- #1
kent davidge
- 933
- 56
It's known that the Lagragian is invariant when one adds to it a total time derivative of a function. I was thinking about the possible forms of this function. Could it be a function of ##x,\dot x, \lambda##? Or it should be a function of ##x## only?
Here ##x: \ \{x^\mu (\lambda) \}## are the coordinates, ##\dot x: \ \{ dx^\mu/d\lambda \}## the corresponding "velocities" and ##\lambda## the parameter.
I think my question is due to my lack of knowledge on variation, because:
if one considers ##\delta \int (df / d\lambda) d\lambda = \delta (f|_{\lambda_f} - f|_{\lambda_i})## then it seems that ##f## can be a function of the three quantities I mentioned above, because the variation will be zero anyways. But if the correct way is to put the ##\delta## symbol inside the integral first, then the variation would vanish only if ##f = f(x)##.
Here ##x: \ \{x^\mu (\lambda) \}## are the coordinates, ##\dot x: \ \{ dx^\mu/d\lambda \}## the corresponding "velocities" and ##\lambda## the parameter.
I think my question is due to my lack of knowledge on variation, because:
if one considers ##\delta \int (df / d\lambda) d\lambda = \delta (f|_{\lambda_f} - f|_{\lambda_i})## then it seems that ##f## can be a function of the three quantities I mentioned above, because the variation will be zero anyways. But if the correct way is to put the ##\delta## symbol inside the integral first, then the variation would vanish only if ##f = f(x)##.