Sorry about the endless stream of questions about Lagrangians. I am actually beginning to detest them a bit;p(adsbygoogle = window.adsbygoogle || []).push({});

Anyway, if we have a Lagrangian in three dimensional space:

[tex]L=\frac{1}{2}m\dot{\vec{x}}^{2}+e\vec{A}.\dot{\vec{x}}[/tex]

where [tex]A_{i}=\epsilon_{ijk}B_{j}x_{k}[/tex] and B is just a constant (magnetic field).

The question is to find the equations of motion. To do this we can just use the Euler-Lagrange equations thus:

[tex]\frac{{\partial}L}{{\partial}q^{i}}-\frac{d}{dt}\frac{{\partial}L}{{\partial}{\dot{q}}^{i}}=0[/tex]

Now,

[tex]\frac{{\partial}L}{{\partial}x^{i}}=0[/tex]

for all i.

So the EOM are given by

[tex]\frac{d}{dt}\frac{{\partial}L}{{\partial}{\dot{x}}^{i}}=0[/tex]

Now

[tex]\frac{{\partial}L}{{\partial}{\dot{x}}^{i}}=m\dot{x}^{i}+eA_{i}[/tex]

so d/dt of this should just be:

[tex]\frac{d}{dt}\frac{{\partial}L}{{\partial}{\dot{x}}^{i}}=m\ddot{x}^{i}+e\dot{A}_{i}[/tex]

Since [tex]A_{i}=\epsilon_{ijk}B_{j}x_{k}\Rightarrow\dot{A}_{i}=\epsilon_{ijk}B_{j}\dot{x}_{k}[/tex]

so we should get this:

[tex]\frac{d}{dt}\frac{{\partial}L}{{\partial}{\dot{x}}^{i}}=m\ddot{x}^{i}+e\epsilon_{ijk}B_{j}\dot{x}_{k}[/tex]

but according to the solution to the problem it should be:

[tex]\frac{d}{dt}\frac{{\partial}L}{{\partial}{\dot{x}}^{i}}=m\ddot{x}^{i}+2e\epsilon_{kji}B_{j}\dot{x}_{k}[/tex]

Note the factor of 2.

I posted a similar problem and I was lead to believe that I need to differentiate wrt another superscript - I am not sure if this is at all relavent here since we are differentiating wrt time.

Would be really grateful if someone could offer me some advice...

Thanks..

**Physics Forums - The Fusion of Science and Community**

# Another question on Lagrangians (sorry)

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Another question on Lagrangians (sorry)

Loading...

**Physics Forums - The Fusion of Science and Community**