Two Spaceships question

[swansont]

oh yeah ^^;

ok' date=' so then from the shooter frame, the distance that the target will travel in the right amount of time from the rest frame.

 

what if he didn't know he was moving though? would he be able to shoot in the right place?

[/quote']

 

 

He doesn't know that he's moving. As far as he's concerned, he and the other ship are at rest, and the rest of the universe is moving. Since the target isn't moving, he aims straight at it, and hits.

[Jacques]

He doesn't know that he's moving.

Because he is blind !

 

he and the other ship are at rest

That the principel of relativity.

 

The hard part in relativity is to leave the notion of absolute space behind us.

No reason to have an absolute space.

 

But here we have a prefered frame: the frame of the light source. When the source emit the photons, they have the momentum of the electrons of the laser in the ship. Is it a way we can rationalize the surprising solution of relativity to this problem ?

[iglak]

why does it hit?

that doesn't make any sense to me.

unless the movement of the spaceship is added to the movement of the laser, but that doesn't work according to all of those "what would happen if a car moving at c turned on it's headlights?" threads.

 

and how is it possible that it'd appear that he's firing ahead of him from a rest frame if from his frame he's firing behind him? to my knowledge, that's not explained in relativity.

[swansont]

why does it hit?

that doesn't make any sense to me.

 

 

If you've ever shot at a target: did you take into account the speed of the rotation of the earth (0.46 km/s at the equator)? The orbital speed of the earth (30 km/s)? Why not?

[insane_alien]

when the laser is fired directly towards the second ship it still has the velocity of the ship (1000km/s i think but it is irrelevant) and it is also moving towards the second ship

at what would appear to be c from the ships due to time dilation and RELATIVITY but to an outside observer where he/she/it sees the ship moving and the light going in a diagonal line at c. this is a crap explanation but its the best i can do at my current state of hungoverness.

[calbiterol]

I wonder if the way you read the question makes any differnce, but it's quite possible it doesn't. When I originally read the situation, I thought the spaceships were travelling parallel, one behind the other (essentially the same line), and in that case, it really wouldn't matter where the laser was aimed, as long as it is dead ahead. Make sense? But I don't know about sideways... I would think you would have to accomodate for the fact that it takes light time to travel the distance, but then again, it's like they're stationary. Ahh, light bulb. Anyways, I just thought it was interesting that I read the question one way, and from the looks of it, everyone else read it differently.

[Jacques]

But I don't know about sideways...

That what the question is all about

[calbiterol]

I don't think it was ever specified which one, but it definitely seems like everyone else took it that way. No matter, they are at V=0 WRT each other and no "leading" of the target (the other ship) is needed.

[pljames]

Two Spaceships Head out of the solar system... They position themselves exactly 1 million km apart and then accelerate, in the same direction, to a velocity of 1000 km/s. They maintain this velocity until they leave the galaxy (Yes I know this would take a long time). Once out of the galaxy they both coast at this velocity for the duration of my question.

 

[/i][/b]Spaceship (a) is in what relevant position/frame to spaceship (b). Position = side,top,bottom, ahead or behind? pljames

[losfomot]

 

[/i][/b]Spaceship (a) is in what relevant position/frame to spaceship (b). Position = side' date='top,bottom, ahead or behind? pljames[/quote']

 

I looked back at my original question, and it seems obvious to me. They are travelling PARALLEL to eachother. Parallel does not usually mean one in front of the other. Two seperate paths that are parallel to eachother... Here, I will try to illustrate it:

 

^^^^^^^^^^^^^^^^^^

A --One Million Km apart -- B

 

where ^ = direction of travel

[calbiterol]

Yes, but parallel lines can be the same line. The definition of parallel means they have the same slope. Two of the same lines have the same slope.

 

But that's overanalyzing it.

[eon_rider]

very cool thread....sorry for butting in, but I'm learning alot...

 

Thanks

[Halucigenia]

Hey, I like the idea of these text diagrams, it reminds me of the good? old days of computing.

 

Anyway since there still seems to be some confusion about these 2 spaceships and their laser beam experiment I will try to explain with the aid of some text diagrams.

 

Any of you who are more knowledgeable than me on the subject are welcome to point out if I have made any mistakes.

but here is how I understand the situation.

 

A = spaceship A

 

B = spaceship B

 

--- = laser beam

 

> & < = direction of beam

 

(Diagrams are on X Y spatial axes)

 

 

A--<-->--B

 

observer at rest

WRT A&B

 

 

To clarify a bit let's split the diagram into 3 times, all with the observer(O) at rest WRT A&B

t1 when A fires laser at B, t2 when B sees the laser beam hit B and t3 when A sees that the laser beam has hit B

(time increasing up the page, which I think is the correct convention)

 

t3 A<-------B

 

 

t2 A------->B

 

 

t1 A--> B

 

As A, B and the observer (O) are all at rest WRT each other there are no angles involved

A just fires directly at B and hits the target.

No relativity involved at all here.

 

Now for the difficult bit, the view from an observer (O') at rest WRT to the star by which A and B calculate their velocity. (this is where the text diagrams idea breaks down, so I will link to images instead)

 

\ and / = laser beam

| = path of A & B

 

 

Now this seems odd because the observer sees A shoot at an angle to hit B.

However, think of it like this - 2 kids playing catch with a ball on a train, to them the ball is going straight back and forwards.

But to an observer it will look like the ball has a zig zag path

 

Now for the really hard bit - what about the speed of light, how come both the observers see light traveling at the same speed.

Well, for the observer (O) at rest WRT A & B there is no problem, it's as if the only thing that is moving is the laser beam.

(and that distant star, but all distant stars appear to be moving and some in different directions, so that can't be affecting anything)

 

For the observer (O') at rest WRT to the star by which A and B calculate their velocity, it is a different matter, now we have to invoke relativity.

Because A, B and O appear to him to be moving, their clocks appear to be going slower, so their calculation of the speed of light emitted by the laser appears to be wrong.

To this observer, O', the time taken for the laser light to travel between A and B, what appears to him to be a longer path than the other observer O would see, is still consistent with his calculation of C.

 

Now for the realy realy hard bit attempting to create text spacetime diagrams of the 2 scenarios.(If you don't know what a specetime diagram is, google it)

 

 

observer at rest

WRT A&B

 

Surprisingly this looks just like the observer O' diagram above, but remember that the y axis is time.

 

For the spacetime diagram from the perspective of O' imagine the A and B worldlines tilted, but the Laser lines remaining at the same angle (the angle should always be 45 deg on a spacetime diagram). What would happen is that the distance (on the diagram) between the points at t1, and t2 would have to stretch to accommodate this.

Anyway what you should see is that the times between t1 abd t2 would be greater, according to O', leaving the speed of light the same - i.e. always = to C.

 

 

observer at rest

WRT the star.

 

I hope that this helps, it was fun to try and do.