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Thread: What will I see if the mirror will move very fast?

  1. #1 What will I see if the mirror will move very fast? 
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    Hi. I have a question that gives me a very big headache for a long time; hope i will find an answer here:

    Lets pretend that I am standing infront of a "mirror train"; A very-very long train which is actually one long mirror.

    The train is staionary and obviously I can see my image as it is reflected to me.

    Now, the train starts moving and it accelerates to "siginificant" speeds.


    And the question is - WHAT WILL I SEE NOW?


    Another way of putting it - if I shoot a single photon verticaaly to the very fast moving mirror - what will happen too it?

    What proof you can gice your answer?


    Thank you for you wise answers!
     

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    Hi chaimc, welcome to the forum. This is a very good question and I am sure will generate lots of responses. My initial thought would be that it should be equivalent to you flying fast past a stationary mirror. That being the case you should observe no difference. But I am not sure.
     

  3. #3 I dont think so... 
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    Quote Originally Posted by Jilan View Post
    Hi chaimc, welcome to the forum. This is a very good question and I am sure will generate lots of responses. My initial thought would be that it should be equivalent to you flying fast past a stationary mirror. That being the case you should observe no difference. But I am not sure.
    Hi Jilan

    1- What do you think will I will see if I fly-by a stationary mirror very fast? Ofcourse it depends on my speed and distance from the mirror. If I move very-very fast and with a significant distance from the mirro, I wont see myself.... Dont you think so?

    2- Therefore my original question scenario is not the same. What will happen then do you think aacording to my original scenario?


    Thx Chaim
     

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    Chaim, , No I,think you will see yourself well however fast you are going. Although if you are a significant distance from the mirror you may look a bit younger!
    The original scenario is equivalent to the second one by the principle of relativity (who is to say whether you or the mirror is stationary?)
     

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    Cheers and Welcome to chaimc!

    In regard to your 1st. question..."what will I see?" You didn't specify an exact velocity, so I'll assume you mean very fast (a relativistic speed) and the answer is...you will see the train

    just as you saw it when stationary, or if you are the one moving, it will appear the same. Very tricky, yes? (this is same answer as Jilan gave you, just worded a little different)

    (Thanks for reading!)
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  6. #6  
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    Quote Originally Posted by chaimc View Post
    Hi. I have a question that gives me a very big headache for a long time; hope i will find an answer here:

    Lets pretend that I am standing infront of a "mirror train"; A very-very long train which is actually one long mirror.

    The train is staionary and obviously I can see my image as it is reflected to me.

    Now, the train starts moving and it accelerates to "siginificant" speeds.


    And the question is - WHAT WILL I SEE NOW?


    Another way of putting it - if I shoot a single photon verticaaly to the very fast moving mirror - what will happen too it?

    What proof you can gice your answer?


    Thank you for you wise answers!
    This is a very good question, the answer is far from trivial, it is treated in great detail by W. Pauli in his excellent book "Theory of relativity". (pages 96-97). Get the book, it is well-worth the few dollars it costs.

    For your case (simpler than the one treated in the book), your image will appear color shifted towards the blue spectrum. The reason is that light reflecting off you gets blue shifted when it hits the incoming mirror and it gets blue shifted again when it gets reflected from the mirror back into your eye. So, you get a double blue-shift effect:


    If you do the experiment with the mirror receding from you, you will get a double red-shift effect:





    If the relative speed between you and the mirror v, is a significant fraction of light speed,c, the effect is noticeable.
    There are many cool websites that show simulations of the above. This one is one of my favorites.
     

  7. #7  
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    Quote Originally Posted by Jilan View Post
    That being the case you should observe no difference. But I am not sure.
    If you are not sure, you shouldn't confuse the others: your answer is incorrect. See above for the correct answer.
     

  8. #8  
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    Quote Originally Posted by Gerry Nightingale View Post

    just as you saw it when stationary, or if you are the one moving, it will appear the same. Very tricky, yes? (this is same answer as Jilan gave you, just worded a little different)
    Yes, both of you posted the same mistake. Not surprising, given the principals.
     

  9. #9  
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    Quote Originally Posted by x0x View Post
    This is a very good question, the answer is far from trivial, it is treated in great detail by W. Pauli in his excellent book "Theory of relativity". (pages 96-97). Get the book, it is well-worth the few dollars it costs.

    For your case (simpler than the one treated in the book), your image will appear color shifted towards the blue spectrum. The reason is that light reflecting off you gets blue shifted when it hits the incoming mirror and it gets blue shifted again when it gets reflected from the mirror back into your eye. So, you get a double blue-shift effect:


    If you do the experiment with the mirror receding from you, you will get a double red-shift effect:





    If the relative speed between you and the mirror v, is a significant fraction of light speed,c, the effect is noticeable.
    There are many cool websites that show simulations of the above. This one is one of my favorites.
    I interpret the opening post as the mirror moving transverse to the line of sight, in which case Jilan's approach and answer are correct.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
     

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    Quote Originally Posted by KJW View Post
    I interpret the opening post as the mirror moving transverse to the line of sight, in which case Jilan's approach and answer are correct.

    I interpret it as the train coming towards the observer.
    If the train is moving transverse, then you (and Jilan) are STILL wrong, you would get a transverse Doppler effect instead of the longitudinal Doppler effect. This means that you would be going through three stages:

    1. Blueshift (train approaching)
    2. No shift (at a critical angle corresponding to transverse Doppler effect)
    3. Redshift (train receding)

    In addition to the above, you would be getting some serious distortions due to the Terrell-Penrose effect. See this very good paper on the subject. I am trying to find another website that illustrates this very well. Found it.
     

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    In reply to #8 post.

    NO...no mistake. I answered the question as it was written..."what would I see if"...and the answer I wrote is valid. If the "observer" is static in relation to the mirror-train and parallel

    w/it, the image the observer "sees" will be the same as if the mirror were not in motion ( no "frame-dragging" effect, as the distance of the train and observer are still relative to each other)

    The same aspect will be maintained if the "observer" is moving and the mirror-train is static. (Einstein pondered this same problem while shaving...and his conclusions are still valid)

    ......

    You have added more to the problem than was asked, and thus changed the original query. (c is still c, no mention was made of the "distance of observation" being a variable)
     

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    Thx Jilan an Gerry and others ( I am responding to all in one post I hope no one gets offended by it....):

    I am truly learning from your answers.

    1- Interesting thing is that none of you actually mentioned the angels of reflection.... Is that so obvious? I mean are you sure the single photon will bounce back in 90 degs regardless of the motion of the mirror? that is actually the question...

    2- Let me rephrase it. Lets pretend that I have a device that can "shoot" a single photon on my request ( is there such a device...?). If I shoot it vertically - will it bounce back to the device if the mirror is moving?

    3- The distances are "not relativitic".

    thans again!
     

  13. #13  
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    Quote Originally Posted by chaimc View Post
    Thx Jilan an Gerry and others ( I am responding to all in one post I hope no one gets offended by it....):

    I am truly learning from your answers.

    1- Interesting thing is that none of you actually mentioned the angels of reflection.... Is that so obvious? I mean are you sure the single photon will bounce back in 90 degs regardless of the motion of the mirror? that is actually the question...
    Actually, I did.

    2- Let me rephrase it. Lets pretend that I have a device that can "shoot" a single photon on my request ( is there such a device...?). If I shoot it vertically - will it bounce back to the device if the mirror is moving?

    !
    Only if the device is moving with the mirror. Otherwise, NOT. You get something called "aberration".
     

  14. #14  
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    Quote Originally Posted by x0x View Post
    If the train is moving transverse, then you (and Jilan) are STILL wrong, you would get a transverse Doppler effect instead of the longitudinal Doppler effect.
    There is no transverse Doppler effect in this case. It is a mirror that is simply reflecting the light from the observer back to the observer. It is not emitting its own light. The light from the moving observer is blueshifted, strikes the stationary mirror at an angle and reflected to the location the moving observer will be, who sees this light redshifted cancelling the original blueshift.
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    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
     

  15. #15  
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    Quote Originally Posted by KJW View Post
    There is no transverse Doppler effect in this case. It is a mirror that is simply reflecting the light from the observer back to the observer. It is not emitting its own light. The light from the moving observer is blueshifted, strikes the stationary mirror at an angle and reflected to the location the moving observer will be, who sees this light redshifted cancelling the original blueshift.
    You didn't look at the websites I recommended, did you?
    The light comes from a source, hits the object, goes to the mirror and comes back. Depending where the light source is, you get a doubling of the blueshift, a doubling of the redshift, etc, exactly like in the simpler case I explained in the first post. In ADDITION, you get severe distortions through the Terrell effect.
     

