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The doppler radar trap paradox on the path to gravity.

  1. May 8, 2009 #1
    Hello All,

    I have been challenged by a friend to look into Gravitational waves and some questions he has about them -- but I have always been a slow starter. Took physics with SR in college, got my BSEE, happily can build analog circuitry of all kinds -- but found that certain questions about SR never did get answered by the Prof and have gnawed at me ever since. Now I accept SR by and large -- but don't really accept GR yet -- so gravity waves are still in the Poincaire electrical charge theory realm (and Feynman had some speculation too, if I remember correctly) for me. And I figure, I best resolve the simple issues first before tackling the bigger ones ... a physics oriented crowd has perhaps studied some experiment that I haven't and might already know which of a set of answers is a dead end... hopefully this thread will grow in knowledge and build toward my friends questions -- which intrigue me, and I hope -- you.

    So here is a simple gedanken with inertial frames to begin the journey toward gravity. I am not good with Minowski 4 vectors, so -- if you don't use them thanks! -- but if you do, please be very clear what they mean.

    I am envisioning a radar trap with a policeman. We can ignore the rotation of the earth and gravity here. I am also envisioning a craft passing the policeman at c/2. Since we are with the policeman at the moment, he is our inertial frame. The craft is another inertial frame -- not accelerating nor decelerating. There is sufficient rest mass with both the policeman and the craft so that the radar gun he fires at the craft does not change the speed between them measurably.

    Please correct any assertion that is clearly contradictory:

    1. Since there are only two objects -- the policeman and the craft -- the speed measured by one ought to be the same as the speed observed by the other. There is no third body issue.

    2. The policeman is firing a low wattage continuous wave radar gun. Eg: let's say 300MHz in his frame -- and he is aiming it at the craft passing him. At T=0 the craft is touching the gun, and thereafter it recedes directly away from the gun at a constant velocity V.

    3. Older radar guns work on the Doppler effect (red shift/blue shift). Our police man is using one -- eg: Essentially, he is measuring the beat frequency of his outgoing wave with the incoming one -- and using that to determine the crafts speed. The craft -- for its part -- is made of a superconductor, and is acting as a perfect mirror nicely aimed to give the policeman an accurate reading.

    4. Einstein believed in "photons" and the photoelectric effect. Hence, we may view (consistently for his thinking) the police man as ejecting photons with wavelength 1 meter, frequency 300Mhz (assuming c=3x10**8 exactly, otherwise -- scale accordingly for nice math.).

    5. Using my old EE books on radiation, by Dr. Aziz Inan -- we can view this as a simplification of Maxwell's equation. We can treat the outgoing and reflected waves as a plane wave in the x direction. The boundary condition is that the perfect conductior has no E-field inside it (charge neutral superconductor) and its location is x=v*t. The resulting TE (transverse electric) then is zero at x=vt. When I solve Maxwell's equations this way, I arrive at a standing wave. The returning wave has the same amplitude as the outgoing one, and is 100MHz.

    Now; Here's where my choose your poison/paradox/irritation begins.
    I naively thought, since the original wave had a length 1M in the policeman's reference frame -- in the crafts frame it ought to have one of 1M / L, where L=sqrt( 1- v**2/c**2). Now that is clearly wrong. When I find the second place away from the reflector where no E-field exists by Maxwell's equations -- I come to the conclusion it is at 1.5Meters from the reflector in the police frame. now 1.5Meters/L = 1.7329Meter'.
    Where the ' means the craft's meter stick. That answer looks right.

    So, here's the first question: Since Einstein viewed photons as having energy E=hf; What is the proper way to balance Energy for a *single* photon? Upon reflection from a lossless mirror which has sufficient rest mass such that acceleration is negligible -- Maxwell's equations predict the reflected wave will have a lower frequency in the police man's reference frame. eg: the photon lost energy upon a lossless mirror. The loss depends on the velocity of the mirror, and nothing else.

    In the reference frame of the craft -- the light does not know who (police or craft) is moving -- so being a good mirror -- the light is reflected and is the same color to the observer on the craft. ergo: No energy loss. The perspectives are at odds to each other -- so rather than believe the police unconditionally -- tear these to shreds:

    1. proposition: Photons do not really have E=hf energy each -- that is just the energy they release to an electron during a reaction ejecting an electron from a photoconductor. Actual photons may have FRACTIONAL amounts more or less than hf.

