Calculating Time Dilation: QW Space Station

[kingyof2thejring]

QW the space station is moving at a speed of 0.01 c relative to the earth. According to the clock on Earth , how much longer than one year would have passed when exactly one year passed in the rest frame of a space station?
How do i solve this problem?
thanks in advance

 

[Hootenanny]

HINT: Use the Lorentz transformation for time.

 

[nrqed]

QW the space station is moving at a speed of 0.01 c relative to the earth. According to the clock on Earth , how much longer than one year would have passed when exactly one year passed in the rest frame of a space station?
How do i solve this problem?
thanks in advance

As Hootenanny said you may use the Lorentz transformations. Or, in this case, you may as well use directly \Delta t = \gamma \Delta t' (which does follow from the Lorentz transformations).

Patrick

 

[kingyof2thejring]

yeh
if we define frames of reference,
S: earth
S': space-station
then S' moves at 0.01 c w.r.t. S

start S: x_1=0 t_1=0
S' x'_2=0 t'_2=0
end S: x_1=? t_1=?
S' x'_2=? t'_2=365*24*60

t'_2-t'_1=gamma[(t_2-t_1)-v(x_2-x_1)/c^2
then what happens?

 

[kingyof2thejring]

\Delta t = \gamma \Delta t'
how does this work?

 

[Doc Al]

\Delta t = \gamma \Delta t'
how does this work?

This is the infamous "time dilation" formula, which gives you the time measured by "stationary" clocks (\Delta t) in terms of the time measured by a moving clock (\Delta t'). In your problem, the space station clock is the moving clock.

In your LT calculation, use the version that gives unprimed coordinates in terms of primed coordinates. Then you'll get the same time dilation formula.

 

[Hootenanny]

\Delta t = \gamma \Delta t'
how does this work?

This is called time dilation and as Pat said, follows direction from the Lorentz transformations. You apply this formula in exactly the same way you would a Lorentz transformation, set up your reference frames exactly as before. I.e. S frame is the Earth frame and S' is the space station's rest frame traveling at \beta = 0.01 wrt S. Therefore \Delta t will be the time period as measure on Earth and \Delta t' will be the time period as measure in the S' frame.

Edit: That's twice in two days Doc's beat me to it