  16. #16  
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    Quote Originally Posted by x0x View Post
    You didn't look at the websites I recommended, did you?
    No, I did not because I did not need to. I confidently worked it out for myself.


    Quote Originally Posted by x0x View Post
    The light comes from a source, hits the object, goes to the mirror and comes back. Depending where the light source is, you get a doubling of the blueshift, a doubling of the redshift, etc, exactly like in the simpler case I explained in the first post. In ADDITION, you get severe distortions through the Terrell effect.
    By considering a moving observer relative to a stationary mirror, comparing this to a moving observer relative to a moving mirror, which is the same as a stationary observer relative to a stationary mirror, it becomes clear that the transverse motion of the mirror has no effect on the image the observer sees of himself.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
     

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    Dear X0X

    You were answring as if I was moving away/towards the mirror. is that correct?

    the scenario I am after is I am staionary and the mirror moves sideways.

    See #12.
     

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    Quote Originally Posted by x0x View Post
    I interpret it as the train coming towards the observer.
    If the train is moving transverse, then you (and Jilan) are STILL wrong, you would get a transverse Doppler effect instead of the longitudinal Doppler effect. This means that you would be going through three stages:

    1. Blueshift (train approaching)
    2. No shift (at a critical angle corresponding to transverse Doppler effect)
    3. Redshift (train receding)
    The transverse Doppler shift applies to moving sources not moving mirrors. Anyway the experimental evidence says there will be no change in the frequency.
    Reflection from a transversely moving mirror-experimental details - Abstract - Journal of Physics E: Scientific Instruments - IOPscience
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  19. #19  
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    Quote Originally Posted by KJW View Post
    No, I did not because I did not need to. I confidently worked it out for myself.
    This is really bad, I'll give it one more try and I'll give up.



    By considering a moving observer relative to a stationary mirror, comparing this to a moving observer relative to a moving mirror, which is the same as a stationary observer relative to a stationary mirror, it becomes clear that the transverse motion of the mirror has no effect on the image the observer sees of himself.
    But this is not what happens. Light strikes the observer (you). The object reflects the light back in all directions in the form of a spherical front (it is called diffuse reflection). The spherical front encounters the incoming mirror at some angle. The incoming mirror reflects the light back towards you. You get one blue shift in the path light from you to the mirror.
    The light is reflected by the mirror back towards the observer. You get an additional blueshift.
    When the train passed you, you may get blueshift followed by redshift or redshift followed by redshift, depending on the position of the light source in the scene.
    In ADDITION, in ALL cases, because the ray paths light source-observer-mirror-observer are unequal in length, you get a (severe) Terrell effect, so, your mirror image will appear twisted.
    This is much more complicated than you are thinking. I wrote quite a few simulators that deal with these effects, you are grossly oversimplifying.
     

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    Quote Originally Posted by Jilan View Post
    The transverse Doppler shift applies to moving sources not moving mirrors.
    A moving mirror constitutes a (secondary) light source. See above.

    I am very familiar with the paper. You obviously do not understand how the conditions in the respective paper differ from what is being discussed. In the paper, you get a cancellation of the redshift Doppler followed by a blueshift Doppler (and vice-versa).
    This is not what happens when you look at yourself in a moving mirror.
     

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    Quote Originally Posted by chaimc View Post
    Dear X0X

    You were answring as if I was moving away/towards the mirror. is that correct?

    the scenario I am after is I am staionary and the mirror moves sideways.

    See #12.
    I answered all the situations.
     

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    Quote Originally Posted by x0x View Post
    I answered all the situations.
    Sorry, i cant find where you answered about the mirror moving sideways. In which #post?
     

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    Quote Originally Posted by chaimc View Post
    Sorry, i cant find where you answered about the mirror moving sideways. In which #post?
    Post 10 and the followups to KJW and Jilian.
     

  24. #24  
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    Quote Originally Posted by x0x View Post
    The spherical front encounters the incoming mirror at some angle. The incoming mirror reflects the light back towards you.
    The mirror is not incoming. It's motion is transverse.


    Quote Originally Posted by x0x View Post
    This is much more complicated than you are thinking.
    It is not complicated. It is a fairly straightforward application of the principle of special relativity.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
     

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    Quote Originally Posted by x0x
    I answered all the situations.
    Remember that my understanding of mirrors boils down to crude Feynman lecture for layman.

    Does the process of reflection itself does not imply that photon reflected are not the same that those incoming ?

    Can I deduce that there is at least some kind of momentum exchange between mirror and light (thus slowing the mirror and speeding up the light (arrrgg ;-))
    I mean the incoming light from the left (I face the mirror) is redshifted (lower momentum) the other blue shifted (greater momentum) (that is how the mirror would 'see' me, like a little rainbow)
    And the same process happens again when photon leave the mirror, but inverted, my right arm photon is blue shifted when coming to the mirror, but bounced red shifted in the other path (OK there is not one photon but that would be the net result while adding all the contributing paths)

    I also think the reflection will also be left-right distorted (in amplitude of frequencies) because various wavelength are not reflected equally (no mirror is perfect)
    I would imagine a mirror speeding near light speed would only reflect in the visible range a small band around my middle axis.
     

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    Quote Originally Posted by KJW View Post
    The mirror is not incoming. It's motion is transverse.
    You are standing on the platform, the train with the mirror is moving on the tracks. Its motion is transverse to you. The effects that I am trying (and failing) to explain to you are happening because the train:

    -approaches you
    -is right in front of you
    -recedes from you

    Since the whole train is a giant mirror, you get different images depending on your direction of gazing. Regardless of your direction of gazing, your image will be severely distorted, both from a geometry and from a color point of view.

    It is not complicated. It is a fairly straightforward application of the principle of special relativity.
    It is not the way you think it is, it is far more complicated.
     

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    Quote Originally Posted by x0x View Post
    You are standing on the platform, the train with the mirror is moving on the tracks. Its motion is transverse to you. The effects that I am trying (and failing) to explain to you are happening because the train:

    -approaches you
    -is right in front of you
    -recedes from you

    Since the whole train is a giant mirror, you get different images depending on your direction of gazing. Regardless of your direction of gazing, your image will be severely distorted, both from a geometry and from a color point of view.



    It is not the way you think it is, it is far more complicated.
    **********

    Pls, as I am not a physicist I am doing my best to understand your wise answers....

    Can someone describe what will happen in this scenario:

    Lets pretend that I have a device that can "shoot" a single photon on my request ( is there such a device...?). If I shoot it vertically - will it bounce back to the device if the mirror is moving?
     

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    Quote Originally Posted by chaimc View Post
    **********

    Pls, as I am not a physicist I am doing my best to understand your wise answers....

    Can someone describe what will happen in this scenario:

    Lets pretend that I have a device that can "shoot" a single photon on my request ( is there such a device...?). If I shoot it vertically - will it bounce back to the device if the mirror is moving?
    ONLY IF the device is moving with the mirror.
     

  29. #29  
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    [QUOTE=Boing3000;19340]Remember that my understanding of mirrors boils down to crude Feynman lecture for layman.

    Does the process of reflection itself does not imply that photon reflected are not the same that those incoming ?

    Can I deduce that there is at least some kind of momentum exchange between mirror and light[quote]

    Yes.


    (thus slowing the mirror
    Yes, it is called "radiation pressure".

    and speeding up the light (arrrgg ;-))
    Absolutely NOT!

    I mean the incoming light from the left (I face the mirror) is redshifted (lower momentum) the other blue shifted (greater momentum) (that is how the mirror would 'see' me, like a little rainbow)
    I do not understand what you are saying here but I can tell you that the frequency is shifted UNDER CERTAIN circumstances.



    I also think the reflection will also be left-right distorted (in amplitude of frequencies) because various wavelength are not reflected equally (no mirror is perfect)
    Maybe. The main distortion is due the Terrell-Penrose effect.

    I would imagine a mirror speeding near light speed would only reflect in the visible range a small band around my middle axis.
    No, it reflects all frequencies. And it shifts them.
     

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    In reply to "whomever'.

    Congrats! I did not think I would live long enough to see Einstein's observations repudiated!

    There is NOTHING here that is "far more complicated than you think". In the instance of the mirror-train/observer, all the conditions are assumed as relative absolutes. The only change from

    a rest-state is the velocity of either object, w/ one them remaining static. In this instance, the velocity component acts upon the matter of the train, NOT the aspect of reflected light.

    (I am assuming the train is not capable of FTL, in which case "all bets are off")

    ......

