The doppler radar trap paradox on the path to gravity.

[andrewr]

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?

 

[Ich]

welcome to PF,

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?

Adjust the frequency.

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

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.)

 

[Dale]

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.

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.

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

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.

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?

None of these are needed. There is no paradox here to resolve, just an accounting error.

 

[Ich]

The photon's energy decreases in all inertial frames and the mirror's KE increases in all inertial frames.

No. KE is frame dependent, and does not transform linearly.

 

[Dale]

No. KE is frame dependent, and does not transform linearly.

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.

 

[andrewr]

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.)

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.

 

[andrewr]

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.)

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.
Hmm...BTW. Thanks for the welcome -- I'm glad to find people actually interested in thinking a bit.

 

[Dale]

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" , which I find to be one of the most useful concepts of SR.

 

[andrewr]

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.

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.
QED.

 

[Dale]

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.

 

[andrewr]

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.

Dale,
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.

 

[Phrak]

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.

 

[andrewr]

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. He sees beats on the cable. The transmitted and reflected signal are no longer equal in frequency. The reflected energy is not equal to the transmitted energy.

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 separated 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 traveling 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.

 

[cesiumfrog]

The issue I am having probably centers around the idea of instantaneous action at a distance.

Please be direct.

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? [...waffle...] 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.

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

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 realized), 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.

 

[Dale]

Energy is conserved in Maxwell's equation -- if energy is dissipated any OTHER way maxwells equation is WRONG.

What do you mean by this?

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

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.

 

[andrewr]

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.

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.

 

[andrewr]

What do you mean by this?

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.

 

[andrewr]

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.

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 traveling 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 ---

 

[andrewr]

-- 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.

 

[andrewr]

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).

 

[Dale]

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.

I still don't know what you mean by the "outside" of the experiment or "leaking" as you said in other posts. In any case, there is no basic discrepancy between the photon or Maxwell description here, nor is there any even apparent paradox here, just stubborn accounting errors on your part.

Let's consider the following scenario and carefully account for the energy and momentum:
Without loss of generality we will use units where the speed of light is 1 and the vacuum permitivity is also 1. Consider a brief pulse of light traveling through free space initially in the -x direction towards a perfect mirror located at the origin. The mirror is equipped with a stabilizer which will provide any force necessary to prevent it from accelerating. The pulse is timed such that the center of the pulse reaches the mirror at t=0. For purposes of the Maxwell description this pulse of light will be a rectangular pulse of unit amplitude and unit duration with the E field of the pulse pointing in the -y direction. For purposes of the photon description the pulse will be a single photon with unit energy. We will examine the situation in the reference frame described above and also in the primed reference frame where the mirror is moving in the positive x' direction at 0.6 c.

Photon approach:
The initial four-momentum is
p0 = (1,-1,0,0)
The final four-momentum is
p1 = (1,1,0,0)
The change in four-momentum is
dp = (0,2,0,0)
By conservation of four-momentum the stabilizer must provide an impulse of
ps = (0,2,0,0)
which does no work. This is consistent with the fact that the KE of the mirror is unchanged and the energy of both the incident and reflected photons are equal.

Maxwell approach:
The E-field of the incoming pulse is (0,-1,0) as defined above and therefore the B-field is (0,0,1) in our units. The energy density is (E.E)/2 + (B.B)/2 = 1 and since the pulse is of unit duration it is also of unit length. Therefore, integrating over all space we get that the pulse has total energy of 1. The momentum density is ExB, and so integrating over all space gives a total initial momentum of (-1,0,0).
The E-field of the outgoing pulse is (0,1,0) and the B-field is (0,0,1). The total energy is again 1, but the total final momentum is (1,0,0).
The change in energy is
de = 0
The change in momentum is
dp = (2,0,0)
So by conservation of momentum the impulse provided by the stabilizer must also be
ps = (2,0,0)
Since the impulse acts over a unit time the force required is
f = dp/dt = (2,0,0)
The mirror does not move so the work done by the stabilizer is
w = f.d = 0. This is again consistent with the fact that the KE of the mirror is unchanged and that the energy of the reflected and incident pulse are equal. This also agrees with the photon description above.

We will now examine the situation in the primed frame where the mirror is moving in the positive x direction at a speed of 0.6 c which corresponds to a time dilation factor of γ = 1.25. The Doppler shift at that speed is a factor of 2.

Photon approach:
The initial four-momentum is redshifted by a factor of 2
p0' = (0.5,-0.5,0,0)
The final four-momentum is blueshifted by a factor of 2
p1' = (2,2,0,0)
The change in four-momentum is therefore
dp' = (1.5,2.5,0,0)
By conservation of four-momentum the stabilizer must provide an impulse of
ps' = (1.5,2.5,0,0)
which does 1.5 units of work. This is consistent with the fact that the KE of the mirror is unchanged and the energy of the reflected photon is higher, corresponding to a blueshift. All of the energy supplied by the stabilizer goes into the reflected photon.

Maxwell approach:
Because of the Doppler effect the E-field of the incoming pulse is (0,-0.5,0) and the B-field is (0,0,0.5). The energy density is (E.E)/2 + (B.B)/2 = 0.25 and due to the Doppler shift the pulse is of duration 2 and length 2. Therefore, integrating over all space we get that the incoming pulse has total energy of 0.5. The momentum density is ExB, and so integrating over all space gives a total initial momentum of (-0.5,0,0).
The E-field of the outgoing pulse is (0,2,0) and the B-field is (0,0,2). The energy density is 4 and due to the Doppler shift the pulse is of length 0.5. The total energy is therefore 2, and the total final momentum is (2,0,0).
The change in the energy of the light pulse is
de' = 1.5
The change in the momentum of the light pulse is
dp' = (2.5,0,0)
So by conservation of momentum the impulse provided by the stabilizer must also be
ps' = (2.5,0,0)
Since the impulse acts over a time of 1.25 the force required is
f = dp'/dt' = (2,0,0)
The mirror moves over a distance of 0.75 so the work done by the stabilizer is
w = f.d = 1.5. This is again consistent with the fact that the KE of the mirror is unchanged and that the energy of the reflected pulse is greater than the energy of the incident pulse by 1.5. This also agrees with the photon description above.

So you see, there is no mystery about the energy and the momentum in either frame using either approach. Everything agrees completely and both energy and momentum are properly conserved in each frame using each method.

 

[andrewr]

I still don't know what you mean by the "outside" of the experiment or "leaking" as you said in other posts. In any case, there is no basic discrepancy between the photon or Maxwell description here, nor is there any even apparent paradox here, just stubborn accounting errors on your part.

Dale, the very fact that you use a square wave -- whose frequency is unknown -- means you haven't a clue about the setup of the problem I am posing. The error I am speaking about comes from the fact that an item with a strict frequency is a sinusoidal source (CW). Such a source has sinusoidal E and H components. Yes, they are in phase, as you show in your square wave pulse -- but they are sine-waves. When we speak of E=hf, or hν, we are talking about a nearly strict frequency.

A forward and reverse traveling wave are computed such that the boundary condition of perfect reflection occurs at the mirror. ( E field = 0; B field may be nonzero ). There may be other boundary conditions associated with the derivatives found in Maxwell's equations.

We need to know the waveform before reflection and after reflection *in the same frame* in order to check energy conservation. (If I don't get an exact balance, that's O.K., but surely more than 50% of the Energy better show up!)

As to the big deal you are making about the momentum transfer...
A Nichols radiometer shows quite conclusively that radiation pressure is quite small.

If I shine a 10W (10 Joule/second) laser on a silvered mirror attached to a small RC car motor -- there will not be enough force to turn the motor let alone generate even 1/2 watt in a vacuum.
I have never seen an equation which says radiation pressure INCREASED as something ran away from light -- in fact, I rather think it decreases. So the error you seem worried about ought to be LESS than that of a mirror which is NOT moving.

(Air is a different matter...) There is no need to worry about energy/momentum being transferred by light to the mirror as the "loss" I am speaking of.

A policeman typically uses a radar gun with less than 1KW of power... *much less*
So, let's say 1400 watts/m**2 to be conservative. (Enough warm up lots of things...)
In that case, the radiation pressure would be: 4.67 x 10**-6 N/m**2. A *VERY* tiny pressure indeed.
Of the Energy arriving, only 1 part in c (3x10**8), is turned into a Newton/meter**2.
A mirror weighing, say, 1kg total, and being 1Meter**2, (from rest) would accelerate at 9.1 micro meters/second **2.
The energy adsorbed by it's acceleration would be less than (classic works fine here...) E(joules) = 1/2 * 1Kg * ((1 x 10**-5) * t)**2
In other words, for a 1 second experiment (which is plenty long to get the results I am after),
the energy loss from a 1000+ watt * 1 second = 1,000 Joule energy transfer -- would be less than a nano Joule.

