00:01
We need to find the time, or show that the time difference in their arrival, show that when e is much, much greater than mc squared, the speed of u is given by the expression in your textbook.
00:14
And using the result of b, compute u1 minus u2 for the energies in mass above and calculate delta t for x is equal to 1 .710 to the 5th light years.
00:27
Okay, so starting with part a, we know that t1 is x over u1, t2 is x over u2.
00:49
So delta t, which is t2 minus t1, is equal to x over u2 minus u1, so getting like denominators u1, 1 minus u2 over u1, u2, which is x delta x over approximately c squared.
01:20
So that is for part a.
01:25
For part b, we know e is equal to in c squared over the square root of 1 minus u squared over c squared.
01:41
So squaring both sides and moving the denominator over, we get 1 minus u squared over c squared is equal to mc squared over e squared.
01:59
So we have u over c is equal to 1 minus mc squared over e squared to the 1 half power.
02:15
So we're going to expand the right hand side and keep only the first term.
02:22
So this is approximately 1 minus 1 half mc squared over e squared.
02:35
So that's u over c...