Naked Science Forum
On the Lighter Side => That CAN'T be true! => Topic started by: Fruityloop on 14/08/2016 00:28:40

[G] [A1 lightsecondB]
> 0.8c
George (designated by the G) is watching his friends Sue (location A) and Sally (location B) zoom away from him at 0.8c. His friends are located at both ends of a spaceship 1 lightsecond long. From George's point of view, the ship is only 0.6 lightseconds long. Sue sends a light beam to Sally and it only takes 1 second to get from Sue to Sally. According to George, because Sally is running away from the light, the trip actually takes 0.6/0.2 = 3 seconds. So 1 second for Sue and Sally is equal to 3 seconds for George. Sally decides to send a beam of light to her friend Sue. Once again, it takes 1 second to get from Sally to Sue. Because Sue is racing towards the beam of light, according to George the trip takes 0.6/1.8 = 1/3 seconds. So 1 second for Sue and Sally is equal to 1/3 seconds for George.
So which is it?
Is 1 second for Sue and Sally equal to 3 seconds for George or 1/3 seconds for George?

[G] [A1 lightsecondB]
> 0.8c
George (designated by the G) is watching his friends Sue (location A) and Sally (location B) zoom away from him at 0.8c. His friends are located at both ends of a spaceship 1 lightsecond long. From George's point of view, the ship is only 0.6 lightseconds long. Sue sends a light beam to Sally and it only takes 1 second to get from Sue to Sally. According to George, because Sally is running away from the light, the trip actually takes 0.6/0.2 = 3 seconds. So 1 second for Sue and Sally is equal to 3 seconds for George. Sally decides to send a beam of light to her friend Sue. Once again, it takes 1 second to get from Sally to Sue. Because Sue is racing towards the beam of light, according to George the trip takes 0.6/1.8 = 1/3 seconds. So 1 second for Sue and Sally is equal to 1/3 seconds for George.
So which is it?
Is 1 second for Sue and Sally equal to 3 seconds for George or 1/3 seconds for George?
MODS  I think I can reply in this section?
1_{s}=1035mph

MODS  I think I can reply in this section?
1_{s}=1035mph
Yes you can.
In general our policy is not to provide answers to homework questions, but just to provide hints.
In this case the question doesn't make any sense and the OP may have copied it down incorrectly.

Because Sue is racing towards the beam of light
No she isn't. You said so.
His friends are located at both ends of a spaceship 1 lightsecond long.
i.e. not moving with respect to each other.

Because Sue is racing towards the beam of light
I thought it was obvious that I meant from George's point of view.
The question makes perfect sense, I'm not sure what's confusing about it.

Because Sue is racing towards the beam of light
I thought it was obvious that I meant from George's point of view.
The question makes perfect sense, I'm not sure what's confusing about it.
You are saying Sue and Sally are standing in a spaceship that is 1 light second long, Sue is standing at the front of the spaceship and Sally stands at the rear, the spaceship travels away from George, both Sally and Sue move in unison with each other relative to George who observes the event?

Because Sue is racing towards the beam of light
I thought it was obvious that I meant from George's point of view.
The question makes perfect sense, I'm not sure what's confusing about it.
George (designated by the G) is watching his friends Sue (location A) and Sally (location B) zoom away from him at 0.8c. His friends are located at both ends of a spaceship 1 lightsecond long.
Nothing confusing about the question, provided you read what it says. Nobody is travelling towards anyone or anything.
The signals take one second to travel in either direction between Sa and Su. If you know how far away from G they are, you can calculate how long it takes for the information about each signal to reach G.

OK. I think I see where the confusion is. How long it takes for the light from the ship to reach George is irrelevant to the question. Perhaps I shouldn't have used the word 'watching'.

Far from irrelevant, but significant. If they are 1 ls away when Sa transmits then you and Su will receive the signal at the same time, but Su will be 1.8 ls away by the time she receives it, so you won't know she has received it until 2.8 ls after it was sent. And vice versa.

I think the opening post is being completely misunderstood. The length of time for light to pass between Sue and Sally in their own frame of reference is being compared to how long it takes to pass between Sue and Sally from George's frame of reference. How far away George is has nothing to do with this situation.

The difference will always be 1.8 seconds if they are far enough that tan x ~ sin x ~ x where x is the angle subtended at G by AB.

So if the angle is 0 then tan 0 = sin 0 = 0 and for George it must be 1.8 seconds? Why is that? Maybe you can explain. I don't see what is wrong with either the math or the logic in the opening post.