{\mu}=0$

$\implies (m^{2}+\partial_{\mu}\partial^{\mu})C^{\nu}-\partial^{\nu}(\partial_{\mu}C^{\mu})=0$

So, for $\partial_{\mu}C^{\mu}=0$ to hold when $m\neq 0$, we must have $(m^{2}+\partial_{\mu}\partial^{\mu})C^{\nu}=0$.

But why is $(m^{2}+\partial_{\mu}\partial^{\mu})C^{\nu}=0$ when $m \neq 0$? Surely $C_{\mu}$ is not the Klein-Gordon field, but rather a massive vector field!?

4. Apr 28, 2016

### CAF123

That's not the case - instead of multiplying it all out, contract the equation of motion with $\partial_{\nu}$ so that you get $$m^2 \partial_{\nu} C^{\nu} + \partial_{\nu} \partial_{\mu} F^{\mu \nu} = 0$$ Using properties of $F^{\mu \nu}$ deduce this second term is zero.

5. Apr 29, 2016

### spaghetti3451

$m^{2}C^{\nu}+\partial_{\mu}F^{\mu\nu}=0$

$\implies m^{2}C_{\nu}C^{\nu}+\partial_{\nu}\partial_{\mu}F^{\mu\nu}=0$

$\implies m^{2}C_{\nu}C^{\nu}+\partial_{\mu}\partial_{\nu}F^{\nu\mu}=0$, where we have relabeled indices $\mu$ and $\nu$ in the second term

$\implies m^{2}C_{\nu}C^{\nu}-\partial_{\nu}\partial_{\mu}F^{\mu\nu}=0$, where we have used the commutativity $\partial_{\mu}\partial_{\nu}=\partial_{\nu}\partial_{\mu}$ of partial derivatives and the antisymmetry $F^{\mu\nu}=-F^{\nu\mu}$ of the electromagnetic field strength tensor

So, we have both $m^{2}C_{\nu}C^{\nu}+\partial_{\nu}\partial_{\mu}F^{\mu\nu}=0$ and $m^{2}C_{\nu}C^{\nu}-\partial_{\nu}\partial_{\mu}F^{\mu\nu}=0$.

Adding the equations gives us $2m^{2}C_{\nu}C^{\nu}=0$ so that $C_{\mu}C^{\mu}=0$ when $m\neq 0$.

Am I correct?

6. Apr 29, 2016

### CAF123

Yup, the argument is correct but the term proportional to m^2 should be $\partial_{\nu} C^{\nu}$ not $C_{\nu}C^{\nu}$. The argument is completely general - whenever you have a symmetric index pair (here $\partial_{\mu} \partial_{\nu}$) contracting with an antisymmetric one (here $F^{\mu \nu}$) their contraction always vanishes.

7. May 1, 2016

### spaghetti3451

Thanks!

Now for the next part of the question.

Further show that $C_0$ can be eliminated completely in terms of the other fields by $\partial_{i}\partial^{i}C_{0}+m^{2}C_{0}=\partial^{i}C_{i}.$

Starting with the equation of motion $m^{2}C^{\nu}+\partial_{\mu}F^{\mu\nu}=0$ and setting $\nu=0$, we have

$m^{2}C^{0}+\partial_{\mu}F^{\mu 0}=0$

$\implies m^{2}C^{0}+\partial_{\mu}(\partial^{\mu}C^{0}-\partial^{0}C^{\mu})=0$

$\implies m^{2}C^{0}+\partial_{\mu}\partial^{\mu}C^{0}-\partial_{\mu}\partial^{0}C^{\mu}=0$

$\implies m^{2}C^{0}+\partial_{0}\partial^{0}C^{0}+\partial_{i}\partial^{i}C^{0}-\partial_{0}\partial^{0}C^{0}-\partial_{i}\partial^{0}C^{i}=0$

$\implies m^{2}C^{0}+\partial_{i}\partial^{i}C^{0}-\partial_{i}\partial^{0}C^{i}=0$

$\implies m^{2}C^{0}+\partial_{i}\partial^{i}C^{0}-\partial_{i}\dot{C}^{i}=0$

$\implies \partial_{i}\partial^{i}C^{0}+m^{2}C^{0}=\partial_{i}{\dot{C}}^{i}$

$\implies \partial_{i}\partial^{i}C_{0}+m^{2}C_{0}=\partial_{i}\dot{C}_{i}$.

Are my steps correct?

8. May 3, 2016

### CAF123

Correct up to the last line - assuming your metric is (1,-1,-1,-1), for the spatial components part of the four vectors $X^{\mu}$ you have $X_i = -X^i$.

9. May 6, 2016

### spaghetti3451

Thanks! In that case, I have

$\partial_{i}\partial^{i}C^{0}+m^{2}C^{0}=\partial_{i}{\dot{C}}^{i}$

$\implies \partial_{i}\partial^{i}C_{0}+m^{2}C_{0}=\partial^{i}{\dot{C}}_{i}$

Let me try to solve the remaining parts of the question.

Construct the canonical momenta $\Pi_{i}$ conjugate to $C_{i}, i=1,2,3,$ and show that the canonical momenta conjugate to $C_0$ is vanishing.

$\Pi_{i}=\frac{\partial \mathcal{L}}{\partial \dot{C}_{i}}=\frac{\partial}{\partial \dot{C}_{i}}\Big(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m^{2}C_{\mu}C^{\mu}\Big)$

$=\frac{\partial}{\partial \dot{C}_{i}}\Big(-\frac{1}{2}(\partial_{\mu}C_{\nu})^{2}+\frac{1}{2}(\partial_{\mu}C^{\mu})^{2}\Big)$, where the two Lagrangians are equivalent upto a total derivative and I left out $\frac{1}{2}m^{2}C_{\mu}C^{\mu}$

$=\frac{\partial}{\partial \dot{C}_{i}}\Big(-\frac{1}{2}(\partial_{0}C_{\nu})(\partial^{0}C^{\nu})+\frac{1}{2}(\partial_{0}C^{0}+\partial_{j}C^{j})(\partial_{0}C^{0}+\partial_{j}C^{j})\Big)$

$=\frac{\partial}{\partial \dot{C}_{i}}\Big(-\frac{1}{2}(\partial_{0}C_{j})(\partial^{0}C^{j})+(\partial_{0}C^{0})(\partial_{j}C^{j})\Big)$

$=(\partial_{0}C_{i})+\delta_{0i}(\partial_{j}C^{j})$.

What do you think?