Relativity homework help?

Question

Two trains move at right angle to each other, at equal speeds v1=v2=v. Pwo passenges, one on each train fire photonic blast guns in horizontal directions perpendicular to their tracks.
As it turns out both photonic blasts went in the same direction.

What is speed v?

Answer 1

This is how it goes:
- First imagine that train A is traveling rightwards (East) along the x-axis with speed v, and fires his light-beam along the y-axis (North). From his point of view, the beam goes straight north. But now look at it from the point of view of someone standing on the ground. Since the guy in train A is going east, and the spot of light did not appear to HIM to be going east or west, then from the groundsman's point of view, that spot has moved as far east as has train 1. This means that the beam of light has x-velocity = v.
- But if the light-beam has x-velocity = v, what is its y-velocity, as seen from the ground? It cannot be what Mr. A sees, which is speed c; or else, by Pythagoras's theorem, the total speed of light (as seen on the ground) would be sqrt(c^2 + v^2) which is definitely > c. In fact, that is the clue: in order for BOTH Pythagoras' theorem AND the constancy of the speed of light to be true simultaneously, the y-velocity must be sqrt(c^2 - v^2). So, the velocity of the beam from train A is: (v, sqrt(c^2 - v^2)
- For train B, traveling northward along the y-axis and shooting eastward along the x-axis, everything works similarly, but x and y change roles. So the velocity of the beam from train B is: (sqrt(c^2 - v^2), v), also as seen from the ground.
- So now if these two beams to be parallel, they have the same velocity. Equating these two velocities, we see:
v = sqrt(c^2 - v^2) , which immediately leads to
v^2 = c^2/2 , or v = c/sqrt(2).

So that's it: v = c/sqrt(2)

Answer 2

Acording to Einstein Fisrt postulate of relativity the Speed of the photonic blast(speed of light) is the same regardless of the motion of the source.

Answer 3

Well from what I recall the universal speed limit is the speed of light, therefore the photonic blasts would not be able to go any faster than the speed of light.

Just a guess but.. seems logical to me.

Answer 4

let train a be traveling in y direction, train b in x
photon from a train travels in + x direction.
photon from b train travels in + y direction.

draw it out if you need to....

Pa = photon from train a
Pb = photon from train b
c = velocity of photon train a = velocity photon train b

the direction Pa travels is the vector from it's velocity imparted by the train, v and it's velocity from the gun, c

let Θ = angle from x axis to direction Pa is traveling
= angle from x axis to direction Pb is traveling (your problem definition)

(1) for train a, tan Θ = v/c

(2) for b train, tan Θ = c/v

substituting, (1) into (2), v/c = c/v ---> v^2 = c^2

in which case, v = c

the answer is the train must be traveling at the same speed the photon or bullet is fired.

Answer 5

Einstein used Pythagoras with that light clock thingy. I would use it here mate.

Answer 6

God that's fricking hard. Maybe it's c over square root 2 or something. Maybe just c. I bet it's c.

Answer 7

I suppose v = speed of light.