Lightspeed.

You are in a spaceship that is not moving in any direction.
You are facing forward, so in the direction of the ships nose (and direction of flight when it moves).
In front of you, a couple of meters away, there is a mirror attached to the wall.
You have a flashlight in your hand. You turn it on and point it at the mirror.
The light takes t to travel from the flashlight to the mirror and back to you again.

The spaceship now accelerates to 50% c and then maintains a constant speed.
You are still in the same spot when you turn the flashlight on and point it at the mirror once again.

How long does the light take to travel from the flashlight to the mirror and back to you again ?
- t
- 1.5t<
- 2t

And why ?

Also, what would prevent light to travel at c relative to the flashlight ?

 

Log in or Sign up to hide all adverts.

-=-

If the speed of the ship is X, everything in the ship is moving at the speed of X. A ball sitting in my hand is moving at the speed of X.
If I throw the ball at the mirror, the ball must move faster than X or it would never reach the mirror.

 

Log in or Sign up to hide all adverts.

Yes, but here we're dealing with a maximum speed; c.

 

Log in or Sign up to hide all adverts.

I'd say 0.75t. You flip the light on at the same point, when the light reaches the mirror, you are 0.5 the way there. On the lights return trip, you meet it again 0.5 the remaining distance.

 

If the ball cannot move faster than the mirror, it will never hit it.

 

So ?
We're talking about light..
And the ship is going half the speed of light (c).

 

Huh? How do you figure the light completes the 'course' faster than when the spacecraft isn't moving ?

 

it takes t in both cases at c is relative to the observer and the observer is not moving relative to themselves.

 

Yes, that's according to the theory.

But now imagine that the mirror is replaced by a window.
One light year ahead is a stationary mirror floating in space.
So when the ship was not moving the light would take one year to arrive at the mirror.
But now that the spaceship is moving at 50% c and the light moves at c relative to the observer, the light will arrive at the mirror in two-thirds of that time (!).
The light will now return to the observer half way through the journey to the mirror. But taking the maximum speed of light, the light and the observer should meet up at 2/3 of the way and not at 1/2 of the way.
This seems to break some rules..
So what am I doing wrong ?

 

Well, you meant to say: "the light will take ONE YEAR," not "one LIGHT year." heh.

But seriously, the light will take one year to reach the mirror: at the point at which you start, the light will take one year (because the mirror is one light year away); at the midway point (one half year has passed), the light from that point even will take only a half-year (because the distance between you and the mirroris one half of a light year) PLUS the half-year which you've spent traveling. So either way, the light will take one year to reach the mirror, irrespective of whether you are stationary or traveling.

 

Oh lol, thanks for pointing that out

Please Register or Log in to view the hidden image!

How can that be ? We just established that the light is traveling at 1.5c relative to the mirror out in space..

Edit: I see you didn't account for the speed of the ship during the time the light spent traveling from the mirror back to the ship. But that's beside the point anyway.

 

Edit #2: The following DOES NOT MAKE SENSE. I don't know what the hell I was thinking.

The speed at which you are traveling is inconsequential, the light still travels at the speed of light. The speed of light is the speed of light. If the mirror were 10 km away and you threw a ball at 10 km/hour from a stationary position, it would take 1 hour to reach the mirror; if you were moving forward 10 km/hour (or 5 even), the ball would still be traveling 10 km/hour relative to you. So at the starting position, the ball takes 1 hour to reach the mirror irrespective of whether you are stationary or moving.

Edit: I don't think I explained that well. Think about it this way: if you a standing still, and throw a ball at a speed of 10 km/hour, it moves at 10 km/hour; but if you are running at 10 km/hour and throw the ball at 10 km/hour, you will both be moving 10 km/hour--and the ball will never leave your side. So actually it's NOT relative to you (relative to you, the ball is not moving--relative to a stationary position, the ball is moving 10 km/hour).

 

I don't understand. The observer is traveling at 0.5c and the light is traveling at 1c relative to the observer.
Then the light must be traveling at 1.5c relative to the mirror, because the ship is traveling at 0.5c relative to the mirror.

Please Register or Log in to view the hidden image!

The only way I can see this working is if the light in the second example from the OP is traveling at 0.5c relative to the observer (and obviously at 1c relative to any stationary objects).
This then means that time has slowed down for the observer making it appear (to the observer) that the light is traveling at 1c relative to him.
The light from the example in post 9 will then be traveling at 1c relative to the mirror out in space..
This seems to prevent any contradictions. What do you think ?

 

That means that you agree with what I wrote in my previous post, right ?
But it is apparently contradicting theory though..

 

First I am not expert at relativity, but I will do my best.

I am also unsure what you mean by

"The light will now return to the observer half way through the journey to the mirror. But taking the maximum speed of light, the light and the observer should meet up at 2/3 of the way and not at 1/2 of the way."

 

First I am not expert at relativity, but I will do my best.

I am also unsure what you mean by

"The light will now return to the observer half way through the journey to the mirror. But taking the maximum speed of light, the light and the observer should meet up at 2/3 of the way and not at 1/2 of the way."

 

I just put that in there to make it clear that we still have an observer for the effect.

If the light is traveling at 1c relative to the mirror out in space by the time the ship is halfway through the journey the light will have reached the mirror.
The light travels twice as fast as the ship, so it will travel 2/3 of the remaining half of the distance back to the ship in the same time that the ship travels 1/3 of the remaining half of the distance.
So the light and the ship meet up again at 4/6 (=2/3) of the original distance.
Did that help ?

 

already addressed--nevermind.

 

Yes that post makes sense to me. I have been trying to work this out with my son. I believe the light would meet you at 2/3 of the way there yes.

 

What bit ? Or did you delete the post ?