Lorentz Invariance of the Euler-Lagrange equation for fields

Question

Given an Lorentz invariant Lagrangian density $L$ of a Lorentz invariant scalar field $\phi$,

How does one show that the following term in the Euler-Lagrange equation is invariant under Lorentz transformations?

$$\frac{\partial }{\partial x^{\mu}}\left(\frac{\partial L }{\partial \left(\frac{\partial \phi}{\partial x^{\mu}}\right)}\right) \tag{1}$$

I was thinking something like:

$$\frac{\partial x'^{\nu} }{\partial x^{\mu}}\frac{\partial }{\partial x'^{\nu}}\left(\frac{\partial L }{\partial \left(\frac{\partial x'^{\nu} }{\partial x^{\mu}}\frac{\partial \phi}{\partial x'^{\nu}}\right)}\right) \tag{2}$$

But I'm stuck...

Maybe there's some way of applying the chain rule to that derivative of the Lagrangian density?

Note you can get larger nested brackets in Mathjax using the \left and \right modifiers on bracket pairs (you have to match left and right pairs). — StephenG - Help Ukraine, Apr 03 '20 at 00:46
Covariance under Lorentz transformations and general coordinate transformations are not the same. — Qmechanic, Apr 03 '20 at 14:52
Related: https://physics.stackexchange.com/q/144389/2451 , https://physics.stackexchange.com/q/118768/2451 — Qmechanic, Apr 03 '20 at 14:54

score -1 · Accepted Answer · answered Apr 03 '20 at 18:39

Nevermind, I've got it already:

$L$ ad $\phi$ are invariants so,

$L\left(\phi(x),\frac{\partial \phi(x)}{\partial x},g(x)\right)=L'\left(\phi'(x'),\frac{\partial \phi'(x')}{\partial x'},g'(x')\right)\tag1$

where $g$ is the metric tensor, and

$\phi(x)=\phi'(x')\tag2$

The derivative of a scalar transforms as a rank 1 covariant tensor:

$\frac{\partial \phi}{\partial x^{\mu}}=\frac{\partial \phi'}{\partial x'^{\nu}}\frac{\partial x'^{\nu}}{\partial x^{\mu}}\tag3$

$\frac{\partial \phi'}{\partial x'^{\nu}}\left(\frac{\partial \phi}{\partial x} \right )=\frac{\partial \phi}{\partial x^{\mu}}\frac{\partial x^{\mu}}{\partial x'^{\nu}}\tag4$

We can show that $\frac{\partial L }{\partial \left(\frac{\partial \phi}{\partial x^{\mu}}\right)}$ transform as a rank 1 contravariant tensor as follows,

Using (1), and the fact that the derivatives of $\phi'$ w.r.t. the $x'$ can be written as a function of the derivatives of $\phi$ w.r.t. the $x$ as shown in (4):

$\frac{\partial L\left(\phi,\frac{\partial \phi}{\partial x},g\right)}{\partial \left( \frac{\partial\phi }{\partial x^{\mu}}\right )}= \frac{\partial L'\left(\phi',\frac{\partial \phi'}{\partial x'}(\frac{\partial \phi}{\partial x}),g'\right)}{\partial \left( \frac{\partial\phi }{\partial x^{\mu}}\right )}\tag5$

Chain Rule on the last term, then use (4):

$\frac{\partial L'}{\partial \left( \frac{\partial\phi '}{\partial x'^{\nu}}\right )} \frac{\partial (\frac{\partial \phi'}{\partial x'^{\nu}})}{\partial (\frac{\partial \phi}{\partial x^{\mu}})}=\frac{\partial L'}{\partial \left( \frac{\partial\phi '}{\partial x'^{\nu}}\right )} \frac{\partial (\frac{\partial \phi}{\partial x^{\gamma}}\frac{\partial x^{\gamma }}{\partial x'^{\nu}})}{\partial (\frac{\partial \phi}{\partial x^{\mu}})}\tag6$

And Finally this last term is:

$\frac{\partial L'}{\partial \left( \frac{\partial\phi '}{\partial x'^{\nu}}\right )} \frac{\partial x^{\mu}}{\partial x'^{\nu}}\tag7$

Lorentz Invariance of the Euler-Lagrange equation for fields

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