Particle´s acceleration respect two inertial frames

[Aler93]

Homework Statement

System S' moves with constant speed v=(vx,0,0) respect to the system S. On the S' system a particle moves with a constant acceleration a=(ax,ay,az).
What is the acceleration a'=(ax',ay',az') measured from the system S?.

Homework Equations

Lorentz transformation

The Attempt at a Solution

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I use the velocity-addition formula and find the ecuations of the velocity transformation, I think maybe a derivation respect to proper time of the velocity transformation ux',uy',uz' give the aceleration of the particle, but I am not really sure how to do it.

 

[TSny]

Homework Statement

System S' moves with constant speed v=(vx,0,0) respect to the system S. On the S' system a particle moves with a constant acceleration a=(ax,ay,az).
What is the acceleration a'=(ax',ay',az') measured from the system S?.

The notation is confusing. You have the unprimed a for the acceleration as measured in the primed frame S, and you have the primed a' for the acceleration as measured in the unprimed frame S.

 

[Aler93]

The notation is confusing. You have the unprimed a for the acceleration as measured in the primed frame S, and you have the primed a' for the acceleration as measured in the unprimed frame S.

I know, but I´m using the same notation that it´s used in the problem. Solving the problem I´m finding the unprimed acceleration as its seen from the system S.

 

[TSny]

I know, but I´m using the same notation that it´s used in the problem. Solving the problem I´m finding the unprimed acceleration as its seen from the system S.

Well, it's confusing to me. But let's see how it goes.

I agree that starting with the velocity transformation formula is a good idea. I'm not sure why you would want to differentiate with respect to proper time. The acceleration of the particle in frame S would be the derivative of the velocity of the particle in frame S with respect to time as measured in frame S. Similarly for frame S'.

 

[Aler93]

Well, it's confusing to me. But let's see how it goes.

I agree that starting with the velocity transformation formula is a good idea. I'm not sure why you would want to differentiate with respect to proper time. The acceleration of the particle in frame S would be the derivative of the velocity of the particle in frame S with respect to time as measured in frame S. Similarly for frame S'.

I talk with the teacher and he made some corrections:
System S' moves with constant speed v=(vx,0,0) respect to the system S. On the S' system a particle moves with a constant acceleration a'=(ax',ay',az').
What is the acceleration a'=(ax,ay,az) measured from the system S?.

In this case I used the inverse lorentz transformation cause its from the S system and obtain something like

[ tex ]ux=\frac{u_x'+v}{1+frac{u_x' v}{c^2}}[ /tex ]

and

uy=\frac{uy'}{ \gamma(1+ ux'v/c^2)}

To find the acelleration we derive respect to dt, as you said, I was a bit confused, but it depents of proper time τ

a=du/dt=du/dτ dτ/dt

but how i use this equation with the velocitiy formulas we found before?

 

[TSny]

$$u_x=\frac{u_x'+v}{1+\frac{u_x' v}{c^2}}$$
$$u_y=\frac{u_y'}{ \gamma(1+ u_x'v/c^2)}$$To find the acelleration we derive respect to dt, as you said, I was a bit confused, but it depents of proper time τ

a=du/dt=du/dτ dτ/dt

but how i use this equation with the velocitiy formulas we found before?

When finding $a_x = \frac{du_x}{dt}$, you can work with differentials. Thus $a_x$ equals the differential $du_x$ divided by the differential $dt$.

For $du_x$, use your formula above. For $dt$, use the appropriate Lorentz transformation equation.

 

[Aler93]

When finding $a_x = \frac{du_x}{dt}$, you can work with differentials. Thus $a_x$ equals the differential $du_x$ divided by the differential $dt$.

For $du_x$, use your formula above. For $dt$, use the appropriate Lorentz transformation equation.

When finding $a_x = \frac{du_x}{dt}$, you can work with differentials. Thus $a_x$ equals the differential $du_x$ divided by the differential $dt$.

For $du_x$, use your formula above. For $dt$, use the appropriate Lorentz transformation equation.

Thanks, I was complicating myself, I finally obtained the three acceleration transformations.

Thanks for your help!