    Since all of the available light has remained at a constant value, then the qualities of the reflected light will retain the same values they had initially, prior to the movement of either object.

    (the light in the FoR is present as "there"...it is not "racing" back and forth. c has remained c, so there is no "shift" aspect...unless of course, you wish to apply TFOLZOs' "DET" to the

    train/observer. In that event, as the train gained greater relativistic speed, yes, the observer would see "red-shift" only)

    ......

    The only way for light to "shift" in the train/observer scenario is if the train is considered "too fast for some frequencies of light spectra to keep up" and therefore at some point only

    the "fast red" photons are being reflected in straight line-of-sight to the observer...all the other photons are undergoing "angular deflection" since they "cannot keep up w/ the train".

    Sorry...I can't buy this as a true FoR condition.


    (Thanks for reading!)
     

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    Quote Originally Posted by x0x View Post
    A moving mirror constitutes a (secondary) light source. See above.



    I am very familiar with the paper. You obviously do not understand how the conditions in the respective paper differ from what is being discussed. In the paper, you get a cancellation of the redshift Doppler followed by a blueshift Doppler (and vice-versa).
    This is not what happens when you look at yourself in a moving mirror.
    Would you care to illuminate us as to why the experiment is different to what is being discussed here? Are there different types of transversely moving mirrors?
     

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    In further reply to my #30 post.

    If in the "mirror-train/observer" FoR it is a "given" that all of the light present is "within the frame as a constant function" then the shift of frequency would not be a factor. The only

    parameter that changed is the relative velocity of the observer or train...everything else in the FoR remains as a constant.

    ......

    Their is no "coming toward" or "away from" w/respect to the line-of-sight of the observer, as there is no change in distance between observer and train.

    (if I'm wrong here, I would sure like to know "how?" Seriously...I might be missing something, but what?)


    (Thanks for reading!)
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    Quote Originally Posted by x0x
    Quote Originally Posted by me
    and speeding up the light (arrrgg ;-))
    Absolutely NOT!
    It was a joke. Gotcha !

    Quote Originally Posted by x0x
    I do not understand what you are saying here but I can tell you that the frequency is shifted UNDER CERTAIN circumstances.
    My bad. I was mind guessing one of those circumstance. I have made a image of the situation (in the hope to be able to link it (the site is sooo broken here is the url: i.imgur.com/iPNh4HQ.gif)

    The situation is drawn from above. The ray from my left arm L-B is received red-shifted because also light travel at fixed c, a part of its space component (the red arrow L-C is catching up with the moving B (bouncing point) thus the mirror see it as right shifted. (likewise R-B (right arm) is seen blue shifted.
    That is how the mirror sees me (rainbow like distorted)


    Quote Originally Posted by x0x
    Maybe. The main distortion is due the Terrell-Penrose effect.
    Thanks for the information. The wiki aerticle is quite succint. Maybe my hypothesis about shifted frequancies is wrong. The mirror also see my left arm elongated and my right arm shortened. If I add same kind of distortion on the up-down axis, I kind of get the is more circumstance to it.
    Like the light from my foot should bounce at B but then got momentum dragged to the right when coming back from the mirror, more so (or less so) then my nostril because the angle is different.

    [QUOTE=x0x]No, it reflects all frequencies. And it shifts them.[QUOTE]
    Even radio wave and x-Rays ? My hypothesis about Doppler shift would lead to relativistic mirror only able to get the center part of the ME (where the red/blue arrow are very short).
    And then shift them so badly again upon reflection that my eyes would certainly not be able to see me anymore
     

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    Pls, as I am not a physicist I am doing my best to understand your wise answers....

    Can someone describe what will happen in this scenario:

    Lets pretend that I have a device that can "shoot" a single photon on my request ( is there such a device...?). If I shoot it vertically - will it bounce back vertically to the device if the mirror is moving? Meaning, will the angle of the reflection of the single photon (shot from a staionary post) will ber affected by the moving mirror?

    Thank you/
     

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    Quote Originally Posted by chaimc View Post
    Pls, as I am not a physicist I am doing my best to understand your wise answers....

    Can someone describe what will happen in this scenario:

    Lets pretend that I have a device that can "shoot" a single photon on my request ( is there such a device...?). If I shoot it vertically - will it bounce back vertically to the device if the mirror is moving? Meaning, will the angle of the reflection of the single photon (shot from a staionary post) will ber affected by the moving mirror?

    Thank you/
    IF the mirror moves as one body with the source, then the photon will return vertically to the source.
    If the mirror is moving with respect to the source, then the photon will NOT return to the source/
     

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    Quote Originally Posted by Jilan View Post
    Would you care to illuminate us as to why the experiment is different to what is being discussed here? Are there different types of transversely moving mirrors?
    The effect is dependent of how the rays are sent into the mirror:
    - If you send a ray "chasing" the mirror and you try to catch its reflection while the mirror is approaching, you get ZERO Doppler effect (due to cancellation):




    Since it follows that

    - If you send a ray "chasing" the mirror and you try to catch its reflection while the mirror is receding, you get DOUBLE Doppler effect (the case being discussed in this thread).






    - If you send a ray "into" the mirror and you try to catch its reflection while the mirror is approaching, you get DOUBLE Doppler effect (the case being discussed in this thread).
    - If you send a ray "into" the mirror and you try to catch its reflection while the mirror is receding, you get ZERO Doppler effect (due to cancellation).

    You simply read the title of the paper, did not read the inside and concluded that the scenarios are the same. I cannot teach people like you physics, you would have to want to LEARN and you are more preoccupied with pushing your agenda. But you can continue taking your cues from Gerry and Farsight.
    Last edited by x0x; 09-02-2014 at 10:08 PM.
     

  37. #37  
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    Quote Originally Posted by Boing3000 View Post
    It was a joke. Gotcha !


    My bad. I was mind guessing one of those circumstance. I have made a image of the situation (in the hope to be able to link it (the site is sooo broken here is the url: i.imgur.com/iPNh4HQ.gif)

    The situation is drawn from above. The ray from my left arm L-B is received red-shifted because also light travel at fixed c, a part of its space component (the red arrow L-C is catching up with the moving B (bouncing point) thus the mirror see it as right shifted. (likewise R-B (right arm) is seen blue shifted.
    That is how the mirror sees me (rainbow like distorted)
    No, the shift is explained by the general formulas of the relativistic Doppler effect. As such, what matters is just the relative speed and the angle between the velocity (of relative motion) and the line of sight.


    Thanks for the information. The wiki aerticle is quite succint. Maybe my hypothesis about shifted frequancies is wrong. The mirror also see my left arm elongated and my right arm shortened. If I add same kind of distortion on the up-down axis, I kind of get the is more circumstance to it.
    Yes.




    [quote]
    [QUOTE=x0x]No, it reflects all frequencies. And it shifts them.
    Even radio wave and x-Rays ? My hypothesis about Doppler shift would lead to relativistic mirror only able to get the center part of the ME (where the red/blue arrow are very short).
    And then shift them so badly again upon reflection that my eyes would certainly not be able to see me anymore
    All frequencies get shifted, the fact that some get shifted into the invisible spectrum is not relevant.
     

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    Quote Originally Posted by x0x
    No, the shift is explained by the general formulas of the relativistic Doppler effect. As such, what matters is just the relative speed and the angle between the velocity (of relative motion) and the line of sight.
    OK got it. The only angle which count is C-B-Mirror. Incidentally the effect is similar, if I look my left arm B moves to the left (the line of sight), and that's why my left arm is different that my right. I have improved my drawing, if that help chaimc.

    Quote Originally Posted by x0x
    All frequencies get shifted, the fact that some get shifted into the invisible spectrum is not relevant.
    Well it is not irrelevant if the question is to explain what will be reflected by the mirror.
    Or are you saying that the Terrel-Penrose effect only take place after the light is reflected (and not also because the incoming light is already too distorted, from the mirror FoR).

    The transversal experiment Jilan mention was probably done with a rotating mirror tube. A rotating mirror disk must be quite fun too.
     

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    Quote Originally Posted by Boing3000 View Post
    OK got it. The only angle which count is C-B-Mirror. Incidentally the effect is similar, if I look my left arm B moves to the left (the line of sight), and that's why my left arm is different that my right. I have improved my drawing, if that help chaimc.
    It is better but you have the redshift and the blueshift swapped wrt the actual situation.

    Well it is not irrelevant if the question is to explain what will be reflected by the mirror.
    Once again, all frequencies are reflected by the mirror. You may not see some (or any ) of them due to the Doppler shift but they are all there.



