# Two similar proofs that $\frac{\partial}{\partial x'^\mu}={\Lambda_\mu}^\nu\frac{\partial}{\partial x^\nu}$, which one is correct?

Using the chain rule, show that the derivative transforms as $$\frac{\partial}{\partial x^\mu}\to\frac{\partial}{\partial x'^\mu}={\Lambda_\mu}^\nu\frac{\partial}{\partial x^\nu}\tag{A}$$

This is the way I would do this, firstly, by the chain rule $$\frac{\partial}{\partial x'^\mu}=\frac{\partial x^\nu}{\partial x'^\mu}\frac{\partial }{\partial x^\nu}\tag{1}$$ and noting that contravariant vectors transform as $$x^\mu \to x'^\mu={\Lambda^\mu}_\nu x^\nu$$, then the inverse Lorentz transformation for contravariant vectors is $$x^\mu={\Lambda_\nu}^\mu x'^\nu$$. Relabelling $$\mu \to \nu$$ and $$\nu \to \mu$$, then this inverse Lorentz transform can be written as $$x^\nu={\Lambda_\mu}^\nu x'^\mu$$. Then $$\frac{\partial x^\nu}{\partial x'^\mu}=\frac{\partial\left({\Lambda_\mu}^\nu x'^\mu\right)}{\partial x'^\mu}\stackrel{\color{red}{?}}={\Lambda_\mu}^\nu\frac{\partial x'^\mu}{\partial x'^\mu}={\Lambda_\mu}^\nu\times 1\tag{2}$$ now by direct substitution into $$(1)$$, the result, $$(\mathrm{A})$$ follows immediately. I put a red question mark over the part of my proof for which I am unsure about. Since I am assuming the $$4\times 4$$ boost matrix, $${\Lambda_\mu}^\nu$$ is constant as far as the differentiation is concerned.

Now, the proof given by the author is almost identical to mine, (except for one small change which I don't fully understand the need for):

The chain rule gives $$\frac{\partial}{\partial x'^\mu}=\frac{\partial x^\nu}{\partial x'^\mu}\frac{\partial }{\partial x^\nu}$$ Using $$x^\nu={\Lambda_\rho}^\nu x'^\rho$$ $$\frac{\partial x^\nu}{\partial x'^\mu}=\frac{\partial}{\partial x'^\mu}{\Lambda_\rho}^\nu x'^\rho\quad{\color{blue}{=}}\quad{\Lambda_\rho}^\nu\delta_\mu^\rho={\Lambda_\mu}^\nu\tag{3}$$ So $$\frac{\partial}{\partial x'^\mu}={\Lambda_\mu}^\nu\frac{\partial}{\partial x^\nu}$$

Where I note that after blue equality in eqn. $$(3)$$ the author has used the fact that $$\frac{\partial x'^\rho}{\partial x'^\mu}=\delta_\mu^\rho \tag{4}$$

But why did the author need to use $$(4)$$? This makes me believe that the manipulation I used after the question-marked equality for eqn. $$(2)$$ is not valid. Put another way, is differentiating wrt. to a set of coordinates, $$x'^\mu$$, in a new (primed) frame legitimate?

• I apologize that the horizontal spacing between the indices for the Lorentz boost matrices is too large as I had to use \quad as I cannot seem to get \, and \; as described in the MathJax tutorial to render correctly. Commented Jan 18 at 15:36
• A better way to typeset indices of tensor expressions is like this: Lambda_\mu {}^\nu will render to $\Lambda_\mu {}^\nu$.
– jd27
Commented Jan 18 at 15:50

$$\frac{\partial x^\nu}{\partial x'^\mu}=\frac{\partial\left({\Lambda_\mu}^\nu x'^\mu\right)}{\partial x'^\mu}\stackrel{\color{red}{?}}={\Lambda_\mu}^\nu\frac{\partial x'^\mu}{\partial x'^\mu}={\Lambda_\mu}^\nu\times 1\tag{2}$$

Unfortunately, there are two mistakes in this equation.