Just curious,
Now, what will be the waveshape of the returning rectangular pulse you calculated? How long will the pulse be (duration) -- and how large an E and B field? What boundary conditions did you use with Maxwell's equations to arrive at that?

 

[Ich]

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.

Read http://www.fourmilab.ch/etexts/einstein/specrel/www/" :

It is remarkable that the energy and the frequency of a light complex vary with the state of motion of the observer in accordance with the same law.

That should reconcile the alleged photon/Maxwell discrepancy.

Another point: If you transfer a small momentum to an object at rest, you won't significantly change its energy. That's different for a moving object. If you try to follow that reasoning, use inertial frames for the calculations.

 

[Dale]

Dale, the very fact that you use a square wave -- whose frequency is unknown -- means you haven't a clue about the setup of the problem I am posing. The error I am speaking about comes from the fact that an item with a strict frequency is a sinusoidal source (CW). Such a source has sinusoidal E and H components. Yes, they are in phase, as you show in your square wave pulse -- but they are sine-waves. When we speak of E=hf, or hν, we are talking about a nearly strict frequency.

A square wave can easily be decomposed into a sum of CW sinusoids with Fourier analysis (sinc function in frequency domain) if you feel it is important, and since Maxwell's equations are linear the principle of superposition applies. Also, by the Heisenberg uncertainty principle a photon does not have a single energy (frequency) nor a definite duration, they have some finite natural linewidth. Thus a pulse with a spread of frequencies, such as what I proposed, is a much better classical representation of a photon than a CW sinusoid. Also, a CW sinusoid, by itself, has infinite energy which a photon does not, which is another reason that a CW is not a good approximation of a photon.

We need to know the waveform before reflection and after reflection *in the same frame* in order to check energy conservation. (If I don't get an exact balance, that's O.K., but surely more than 50% of the Energy better show up!)

The waveforms both before and after reflection are rectangular pulses (f(x) = A if |x|<w/2, 0 otherwise), the incident pulse traveling towards (f(x+ct)) and the reflected pulse traveling away (f(x-ct)) from the mirror. The pulse widths (w) and amplitudes (A) are given above. You can easily check that the waves satisfy Maxwell's equations for free space plus the usual reflecting boundary condition.

andrewr, you are making irrelevant objections. If you wish to remain confused that is certainly your choice, but your question has been conclusively answered several times now. There appears to be nothing more I can do for you.

 

[andrewr]

If you wish to remain confused that is certainly your choice, but your question has been conclusively answered several times now. There appears to be nothing more I can do for you.

I think there was nothing you could do in the first place except to try and exchange one problem for a different one.
I don't think you know how to solve Maxwell's equations. ( or don't care to do the hard parts ).

A rectangular pulse reflected off of a moving target has a different length. Why isn't that in your square pulse response?
(you don't actually want to work out the whole problem?)

If A CW wave of a given frequency has "infinite energy" -- it only has finite power. That power *will* be the same as many photons overlapping and having their wave patterns merged. I solve QM problems for silicon that way -- I see no reason why it can't apply here where dispersion is not a problem.

 

[andrewr]

Read http://www.fourmilab.ch/etexts/einstein/specrel/www/" :

That should reconcile the alleged photon/Maxwell discrepancy.

Another point: If you transfer a small momentum to an object at rest, you won't significantly change its energy. That's different for a moving object. If you try to follow that reasoning, use inertial frames for the calculations.

No, unfortuneately it doesn't. Einstein isn't comparing a sinusoidal wave against E=hf when he does the problem. I have no way of comparing amplitudes and duration with that method.
Thanks, though, it is interesting.

Let me shift this argument a bit -- by first going over the relativity and frame arguments.
I will note that it isn't proper to make the second frame "free floating" as it wasn't in my original problem -- but I think there may be enough information here to at least track down some of the issues I am facing.

--more--

 

[andrewr]

Read http://www.fourmilab.ch/etexts/einstein/specrel/www/" :
Another point: If you transfer a small momentum to an object at rest, you won't significantly change its energy. That's different for a moving object. If you try to follow that reasoning, use inertial frames for the calculations.

I am glad that the relativistic Doppler shift was included in that link -- it verifies the wavelength I computed for the reflection using Maxwell's equations.

Since the lab frame uses the relativistic formula twice over, that also verifies that a 300MHz
transmitted wave will become a 100MHz reflected wave as predicted.

Perhaps, then, if you think I am focusing on the problem wrong -- let me lay out some points, and you tell me which is wrong and shown by any experiment. I'll stick to the relativity first -- and we can build on that step by step. (It may be enough by itself.)

E = hν

When transmitted, and transmission is at 300MHz, a *single* photon at that frequency has:
Eph = ~6.63 x 10**-34 Joules/Hz * 300MHz = ~198.9 x 10**-27 Joules

Since Maxwell predicts a momentum transfer of p=2E/c for a *totally* reflected photon;
The max momentum we are talking about must be Pph = 2 * 198.9 x 10**-27 Joules / 3.0 x 10**8 m/s
Or simply: Pph ~= 1.326 x 10**-33 Newton*seconds or kg*m/s for each photon[/color]

Since the lab frame and mirror are the only two reference frames in the experiment I don't think it possible that they would measure/calculate the velocity between them that is different from each other. That is, can the lab frame measure a velocity which is different from the mirror frame? Or will they *necessarily* agree on how fast they move apart?
(I think it is the latter, but I am asking the question -- is there an ether -- I am holding no.)

Given that they can *not* measure different speeds, then it would follow that any event that happens in one frame to accelerate it would necessarily be registered in the other frame as a velocity change. That velocity will be agreed to by BOTH frames.

Ok, then let's examine a police man with a reflector at rest to his frame first. The rest frame of the police man is so massive ( a million Kg or infinite), that no velocity change can be detected for the number of photons involved. Yet we ignore gravity... (You can rework these assumptions if they make a difference -- but gravity is not part of SRT's domain.)

A 1 photon packet is released and reflected. The mirror accelerates, and both the police man and the reflector register the same velocity change. The velocity is non-relatavistic, and we may use correspondence principle.

The momentum carried by a single photon of 300Mhz light is:

p = 2 * 198.9 x 10**-27 Joules / 3.0 x 10**8 m/s = 1.326 * 10**-33m/s.
In classical physics, momentum / mass = velocity; so: 1Kg --> 1.326 * 10**-33 m/s

But, in general, E = 1/2*m*v**2 would give: 0.5 * 1Kg * ( (Eph/c) / 1Kg ) **2
Or, 0.5 * Eph**2 / m0

From which it is clear that the larger the mass of the mirror, the less energy is transferred by a photon reflection off a mirror. The result is a function of the photon Energy**2.

Thus, I would expect LESS energy to be transferred as the velocity of the mirror increased as well.
Eg: In the frame of reference that the photon is emitted in, the mirror which is moving must have more mass.
If this is wrong -- let it go for now, and we will ascertain the correct/falsity a little bit later.
I am after the rest frame balance first.

So:
Whatever velocity change is registered in the moving mirror frame must *also* register in the stationary police frame -- and it must be identical.

For conservation of energy to hold, the energy NOT returned in a photon must exactly be offset by the amount which the other body adsorbs. Otherwise we have a creation or deletion of energy.

So, since the mirror gains KE from rest, where KE = m0*c**2*( γ - 1 )

And ... The reflected photon must have Eref = Eph - KE = initial energy - transferred energy

But we only have momentum of the photon to work with from Maxwell, so translating momentum into KE is done this way:
(TE=total energy of object)

KE = TE - m0*c**2
KE = sqrt( p**2 * c**2 + m0**2 * c**4 ) - m0*c**2

Since the momentum gained is: 2*Eph/c according to Maxwell:

KE = sqrt( 4 * Eph**2 / c**2 * c**2 + m0**2 * c**4 ) - m0*c**2
KE = m0*c**2 * ( sqrt( 4 * Eph**2 / (m0**2*c**4) + 1 ) - 1 )

Which by inspection of the original KE = m0*c**2*( γ - 1 ) , means that:
γ = sqrt( 4*Eph**2 / (m0**2 * c**4) + 1 )

calling: α = m0*c**2

1/sqrt( 1 - (v/c)**2 ) = sqrt( 4*Eph**2 / α**2 + 1 )
1/( 1 - (v/c)**2 ) = 4*Eph**2 / α**2 + 1
1 - (v/c)**2 = 1 / ( 4*Eph**2 / α**2 + 1 )
(v/c)**2 = 1 - 1 / ( 4*Eph**2 / α**2 + 1 )
(v/c)**2 = ( 4*Eph**2 / α**2 + 1 - 1 ) / ( 4*Eph**2 / α**2 + 1 )
(v/c)**2 = ( 4*Eph**2 / α**2 ) / ( 4*Eph**2 / α**2 + 1 )
v**2/c**2 = ( 4*Eph**2 ) / ( 4*Eph**2 + α**2 )
v**2 = c**2 * ( 4*Eph**2 ) / ( 4*Eph**2 + α**2 )
v**2 = c**2 / ( 1 + α**2/(4*Eph**2) )

v = c / sqrt( 1 + 0.25*(α/Eph)**2 ) )
v = 1/sqrt( 1/c**2 + 0.25*(m0*c**2/Eph)**2/c**2 )
v = 1/sqrt( 1/c**2 + (m0*c/(Eph*2) )**2 )

This agrees qualitatively with the classical formula. As the mass becomes larger, the velocity drops & energy transfer drops.
v ~= 1.326 * 10**-33

KE( mirror ) ~= m0*c**2*( γ-1.0) = 879.138 * 10**-69 Joules.
Photon return energy ~= 198.9 x 10**-27 Joules - 879.138 * 10**-69 Joules
Δf=( Eph - KE )/h - 300MHz = -1.326 x 10**-33 Hz.
Δf=((c - v)/(c + v) - 1)*f0= -2.652x10**-33 Hz.

