On Thu, Feb 27, 2014 at 4:49 PM, Jesse Mazer <[email protected]> wrote:
> > > A simple example: say in Alice's rest frame, there are two markers at rest > in this frame 20 light-years apart, and Bob moves inertially from one > marker to the other a velocity of 0.8c in this frame. What is the proper > time on Bob's worldline between passing the first marker and passing the > second? In Alice's frame we could calculate this by first noting it should > take 20/0.8 = 25 years of coordinate time in this frame for Bob to get from > one to the other, and then the time dilation equation tells us that if he's > moving at 0.8c his clock should be slowed by a factor of sqrt(1 - 0.8^2) = > 0.6 in this frame, so Bob's own clock should tick forward by 25*0.6 = 15 > years between passing the first marker and the second. That is BOB'S PROPER > TIME, AS CALCULATED IN ALICE'S REST FRAME. > > You could of course calculate the proper time in Bob's rest frame too. In > this case, you have to take into account length contraction--the markers > are moving at 0.8c relative to Bob's frame, so if the distance between them > was 20 light-years in their own rest frame, in Bob's frame the distance > between them is shortened by a factor of sqrt(1 - 0.8^2) = 0.6, so in Bob's > frame the second marker is 20*0.6 = 12 light-years away at the moment he is > passing the first marker. Thus, if the second marker is moving towards him > at 0.8c, it will take 12/0.8 = 15 years of coordinate time in this frame to > reach him after the first marker passed him. And since he is at rest in > this frame, his clock ticks at the same rate as coordinate time, so his > clock should also tick foward by 15 years between passing the first marker > and passing the second. That is BOB's PROPER TIME, AS CALCULATED IN BOB'S > REST FRAME, and you can see that we get exactly the same answer as when we > calculated his proper time using Alice's rest frame. > > Incidentally, for two events E1 and E2 on the worldline of an inertial clock (like the events of Bob passing each marker), there is also a simple formula for calculating the proper time the clock ticks between those events, using the coordinates of any frame you like. That is: (proper time between E1 and E2)^2 = (coordinate time between E1 and E2)^2 - (1/c^2)*(coordinate distance between E1 and E2)^2 Or using more common notation, dtau^2 = dt^2 - (1/c^2)*dx^2 If you use units where c = 1, like years for time and light-years for distance, this reduces to: dtau^2 = dt^2 - dx^2 For example, in Alice's frame we have dt = 25 years, and dx = 20 light-years, so dtau^2 = 25^2 - 20^2 = 625 - 400 = 225, so dtau is the square root of 225, or 15. Likewise, in Bob's frame we have dt = 15 years (which you could derive using the Lorentz transformation if you knew the coordinates of passing each marker in Alice's frame), and dx = 0, which again gives dtau^2 = 225 and therefore dtau = 15. This formula is the spacetime analogue of the Pythagorean formula in Euclidean geometry, which tells you that if you have a line segment that has some length ds that you want to calculate, then if you use any cartesian coordinate system to define the x and y coordinates of its endpoints, so you can find dx and dy between the endpoints, then ds^2 = dx^2 + dy^2. Just as the length of a line segment will have an answer that is the same regardless of how you orient your Cartesian coordinate axes (dx and dy may change depending on the axes, but ds will always be the same), so the proper time between two events on a worldline has an answer that is the same regardless of what inertial frame you use (dx and dt can vary, but dtau will always be the same)--both are coordinate-independent quantities, and both are understood to reflect the "geometry" of the space/spacetime in which they are defined. Jesse -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
