Velocity of Particle vs Inertial Frame Velocity

[Phillipdanbury]

Hello all,

This post is in reference to a previous homework post, found here:
https://www.physicsforums.com/threads/show-that-f-gamma-3-ma.338744/

That thread is closed to further replies. Probably because it's nearly 10 years old.

That thread is about deriving relativistic force from the derivative of relativistic momentum specifically when the force is parallel with the velocity. The OP must prove that F = (d/dt) (γmv) = γ³ma.

I followed that thread fine and I see where the answer came from. However, when I first attempted the problem (before I found my way here), I was assuming that the velocity of the particle was not necessarily the same as the velocity of the inertial frame buried inside of the lorentz factor γ . Feedback from the community suggested that they are necessarily the same and, indeed, the proof would be impossible if they were not the same.

Why are they the same?

I thought perhaps they are the same because we assume that the velocity vector is the reference frame. In other words, we are traveling on the back of the particle moving at velocity v. Is there a simpler way to think about the problem?

 

[Orodruin]

The gamma factor is not associated to any inertial frame. It is an integral part of the relativistic definition of momentum.

 

[Herman Trivilino]

I followed that thread fine and I see where the answer came from. However, when I first attempted the problem (before I found my way here), I was assuming that the velocity of the particle was not necessarily the same as the velocity of the inertial frame buried inside of the lorentz factor γ .

It's the same $v$ in both relations: $p=\gamma mv$ and $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$

What is it that makes you think they are defined differently?

 

[Sorcerer]

It's the same $v$ in both relations: $p=\gamma mv$ and $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$

What is it that makes you think they are defined differently?

Is there not a case where two coordinate systems S and S' move with respect to each other with their x-axes coinciding at speed v, who both watch an object move with speed u in S and u' in S'?

In that case. wouldn't you have γ(v) for the speed between the two systems S and S' and then γ(u) and γ(u') for what observers at rest in S and S' measure for the moving object?

 

[Herman Trivilino]

Is there not a case where two coordinate systems S and S' move with respect to each other with their x-axes coinciding at speed v, who both watch an object move with speed u in S and u' in S'?

Yes, but the case being discussed here is not one of them.

 

[Sorcerer]

Yes, but the case being discussed here is not one of them.

Oh, sorry, I thought you were talking generally. Honestly I've imbibed a bit too much of a certain magical brain cell killing Sorerer's potion tonight.

 

[Phillipdanbury]

Is there not a case where two coordinate systems S and S' move with respect to each other with their x-axes coinciding at speed v, who both watch an object move with speed u in S and u' in S'?

In that case. wouldn't you have γ(v) for the speed between the two systems S and S' and then γ(u) and γ(u') for what observers at rest in S and S' measure for the moving object?

Yes, but the case being discussed here is not one of them.

Well yes actually, that is exactly the type of example I was thinking. In my case above, I don't see why the velocities must be the same. Is my case special in some other way that I'm not seeing?

The gamma factor is not associated to any inertial frame. It is an integral part of the relativistic definition of momentum.

I don't really know what this means. I'm 5 weeks through my first course in non-classical physics.

 

[Ibix]

The point is that only one frame is being considered here, the one in which the particle is moving at $v$ and has momentum $\gamma mv$. You don't need any other frame, so it's not clear to me why you think there's a second one or how you intend to use it.

 

[Sorcerer]

Well yes actually, that is exactly the type of example I was thinking. In my case above, I don't see why the velocities must be the same. Is my case special in some other way that I'm not seeing?
I don't really know what this means. I'm 5 weeks through my first course in non-classical physics.

Your case, IMHO, is like choosing to do a free body diagram of a box being dragged in which the coordinates you choose have axes that are not parallel to any force involved or any direction of motion.

What I mean is, why not just choose your moving reference frame so that γ(v) = γ(u)?

Of course if you have more objects involved you can’t do that, but for the derivation in question it seems simpler.

 

[Herman Trivilino]

Well yes actually, that is exactly the type of example I was thinking. In my case above, I don't see why the velocities must be the same. Is my case special in some other way that I'm not seeing?

One frame of reference is the rest frame of the laboratory. In this frame the speed of the particle is $v$.

Another frame of reference is the rest frame of the particle. This frame moves with (the same) speed $v$ relative to the laboratory.

 

[Herman Trivilino]

I thought perhaps they are the same because we assume that the velocity vector is the reference frame.

Huhh? A velocity vector can't be a reference frame.

In other words, we are traveling on the back of the particle moving at velocity v. Is there a simpler way to think about the problem?

The rest frame of the particle is the frame of reference in which the particle is at rest. That particle can be located anywhere in that reference frame. If you are also at rest in that reference frame then you are traveling along with the particle, but that doesn't tell you where you are relative to the particle, just that the particle maintains its position relative to you.