Are You Struggling to Understand Reference Frames in Relativistic Dynamics?

[kreil]

I'm having trouble getting all of the relativistic dynamics information straight in my head. In particular, I'm having trouble with reference frames. It seems that when I am trying to solve problems, even simple ones, I am struggling with what each observer in the different frames should be seeing.

For example, let's say in the rest frame O an observer sees a particle moving with velocity V. The particle has a four velocity given by (note: I use c=1 units everywhere)

$$\bar u = \frac{d \vec x}{d \tau} = (\gamma, \gamma \vec V)$$

and the four force on the particle would be

$$\bar f = \gamma (\vec F \cdot \vec V, \vec F)$$

Now let's say an observer in the frame O', moving with velocity v along the x-axis of O, observes the particle. From the velocity law, he sees the particle moving at speed:

$$V^{x'} = \frac{V^x-v}{1-vV^x}$$

where $V^x$ is the observed velocity in O.Question: Is it true that the particle's four velocity and four force are transformed to use this $V^{x'}$ ? Why or Why not?

Question: Is the velocity in the Lorentz Factor $\gamma$ the velocity of the particle in the rest frame V, or the relative velocity between the frames v?

Question: Is it correct to say that for any problem where a particle's velocity is being observed there are 3 reference frames to worry about: the rest frame O, the moving frame O', and the frame of the particle? If it is relevant, I encountered these thought problems while trying to find the four force and four velocity components for a particle moving in the Lorentz Force Field (Hartle problem 5.21),

$$\vec F = \frac{d \vec p}{dt} = q(\vec E + \vec V \times \vec B)$$

which is invariant between inertial frames.

I realize that this is all quite basic stuff math wise, but the thought process that leads to the correct calculations is giving me trouble, and it is crucial I get it solidified into my head correctly.

 

[vela]

Question: Is it true that the particle's four velocity and four force are transformed to use this $V^{x'}$ ? Why or Why not?

Yes, of course. In O', the only thing you see is the particle's three-velocity V^x'. The four-velocity and four-force aren't going to depend on quantities observed in a different frame. After all, there would be an infinite number of frames you could choose from, right? Which one would you pick?

Question: Is the velocity in the Lorentz Factor $\gamma$ the velocity of the particle in the rest frame V, or the relative velocity between the frames v?

If you're calculating the four-force or four-velocity in O', you'd use neither. You'd use the three-velocity of the particle as observed in O'. Is this what you were asking about? Your question seemed a bit ambiguous.

Question: Is it correct to say that for any problem where a particle's velocity is being observed there are 3 reference frames to worry about: the rest frame O, the moving frame O', and the frame of the particle?

Not for any problem. In particle physics, sometimes you can do an entire problem in the lab frame; sometimes, it's convenient to use a center-of-mass frame, where none of the particles is at rest; sometimes, you find the particle's rest frame a convenient choice. If you're working with massless particles, like photons, there is no rest frame (of the particle).

If it is relevant, I encountered these thought problems while trying to find the four force and four velocity components for a particle moving in the Lorentz Force Field (Hartle problem 5.21),

$$\vec F = \frac{d \vec p}{dt} = q(\vec E + \vec V \times \vec B)$$

which is invariant between inertial frames.

I realize that this is all quite basic stuff math wise, but the thought process that leads to the correct calculations is giving me trouble, and it is crucial I get it solidified into my head correctly.

 

[kreil]

The lorentz factor is,

$$\gamma = \frac{1}{\sqrt{1-v^2}}$$ in units of c=1.

My question related to the fact that many things get multiplied by this gamma factor: Lorentz Coordinate Transformations, four-velocity, etc.. but is the velocity in gamma always the same? i.e. Are we using the velocity between frames in gamma for the Lorentz Transformations, but the three-velocity in gamma when calculating the four-velocity?

 

[vela]

No, the appropriate velocity in γ depends on the context. For the four-velocity, it is the velocity of the particle. For a Lorentz boost, it's the velocity of the moving frame.

 

[kreil]

Thanks vela, you've been helpful as usual!

 

[kreil]

Just to make sure I understand, is this correct?

"Assume that in all frames the force on a charged particle is given by the lorentz force law (given above). Consider a frame moving with speed v along the x-axis relative to the lab frame. Find the components of the 4-force in terms of E, B and the components of the 4 velocity."Assume the particles velocity is $V'=\frac{V-v}{1-Vv}$ as observed from the moving frame. Then,
$$\bar f' = \gamma (\vec F \cdot \vec V', \vec F)$$

$$\vec F \cdot \vec V' = q(\vec E + \vec V' \times \vec B) \cdot \vec V'=q|E||V'|cos 0 =qEV'$$

because V'xB is perpendicular to V' their dot product is zero.

$$\bar f' = \frac{q}{\sqrt{1-\vec V'^2}} \left [ EV', (\vec E + \vec V' \times \vec B) \right ]$$

and since,

$$\bar u' = \frac{1}{\sqrt{1-\vec V'^2}} ( 1, \vec V' ) = ( \vec u_{t'}, \vec u_{x'})$$

we get:

$$\bar f' = q\left [ E \vec u_{x'}, (\vec E \vec u_{t'} + \vec u_{x'} \times \vec B) \right ]$$

 

[vela]

You're assuming the electric field and velocity point in the same direction. If you don't assume that, the best you can do is say that

$$\vec{F}\cdot\vec{V}' = q\vec{E}\cdot\vec{V}'$$