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Bullet vs Photon: speed of light, momentum and reference frames

  1. Mar 22, 2010 #1
    Code (Text):

            moving   |
           observer  |
                     | ---> v (moving speed train)
      photon         |

                    ^       moving   |
                ^   ^      observer  |
         d1 ^       ^                | ---> v
        ^           ^d2              |
    ^               ^________________|

    --- standing observers see d1, passengers see d2
    However, if speed of light is to be constant in each and every reference frame and independent of motion of the light source, which means no horizontal momentum for this photon, then it must miss the target. If it was a bullet then it would have this horizontal momentum and it would hit the target, but photon, as fast as it is, actually has to miss it, right? How can it "know" it has to move horizontally (as well) if it can not be impacted with this sideways momentum like a bullet would? Solution, then...

    Code (Text):

            moving   |
           observer  |
                     | ---> v
     photon          |

    ^                          moving  |
    ^                         observer |
    ^d1                                | ---> v
    ^                    d1=d2         |
    ^                 . _______________|

    --- standing observers see d1, passengers see d1
    ...and so the speed of light is constant and same in all frames. Except that this looks a bit strange, so is this really what experiments measure? Is there any such experiment that tries to measure some lateral momentum of photons like in the scenario above? What is the speed of light relative to?
  2. jcsd
  3. Mar 22, 2010 #2
    Your first drawing is the correct one. Note that we always talk about the constant speed of light and not the constant velocity of light. When the source is moving the direction the photon is emitted in can be frame dependent, but the speed is always constant. (Speed says nothing about direction, it is just a magnitude and not a vector like velocity.)

    Welcome to PF by the way and congratulations on finding out how to post ascii art in a tidy format using the code insert facility on your very first post.
    Last edited: Mar 22, 2010
  4. Mar 23, 2010 #3
    Speed is not the same nor constant for all reference frames if the 1st diagram is true as that photon would travel two different distances in the same time, which means different speed. That also implies photon would have this lateral momentum, like a bullet would, so can you point some experiments or other reference about this photon 'lateral momentum'?
  5. Mar 23, 2010 #4


    Staff: Mentor

    You are correct to notice this problem, with the key (incorrect) assumption in bold. In order for the speed to be the same and the distance to be different the time must also be different. Different frames measure different distances and different times such that the ratio is the same (for light).
  6. Mar 24, 2010 #5
    Momentum. "Lateral momentum", in example above.

    - We agree a bullet would hit the target because of this sideways momentum.
    - I thought SR says photons are independent of the motion of the light source.

    So, I want to confirm two things, 1.) what SR actually says about this "lateral momentum" and 2.) whether there are any actual experiments as in diagram above - on some YouTube video, some paper or online article.

    I think this would be easy experiment, we just need a very, very thin laser beam, or even better - a single photon emitter that in stationary reference frame always hits one and the same spot, very, very tiny spot, and that it hits no where else. This is the only precision required to perform this experiment, then we just sit in a plane and see if photons would still hit exactly the same spot, and maybe even measure if there was any angle in impact other than 90 degrees. Is there any such experiment?
  7. Mar 24, 2010 #6


    Staff: Mentor

    As has already been pointed out to you SR says that the SPEED of light is independent of the emitter. Velocity is a vector quantity meaning that it has a direction as well as a magnitude (the magnitude of velocity is speed). So the movement of a source can indeed impart "lateral momentum" to the light, but cannot change the speed of light.
  8. Mar 24, 2010 #7
    I'm talking about magnitudes too, distance and time: speed = distance/time. "Independent of emitter movement" reads to me "no change in momentum imparted on photons due to change in velocity of the emitter, what so ever". How can momentum transfer work if photons are supposed to have no mass? -- What is the speed of light relative to? Does absolute speed not require absolute reference frame?

