2

For instance, when differentiating four-vectors the result is straightforward: $$\frac{\partial x^\mu}{\partial x^\nu}=\delta_\nu^\mu$$ as the derivative is only non-zero when the Lorentz indices match. Here $\mu, \nu = 0,1,2,3$.

But when differentiating type $(0,2)$ tensors, for example, $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_c^\mu\delta_d^\nu\tag{1}$$

I note that the order of the indices in $(1)$ (going from left to right) is preserved; $\mu$ is 'paired' with $\mathrm{c}$, and $\nu$ is paired with $\mathrm{d}$.

My question is very simple, am I allowed to commute the two contravariant or covariant derivatives in $(1)$?

Put simply, is it true that $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_c^\mu\delta_d^\nu=\delta_c^\color{red}{\nu}\delta_d^\color{red}{\mu}=\delta_\color{blue}{d}^\mu\delta_\color{blue}{c}^\nu?$$

My reason for asking is that since the condition for $(1)$ to be non-zero is that all the Lorentz indices must be equal, then what is to stop me from interchanging two of the contravariant (or covariant) indices?

Update:

After writing this post another question came to mind, namely, why do we need a product of Kronecker deltas and why not write $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_{cd}^{\mu\nu}$$ instead of eqn. $(1)$? I had this idea after looking at this Wikipedia page on the generalized Kronecker delta.

  • 1
    Consider a special case, say, $\mu = c = 1$, $\nu = d = 2$. Then $\delta_c^\mu \delta_d^\nu = 1$ but $\delta_c^\nu \delta_d^\mu = 0$. Hence they cannot possibly be equal in general. – ummg Feb 28 '24 at 20:19
  • @ummg Thanks for the confirmation, why then is $(1)$ written the way it is? What I mean by that is, while I now acknowledge that generally $\delta_c^\mu\delta_d^\nu\ne\delta_c^\color{red}{\nu}\delta_d^\color{red}{\mu}\ne\delta_\color{blue}{d}^\mu\delta_\color{blue}{c}^\nu$, why not write $\dfrac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_c^\nu\delta_d^\color{red}{\mu}$ instead? In other words, does $\mu$ have to be 'paired' with $\mathrm{c}$, and $\nu$ paired with $\mathrm{d}$? –  Feb 28 '24 at 20:27
  • In terms of Iverson brackets, the derivative is obviously $[c=\mu\land d=\nu]=[c=\mu][d=\nu]$, which you can rewrite with Kronecker deltas. – J.G. Feb 28 '24 at 20:53
  • @J.G. Hi there, I'm sorry that answer was a little too technical for my level. A simple yes or no to is $\dfrac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_c^\nu\delta_d^\color{red}{\mu}$ true would be great? Also, this may seem like a stupid question but I heard the generalized Kronecker delta can be written in any number of dimensions so my last question is why not write $\dfrac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_{cd}^{\nu\mu}$? Many thanks! –  Feb 28 '24 at 21:14
  • The GKD assumes your problem has certain antisymmetries this one doesn't. When we say $\frac{\partial x^\mu}{\partial x^\nu}=\delta^\mu_\nu$ we mean $x^\mu$ doesn't change in response to $x^\nu$ changing unless $\mu=\nu$. Similarly, $\partial_cA_d$ doesn't change because of $\partial_\mu A_\nu$ unless $c=\mu\land d=\nu$. – J.G. Feb 28 '24 at 21:22
  • @J.G. Hi there, I'm sorry I still don't understand. Is $\dfrac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_c^\nu\delta_d^\color{red}{\mu}$? Only a yes or no is required. Moreover, Why is $\dfrac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}$ not being written as $\delta_{cd}^{\nu\mu}$? I don't understand statements like $c=\mu\land d=\nu$, or GKD etc. –  Feb 28 '24 at 21:49
  • 1
    $\land$ means and, and GKD is generalized Kronecker delta. And no, the red version isn't right. You definitely need a $\delta_c^\mu$ factor. Your proposed notation is defined to be antisymmetric for concision when that's needed, but for this you just multiply two Kronecker deltas. – J.G. Feb 28 '24 at 22:20
  • @digital Can you say something about the context where you have encountered these derivatives? Are you working with a Lagrangian? In that case I might be able to give some more insight. – ummg Feb 29 '24 at 20:42
  • @ummg Hi, sorry for the slow reply, not been here for a while, the context is in computing Euler-Lagrange equations to get the equation of motion for the Maxwell Lagrangian, $\mathcal{L}=-\frac14F^{\mu\nu}F_{\mu\nu}-j^\mu A_\mu$, where $F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$, $j^\mu$ is the external current and $A^\mu$ is the four-potential. –  Mar 07 '24 at 22:38

1 Answers1

0

Suppose for a contradiction that $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_c^\color{red}{\nu}\delta_d^\color{red}{\mu}.$$ Take $c=\mu=0$ and $d=\nu=2$ in such equation. Then we have $$\frac{\partial\left(\partial_0A_2\right)}{\partial\left(\partial_0 A_2\right)}=\delta_0^\color{red}{2}\delta_2^\color{red}{0}=0.$$ This is a contradiction, because $$\frac{\partial\left(\partial_0A_2\right)}{\partial\left(\partial_0 A_2\right)}=1.$$ Therefore, $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}\neq\delta_c^\color{red}{\nu}\delta_d^\color{red}{\mu}.$$ Now, suppose for a contradiction that $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_\color{blue}{d}^\mu\delta_\color{blue}{c}^\nu.$$ Take $c=\mu=0$ and $d=\nu=2$ in such equation. Then we have $$\frac{\partial\left(\partial_0A_2\right)}{\partial\left(\partial_0 A_2\right)}=\delta_2^\color{blue}{0}\delta_0^\color{blue}{2}=0.$$ This is also a contradiction, because $$\frac{\partial\left(\partial_0A_2\right)}{\partial\left(\partial_0 A_2\right)}=1.$$ Therefore, $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}\neq\delta_\color{blue}{d}^\mu\delta_\color{blue}{c}^\nu.$$ It is worth point out that $\delta_c^\color{red}{\nu}\delta_d^\color{red}{\mu}=\delta_\color{blue}{d}^\mu\delta_\color{blue}{c}^\nu$. However, $\delta_{c}^\mu\delta_{d}^\nu\neq\delta_c^\color{red}{\nu}\delta_d^\color{red}{\mu}$.

In conclusion, we must have $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_c^\mu\delta_d^\nu.$$

QuantumSuperfield
  • 2,751
  • Very good, thanks. So the reason why $\mu$ is 'paired' with $\mathrm{c}$, and $\nu$ is paired with $\mathrm{d}$ is something do with the $\mathrm{c}$ and $\mu$ being indices that come first - before the $\mathrm{d}$ and $\nu$ in $\dfrac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}$? –  Feb 28 '24 at 22:42
  • 1
    No. Note that \begin{equation} \frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\begin{cases} 1 , \text{if} \ c=\mu \ \text{and} \ d=\nu\ 0 , \text{all the other cases} \end{cases} \end{equation} and \begin{equation} \delta_c^\mu\delta_d^\nu=\begin{cases} 1 , \text{if} \ c=\mu \ \text{and} \ d=\nu\ 0 , \text{all the other cases} \end{cases} \end{equation} Therefore, $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_c^\mu\delta_d^\nu.$$ – QuantumSuperfield Feb 28 '24 at 23:00