Hurkyl said:
Ok, let us put coordinates on this system in order to solve it. (Things are much clearer with coordinates, which is probably one of the reasons why anti-SR crackpots tend to dislike them)
In the rest frame of C, let us put the event where A, B, and C are all at the same place at the origin of the coordinate system. Let the coordinate axes x, y, and z be right, up, and out of the page respectively, and let t be the coordinate time.
The worldlines of A and B are given parametrically by (being sloppy and parametrizing by t):
A: (t, 0, 0.7ct, 0)
B: (t, 0.7ct, 0, 0)
Letting t', x', y', and z' be the coordinates in A's reference frame, the Lorentz transform from C's frame to A's frame gives the relations:
x' = x
y' = γ(y - 0.7ct)
z' = z
t' = γ(t - 0.7y/c)
where γ := 1/√(1-0.72) = 1/√0.51
So, the worldlines are given parametrically in A's frame by (yes, I mean to parametrize by t and not t'):
A: (γ(t - 0.7 (0.7ct)/c), 0, γ(0.7ct - 0.7ct), 0)
A: (γ 0.51 t, 0, 0, 0)
B: (γ(t - 0.7 0/c), 0.7ct, γ(0-0.7ct), 0)
B: (γt, 0.7ct, -γ0.7ct, 0)
(I only did A as a sanity check)
So, we have that the coordinate velocity of B in A's frame is given by
v = (0.7c √0.51, -0.7c, 0)
So, unless I've made a mistake along the way, B's speed relative to A is 0.7 c √1.51.
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No, you didn't make a visible mistake along the way, and the reason you didn't is because the initial speed was taken as .7c. Let it be c, and redo the calculation.
Kind regards,
StarThrower
Here, I will start you off:
GOAL: Compute the relative speed of B and A.
As Hurkyl did, start off viewing the motion of A,B in the rest frame of C.
Let there be a clock at rest (anywhere in this frame, it doesn't matter), whose readings are denoted by the letter t. At the beginning of this event, the clock reads zero. As time goes foward, this clock's readings increase numerically. And let us presume that the clock ticks so fast, that whenever point A or point B changes coordinates in this frame, this clock has changed readings. Furthermore, let the tick rate of this clock be constant. This having been said...
At any moment in time t in this frame, the y coordinate of point A is ct on the y axis, x coordinate = 0, and z coordinate =0.
Simultaneously in this frame, the x coordinate of B in this frame is ct, y coordinate =0, and z coordinate =0.
Or using coordinates so as to not be accused of being an anti-SR crackpot, we have:
A: (t, 0, ct, 0)
B: (t, ct, 0, 0)
Now, we are interested in viewing the motion of B in the rest frame of A. Let a coordinate system have been set up, whose origin is the point A. Let the coordinate axes x`, y`, and z` be right, up, and out of the page respectively, and let t` be the coordinate time. The motion of B is going to be diagonally down, and to the right, placing B in the fourth quadrant of this frame FA, once t`>0.
There is a clock at rest in this frame, whose rest rate matches that of the other clock. In fact, this clock is constructed identically to the other clock. When the event begins, this clock reads zero too. Hence, if t=0 then t`=0.
The Lorentz transformations from rest frame C, to rest frame A give:
x' = x
y' = γ(y - ct)
z' = z
t' = γ(t - y/c)
where
γ := 1/√(1-1) = 1/0
Oops, I guess I can't use the Lorentz tranformations. Let me try the Galilean transformations instead... etc. etc.
Lets check out the location of B in FA, when t` = 1.
We have a clock at rest in frame A, which we shall call clock A. In the time it took clock A to go from zero, to 1, the point C moved downwards, onto the negative y` axis. This distance down it moved (in this rest frame FA), is given by ct`. Now, the y coordinate of A in reference frame C, at any moment in time according to clock C is ct, and the y` coordinate of C in reference frame A according to clock A is -ct`. Separation distance must be equivalent, hence we must have |-ct`| = |ct| which implies t=t`. In other words, the current coordinates of point C in frame A are: (1, 0, -c, 0), and clock C also reads 1.
Using the galilean tranformation we have t = t`, hence t = 1.
In this time, the point B has moved onto the positive x-axis of frame C, at an x coordinate of ct or ct` (take your pick). This places point B in the fourth quadrant of the x`,y` plane, at a location with the following coordinates in frame A:
(t`,x`,y`,z`) = (1,c,c,0)
At this point in the argument c is in meters, because c = 299792458 m/s, but t` = one second in frame A= one second in frame C = t, hence the distance in meters is 299792458 meters.
The hypotenuse of this triangle can be computed using the pythagorean theorem. It corresponds to the distance traveled by point B in frame A, in the time it took clock A to go from zero to one. Using the pythagorean theorem we have:
D^2 = c^2 + c^2 = 2c^2
From which we infer that the distance traveled is:
D = (\sqrt{2}) c = (\sqrt{2}) (299792458) meters
Hence, the speed of B relative to A, in the rest frame of A is:
|\vec{V}| = (\sqrt{2}) (299792458) meters/second
Because it did so in one tick of clock A, and I recently stipulated that the unit of clock A is the second.
Let objects A,B be photons. Thus, the relative speed of these two photons exceeds the speed of light in the rest frame of A. Thus, if the fundamental postulate of the special theory of relativity is true, then FA is not an inertial reference frame. FC is an inertial reference frame by stipulation, and FA is moving relative to FC at a constant speed. Therefore, FA is an inertial reference frame. Therefore, FA is and isn't an inertial reference frame, which is impossible. Therefore, the fundamental postulate of SR is false.
Kind regards,
StarThrower
