Question about Lagrangian in electromagnetic interaction

[mings6]

Sorry for a naive question.

In EM textbook and QM path integral textbook, the action and Lagrangian in electromagnetic interaction are

S = L dt = e(\phi – A v) dt ---equ.(1)

But in QFT textbook, the action and Lagrangian density are

S = L d^4x = A J d^4x ---equ.(2)

As I understand, in equ.(2), J = \rho U = \rho \gamma V
In which \rho is density, U is the 4-velocity=dx/d\tau, and V is the common velocity=dx/dt, \gamma is \sqrt (1-v^2/c^2).

So equ.(2) will have a factor of \gamma, but equ.(1) does not have.

So where is my mistake?

 

[bapowell]

Sorry for a naive question.

In EM textbook and QM path integral textbook, the action and Lagrangian in electromagnetic interaction are

S = L dt = e(\phi – A v) dt ---equ.(1)

But in QFT textbook, the action and Lagrangian density are

S = L d^4x = A J d^4x ---equ.(2)

As I understand, in equ.(2), J = \rho U = \rho \gamma V
In which \rho is density, U is the 4-velocity=dx/d\tau, and V is the common velocity=dx/dt, \gamma is \sqrt (1-v^2/c^2).

So equ.(2) will have a factor of \gamma, but equ.(1) does not have.

So where is my mistake?

The 4-current is [itex]J^\alpha = \rho_0 u^\alpha[/itex], where [itex]\rho_0[/itex] is related to the charge density by [itex]\rho_0 = \rho/\gamma[/itex] and [itex]u^\alpha[/itex] is the 4-velocity, [itex]u^\alpha = \gamma(c,{\bf v})[/itex]. You'll find that the [itex]\gamma[/itex]'s cancel.