Muon Decay at t=2.2x10^-6s in Frame s': Coordinates and Time

[s7b]

consider two frames s and s' where s' is moving with velocity vector (0.9c,0,0) with respect to s frame. At time t=t'=0 the origin of the two frames coincide. At time t=2.2x10^-6s a muon decays at coordinates (100,0.8,1.0) meters in the s frame.

a) At what time and coordinates does the muon decay in frame s''?


Attempt:

For this question I used the formula for time dilation to find the time the muon decays in the frame s''.

I used t=(gamma)(proper time) where the proper time is the 2.2x10^-6

How do I go about finding the new co-ordinates??

 

[granpa]

https://www.physicsforums.com/showthread.php?t=257689

 

[baba89]

To find the coordinates of the muon decay in frame s'', we can use the Lorentz transformation equations. These equations describe how coordinates and time are related between two different frames of reference in special relativity.

In this case, we are given the velocity vector (0.9c,0,0) of frame s' with respect to frame s. We can use this to calculate the Lorentz factor, gamma, as follows:

gamma = 1/sqrt(1-(v^2/c^2)) = 1/sqrt(1-(0.9c)^2/c^2) = 1/sqrt(0.19) = 1.051

Next, we can use the Lorentz transformation equations for position to find the coordinates (x'', y'', z'') in frame s'':

x'' = gamma(x' + vt')
y'' = y'
z'' = z'

Where x', y', and z' are the coordinates in frame s' and t' is the time in frame s'.

Plugging in the given values, we get:

x'' = 1.051(100 + (0.9c)(2.2x10^-6)) = 105.1 meters
y'' = 0.8 meters
z'' = 1.0 meters

Therefore, in frame s'', the muon decays at coordinates (105.1, 0.8, 1.0) meters at the same time t''= 2.2x10^-6 seconds. This shows that the coordinates and time of the muon decay are different in different frames of reference, as predicted by special relativity.