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In section 6.2 (page 128) of David Tong's Lectures on QFT, Gauss' law is derived for the free Maxwell theory. The result of computing the Hamiltonian of the theory is (eq. 6.17),

$$H = \int d^{3} x \frac{1}{2} \vec{E} \cdot \vec{E}+\frac{1}{2} \vec{B} \cdot \vec{B}-A_{0}(\nabla \cdot \vec{E}).\tag{6.17}$$

I am comfortable reproducing this computation, but I do not understand what is meant by the claim:

"So $A_0$ acts as a Lagrange multiplier which imposes Gauss' law".

Earlier in eq. (6.10), it is established that $A_0$ is not an independent physical degree of freedom, and I'm assuming this fact is somehow operative in the Gauss's law statement. It's not clear to me though.

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    $\begingroup$ Going to Hamiltonian formalism, you actually eliminate the $A_0$ component of the field. However, being fixed to some value it has to still obey the equations of motion. The equation of motion, obtained from variation of $A_0$ is : $\partial_i F^{0 i} = 0 \Rightarrow \partial_i E_i = 0$, which is a Gauss law. $\endgroup$ May 31, 2020 at 8:52
  • $\begingroup$ the Gauss law is not derived on p128 from 6.17. the gauss law is a consequence of the equations of motion as it is written earlier in 6.8. so the constraint is a secondary constraint. $\endgroup$
    – Naima
    Jun 23, 2020 at 6:09

1 Answer 1

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  1. The Hamiltonian formulation$^1$ (6.17) has 7 independent fields $A_0$, $\vec{A}$ & $\vec{E}$. In particular $A_0$ acts as a Lagrange multiplier in the corresponding (Hamiltonian) Lagrangian.

  2. In eq. (6.10) Tong apply an EOM to detemine $A_0$. This does not contradict the fact that $A_0$ is an independent field prior to the use of EOMs.

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$^1$ Tong is here discussing the situation prior to any gauge-fixing. Note that in the Hamiltonian formulation (6.17) the momentum $E^0$ for $A_0$ has been eliminated by a primary constraint $E^0\approx 0$.

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