Special relativity equation question

[Alex Pavel]

Good evening. I think I've got it as far as time dilation is concerned. Could someone please let me know if this calculation is correct?

- From frame C( stationary camera) Bob is moving at 50,000KM/second parallel to Alice's path
- Bob is also 50,000KM behind Alice on the parallel path
- Bob's path is 299,972 kilometers (1 light second) north of Alice's
- Knowing this, Alice knows that if she fires a laser bolt straight north, Bob will run into it.
- Alice fires the shot. Her camera (C) remains on the spot, facing North. She does not alter her vector
- When the bolt hits Bob, the camera records the time as one second after the shot is taken
- When the bolt hits Bob, Alice's clock shows 0.972 seconds since the shot
- Bob's clock also shows 0.972 seconds since the shot
- I put H (hypotenuse) in for reference but i do not think we need it, even though this is the path Alice 'sees'.
- Key issue is the camera sees path Y and it takes one second to reach the target in that frame.
- What does Alice's clock show if she triggered the shot at 0? Is it the same as Bob's?

Thanks!

 

[mfb]

Knowing this, Alice knows that if she fires a laser bolt straight north, Bob will run into it.

North in the reference frame of C, but not in the reference frame of A. Alice shoots slightly backwards.

Bob's clock also shows 0.972 seconds since the shot

Only if you define "the shot" in the reference frame of C. Otherwise B will start his clock at a different time than C.

When the bolt hits Bob, Alice's clock shows 0.972 seconds since the shot

Same problem here.

Relativity of simultaneity is often overlooked.

 

[Alex Pavel]

Thanks! I am working to understand simultaneity.

- Let's say a train is passing by Bob, heading east at a relative speed of 50,000 KM/Second.
- Bob has a laser rifle pointing geometrically perpendicular to the path of the train.
- He is not aiming at a specific target, rather, he is simply shooting due North relative to the train's eastern route.
- The shot whizzes right over Alice's head, who is at the southern window
- The northern wall of the train is 29,972 KM from the window (train is 1 light second wide)
- On Bob's clock, the shot hits the north wall at one second
- Relative to the entry point of the bolt, the bolt strikes the north wall 50,000 KM west
- Relative to Bob, the bolt strikes the northern wall at 0 (west or east)
- What time is it on Alice's clock when the bolt hits the north wall, taking the entry as 0 on her clock?

When I have this answer I think I will have this down, finally.

 

[PeterDonis]

I am working to understand simultaneity.

Trying to do this without using spacetime diagrams is, IMO, a bad idea. You are drawing diagrams, but they are diagrams in space only. But the whole issue in relativity is that "space" is not an entity by itself; only spacetime is. So to really see what's going on, you need to draw diagrams that explicitly represent time as well as space--i.e., spacetime diagrams.

 

[mfb]

Spacetime diagrams with two spatial dimensions can be a bit tricky to draw (but it is possible).
Find the point in the train where it hits as seen by Alice, the time is then easy to find.

 

[Ibix]

geometrically perpendicular

Is there any other kind of perpendicular?

When I have this answer I think I will have this down, finally.

Write down the coordinates of events in Bob's frame. Apply the Lorentz transforms. When the results stop surprising you, you understand it.

According to Bob, the laser fires at x=0, y=0, t=0. It hits at x=0, y=1, t=1 (distances in light seconds). The train is moving at v=c/6. You just need to feed that velocity and those two sets of coordinates into the Lorentz transforms and you'll get the positions and times of the events according to Alice.

 

[Alex Pavel]

I think I have it.

The bolt hits the far wall for Bob at 1 second, X=0, Y=299972, and Z = 0.

Alice in the train going 50,000 / sec when the bolt hits:
X = -50709
Y = 29972
Z = 0
T = 1.01418

So, is it true when Alice is 50,000KM away from Bob from Bobs perspective.

However, when the bolt hits, Alice is 50,709 away from her perspective, and it happened 1.014 seconds ago?

And this is clearly not due to the 'catch up' time needed for the light information to reach Alice.

