# is action integral Lorentz invariant?

I need to find the Lagrangian for charged particles in EM fields considering relativistic effects. Is action integral Lorentz invariant. $$A = \int_{t_1}^{t_2} L (q_i, \dot q_i, t) dt$$

According to my note

According to the first postulate of special relativity, the action integral $A$ must be invariant because the equation of motion is determined by extreme condition $\delta A = 0$.

I do not understand how does this make $A$ invariant.

The action is obviously not invariant because energy is different in different frames. What's invariant is the trajectory that makes the action stationary. Specifically, transforming frames will add a total derivative to the Lagrangian, thereby adding a constant to the action. See, for example, the first few pages of Mechanics by Landau and Lifshitz.

• Although this is certainly true, I think it might help to clear up a bit of confusion for the OP if, perhaps, you were to elaborate a bit more on the relationship between the "first postulate" and the fact that "what's invariant is the trajectory which makes the action stationary." Incidentally, I find this a bit of a confusing way of saying what I think you mean since a stationary point of the action is not mapped to itself under Lorentz transformation but is nonetheless mapped into another stationary point of the action. Commented Jan 13, 2014 at 6:59
• I'm not sure what the "first postulate" is, sorry. I'm saying that there is some $\mathbf{x}(\tau)$ which is the trajectory. It's a four-vector position as a function of proper time and is invariant. If you find it in two different frames, you'll get two different sets of coordinates describing the same $\mathbf{x}(\tau)$. Commented Jan 13, 2014 at 7:05
• I'm guessing "first postulate" here means something like "laws of physics are form invariant under change of inertial frame." I see, I suppose that in common physics terminology you're thinking of Poincare transformations as "passive," namely the curve is just a set of points on the spacetime manifold, and changing frames corresponds to choosing different local coordinates, but keeps the curve unaltered. In this case, I think the first postulate would be stated as something like "laws of physics can be written an a coordinate-free way." and presumably, this restricts the action similarly. Commented Jan 13, 2014 at 7:49

In general, invariance of the action under some transformations, such as Lorenz rotations, is an extra condition that we impose on a theory. This is not related to the extreme condition that just gives trajectories along which the classical evolution goes.

In your case, I guess, there is some confusion in formulation of the idea. According to the 1st postulate of special relativity, all physical laws are the same in any inertial frame (related by a Lorenz transformation). To realise this theoretically, we need to make our Lagrangian invariant under such transformations. Then it will give the same physics in all frames.

Hence, one should impose some extra condition to satisfy the 1st postulate.

• It is not correct to say that to realize the 1st postulate in the form you state, the Lagrangian must be Lorentz-invariant. What matters is that the equations of motion must be Lorentz-invariant, and this allows for the Lagrangian to change by a total time derivative. Commented Jan 13, 2014 at 20:35
• Yes, that's true, I usually forget about this extra total derivative if one is not interested in boundary effects. Commented Jan 14, 2014 at 0:18