Photon path .. ALT Junior Member Posts: 45 Threads: 2 Joined: May 2016 Reputation: 0 05-10-2016, 12:30 PM Hi all. I have no theory or hypothesis - I just want to ask a question. I thought this the best place to do it, but please move if needed. I am curious about the trajectory of a photon. Not a group of photons, or a beam of light, but to keep it simple, one photon. I understand that a photon is a mass-less particle, and I understand that, being light, regardless of the speed of travel of the photon emitter., the photon will travel at c. But there is another issue I have tried to explore and understand, and having raised it on various science forums has resulted in termination of the subject thread, or derailment of it, after what seemed to me, much obfuscation. So I'll see if you guys can help. But first, let me back up a bit. Imagine I am standing on a wide, vast, flat, open plain. In front of me, at some convenient distance, is a rail track. On the track, mounted on a carriage and perpendicular to the horizontal plain, i.e., pointing vertically upwards, is a rifle. The carriage with the fixed rifle is travelling from my left to my right; L > R. Directly above the rifle barrel and rectilinear to it, at some convenient distance, is affixed a target. A bullet is fired. Does the bullet hit the target ? Or does it miss it, given that the whole assembly is moving at speed L > R, and the target would have moved away to the right, before being hit. We al agree that (ignoring air resistance, etc.) the bullet would hit the target, as it is also moving L > R along with the gun and the whole assembly. Simple. And I apologise for having bored you with this so far, but I needed to put it like that, so you understand where I'm coming from, and what I ask next. Now imagine the same setting, but instead of a rifle we have a photon emitter, and scale the speeds, etc., to suitable measurements for this to be considered. The emitter. fires one photon. Does the photon hit the target (now a photoelectric receptor) or does it miss it ? It would hit it if it was influenced by the transverse motion of the emitter. It would miss it if it wasn't so influenced. There can only be two possible answers. Which is it ? Thanks. Schmelzer Administrator Posts: 215 Threads: 31 Joined: Dec 2015 Reputation: 0 05-10-2016, 03:11 PM Very good question. The answer is given by the equivalence principle - the moving train is unable to identify the own movement, but if the target would be missed only if it moves, this would allow to detect the own movement. But at a first look, another, wrong answer comes to mind: the light ray moves from the same event, in the same direction - orthogonal - with the same velocity c. But the target moves. So, the light ray can hit only one of the two different targets. What is wrong here? We have to add the velocities. Velocity of the light ray from the train at rest in y direction would be (0,c,0). Once the train moves in x direction with some (v,0,0), we have to add the two velocities. With the relativistic formula, of course. Which gives $$(v,c\sqrt{1-v^2/c^2},0)$$. This is also a light ray with velocity c. But in a different direction. So, it will hit a different target. rjbeery Junior Member Posts: 2 Threads: 1 Joined: May 2016 Reputation: 0 05-10-2016, 10:37 PM (05-10-2016, 12:30 PM)ALT Wrote: Hi all. I have no theory or hypothesis - I just want to ask a question. I thought this the best place to do it, but please move if needed. I am curious about the trajectory of a photon. Not a group of photons, or a beam of light, but to keep it simple, one photon. I understand that a photon is a mass-less particle, and I understand that, being light, regardless of the speed of travel of the photon emitter., the photon will travel at c. But there is another issue I have tried to explore and understand, and having raised it on various science forums has resulted in termination of the subject thread, or derailment of it, after what seemed to me, much obfuscation. So I'll see if you guys can help. But first, let me back up a bit. Imagine I am standing on a wide, vast, flat, open plain. In front of me, at some convenient distance, is a rail track. On the track, mounted on a carriage and perpendicular to the horizontal plain, i.e., pointing vertically upwards, is a rifle. The carriage with the fixed rifle is travelling from my left to my right; L > R. Directly above the rifle barrel and rectilinear to it, at some convenient distance, is affixed a target. A bullet is fired. Does the bullet hit the target ? Or does it miss it, given that the whole assembly is moving at speed L > R, and the target would have moved away to the right, before being hit. We al agree that (ignoring air resistance, etc.) the bullet would hit the target, as it is also moving L > R along with the gun and the whole assembly. Simple. And I apologise for having bored you with this so far, but I needed to put it like that, so you understand where I'm coming from, and what I ask next. Now imagine the same setting, but instead of a rifle we have a photon emitter, and scale the speeds, etc., to suitable measurements for this to be considered. The emitter. fires one photon. Does the photon hit the target (now a photoelectric receptor) or does it miss it ? It would hit it if it was influenced by the transverse motion of the emitter. It would miss it if it wasn't so influenced. There can only be two possible answers. Which is it ? Thanks. Take your photon emitter and place the rifle barrel above it so that the photon travels through it.  