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Can we see light if we move our eyes at speed of light?

  1. Sep 16, 2012 #1
    Hello,
    We know we can not see light.It is only the effect of light that we actually see .
    Is it because of the fact that our eyes cant move at a speed of light? or its just basically the nature of light that we can never see it..?
    If I move my eyeballs at 'c', would I be able to see light?


    Remember: We theoretically assume that movement of eye-balls at the speed of light is possible. The discussion is not about achieving speed of light.,etc.
     
    Last edited: Sep 16, 2012
  2. jcsd
  3. Sep 16, 2012 #2
    actually, i saw a video on youtube. It was about a camera that can capture the movement of a light through a coke bottle,

    so, if you have that kind of eyes, then you might able to see the light just like a glowing object..
     
  4. Sep 16, 2012 #3

    russ_watters

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    Welcome to PF!

    Of course we can see light! Light is what we see -- that's basically a tautology. We see light that hits our eyes and can't see light that doesn't hit our eyes, so moving doesn't have any impact on whether we can see light that doesn't hit our eyes.

    This is like asking if one were to walk next to a river at the same speed as the river flows, would you get wet?
     
  5. Sep 16, 2012 #4

    Dale

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    Under which theory can we theoretically assume that? We cannot assume it under special relativity.
     
  6. Sep 16, 2012 #5

    HallsofIvy

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    What, exactly, do you mean by "see" here? Many would argue that light is the only thing we see because it is the only thing that affects the retinas of or eyes. When we say we see objects we really mean we see images of them in the light- we see the light coming from the objects.

    While we can "theoretically assume" anything we want, even that relativity is NOT true, it makes no sense to assume that relativity is NOT true and then ask "what would relativity say about this situation". And that is what you doing here. What you could do is ask "suppose we are moving at 99.9% the speed of light (relative to some frame of reference), what would we see? And the answer to that is that we would exactly what we would see at any speed relative to that frame of reference- we would see light coming toward us at "c". The speed of light is the one constant speed in relativity.
     
    Last edited by a moderator: Sep 16, 2012
  7. Sep 16, 2012 #6

    ghwellsjr

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    It's true that when we emit a flash of light, we cannot see the light as it is traveling away from us. Is that what you are talking about?

    If so, then we can put a reflector at some distant point and wait for the light to bounce off of it and return back to us. The net result is that we can see that the light reached the reflector but because of the additional time that it took to return, we can't really say when it hit the reflector. However, Einstein came up with the idea that we can assume that the light hit the reflector half way during its round trip. Is this the issue that you are concerned with?
     
  8. Sep 16, 2012 #7
    Though if light is moving in air, so at <c, you could catch up with it if you run fast enough.
     
  9. Sep 16, 2012 #8

    HallsofIvy

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    And if you can run fast enough, be sure to enter the Olympics!
     
  10. Sep 16, 2012 #9

    ghwellsjr

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    Long before you can run that fast in air, you will burn up.
     
  11. Sep 17, 2012 #10
    Sorry could you link that vid?
     
  12. Sep 17, 2012 #11
    Hello joemmonster, can you send me the link of that video please
     
  13. Sep 17, 2012 #12
    Hi russ_watters,
    Well I partly agree with that..
    Now consider the situation below:
    I am standing at (0,-2) and facing X-axis....
    A light ray is moving along positive X-axis direction and is at the origin(0,0) at present
    (i mean its nearby to me)...
    ....There is no object anywhere...Its just the single ray of light...
    Can I still see that..?

    Assume linear propogation as I said single light ray...Dont consider huygen's wave or imaginary spheres.....
     
  14. Sep 17, 2012 #13

    hi ghwellsjr,
    Thanx 4 the reply...
    Actually I am afraid its not about that....This is the situation I was talking of:
    I am standing at (0,-2) and facing X-axis....
    A light ray is moving along positive X-axis direction and is at the origin(0,0) at present
    (i mean its nearby to me)...
    ....There is no object anywhere...Its just the single ray of light...
    Can I still see that..?

    Assume linear propogation as I said single light ray(like a LASER source,,etc)...Dont consider huygen's wave or imaginary spheres.....
     
  15. Sep 17, 2012 #14


    HI HallsofIvy,
    Actually thats not exactly my doubt..Yeah,ofcourse wen i say 'assuming speed of 'c', imean 0.99c or somthing like that......

