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Author Topic: What would happen to light inside a tube which was accelerating?  (Read 3127 times)

Offline thebrain13

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Heres a thought experiment i thought of that i was wondering what the result would be. What if you had a tube in outerspace, with rockets positioned outside and directed perpendicular to the tube. you then accelerate the tube to say .1c relative to a light source that emits a beam parallel and inside of the tube. The tube is accelerated in the direction perpendicular to the length of the tube. say it was a mile long and about an inch in diameter. would the light make it through or would it get intercepted somewhere inside of it?

[mod - subject edited to make it a question - please endeavour to do this in future to remain consistent with the forum standard, thanks. CS]
« Last Edit: 08/06/2008 22:07:40 by chris »


 

Offline lightarrow

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Heres a thought experiment i thought of that i was wondering what the result would be. What if you had a tube in outerspace, with rockets positioned outside and directed perpendicular to the tube. you then accelerate the tube to say .1c relative to a light source that emits a beam parallel and inside of the tube. The tube is accelerated in the direction perpendicular to the length of the tube. say it was a mile long and about an inch in diameter. would the light make it through or would it get intercepted somewhere inside of it?
But you mean that the tube moves perpendicular to the light beam the source of which is stationary? Then the light will be intercepted.

The condition for light to go along all the tube is:

v/c < d/L

v = tube's speed
d = tube's diametre
L = tube's lenght

With your data: v/c = 0.1; d/L = 1.59*10-5 = 0.0000159
and 0.1 is clearly not < 0.0000159.
You can also see that light will be intercepted after have travelled 160 m along the tube (160 m = 0.1 mile).
« Last Edit: 08/06/2008 14:10:18 by lightarrow »
 

Offline graham.d

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The problem is made a little obscure by you having the tube "accelerated" with respect to the light source and referring to a "beam" of light. I guess, like lightarrow, that you mean the tube to be passing the beam at a constant speed having previously accelerated to 0.1c from a state where it would previously have been aligned with the beam.

If the light source had been inside the tube then the beam would pass through the tube as long as the tube is not accelerating (at its initial or final speeds), but may be intercepted by the sides of the tube during the period of acceleration, depending on how much the acceleration is. This acceleration would then be indistinguishable from a gravitational field inside the tube, which would bend the light towards the side. It may or may not be sufficient to prevent the beam exiting the tube.
 

Offline LeeE

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I seem to recall reading somewhere that this is a factor that has to be taken into consideration in the design of terrestrial based solar telescopes.  Because they generally use very long focal lengths the distance between the optics is great enough that an offset in the targetting has to be made to allow for either the Earth's rotation or it's passage along it's orbit - I can't remember exactly which, and can't find any references to this specific aspect of their designs either, so I might be mis-remembering all this.
 

Offline Soul Surfer

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The process is known as the abberration of light and it was discovered by Bradley even before the velocity of light was measured properly. It causes stars to move their positions in the sky very slightly!.

see   http://en.wikipedia.org/wiki/Aberration_of_light
 

Offline LeeE

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Yup - that'll be it:)
 

Offline lightarrow

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The process is known as the abberration of light and it was discovered by Bradley even before the velocity of light was measured properly. It causes stars to move their positions in the sky very slightly!.

see   http://en.wikipedia.org/wiki/Aberration_of_light

Right. In order for the beam of light to pass through the tube regardless of its lenght, the tube must be tilted towards the source of light by an anle θ such that sinθ = v/c. In this case sinθ = 0.1, so θ ≈ 0.1rad ≈ 5.73.
 

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