I carried the math out to 1000 decimal places and rounded off only in the answers I wrote down. I kept around 20 significant digits.
bc under Linux is awesome!

Do you see any mistakes in this analysis so far?
If so, would you kindly point them out?

From the relativity values, I have computed:

1) The velocity of the Mirror and its KE.
2) The energy of the recoil photon as Eph - KE.
3) The frequency of the recoil photon from two ways
a) from relativistic Doppler shift formula based on the velocity and source frequency,
b) from the recoil Energy (conservation of energy argument) and E=hf.

The frequency does not match. Doppler shift and relativity seem to be a bit odd.

 

[Dale]

A rectangular pulse reflected off of a moving target has a different length. Why isn't that in your square pulse response?

It is if you had bothered to actually read:

We will now examine the situation in the primed frame where the mirror is moving in the positive x direction at a speed of 0.6 c which corresponds to a time dilation factor of γ = 1.25. The Doppler shift at that speed is a factor of 2.

...

Maxwell approach:
Because of the Doppler effect the E-field of the incoming pulse is (0,-0.5,0) and the B-field is (0,0,0.5). The energy density is (E.E)/2 + (B.B)/2 = 0.25 and due to the Doppler shift the pulse is of duration 2 and length 2. Therefore, integrating over all space we get that the incoming pulse has total energy of 0.5. The momentum density is ExB, and so integrating over all space gives a total initial momentum of (-0.5,0,0).
The E-field of the outgoing pulse is (0,2,0) and the B-field is (0,0,2). The energy density is 4 and due to the Doppler shift the pulse is of length 0.5. The total energy is therefore 2, and the total final momentum is (2,0,0).

If A CW wave of a given frequency has "infinite energy" -- it only has finite power. That power *will* be the same as many photons overlapping and having their wave patterns merged. I solve QM problems for silicon that way -- I see no reason why it can't apply here where dispersion is not a problem.

Now you are the one changing the problem. In any case the correct answer is the same, but you have to state everything in terms of power instead of energy.

A radiation of a given power exerts a given pressure on a mirror which is equal to the change in momentum of the light over a unit time. To keep your mirror from accelerating this requires a constant force. In the rest frame of the mirror this force does no work, consistent with the fact that the frequency of the incoming and outgoing waves are the same and their power is the same. In a frame where the mirror is moving towards the light that force does work, this is consistent with the fact that the frequency of the outgoing waves are blueshifted and their power is higher. No missing power.

 

[Ich]

No, unfortuneately it doesn't. Einstein isn't comparing a sinusoidal wave against E=hf when he does the problem. I have no way of comparing amplitudes and duration with that method.

Sorry, I don't understand. Einstein shows immediately before §8 how classical amplitudes transform, and in §8 derives E=const.*f for a classical wave packet. He proposed the photon idea three months before, so it's clear that he identifies it with a fundamental "light complex". His comment I quoted before, "It is remarkable that the energy and the frequency of a light complex vary with the state of motion of the observer in accordance with the same law", shows his satisfaction that the classical approach is fully consistent with the photon view.

Now I'll do the energy calculations in both frames. I regard the limit where the photon energy is very small compared to the mass of the wall.

In the wall's frame:
E_{in}=E_{out}=hf \sqrt{\frac{1-v}{1+v}}
\Delta E_{photon} = \Delta E_{wall} = 0
The change of the wall's energy goes to zero compared with the photon energy.
In the emitter/receiver frame:
E_{in}=hf
E_{out}=hf \frac{1-v}{1+v}
\Delta E_{photon} = -hf(1-\frac{1-v}{1+v})=-hf \frac{2v}{1+v}
The transferred momentum is accordingly
\Delta p = (E_{in} +E_{out})/c=hf \frac{2}{c(1+v)}
The change in the wall's Energy is
\Delta E_{wall} = \frac{\partial E_{wall}}{\partial p} \Delta p = E_{wall}/p \Delta p = v \Delta p = hf \frac{2v}{1+v}
So the wall gains exactly the energy that the photon loses.

 

[andrewr]

Yeah, I didn't read that carefully. I hit apparent mistakes early and quit.
Let me understand what you mean.
(ellipsis mine).

Maxwell approach:
The E-field of the incoming pulse is (0,-1,0) as defined above and therefore the B-field is (0,0,1) in our units. The energy density is (E.E)/2 + (B.B)/2 = 1 and since the pulse is of unit duration it is also of unit length. Therefore, integrating over all space we get that the pulse has total energy of 1. The momentum density is ExB, and so integrating over all space gives a total initial momentum of (-1,0,0).
The E-field of the outgoing pulse is (0,1,0) and the B-field is (0,0,1). The total energy is again 1, but the total final momentum is (1,0,0).

The change in energy is
de = 0
[/color]
...
The mirror does not move so the work done by the stabilizer is
w = f.d = 0. This is again consistent with the fact that the KE of the mirror is unchanged and that the energy of the reflected and incident pulse are equal. This also agrees with the photon description above.

In this part of maxwell's eqn, you do not show (even under the ...) the reflected wave but claim the energy change is zero. I agree.

We will now examine the situation in the primed frame where the mirror is moving in the positive x direction at a speed of 0.6 c which corresponds to a time dilation factor of γ = 1.25. The Doppler shift at that speed is a factor of 2.

Photon approach:
The initial four-momentum is redshifted by a factor of 2
p0' = (0.5,-0.5,0,0)
The final four-momentum is blueshifted by a factor of 2
p1' = (2,2,0,0)
The change in four-momentum is therefore
dp' = (1.5,2.5,0,0)
By conservation of four-momentum the stabilizer must provide an impulse of
ps' = (1.5,2.5,0,0)
which does 1.5 units of work. This is consistent with the fact that the KE of the mirror is unchanged and the energy of the reflected photon is higher, corresponding to a blueshift. All of the energy supplied by the stabilizer goes into the reflected photon.

Why this is red and then blue shifted I have no idea. The mirror is moving APART, so the doppler shift would red-shift twice. The frame of reference would first be the police, then a lower f at the receding mirror, and then lower yet when returning to the original frame.
The non-relativistic Doppler shift is f0*(c-v)/(c+v) -- and that is the square of the relatavistic formula showing a double red shift. Ich shows that correctly in his post.

I asked in an early post that if you used the minkowski type objects, that you be very clear what they mean -- as in why. Is this blue shift you speak of found only in four vectors?

This next section is the one I overlooked -- as I did not see in the original section on maxwell the expanded waveform.

Maxwell approach:
Because of the Doppler effect the E-field of the incoming pulse is (0,-0.5,0) and the B-field is (0,0,0.5). The energy density is (E.E)/2 + (B.B)/2 = 0.25 and due to the Doppler shift the pulse is of duration 2 and length 2. Therefore, integrating over all space we get that the incoming pulse has total energy of 0.5. The momentum density is ExB, and so integrating over all space gives a total initial momentum of (-0.5,0,0).

O.K.

The E-field of the outgoing pulse is (0,2,0) and the B-field is (0,0,2). The energy density is 4 and due to the Doppler shift the pulse is of length 0.5. The total energy is therefore 2, and the total final momentum is (2,0,0).

I don't follow this.
When you solved Maxwell's equation, what boundary conditions did you use in order to compute the amplitude of the returning wave. Did you do them with respect to the derivatives, or just the amplitude. Eg: in a transmission line with a non-moving termination, voltage and current continuity across the short is what results in reflection. Derivatives are ignored as the short isn't moving.
How did you get that answer?

So you see, there is no mystery about the energy and the momentum in either frame using either approach. Everything agrees completely and both energy and momentum are properly conserved in each frame using each method.

Again, there is no sine wave. Please do it with a cosine wave or a wave packet (sinc pulse) if you prefer. I'd like to see your boundary conditions when you solve the wave equation and how the amplitude of the individual sine waves translate.

