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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?

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    $\begingroup$ Invariant under what? $\endgroup$
    – J. Murray
    Apr 3 '20 at 0:45
  • $\begingroup$ 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). $\endgroup$
    – StephenG
    Apr 3 '20 at 0:46
  • $\begingroup$ Invariant under general coordinate transformations $\endgroup$ Apr 3 '20 at 0:54
  • $\begingroup$ Related physics.stackexchange.com/q/506259 $\endgroup$
    – SRS
    Apr 3 '20 at 14:26
  • $\begingroup$ Covariance under Lorentz transformations and general coordinate transformations are not the same. $\endgroup$
    – Qmechanic
    Apr 3 '20 at 14:52
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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:

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

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