# Finding the equation of motion of a given Lagrangian

1. Jan 13, 2016

### spaghetti3451

1. The problem statement, all variables and given/known data

Given the Lagrangian $\mathcal{L} = \frac{1}{2}(\partial_{\mu}\Phi)^{2}+\frac{1}{2}\Phi^{2}-\frac{1}{2}\Phi^{3}+\frac{\alpha}{8}\Phi^{4}$, where $\Phi=\Phi(x)$, find the equation of motion of the system. Assume that the field $\Phi$ is spherically symmetric, i.e. $\Phi = \Phi(r)$.

2. Relevant equations

The Euler-Lagrange equation of motion for a scalar field

3. The attempt at a solution

$S = \int d^{4}x \bigg(\frac{1}{2}(\partial_{\mu}\Phi)^{2}+\frac{1}{2}\Phi^{2}-\frac{1}{2}\Phi^{3}+\frac{\alpha}{8}\Phi^{4}\bigg)$

The spherical volume element in $4-$dimensional Euclidean space is $d^{4}x = r^{3}\ \text{sin}^{2}(\phi_{1})\ \text{sin}(\phi_{2})\ dr\ d\phi_{1}\ d\phi_{2}\ d\phi_{3}$.

Therefore, $S = \int d^{4}x \bigg(\frac{1}{2}(\partial_{\mu}\Phi)^{2}+\frac{1}{2}\Phi^{2}-\frac{1}{2}\Phi^{3}+\frac{\alpha}{8}\Phi^{4}\bigg)$

$=\int_{0}^{2\pi}d\phi_{3}\ \int_{0}^{\pi}d\phi_{2}\ \text{sin}(\phi_{2}) \int_{0}^{\pi}d\phi_{1}\ \text{sin}^{2}(\phi_{1}) \int_{0}^{r}r'^{3}\ dr'\ \bigg(\frac{1}{2}(\partial_{r'}\Phi)^{2}+\frac{1}{2}\Phi^{2}-\frac{1}{2}\Phi^{3}+\frac{\alpha}{8}\Phi^{4}\bigg)$

$=(2\pi)(2)\bigg(\frac{\pi}{2}\bigg)\int_{0}^{r}r'^{3}\ dr'\ \bigg(\frac{1}{2}(\partial_{r'}\Phi)^{2}+\frac{1}{2}\Phi^{2}-\frac{1}{2}\Phi^{3}+\frac{\alpha}{8}\Phi^{4}\bigg)$.

Am I correct so far?

2. Jan 13, 2016

### nrqed

Why do you bother with the volume element? Just apply the E-L equation to the Lagrangian density an you will directly get the equation of motion.

3. Jan 13, 2016

### spaghetti3451

If I don't bother with the volume element, won't the crucial factor of $r'^{2}$ in the Lagrangian density be missing?

4. Jan 14, 2016

### nrqed

Hi,

The least action principle leads to the Euler-Lagrange equations which are conditions to what is integrated over all space. So there is no need to do anything with the volume element, $d^4x$.

The E-L equations are

$\frac{\partial {\cal{L}}}{\partial \phi} - \partial_\mu \frac{\partial {\cal{ L}}}{\partial (\partial_\mu \phi)} = 0$

5. Jan 14, 2016

### spaghetti3451

I understand that.

But wouldn't the equation of motion take a simpler form in spherical polar coordinates than in Cartesian coordinates, due to the radial symmetry of the potential?

6. Jan 21, 2016

### kunyao

Yes, the equation of motion will take a simpler form in spherical polar coordinates. But as nrqed said, you should first derive the Euler-Lagrange equation and then impose the spherical symmetry (actually this just means to replace the d'Alembert operator by a double derivative with respect to "r").