    The transversal experiment Jilan mention was probably done with a rotating mirror tube. A rotating mirror disk must be quite fun too.
    She didn't read the paper.
     

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    In reply to post # 36, re: your comments.

    Your "Wiki" quotes of "cut & paste" are very nice.

    As for "TEACHING?" anyone physics, at least w/ regard to "Jilan"...your own knowledge of theory is lacking. (C'mon now...fess up! I'm seriously starting to think you are rpenner! (hates

    girls...loves overly elaborate maths)

    ......

    You think references to "links" means YOU created all of the examples? In what manner? You "teach?" This is actually kinda' funny...I asked for an explanation of why "Mercury does not

    fly off into space, or fall into the Sun?" And you couldn't come up w/ anything except references to "orbit functions"...and that is NOT what I asked you!

    How you can imply you "know physics" and NOT KNOW the answers to a simple question of gravity is "beyond the pale".

    (this what leads to the "rpenner" thought...he couldn't answer either, for all his elaborate "superfluous learnedness"...maybe he should've asked his "daddy"?)

    ......

    All of your knowledge is the work of OTHERS, not your own...you cannot even match me, never mind Jilan!!!


    (Thanks for reading!) or, should I use a numerical sequence...so you can understand English as a "number".
     

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    Quote Originally Posted by Gerry Nightingale View Post

    As for "TEACHING?" anyone physics, at least w/ regard to "Jilan"...your own knowledge of theory is lacking. (C'mon now...fess up! I'm seriously starting to think you are rpenner! (hates

    girls...loves overly elaborate maths)
    ...despises idiots....
     

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    In reply to post #41, re: my #40 post.

    Yes! I thought so...you gave the game away! (go back to your little playground at "phys" where you can impress "Doughnut" and "Brucie" and thrall them w/ your acumen of all things in

    physics theory...and tell your "daddy" how mean I am!) Oh...I forgot! Did you like my message I left on the "other" site? I thought it described you perfectly!

    ......

    Your words are weak, as well as your grasp of theory...the only thing you despise more than yourself is Einstein! (you are STILL trying to find some way to deny GR in every single post reply

    concerning light theory. Every time. (I don't know which of you is more obsessed w/ A.E. You, or "T" in terms of denial)
     

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    Quote Originally Posted by x0x View Post
    IF the mirror moves as one body with the source, then the photon will return vertically to the source.
    If the mirror is moving with respect to the source, then the photon will NOT return to the source/

    As I said, i am an idiot who is trying to understand and i thank you for your effort... hahah

    In the moving mirror scenario, what is the cause of he shifting of the light? Is it because the time elapsed between the absorption abd reflection? Is it because "quantum things"?

    light (as I see it ....) is an independant entity. This is why i expected that the light will bounce back vertically regardless of the motion of the mirror...

    Thx.
     

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    In reply to chaimc, re: your # 43 post.

    No, you are not an "idiot"...an idiot would not be able to think of light as an "independent entity" and you are correct to think so.

    (thanks for reading!)
     

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    Quote Originally Posted by Gerry Nightingale View Post
    In reply to chaimc, re: your # 43 post.

    No, you are not an "idiot"...an idiot would not be able to think of light as an "independent entity" and you are correct to think so.

    (thanks for reading!)

    Gerry, thanks for the compliment.

    So, is there a base to expect that what I presumed is correct? Light's angle of reflection is independent of the motion of the mirror?

    Thnks for answering.
     

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    In reply to chaimc, re: your #45 post.

    You are welcome.

    My answer to your question of angle of reflection has more to do w/ your position as observer, rather than the motion or lack of motion of the mirror. The reflected light possesses all of

    same qualities it had before being reflected, although a very small amount of energy was imparted to the surface of the mirror by the light.

    Yes, since the reflected light is independent of the mirror...the mirror cannot dictate "angle" to light whether or not the "mirror is moving". Assuming "light" is always traveling in a straight-line,

    then "angle is observer-dependent".

    (Jilan can explain this MUCH better I can...she has years of formal education in physics. I do not have any...my study was in emergency medicine)

    "Physicist" can help also...but you might receive a TON of maths to go w/it!


    (Thanks for reading!)
     

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    Quote Originally Posted by Gerry Nightingale View Post
    In reply to chaimc, re: your #45 post.

    You are welcome.

    My answer to your question of angle of reflection has more to do w/ your position as observer, rather than the motion or lack of motion of the mirror. The reflected light possesses all of

    same qualities it had before being reflected, although a very small amount of energy was imparted to the surface of the mirror by the light.

    Yes, since the reflected light is independent of the mirror...the mirror cannot dictate "angle" to light whether or not the "mirror is moving". Assuming "light" is always traveling in a straight-line,

    then "angle is observer-dependent".

    (Jilan can explain this MUCH better I can...she has years of formal education in physics. I do not have any...my study was in emergency medicine)

    "Physicist" can help also...but you might receive a TON of maths to go w/it!


    (Thanks for reading!)

    Hi Gerry

    Until I digest what i learnt so far, (maybe Jilan and others will elighten me even more as you suggested...) now i understand even more clearly why Newton named the light phenomena as "spectrum", demons....

    Thanks,
     

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    Quote Originally Posted by x0x
    Once again, all frequencies are reflected by the mirror.
    That settles it then. In a transverse mirror frame, light is normal, and received as usual. Although I cannot understand why, each "Feynman path" of light have different angle, and a different transverse momentum (but not Doppler shift, says you )

    What in the case of a receding mirror ? Surely my image incoming photon would be so red shifted, that they would be radio wave in the mirror FoR and goes trough it ?

    Quote Originally Posted by x0x
    She didn't read the paper.
    No, nor I. That would be 33$ , and there is no chance I would by something out of my league of comprehension (I am not that stupid)
    If you have, what was the basis setup ? A disk, a tube ? surely not a fast moving train.

    Bonus question, is there a legal path to get to read more than abstract of such articles without being racketed ?
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    Quote Originally Posted by chaimc
    As I said, i am an idiot who is trying to understand and i thank you for your effort... hahah
    No your are not. An idiot is someone that is NOT trying to understand things. You are on the right path to enlightenment
    A word of caution, some idiot have responded to you (to be clear Jilan is not one of them)

    Quote Originally Posted by chaims
    In the moving mirror scenario, what is the cause of he shifting of the light? Is it because the time elapsed between the absorption abd reflection? Is it because "quantum things"?
    I strongly urge you to read Feynman lecture on QED (or all of them really). It is fun, easy, and mind blowing.
    The only Quantum thing here is the explanation of how mirror works in general (moving or not)

    Quote Originally Posted by chaims
    light (as I see it ....) is an independant entity.
    What do you mean by that ? Light is clearly part of the observable universe, with a lot of properties. It is not independent of anything, it interact quite extensively with other "things".

    Quote Originally Posted by chaims
    This is why i expected that the light will bounce back vertically regardless of the motion of the mirror...
    Things that "bounce" do bounce. You'll have the instinctive feelings that it is because some momentum is conserve (like billiard ball bouncing, or waves bouncing in your bathtub).
    Now re-examine your expectation. A bounce is a interaction between something and something else. Whatever the process, if you know that this interaction involves exchange of momentum, you know already that there is a great deal of chance that a moving(high momentum) mirror will not "bonce" the like a "at-rest" one.

    Although, as Feynman will told you, there is no reason nature behave like your gut feeling tells you. You then make experiment to test this momentum exchange (that have been done over and over again). Or the doppler shift, etc etc..

    You experiment is one of those thing, a very smart things to imagine.
     

  50. #50  
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    Quote Originally Posted by x0x View Post
    Quote Originally Posted by KJW View Post
    Quote Originally Posted by x0x View Post
    You didn't look at the websites I recommended, did you?
    No, I did not because I did not need to. I confidently worked it out for myself.
    This is really bad, I'll give it one more try and I'll give up.
    I'll put it another way. I regarded the websites you recommended as a red herring.


    Quote Originally Posted by x0x View Post
    Quote Originally Posted by KJW View Post
    The mirror is not incoming. It's motion is transverse.
    You are standing on the platform, the train with the mirror is moving on the tracks. Its motion is transverse to you. The effects that I am trying (and failing) to explain to you are happening because the train:

    -approaches you
    -is right in front of you
    -recedes from you

    Since the whole train is a giant mirror, you get different images depending on your direction of gazing. Regardless of your direction of gazing, your image will be severely distorted, both from a geometry and from a color point of view.
    This is incorrect. The motion of the mirror is parallel to its surface. Thus, the reflection plane is the same whether the mirror is moving or not. Therefore, the image the person standing in front of the mirror sees will be same regardless of whether the mirror is moving or not.

