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?

  • 1
    $\begingroup$ Invariant under what? $\endgroup$
    – J. Murray
    Commented Apr 3, 2020 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). $\endgroup$ Commented Apr 3, 2020 at 0:46
  • $\begingroup$ Invariant under general coordinate transformations $\endgroup$
    – Quanta
    Commented Apr 3, 2020 at 0:54
  • $\begingroup$ Related physics.stackexchange.com/q/506259 $\endgroup$
    – SRS
    Commented Apr 3, 2020 at 14:26
  • $\begingroup$ Covariance under Lorentz transformations and general coordinate transformations are not the same. $\endgroup$
    – Qmechanic
    Commented Apr 3, 2020 at 14:52

1 Answer 1


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


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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.