# Covariance of Euler-Lagrange equations for fields

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• topolosaurus
In summary, the Euler-Lagrange equations for fields are not derived from a variational principle like for particles, but from the principle of least action. However, they still possess the important property of covariance under coordinate transformations, which is necessary to preserve their physical meaning. This can be shown by considering the transformation of the action functional, which involves the Jacobian of the coordinate transformation.
topolosaurus
$$\newcommand{\dep}[1]{\partial_{#1}} \newcommand{\parcial}[2]{\frac{\partial{#1}}{\partial{#2}}} \renewcommand{\d}{\text{d}} \newcommand{\ddt}{\frac{\text{d}}{\text{d}t}} \newcommand{\ppartial}[3]{\frac{\partial^2{#1}}{\partial{#2}\partial{#2}}}$$

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.

First of all, it is important to note that the Euler-Lagrange equations for fields are not derived from a variational principle in the same way as for particles. Instead, they are derived from the principle of least action, which states that the action functional S[\phi] is stationary (not necessarily minimal) under small variations of the field \phi. Therefore, the equations you have derived for fields are not the classical Euler-Lagrange equations, but rather the field equations.

That being said, the covariance of the field equations under coordinate transformations is an important property and can be seen as a consequence of the principle of general covariance in general relativity. In fact, the field equations are required to be covariant under any coordinate transformation in order to preserve the physical meaning of the equations.

To show this, one can follow a similar approach as you have done for the particle case. However, instead of considering the transformation of the field derivatives, one must consider the transformation of the action functional S[\phi] itself. This can be done using the chain rule and the transformation properties of the field, and will yield the desired result that the field equations are covariant under coordinate transformations.

It is true that the presence of the Jacobian in the transformation of the action functional can complicate things, but it is necessary in order to preserve the physical meaning of the equations. In fact, this is a fundamental concept in differential geometry, where the transformation of tensors involves the Jacobian of the coordinate transformation.

In conclusion, the covariance of the field equations under coordinate transformations is an important property that must be considered when dealing with field theories. It can be shown that the equations are indeed covariant, but it requires a different approach than for the particle case.

## 1. What is the covariance of Euler-Lagrange equations for fields?

The covariance of Euler-Lagrange equations for fields refers to the invariance of these equations under a change of coordinates. This means that the equations will have the same form and solutions regardless of the coordinate system used.

## 2. Why is covariance important in the context of Euler-Lagrange equations for fields?

Covariance is important because it ensures that the equations are consistent and applicable in any coordinate system, making them more general and applicable to a wider range of physical systems.

## 3. How does covariance impact the solutions of Euler-Lagrange equations for fields?

Covariance does not affect the solutions of the equations themselves, but it allows for the same solutions to be found in different coordinate systems. This allows for a deeper understanding of the underlying physical principles and relationships.

## 4. What is the relationship between covariance and symmetry in Euler-Lagrange equations for fields?

Covariance and symmetry are closely related in the context of Euler-Lagrange equations for fields. Symmetry refers to the invariance of a physical system under certain transformations, and covariance ensures that the equations governing this system are also invariant under these transformations.

## 5. How is covariance applied in practical applications of Euler-Lagrange equations for fields?

In practical applications, covariance is used to simplify and generalize the equations by choosing an appropriate coordinate system that makes the problem easier to solve. It also allows for the equations to be applied in a wider range of physical systems with different coordinate systems.

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