    This reasoning is about geometric mirrors. The reasoning in earlier posts is about physical mirrors and shows that the independence of the motion is also true for physical mirrors.
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    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
     

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    Quote Originally Posted by KJW
    This reasoning is about geometric mirrors.
    Ok, I am lost now. If the mirror momentum is different, and there is momentum exchange between light and mirror, how reflection would not change ?

    Should I conclude that those kind of wiki page are wrong ?
     

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    Quote Originally Posted by Boing3000 View Post
    momentum exchange between light and mirror
    I think in this problem, we can regard the momentum exchange between light and mirror as insignificant. It is not part of any argument used in this thread.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
     

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    Quote Originally Posted by KJM
    I think in this problem, we can regard the momentum exchange between light and mirror as insignificant. It is not part of any argument used in this thread
    So much for my arguments. The momentum of a mirror at relativistic speed is insignificant.

    So your point is that in that article
    For images of passing objects, the apparent contraction of distances between points on the object's transverse surface could then be interpreted as being due to an apparent change in viewing angle, and the image of the object could be interpreted as appearing instead to be rotated
    {bold added} the mirror surface is not parallel to the moving direction ? Like a train but with mirror disposed like a jigsaw (as seen from above) ?

    What about the real experiment, I bet the mirror was circular (a tube), thus also "parallel" to its surface. Will it also be reflection neutral, even so its surface is no more following the geodesic (accelerating) ?
     

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    Quote Originally Posted by KJW View Post
    I'll put it another way. I regarded the websites you recommended as a red herring.
    That's bad, you could have learned how the problem is solved. I have also recommended a very good paper on the subject.



    This is incorrect. The motion of the mirror is parallel to its surface. Thus, the reflection plane is the same whether the mirror is moving or not.
    This is , of course, false, the reflection plane being the same is a condition insufficient to insure that there is no frequency shift. I provided the math that proves your claim false, it is not very complicated. There are cases when your claim is true and there are cases when your claim is false.
    IN ADDITION, there is ALWAYS distortion due to the Terrell effect, an effect that you have not addressed in your debate.

    Therefore, the image the person standing in front of the mirror sees will be same regardless of whether the mirror is moving or not.
    This is a wide sweeping claim and it is false. The math disproving your claim is very simple. You are missing a fundamental point: the image in the mirror is formed by the integration of the rays of light of equal time path. As such, it corresponds to different instances of the object in motion when creating the reflection image. This is what generates the twisting (Terrell) effect as well as the changes in frequency.
    You are absolutely right when you talk about the reflection of ONE individual ray (no change in frequency), you are wrong when you try to apply that to the creation of a mirror image.
    Last edited by x0x; 09-03-2014 at 04:46 PM.
     

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    Quote Originally Posted by Boing3000 View Post
    So much for my arguments. The momentum of a mirror at relativistic speed is insignificant.

    So your point is that in that article
    {bold added} the mirror surface is not parallel to the moving direction ? Like a train but with mirror disposed like a jigsaw (as seen from above) ?
    KJW has never addressed the Terrell effect in his posts. The Terrell effect is always present, even if the mirror moves parallel to itself.
     

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    In reply to AIP's this thread, re: light.

    So...as is true on every physics site, a "thread" is made in into a war of interpretations regarding the properties of light. And of course, the "He-Man Einstein Haters Club" always starts

    off with "Einstein's Relativity is wrong! Here are the MATHS to prove it!!!" (this is the singular worst thing to ever happen in physics theory..."numbers over reality and logic")

    ......

    Stating that a source dictates to an emitted factor is analogous to declaring "photons are actually tiny yo-yo's"...!


    (Thanks for reading!)
     

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    Quote Originally Posted by KJW View Post
    I'll put it another way. I regarded the websites you recommended as a red herring.




    This is incorrect. The motion of the mirror is parallel to its surface. Thus, the reflection plane is the same whether the mirror is moving or not. Therefore, the image the person standing in front of the mirror sees will be same regardless of whether the mirror is moving or not.

    This is what I am looking after!!! If I am asked to prove it - how do i do it??

    This reasoning is about geometric mirrors. The reasoning in earlier posts is about physical mirrors and shows that the independence of the motion is also true for physical mirrors.
    ooops... What is geometric mirror and what is physical mirror?

    Thx !!
     

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    Well, I didnt expect to cause such riots with this question. I LOVE IT....

    As i havent come to full digestion of the issue yet, I came with another similiar but different experiment, maybe it will narrow the argument to a firmer conclusion.

    In this scenario I am not standing in front of a MIRROR TRAIN. i an standing in front of a CYLINDER MIRROR. As the mirror is a cylinder what i see of myself is a narrow "slice" of my image. Now' the cylinder mirror begins spinning.... what will I see now?

    To take it to a higher level of precision istead of "fat-me" I will put infront of the revolving mirror something narrow - a broom stick' or whatever.

    Level #3: The cylider mirror is behing a black wall and there is only a narrow slot which I can see myself through. A portion of light enters the mirror through the slot and escapes back to me throgh that same slot (or keyhole maybe?).

    The question to "level 3": will the light will always escape or maybe in "very high speed", due to a shift, the light will not escape and will be absorbed by the inner wall?

    (what have I done now....?)

    ....does it help?

    THANK YOU ALL FOR YOU WISE ANSWERS!
     

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    Well, I didnt expect to cause such riots with this question. I LOVE IT....

    As i havent come to full digestion of the issue yet, I came with another similiar but different experiment, maybe it will narrow the argument to a firmer conclusion.

    In this scenario I am not standing in front of a MIRROR TRAIN. i an standing in front of a CYLINDER MIRROR. As the mirror is a cylinder what i see of myself is a narrow "slice" of my image. Now' the cylinder mirror begins spinning.... what will I see now?

    To take it to a higher level of precision istead of "fat-me" I will put infront of the revolving mirror something narrow - a broom stick' or whatever.

    Level #3: The cylider mirror is behing a black wall and there is only a narrow slot which I can see myself through. A portion of light enters the mirror through the slot and escapes back to me throgh that same slot (or keyhole maybe?).

    The question to "level 3": will the light will always escape or maybe in "very high speed", due to a shift, the light will not escape and will be absorbed by the inner wall?

    (what have I done now....?)

    ....does it help?

    THANK YOU ALL FOR YOU WISE ANSWERS!
     

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    Quote Originally Posted by chaimc
    Well, I didnt expect to cause such riots with this question.
    There is no riot here, just point of view trying to converge

    Quote Originally Posted by chaimc
    In this scenario I am not standing in front of a MIRROR TRAIN. i an standing in front of a CYLINDER MIRROR
    I came with that idea at post #38.

    The case of a mirror disk (spinning) must be quite complicated because every part of the mirror is moving at different speed AND direction.
    Now what if that disk is fixed on the wheel of your train ?

    Seriously though, over-complicating things won't add much to the comprehension of your first suggestion. Just the opposite.

    And all those experimental idea are quite evident, if you come to think of it (like every shape with a symmetrical axis (sphere, cones, etc..).

    So lets wait for the real physicists to settle the case (I am leaning toward distortion at the moment)
     

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    Quote Originally Posted by Gerry Nightingale View Post
    "numbers over reality and logic"
    Nope.

    The logic of mathematics applied to empirical measurements, over peoples personal ideas about what reality should mean.
    ......

    Quote Originally Posted by Gerry Nightingale View Post
    Stating that a source dictates to an emitted factor is analogous to declaring "photons are actually tiny yo-yo's"...!
    Only if your analogy is based in a frame of reference that does not exist. Nobody in any frame of reference would ever measure photons to be acting analogously to tiny yo-yo's.

    Your idea of "reality" is unphysical.
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    Quote Originally Posted by SpeedFreek View Post

    Your idea of "reality" is unphysical.
    Why are Gerry and TFOLZO allowed to post in the main forum? They as bad as Farsight (who is no longer allowed, thank goodness).
     

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    Quote Originally Posted by chaimc View Post
    Well, I didnt expect to cause such riots with this question. I LOVE IT....
    Chaimc, I'm loving it too, it was a great question. Let's see if we can summarise where we are. It would appear that for a single photon there should be no doppler shift (any blue shift at the mirror being later cancelled out by redshift back at the observer due the angle of incidence and reflection being equal). Also the image of your face might look like a multiple projected images if you were able to change your expression at relativistic speeds (blinks very fast)

    If everyone agrees we can move onto your second scenario otherwise I would suggest that the other question should form another thread as that would mean we haven't bottomed this one out yet.
     