Clearly the photon must move an an angle from the emitter to the target in order to traverse the rifle barrel.  How the person standing in the plain reconciles this is not immediately clear because it depends on the specific construction of the emitter.  Perhaps the emitter appears tilted? secur Member Posts: 165 Threads: 0 Joined: May 2016 Reputation: 0 05-10-2016, 11:33 PM hi rjbeery, If I understand you correctly, no, the rifle barrel wouldn't appear tilted. First note that to visualize this at all we must imagine the train is traveling very fast. As the photon travels up the rifle barrel, it moves sideways at just the right speed (from pov of person on plain, of course) so the photon, which is moving on a diagonal, is always within the barrel. But the barrel always appears upright. Note, of course, he can't actually see the photon in flight! But we can imagine instead a laser beam, say, going through fog, filmed by a super-high speed camera; so you can actually see the progress of the light pulse. Consider the following gedanken. Suppose the rifle barrel is a long transparent tube (filled with diffuse fog) with pulsed lasers at both ends, pointing at each other. Then person on plain sees two pulses travel diagonally along the barrel towards each other. Clearly the barrel can't tilt to accommodate their passage, it would have to tilt in two directions at once! Instead it moves sideways, remaining upright, at the right speed so each pulse remains centered in the barrel as it travels. secur Member Posts: 165 Threads: 0 Joined: May 2016 Reputation: 0 05-11-2016, 01:27 AM ALT: It would hit it if it was influenced by the transverse motion of the emitter. It would miss it if it wasn't so influenced. There can only be two possible answers. Which is it ? BTW ALT note that actually, it's not one of these two alternatives! The photon is influenced by the transverse motion. But (as Schmelzer explained) the vector addition is relativistic, so it winds up traveling orthogonally towards the target slower by sqrt(1 - v^2/c^2). So it will miss, by falling short. The reason the rifle bullet does hit the target: vector addition is also relativistic in this case but the speeds are so slow that sqrt(1 - v^2/c^2) is negligible. ALT Junior Member Posts: 45 Threads: 2 Joined: May 2016 Reputation: 0 05-11-2016, 01:59 AM (This post was last modified: 05-11-2016, 02:25 AM by ALT. Edit Reason: edit 1st line. ) (05-10-2016, 03:11 PM)Schmelzer Wrote: Very good question.  The answer is given by the equivalence principle - the moving train is unable to identify the own movement, but if the target would be missed only if it moves, this would allow to detect the own movement.   But at a first look, another, wrong answer comes to mind:  the light ray moves from the same event, in the same direction - orthogonal - with the same velocity c.  But the target moves.  So, the light ray can hit only one of the two different targets.  What is wrong here?   We have to add the velocities.  Velocity of the light ray from the train at rest in y direction would be (0,c,0).  Once the train moves in x direction with some (v,0,0), we have to add the two velocities.  With the relativistic formula, of course.  Which gives $$(v,c\sqrt{1-v^2/c^2},0)$$.   This is also a light ray with velocity c.  But in a different direction.  So, it will hit a different target. Hi Schmelzer .. I am a little bewildered about this thread being moved to Special Relativity, not because it isn't the appropriate thing to do, but because I know as much about SR as I do about brain surgery.  No matter - just as long as y'all understand this. Your 1st para; I understand what you are saying and tend to agree. Your 2nd para; Yes .. Your 3rd; The math is completely beyond me (it really is - you wouldn't even call me a layman) but I agree with your conclusion - it will hit a different target, ie, miss the target directly above it. I will have something very important (to me at least) to say about that soon, but first, I want to deal with the other replies, and also stew on it all for a while. Thanks. (05-10-2016, 10:37 PM)rjbeery Wrote: (05-10-2016, 12:30 PM)ALT Wrote: Hi all. I have no theory or hypothesis - I just want to ask a question. I thought this the best place to do it, but please move if needed. I am curious about the trajectory of a photon. Not a group of photons, or a beam of light, but to keep it simple, one photon. I understand that a photon is a mass-less particle, and I understand that, being light, regardless of the speed of travel of the photon emitter., the photon will travel at c. But there is another issue I have tried to explore and understand, and having raised it on various science forums has resulted in termination of the subject thread, or derailment of it, after what seemed to me, much obfuscation. So I'll see if you guys can help. But first, let me back up a bit. Imagine I am standing on a wide, vast, flat, open plain. In front of me, at some convenient distance, is a rail track. On the track, mounted on a carriage and perpendicular to the horizontal plain, i.e., pointing vertically upwards, is a rifle. The carriage with the fixed rifle is travelling from my left to my