It is if you had bothered to actually read:
Now you are the one changing the problem. In any case the correct answer is the same, but you have to state everything in terms of power instead of energy.

No, I did not change the problem. You did. Mine was based on a sine wave which can be checked for boundary conditions using Maxwell's equations. For your solution to work, you would need to show how E field cancels at the location of the mirror. A square wave with different amplitudes on a boundary does not cancel, so there is an E-field at the mirror.
A sine wave has an amplitude which goes to zero at certain points, and when the waveform is modulated can produce a solution which has different forward and backward waves -- esp. if the argument is nonlinear in time -- something that the standard wave solutions do not take into consideration.

What were your boundary conditions?
Those might solve my problem as I have questions about the proper ones -- !

Again. In my original problem:
1. the mirror was traveling away -- so there is double redshift.
2. I used a sinewave source.

Neither of these did you do in your problem.
Also; if you are now trying to say that the problem can't be done with individual photons -- assign a wave packet to them -- then you appear to invoke the wave particle duality paradox. I am going to laugh at you if , in telling me that what I am doing is not a paradox at the start of the thread by invoking another paradox. Of COURSE quantum particles can violate conservation of energy - ... - for a short time.

 

[andrewr]

Sorry, I don't understand. Einstein shows immediately before §8 how classical amplitudes transform, and in §8 derives E=const.*f for a classical wave packet. He proposed the photon idea three months before, so it's clear that he identifies it with a fundamental "light complex". His comment I quoted before, "It is remarkable that the energy and the frequency of a light complex vary with the state of motion of the observer in accordance with the same law", shows his satisfaction that the classical approach is fully consistent with the photon view.

Hmmm... perhaps if you had told me to read the section before §8 I would have read it.
I'll go look.

Now I'll do the energy calculations in both frames. I regard the limit where the photon energy is very small compared to the mass of the wall.

In your workup -- you end up simply assigning the energy change to the proper item.
On cursory glance, I don't see two different ways of calculating the Energy so that E=hf can be compared -- but perhaps I missed something.

When I attempted the purely relativistic approach, above, for a free mirror -- as I understood you wanted me to do that -- I do not arrive at the same energy change when using the photon calculation from energy only consideration vs. Doppler shift.

Do you see a mistake in my attempt someplace? (I do not, and suspect I might have one -- though it baffles me.).
I'll read the section you mention on Einstein previous to the section I was told to read before -- but if he is using a "classic" wave packet, the fact that all sine waves in a wave packet will stretch linearly still makes me think I will be forced to do some kind of fixup that splits photons with Maxwell's equations and E=hν in order for it to work.

Thanks, Ich, I really appreciate the help.
I'll answer your post in detail a bit later when I have a chance to adsorb it more.

 

[andrewr]

Oh, and Dale, I most certainly HAVE been talking about power all along.

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.
QED.

And to show that I am averaging photons over time for power, note:

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[/color]power[/color] reflected -- eg:
power transmited*transmission time = power reflected * (transmission time + delay until coax is empty of radiation)..

See the above analysis.

Power is not conserved, but total energy is.

A delay line, which space is, has capacitance and inductance per distance. These capacitance and inductance elements are NOT carried along with the energy. They are not Ether. So, The energy one pours into any differential element actually exists at that point (whatever that point is philosophically). When a perfect reflector is modeled in Maxwell's equations, and it moves, what is happening is that more differential elements of inductance and capacitance are being exposed. There is *no* "motion" of these elements, as there is no ether. On average, when doing the idealized calculation for a transmission line -- the energy must spread out *in some way* to fill the extra space -- eg: energy storage capacitance and inductance per unit length. From experiments done in the lab with a fast 'scope and a movable short -- done multiple times to cancel noise -- I am pretty certain that a moving short does not adsorb any more energy than a stationary one. -- even if the short is made to change position by oscillation in the low MHz region -- which is the fastest I can make it move.

I am not doing all of this, here, to grind some pet theory -- I am trying to reconcile experiment with maxwell's equation and relativity *both*.
I hope by doing this, I will be able to begin to understand the general theory of relativity and a particularly tricky problem that has been presented to me.

 

[andrewr]

Sorry, I don't understand. Einstein shows immediately before §8 how classical amplitudes transform,

Yes. I see in section §7 what you are speaking about, now that I have read that section too.
For all practical purposes in this discussion, we are speaking about the simplified φ=0 in the 1905 document.

As a side note; Einstein does not appear to do a derivation from Maxwell. He does *not* convert the amplitude of the field -- rather he does the conversion of A**2. When one works with Maxwell's equations, one does not have A**2 as part of the wave equation.

I don't know how to convert to A**2 in Maxwell's equation, as the square of the individual components of the wave equation can-not be squared independently without changing the character of the equation -- and the meaning of the boundary conditions.

This doesn't really help me understand what is going on as an Engineering student familiar with Maxwell's equations -- which is most of the point of doing the exercise, and why am am somewhat lost here.

and in §8 derives E=const.*f for a classical wave packet.

Yes, but he transformed the intensity**2 in the first place, which is identified with energy ?
If the transformation truly predicted the change of energy with frequency, then he ought to have been able to derive plank's constant -- for E=m*c**2, links the other terms together, giving a scale.
I am not aware of Plank's constant being derivable from the speed of light; so I tend to wonder if the coincidence has to do with the way he transformed amplitude squared in the first place. Or else, it may be that we shall have an exact value for the constant soon in terms of c...

He proposed the photon idea three months before, so it's clear that he identifies it with a fundamental "light complex". His comment I quoted before, "It is remarkable that the energy and the frequency of a light complex vary with the state of motion of the observer in accordance with the same law", shows his satisfaction that the classical approach is fully consistent with the photon view.

That may well be -- I just wish he had done it from the classical equation!
It also makes me wonder why my calculation came out with two different answers using the principle equations that are taught in college physics courses including relativity, which I took. They did not do the four vectors, but only the way momentum and energy and mass are related -- and I understand that well enough to be able to at least make calculations. (even though they don't agree with E=hf based on the values I have).

I'll get to your analysis here in just a bit ... I haven't quite got everything straight in my mind.

 

[Dale]

In this part of maxwell's eqn, you do not show (even under the ...) the reflected wave but claim the energy change is zero. I agree.

I am sorry that you are finding it confusing. The information is there, but because I was having to work the same problem 4 different ways I necessarily did a very sketchy outline and expected the readers to fill in the details. I gave the formula for energy density (E.E)/2 + (B.B)/2. When you integrate energy density over space, as I mentioned in the next sentence, then you get energy of the incident pulse. Then I did the same for the reflected pulse. Since they got the same energy the difference in energy is obviously 0.

Why this is red and then blue shifted I have no idea. The mirror is moving APART, so the doppler shift would red-shift twice. The frame of reference would first be the police, then a lower f at the receding mirror, and then lower yet when returning to the original frame.
The non-relativistic Doppler shift is f0*(c-v)/(c+v) -- and that is the square of the relatavistic formula showing a double red shift. Ich shows that correctly in his post.

This is not a big deal. I worked the problem with the mirror going into the light so that the stabilizer would do work. It is a simple matter to reverse the sign of the velocity, get the mirror moving in the opposite direction, and have work be done on the stabilizer. However, in that case it would be blueshifted and then redshifted by my usage.

Remember, I am starting with the reference frame where the mirror is at rest, this is my base reference frame. In that frame there is no Doppler shift from the reflection, the radiation is the same base frequency on the incident and the reflected wave. (note that this is not necessarily the same frequency as that emitted by the source of the wave since I never specified the source's velocity). If the mirror moves away from the incident light pulse (-x' direction) in the primed frame frame then in the mirror's frame the primed frame is moving towards the incident pulse (+x direction). Thus the initial blueshift if I reversed the sign of the velocity. If you are defining the "base" frequency as the frequency of the source rather than the frequency in the mirror's frame that is fine too and depending on the source's velocity in the primed frame you could get either a redshift or a blueshift. It is not an important detail as long as everyone is clear about what they are doing, I hope this clarifies what I did.

I asked in an early post that if you used the minkowski type objects, that you be very clear what they mean -- as in why.

I posted a link to the Wikipedia page on four-momentum earlier. It is a very useful concept. It is a vector formed by (E/c, px, py, pz) where E is the total energy (rest+kinetic) of the particle, px is the particle's momentum in the x direction and so on. The conservation of four-momentum embodies in one nice neat package the classical principles of conservation of energy and momentum. The four-momentum transforms like any other four-vector, which makes it particularly convenient to work with for problems like this.

I don't follow this.
When you solved Maxwell's equation, what boundary conditions did you use in order to compute the amplitude of the returning wave. Did you do them with respect to the derivatives, or just the amplitude. Eg: in a transmission line with a non-moving termination, voltage and current continuity across the short is what results in reflection. Derivatives are ignored as the short isn't moving.
How did you get that answer?