So, if Alice wants to bring the train to a stop so she can go get Bob arrested, what decrease in velocity is needed from her perspective?

I suppose she would need to apply enough braking power to reduce speed by 50,709 from her reckoning?

 

[mfb]

However, when the bolt hits, Alice is 50,709 away from her perspective

Alice is 1.014 light seconds away. That's what you just calculated because there was a light beam going from Alice to the other wall (a bit behind her).

So, if Alice wants to bring the train to a stop so she can go get Bob arrested, what decrease in velocity is needed from her perspective?

She has to stop the train - both agree that it travels at 50,000 km/s.

 

[Alex Pavel]

Thanks! These concepts are slowly sinking into my mind.

 

[Alex Pavel]

Is there any other kind of perpendicular?

Is there some sort of different measure for perpendicular in different reference frames where spacetime becomes curved?

 

[mfb]

This is special relativity, no curved spacetime.

Even in General Relativity, locally you always see the spacetime of SR - you can always define "perpendicular" in your reference frame in a clear way. Different observers will measure different angles, however.

 

[Ibix]

Different observers will measure different angles, however.

Depends what they're measuring, surely. The laser will be perpendicular to the track in all frames. Its beam won't.

 

[Alex Pavel]

So even with the phenomenon of length contraction, curved space time, etc, perpendicular remains perpendicular. Even in the case of, say, the aberration of light when you view through a telescope, you will see the object in a different place than it really is, the moment its light reaches the lens. Because the speed of light is finite, we are always somewhat 'seeing the past'. Therefore, even something that appears optically perpendicular is not necessarily considered geometrically perpendicular, depending on velocity.

 

[mfb]

Depends what they're measuring, surely. The laser will be perpendicular to the track in all frames.

Not in all frames. The angle changes in frames moving relative to laser and track but with the direction of motion not aligned with either of them.

So even with the phenomenon of length contraction, curved space time, etc, perpendicular remains perpendicular.

No, not in general.

 

[Alex Pavel]

So, my math shows that if I am in the middle of a train moving at .6C, and I shine a flashlight in both directions, the light hits both the front and back simultaneously by my perspective? Even if one end is coming toward me and the other end is moving away? Is this correct or have I missed something?

 

[Ibix]

So, my math shows that if I am in the middle of a train moving at .6C, and I shine a flashlight in both directions, the light hits both the front and back simultaneously by my perspective?

From your perspective, you aren't moving and neither is either end of the train, so your maths is correct. However this:

Even if one end is coming toward me and the other end is moving away?

...doesn't make sense because the ends of the carriage aren't moving from your perspective.

 

[PeroK]

So, my math shows that if I am in the middle of a train moving at .6C, and I shine a flashlight in both directions, the light hits both the front and back simultaneously by my perspective? Even if one end is coming toward me and the other end is moving away? Is this correct or have I missed something?

Imagine you are on an aeroplane with two friends, traveling at $1000 km/h$. You are in the middle of the plane, one friend at the front another at the back.

Can you tell which direction the plane is moving?

If you toss a coin normally, does it go straight up and down relative to you?

If you throw something to your friend at the front, does it never reach them because you can't throw it at $1000km/h+$?

If you throw something to your friend at the back, does it kill or seriously injure them because they are moving towards it at $1000km/h$?

 

[Alex Pavel]

From your perspective, you aren't moving and neither is either end of the train, so your maths is correct. However this:

...doesn't make sense because the ends of the carriage aren't moving from your perspective.

Therefore, no matter my velocity, if i shine a light on something, regardless of direction, it reaches the target at (c/distance). Hard to process in our primitive brains!

 

[Ibix]

Therefore, no matter my velocity, if i shine a light on something, regardless of direction, it reaches the target at (c/distance).

distance/c, I think. But this is true in any frame. The special thing about the rest frame is that the distance light travels is half the rest length of the carriage in both directions. In any other frame it's shorter in one direction than the other because the carriage ends are moving.

 

[mfb]

Therefore, no matter my velocity, if i shine a light on something, regardless of direction, it reaches the target at (c/distance).

If "distance" is the distance to you at the time it hits the target: yes.