# Lorentz Invariance of the Euler-Lagrange equation for fields

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?

• Invariant under what? Apr 3 '20 at 0:45
• 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). Apr 3 '20 at 0:46
• Invariant under general coordinate transformations Apr 3 '20 at 0:54
• – SRS
Apr 3 '20 at 14:26
• Covariance under Lorentz transformations and general coordinate transformations are not the same. Apr 3 '20 at 14:52

$$L$$ 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:

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