Harmonic oscillator onboard a relativistic rocket

[Methavix]

Homework Statement

We have a first frame S (named coordinate frame) at rest on the Earth (we suppose a non-rotating earht and this frame as a perfect inertial frame) and a second frame S' (named proper frame) moving in respect to the first with a constant relativistic velocity V along the x-axis.
The second frame is fixed to a relativistic rocket where onboard is placed a non-relativistic oscillator: mass-spring-damper system. The oscillator swings classically (that is at non relativistic velocities) and I want to write its equations of motion (and their solutions) in the proper frame and in the coordinate frame, also to find the tranformation of the spring constant and the damping factor from a frame to the other.

Homework Equations

I know the equation of motion in the proper frame:

m'*d2x'/dt'2+b'*dx'/dt'+k'*x'=0

where x' and t' are the coordinates in the S' frame, b' is the damping factor and k' is the spring constant in the same frame.
Besides I define:

gamma = 1/sqrt[1-(V/c)^2]
d2x'/dt'2 = a'
dx'/dt' = u'

The Attempt at a Solution

I can directly transform all quantities from this equation :

(m/gamma)*(a*gamma^3)+(b')*[(u-V)/(1-u*V/c^2)]+(k')*(x*gamma) = 0

but i cannot transform b' and k'.
Besides, being u' not relativistic can I do anyway this calculation?

thanks

 

[AlephZero]

Some other ways to get a handle on the other transformations:

1. If the oscillator is underdamped, it is effectively a clock so its oscillation frequency must transform correctly.

2. The 3 terms mx" bx' and kx have the same units (force) so if you can transform one of them the transformation for the other two must be the same.

3. Some features of the motion would be observed the same in both frames - e.g. both observers would see the same number of oscillations for the amplitude of an underdamped oscillator to reduce by half.

 

[Methavix]

1. If the oscillator is underdamped, it is effectively a clock so its oscillation frequency must transform correctly.

what does it mean? i mean, how can i transform the oscillation frequency? is there a known law for it?

2. The 3 terms mx" bx' and kx have the same units (force) so if you can transform one of them the transformation for the other two must be the same.

but i can't do it since gamma has not unit of measurement.
if i write this known transformation:

m'*a' = m*a* gamma^2 --> means that i have to transform this quantity as follows b'*u' = b*u*gamma^2 ?
besides, I found m'*a' = m*a* gamma^2 but F=F' if they are in the x direction, right? is it in disagree with my result?

3. Some features of the motion would be observed the same in both frames - e.g. both observers would see the same number of oscillations for the amplitude of an underdamped oscillator to reduce by half.

sorry, but what do you mean?

thanks
luca

 

[AlephZero]

The general point I'm making is that the solutions of the equation (i.e. what you could observe about the system from the other reference frame) also have to transform correctly.

If b = 0, there is a solution x = A sin wt where w^2 = k/m. That could be used as a clock sending a "tick" at intervals 2 pi/w. If you observe the clock from the other system you know how time is tranformed.

For a lightly damped system a solution is of the form Ae^{-pt}cos wt. (This is not the same w as in the undamped solution of course). The decay in amplitude in one cycle is e^{-2 pi p/w}. That amount of decay must be the same when observed from the other system so p/w must be the same in both systems.

 

[tim_lou]

get a four vector solution to the oscillator in the rocket's frame (x,t) and smack a Lorentz transformation matrix in front of it?

 

[Methavix]

The general point I'm making is that the solutions of the equation (i.e. what you could observe about the system from the other reference frame) also have to transform correctly.

If b = 0, there is a solution x = A sin wt where w^2 = k/m. That could be used as a clock sending a "tick" at intervals 2 pi/w. If you observe the clock from the other system you know how time is tranformed.

For a lightly damped system a solution is of the form Ae^{-pt}cos wt. (This is not the same w as in the undamped solution of course). The decay in amplitude in one cycle is e^{-2 pi p/w}. That amount of decay must be the same when observed from the other system so p/w must be the same in both systems.

ok, if we suppose to have an undamped system the solution in the rocket's frame is (as you write):

x'(t') = x'(0)*cos[sqrt(k'/m')*t']

but the transformation i have to do is:

x'(t') = (x(t)-V*t)*gamma

m' = m*gamma*[1-(u(t)*V)/c^2]

t' = gamma*[t-(x(t)*V)/c^2]

right? because the mass is moving in respect to the rocket frame with a velocity u'(t'), but on the other hand u' is non-relativistic... so I'm not sure...
if these transformations are correct, we yield:

(x(t)-V*t)*gamma = x(0)*gamma*cos[sqrt(k'/(m*gamma*(1-(u(t)*V)/c^2)))*gamma*(t-(x(t)*V)/c^2)]

if i did right, what can i do to continue?
thanks