right; L > R. Directly above the rifle barrel and rectilinear to it, at some convenient distance, is affixed a target. A bullet is fired. Does the bullet hit the target ? Or does it miss it, given that the whole assembly is moving at speed L > R, and the target would have moved away to the right, before being hit. We al agree that (ignoring air resistance, etc.) the bullet would hit the target, as it is also moving L > R along with the gun and the whole assembly. Simple. And I apologise for having bored you with this so far, but I needed to put it like that, so you understand where I'm coming from, and what I ask next. Now imagine the same setting, but instead of a rifle we have a photon emitter, and scale the speeds, etc., to suitable measurements for this to be considered. The emitter. fires one photon. Does the photon hit the target (now a photoelectric receptor) or does it miss it ? It would hit it if it was influenced by the transverse motion of the emitter. It would miss it if it wasn't so influenced. There can only be two possible answers. Which is it ? Thanks. Take your photon emitter and place the rifle barrel above it so that the photon travels through it.  Clearly the photon must move an an angle from the emitter to the target in order to traverse the rifle barrel.  How the person standing in the plain reconciles this is not immediately clear because it depends on the specific construction of the emitter.  Perhaps the emitter appears tilted?Hi rjbeery and thank you for the contribution. Placing a barrel above the photon emitter was not part of my thought experiment. So I don't want to complicate it with that. Also, observer dependent aspects too, was not part of it. The emitter is perpendicular to the track. Whether it appears so or not is irrelevant to the result. In other places, much circuitous argument, and and even more obfuscation transpired by bringing in various observers, etc. Therefore, I will now obviate the possibility of this happening here, with a slight amendment .. There is no observer. The hit or miss is recorded by computer and the result is subsequently transmitted to me in Sydney and my friend in Athens (and my mistress waiting for me in the Bahamas). We all get the same info. (05-10-2016, 11:33 PM)secur Wrote: hi rjbeery, If I understand you correctly, no, the rifle barrel wouldn't appear tilted. First note that to visualize this at all we must imagine the train is traveling very fast. As the photon travels up the rifle barrel, it moves sideways at just the right speed (from pov of person on plain, of course) so the photon, which is moving on a diagonal, is always within the barrel. But the barrel always appears upright. Note, of course, he can't actually see the photon in flight! But we can imagine instead a laser beam, say, going through fog, filmed by a super-high speed camera; so you can actually see the progress of the light pulse. Consider the following gedanken. Suppose the rifle barrel is a long transparent tube (filled with diffuse fog) with pulsed lasers at both ends, pointing at each other. Then person on plain sees two pulses travel diagonally along the barrel towards each other. Clearly the barrel can't tilt to accommodate their passage, it would have to tilt in two directions at once! Instead it moves sideways, remaining upright, at the right speed so each pulse remains centered in the barrel as it travels.Hi secur. As stated above, the photon is NOT travelling up any rifle barrel first. It is emitted from the tip of a photon emitter. Considering measurements, tolerances, etc, please note this is just a thought experiment, although it certainly is performable and measurable with today’s technology - search .'ring laser gyroscope' for instance. And note, to keep it simple, there is now NO observer. See above post. (05-11-2016, 01:27 AM)secur Wrote: ALT: It would hit it if it was influenced by the transverse motion of the emitter. It would miss it if it wasn't so influenced. There can only be two possible answers. Which is it ? BTW ALT note that actually, it's not one of these two alternatives! The photon is influenced by the transverse motion. But (as Schmelzer explained) the vector addition is relativistic, so it winds up traveling orthogonally towards the target slower by sqrt(1 - v^2/c^2). So it will miss, by falling short. The reason the rifle bullet does hit the target: vector addition is also relativistic in this case but the speeds are so slow that sqrt(1 - v^2/c^2) is negligible. The science and math is beyond me, but your conclusion is clear. It will MISS. It will fall short. It IS influenced by the transverse motion. VERY interesting ! secur Member Posts: 165 Threads: 0 Joined: May 2016 Reputation: 0 05-11-2016, 04:32 AM (This post was last modified: 05-11-2016, 04:36 AM by secur.) ALT wrote: ... your conclusion ... Please note, although I understand the math just fine I'm not a physicist. So it's not actually "my" conclusion; rather, all I intended to do was clarify Schmelzer's conclusion. Although his math clearly says the photon IS influenced by the transverse motion, he didn't say so explicitly and I thought a layman might miss it. But I should have left it at that, and not said it misses the target. I meant to just repeat his statement that it would "hit a different target". But I should have thought twice. Consider: if you're moving with the train, to