My question is in three parts and asks for a relation between two space time intervals, an explanation of length contraction and time dilation, then an application of proper time.
i) Two inertial frames O and O' are in standard configuration. Write down the two equations relating spacetime intervals Delta x' and Delta t' in O' to the corresponding intervals Delta x and Delta t in O. Then write the corresponding two inverse relations for Delta x and Delta t in terms of Delta x' and Delta t', explain any special symbols you use.
ii) Now consider these four relationships. Explain how the length contraction relationship follows immediately from one of them, and the time dilation relationship follows immediately from another. Include all essential steps in the argument, in particular stating the reasons for your choice of equation.
iii) The time dilation relationship just derived applies to uniform motion. Briefly explain how the concept of proper time permits the time dilation relationship to be applied to, the half-life of a muon circulating in a magnetic storage ring.© BrainMass Inc. brainmass.com July 16, 2018, 6:19 am ad1c9bdddf
We can use the equations for the Lorentz transformation:
x' = gamma[x - v t]
t' = gamma[t - v x/c^2]
Here it is assumed that at t = 0 the observer O' is at x=0 (his clock is set to zero at that time), and that O' is moving in the x-direction with velocity v. Suppose that some event happens at position x and time t in the inertial frame of observer O, then that same event happens at position x' at time t' in the inertial frame of observer O'. The symbol gamma is defined as:
gamma = 1/sqrt[1 - v^2/c^2]
And c is the speed of light.
Suppose we have two events at (x,t) and at (x + Delta x, t + Delta t) then from the above Lorentz transformation equations it follows that:
Delta x' = gamma[Delta x - v Delta t] (1)
Delta t' = gamma[Delta t - v/c^2 Delta x] (2)
where Delta x' and Delta t' are the intervals between the two events in the inertial frame of observer O'. To obtain the inverse relation, note that observer O' sees observer O moving in the opposite direction, i.e. with velocity -v. This means that dx and dt are given in terms of Delta x' and Delta t' in just the same way as Delta x' and Delta t' are given in terms of Delta x and Delta t, but with v replaced by -v (note that gamma stays the same under the interchange v --> -v):
Delta x = gamma[Delta x' + v Delta t'] (3)
Delta t = gamma[Delta t' + v/c^2 Delta x'] (4)
We can, of course, derive (3) and (4) directly from (1) and (2) without using the symmetry between ...
A detailed solution is given