First mistake, this expression $$\frac{\partial\left({\Lambda_\mu}^\nu x'^\mu\right)}{\partial x'^\mu}$$ is wrong, because the index $$\mu$$ appears three times. You should have changed $$\mu$$ to another letter, say, $$\rho$$. Thus, you would obtain $$\frac{\partial x^\nu}{\partial x'^\mu}=\frac{\partial\left({\Lambda_\rho}^\nu x'^\rho\right)}{\partial x'^\mu}={\Lambda_\rho}^\nu\frac{\partial x'^\rho}{\partial x'^\mu}$$

Second mistake, this expression $$\frac{\partial x'^\mu}{\partial x'^\mu}=1$$ is wrong. The correct result is $$\frac{\partial x'^\mu}{\partial x'^\mu}=\frac{\partial x'^0}{\partial x'^0}+\frac{\partial x'^1}{\partial x'^1}+\frac{\partial x'^2}{\partial x'^2}+\frac{\partial x'^3}{\partial x'^3}=4,$$ because the convention that repeated indices imply the summation is to be done.

But why did the author need to use $$(4)$$?

Because $$x'^0,x'^1,x'^2,x'^3$$ are independent variables. Thus, for example, $$\frac{\partial x'^0}{\partial x'^0}=1 \ ; \frac{\partial x'^0}{\partial x'^1}=0; \frac{\partial x'^0}{\partial x'^2}=0; \frac{\partial x'^0}{\partial x'^3}=0,$$ and so on. Therefore, $$\frac{\partial x'^\rho}{\partial x'^\mu}=\delta^\rho_\mu=\begin{cases}1, \text{if} \ \ \rho=\mu\\0, \text{if} \ \ \rho\ne\mu\end{cases}$$ is correct.

• Thanks for your answer, in the last expression for the case that $\rho=\mu$, did you mean $4$ instead of $1$? As $\delta_\mu^\mu=4$ due to the summation convention? Thank you also for editing my question, looks much better now. Commented Jan 18 at 23:59
• Thanks, I think I understand you, it's a subtlety, like an 'if....then' type logical statement where the order matters. So if $\rho=\mu$ then $\delta^\mu_\mu=4$, but if $\delta_\mu^\rho=1$ then $\rho=\mu$. That's my interpretation anyway, or maybe I'm just overthinking it. Commented Jan 19 at 0:55
• @SiriusBlack Let $M$ be a matrix $4\times 4$ whose entries are $\displaystyle\frac{\partial x'^\rho}{\partial x'^\mu}$ and $I$ be the identity matrix $4\times4$ whose entries are $\delta^\rho_\mu$. Then the equation $$\frac{\partial x'^\rho}{\partial x'^\mu}=\delta^\rho_\mu$$ means that $$M=I.$$ Commented Jan 19 at 2:04
• @SiriusBlack If $M=I$, then $\operatorname{Tr}(M)=\operatorname{Tr}(I)=4$, where $\operatorname{Tr}$ is the trace. This means that if $$\frac{\partial x'^\rho}{\partial x'^\mu}=\delta^\rho_\mu$$, then $$\frac{\partial x'^\mu}{\partial x'^\mu}=\delta^\mu_\mu=4.$$ Commented Jan 19 at 2:04
• That I will, I just would like to read this answer/comments a bit more first to try to understand it better if that's okay? Don't worry, I have never left a question I asked with an answer unaccepted. As this $\delta_\mu^\mu=1 \quad \text{or} \quad 4 \quad$ is quite confusing, as one could argue that the upper statement in the last expression of your answer; $\delta^\rho_\mu=1 \quad \text{if} \quad \rho=\mu\quad$ actually means we have $\delta_\mu^\mu$ and as you've said, that is equal to $4$ (and not $1$) due to the repeated index - Einstein summation convention. Commented Jan 19 at 10:39