If n is a unit vector normal to the surface of the conductor then the usual boundary conditions for a stationary conductor are nxE=0 and n.B=0 on the surface of the conductor. You can verify that the sum of the incident and reflected pulses satisfy this boundary condition at x=0 in the unprimed frame.

Again, there is no sine wave. Please do it with a cosine wave or a wave packet (sinc pulse) if you prefer. I'd like to see your boundary conditions when you solve the wave equation and how the amplitude of the individual sine waves translate.

I am not interested in doing that, it is a really unimportant detail that doesn't change any of the points that I have made (due to the linearity of Maxwell's equations) and serves only to unnecessarily complicate the math. The amplitudes transform by the Doppler factor, if you have a redshift of 2 then your amplitude is half and your duration and length are doubled. If you want more details about how the electric and magnetic fields transform see the last set of equations in section 6 of Einstein's paper that Ich linked to earlier.

For your solution to work, you would need to show how E field cancels at the location of the mirror. A square wave with different amplitudes on a boundary does not cancel, so there is an E-field at the mirror.

In the mirror's frame the electric field has the same amplitude and opposite direction and therefore cancels at the mirror. This satisfies the boundary conditions.

if you are now trying to say that the problem can't be done with individual photons

I don't know how you would possibly come to that conclusion when I worked the problem in both frames using individual photons. I even put them in sections that I clearly labeled "Photon approach". All I said is that a continuous sine wave is a very poor classical approximation to a single photon. A rectangular pulse is not only mathematically more convenient, but is also a much better analogy.

 

[andrewr]

I am sorry that you are finding it confusing. The information is there, but because I was having to work the same problem 4 different ways I necessarily did a very sketchy outline and expected the readers to fill in the details. I gave the formula for energy density (E.E)/2 + (B.B)/2. When you integrate energy density over space, as I mentioned in the next sentence, then you get energy of the incident pulse. Then I did the same for the reflected pulse. Since they got the same energy the difference in energy is obviously 0.

Dale, as far as I know maxwell's equation (for light's pressure or momentum) calculates the momentum of a wave incident on a reflector based on a sinusoidal wave -- as my professor shows in his book which I already mentioned. If you do not wish to rework the problem without changing frames -- and NOT introducing yet a potential "third" frame or a condition which upsets the problem I originally posed by causing extra transference of energy which is unaccounted for explicitly *a light source not in either frame of reference?!* -- you will only cause confusion.

If you do not feel like doing Maxwell's equations in a single frame of reference so that all E and H field boundaries can be checked independent of changing frames of reference (obscurely), I cordially thank you -- but ask that you do not post any more in this thread. Maxwell's equations do not require correction by frame changes to work -- and this is a source of confusion that I find unnecessary.

The conservation of four-momentum embodies in one nice neat package the classical principles of conservation of energy and momentum. The four-momentum transforms like any other four-vector, which makes it particularly convenient to work with for problems like this.

I already worked out the problem for a free mirror for ICH -- which should have save everyone a lot of work and study as even that example shows an error.
Relativity did not help me reconcile the mistake. I fail to see that four vectors is going to do anything except increase the number of potential misinterpretations I might have -- eg: as the information is unfamiliar -- which is why I asked people not use them if possible in the first place.

If n is a unit vector normal to the surface of the conductor then the usual boundary conditions for a stationary conductor are nxE=0 and n.B=0 on the surface of the conductor. You can verify that the sum of the incident and reflected pulses satisfy this boundary condition at x=0 in the unprimed frame.

Not according to what I learned -- in full disclosure. Even if one takes the infintesimal surface, the H field does not have to be zero (or B, if it has the same meaning in your eyes). In any event, in the earlier work you gave -- I do not see that it is possible for E and H to both be zero in the conductor. I suspect, that if your above statement is true -- it is only a partial boundary condition; and the change to B is to obscure things yet more.

The choice of the normal is arbitrary and hides important information at that point. eg: you are choosing the normal at right angles to the information required to solve the problem.

The E field of a square wave at 90 degrees to the direction of travel along the proper axis ( not the cross product ) is set to zero on a boundary. Therefore, it is impossible to have TWO E-fields of differing strengths superposed at the reflection point without having a non-zero E-Field. Even more importantly, the continuity of the E field is violated at a point if that happens -- which is not permitted according to my class on Maxwell's equations.

I don't know how you would possibly come to that conclusion when I worked the problem in both frames using individual photons. I even put them in sections that I clearly labeled "Photon approach". All I said is that a continuous sine wave is a very poor classical approximation to a single photon. A rectangular pulse is not only mathematically more convenient, but is also a much better analogy.

Actually, I re-read your earlier posts -- and it looks to me that you were being rude, and claiming I am changing the problem by using the power average -- even though that has been in the thread even since the first page. Therefore, I have asked a moderator to look over what is going on -- to give you the benefit of the doubt concerning the credibility of your final post. But I would cordially invite you to end the dialog here. Thank you.

 

[Dale]

Dale, as far as I know maxwell's equation (for light's pressure or momentum) calculates the momentum of a wave incident on a reflector based on a sinusoidal wave -- as my professor shows in his book which I already mentioned.

The momentum of a EM field is certainly not restricted to the special case of a sinusoidal (plane) wave. In fact, even static fields that do not change in time can carry momentum. The momentum density is:
p/V = 1/c² ExB

If you do not feel like doing Maxwell's equations in a single frame of reference so that all E and H field boundaries can be checked independent of changing frames of reference (obscurely), I cordially thank you -- but ask that you do not post any more in this thread. Maxwell's equations do not require correction by frame changes to work -- and this is a source of confusion that I find unnecessary.

The transformations are well known and essential to the topic of this thread (since you are interested more than one frame) so I don't really see the problem here. One of the practical reasons for learning how to do transformations is exactly for this kind of situation, so you can transform it to a simpler coordinate system and solve an equivalent but mathematically easier problem.

If you know the correct boundary conditions for a moving conductor then I would be glad to check and see if my solution satisfies it, but since it satisfies the appropriate boundary condition in one frame then it must do so in all other frames also.

I fail to see that four vectors is going to do anything except increase the number of potential misinterpretations I might have -- eg: as the information is unfamiliar -- which is why I asked people not use them if possible in the first place.

I strongly recommend that you make the effort to learn four-vectors. Minkowski geometry (of which four-vectors are an essential part) is the mathematical framework of SR, it vastly simplifies and clarifies SR.

I suspect, that if your above statement is true -- it is only a partial boundary condition;

I don't know what you mean by a partial boundary condition. The only boundary conditions I know about for the surface of a perfect conductor at rest are nxE=0 and n.B=0.

The choice of the normal is arbitrary and hides important information at that point. eg: you are choosing the normal at right angles to the information required to solve the problem.

The normal is not arbitrary at all, it is completely determined by the geometry of the mirror. In this case the mirror is the plane x=0, so the normal is unambiguously (1,0,0).

it looks to me that you were being rude, and claiming I am changing the problem by using the power average -- even though that has been in the thread even since the first page. Therefore, I have asked a moderator to look over what is going on

oh come on. You said the exact same thing to me earlier plus several other minor rude things. You can go ahead and complain to a moderator but nobody is going to censure me for my conduct here. This has been a reasonably tame thread overall.

 

[andrewr]

oh come on. You said the exact same thing to me earlier plus several other minor rude things. You can go ahead and complain to a moderator but nobody is going to censure me for my conduct here. This has been a reasonably tame thread overall.

I didn't ask them to censer you -- I asked them to check the validity of your last statement.
I also asked you, as a courtesy, to leave the thread if you do not intend on answering me fully and non evasively on the grounds which I have asked. There are several options.
The fact that you can point to equally curt comments on my part in response to your comments -- is all the more reason not to progress farther.

Thank you.

 

[Dale]

I also asked you, as a courtesy, to leave the thread if you do not intend on answering me fully and non evasively on the grounds which I have asked.

I have answered you fully and non evasively from the beginning. There is no paradox here, there is no missing energy, it does not matter if you use the concept of photons or waves, and it does not matter what frame you choose for the analysis. Although I made a good-faith effort to answer the question on your "grounds", the fact is that the conservation of energy and momentum are general features of the laws themselves and therefore hold in every situation that the laws govern regardless of the details. So your "grounds" are neither relevant nor important, they are just a poor excuse for you to ignore an answer you don't like.

 

[andrewr]

I have answered you fully and non evasively from the beginning. There is no paradox here, there is no missing energy, it does not matter if you use the concept of photons or waves, and it does not matter what frame you choose for the analysis. Although I made a good-faith effort to answer the question on your "grounds", the fact is that the conservation of energy and momentum are general features of the laws themselves and therefore hold in every situation that the laws govern regardless of the details. So your "grounds" are neither relevant nor important, they are just a poor excuse for you to ignore an answer you don't like.