    2. proposition. Photons are not really individual (a variation of 1) -- if we take the extra time delay from when a the police gun is shut off to when the last of the light arrives back at the gun -- we will find that the lower energy returning photons last for just enough extra time that a the energy (assuming a CW wave and ignoring photons) balances out. Low power x longer time = higher powe x shorter time. (In fact it does balance out.)

    3. What would you propose that completely explains the reconciling of the policeman's view -- and the crafts view?
    Last edited: May 8, 2009
  2. jcsd
  3. May 8, 2009 #2


    User Avatar
    Science Advisor

    welcome to PF,

    Adjust the frequency.
    Before jumping to conclusions concerning quantum physics and relativity, make sure you understand and resolve the exact same contradiction that arises in classical mechanics. Use perfectly elastic rubber balls, bouncing from a moving surface. In one frame, they don't lose energy. In another, they do. Where did the energy go?
    (Hint: the velocity of the wall will change, and the effect of that change is not negligible.)
  4. May 8, 2009 #3


    Staff: Mentor

    The lost energy goes into increased KE of the mirror. If you are neglecting the acceleration then you are also neglecting the change in energy. There is nothing wrong with that since it is a small effect, but neglecting one object's change in KE and then finding that energy is not conserved is hardly paradoxical.

    No, they are not at odds to each other. The photon's energy decreases in all inertial frames and the mirror's KE increases in all inertial frames. Again, purposely neglecting something does not make for a paradox.

    None of these are needed. There is no paradox here to resolve, just an accounting error.
  5. May 8, 2009 #4


    User Avatar
    Science Advisor

    No. KE is frame dependent, and does not transform linearly.
  6. May 8, 2009 #5


    Staff: Mentor

    D'oh! You are right. In frames where the mirror is traveling towards the photon the KE of the mirror decreases and the energy of the photon increases! Of course, energy is still conserved in all frames, but the amount and direction of the energy transfer is frame dependent.
  7. May 8, 2009 #6
    If the wall were free, yes its velocity would change.

    I accept most of your answer -- although I find it quite unsettling for the kinds of things I am exploring -- and am not sure it really applies. (My fault for not noticing the obvious?)

    BTW: I am not trying to disprove SR in favor of "classical physics" -- and mentioning that the same problem exists in classical physics doesn't really solve my problem -- it doubles it.

    It is exactly this kind of assumption that I am often forced to make in EM calculations, for say -- transmission lines. A shorted transmission line does *NOT* transmit any energy beyond the short -- nor does the short adsorb any energy. All energy is reflected at the short circuit.

    Or again...
    If the wall, or mirror, were, say -- the end of the universe (a transmission line which is "open" so to speak) -- the energy loss would still exist, although no energy could transmit beyond the edge....
    In that case -- as in the case of a mirror -- 100% reflection would still occur. (The sign of the reflection would simply change from inverted to non-inverted.). The experimenter can not know whether his photon hits the edge of the universe or not -- so, this seems to imply, that an assumption is made every time a physics equation is calculated about the nature of the reflector. (It was my intent to avoid the assumption if possible -- and that is also the reason I tried to avoid any third body issues/calculations where a rest mass was concerned.)

    Your solution requires an assumption that I am trying to pin down the meaning of. I am beginning to think it is elusive because something is swept under the rug in my EE classes (so to speak.).

    So, to be perhaps hair splitting:

    In a frame of reference where the wall is NOT moving, one does not say "the wall accelerates" and that "acceleration is not negligible". I see it commonly stated/assumed that a wall does not accelerate, and the velocity change is in fact ignored. ERGO: the energy of the ball is conserved and the wall gains and looses no energy -- the way (I presumed) this was known, is that the wall does not accelerate. But now, I am thinking, it is simply that the WALL's frame of reference is preferred -- because the mathematics becomes subject to roundoff error, otherwise...

    Given a transmission line (an exact analogy to the mirror and the police-man with a SINGLE frequency CW signal -- eg: non-dispersive) where a reflector moves along the transmission line with time -- lengthening it by exposing more of it (a sliding short with exposed conductors) -- as an EE I would view it as mathematically identical to the photon/speed trap problem. According to all calculations (and the poynting vector) -- 100% of the energy *IS* reflected at the mirror/short.