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    Quote Originally Posted by Jilan View Post
    Chaimc, I'm loving it too, it was a great question. Let's see if we can summarise where we are. It would appear that for a single photon there should be no doppler shift (any blue shift at the mirror being later cancelled out by redshift back at the observer due the angle of incidence and reflection being equal).
    The above is true only if there is no movement between the light source and the plane of the mirror.

    Also the image of your face might look like a multiple projected images if you were able to change your expression at relativistic speeds (blinks very fast)
    You didn't get this one quite right , either. The reflected image is a composite of the reflections of different points on the object (you can see both the front and the back of the object). This is what the Terrell effect is all about.

    If everyone agrees we can move onto your second scenario otherwise I would suggest that the other question should form another thread as that would mean we haven't bottomed this one out yet.
    Rotating mirrors, by virtue of changing their plane of motion, will surely exhibit both frequency and shape distortion.
     

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    Quote Originally Posted by x0x View Post
    The above is true only if there is no movement between the light source and the plane of the mirror.



    You didn't get this one quite right , either. The reflected image is a composite of the reflections of different points on the object (you can see both the front and the back of the object). This is what the Terrell effect is all about.



    Rotating mirrors, by virtue of changing their plane of motion, will surely exhibit both frequency and shape distortion.
    OK, let's restate the situation. There will be no distortion, Doppler effect etc unless you can blink relativistically and your face is really really wide. Would you be happy with that? I am keen to establish common ground here.
     

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    Quote Originally Posted by Jilan View Post
    OK, let's restate the situation. There will be no distortion, Doppler effect etc unless you can blink relativistically and your face is really really wide. Would you be happy with that? I am keen to establish common ground here.
    When you get petulant, your posts get at the same level as the ones made by Gerry and TFOLZO.
     

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    This is not petulance, this is me trying to save your face. If you don't realise it then you are even less smart than I have you down for.
     

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    Quote Originally Posted by Jilan View Post
    This is not petulance, this is me trying to save your face.
    You can't because you don't understand an iota of physics. You are at the same level as Gerry, TFOLZO and your idol, Farsight.
     

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    In reply to SpeedFreek, re: your #61 post.

    In what is my understanding "unphysical?" If a photon is emitted, then it is no longer part of the entity that emitted it. If a photon is considered in some metaphysical sense to still

    be dependent upon the source, then it is easy enough to consider it a "yo-yo" , whether there is a "quantum string" attached or no.

    ......

    In what manner did I present a "personal idea?" w/ regard to the reality of light? My personal ideas I post in "alternate theories", not as "actual accepted facts".

    ......

    I can "prove" almost anything w/the application of mathematical theory, and so can you...there are many who do nothing else! I disagree w/many areas of what to me a sheer conjecture, and

    you don't like it because I point out that a "given" supposition may not be correct, even if the "numbers" say it is.

    All I'm implying is that many things assumed to be true in the past were in fact never true, even though a mathematical theorem said a "thing was true".

    Why put me under a broad-spectrum indictment of "you deny all things mathematical" when in fact I do not. (at least this is my impression from your post reply #61)


    (Thanks for reading!)
     

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    Quote Originally Posted by Boing3000 View Post
    There is no riot here, just point of view trying to converge


    I came with that idea at post #38.


    ... indeed you did. Sorry...


    The case of a mirror disk (spinning) must be quite complicated because every part of the mirror is moving at different speed AND direction.
    Now what if that disk is fixed on the wheel of your train ?

    Seriously though, over-complicating things won't add much to the comprehension of your first suggestion. Just the opposite.

    And all those experimental idea are quite evident, if you come to think of it (like every shape with a symmetrical axis (sphere, cones, etc..).

    So lets wait for the real physicists to settle the case (I am leaning toward distortion at the moment)
    OK
     

  72. #72  
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    Quote Originally Posted by x0x View Post
    Quote Originally Posted by KJW View Post
    Therefore, the image the person standing in front of the mirror sees will be same regardless of whether the mirror is moving or not.
    This is a wide sweeping claim and it is false.
    It's true and it can be proven quite simply. I already provided enough information to piece together a proof. But let's be clear about this. It is sufficient that for a pulse of light reflecting off a mirror at an arbitrary given angle, the direction and frequency (or length) of the reflection is independent of the velocity component parallel to the surface of the mirror. But this can be proven by choosing the inertial frame of reference in which the mirror is at rest. Because the mirror is at rest, its reflection properties are indisputably straightforward. The angle of reflection is equal to the angle of incidence and the frequency of the reflected pulse is equal to the frequency of the incident pulse. Returning to an inertial frame of reference in which the mirror is in motion parallel to its surface, the reflection properties established in the rest frame are maintained in the moving frame. Note that velocity component of the light parallel to the mirror is unchanged by the reflection.


    Quote Originally Posted by x0x View Post
    an effect that you have not addressed in your debate
    This is a red herring.


    Quote Originally Posted by x0x View Post
    I provided the math that proves your claim false
    ...
    The math disproving your claim
    The maths doesn't prove anything because the reasoning behind its application is wrong. You need to specifically address the proof that I have presented.


    Quote Originally Posted by x0x View Post
    You are missing a fundamental point: the image in the mirror is formed by the integration of the rays of light of equal time path. As such, it corresponds to different instances of the object in motion when creating the reflection image.
    Why would this be different for a moving mirror compared to a stationary mirror?
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  73. #73  
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    Quote Originally Posted by KJW View Post
    It's true and it can be proven quite simply. I already provided enough information to piece together a proof. But let's be clear about this. It is sufficient that for a pulse of light reflecting off a mirror at an arbitrary given angle, the direction and frequency (or length) of the reflection is independent of the velocity component parallel to the surface of the mirror.
    No one claimed differently.



    But this can be proven by choosing the inertial frame of reference in which the mirror is at rest. Because the mirror is at rest, its reflection properties are indisputably straightforward. The angle of reflection is equal to the angle of incidence
    No one claimed any different


    and the frequency of the reflected pulse is equal to the frequency of the incident pulse.
    Not if the light source is moving wrt the plane of the mirror. This is the part that you keep missing.
    You are trying to treat this problem strictly as a relativistic physics problem but image formation is a relativistic raytracing problem, so you need to learn about how the motion of objects, mirrors AND light sources influences the final image when it comes to frequency shift.
    In addition, you continue to ignore the geometric distortion.



    Returning to an inertial frame of reference in which the mirror is in motion parallel to its surface, the reflection properties established in the rest frame are maintained in the moving frame. Note that velocity component of the light parallel to the mirror is unchanged by the reflection.
    No one claims any different. Except that you keep missing some key facts (see above).



    This is a red herring.
    No, it isn't, there is ample literature on the subject that you refuse the learn from. I could compile a list of references that refutes your rather narrow understanding on the issue.



    The maths doesn't prove anything because the reasoning behind its application is wrong. You need to specifically address the proof that I have presented.
    Actually, I did address it. Unfortunately, you clearly do not understand the issue.



    Why would this be different for a moving mirror compared to a stationary mirror?

    Simple: because the color of the object is function of the spectrum of the light source. Raytracing teaches you that.
    The motion of the source wrt to the mirror changes the spectrum of the light source (shifts it up or down) as I have already showed you. Hence, it changes the color of the object.

    Now, you are an intelligent guy, very knowledgeable . You seem nevertheless more intent on being right than learning. This is an opportunity for you to learn something new, like the Terrel effect (that you have steadfastedly refused to address), like the realistic way reflections are dealt with in relativistic raytracing.
     

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    Quote Originally Posted by x0x View Post
    Quote Originally Posted by KJW View Post
    It's true and it can be proven quite simply. I already provided enough information to piece together a proof. But let's be clear about this. It is sufficient that for a pulse of light reflecting off a mirror at an arbitrary given angle, the direction and frequency (or length) of the reflection is independent of the velocity component parallel to the surface of the mirror.
    No one claimed differently.


    Quote Originally Posted by KJW View Post
    But this can be proven by choosing the inertial frame of reference in which the mirror is at rest. Because the mirror is at rest, its reflection properties are indisputably straightforward. The angle of reflection is equal to the angle of incidence
    No one claimed any different
    Except you contradicted yourself with this statement:

    Quote Originally Posted by x0x View Post
    In addition, you continue to ignore the geometric distortion.
    The above two points prove that motion of the mirror parallel to its surface does not cause geometric distortion.

    I should remark that if I stand in front of the mirror with outstretched arms, the image of me that I see is not moving. Therefore, one cannot attribute to the image any relativistic effects that one would attribute to a moving object. It appears to me that you are conflating the motion of the mirror with the motion of the images.


