Lorentz Force Law derivation using the Lagrangian

  • Thread starter Elvex
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  • #1
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* implies time derivative, _i and _j are indices.


Homework Statement


Consider a Particle of mass m and charge q acted on by electric and magnetic fields E(x,t) and B(x,t). these fields can be described in therms of the scalar and vector potentials PHI(x,t) and A(x,t) for which E= - gradPHI and B = CurlA.
Show that the Lagrangian equations of motion for this Lagrangian give back the Lorentz Force Law.


Homework Equations


L= (1/2)mx*^2 - qPHI + qx*.A

Lorentz Force Law: mx**= qE + qx* x B

Lagrange Equation of Motion: (d/dt)(dL/dx*)= dL/dx


The Attempt at a Solution



It was easy for me to derive the qE part of the force law, but I'm having problems with qx* x B.
First, I used einstein summation notation to express x* X Curl(A), a double cross product and I got two terms,

x*_i (dA_i / dx_j) e_j - x*_j (dA_i / dx_j) e_i

There's a chance this is wrong, I haven't used summation notation in a while.

Now when I take dL/dx for the term with A in the lagrangian. I only get that first term from the double cross product.

I'm not really sure what to do in taking the derivative of x* . A

would that be... x*. gradA ?

Basically, I don't know how to get the qx* x B term out of taking derivatives of the Lagrangian.
 

Answers and Replies

  • #2
dextercioby
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The problem's not too simple. You need to use the identity

[tex] \nabla (\overrightarrow{a}\cdot \overrightarrow{b})=(\overrightarrow{a}\cdot
\nabla )\overrightarrow{b}+(\overrightarrow{b}\cdot \nabla )\overrightarrow{a}+\overrightarrow{a}\times (\nabla \times \overrightarrow{b})+
\overrightarrow{b}\times (\nabla \times \overrightarrow{a}) [/tex]
 

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