# I Covariance of Euler-Lagrange equations for fields

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1. Aug 4, 2017

### topolosaurus


I haven't found this problem solved around maybe because it's a simple thing that I'm not yet able to see.
I have studied a bit of calculus of variations and I have succesfully derived Euler equations for the case of particles (i.e. for curves $q: [ a , b ] \rightarrow \mathbb{R}^n$) when I have a lagrangian and action of the form:

$$S[q] = \int_a^b L(t,q,\dot{q}) \d t$$

This yields the classical Euler-Lagrange equations

$$0 = \parcial{L}{q^k}-\ddt\parcial{L}{\dot{q}^k}$$

I have been able to show that these equations share the same form for a change of coordinates in configuration space $q^\mu \rightarrow x^\mu$ which means that if E-L equations for $q$ holds then:

$$0 = \parcial{L}{x^k} - \ddt \parcial{L}{\dot{x}^k}$$
This also happens if $t$ is involved in the coordinate transformation. Now I want to check that this covariance holds also for Euler Lagrange equations of fields, in which case we have:

• $q^\mu, x^\mu$ are now space-time coordinates
• $\phi$ is the scalar field involved

Now the action over the field is

$$S[\phi] = \int_\Omega \mathcal{L}(x,\phi,\dep{\mu}\phi) \d^4 x$$

The derived Euler Lagrange equations are:

$$0 = \parcial{\mathcal{L}}{\phi} - \dep{\mu}\left(\parcial{\mathcal{L}}{(\dep{\mu}\phi)}\right)$$
For convenience I shall write out explicitly partial derivatives

$$0 = \parcial{\mathcal{L}}{\phi} - \parcial{}{x^\mu} \left( \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)} \right)$$

Now consider a space-time transformation $x^\mu \rightarrow q^\mu$ which in principle is not Lorentz transformation. The transformation of the derivatives of the field follow simply from the chain rule:
$$\parcial{\phi}{x^\mu} = \parcial{\phi}{q^\nu} \parcial{q^\nu}{x^\mu}$$
Hence:
$$\parcial{}{\left(\parcial{\phi}{q^\sigma}\right)}\left(\mathcal{L}(\phi,\parcial{\phi}{x^\mu}) \right) = \parcial{}{\left( \parcial{\phi}{q^\sigma}\right)} \left(\mathcal{L}(\phi,\parcial{\phi}{q^\nu}\parcial{q^\nu}{x^\mu}) \right) = \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)} \parcial{}{\left(\parcial{\phi}{q^\sigma}\right)} \left( \parcial{\phi}{q^\nu}\parcial{q^\nu}{x^\mu} \right) = \parcial{\mathcal{L}}{\left(\parcial{\phi}{x^\mu}\right)} \parcial{q^\sigma}{x^\mu}$$
This yields:
\begin{aligned} \parcial{}{q^\sigma} \left( \parcial{\mathcal{L}}{\left( \parcial{\phi}{q^\sigma}\right)} \right) && = && \parcial{}{q^\sigma} \left( \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)} \right) \parcial{q^\sigma}{x^\mu} + \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)} \parcial{}{x^\alpha} \left( \parcial{q^\sigma}{x^\mu} \right) \parcial{x^\alpha}{q^\sigma}\\ && = && \parcial{}{x^\alpha} \left( \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)} \right) \parcial{q^\sigma}{x^\mu} \parcial{x^\alpha}{q^\sigma} + \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)} \ppartial{q^\sigma}{x^\alpha}{x^\mu} \parcial{x^\alpha}{q^\sigma}\\ && = && \parcial{}{x^\alpha} \left( \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)} \right) \delta_\mu^\alpha + \parcial{\mathcal{L}}{\left( \parcial{\phi}{x^\mu}\right)} \ppartial{q^\sigma}{x^\alpha}{x^\mu} \parcial{x^\alpha}{q^\sigma}\\ \end{aligned}
This and the fact that $\parcial{\mathcal{L}}{\phi}$ is independent of coordinates yields that the latter extra term in Euler Lagrange equations for $q$ coordinates. Moreover, I tend not to trust second derivatives expressions of the coordinates to be covariant from differential geometry (I may be wrong for a general case though). Any ideas on how to continue from this point or possible mistakes?

One possible thing I could be stepping on is the fact that lagrangian is a density and maybe I have to consider the jacobian of the transformation as a weight. However this seems a very big complication to treat a 4x4 determinant expression and could not be generalized (or at least have an equal treatment) for higher dimensions.

2. Aug 9, 2017

### PF_Help_Bot

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