- #1

- 45

- 0

## Homework Statement

Show that the Lorentz force law follows from the following variational principle:

[tex]S=\frac{m}{2}\int\eta_{\mu\nu}u^\mu u^\nu ds-q\int A_\mu u^\mu ds[/tex]

## Homework Equations

Definition of Field Strength Tensor

Integration by Parts

Chain Rule & Product Rule for Derivatives

## The Attempt at a Solution

In order to find the force equation, we need to vary the action and find when [itex]\delta S=0[/itex].

So, we begin by varying the action integral,

[tex]\delta S=\frac{m}{2}\int\eta_{\mu\nu}\delta(u^\mu u^\nu)ds-q\int\delta(A_\mu u^\mu)ds[/tex]

Then apply product rule to expand all the terms,

[tex]\delta S=m\int\eta_{\mu\nu}u^\nu\frac{d(\delta x^\mu)}{ds}ds-q\int\left(\partial_\nu A_\mu u^\mu\delta x^\nu+A_\mu\frac{d(\delta x^\mu)}{ds}\right)ds[/tex]

Simplified and renamed indices in the second term,

[tex]\delta S=m\int u_\mu\frac{d(\delta x^\mu)}{ds}ds-q\int\left(\partial_\mu A_\nu u^\nu\delta x^\mu+A_\mu\frac{d(\delta x^\mu)}{ds}\right)ds[/tex]

Reverse integration by parts to get the variation out of the derivatives,

[tex]\delta S=m\int\left(\frac{d}{ds}\left(u_\mu\delta x^\mu\right)-\frac{du_\mu}{ds}\delta x^\mu\right)ds-q\int\left(\partial_\mu A_\nu u^\nu\delta x^\mu+\frac{d}{ds}\left( A_\mu\delta x^\mu\right)-\frac{dA_\mu}{ds}\delta x^\mu\right)ds[/tex]

Grouping like terms we have,

[tex]\delta S=\int\left(\frac{d}{ds}\left(mu_\mu\delta x^\mu-qA_\mu\delta x^\mu\right)\right)ds+\int \left( q\left(\frac{dA_\mu}{ds}\delta x^\mu-\partial_\mu A_\nu u^\nu\delta x^\mu\right)-m\frac{du_\mu}{ds}\delta x^\mu\right)ds[/tex]

After integration, the first term should be zero since the variations must vanish at the endpoints. Also, using the chain rule on the dA/ds derivative gives me

[tex]\delta S=\int \left( q\left(\partial_\nu A_\mu u^\nu-\partial_\mu A_\nu u^\nu\right)-m\frac{du_\mu}{ds}\right)\delta x^\mu ds[/tex]

Rearranged terms to simplify further,

[tex]\delta S=\int \left( -q\left(\partial_\mu A_\nu-\partial_\nu A_\mu\right)u^\nu -m\frac{du_\mu}{ds}\right)\delta x^\mu ds[/tex]

Then I use the definition of the field strength tensor to get,

[tex]\delta S=\int \left( -qF_{\mu\nu}u^\nu -m\frac{du_\mu}{ds}\right)\delta x^\mu ds[/tex]

Then [itex]\delta S=0[/itex] for any arbitrary variation [itex]\delta x^\mu[/itex] only if the integrand itself is zero, so

[tex]qF_{\mu\nu}u^\nu+m\frac{du_\mu}{ds}=0[/tex]

Rearranging the expression gives

[tex]\frac{du_\mu}{ds}=-\frac{q}{m}F_{\mu\nu}u^\nu[/tex]

This is the correct form for the Lorentz force law, however I have an additional minus sign. It's probably a trivial problem, but I've double checked all the work several times and I can't figure out where it's coming from.

Last edited: