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rsl12 · 10-25-2005, 08:28 AM

stingc and knifemissile have already summed it up nicely. I will offer then just these two useful illustrations for time dilation and length contraction. The illustrations follow from the two facts described by knifemissile (i.e., physics works no matter from where you are looking, and light always travels at speed c).

TIME DILATION
Ok, you're sitting on your front porch, and your friend is driving past in his very fast car (at a constant velocity), and he is holding up to the window a see-through box containing a light beam, which is he has managed to trap between two perfectly polished mirrors (parallel to the floor). To your friend, the light beam is bouncing straight up and down, and the time it takes for the beam to go from one mirror to the next is d/c, where d is the distance between the mirrors.

To you, however, the light beam isn't going straight up and down--it's going ziggy zaggy because the car is moving! (Pretend you're in dark dark space, and all you can see is the light beam, like a flashlight waving around in the dark. You wouldn't see it go up and down if your friend is travelling past you--you'd see it go up-left, down-left, up-left, etc.) Nevertheless, light still travels at the speed of light, despite the fact that the distance it must travel is longer (the vertical distance between mirrors is the same, but now there's a horizontal component to the movement), so the time it takes for the light to go from one mirror to the next is longer for you than it is for your friend (i.e., for you, it appears that time is moving slower in the car!). Using some pythagorean theorem and rearranging terms using simple algebra, you can figure out that the time it takes is d/(c^2-v^2)^0.5, where d is the distance between mirrors and v is the speed of the car. (Note, however, that if you were holding the same box with the same light beam, your friend would see it moving ziggy zaggy. Therefore, to your friend, you are the one moving slower through time! But that's another matter altogether.)

LENGTH CONTRACTION
This is a bit trickier. Okay, now let's say that your friend is still in the car with the light beam in a box, but that the box now has the mirrors perpendicular to the floor (i.e., the light beam travels parallel to the floor, going back and forth in the same direction as the car's movement). The time it takes for the light beam to go from one mirror to the next is, for your friend, still d/c, or 2d/c to do one complete lap. For you, however, the light is travelling forward 2 steps, then going back one (I'm talking figuratively now) since the car and the mirrors are in motion. How long does it take to make one complete lap? Well, since TIME DILATION is occuring, we know that the time should be 2d/(c^2-v^2)^0.5. However, we also know that you can calculate the time by figuring out the time it takes to get to the 1st mirror, or d/(c-v), and the time it takes to get back, which is d/(c+v). So the time it takes is the sum of these. Using the fact that (c-v)(c+v)=(c^2-v^2), we can figure this time to be 2dc/(c^2-v^2). So if we set these two methods of calculating time to be equal, we have

2d/(c^2-v^2)^0.5=2dc/(c^2-v^2)

how is it possible? the left side is *almost* identical to the right side, except that the right is multiplied by a peksy term of c/(c^2-v^2)^0.5!!

The answer is that the 'd' in the left side is *not* the same as the 'd' on the right side. The 'd' on the left was calculated using the distance between the mirrors when the mirrors were parallel to the floor (i.e., the same distance that your friend measures between the mirrors), whereas the 'd' on the right is the distance you observe between the mirrors when the mirrors are *perpendicular* to the floor! Let's call this second distance d prime (d'). Therefore the relationship between d and d' is:

d = d'c/(c^2-v^2)^0.5

The lengths of things aligned with the direction of motion appear *shorter* to you!

10-25-2005, 08:28 AM	#6 (permalink)
rsl12 On the lam Location: northern va	stingc and knifemissile have already summed it up nicely. I will offer then just these two useful illustrations for time dilation and length contraction. The illustrations follow from the two facts described by knifemissile (i.e., physics works no matter from where you are looking, and light always travels at speed c). TIME DILATION Ok, you're sitting on your front porch, and your friend is driving past in his very fast car (at a constant velocity), and he is holding up to the window a see-through box containing a light beam, which is he has managed to trap between two perfectly polished mirrors (parallel to the floor). To your friend, the light beam is bouncing straight up and down, and the time it takes for the beam to go from one mirror to the next is d/c, where d is the distance between the mirrors. To you, however, the light beam isn't going straight up and down--it's going ziggy zaggy because the car is moving! (Pretend you're in dark dark space, and all you can see is the light beam, like a flashlight waving around in the dark. You wouldn't see it go up and down if your friend is travelling past you--you'd see it go up-left, down-left, up-left, etc.) Nevertheless, light still travels at the speed of light, despite the fact that the distance it must travel is longer (the vertical distance between mirrors is the same, but now there's a horizontal component to the movement), so the time it takes for the light to go from one mirror to the next is longer for you than it is for your friend (i.e., for you, it appears that time is moving slower in the car!). Using some pythagorean theorem and rearranging terms using simple algebra, you can figure out that the time it takes is d/(c^2-v^2)^0.5, where d is the distance between mirrors and v is the speed of the car. (Note, however, that if you were holding the same box with the same light beam, your friend would see it moving ziggy zaggy. Therefore, to your friend, you are the one moving slower through time! But that's another matter altogether.) LENGTH CONTRACTION This is a bit trickier. Okay, now let's say that your friend is still in the car with the light beam in a box, but that the box now has the mirrors perpendicular to the floor (i.e., the light beam travels parallel to the floor, going back and forth in the same direction as the car's movement). The time it takes for the light beam to go from one mirror to the next is, for your friend, still d/c, or 2d/c to do one complete lap. For you, however, the light is travelling forward 2 steps, then going back one (I'm talking figuratively now) since the car and the mirrors are in motion. How long does it take to make one complete lap? Well, since TIME DILATION is occuring, we know that the time should be 2d/(c^2-v^2)^0.5. However, we also know that you can calculate the time by figuring out the time it takes to get to the 1st mirror, or d/(c-v), and the time it takes to get back, which is d/(c+v). So the time it takes is the sum of these. Using the fact that (c-v)(c+v)=(c^2-v^2), we can figure this time to be 2dc/(c^2-v^2). So if we set these two methods of calculating time to be equal, we have 2d/(c^2-v^2)^0.5=2dc/(c^2-v^2) how is it possible? the left side is almost identical to the right side, except that the right is multiplied by a peksy term of c/(c^2-v^2)^0.5!! The answer is that the 'd' in the left side is not the same as the 'd' on the right side. The 'd' on the left was calculated using the distance between the mirrors when the mirrors were parallel to the floor (i.e., the same distance that your friend measures between the mirrors), whereas the 'd' on the right is the distance you observe between the mirrors when the mirrors are perpendicular to the floor! Let's call this second distance d prime (d'). Therefore the relationship between d and d' is: d = d'c/(c^2-v^2)^0.5 The lengths of things aligned with the direction of motion appear shorter to you! __________________ oh baby oh baby, i like gravy. Last edited by rsl12; 10-25-2005 at 09:29 AM..