Physicist1231 said:
Jesse,
You need to have rigid space and time to properly calculate any movement ever.
No you don't, you can just calculate everything from the perspective of an inertial frame, without having to say anything about whether that frame is in absolute motion or is at absolute rest.
Physicist1231 said:
I will create a scenario and ask a few questions about the results. You solve for me using Relativity. Then I will solve using NPs.
Sure, but you didn't answer my question, what does "NPs" mean? Presumably the "N" and "P" stand for something?
Physicist1231 said:
Case in point.
Time:
T=0
Positions:
A light source at 0,0,0
Body A, B and C are at coord 20,0,0
Velocities:
Body A remains at 20,0,0
Body B has a speed of .5c,0,0
Body C has a speed of 0,.5c,0
Light is emitted from the source at T=0 and at this time Body B and C start their motion.
Questions:
At what point in time does do each body percieve that light? Please have separate answers for all three and explain the math.
At what coordinates is each body when it perceives the light? Again please have separate answers and the math.
The math is easier if we put the origin at the point where all three coincide (the Lorentz transformation assumes that the spatial origins of each frame coincide at t=0 in each frame, if you drop this assumption you need the
Poincaré transformation), and hopefully you agree that where we put the origin is just a matter of convention, so let's assume that in A's rest frame, A is at x=y=z=0, and at t=0 we assume B and C are also at that position (and let's assume the clocks of all three are set to read 0 at that moment), while the light source is at x=-20, y=0, z=0, and at t=0 in this frame the source sends out a flash. And as you said before, in this frame B has a velocity of 0.5c in the x-direction, and C has a velocity of 0.5c in the y-direction. OK so far?
Let's first find the answer just using A's frame, which is pretty easy. Here a ray of light traveling along the x-axis (which will first pass A, then B) has position as a function of time given by x(t)=1c*t - 20, so it will obviously pass A 20 seconds later at t=20, while B has position as a function of time given by x(t)=0.5c*t, so the light will reach B when 1c*t - 20 = 0.5c*t, or when t=20/0.5=40. Finally, the path of the ray of light that meets C after some time t will be the hypotenuse of a right triangle whose horizontal side has length 20 and whose vertical side has length 0.5c*t, with the hypotenuse itself having length 1c*t since this ray must move at 1c as well. So we have (20)
2 + (0.5c*t)
2 = (1c*t)
2, solving for t gives t=\sqrt{400/0.75} = 23.094010767585. So now we know that the coordinate time of the light reaching B is t=40 and the coordinate time of the light reaching C is t=23.094010767585, but to find the actual times on the clocks of B and C we have to consider the fact that both clocks are running slow by a factor of \sqrt{1 - 0.5c^2/c^2} = 0.866025403784439 in this frame, so when the light reaches B its clock reads 40*0.866025403784439 = 34.6410161513776, and when the light reaches C its clock reads 23.094010767585*0.866025403784439 = 20. And of course A is at rest in this frame, so when the light reaches it at t=20 it reads a time of 20 as well. So, to sum up, when we calculate everything in A's frame we find the following:
--A's clock reads 20 when the light reaches it
--B's clock reads 34.6410161513776 when the light reaches it
--C's clock reads 20 when the light reaches it
Now we can double-check that everything is consistent by analyzing things from the perspective of B's frame, and from C's frame. To save some time I'll just figure out the time on B's clock using B's frame, and the time on C's clock using C's frame, though I could also figure out the time on A and C's clock using B's frame or figure out the time on A and B's clock using C's frame if you really need to see those calculations.
Let's start with B's frame. If we denote the coordinates in B's frame using symbols x',y',z',t', then according to the Lorentz transformation the relation between these coordinates and the coordinates of A's frame (x,y,z,t) is given by:
x' = gamma*(x - vt)
y' = y
z' = z
t' = gamma*(t - vx/c^2)
where v=0.5c, and gamma = 1/sqrt(1 - v^2/c^2) = 1.15470053837925. We know that in A's frame the coordinates of the emitter sending out the flash were x=-20, y=0, z=0, t=0, so in B's frame the coordinates of the flash being sent out are:
x' = 1.15470053837925*(-20) = -23.094010767585
y' = 0
z' = 0
t' = 1.15470053837925*(-0.5*-20/c) = 11.547005383792
B is at rest at the origin in this frame, so naturally if the flash was sent out from position x'=-23.094010767585 and a ray moves towards the origin at 1c, it takes a time of 23.094010767585 for the light to reach B. But the flash wasn't sent out until a time of t'=11.547005383792 in this frame, so naturally it won't reach B until a time of t'=11.547005383792+23.094010767585=34.641016151377. And since B is at rest in this frame, its clock will also read 34.641016151377 when the light reaches it, which is the same (aside from a difference in the last decimal place due to roundoff error) as what we predicted when we calculated things in A's frame.
Now let's figure out what C's clock should read when the light hits it, using C's frame. In A's frame C is moving along the y-axis, so if C uses coordinates x'',y'',z'',t'' then according to the Lorentz transformation these coordinates are related to A's x,y,z,t coordinates by:
x'' = x
y'' = gamma*(y - vt)
z'' = z
t'' = gamma*(t - vy/c^2)
And again we have v=0.5c, and gamma = 1/sqrt(1 - v^2/c^2) = 1.15470053837925. So if we want to know the coordinates of the source sending the flash in C's frame, we take the coordinates in A's frame, x=-20, y=0, z=0, t=0, and plug them into the transformation:
x'' = -20
y'' = 0
z'' = 0
t'' = 0
Very simple in this case! And if light is sent from position x''=-20 at t''=0, while C is at rest at the origin, then obviously if the light moves at 1c in C's frame we will conclude the light hits C at t''=20, and C's clock is keeping pace with coordinate time in this frame so we conclude that C's clock reads 20 at the time the light hits it. Again, this is exactly the same as what was calculated using A's frame.
