dubiousraves said:
Thanks. I am aware of the time dilation equation. However, if you say the distance between the Earth and the star contracts from the astronaut's reference frame, is there not a formula that would express that exact contraction?
The formula for Length Contraction is the inverse of the one for Time Dilation, but those are just short-cuts that can get you into trouble. The equations that will never get you into trouble are called the
Lorentz Transformation process. I like to use units for a simplified version of those equations where c has the value of 1, for example, 1 foot per nanosecond. I also like to express velocity as a fraction of c which we call beta and assign the Greek letter beta (β).
So let's say we have a rod of length 5 feet laying motionless along the x-axis. A spacetime diagram showing this simple scenario is:
The blue and red lines represent the two ends of the rod. The rod is spread out all along the space between it's two endpoints but we don't care about that, we only care about the endpoints that are used in determining its length. I'm only showing the positions of the endpoints for just a short time but we understand that the lines really extend upward and downward. Now let's see what happens if we transform the coordindates of this scenario into one that is moving at 0.6 toward the left. In this new diagram, the rod will be moving to the right at 0.6c and we can see how long it is.
The first thing we have to do when using the Lorentz Transformation is calculate the value of gamma, the Lorentz factor (the same value used to determine Time Dilation), as a function of β, in this case -0.6:
γ = 1/√(1-β
2) = 1/√(1-(-0.6)
2) = 1/√(1-0.36) = 1/√(0.64) = 1/0.8 = 1.25
Now we take the value of the t and x coordinates of each event (dot) in the diagram and calculate new values for the new diagram. The two equations for the new primed values are:
t' = γ(t-βx)
x' = γ(x-βt)
We can take another shortcut and just calculate the two endpoints of each of the two worldlines and fill in the extra events (dots) proportionally. So for the top of the blue line, x=0 and t=4:
t' = 1.25(4-(-0.6)*0) = 1.25(4) = 5
x' = 1.25(0-(-0.6)*4) = 1.25(2.4) = 3
For the bottom end of the blue line we have x=0 and t=0:
t' = 1.25(0-(-0.6)*0) = 1.25(0) = 0
x' = 1.25(0-(-0.6)*0) = 1.25(0) = 0
For the top of the red line we have x=5 and t=4:
t' = 1.25(4-(-0.6)*5) = 1.25(4+3) = 1.25(7) = 8.75
x' = 1.25(5-(-0.6)*4) = 1.25(5+2.4) = 1.25(7.4) = 9.25
For the bottom of the red line we have x=5 and t=0:
t' = 1.25(0-(-0.6)*5) = 1.25(3) = 3.75
x' = 1.25(5-(-0.6)*0) = 1.25(5) = 6.25
From those calculations and by interpolating (or by more calculations) we can make this diagram:
Now it's important to know that when we want to determine the length of an object (or the space between objects), we have to do it along a line where time is constant. So look at the horizontal grid line where time is 5 nanoseconds. The top of the blue line is on that grid line at x=3 feet and immediately to its right is the second dot up on the red line with an x coordinate value of 7 feet. The difference between these to events (dots) is 4 feet. This is the contracted length of the rod from its Proper Length in its rest frame divided by gamma (5/1.25=4).
You can also see the Time Dilation in this diagram where the dots representing 1 nsec intervals of time in the rest frame of the rod are now stretched out to 1.25 nsec of Coordinate Time.
The other important feature of Special Relativity call Relativity of Simultaneity is also shown where events that were simultaneous in the first frame occur at different times in the transformed frame.