    Now, has anyone seen a case where the force on such a short circuit is shown to be non-zero?
    (superconducting transmission line and short circuit.)? My gut reaction, having just toyed with the equations, is that the poynting vector shows 100% energy reflection at the superconducting short circuit. ( Perhaps I have made a mistake...? )

    If not, is there a difference when computing photons vs. an EM wave on a transmission line?
    What is this difference?

    To be clear: I am not in favor of classical physics over SR -- I am trying to grasp where the assumptions fail -- and *why*. -- see possibility 3. I am a slow starter, as I said.
  8. May 8, 2009 #7
    Not to be asanine, but since I am after a subtle point -- I wish to make sure we're comparing like to like by enforcing a bit of narrow mindedness.

    Here's a simple analysis of your classical problem.
    Correspondence principle -- nonrelativistic speeds:

    Initially, from the frame of reference of the ball, the wall is approaching at a given velocity v.
    After impact the wall recedes at the same velocity. The magnitude of the wall's velocity as measured from the ball is constant w/ a sign change, ergo the energy which is related to the *square* of that speed as mass does not change classicly with speed. eg: m*v**2 = m*(-v)**2: eg: even In the limit as mass approaches infinite, the equation is exactly balanced such that regardless of the mass of the wall and the ball -- the approach speed and the leaving speed are identical and opposite. Therefore, energy is *always* conserved -- strictly. Momentum is a different issue, but I was not considering momentum in my original problem either. The magnitude of momentum is clearly conserved as well -- but the direction is not.

    In any elastic collision between two masses, the relative speed (the only physically measurable speed) in a strict two body problem -- is conserved regardless of the individual masses. The ball and the wall may be considered inertial before and after the collision, with the non-inertial consideration limited to a point discontinuity.

    From the perspective of the wall, the exact same is true. No imbalance in the conservation of energy is detected. Even if the wall is not infinitely massive, and the attraction between ball and wall are known -- working the problem out from either reference frame will produce the same result classically.

    No paradox is noted.

    I do see that if I choose a third reference frame, the mass of the wall must be known in order to solve the problem, and that (in that case which I purposely avoided) the wall becomes important as a sink of energy.

    So, it would seem, that even a photon of light must indroduce a three body problem, even though the photon has zero rest mass in SR -- and the "Ether" is supposedly irrelevant as a reference frame.
    Energy of a photon, then, still must be an incomplete description of that photon, then, somehow.

    BTW. Thanks for the welcome -- I'm glad to find people actually interested in thinking a bit.
    Last edited: May 8, 2009
  9. May 8, 2009 #8


    Staff: Mentor

    You are really over-analyzing this, there is nothing complicated here.

    Just write down the expression for the initial momentum of your mirror for some arbitrary velocity and write down the expression for the initial momentum of your light pulse. You know the mirror's velocity so you know the Doppler shift of the light pulse and therefore the change in energy and momentum for the light.

    Use conservation of momentum to determine the mirror's new velocity and then you can check to make sure that energy is conserved for any arbitrary mirror velocity.

    Btw, you might be interested in the http://en.wikipedia.org/wiki/Four-momentum" [Broken], which I find to be one of the most useful concepts of SR.
    Last edited by a moderator: May 4, 2017
  10. May 8, 2009 #9
    Ηi Dale!

    A paradox is an apparent (even if illusionary) contradiction. The twin paradox is a paradox precisely because when looked at in the way presented -- it does not have a solution unless further information is used to break they symmetry. eg: one person "feels" the acceleration... etc. I don't mean to imply that there is no solution to the problem I am presenting -- I mean that it causes certain irritations because assumptions in EM waves class leads to the paradox, and it is NOT self evident how to resolve the paradox to me -- I expect there is a solution.

    The issue I am having probably centers around the idea of instantaneous action at a distance.
    I defined the problem simply as a starting point of a discussion, not as an all encapsulating description of an irrefutable contradiction. Please see my posts to the German "I" ;) not, I hope, a synonym for the greek Ego. ηγω.

