Variational principle & lorentz force law

  • Thread starter Brian-san
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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:

Answers and Replies

  • #2
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4
Perhaps the problem is this:

On the right hand side you have [tex]u^\nu[/tex]
On the left hand side [tex]u_\mu[/tex]

The relation between the two is given by [tex]\eta_{\mu\nu}[/tex] and that depends on your signature convention. This signature is expressed in the sign of the first term of your action principle.
 
  • #3
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I checked my notes and it's supposed to come out to
[tex]\frac{du_\mu}{ds}=\frac{q}{m}F_{\mu\nu}u^\nu[/tex]

We've been using the convention that the spatial terms have the minus signs in the Minkowski metric (+---). In an earlier step I used [itex]u_\mu=\eta_{\mu\nu}u^\nu[/itex] to get the lower mu index in the first term with the mass. I can't find anything that says how the signature affects the process of raising/lowering indices though.
 
  • #4
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4
Question is why do you have minus sign in front of the second (or plus in front of the first) term in:

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

Landau and Lifgarbagez has this action

em.jpg


The first term is [tex]\sqrt{\eta_{\mu\nu}u^{\mu}u^{\nu}}\,ds[/tex] but it should not matter.
 
Last edited:

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