How does light know which path to take ?

  • #1

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This is a two part question. Part 1: I am embarrassed to ask but if someone could please help me. I was thinking about a train car moving at a constant speed which had a simple apparatus onboard. A light source on a pole which emits light straight down only (say a laser of some kind). Below the light is a light sensor which can detect where the light falls (i.e. CCD sensor).

My question: Do the photons of light hit the sensor directly under the source or do they hit the sensor offset some horizontal distance. (see pic)
 

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  • #2
russ_watters
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Light doesn't "know" anything. It goes straight down because the laser is pointed straight down and the laws of physics say that motion is relative. Note, so far this has nothing to do with Relativity or light, as you could run the same experiment with a tennis ball and get the same result, as Galileo predicted.
 
  • #3
Light doesn't "know" anything. It goes straight down because the laser is pointed straight down and the laws of physics say that motion is relative. Note, so far this has nothing to do with Relativity or light, as you could run the same experiment with a tennis ball and get the same result, as Galileo predicted.
I guess my title was wrong for part 1 of my question, but not part 2 which is dependant on the answer to part 1. But the question is still the question. Thankyou for your reply.

I would like confirmation of this though, is this correct that the light travels in a 'angled' path if viewed from an observer on the ground as the train passes by. If so then part 2 of question will follow soon. Thank you for your replies.
 
  • #4
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  • #5
Yes. That is called "aberration".

http://en.wikipedia.org/wiki/Aberration_of_light
Sorry, I should have been more explicit please look at picture. I realize that someone on the ground would not actually even see the light pulse or photons hitting sensor, so I think aberration is not involved here. What I meant is from the observer's perspective if he was informed that the sensor had detected the light directly inline vertically from the source and not offset horizontally then the observer would conclude that the light had travelled along a path with both vertical and horizontal components since the train had travelled some distance in the time it took for the light to have reached the sensor.

So is this correct though that in the attached picture it would be case T1a not T1b. ?
 
  • #8
russ_watters
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Sorry, I should have been more explicit please look at picture. I realize that someone on the ground would not actually even see the light pulse or photons hitting sensor...
The cool thing about a laser in air is that a 3rd party can actually watch the beam since some of it is scattered. So someone on the ground can see the beam extending from the laser to the target.
 
  • #9
Thanks again for the links, it seems there is consensus on this that T1a is what happens.
So part 2 of the questions becomes the following. In the trainb.jpg picture I have depicted a scenario where the train car is travelling right past a light pole with the exact same type light source. Both the light pole and train apparatus fire off a photon of light at the precise moment the two light sources graze each other. So at an instant in time just as the photons have left the light sources it would seem that one of them is travelling slightly faster than the other (in order to be able cover the extra horizontal distance) and it for some reason knows to travel at a slight angle. Both of these photons of light are travelling in the same space and thru the same medium, and a moment later these photons know to keep exhibiting these behavior differences. I need some help understanding what it is about the train's photon, where is this information 'stored' ? Do we have experiments or some way of observing sub atomic difference between the two photons that causes one to behave differently than the other ?

Theoretically, if the observer on the ground knew the train's schedule could he not snag the train's photon on some sensor by quickly sliding it into it's projected path, and do the same with the ground based light pole, and try to figure out what is different between them ? But then the same could be said for an observer on the train, except the measured differences would supposedly by inverted, that is the ground based light pole's photon would be the faster one. I would really appreciate some guidance in understanding how one photo emitted from exactly same type source 'knows 'that it should behave differently than the other photon.
 

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  • #10
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at an instant in time just as the photons have left the light sources it would seem that one of them is travelling slightly faster than the other (in order to be able cover the extra horizontal distance)
No, they both always travel at c in all reference frames.

and it for some reason knows to travel at a slight angle. Both of these photons of light are travelling in the same space and thru the same medium, and a moment later these photons know to keep exhibiting these behavior differences. I need some help understanding what it is about the train's photon, where is this information 'stored' ?
In the momentum. In the rest frame of the emitter the light has a given momentum. Simply transform that momentum into any other frame to determine the momentum of the light in the other frame.
 
  • #11
JesseM
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No, they both always travel at c in all reference frames.
Yes, and given his username, Tim Edilation may be interested to know that if the source is moving along with the train, this implies that the light must take longer to get from the source to the floor of the train as seen in the ground frame than in the train frame (because in the train frame it goes straight down while in the ground frame it goes diagonally, so if it goes at c in both frames it must take longer in the ground frame), and that this is a good way of deriving the existence of time dilation in relativity. See the light clock for more info.
 
  • #12
No, they both always travel at c in all reference frames.

In the momentum. In the rest frame of the emitter the light has a given momentum. Simply transform that momentum into any other frame to determine the momentum of the light in the other frame.
Ok thanks, all this time I did not know that a moving emitter transfers it's momentum to the light being emitted. I thought for instance if an emitter was moving v in x-axis direction, and fired a photon perpendicular say exactly in direction of y when it was at say x coordinate 10, that the when the photon arrived some time later at some distance y away it's x-coordinate would still be 10 not 10 + v*time. Really. Now that I write it out it sounds correct. Are you sure the light photon would continue moving in x direction because the emitter was moving in the x direction ?
 