    Code (Text):

                  Energy Bar: X....|..a.|....|...b|....|
                              o  dA  o        o              
                              o   o     o dB                
                              o o  o              
    ===[o]v--> ===============-===================Track ===
    Photon Emitter            ^Trigger
    0. Everything's static relative to ground except photon emitter
    1. Cart is set stationary over trigger to confirm 90 degree angle
    2. Cart with constant velocity vA scores +a energy pts, distance dA
    3. Cart with constant velocity vB scores +b energy pts, distance dB

    Static timer records difference in "hit time" and "trigger time", dtA & dtB.
    Will the timer measure the same delta-time for both dA and dB trajectory?

    If the time it takes photons to hit the target does not change with the velocity of the emitter, then something is wrong because the distance would obviously change and then this speed would not only vary but would exceed the "c". What is wrong?
  9. Mar 24, 2010 #8


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    Gold Member

    You're probably thinking of momentum in the p = mv sense. For photons, this is not the expression for momentum. A photon of energy E has a momentum E/c.

    Only if you use the classical ideas of how to transform from one reference frame to another. In relativity, the transformation from one frame to another is different, and this transformation preserves the speed c, i.e. if the speed of something is c in one frame, it is c in every other frame.
  10. Mar 24, 2010 #9


    Staff: Mentor

    This is simply wrong. The SPEED of light is independent of the emitter movement, but that statement does not restrict the momentum in any way. As dx pointed out the magnitude of the momentum of a photon is independent of its speed (which is constant) and is only dependent on its energy. Because the motion of the source can affect the energy (Doppler shift) it can affect the magnitude of the momentum.
  11. Mar 24, 2010 #10
    I should have said, better part of my understanding about this comes from cartoons, so I made interpretation as best I could from what I managed to gather, but none of it was about anything like these examples, and Google didn't explain anything about photon momentum and moving emitters, so I still do not know what's simple about it. Ok, I hear you, I have no problem with momentum and that SPEED is supposed to be constant and same for all reference frames. Photons having momentum makes more sense to me anyway, but then there is everything else. By the way, how do you know what you're saying is correct, are there any actual experiment similar to those diagrams?

    Al right, speed is constant and energy changes, so that means that mass or frequency can change, right? Though that still does not make sense to me, so can you please explain the second diagram, why would "time from emission to impact" (dTA & dTB) change as velocity of the emitter changes? That's what you saying, the delta-time will be different for different distances dA and dB, right?

    Code (Text):

    FRAME 1: target IS stationary:                      
    =========================[X....|.A..|....|..B.|]=====Track ===
                              o    o         o
                              o   o dA    o
                              o  o     o              
                              o o   o  dB                
        |                     oo o              dtA=dtB?  
        |                     o                            
    ===[o]v-----> ============-===========================Track ===
    90deg Emitter             ^Trigger

    FRAME 2: emitter is "stationary":
    =========== <-----[X....|.A..|....|..B.|]=============Track ===
                              o dA=dB
                              o                  dtA=dtB        
    =========================[|]===========================Track ===
                              90deg Emmiter

    - "Flight time" (dtA&dTB) differs for dA & dB in frame one?
    - How do photons "know" whether emitter or target is moving?
  12. Mar 24, 2010 #11


    Staff: Mentor

    Yes, there are over 100 years worth of experiments: http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

    Yes, the speed of light is c, so dTA=dA/c and dTB=dB/c. Therefore, if dA is different from dB then dTA must also be different from dTB.

    I am not sure I understand what you are trying to draw here. In the first drawing are you showing frame 1's analysis of three different flashes of light emitted at the same time in 3 different directions and striking 3 different targets. If so, then why is there only one flash in one direction in frame 2's analysis.
  13. Mar 24, 2010 #12
    I've seen that page, but I do not recognize any experiments are similar to this. I understand you are interpreting mainstream and accepted theory, me too, but I do not see what part of the theory address this issue and what effect are we talking about - "simultaneity"?

    Two new diagrams represent three separate measurements. There is only one photon vertically emitted per trial. Vertical path is obtained when both target and emitter are stationary relative to ground, other two distances are obtained when emitter is moving with some constant speed. The faster it moves sideways, the more it will miss stationary target located right in front of it. Top diagram is the same as the very 1st diagram except the target is now stationary on the ground. The diagram below is this same situation but in reference frame of moving emitter, so something is wrong.