    OK. Let me then, push the wave/maxwell equation for a moment. If, as you say, the energy is transferred to the mirror -- then why is it that the time average power computed for the returning wave EXACTLY balances out?

    A quick and crude calculation I emailed to a friend is attached:

    My question to you, is that since the wave equation (not the photon based one) predicts all energy is in fact returned -- how is it that you say the mirror is necessarily accelerated?
    Also, the problem EXPLICITLY sets the location of the mirror to x=vt; therefore, the form of the solution which follows (missing only a scalar constant to match it in terms of volts/meter) ought to take into account the fact that the mirror does not accelerate. What you appear to argue, then, is that something is wrong with Maxwell's equation -- for I asked the question thusly: given a wave which hits a mirror that is NOT accelerating, what wave must return in order to solve Maxwell's equation.
    I would have expected the solution to take into account any implied forces or masses which prevent the mirror from accelerating. The problem is: since the mirror does NOT accelerate, and the returning wave from a NON accelerating mirror is a wave with lower energy -- why do you insist that the mirror "must" have accelerated. (This kind of thinking is irritating, and I apologize -- I am doing this as an exercise to get to the bottom of a certain issue that cropped up from my EM classes. But realize, that if there is a counterbalancing force which prevented the acceleration -- I am fishing for why Maxwell's equation does not compensate for that -- and if Maxwell's equation DID -- why is it that the reflected wave does not appear to agree with a wave packet/photon type of approach.)

    ;------------------------- quick and crude calculation --------------------------
    Assume a mirror which at t=0 is touching a microwave antenna. It is
    moving away from the antenna at velocity v (a constant), the antenna is
    radiating source energy of radian frequency "ωs" at the mirror.
    ... and of course, the speed of light is just 'c'.

    In this case, one can set up a standing wave equation, satisfying
    Maxwell's eqns. And knowing 100% reflection occurs:
    Efield = cos( ωs - ωs/c * x ) + cos( ωr + ωr/c * x + α )
    Knowing that the E-field is 0 inside a conductor (eg: the mirror), and
    that the mirror is moving with velocity v, the above equation is zero at
    x = v*t; e.g. the mirror. The answer pops out:
    ωr = ωs * (c - v)/(c + v). for example, if v=c/2, and ωs = 2*π*300Mhz
    (wavlength=1Meter) the returning frequency ωr = ωs*(.5/1.5) = ωs*(1/3).
    So, to the observer standing at the transmitter, the photons coming back
    have 1/3 the energy (2/3'rds lost!).
    However, when one works out how long until the light stops returning,
    assuming the transmiter turned off after T seconds (the mirror being T*v
    meters away):

    The distance the mirror travels until the last of the wave hits it:
    v*T + v*Td = c*Td
    So, Td = v/(c-v)*T
    The total time, then, is the "on" time + last of outward going light
    making a round trip.
    T + 2*Td:
    So the total time receiving is just: T + T*2*v/(c-v)
    = T * ( c + v )/( c - v).
    The equation is scaled exactly opposite to the loss of energy.
    loss of energy scalar = (c-v)/(c+v)
    gain of time scalar = (c+v)/(c-v)
    multiply together, answer is 1. Therefore, energy is conserved.
  11. May 8, 2009 #10


    Staff: Mentor

    If the mirror does not accelerate then your mirror/light system cannot be isolated and whatever external force is keeping it from accelerating can transfer energy to or from the system. So again, no problem.

    If you neglect something (either the acceleration of the mirror or the external force preventing it from accelerating) then obviously you can get these accounting errors you are worried about.
    Last edited: May 8, 2009
  12. May 8, 2009 #11
    Energy is conserved in Maxwell's equation -- if energy is dissipated any OTHER way maxwells equation is WRONG.

    Maxwell's eqations, I was taught, are correct on *average* -- eg: an expectation value.

    If so, then all the energy is accounted for in the wave itself. NO ENERGY IS REQUIRED ANYWHERE ELSE. It does NOT flow around the mirror, or any other way in the problem I have set up.

    Do you have any proof that the energy MUST flow around it somehow?

    Again: It balances according to Maxwell's equations but NOT when individual photons are imagined.

    Why do these complemetary methods predict different things must be happening?