  • #13
Yes, and given his username, Tim Edilation may be interested to know that if the source is moving along with the train, this implies that the light must take longer to get from the source to the floor of the train as seen in the ground frame than in the train frame (because in the train frame it goes straight down while in the ground frame it goes diagonally, so if it goes at c in both frames it must take longer in the ground frame), and that this is a good way of deriving the existence of time dilation in relativity. See the light clock for more info.
Thanks for the link, am reading it now.
 
  • #14
JesseM
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Ok thanks, all this time I did not know that a moving emitter transfers it's momentum to the light being emitted. I thought for instance if an emitter was moving v in x-axis direction, and fired a photon perpendicular say exactly in direction of y when it was at say x coordinate 10, that the when the photon arrived some time later at some distance y away it's x-coordinate would still be 10 not 10 + v*time. Really. Now that I write it out it sounds correct. Are you sure the light photon would continue moving in x direction because the emitter was moving in the x direction ?
Yes, remember that "moving" and "at rest" have no absolute meaning in relativity, they can only be defined relative to a particular arbitrary choice of inertial frame, and the first postulate of relativity says that the laws of physics work exactly the same in every inertial frame. So if I have an emitter which is "at rest" relative to my own frame, and I see that the beam is emitted parallel to the direction of the emitter in my frame, then it must likewise be true in your frame that if you have an emitter "at rest" relative to your frame, the beam must be emitted parallel to the direction of the emitter in your frame. So if your emitter is hanging straight down from the ceiling of a room at rest in your frame, the spot from the beam on the room's floor will necessarily be directly under the emitter--and that must mean that your beam moves at a diagonal as viewed in my frame!
 
  • #15
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The cool thing about a laser in air is that a 3rd party can actually watch the beam since some of it is scattered. So someone on the ground can see the beam extending from the laser to the target.
Hey Russ,
Can you please refer me to a link that shows actual experimental pictures of light aberration, as seen from a 3rd party standing outside the moving light apparatus?

Thanks,
Roi.
 
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  • #16
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Are you sure the light photon would continue moving in x direction because the emitter was moving in the x direction ?
Yes. I am sure. I can post the math if you like.
 
  • #17
Yes. I am sure. I can post the math if you like.
Thanks for the offer, but it is not the math (i am capable of the simple math involved) but the notion that light maintains the momentum of the emitter. I am willing to admit I was wrong all these years, but just one last confirmation. I made a pic called trainc.jpg which is plan view of a moving train whose emitter shoots out a pulse towards the east. I am still stuck on the notion that the light will not hit a wall some distance away at exactly the north coordinate that the train was at when it fired the photon. I will go away after this but one last confirmation, The light will hit the wall farther north then when the photon was fired, actually it will hit the wall at the current north position of the train (since north velocity is imparted to the photon) Really ? I thought it would have hit the wall at the dashed line. Is everyone here sure the photon keeps the northerly momentum from the train's emitterr ?
 
  • #18
Yes. I am sure. I can post the math if you like.
whoops forgot picture
 
  • #19

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  • #20
JesseM
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Thanks for the offer, but it is not the math (i am capable of the simple math involved) but the notion that light maintains the momentum of the emitter. I am willing to admit I was wrong all these years, but just one last confirmation. I made a pic called trainc.jpg which is plan view of a moving train whose emitter shoots out a pulse towards the east. I am still stuck on the notion that the light will not hit a wall some distance away at exactly the north coordinate that the train was at when it fired the photon. I will go away after this but one last confirmation, The light will hit the wall farther north then when the photon was fired, actually it will hit the wall at the current north position of the train (since north velocity is imparted to the photon) Really ? I thought it would have hit the wall at the dashed line. Is everyone here sure the photon keeps the northerly momentum from the train's emitterr ?
Yes, did you understand my point in post #14 about all velocities being relative? If so, just consider things from the perspective of a frame where the train is at rest and the wall is moving, this is also the rest frame of the emitter so naturally the laser will travel in a purely easterly direction in this frame, not in the wall's frame.
 
  • #21
Yes, did you understand my point in post #14 about all velocities being relative? If so, just consider things from the perspective of a frame where the train is at rest and the wall is moving, this is also the rest frame of the emitter so naturally the laser will travel in a purely easterly direction in this frame, not in the wall's frame.
Thank you for sticking with it. I appreciate it. I get it now. Thanks
 
  • #22
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The light will hit the wall farther north then when the photon was fired, actually it will hit the wall at the current north position of the train (since north velocity is imparted to the photon) Really ?
Really, really.

If you can do the math, then you may want to work a problem on your own just to convince yourself that it is correct, even if you don't have an intuitive "feel" for it yet.
 
  • #23
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The cool thing about a laser in air is that a 3rd party can actually watch the beam since some of it is scattered. So someone on the ground can see the beam extending from the laser to the target.
Hey Russ,
Can you please refer me to a link that shows actual experimental pictures of light aberration, as seen from a 3rd party standing outside the moving light apparatus?

Thanks,
Roi.
 
  • #24
JesseM
Science Advisor
8,496
12
Hey Russ,
Can you please refer me to a link that shows actual experimental pictures of light aberration, as seen from a 3rd party standing outside the moving light apparatus?

Thanks,
Roi.
You'd need a source traveling at a significant fraction of light speed to actually see the change in angle visibly, which isn't possible in a lab! Russ didn't say specifically that aberration could be observed this way, just that lasers can be "visible" in some experiments. But for something in the ballpark of what you're asking you could consider the aberration of lasers aimed at orbiting reflectors on satellites (see the bottom of this page) or at the moon (see here).
 

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