# Light beam collision

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Just thinking here.

If two moving, and opposite, sources emitted a radio signal from distances x and y. Would you always see the light beams meet at $\frac{x+y}{2}$?

If you were in-between the sources and nearer to one?

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I think that your own velocity will also make a difference, so it might be that there is not enough information in the question for a simple answer.

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First, let's say I'm stationary.

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As I understand it, under those conditions the answer is no. However there is the small issue of observers in different reference frames seeing a different point of collision than the one you see. And you'd both be "right".

Is this video essentially describing your scenario?

Edited by Sayonara³
multiple post merged

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Stationary relative to whom/what?

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As I understand it, under those conditions the answer is no.

Did you mean yes?

Kinda' like that video but with only one observer.

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I think I understand your setup.

What makes you think the answer is yes?

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As long as you're stationary the beams will always travel towards each other at c, so they'd always meet at $\frac{x+y}{2}$. The sources are like expanding circles.

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Right, the beams of light meet at that point because the speed of light is constant in all inertial frames. But is that necessarily what you see as an observer (assuming you can perceive the collision, of course)? I think I might have confused myself on this one. Damn you Einstein.

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As long as you're stationary the beams will always travel towards each other at c, so they'd always meet at $\frac{x+y}{2}$. The sources are like expanding circles.

The beams will travel toward any inertial-frame observer at c. You'd actually see the two beams approaching each other at 2c.

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QUESTION

how can one see beams moving ?

if i shine avery powerful beam into direction x ( deep space) .

and an observer perpendicular to me is at point y .

would he see the beam propergating outward to x .

I dont see how this is possible, unless there where atoms in the beams path reflecting them my way.

But what if there were no atoms to reflect the beam? You would not see it , correct?

Has there been any experiment done , where the propergating , beam

has been captured . Say on video , like a beam reflecting back from the moon for instance?

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QUESTION

how can one see beams moving ?

if i shine avery powerful beam into direction x ( deep space) .

and an observer perpendicular to me is at point y .

would he see the beam propergating outward to x .

I dont see how this is possible, unless there where atoms in the beams path reflecting them my way.

But what if there were no atoms to reflect the beam? You would not see it , correct?

Has there been any experiment done , where the propergating , beam

has been captured . Say on video , like a beam reflecting back from the moon for instance?

Bouncing lasers off of the moon and measuring time to do so is how we get the most accurate measurements of the moon and its orbit.

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QUESTION

how can one see beams moving ?

if i shine avery powerful beam into direction x ( deep space) .

and an observer perpendicular to me is at point y .

would he see the beam propergating outward to x .

I dont see how this is possible, unless there where atoms in the beams path reflecting them my way.

But what if there were no atoms to reflect the beam? You would not see it , correct?

Has there been any experiment done , where the propergating , beam

has been captured . Say on video , like a beam reflecting back from the moon for instance?

That's correct. If there's nothing to scatter the light, you won't physically see it except by being along its path. In these discussions, however, "see" is not always meant to be taken literally. See, observe, measure are some of the words that are used in this way; essentially it is being assumed you've set up an experiment to take all of the data you'd need to learn what happened.

In this case, it would be a measurement of the time it takes for the light to travel the paths and the distance of those paths. I would conclude that each beam traveled at c.

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