My grounds for not attempting four vectors is personal, associated with medication I have to take for anxiety.
I don't know why you are choosing to be antagonistic with me.
If you are so angry: I have shown a relativity calculation of doppler and NON doppler for a photon.
They DO NOT AGREE -- you can prove me wrong by just finding the mistake.
But you prefer to belittle me in an area I admit I have no strength in.
I thought that kind of act was usually attributed to the Pope by physics types.

Hey, and "Doppler" is not "Einsten" -- Gee: Einstein didn't derive Doppler's formula -- he inherited it.
Well; now I never intimated you were evil by saying you wanted to stay in the dark -- but that's how you started baiting and bullying me. What gives, do I have -- treat me like garbage written on me?
I thought this was science, mr. advisor to look up to.

So, How do you use your pretty shortcut to DERIVE dopplers formula?
I come up with different values of Energy do not scale linearly with momentum transferred.
I showed the math.

I didn't see you correct me -- I asked. No, you try and give me a "shortcut" which can never prove the thing upon which it is based. YOU DEMAND FAITH OF ME?
Forget it. Doppler didn't have RELATIVITY to derive his formula. Einstein just uses it in section 8 of his 1905 paper -- without proof.
So, How do we know Doppler's formula works ENERGY wise without doing the boundary conditions?

I don't know -- so I ask the experts.

You have a badge so the board will let you get away with anything?
If the physics board allows bullies like you -- the world can sure use another Pope to balance it out.
God bless!

 

[sylas]

My grounds for not attempting four vectors is personal, associated with medication I have to take for anxiety.

In which case you should probably bear in mind that you are prone to anxiety and perhaps likely to perceive aggression that isn't there.

I don't know why you are choosing to be antagonistic with me.

I'm just an onlooker here, but sometimes an extra word from outside can help. This is a case of seeing stuff that isn't there. Relax. No offense intended by me either. Just advice. You're getting good information, and you appear to be seeing aggression in what is simply direct substantive corrections, given fairly and dispassionately. Settle down.

Cheers -- sylas

 

[jtbell]

My grounds for not attempting four vectors is personal, associated with medication I have to take for anxiety.

 

[Dale]

My grounds for not attempting four vectors is personal, associated with medication I have to take for anxiety.

I am sorry to hear that you have such severe anxiety, but it does explain a lot about this conversation. Anyway, I will take your comments with an extra "grain of salt", and I would encourage you to do the same with me. That should help to defuse any emotional component here.

However, if you are anxious about relativity then I would recommend learning four-vectors even more. I struggled with relativity for about 7 years until I accidentally stumbled on the concept of four-vectors and Minkowski geometry. Suddenly all of these separate concepts that just didn't make sense crystalized into one clear and simple framework of understanding. Imagine someone coming to you trying to do classical physics without regular three-vectors or calculus; to me that is the closest analogy to trying to do relativity without four-vectors or Minkowski geometry.

I have shown a relativity calculation of doppler and NON doppler for a photon.
They DO NOT AGREE -- you can prove me wrong by just finding the mistake.
...
I come up with different values of Energy do not scale linearly with momentum transferred.
I showed the math.

I didn't see you correct me -- I asked.

Which post(s) are you referring to? And what specifically is the disagreement you computed?

EDIT: I guess you mean this?

Δf=( Eph - KE )/h - 300MHz = -1.326 x 10**-33 Hz.
Δf=((c - v)/(c + v) - 1)*f0= -2.652x10**-33 Hz.

It is pretty clearly missing a factor of 2 somewhere. I will look for it.

 

[andrewr]

I'll stick to the relativity first -- and we can build on that step by step. (It may be enough by itself.)

E = hν

When transmitted, and transmission is at 300MHz, a *single* photon at that frequency has:
Eph = ~6.63 x 10**-34 Joules/Hz * 300MHz = ~198.9 x 10**-27 Joules

Since Maxwell predicts a momentum transfer of p=2E/c for a *totally* reflected photon;
The max momentum we are talking about must be Pph = 2 * 198.9 x 10**-27 Joules / 3.0 x 10**8 m/s
Or simply: Pph ~= 1.326 x 10**-33 Newton*seconds or kg*m/s for each photon[/color]

... [ please see actual post for derivation ] ...

v = 1/sqrt( 1/c**2 + (m0*c/(Eph*2) )**2 )

This agrees qualitatively with the classical formula. As the mass becomes larger, the velocity drops & energy transfer drops.[/color]
v ~= 1.326 * 10**-33

KE( mirror ) ~= m0*c**2*( γ-1.0) = 879.138 * 10**-69 Joules.
Photon return energy ~= 198.9 x 10**-27 Joules - 879.138 * 10**-69 Joules
Δf=( Eph - KE )/h - 300MHz = -1.326 x 10**-33 Hz.
Δf=((c - v)/(c + v) - 1)*f0= -2.652x10**-33 Hz.

I carried the math out to 1000 decimal places and rounded off only in the answers I wrote down. I kept around 20 significant digits.
bc under Linux is awesome!

Do you see any mistakes in this analysis so far?
If so, would you kindly point them out?

From the relativity values, I have computed:

1) The velocity of the Mirror and its KE.
2) The energy of the recoil photon as Eph - KE.
3) The frequency of the recoil photon from two ways
a) from relativistic Doppler shift formula based on the velocity and source frequency,
b) from the recoil Energy (conservation of energy argument) and E=hf.

The frequency does not match. Doppler shift and relativity seem to be a bit odd.

The setup was a free floating mirror in the rest frame -- as ICH seemed to recommend -- in order to get used to what reference frames required. The point was just to verify that (numerically) Einstein's equations gave a consistent result. I note that the numeric result differs far beyond anything that round-off error could possibly explain. Unless I made a typo, and I tried to double and triple check -- I can't explain it.

I wanted to start here because I understand the equation, I don't feel disoriented by new math, and it could quickly get to the root of my issue (if I haven't just made a dumb math mistake).
I only have about 4 good hours a day where I don't shake and can concentrate enough to attempt complex tasks. This derivation (though easy) took me over six hours. Learning a totally new framework -- right now -- is far beyond my ability. No offense -- but I am going to have to take things in very small steps. As I said at the thread start -- I am a slow starter.


Also notice -- changing the value of E will change the amount of energy transferred to the stationary mirror.
Therefore, two photons with energy E/2 each will have a different effect than one photon of energy E.
[/color]

On the Maxwell Eqn. side --
In Dr. Aziz Inan's class I was taught to do the boundary condition of a shorted reflector by setting the E field to zero at the boundary. Or as he puts it:

Engineering electromagnetics: Umran S. Inan, Aziz S. Inan.
Addison Wesley, (c)1999
Chapter 8: Reflection and transmission of Waves at Planar Interfaces

p.693 -- Normal Incidence on a Perfect Conductor
The boundary condition on the surface of the conductor requires the total tangential electric field to vanish[/color] ... since the electric field inside medium 2 (perfect conductor) must be zero.

In the work he shows that the tangential electric field is that which is perpendicular to the direction of travel. The H-field, by contrast, typically doubles. That corresponds to current flow inside a conductor -- but no voltage drop across the current flow -- as it is a perfect conductor.

Dr. Inan does not use the normal anywhere in the boundary conditions.
In addition, I note:
The value of the E field along the normal -- eg: what one gets when they do a dot product of the normal to the conductor surface with the E-field of the incoming wave -- will not guarantee the boundary condition Dr. Inan requires is satisfied.

Any equation which has a boundary condition for an arbitrary velocity, must (by correspondence) degenerate into the boundary condition already stated for a non-moving short circuit by Dr. Inan.

When I check a second author:
Who is very thorough in showing how to get Maxwell's equations from gauss's law, etc.
William Hayt, Jr.
Engineering Electromagnetics -- 1981, McGraw Hill
pp. 405 & 6, sec 11.6, Reflection of Uniform Plane Waves
Both authors have the wave propagating in z, not x -- so rotation of axes is required to make my problem solve.

He says:"
Now we must try to satisfy the boundary conditions at z=0 with these assumed fields. Ex is a tangential field;[/color]therefore the E fields in regions 1 and 2 must be equal[/color] at z=0. ..."

From which, the tangent to the direction of propagation is clearly 90 degrees. Or parallel to the E field vector of the traveling wave.
He shows that this is a continuity requirement:

"therefore the E fields in regions 1 and 2 must be equat at z=0."

Which means one can not have two values of E at anyone point and E, when taking a limit from any direction -- must have the same value from a left sided limit or a right sided limit. Eg: if inside the conductior has no E field, then just outside (differentially) there must also be NO E field.

That is standard engineering approach to solving Maxwell's equations at a boundary for a plane wave.

