Question about velocity-dependent Lagrangian involving magnetic fields

  • Thread starter anton01
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Homework Statement


The Lagrange method does work for some velocity dependent Lagrangian. A very important
case is a charged particle moving in a magnetic field. The magnetic field can be represented as a "curl" of a vector potential ∇B = ∇xA . A uniform magnetic field B0 corresponds to a
vector potential ∇A = 1/2 B0 x r.

(a) Check that B0 = ∇xA
(b) From the Lagrangian
[itex]\frac{1}{2}mv^{2}+e\overline{v}.\overline{A}[/itex]
show that the EOM derived are identical to the classical Newton's law with the Lorentz
force F = ev x B .

Homework Equations


Euler-Lagrange equations:
[itex]\frac{d}{dt}(\frac{\partial L}{\partial \dot{s_{j}}})[/itex]=[itex]\frac{\partial L}{\partial s_{j}}[/itex]

Triple product:
a.(b x c)=b.(c x a)
a x (b x c)=(a.c)b - (a.b)c


The Attempt at a Solution



For a, I have tried using the triple product:
∇ x A = 1/2 ∇x(Bo x r) = 1/2 [Bo(∇.r)-(∇.Bo)r]
Since Bo is uniform, its divergence is zero and so:
∇ x A =1/2 Bo(∇.r)
From here, I guess ∇.r=2 for the proof to work, but I really don't see it.

For part b, I am not even sure where to start it. As in how can I apply the E-L equations to it? What scares me the most is how exactly do I take the derivatives on a Lagrangian involving vectors.

Any help is welcome, thank you!
 

Answers and Replies

  • #2
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(a) Looks like you have ∇.B where you should have B.∇ in your vector identity.

(b) The Lagrangian is still a scalar - it's just that it's been written with one piece that's a scalar product of two vectors. So just expand out that product and do Euler-Lagrange stuff as usual.
 

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