    This is the situation I was talking of:
    Consider the co-ordinate system...
    .I am standing at (0,-2) and facing X-axis....
    A light ray is moving along positive X-axis direction and is at the origin(0,0) at present
    (i mean its nearby to me)...
    ....There is no object anywhere to reflect it.etc...Its just the single ray of light...
    Can I still see that..?

    Assume linear propogation as I said single light ray(like a LASER source,,etc)...Dont consider huygen's wave or imaginary spheres.....
     
  16. Sep 17, 2012 #15
    When I say moving at 'c',obviously i mean 0.99c...NOt breaking any relativity laws..
     
    Last edited: Sep 17, 2012
  17. Sep 17, 2012 #16

    mfb

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    Cola bottle + light video together with the corresponding TED talk.

    Light is moving away from you? No, you cannot see it.
    Why did you ask this in 3 posts in a row?
    That is not obvious at all, as it is a completely different situation.
     
  18. Sep 17, 2012 #17
    --------------------------------------------
    Well I had 2 reply differently to different answers.So asked in 3 posts..Could hav written one instead...Anyways,....so u say we cant see that light ray...

    (yup ..by 'obviously' , I meant it is understood that you cant move your eyeballs at more than 'c'.So it shud b a bit less than 'c'..Thats wat i meant...)

    Anyways, thanx 4 d answer bro..
     
  19. Sep 17, 2012 #18

    Dale

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    No. You only ever see light that is moving towards you, you can never see light which is moving away from you.
     
  20. Sep 17, 2012 #19

    HallsofIvy

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    Then say "near c" !

    But you say "0.99c or something like that" relative to what frame of reference. In any case, as several people have said, the speed of light, relative to any frame of reference is "c". How you are moving, relative to anything, is irrelevant to light.

    Once again the ONLY light you "see" is light that goes into your eye and to your retina. If the light was " moving along positive X-axis direction" and you are at (-2, 0), it is moving away from you and cannot go into your eye. You cannot see it at all.

    You may be thinking of a situation in which a person in front of you is shining a flash light and you see a beam of light going out from the flash light. What you are seeing is light that has reflected of water globules, dust, etc. in the air and has been reflected back to your eyes. If this were done in a perfect vacuum, you would see nothing at all.
     
    Last edited by a moderator: Sep 17, 2012
  21. Sep 17, 2012 #20

    ghwellsjr

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    Your explanation is very confusing. What does (0,-2) stand for?

    What does it mean that you are facing the X-axis?

    But as I answered before: "we cannot see the light as it is traveling away from us" and then you said that's not what you are asking about and then you asked if you can see a single ray of light which I presume is either traveling away from you or traveling across your field of view, and it is not hitting any object so that it cannot reflect off of anything, but even if it is traveling toward you, you will not see it unless it enters your eye.

    Somehow, I don't think I have answered your question but that is because I don't understand your question. You have to be very precise when you ask a question like this. Assume I'm really stupid and have no idea what you are talking about. Don't take anything for granted.

    Please note in the Coke bottle video that they showed a ray of light coming from a laser and they pointed out that the only reason you could see it is because it is reflecting off of particles in the air. At first this was a constant ray of light so you could see it as a steady beam. But then they changed it to a very short pulse so that you could see its progress as if it were a bullet. That's why I used the word "flash of light". You need to do something similar in your explanation. Are you talking about a constant ray of light or a short burst of light?
     
  22. Sep 17, 2012 #21

    Dale

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    Whether or not you see something is pretty straight forward: if the light hits your retina you see it. (Assuming no health issues)
     
  23. Sep 17, 2012 #22
    lol,

    All else is "elsewhere"
     
  24. Sep 17, 2012 #23
    hello george,
    As u hav assumed ,the light is traveling across my field of view, and it is not hitting any object ,in my question .And it is a flash of light.
    By giving those numbers,I meant the relative positions on the XY-coordinates.

    Anyway, I agree that it is difficult to explain the question. The point is that in physics, a simple illustrative diagram makes a perfect sense and explains a whole lot .

    I got the answer and my doubt got cleared....
    Thanks for your answer man...
    -Aditya
     
  25. Sep 17, 2012 #24
    health issues...haha..lol!!

    nyway thanx 4 d reply...!!
     
