Covariant form of the Lagrangian for Lorentz force.

[Peeter]

If I use the following Lagrangian:

$$\mathcal{L} = \frac{1}{2} m v^2 + e A \cdot v/c = \frac{1}{2} m \dot{x}_\mu \dot{x}^\mu + e A_\nu \dot{x}^\nu /c$$

I can arrive at the Lorentz force equation in tensor form:

$$m \ddot{x}_\mu &= (q/c) F_{\mu\beta} \dot{x}^\beta$$

details offline here:
http://www.geocities.com/peeter_joot/lut/maxwell_tensor_lagrangian.pdf


However, reading a translation of the DeBroglie thesis he appears to use the same Lagrangian, but it differs by a factor of two in the v^2 term

http://www.nonloco-physics.000freehosting.com/ldb_the.pdf

(equation 2.3.5 on page 26 of the pdf)

I haven't gotten far enough that I see what he does with this, but would like to understand why my previous calculation appears to be off by a factor of two before continuing.

Does anybody know of the correct covariant Lagrangian to arrive at the Lorentz force equation? I started with the non-covariant form in Goldstein and got the equation above with a bit of guess work.

 

[gtoguy97]

I am aware of the equations in Landau and Lifshitz but they don't appear to provide a starting Lagrangian.The discrepancy between the two Lagrangians is perplexing. It appears that the DeBroglie thesis Lagrangian is missing a factor of two in the velocity term, and I'm not sure why this would be the case. Does anyone have any ideas?

 

[astrokreng]

I can confirm that the covariant form of the Lagrangian for the Lorentz force is indeed given by the equation provided in the content. The factor of two difference in the v^2 term in the DeBroglie thesis may be due to a different convention or notation being used. It is important to note that the Lagrangian is not unique and different forms may be used to describe the same physical system. As long as the resulting equations of motion are consistent with the Lorentz force equation, the specific form of the Lagrangian does not affect the results. It is also possible that DeBroglie may be considering a slightly different system or including additional terms in the Lagrangian. Further investigation and comparison with experimental results may be necessary to fully understand the difference in the v^2 term.