Relativistic Action: Lorentz Transform to Get Action A

[actionintegral]

I am at rest. I see an object with action A. Someone moves by with
velocity v. How should I use the Lorentz transform to get the action A' in his frame?

 

[coalquay404]

What do you mean by "I see an object with action A"? This doesn't make much sense to me.

 

[Jheriko]

Do you mean an "I see an object which follows a path that minimises the action integral A"?

 

[actionintegral]

Do you mean an "I see an object which follows a path that minimises the action integral A"?

Not really - I am trying to treat action on it's own merit. Let's keep the path (dt) or (dx) infinitesimally small.

 

[coalquay404]

I'm still confused. The action is just a number formed from an integral of some quantity:

$$S = \int L(q,\dot{q}).$$

If you're interested in an action which is appropriate for, say, a relativistic point particle of mass $m$, then the obvious candidate is

$$S = -m\int d\tau$$

where $d\tau$ is the infinitesimal proper time along the world-line. The point here is that Lorentz invariance is built in to the action from the start since $d\tau$ is a Lorentz invariant. Because of this you're guaranteed that the equation of motion that you get from varying this action will automatically be Lorentz invariant also.

Apologies if I'm missing something here, but the question seems trivial.

 

[actionintegral]

the obvious candidate is
...
the question seems trivial.

What I am missing is the following: I am at rest. I see an object with energy E. I watch it for time dt. I calculate E*T. A moving person will see the same object with energy E' and the time interval in question is dt'. It is not obvious to me that Edt = E'dt' but I will try to convince myself of that.

 

[Los Bobos]

What I am missing is the following: I am at rest. I see an object with energy E. I watch it for time dt. I calculate E*T. A moving person will see the same object with energy E' and the time interval in question is dt'. It is not obvious to me that Edt = E'dt' but I will try to convince myself of that.

Is it possible, that you are confusing the coordinate time t and the proper time $$\tau$$, which is by definition Lorentz invariant (as coalquay404 explained) as it is the square root of the spacetime interval?