    If it was a bullet its "straight towards the target" speed and therefore this delta-time would not change regardless of how fast the train is moving sideways, but the impact energy on the *stationary* target will actually be higher as emitter is moving faster. So, the same thing happens with photons too, except this momentum transfer works around velocity and mass, so we expect it to change only in frequency and blue-shift, so to increase energy of impact at stationary target the faster the emitter is moving sideways relative to it, just like a bullet?

    I still do not see how any of that connects to the "total speed" as we are shooting photons directly in front, perpendicularly to the direction of movement, so how can sideways momentum take anything from this "straight ahead" speed photons are supposed to have when being shot directly forward?
    Last edited: Mar 24, 2010
  14. Mar 24, 2010 #13


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    Staff: Mentor

    How is your first diagram different from what the Michelson Morley Experiment was designed to test?
  15. Mar 25, 2010 #14
    I don't know why are you asking me, if you could explain that it was similar then that might answer half of the questions I am asking. I don't see how what I'm talking about has anything to do with any aether wind, and I do not see mention of any momentum in those kinds of experiments.

    The greatest difference I see is that "Michelson-Morley" type of experiments are pretty static in relation to ground, to Earth's gravity field, which is where my 2nd question came from - "what is the speed of light relative to?" - since all the measurements of it seem to be relative to this infamous "laboratory" reference frame, and to me that sounds as we are bound to end up as Earth itself to be our "absolute reference frame", but I'm not talking about any "aether", but simply *gravity field*.

    So, as a side question -- are there any measurements of the speed of light taken far away from gravity fields, is there anything we know of that can guarantee those measurements would be the same as what we measure down on Earth surface? Do we have any measurements about the speed of light taken from some space probes away from planetary gravity fields, on Moon, Mars? Did we check time dilation and other effects with any of those "deep space" spaceships far away from gravity fields?
  16. Mar 25, 2010 #15


    Staff: Mentor

    OK, that is three scenarios with two different reference frames for each for 6 different analyses. That is too much, I don't want to do that many problems, and chances are that the scenario is so complicated that you wouldn't learn anything from it anyway. Please consider what your real confusion is and distill it into the most simple scenario possible to analyze. I will walk you through the equations in detail and then you can repeat that for more complicated scenarios.

    The speed of a bullet does change depending on how fast the train is moving. That is the key difference between bullets and light.

    Yes. The energy and momentum of a blueshifted photon is increased.

    Please correct me if I am wrong, but you seem to think that if the path of a light pulse is perpendicular in one frame it should be perpendicular in another frame. That is not the case, angles of motion are frame variant, almost by definition. Don't forget how a laser works, there is a standing wave resonating inside the laser cavity. That wave is obviously traveling sideways inside the moving resonator. Once the pulse leaves the cavity it simply continues traveling sideways as it did inside the cavity.
  17. Mar 25, 2010 #16
    It's only one scenario, the same one from the beginning only different questions. I'm not talking about directions, primarily I'm concerned with DISTANCE and TIME. We pretty much narrowed everything down, so I just want to make sure I got everything right and then I'll try to make my point again. The most important thing is that we're talking about the same thing.

    Code (Text):

    =TARGET SENSOR=======B ..b1 ..b2 ..b3 ..b4
    E------------------->A --->        

    Photon emitter 'E' moves from left to right with arbitrary constant velocity.
    When E passes over point 'A' it shoots a single photon in the direction of 'B'.

    1.) The faster E moves, the greater energy the sensor will register upon impact?
    2.) The faster E moves, the further away from B(b1, b2, b3..) photon hits the sensor?
    3.) The faster E moves, the more time photon takes to cross AB distance and hits sensor?

    That's what you said so far I believe, and that's all fine except for the third one, how do you explain that and can you please let me know where do you draw your conclusions from so I can read and learn about it too. What effect are we talking about here?
  18. Mar 25, 2010 #17


    Staff: Mentor

    OK, got it.