    Quote Originally Posted by x0x View Post
    Quote Originally Posted by KJW View Post
    and the frequency of the reflected pulse is equal to the frequency of the incident pulse.
    Not if the light source is moving wrt the plane of the mirror.
    The motion of the source is irrelevant. I've already established that the equality of the frequencies of the incident and reflected pulses for the moving mirror is a direct consequence of the equality of the frequencies of the incident and reflected pulses for the non-moving mirror. Note that because the parallel component of the velocity of light is unchanged by the reflection, then the frequencies of the incident and reflected pulses will be equally transformed by the change of reference frame.


    Quote Originally Posted by x0x View Post
    The motion of the source wrt to the mirror changes the spectrum of the light source
    This is irrelevant because we are only concerned with the effect of the motion of the mirror.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
     

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    Quote Originally Posted by KJW View Post
    Except you contradicted yourself with this statement:



    The above two points prove that motion of the mirror parallel to its surface does not cause geometric distortion.
    Your claim is false, it does create a Terrell effect. I already explained to you the reason, the light paths source-mirror-object-mirror-back to object (observer) are unequal, so you end up seeing a composite of your front and of your side view. The faster the motion between the object and the light source, the stronger the effect.

    I should remark that if I stand in front of the mirror with outstretched arms, the image of me that I see is not moving. Therefore, one cannot attribute to the image any relativistic effects that one would attribute to a moving object. It appears to me that you are conflating the motion of the mirror with the motion of the images.
    No, I am not, you simply do not understand the raytracing part.



    The motion of the source is irrelevant.
    False again. The motion of the source alters its spectrum, you stubbornly refuse to learn how images form via raytracing.


    This is irrelevant because we are only concerned with the effect of the motion of the mirror.
    No, we are concerned with the way images form in the REAL world. Objects get illuminated by MOVING sources (moving cars headlights, reflections on chrome parts of other objects moving in the scene, sun moving in the sky , etc). Until you learn that, you will have a crippled understanding of the issue.
     

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    Quote Originally Posted by x0x View Post
    Quote Originally Posted by KJW View Post
    This is irrelevant because we are only concerned with the effect of the motion of the mirror.
    No, we are concerned with the way images form in the real world.
    You might be, but I'm only interested in answering the question asked in the opening post.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
     

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    Quote Originally Posted by KJM
    The motion of the source is irrelevant. I've already established that the equality of the frequencies of the incident and reflected pulses for the moving mirror is a direct consequence of the equality of the frequencies of the incident and reflected pulses for the non-moving mirror
    Inverting the problem and considering the mirror static and the person moving is very interesting (I generally do that plus bring the process to its limit).
    Here is how I picture things.

    I cannot see any difference. Won't I appear totally blue shift, than normal, than red shifted, for each part of the mirror that's gona participate in the reflection (in the static mirror frame) ?
    Won't I in every case appear squeezed along my vertical axis ? I know it is no big deal, because my eyes are squeezed too, but is this length contraction constant or does depend on the angle between the normal to reflection M and "me" ? (If so distortion are an-avoidable)

    Because the speed of light is constant in the mirror frame, thus if I move fast enough a light from my red(and small?) eye may end up in my blue(a little less small?) eye, or miss it altogether.
    I get a feeling that maybe all those effect will cancel out because there is an infinity of M points, but cannot do the math...
    If I move at light speed what can the mirror reflect but a very slim gama ray burst ? And only if I move totally parallel to the mirror and if I look totally along the normal (the question of this post).
    Incidentally no-one can look at a mirror only along the normal. Or then only shave a very narrow part of one chin.

    A third actor "the source" must be added, and even if I hold an electric torch, its rays will never be parallel to the path, but in one case (M, not Mī). And Both its length, speed and emitted light are squeezed slightly different than me (delta in time). Every part of me also is.

    Or not ?
     

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    The photons from your face will be blueshifted when they reach the mirror in the mirror frame. When those photons are reflected in the mirror frame they will have the frequency in the mirror frame. However when the receiver absorbs them they are redshifted back to the original frequency. Image aberration will occur, but only if the source of the image if changing at relativistic speed (blinks very fast).
     

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    Image aberration will occur, but only if the source of the image if changing at relativistic speed (blinks very fast).
    Good point. Not only every part of me is moving differently with regard to the mirror, but every part of me is moving in regards to every part of me (unless I am dead cold). Let's say I am hand-waving like hell (pun intended)
     

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    Quote Originally Posted by KJW View Post
    You might be, but I'm only interested in answering the question asked in the opening post.
    Fine, we are done.
     

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    Quote Originally Posted by Gerry Nightingale View Post
    In reply to SpeedFreek, re: your #61 post.

    In what is my understanding "unphysical?" If a photon is emitted, then it is no longer part of the entity that emitted it. If a photon is considered in some metaphysical sense to still

    be dependent upon the source, then it is easy enough to consider it a "yo-yo" , whether there is a "quantum string" attached or no.
    Show me any way for any observer to make an experiment or observation where a photon acts in a way that can be considered to be like a "yo-yo". In what frame of reference does anyone conclude that a photon acts like this?
     

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    In reply to SpeedFreek, re: your #81 post.

    I didn't set-up the parameters of post #1. To me it is an exercise in logic, and if it is considered that all of the components of the FoR are "relative absolutes" then no "frame-dragging" or

    "chromatic-shifts" are going to occur.

    "XoX" maintains that an "absolute velocity" (even) of the mirror-train is going to exhibit shift w/regard to the perception of the observer as soon as the train moves. I state this could only
    occur IF the distance bet. the observer or train changed in some manner, and it has not changed.

    "XoX" maintains the reflected light will shift...the inherent implication being that "the mirror can be considered as an emitter due to surface of the mirror "throwing back" photons, and in this

    circumstance the light will exhibit shift due to the increasing velocity of the mirror-train"<(I'm paraphrasing here, these are my words, not his)

    My "yo-yo" analogy of the photons is apt, in this instance. To state that the velocity of the mirror dictates how the light will be perceived by the observer as "red in one aspect, blue in the

    other" means the photons THEMSELVES are contiguous with the matter of mirror's surface, and to me such an observation is literally impossible!

    Just how does an observer "see" a photon which has not been emitted or "reflected?" The surface of the mirror is reflecting light that is already in the FoR! Just "how" can light be shifted

    that is already present?!?! The mirror is not creating the photons whether it is moving or static...it is only reflecting that which is already in-place! Observer/Light/ Mirror-train are already

    in the FoR...the only change is the velocity of mirror, relative to it's presumed previous static position.

    ......

    Consider the Sun as an analogue of the "mirror-train"...if an observer looks at it from a static position in space (no aberrational movement) does one aspect of the Sun's edge demonstrate

    a "red-shift" while the opposite edge shows "blue-shift?" According to what I read in XoX's posts, it should demonstrate exactly that!

    I maintain that the Sun exhibits no shift because all of the photons emitted are being emitted at the same instant, regardless of "where they are in relation to the observer".

    I maintain that unless the observer is moving, as well as the mirror-train or the Sun, then no chromatic-shift will be observed regarding the light from the train or the Sun, or even

    reflected light from the Moon!

    ......

    I am not the one who maintains that a photon which has already been reflected or emitted is still subject to the velocity of the emission source!


    (Thanks for reading!)
     

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    Quote Originally Posted by Gerry Nightingale View Post
    Consider the Sun as an analogue of the "mirror-train"...if an observer looks at it from a static position in space (no aberrational movement) does one aspect of the Sun's edge demonstrate

    a "red-shift" while the opposite edge shows "blue-shift?"
    Yes it does, due to the Sun's rotation.
     

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    In reply to SpeedFreek, re: your #83 post.

    As per your statement "yes it does", I stand corrected...and by implication, XoX is now the greatest living person in physics theory!!! "Einstein is DEAD! Long live the new King!"

    (I am not going to argue one iota further w/regard to "chromatic shift" being observer dependent...it's pointless because XoX is now correct in all respects with no margin for error)

    Besides, you're the boss here, and what you say goes and I have no problem with that.


    (Thanks for reading!)
     

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    Don't just take my word for it that a stars rotation causes the light from one side to be redshifted whilst the other side is blueshifted. Do some research. There are many sites and articles out there that explain it, and many are at entry level like this one (made by a friend of mine) or this one

    And I was addressing that point alone, so don't just assume it backs up anything else anyone has stated by implication. Science often finds results that are counter-intuitive.

    As for the "what I say goes" thing, that only applies when I am moderating (where I use coloured text). When it comes to discussing science I can be wrong just like everyone else can be wrong.
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    In reply to SpeedFreek, re: your #85 post.