  26. Sep 18, 2012 #25
    Because i do not know about all the special cases how light behaves when it is not within a vacuum away of gravitational fields i will answer the question only for that special situation in which we assume you are in a perfect vacuum and there is no gravity field present.

    So what we have here is the first postulate (some like to call it axiom) of special relativity.
    Light will ALWAYS travel at C in a vacuum absent of any gravity.

    Therefore, if you hold a flashlight towards any direction. The photons it emits will travel at C away of you. It would be impossible to see those photons unless they hit an object and bounce back. At least that is what one has to assume when he wants to arrive at the formulas Einstein arrived at for special relativity.
    My knowledge about QM is too minimal to even attempt a guess on which scale it would change the formulas if photons would not travel as Einstein has imagined them to travel or as he deliberately postulated to "keep things simple enough" for his first shot.

    Anyway, if we want to arrive at the formulas Einstein arrived at we HAVE TO consider those idealistic photons all traveling at C or at least "lightbeams" traveling at C within a vacuum away of gravity.

    l*gamma = l' with
    gamma = sqrt(1 - (v^2/c^2))
    as an example of one of the formulas you will arrive at if you follow the two postulates for special relativity strictly.
    (There are actually more postulates required but then it gets almost philosophical if we dive that deep)

    And here is the clue. YOU yourself, seen from your own reference system DO NOT MOVE at all. You have to consider yourself standing still, and everything else is moving around you.
    Your eyeballs move at exactly zero m/s within your reference frame. Remember it is called relativity for a reason. Everything moves relative to each other.

    You could see someone ELSE's eyeballs move at CLOSE TO but never reach C, following a photon closely and getting distance to it very very slow.
    The mindbending part here is that the eyeballs of the other guy you see traveling close to C, just a tiny bit slower than the photon they follow, will, when seen from that other guy's perspective(reference system) see that photon traveling at C (and so will everyone else in any inertial reference system - i repeat ALL WILL SEE THAT PHOTON TRAVEL AT C no matter which inertial reference system they are observing it from), and this guy or his eyeballs will have to consider himself NOT moving.
    If your mind did not blow up yet, continue reading...

    This is not easy at all to get your mind around. It took me months. But after drawing this

    http://img204.imageshack.us/img204/2425/89825284.jpg [Broken]

    representing two observers traveling at v= 0.5c RELATIVE to each other (Both consider themselves not moving, but see each other passing at 0.5c) i finally managed to arrive myself at the formulas, using only the two postulates(or axioms if you prefer).

    I now KNOW that Einstein was right first hand, FINALLY!! (As long as the two postulates are valid. It would be "easy" to adjust the formulas if QM told us that photons don't behave as idealistic as assumed in the postulates but nevertheless we should be very close to same formulas still).


    Those are the two postulates required to build the formulas for special relativity. This, maths and maybe some geometry will get you the formulas.

    1. Light(photons? lightbeams?) in a vacuum away of any gravity field will ALWAYS travel at C in all inertial frames of reference.

    2. The laws of physics are the same in all inertial reference systems.



    Or as i would say.. if two reference systems A and B are equal in quality, then there is no reason to consider any of the two special.

    Whatever an observer a within a ref system A sees in ref system B under certain conditions, an observer b in system B will see the same if same conditions in A are met.

    If for example observer a sees a ruler which is 1 meter within b's reference system (not moving relative to b) shrink by 80% to 0.8m, then b will see a different ruler which is 1 meter within a's reference system (not moving relative to a) also shrink by 80% to 0.8m.

    Why? Because neither a's reference system nor b's reference system are special.
    The laws of physics are the same in both reference systems(2nd postulate).
    a sees b moving and considers himself motionless. b sees a moving and considers himself motionless. They move relative to each other at a certain speed. Their relative speed to each other is a common factor which cannot be used to assume either reference system A or B being special in any way.

    Start drawing two diagrams representing two reference systems, and follow strictly the postulates like i did. The yellow lines are lightbeams. As you can see, the diagrams are drawn in a way, so llightbeams moving at C are at 45° always (see Minkowski diagrams)




    After you understood this, your next stop will be the twin paradoxon most likely. The key to this is that acceleration changes the reference system the accelerating twin is in which creates an asymmetrical situation. One twin being in the same reference system at all times, while the other twin is passing through several reference systems by accelerating at times.
     
    Last edited by a moderator: May 6, 2017
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