Physicist1231 said:
do you have the math on that? Cause using Newtonian Physics (seems to be a cusw word on here some times
then that Photon that was emitted would have reached all three individuals at the same time but each person is in a separate location. So thus that one photon is in X (number of reference points) places at the same time.
To understand what's going on here, you have to understand that each observer is using clocks at rest relative to themselves which have been synchronized according the the
Einstein clock synchronization convention--the idea is that each observer defines two of their clocks to be "in sync" if they set a light flash off at the midpoint between the two clocks, and both clocks show the same reading at the moment the light strikes them, because each observer
assumes that the light traveled at the same speed in both directions
relative to themselves. But this necessarily implies that if I have two clocks which
I define to be in sync, in your frame my two clocks must be out-of-sync! After all, suppose I am in a ship that is moving relative to you, with clocks at the front and back of the ship, and I set off a flash at the midpoint of the ship and set both to read the same time when the light strikes them. And suppose at the moment I set off the flash at the middle, the ship is passing by you, and you are right next to the flash as it happens, so in your frame you define your own position to be "the position where the flash occurred". In that case, since the ship is moving forwards in your frame, from your perspective the clock at the front of the ship is moving
away from "the position where the flash occurred", while the clock at the back of the ship is moving
towards that position! So if you assume that light moves at the same speed in both directions in
your frame, then according to your own definition of simultaneity the light must catch up with the back clock before it catches up with the front clock. So if I set both clocks to read the same time when the light hits them, then in your frame the clocks must be out-of-sync, because the back clock will show that time at an earlier moment (in your frame) than the front clock. It works out that if the clocks are synchronized in my frame, and the distance between them in my frame is D, then if the ship is moving at speed v in your frame, then at any given instant in your frame the two clocks will be out-of-sync by an amount vD/c^2.
If you combine this equation for the out-of-syncness of moving clocks with the equation for length contraction, which says that a ruler of length L in my frame is shrunk to a length L*sqrt(1 - v^2/c^2) in your frame, and the equation for time dilation, which says that two ticks of my clock which happen at a time interval of T apart in my frame happen at an interval of T/sqrt(1 - v^2/c^2) in your frame (which is equivalent to saying that in your frame the rate at which my clocks tick is only sqrt(1 - v^2/c^2) the rate of your own clocks), then you can construct a simple example to show how both observers measure the speed of a single light beam to be 1c in their own frames, using their own rulers and synchronized clocks. Here is such an example:
Say there's a ruler that's 50 light-seconds long in its own rest frame, moving at 0.6c in your frame. In this case the relativistic gamma-factor of 1/sqrt(1 - v^2/c^2) is equal to 1.25, so in your frame its length is 50/1.25 = 40 light seconds long. At the front and back of the ruler are clocks which are synchronized in the ruler's rest frame; because of the relativity of simultaneity, this means that in your frame they are out-of-sync, with the front clock's time being behind the back clock's time by vD/c^2 = (0.6c)(50 light-seconds)/c^2 = 30 seconds.
Now, when the back end of the moving ruler is lined up with the x=0 light-second mark of your own ruler (which of course is at rest relative to you), you set up a light flash at that position. Let's say at this moment the clock at the back of the moving ruler (which is right next to the flash as it happens) reads a time of 0 seconds, and since the clock at the front is always behind it by 30 seconds in your frame, then in your frame the clock at the front must read -30 seconds at that moment, and it will be at a position of x=40 light-seconds since the ruler has a length of 40 ls in your frame. 100 seconds later in your frame, the back end will have moved (100 seconds)*(0.6c) = 60 light-seconds along your ruler and be at position x=60 ls, which means the front end will be lined up with the x=100 ls mark on your ruler. Since 100 seconds have passed, if the light beam is moving at c in your frame it must have moved 100 light-seconds in that time, so it will also be at the 100-light-seconds mark on your ruler, just having caught up with the front end of the moving ruler.
Since 100 seconds passed in your frame, then thanks to the slower rate of the clocks on the moving ruler this means 100/1.25 = 80 seconds have passed on the clocks at the front and back of the moving ruler. Since the clock at the back read 0 seconds when the flash was set off, it now reads 80 seconds; and since the clock at the front read -30 seconds, it now reads 50 seconds. And remember, the ruler was 50 light-seconds long in its own rest frame! So in its frame, where the clock at the front is synchronized with the clock at the back, the light flash was set off at the back when the clock there read 0 seconds, and the light beam passed the clock at the front when its time read 50 seconds, so since the ruler is 50-light-seconds long, the beam must have been moving at 50 light-seconds/50 seconds = c as well! So you can see that everything works out--if you measure distances and times with rulers and clocks at rest in my frame, you conclude the light beam moved at 1 c, and if a moving observer measures distance and times with rulers and clocks at rest in his frame, he also concludes the same light beam moved at 1 c.
For symmetry we can also consider that at the moment the flash was set off, there was a second ruler moving at 0.6c which was also 50 ls long in its own rest frame, but with its
front end next to the position the flash was set off, which means at the same moment in your frame the back end will be next to the x=-40 ls mark on your ruler. If we assume the clock at the front end of the moving ruler read a time of 0 seconds when the flash went off, then according to the relativity of simultaneity, the clock at the back end must have read 30 seconds at the same moment in your frame.
25 seconds later in your frame, since the back end is moving at 0.6c it will have moved forward by (25 s)*(0.6c) = 15 light-seconds, so it will have moved from x=-40 to x=-25 along your own ruler. And if the light beam was traveling at c in your frame, then the beam will also be at x=-25 light seconds at that moment. And in 25 seconds, the clocks at the front and back only tick forward by 25/1.25 = 20 seconds, so at this moment in your frame the clock at the front reads 20 seconds while the clock at the back reads 50 seconds. So just like with the first ruler, the clock at the far end of the ruler (the one that wasn't right next to the flash when it happened) reads 50 seconds when the light hits it, and again the ruler is 50 light-seconds long in its own frame, and again in its own frame clocks at either end are defined to be "synchronized" and the clock next to the flash read 0 seconds when it happened. So from this you can hopefully see why an observer at rest relative to these rulers would conclude the light moved at the same speed of c in both directions from the flash, just as was true in your own frame.