    I admit there is an error, but can you show what the error is such that MAXWELLS equations do not say that all the power is reflected back -- or the converse, that in the case of photons -- all energy must be reflected back. One of these two is required for the equation to balance.
    The mirror does not have to be isolated, and I never claimed it was. I simply claimed that it was not accelerating.
  13. May 8, 2009 #12
    A la Maxwell. Take an coaxial cable, stationary in the lab frame. As you assume, one end is shorted and at the other end, you inject a signal. With all the energy reflected from the end, the cable contains a standing wave.

    In an inertial frame in relative motion parallel with the length of the cable, the observer will no longer see a cable containing standing waves. As he travels down the cable he will see alternating nodes and antinodes. The transmitted and reflected signal are no longer equal in frequency. The reflected energy is not equal to the transmitted energy per E=hw.
    Last edited: May 8, 2009
  14. May 8, 2009 #13
    In the example problem I gave, the policeman sees either 1: A 300MHz outgoing wave CW -- and a 100MHz returning wave CW -- (if they are seperated spatially by say a double reflecting 90' mirror or two coaxes side by side) -- or else he sees the superposition of a 100Mhz and 300Mhz wave -- which has standing and travelling components. It is correct that he sees a modulated wave. The modulation is has null spots every 0.75Meter, and repeats. The modulated envelope travels with and is stationary to the reflecting mirror -- which means that the modulation moves TOWARD the mirror in the police man's frame (AKA lab frame.).

    Since the mirror/short is moving, the length of the cable is effectively changing with time -- it is effectively becoming longer. The result is, that should one turn the source off at any time -- the power will still be reflected back for a certain period of time AFTER the transmitter is turned off. (The mirror continues to move with the same velocity). This length of time happens to exactly balance the the change in power reflected -- eg:
    power transmited*transmission time = power reflected * (transmission time + delay until coax is empty of radiation)..

    See the above analysis.

    Power is not conserverd, but total energy is.
  15. May 8, 2009 #14
    Please be direct.
    OK, so you have a mirror that does not accelerate. As you point out, if we reflect a classical EM wave against the mirror, the reflected beam will have lost zero energy (and its wavelength will be unchanged, although its momentum has been reversed). Similarly (as you may not have realised), if we bounce photons of a mirror that does not accelerate then the photons (at least on average) will also not lose any energy.

    Why doesn't it accelerate, when in both cases there is obviously a force against one side of the mirror? There must be a balancing force on the other side. Perhaps a symmetric EM wave (or sequence of photons) on the other side. Perhaps a free mass bounces bounces off of a spring that is attached to the mirror's other side (conserving its energy but changing its momentum to balance the light's). Perhaps the mirror has a rocket engine (converting chemical energy to kinetic energy of the exhaust which again balances the light's momentum).

    Conversely, if the mirror was isolated then the mirror will be accelerated and the light will lose energy, regardless of whether you consider individual photons or a classical EM wave.
    Last edited: May 8, 2009
  16. May 8, 2009 #15


    Staff: Mentor

    What do you mean by this?

    It doesn't matter if you use Maxwell's equations or photons, the result is the same. The light (Maxwell or photons) has momentum which is approximately reversed by the reflection. Conservation of momentum requires the mirror to accelerate (or an external force to act on it). Any energy missing from the light is accounted for by the mirror's acceleration or the force on the mirror.
  17. May 9, 2009 #16
    Reasonable sounding, but still missing the target.

    Have you ever heard of the ultraviolet catastrophe? It is a mis-prediction based on Maxwell's equation. (the wave equation as opposed the schrodinger diffusion equation.). There is all kinds of extra energy predicted that doesn't exist by the Raleigh Jean Law. Plank is the one who remedied it. So, I tend to think it does matter. Plank did not derive his result from Maxwell's equations -- he assumed them empirically. Now I don't remember all the details... but, that's part of why I am trying to work this out.

    The thing is, if Maxwell's equation says that for a given frequency reflected by a superconducting mirror will reflect *all* the energy -- albeit, at a lower frequency(/ies) and a longer duration -- and that exactly balances out, so there is no reason to require extra energy going around the outside as far as I can tell.

