petm1 said:
No I think the math is correct for both length contraction and time dilation, but put them into the equation for the speed of light at the same time and you will change the speed of light. The contraction is (length)/gamma. The dilation is t (gamma) let's call gamma .5 now plug it into the speed of light. (300,000,000/1) X 1 meter/1 second X 1/.5 /1 X .5 now I come up with a number of 1,200,000,000 how about you? I don't disagree with any of SR, I just think that I am not seeing the whoe picture.
Do you understand the concept of the "relativity of simultaneity"? It is assumed that each observer measures the speed of anything, including a light beam, using
multiple clocks which are at rest in their frame, and "synchronized". For example, if an object passes by the 1-meter mark on my ruler and at the moment it passes a clock at the 1-meter mark reads "15 seconds", and then later the object passes the 9-meter mark on my ruler, and at the moment it passes a second clock at the 9-meter mark reads "17 seconds", then I'll conclude that in my frame the object moved 9-1=8 meters in 17-15=2 seconds, for a speed of 8/2=4 meters/second.
So, each observer is using multiple synchronized clocks to measure speed. But--and this is a key point--
different observers disagree on what it means for clocks to be 'synchronized. Two clocks which are synchronized in the frame of an observer at rest relative to them will be out-of-sync in the frame of an observer moving relative to them. The reason for this is that each clock is synchronized using the assumption that light travels at the same speed in all directions in the clocks' rest frame, so I can synchronize two of my own clocks by setting off a flash at the midpoint between them, and setting them to read the same time at the moment the light from the flash reaches each one. But if in your frame my clocks are moving, then one clock will be moving
toward the point where the flash went off, while the other will be moving
away from that point...so if you assume the light moves at the same speed in both directions in
your frame, then in your frame the light must reach the clocks at different times! This means that if I set the clocks to read the same time when the light hits them, in your frame the clocks will be out-of-sync. It works out that if the two clocks are at a distance of x apart in their own rest frame, and they are synchronized in that frame, then if in your frame the clocks are moving at velocity v parallel to the axis that joins them, in your frame the back clock's time will be ahead of the front clocks' time by a factor of vx/c^2.
Once you take time dilation and length contraction
and the relativity of simultaneity into account, you will find that each frame does indeed measure the same speed for a light beam. I've already posted a link to
this thread where I gave an example showing how it works out, but perhaps I should just repost the example here to increase the chance that you'll actually read it:
Suppose, for the sake of making the math a bit easier, that we measure distance in units of "fivers", where a fiver is defined to be the distance light travels in 0.2 seconds (i.e. 1 fiver = 0.2 light-seconds), so that light is defined to have a velocity of 5 fivers/second. Suppose you see a ruler which is moving at a velocity of 3 fivers/second along your x-axis (0.6c). In its own rest frame, this ruler is 40 fivers long; so in your frame its length will appear to be:
40 * \sqrt{1 - v^2/c^2} = 40 * \sqrt{1 - 9/25} = 40 * 0.8 = 32 fivers long. Also, at either end of this ruler is placed a clock; using the time dilation formula, we can see that for every second on your clock, you will only see these clocks ticking 0.8 seconds forward.
Now, say that when t=0 according to your clock, the clock on the left end of the ruler also reads t'=0. At that moment, a light is flashed on at the left end of the ruler, and you observe how long the light pulse takes to catch up with the right end. In your frame, the position of the light pulse along the x-axis at time t will be c*t, while the position of the right end of the ruler at time t will be v*t + 32. So, the light will catch up to the right end when c*t = v*t + 32 which if you solve for t means t = 32/(c - v). Plugging in c = 5 and v = 3, you get a time of 16 seconds, in your frame.
Now the key to understanding how the ruler can also measure this pulse to be moving at c is to realize that different frames have different definitions of what it means for a pair of clocks to be "synchronized". In your frame, when the clock on the left reads t'=0, the clock on the right will not read t'=0; in your frame, the clock on the right is always 4.8 seconds behind the clock on the left (since the clocks are 40 fivers apart in their own rest frame and they are moving at 3 fivers/second in your frame, then plugging that into the formula vx/c^2 gives (3*40)/(5^2) = 120/25 = 4.8), so it will read t'=-4.8. This means that after 16 seconds have passed according to your clocks, only 16*0.8 = 12.8 seconds will have elapsed on the ruler's clocks, which means the clock on the right reads -4.8 + 12.8 = 8 at the moment that the light reaches the right end. So remembering that light was emitted when the clock on the left read t'=0, the light must have taken 8 seconds to cross the ruler in the ruler's own frame; and remembering that the ruler is 40 fivers long in its own frame, the speed of the light pulse is measured to be 40/8 = 5 fivers/second. So, light does indeed have the same speed in both frames.[/quote]
