- #1

topolosaurus

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\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}}}

[/tex]

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 [itex]q: [ a , b ] \rightarrow \mathbb{R}^n[/itex]) when I have a lagrangian and action of the form:

[tex]

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

[/tex]This yields the classical Euler-Lagrange equations[tex]

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

[/tex]

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

[tex]

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

[/tex]

This also happens if [itex]t[/itex] 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:

- [itex]q^\mu, x^\mu[/itex] are now space-time coordinates

- [itex]\phi[/itex] is the scalar field involved

Now the action over the field is

[tex]

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

[/tex]

The derived Euler Lagrange equations are:[tex]

0 = \parcial{\mathcal{L}}{\phi} - \dep{\mu}\left(\parcial{\mathcal{L}}{(\dep{\mu}\phi)}\right)

[/tex]

For convenience I shall write out explicitly partial derivatives

[tex]

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

[/tex]

Now consider a space-time transformation [itex]x^\mu \rightarrow q^\mu[/itex] which in principle is not Lorentz transformation. The transformation of the derivatives of the field follow simply from the chain rule:

[tex]

\parcial{\phi}{x^\mu} = \parcial{\phi}{q^\nu} \parcial{q^\nu}{x^\mu}

[/tex]

Hence:

[tex]

\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}

[/tex]

This yields:

[tex]

\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}

[/tex]

This and the fact that [itex]\parcial{\mathcal{L}}{\phi}[/itex] is independent of coordinates yields that the latter extra term in Euler Lagrange equations for [itex]q[/itex] 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.