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Equations of motion and Hamiltonian density of a massive vector field

  • #1
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Homework Statement



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}##.)

Homework Equations



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?
 

Answers and Replies

  • #2
CAF123
Gold Member
2,894
88
Yes, it's all fine.
 
  • #3
1,344
32
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
CAF123
Gold Member
2,894
88
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##.
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
1,344
32
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.
##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
CAF123
Gold Member
2,894
88
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
1,344
32
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
CAF123
Gold Member
2,894
88
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
1,344
32
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##.
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?
 

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