you emitter and target are stationary. Evidently the photon will go straight across and hit the target. So if you're watching from any other reference frame you have to see the same event happen - although it takes longer. That must be what "a different target" meant. Sorry for the confusion ... I hope you keep digging and make sure to get this right. Also my post about photon travelling along the rifle barrel was to correct rjbeery's idea that the barrel would appear slanted, I realize that's not your question. Schmelzer Administrator Posts: 215 Threads: 31 Joined: Dec 2015 Reputation: 0 05-11-2016, 06:46 AM (05-11-2016, 01:59 AM)ALT Wrote: Hi Schmelzer .. I am a little bewildered about this thread being moved to Special Relativity, not because it isn't the appropriate thing to do, but because I know as much about SR as I do about brain surgery.  No matter - just as long as y'all understand this.The speed of light was involved, the answer required relativistic addition of velocities, a relativistic formula, and that you are a layman does not matter, they also are allowed to have questions about special relativity. (05-11-2016, 01:59 AM)ALT Wrote: (05-10-2016, 03:11 PM)Schmelzer Wrote: But at a first look, another, wrong answer comes to mind:  the light ray moves from the same event, in the same direction - orthogonal - with the same velocity c.  But the target moves.  So, the light ray can hit only one of the two different targets.  What is wrong here?   We have to add the velocities.  Velocity of the light ray from the train at rest in y direction would be (0,c,0).  Once the train moves in x direction with some (v,0,0), we have to add the two velocities.  With the relativistic formula, of course.  Which gives $$(v,c\sqrt{1-v^2/c^2},0)$$.   This is also a light ray with velocity c.  But in a different direction.  So, it will hit a different target.The math is completely beyond me (it really is - you wouldn't even call me a layman) but I agree with your conclusion - it will hit a different target, ie, miss the target directly above it.Ups. I see, you have misunderstood my answer - and it is my fault. So, to correct this misunderstanding: Train at rest, target at rest -> hit. Train moving, target moving (say, target something inside the train) -> hit. Train moving, target at rest (say, a tree outside) -> miss. ALT Junior Member Posts: 45 Threads: 2 Joined: May 2016 Reputation: 0 05-11-2016, 12:24 PM (This post was last modified: 05-11-2016, 12:44 PM by ALT.) (05-11-2016, 04:32 AM)secur Wrote: ALT wrote: ... your conclusion ... Please note, although I understand the math just fine I'm not a physicist. So it's not actually "my" conclusion; rather, all I intended to do was clarify Schmelzer's conclusion. Although his math clearly says the photon IS influenced by the transverse motion, he didn't say so explicitly and I thought a layman might miss it. But I should have left it at that, and not said it misses the target. I meant to just repeat his statement that it would "hit a different target". But I should have thought twice. Consider: if you're moving with the train, to you emitter and target are stationary. Evidently the photon will go straight across and hit the target. So if you're watching from any other reference frame you have to see the same event happen - although it takes longer. That must be what "a different target" meant. Sorry for the confusion ... I hope you keep digging and make sure to get this right. Also my post about photon travelling along the rifle barrel was to correct rjbeery's idea that the barrel would appear slanted, I realize that's not your question. Hi secur, OK, noted about your reply to rjbeery. Confusion .. yes, it can get like that. That's why I try to strip it down to basics. And that's why I've completely removed any observer (see earlier post) - so I won't now comment on your paragraph .. Consider: ... To avoid further confusion I will reply to Schmelzer next, hopefully also addressing your comments above. Oh, BTW, secur, earlier you said; But (as Schmelzer explained) the vector addition is relativistic, so it winds up traveling orthogonally towards the target slower by sqrt(1 - v^2/c^2). So it will miss, by falling short. If I'm reading that correctly, you are saying that it WILL miss the target. And note, I have never said anything about any other target - only the one, orthogonal to and directly above, the emitter. ALT Junior Member Posts: 45 Threads: 2 Joined: May 2016 Reputation: 0 05-11-2016, 01:13 PM Hi Schnelzer, you said; Train at rest, target at rest -> hit. Train moving, target at rest (say, a tree outside) -> miss. Agreed on both counts. Both rather obvious, and not part of my scenario You said; Train moving, target moving (say, target something inside the train) -> hit.  Indeed, this seems to be different to what you said earlier. But probably my fault. So please, forebear with my repitition, look at my OP again to ensure you are familiar with everything I have (and haven't) introduced, and see if I can get it down pat .. 1) You are saying that the photon will hit the receptor directly above it. 2) You are saying (remember, no immediate observer) that the receptor / computer will later report to me a hit. 3) Therefore you are saying that the photon IS influenced by the horizontal (transverse) movement of the photon emitter. « Next Oldest | Next Newest »