    Yes, all correct.

    If you understand 2, then 3 is an inevitable consequence of 2 and the invariance of the speed of light:

    From 2 we can use the Pythagorean theorem to calculate the distance traveled by the light, which clearly increases as E moves faster. Dividing that distance by c gives us the time which therefore also clearly increases as E moves faster.

    I draw my conclusions about the invariance of c from experiment. The evidence on the subject is overwhelming. See sections 3.1-3.4 of http://www.edu-observatory.org/phys...iments.html#Tests_of_Einsteins_two_postulates with special attention to section 3.3 which is directly applicable to your proposed scenario and which also includes a variety of cosmological sources which would not be affected by any laboratory frame "entrainment" issues you worried about earlier.
  19. Mar 25, 2010 #18


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    I'm not sure momentum really matters here. Your first diagram shows a reference frame dependent speed of light based on measuring the speed perpendicular to the motion of the apparatus and comparing it with the speed parallel with the motion of the apparatus. That's exactly the purpose of the MMx. Though the aether wind was assumed before the experiment was conducted, it is really an explanation of why the speed should be frame dependent. All that's really critical to the prediction is the assumption that the speed is frame dependent. Whether they are waves traveling on a medium or particles being shot across a bus really isn't part of the issue.

    A change in momentum might be an implication of that, but that isn't what you are measuring with the experiment.
    Well the earth is rotating, so at the very least you can assume that an experiment measuring the speed of light in an east-west direction will measure a different speed than one measuring it in a north-south direction....if the concept of a frame dependent speed of light is correct.
    Yes we do, but I'm not sure how that is really relevant: gravitational time dilation and special relativistic time dilation are two different things.
  20. Mar 27, 2010 #19
    I'm afraid that's all very different, especially in terms of what is measured and what is calculated. I'm strictly talking about the scenario where TIME and DISTANCE is measured, and velocity is deduced from these two, ALL in one reference frame, not across the interstellar space and between two or more reference frames.

    On closer inspection that web-page made it all even more confusing: - Any ray of light moves in THE “stationary”* system of coordinates with determined velocity c, whether the ray be emitted by a stationary or by a moving body. *Stationary: Let us take a system of coordinates in which the equations of newtonian mechanics hold good. In order to render our presentation more precise and to distinguish this system of coordinates verbally from others that will be introduced hereafter, we call it the “stationary system”.

    MEANING: Photons have constant velocity 'c' relative to Earth...
    ...or whatever planet you're standing on (Newtonian mechanics holds good).

    Einstein does not even mention "all" frames or "any" frames, nor any relations between some different frames, he directly and unambiguously defines the speed of light is relative to THE "static" frame, the Earth. No other frames are mentioned there and what else "where Newtonian mechanics holds good" can mean?

    Yes, but that is not logically (mathematically) possible and I think you might be interpreting SR incorrectly here, there does not seem to be any measurements to DIRECTLY confirm or disprove either really, as far as I see. -- If we take the sensor to move along with the emitter then the photon always takes the same amount of time to reach the target across the distance AB, right? And so, if we were to remove the sensor and let the photon hit the static target at this same distance, then the time can not be different from the impact time if moving target was still there, and this can all be measured in one reference frame.

    It's easy experiment to perform in laboratory if one has a very good stopwatch and laser spinning on a disc with some fixed target on a wall. I think for SR to not fail in this experiment the laser always needs to hit the target in the same spot regardless of its speed, if shot always from the same spot (distance), but that means the 2nd diagram from the beginning is actually the correct one, as I first said, and I think that this is actually what Einstein said.
    Last edited by a moderator: May 4, 2017
  21. Mar 27, 2010 #20
    Do you know some experiment where spinning laser shoots a static dot on a wall? I really do not see any relation of this experiment with any aether and other experiment, I'm not interested in any theory really, I'd rather know what experiments actually measure, than what are they supposed to measure. -- Since you mentioned it, how SR explains two clocks put in airplane, one going E-W and the other W-E, lose/gain different amounts of time?
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