Now, when I am not upset -- I am quite clear headed, but once the adrenaline kicks in -- it takes me a few days to settle down. I can't just "settle down" -- and the medication takes time to work again.
If I am able, I may rejoin you in a few days. Perhaps, if I get lucky -- the math shown by ICH, who is very gentle and nice -- will be something I am able to digest.

Post note addition/edit: The calculation used h=6.63*10**-34 J/Hz; c=3.0 * 10**8 m/s as Einsten indicated h was a constant is verified by his theory, but the value is undetermined by him; so I took him to mean any constant may be permitted for consistency with the mathematics of relativity. However, it is possible that a specific value is required and that would mean h is derivable from relativity. This can be easily checked by repeating the method I used above by solving for the value of h which corrects the error and then retrying with another value of E to see if energy is still conserved.

Also, the units of momentum for the first photon are kg*m/s, not just m/s, and that includes the double momentum for reflection.
The general formula ought to have read: E(mirror) = 0.5 * Eph**2 / (m0*c**2)
The energy transferred is a non-linear function of the energy of the photon.

 

[andrewr]

In which case you should probably bear in mind that you are prone to anxiety and perhaps likely to perceive aggression that isn't there.

Not at all. It is because I have been through abuse that I am sensitive to it in any form.
People often do not recognize what they do; often utilitarianism kicks in where people justify their actions based on the judgment of the ends. eg: you judge that the information given is accurate and sufficient, and therefore seem to believe that it is right to give this information regardless of whether or not the person receiving it is still receptive to it.[/color]

I'm just an onlooker here, but sometimes an extra word from outside can help. This is a case of seeing stuff that isn't there.

That is a judgment that is not verifiable. It is in fact an assumption which is likely half true.

Relax. No offense intended by me either. Just advice.

I believe you.

You're getting good information, and you appear to be seeing aggression in what is simply direct substantive corrections, given fairly and dispassionately. Settle down.

So, now you have judged me in error without proof. In other words, the ends -- correction -- justifies the means --- forcing a particular outlook or way of doing things upon me.

Silas, The mathematics I used for the example to ICH are *EVERY* bit as complete and thorough as four vectors. Four vectors is an extension of the math I gave.

There is no need to force this issue upon me, four vectors are not mandatory to get the correct solution, and no one has the implicit right to force information on someone else which they do not want. In your post you have made the inference that I am ignorant and that Dale is totally right.

Will you apologize if you are proven wrong?

I suggest you work the problem of a square wave pulse that Dale has given and show that the tangential field of the square wave is zero upon a moving conductor[/color] according to his information.

He, somehow, calculates a different amplitude for the forward going square wave and the returning one. The E field, then, is different for the forward and backward wave. Can you show mathematically that fields of differing magnitude can ever cancel to exactly zero? Can you show that any complaint I made against Dale is unfair -- eg: esp: that his boundary conditions are incomplete?

I tell you what. I'll list the snippish comments in the order that they occurred in the thread -- and you can tell me why a correction is justified and is not gratuitous. I will also take each comment and note any double standards in judgment -- eg: who accuses whom of what; and whether or not they themselves are guilty of doing it. Also, and most importantly who admits they are wrong when it becomes obvious -- and who covers it up.[/color]
EDIT: Multiquote will not allow me to take all the pieces together with references, so -- I would have to do it too selectively which is biased;therefore I'll await your objection and answer appropriately.[/color]

I note that a bully is one who uses public opinion, circumstances, or other unstable opportunistic events to inflict damage upon an opponent. I will also note that a bully tends to run and hide, or to change the subject, or obfuscate when called on the matt for their behavior. My wife is an expert -- I don't need any more.

I suggest you go get big calculator (bc) for your PC or for Linux -- and you work the numerical problem I gave earlier. It seems people are slow in making a correction -- and perhaps you can expidite things.

It is free software; and can be easily tested to check its accuracy.

Take me down -- if you are justified. I say -- "I don't know" when I mean "I don't know."
I expect everyone else to give me the same courtesy.

Cheers -- sylas

We'll see.

 

[Dale]

Hi andrewr,

I went through your previous post looking to find where the factor of 2 was. I quickly found this.

The momentum carried by a single photon of 300Mhz light is:

p = 2 * 198.9 x 10**-27 Joules / 3.0 x 10**8 m/s = 1.326 * 10**-33m/s.

The momentum of a single photon is p=E/c. The reason that the momentum transferred is twice that is because the photon bounces back with almost the same momentum but in the opposite direction, a change of ~2p.

However, this factor of 2 was not the cause of the difference that you found between the two methods of calculating Δf. Although your value for p was wrong it appears that you didn't use it anywhere else so the error did not propagate. I went carefully through the remainder of your numbers and they are all correct, Eph, v, KE(mirror), and Δf (energy).

It took me a long time to spot the error because I was looking for a factor of 2. However, the actual problem was just in the formula for the Doppler shift. You had:

Δf=((c - v)/(c + v) - 1)*f0

but the correct formula is:

Δf=(sqrt((c - v)/(c + v)) - 1)*f0

With the correct formula for the Doppler shift there is agreement between the two different methods of calculating Δf. So again, there is no discrepancy between any of the different approaches and energy is always correctly conserved.

 

[Dale]

Dr. Inan does not use the normal anywhere in the boundary conditions.

He uses it explicitly in the title of the section.

p.693 -- Normal Incidence on a Perfect Conductor

The boundary condition on the surface of the conductor requires the total tangential electric field to vanish[/color]

I agree completely. One way of expressing this mathematically is nxE=0, as I said all along.

The H-field, by contrast, typically doubles.

Although I didn't have that as a boundary condition nor a forced constraint you can see for yourself that my solution also has that behavior.

In addition, I note:
The value of the E field along the normal -- eg: what one gets when they do a dot product of the normal to the conductor surface with the E-field of the incoming wave -- will not guarantee the boundary condition Dr. Inan requires is satisfied.

Certainly not, that is why it is nxE=0 and not n.E=0. In fact, n.E=0 would give the opposite behavior as desired.

Which means one can not have two values of E at anyone point and E, when taking a limit from any direction -- must have the same value from a left sided limit or a right sided limit. Eg: if inside the conductior has no E field, then just outside (differentially) there must also be NO E field.

This is not correct. The E-field can be discontinuous wherever there is a charge. For a perfect conductor all charge resides on the surface of the conductor and therefore even differentially just outside the conductor you can have an E-field despite the fact that there is no E-field inside the conductor. That E-field will be purely normal to the surface and will thus have no tangential component.

Look andrewr. You don't like me, I get that. For some reason I make you anxious. But that does not mean I am wrong. There is simply no paradox here, all the energy is accounted for whichever way you work the problem except when you make a mistake such as ignoring the acceleration of the mirror, ignoring the force that keeps the mirror from accelerating, or using the wrong formula. We can keep playing "Where's Waldo" with new mistakes, but that isn't going to change anything. The conservation of energy is a fundamental feature of the laws of special relativity and electromagnetism, so if you apply those laws correctly to any possible scenario you will always find energy is conserved. If you do not find energy conserved then you are 100% guaranteed to have made a mistake.

 

[andrewr]

Dale,

Hi andrewr,
I went through your previous post looking to find where the factor of 2 was. I quickly found this.The momentum of a single photon is p=E/c. The reason that the momentum transferred is twice that is because the photon bounces back with almost the same momentum but in the opposite direction, a change of ~2p.

I know you didn't see this before you wrote.

Also, the units of momentum for the first photon are kg*m/s, not just m/s, and that includes the double momentum for reflection.[/color]
The general formula ought to have read: E(mirror) = 0.5 * Eph**2 / (m0*c**2)
The energy transferred is a non-linear function of the energy of the photon.

However, this factor of 2 was not the cause of the difference that you found between the two methods of calculating Δf. Although your value for p was wrong it appears that you didn't use it anywhere else so the error did not propagate. I went carefully through the remainder of your numbers and they are all correct, Eph, v, KE(mirror), and Δf (energy).

The reason is that I meant the momentum of a single reflected photon.

It took me a long time to spot the error because I was looking for a factor of 2. However, the actual problem was just in the formula for the Doppler shift. You had:

Δf=((c - v)/(c + v) - 1)*f0

My formula IS the correct formula for Doppler shift of a REFLECTED photon.

but the correct formula is:

Δf=(sqrt((c - v)/(c + v)) - 1)*f0

Sorry, that is the change of frames (one way -- not reflected) Doppler shift.
It does not include reflection.
There are TWO doppler shifts in a reflection. Since Doppler shift is multiplicative, you must square the relativistic Doppler shift. This was dealt with and derived at the start of the thread. It is also well known:

eg: http://www.physast.uga.edu/ask_phys_q&a_old.html

* When a policeman measures your speed using radar he is using the Doppler effect to do so. Suppose that the policeman is at rest and you are speeding by. The radar leaves the police car and you receive a Doppler shifted frequency because you are a moving observer. Then you reflect the radar back to him so you become a moving source and there is a Doppler effect again. So this reflection becomes a double Doppler effect.[/color] The policeman now compares the wavelengths of the original radar and that reflected back from you to determine your speed.