    The problem which does *seem* to exist, is that quantum mechanics requires photons -- when detected -- to be found only in one place and to have fixed energy. They are not *smeared* out like maxwell's equation would seem to require.

    Consider: a wave packet, and perhaps a number of sine waves, having the energy of an emitted quantum at 300Mhz -- and a wavelength roughly constant -- when reflected. The mobile mirror stretches the returning wave out in both time and space. The duration of the packet is longer and its amplitude is smaller. When one adds up the components of the reflected photon, they add up to the energy of the *original* photon.

    No matter what the composition of sine waves of the original packet, the mirror would reflect them linearly stretched in time and space. Longer times, wavelengths, and lower frequencies.
    Now photons of *different* frequencies do not have the same energy -- so an *apparent* contradiction may exist -- here's why:

    If it were always something nice like -- the reflected frequecy is 1/3 -- then I could say that three photons of the lower energy existed where only one was before. But the thing is, any velocity is permitted in relativity and it doesn't necessarily work out to a nice number of photon multiplications.

    Secondly, from quantum mechanics, there is the issue of the wave packet. A photon, when it is detected, is detected in only one place. I can only imagine one thing is happening at the mirror when non-nice fractions are set in place by the velocity chosen:

    If a single photon hit the mirror and all the energy was reflected back -- it must split into two photons -- one with more energy, and one with less. Yet, that would suggest one can detect a low freqency wave which directly coincides with the higher frequency wave. It would be as if extra energy were present in the photon (since they coincide). However, in the other reference frame (that of the mirror) only a single photon would ever be detected by any conceivable experiment. (correction: any photoelectric effect experiment which Einstein uses).

    I don't see anything (and I may be blind) in Maxwell's equation which would correspond to this energy splitting. The original packet did not have multiple photons (as we can trace a single one which must split), and the scaled reflection I would expect to have exactly the same composition with a smaller amplitude as the original -- just stretched out in time and space.

    There is no need to *add* energy -- but to reconcile how Maxwell's equation and photons relate to each other. E=hf would seem to require a modification of some kind to maxwell's to correctly identify when photons split (if they do) -- and how much energy to expect in the localized place a photon exists in.
    Last edited: May 9, 2009
  18. May 9, 2009 #17
    Maxwell's equation already accounts for all the energy when NOT using wave packets.
    If there is any extra unbalanced energy going around the "outside" of the experiment -- it is not because of Maxwell equations predictions.

    When I try to add the refinement of the wave packet/quantized photons -- then I have a hard time reconciling where the energy went.
  19. May 9, 2009 #18
    Just curious about the name -- a Battlestar Galactica fan?

    I want to address what you bring up in a bit more detail -- perhaps it will help stimulate some new thought.

    The change of reference frame issue is something I find important -- but also a bit of a false attractor of criticism. According to texts I have read (not analyzed thoroughly) Maxwell's equations are already corrected for relativity (Einstein/Infield).

    There is an issue when travelling along the cable (as you note) where energy "looks" different.
    The issue though, is whether there is a difference in ability to change/interact.

    What I mean is this -- I am looking for reconciliation where computations of an interaction in one frame of reference predict the same chain of events (even though translated through a transform) from a different frame of reference. That is, I want two different observers to say that all events detected in one frame -- are also detected in the other -- just transformed by a predictable amount.

    Now, to the energy accounting problem:
    I am working the original problem out in the lab/(police man) frame -- but noting what happens in the mirror frame as well -- as a sanity check. Using a coax cable as an analogy (with certain defects about propagation speed and invariance not being exactly the same) -- gives *qualitative* ideas about what would happen in the limit where the cable is removed. eg: we are guaranteed that 100% reflection occurs. Therefore 100% of the energy transmitted is returned. The forum member, Ich, noticed that as well -- and rightly pointed out the reference frame issue.

    When I have a problem with Energy, though, it is *NOT* from different frames of reference -- but rather from the same frame of reference. The energy transmitted by the police man is reflected back to him/her. Now, I gave Ich a rather narrow minded analysis of the classical case, to show that (as perverted as it might seem) -- riding along with a ball in an elastic collision (assuming we survived the impact!) would not cause the observer on the ball to note a violation of conservation of Energy.
    (momentum is a different issue.) I would kind of hope Ich would continue the conversation as that member appears to have a pretty good grasp of physical/mathematical relationships.