    Referring me to "Dick and Jane" primers will not convince me that Einstein was wrong...this IS the direct implication of XoX's posts in regard to the "light from the mirror train being reflected".

    If what I wrote concerning the light conditions of the "mirror train and observer" are wrong, and "you" acknowledge that XoX is correct by default, then NO real discussion of the science

    of light is possible. You cannot tell me that a "Star's rotational vectors determine what shift of color an observer will see" and then tell me it does not acknowledge that XoX may or may

    not be "correct"...no, the statement verifies XoX is correct!!! You cannot say "this is correct" and at the same time say "it might be wrong if XoX wrote it". Huh? What?

    Do some "research?" Okay, I'll read more Einstein! Again! Of course, whatever I write regarding physics will be instantly dismissed, if it involves GR or SR as Einstein wrote it!

    Who or from what source do you think I am referencing with when I write anything w/regard to "light theory?" "Star Trek" maybe? No...my knowledge comes from Einstein, not

    "Dr. WHO", however entertaining it may be.

    I am wrong..."Physicist" is wrong..."Jilan" is wrong. Okay, then. That sums it all up. That's three people, two of whom have years of verifiable education in physics theory, and one of whom

    is a verifiable genius<(although he does not think so) whose opinions and educated remarks are "flushed down the toilet" as if they were nothing more than excrement!!!

    So..."Jilan"s education at Oxford concerning physics is NOTHING??? Forty years of reading physics on my own is also nothing? Physicist also?

    Okay...like a guy once said "I don't need a weatherman to tell me which way the wind is blowing".


    (Thanks for reading!)
     

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    I'm not trying to convince anyone that Einstein is wrong and nor is anyone else in this thread, as far as I can tell.

    I was just telling you that you were wrong to assume that the light from one side of the Sun is not redshifted and the light from the other side is not blueshifted.

    You seem to be leaping to conclusions that are unwarranted.
     

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    Yes Gerry, please calm down, Speedfreak is talking about something else altogether.
     

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    In reply to SpeedFreek, re: your #87 post.

    SpeedFreek, what you say regarding chromatic shift due to rotational vectors is true, but NOT for the reason you think it is true. The emitted photons are all the same, yes? At least w/regard

    to their being "packets of energy" does not predicate that some photons have "more" energy than others...this is what "red shift" implies as an inherent condition, that "blue" is weaker than

    red in some unknown fashion, and I refuse to accept this as true! I am stating there are "no color" assignments associated w/photons as part of their intrinsic make-up, either as an

    ab initio state or in transit...they (the photons) are all the same. "Chromatic-scales are in our minds, they are "how our minds see" to establish scale and definition". SpeedFreek, if you can

    perceive light from the Sun's "edges" being "red on one, blue on the other" as they are being emitted, it is because that is "how" our minds perceive it, not a true aspect of light as it

    actually is!

    Our mind's abilities to see light are limited by the speed of chemical reactions that cannot possibly "match" the relativistic speed of light, and therefore all we can see is a "past-tense state"

    of that which was or is current in any given frame-of-reference. <(if you can find a flaw in this statement, I would really like to "see" it, pun intended)

    In terms of the Suns' axial rotation and it's influence on the aspect of observation of "red one side, blue the other" it is a virtual analogy that "one edge is rotating faster than the other!"

    Just how does this mechanism work in relation to a presumed contiguous mass? Because I don't understand this at all.

    ......

    SpeedFreek, it may well be true I'm not the "sharpest knife in the drawer", but I think I'm quite some distance from "dull" or "thick". When I answer something, I automatically

    consider ALL the factors involved, not just the ones which are stated...I answered the original question on this thread to the best of my ability, and if I'm wrong or mistaken, it's okay that

    I am. I won't like it much, but only because my thinking is faulty, not because someone criticized what I wrote.


    (Thanks for reading!)
     

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    Gerry, the colour of light if a function of the receiver as well as the emitter due to the Doppler Effect. Energy is frame dependent.

    Doppler effect - Wikipedia, the free encyclopedia
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    In reply to Jilan, re: your #88 post.

    Jilan, I am "chilled to the max" w/regard to my emotive dissonance...! (I just love using big words! Makes me feel like I'm smarter than I really am!)

    If my "tone" seems hostile, it is not directed to "SpeedFreek" as a reaction to his posts to me...rather, it is my extreme annoyance w/being trolled to the point of exhaustion and seeing

    threads de-railed for an "idiot's delight"<(do you think anyone here will understand the reference? Beside yourself?)

    I "get" what the "gang of three" are attempting to get away with, and I don't like it!


    (Thanks for reading!) Cheerio!
    ps...I know you are trying to "protect me from myself". Thanks.
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    In reply to Jilan, re: your #90 post.

    Of course, "energy is frame dependent" via the mechanisms of the actions of matter. (if you got the impression I'm advocating differently...I'm not.)

    I am only trying to establish that much of what we commonly accept as a "true condition" may have an inherent flaw. "Doppler shift" applies (I think) to any wave-function of what is regarded

    as emitted energy from a definable source, and it may be that "Doppler" is an inherent feature of light that does not require "distance from source" to become manifest.

    In other words, Doppler is "there" at all times, but it requires the distance of the observer to "make it out" visually, or an observer's velocity or an oblique perspective from the source.

    Since c is always c, then "chromatic shift" will always be inherent factor, existing independent of photon c<(at least I think so)


    (Thanks for reading) Ta ra!
     

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    Quote Originally Posted by x0x View Post
    You are standing on the platform, the train with the mirror is moving on the tracks. Its motion is transverse to you. The effects that I am trying (and failing) to explain to you are happening because the train:

    -approaches you
    -is right in front of you
    -recedes from you

    Since the whole train is a giant mirror, you get different images depending on your direction of gazing. Regardless of your direction of gazing, your image will be severely distorted, both from a geometry and from a color point of view.

    It is not the way you think it is, it is far more complicated.
    Question: If the train is so long that one cannot see the front or the back of the train and the train is completely hidden by one seamless mirror facing the observer, would the observer even know that the train is moving and see him/herself just reflected as if the mirror was stationary or would the reflection be distorted?
     

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    Quote Originally Posted by write4u View Post
    Question: If the train is so long that one cannot see the front or the back of the train and the train is completely hidden by one seamless mirror facing the observer, would the observer even know that the train is moving and see him/herself just reflected as if the mirror was stationary or would the reflection be distorted?
    This is exactly what this thread is about.
    You might want to read it from the beginning.
     

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    Quote Originally Posted by Jilan View Post
    This is exactly what this thread is about.
    You might want to read it from the beginning.
    I have and am still waiting for a direct and specific answer to the question of what the observer sees.

    a) does observer see him/herself distorted? yes? no? This was the crux of the original question, IMO
     

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    Quote Originally Posted by write4u View Post
    I have and am still waiting for a direct and specific answer to the question of what the observer sees.

    a) does observer see him/herself distorted? yes? no? This was the crux of the original question, IMO
    I would say not, but we did not reach a consensus on this.
     

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    Quote Originally Posted by Jilan View Post
    I would say not, but we did not reach a consensus on this.
    I agree and that is why reposed the question in a slightly different way. I am actually very interested, because of the inclusion of the mirror, which reflects a reversed image of the original observer.
     

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    I am leaning towards yes. Even with a fixed light source. (Maybe no color aberration, but geometric distortion)
     

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    Quote Originally Posted by Boing3000 View Post
    I am leaning towards yes. Even with a fixed light source. (Maybe no color aberration, but geometric distortion)
    Intuitively I would tend to agree, after all the image from and back to the observer has to travel a very small distance and that takes time, even @ "c". But it gets more complicated than that.

    Next questions; (keeping in mind that the reflected image is reversed), if there is an geometric aberration, how would that appear to the stationary observer? It seems to me that any geometric distortion would apply equally to both sides of the observer. Perhaps the entire image shifts very slightly, but then: (assuming the mirror is moving from left to right relative to the observer), will the observer see himself (and everything else behind him/her) shifted to the left or to the right?
     

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    Consider a person standing in front of a stationary mirror. Suppose an observer moves past this person at relativistic speed in a direction parallel to the mirror surface. In the frame of reference of this observer, light is emitted by the person and is reflected by the moving mirror back to the person. From the Law of Reflection in the person's frame of reference, we deduce that the Law of Reflection also holds for a moving mirror in the observer's frame of reference. If the Law of Reflection holds regardless of the motion of the mirror parallel to its plane, then all rays of light will be unaltered by such motion, and images will not be affected.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
     

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