With the correct formula for the Doppler shift there is agreement between the two different methods of calculating Δf. So again, there is no discrepancy between any of the different approaches and energy is always correctly conserved.

I don't see that. I see that you have substituted half the formula.

He uses it explicitly in the title of the section.
I agree completely. One way of expressing this mathematically is nxE=0, as I said all along.Although

Good; Even though the normal is NOT used in the mathematics -- which triggered my frustration upon seeing the form being unable to be more than my boundary conditions. I concede that your notation with the normal is equivalent to the engineering equations in my book -- When the adrenaline kicked in, I lost concentration -- it is fairly easy to see which post that started at. But, none the less -- my incoherent reasoning had a kernel.
As I noted, that if your conditions were correct...then this follows:[/color]

It still looks to me that your conditions, therefore -- are incomplete -- for a moving target. eg: My boundary conditions were incomplete, as taught by my prof. and yours are equivalent? (I'll demonstrate).

If they are not identical to what my professor claimed, you can expand the reasoning. These boundary conditions are where I suspect the problem lay when I began the thread.

Since the boundary conditions you quote are the same as for a stationary target -- (Correct?) -- then you can't get an answer different from a stationary target with the exception of the boundary condition being at x=vt;
In my original analysis of the sinewave problem, it is abundantly clear that the amplitude does not change -- just as it would not in a stationary conductor.
One can say "A*cos( ωt - ω/c * x ) + B*cos( ωt - ω/c * x + π) = 0" at x=vt or at x=const, and the solution of the equation will produce the same thing.
(The cos I am taking as the E field, the sin would be the H. I take it, when you use B fields you mean -- B = μ0 * H; so these are scalar multiples of each other.

The forward and reverse waves have the same amplitude: A=B -- only the frequency changes.
Now, that isn't correct -- as power is created in such a scenario.
But -- getting the RIGHT answer is not what I am after so much as understanding HOW to get it from Μaxwell's equations -- eg: how other people do it.
I have a limit approach, but that essentially causes some unphysical results which have the right power values.
I am looking for the standard way to solve it. I do not understand what Einstein does, as he appears to add a term which has no physical meaning to me in his derivation... in the 1905 document.
There is the issue of Hall effect, I suppose, as the H field would act differently on moving electrons and protons. That is something I have never seen tried in my engineering texts...

The E-field can be discontinuous wherever there is a charge. For a perfect conductor all charge resides on the surface of the conductor and therefore even differentially just outside the conductor you can have an E-field despite the fact that there is no E-field inside the conductor. That E-field will be purely normal to the surface and will thus have no tangential component.

It still cancels to exactly zero on the boundary, does it not? (at least INSIDE the conductor) -- If so -- how do you get a reverse propagating wave with a different amplitude when using a square wave? (same argument as a sine wave above).
The magnitude of the forward propagating wave and the inverted magnitude of the reverse propagating wave must add up to zero at the conducting boundary[/color].
A step discontinuity to a voltage/E-field would appear to induce infinite current if there were no (in the limit) distance between it and the short.
That is what I meant about continuity -- the poor description of that continuity not withstanding.
My prof states that on a transmission line, the voltage at a boundary (corresponds to E) must be continuous at that point.

Look andrewr. You don't like me, I get that. For some reason I make you anxious. But that does not mean I am wrong.

You have misinterpreted what I have said multiple times and been insulting (continuing to push when I asked you to stop) when the tables were turned. Yes, I don't like you.

There is simply no paradox here, all the energy is accounted for whichever way you work the problem except when you make a mistake such as ignoring the acceleration of the mirror,

Case in point, the original problem defined[/color] no acceleration -- which you originally overlooked. Will you repeat a bad argument?

ignoring the force that keeps the mirror from accelerating, or using the wrong formula. We can keep playing "Where's Waldo" with new mistakes, but that isn't going to change anything. The conservation of energy is a fundamental feature of the laws of special relativity and electromagnetism, so if you apply those laws correctly to any possible scenario you will always find energy is conserved. If you do not find energy conserved then you are 100% guaranteed to have made a mistake.

Please re-read the thread. I made it quite clear to you MULTIPLE times that I believe energy is conserved on average. The free floating mirror problem was a concession -- the disagreement I found was not the original issue. But, since I stumbled across it -- being careful -- I figure we might as well look at it. ESPECIALLY since it isn't relativistic, and can be verified two ways.

In my FIRST post to you I spoke of power and time equating to conservation.
If you don't want to work the problem all the way through, you are welcome to leave.

I fully expect an error of some sort to appear on my part. That was in the original part of the thread. So, I am not fighting you over a real or "average" violation.
Also, 300MHz equates to 1 METER. That is MACROSCOPIC -- and there are waves in the 30CM range that you can actually detect the shape of with a small dipole antenna and a light bulb. This isn't affected appreciably by Heisenburg -- but if it were, you would be actually agreeing that a photon may have a bit more or less energy than the frequency would imply. One of my original answers. ONE THING FOR CERTAIN -- THEY ARE NOT SQUARE WAVES.
An antenna acts as a coherent light source emitter -- just as if a laser were in operation. The wavelength and time at 300MHz are long enough that the E-field can be viewed on an oscilloscope. There are many other inconsistencies that I find in our conversation.

Here, as a nicety -- let me throw out a thought -- the time of collision of the photon at 300MHz is a macroscopic 3ns. I can watch that on an oscilloscope screen.
Therefore, there will be time for acceleration to occur between the first and second halves of the wave. That would mean that the photon hitting the boundary will experience different amounts of shift vs. time. That might have a small effect -- and Einstein seems to ignore that possibility. It may not be big enough -- but it is something to check.

 

[George Jones]

The mirror is moving APART, so the doppler shift would red-shift twice. The frame of reference would first be the police, then a lower f at the receding mirror, and then lower yet when returning to the original frame.
The non-relativistic Doppler shift is f0*(c-v)/(c+v) -- and that is the square of the relatavistic formula showing a double red shift. Ich shows that correctly in his post.

This is what I get for the relativistic elastic collision of a photon with a moving mirror, and I did the calculation in a single reference frame using conservation of energy and spatial momentum. Also, I carried the mass of the mirror through the entire calculation, and only at the end did I take the limit as m \rightarrow \infty.

 

[Dale]

There are TWO doppler shifts in a reflection.

Yes, there are two Doppler shifts in a reflection. But don't forget that the mirror is changing velocity due to the reflection so you can't just use a single velocity, you have to use both velocities. In this case the mirror's initial velocity is 0, so the first Doppler shift drops out and only the second Doppler shift causes any change in frequency. Thus the equation I posted is correct for this situation, the one you posted is for a situation where the velocity does not change.

It still looks to me that your conditions, therefore -- are incomplete -- for a moving target.

Correct, the boundary conditions I cited are only for a stationary perfect conductor. I don't know the conditions for a moving conductor, which is why I worked the problem in the conductor's rest frame and then transformed to the moving frame.

 

[andrewr]

Yes, there are two Doppler shifts in a reflection. But don't forget that the mirror is changing velocity due to the reflection so you can't just use a single velocity, you have to use both velocities. In this case the mirror's initial velocity is 0, so the first Doppler shift drops out and only the second Doppler shift causes any change in frequency. Thus the equation I posted is correct for this situation, the one you posted is for a situation where the velocity does not change.

Actually, I think this has far more implications than you notice.

As a photon, --all the energy of the photon is adsorbed at once -- is the standard quantum claim.
The photoelectric effect is all or nothing.

So, since the transmitter and receiver are by definition (the mirror surface) moving at the same velocity -- the equation I gave is correct *instantaneously*.
The velocity I plugged into it is wrong -- based on the quantum assumption -- but that seemed the more logical choice.

I think Your answer will not be very close should the mirror be moving at say 0.1m/s -- where mine will still be close. The question is, does the increase in velocity always agree with two velocities -- eg:the original before impact, and the increase after impact -- when using single ended Doppler shifts. If not, your solution is rather ad-hoc.

It would seem that the velocity is not constant, and that one truly ought to try an integral of the energy / dt applied to see if the acceleration corrects the issue. If it does, that even deepens the mystery about photons.

Correct, the boundary conditions I cited are only for a stationary perfect conductor. I don't know the conditions for a moving conductor, which is why I worked the problem in the conductor's rest frame and then transformed to the moving frame.

OK, but it leaves a big gap for me in understanding. I can understand analogically, that a delay line exposes inductance and capacitance as it's short moves. Therefore, the inductance freshly exposed has a value affected by the conditions of the conductor which exposed it - that would cause a waveform change, as the energy elements exposed must be traversed forward -- and then reverse. How, exactly, they would change is something of a mystery.