    The thing that is interesting is that there is no accounting error in the picture I gave of the classic system -- it is just a fact of life that when using the ball's reference frame -- which is a translation of a "third" observer's frame who watches two balls collide externally -- that momentum reverses sign for any mass that the observer may believe the wall (or ball) to have. There is NO violation of conservation of momentum going on. It is just exceedingly difficult to realize that from the reference frame of the ball. Conservation of energy is easy to show. However, there is no violaton of either the momentum or the Energy law -- even though momentum appears to be violated (until one includes the momentum of the ball they are riding on which I omitted)

    Dalespam wants to wash the energy issue away by claiming it "leaks" around the outside. Perhaps, if there is a mathematical reason that conservation of energy predicted by the increased time of flight (doppler effect) predicted by delay line & relativity -- I would be interested in indulging Dale.

    My simple analysis goes like this: In classical mechanics, F=ma -- and work (energy) = force(net) x distance.

    If two equal and opposite forces act on each other -- the net force is zero -- and no work (energy) is done/transferred. How can one tell no work was done? Because there is no acceleration. F=ma.

    That is a sufficient condition (classically) to claim, that in the experiment (which is defined with velocity=constant, acceleration=0) that no energy is being transferred to the mirror. (For whatever physical reason).

    --continues in a moment ---
    Last edited: May 9, 2009
  20. May 9, 2009 #19
    -- continued --

    I suppose, it is possible that mass already moving at exactly the same velocity as the mirror could be added to the mirror. But that doesn't in any way affect the fact that Maxwell predicts no energy supply to the mirror; in and of iteself (perhaps I have a mistake in the analysis.).

    What I do see, is that maxwell's predicts points of minimum standing (as in standing wave) energy located a short distance away from the mirror and moving relative to it. Eg: 1.5 meters as measured by the meter stick of the police man -- so that would be 1.7320meter to a person measuring it in the mirror's reference frame.

    Now, I don't have my relativity book handy -- but that (I suppose) is the doppler shift that has been corrected for lorentz contraction for a 300MHz source wave. (Anyone second that opinon/correct me?).

    A second problem which crops up -- is that the naive solution to maxwell's equation using Dr. Inan's simplification of a plane wave soluton -- meets the boundary conditions, but...

    Maxwell's equations have energy which scales as the square of the magnitude of the E field. (Read voltage for a transmission line with metallic conductors). If I plug that in directly, all the sudden energy is created -- Eg: there is no increase in power with frequency in maxwell's equations -- only with voltage/magnetic field or E&B-field. But since the energy/meter (average) is the same for the outgoing and reflected wave -- a longer return time=more energy. Fortunately, the crude calculation I gave earlier is NOT the only possible solution -- and my alternate way predicts a balance in energy as expected.

    (This energy imbalance goes the wrong way for DaleSpam's idea -- more energy comes BACK to the police man than left.)

    This is one of the main issues I am trying to solve. How does the E and B fields translate into more energy as frequency increases, and why.

    So far, I have a solution which agrees with doppler shift in both frames of reference -- it creates a standing wave in the reference frame of the mirror. But there is something missing.

    All the solutions offered so far sort of suggest/hint at the same thing -- certain frames of reference are incapable of balancing energy & momentum.
    I was hoping that in each frame, although the values of energy are different, within a single frame -- energy could be balanced.
    Last edited: May 9, 2009
  21. May 9, 2009 #20
    Ok; I'll put up a bit of info I found in my advanced EE text.
    Apparently, in 1837, Maxwell predicted that EM waves carried momentum.
    From Electromagnetic Waves, Umran S. Inan & Azis S. Inan.

    pp. 76: Radiation pressure by definition is the force per unit unit area exerted by a wave upon a material it is incident on.
    ΔP = Sav * ΔA * Δt / c

    Where Sav is the power from a Poynting vector calculation which gives us watts/meter**2
    The result is that according to maxwell, momentum = Power * time / c;
    or to be more obvious: p = E/c.

    So, the balance of momentum attempted to be transferred is 2*p (to the mirror) since the photon is reflected and not adsorbed. (Inan verfies this on pp.77).
    Last edited: May 9, 2009
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