# Homework Help: Particle´s acceleration respect two inertial frames

Tags:
1. Oct 1, 2016

### Aler93

1. The problem statement, all variables and given/known data
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?.

2. Relevant equations
Lorentz transformation

3. The attempt at a solution

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 im not really sure how to do it.

2. Oct 1, 2016

### TSny

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.

3. Oct 1, 2016

### Aler93

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.

4. Oct 1, 2016

### TSny

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'.

5. Oct 3, 2016

### Aler93

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
Code (Text):
[ tex ]ux=\frac{u_x'+v}{1+frac{u_x' v}{c^2}}[ /tex ]
and
Code (Text):
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?

Last edited: Oct 3, 2016
6. Oct 3, 2016

### TSny

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.

7. Oct 3, 2016

### Aler93

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