Gauge transformations are known to be canonical transformations, see here and here. Using direct conditions to check if they are, however, imposes a constraint on the gauge parameter $\Lambda(t,\mathbf{q},\mathbf{Q})$ that does not appear elsewhere. See below.

A coordinate transform $(\mathbf{q},\mathbf{p},t) \to (\mathbf{Q},\mathbf{P},t)$ is canonical if it satisfies the direct conditions: \begin{equation} \left( \frac{\partial P_m}{\partial p_n} \right)_{\mathbf{q,p}} = \left( \frac{\partial q_n}{\partial Q_m} \right)_{\mathbf{Q,P}} \end{equation} \begin{equation} \left( \frac{\partial P_m}{\partial q_n} \right)_{\mathbf{q,p}} = -\left( \frac{\partial p_n}{\partial Q_m} \right)_{\mathbf{Q,P}} \end{equation}

The gauge transformation is $$t\to t,$$ $$\mathbf{q} \to \mathbf{Q},$$ $$\mathbf{p} \to \mathbf{p} + \frac{q}{c} \nabla \Lambda,$$ for charge $q$ and speed of light $c$.

I've listed the relevant derivatives below:

\begin{equation} \left( \frac{\partial P}{\partial p} \right)_{\mathbf{q,p}} = \frac{\partial }{\partial p} (p+\frac{q}{c} \nabla \Lambda) \big{|}_{\mathbf{q,p}} = 1 \end{equation} \begin{equation} \left( \frac{\partial q}{\partial Q} \right)_{\mathbf{Q,P}} = 1 \end{equation} \begin{equation} \left( \frac{\partial p}{\partial Q} \right)_{\mathbf{Q,P}} = 0 \end{equation} \begin{equation} \left( \frac{\partial P}{\partial q} \right)_{\mathbf{q,p}} = \frac{\partial p}{\partial q} + \frac{q}{c} \frac{\partial \nabla \Lambda}{\partial q} = \frac{q}{c} \frac{\partial \nabla \Lambda}{\partial q} \end{equation} The first direct condition is satisfied but the second implies that \begin{equation} \frac{\partial}{\partial q} \nabla \Lambda = 0. \end{equation} However, $\Lambda$ is supposed to be an arbitrary function of the coordinates.

  • $\begingroup$ Why do you say that $\left(\frac{\partial p}{\partial Q}\right)_{\mathbf Q,\mathbf P}=0$ ? $\endgroup$ – J. Murray Aug 26 at 20:40
  • $\begingroup$ Since $\frac{\partial p}{\partial Q} = \frac{\partial p}{\partial q} = 0$ because $p$ and $q$ are independent. $\endgroup$ – Chamindu Amarasinghe Aug 27 at 17:31
  • $\begingroup$ If that reasoning were true, would that not also imply that $\left(\frac{\partial P}{\partial q}\right)_{\mathbf q,\mathbf p}=0$? $\endgroup$ – J. Murray Aug 27 at 17:45
  • $\begingroup$ I don't think so since we cannot assume that $P$ and $Q$ (or $P$ and $q$ equivalently) are independent. $\endgroup$ – Chamindu Amarasinghe Aug 27 at 17:58

From your definition, $$Q^i(\mathbf q,\mathbf p, t) = q^i$$ $$P_i(\mathbf q, \mathbf p, t) =p_i +\frac{g}{c}\frac{\partial \Lambda(\mathbf q,t)}{\partial q^i}$$

This transformation can be immediately inverted to yield $$ q^i(\mathbf Q,\mathbf P, t) = Q^i$$ $$p_i(\mathbf Q,\mathbf P, t) = P_i - \frac{g}{c}\frac{\partial \Lambda(\mathbf{q}(\mathbf Q,\mathbf P, t),t)}{\partial Q_i}$$ $$ = P_i - \frac{g}{c}\frac{\partial \Lambda(\mathbf Q, t)}{\partial Q_i}$$

Your partial derivatives should follow immediately from there.


$$\frac{\partial P_i}{\partial q^j}= \frac{g}{c} \frac{\partial^2 \Lambda(\mathbf q, t)}{\partial q^i \partial q^j}$$


$$\frac{\partial p_i}{\partial Q^j} = -\frac{g}{c} \frac{\partial^2 \Lambda(\mathbf Q,t)}{\partial Q^i \partial Q^j}$$

You've mentioned several times in the comments that $\mathbf Q = \mathbf q$, so I want to make one additional point. In your original phase space, you label points with $\mathbf q$ and $\mathbf p$. These labels are independent of one another in the sense that there is no functional dependence between them - this should be obvious, as they're simply labels we give to points.

In the transformed phase space, you label points with $\mathbf Q$ and $\mathbf P$. These labels are also independent of one another for precisely the same reason.

However, the transformation takes some ($\mathbf q,\mathbf p$) in the original phase space and maps it to the corresponding ($\mathbf Q,\mathbf P$) in the transformed phase space. In this particular example, $\mathbf Q$ is simply assigned the numerical value of $\mathbf q$, while $\mathbf P$ is assigned the value $\mathbf p + \frac{g}{c}\frac{\partial \Lambda(\mathbf q,t)}{\partial \mathbf q}$.

You should be able to see that $\mathbf P$ is not independent of $\mathbf q$. $\mathbf P$ and $\mathbf Q$ are independent for a given system in the sense that your position does not determine your momentum. However, if you are using a different coordinate system than I am, then my momentum label at a particular time may depend on your position label (and vice-versa for the inverse transformation).

  • $\begingroup$ $\Lambda (\mathbf{Q},t) = \Lambda (\mathbf{q},t)$ is still a function of the coordinate $q$. I don't see how there won't be a constraint on $\Lambda$. $\endgroup$ – Chamindu Amarasinghe Aug 27 at 18:30
  • $\begingroup$ $\Lambda$ is a function which eats a vector and a scalar. You plug $\mathbf q$ and $t$ into it when transforming from $(\mathbf q,\mathbf p, t)\rightarrow (\mathbf Q,\mathbf P,t)$, and you plug $\mathbf Q$ and $t$ into it when transforming back. I'm not sure I understand your objection. $\endgroup$ – J. Murray Aug 27 at 19:11
  • $\begingroup$ Yes, but $\mathbf{q} = \mathbf{Q}$ so $\frac{\partial \Lambda}{\partial q} \neq 0$ which is why the second direct condition does not hold without the constraint that I mentioned. My objection is that the gauge transformation should be canonical without such a constraint. $\endgroup$ – Chamindu Amarasinghe Aug 27 at 19:18
  • $\begingroup$ @ChaminduAmarasinghe Does my edit help? $\endgroup$ – J. Murray Aug 27 at 19:27
  • $\begingroup$ @ChaminduAmarasinghe I added an additional clarification RE: independence of Q and P. $\endgroup$ – J. Murray Aug 27 at 19:42

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