The doppler radar trap paradox on the path to gravity.

[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

... [ 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.
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.


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 ... 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;therefore the E fields in regions 1 and 2 must be equal 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.

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 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.
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.

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

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.
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. 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:

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.
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 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.

 

[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.

This is an odd point, and one I want to look into dearly. What you seem to suggest is that as an object moves faster it becomes more capable of adsorbing energy. I'll work an example -- if I am sane -- tomorrow night. But is *seems* intuitively, that since mass is to appear greater when something moves, that it ought to adsorb less energy from light. It is as if nature were reversing her preferences all the sudden.

Do you happen to know, in section 7 of Einsteins work, what the significance of the "l" is in this:
φ=ω{ t - 1/c( lx + ... )

Eg: when I do a phasor/travelling wave; I do
arg = ω{ t - 1/c * x }

Do you know how to reduce the equation Einstein uses in section 7 so that it represents a plane wave in the x direction? Thanks.

 

[Ich]

Do you know how to reduce the equation Einstein uses in section 7 so that it represents a plane wave in the x direction? Thanks.

Set l=1, m=n=0. These are the components of a unit vector, the direction of the plane wave.

 

[Dale]

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.

Yours was not close, it was off by a factor of ~2. That was kind of the whole point of the exercise.

The formula I cited was for this specific case (v0 = 0). The general one for an isolated mirror reflecting a single pulse is:

f1/f0 = sqrt((c - v0)/(c + v0)) sqrt((c - v1)/(c + v1))

Go ahead and test the general one as much as you like, I tested it in Mathematica with low velocities and relativistic velocities and with low energy photons and high energy photons, and then I verified the equality analytically. I would be glad to post the code if you use Mathematica, but there are four-vectors in the derivation.

 

[andrewr]

Yours was not close, it was off by a factor of ~2. That was kind of the whole point of the exercise.

The formula I cited was for this specific case (v0 = 0). The general one for an isolated mirror reflecting a single pulse is:

f1/f0 = sqrt((c - v0)/(c + v0)) sqrt((c - v1)/(c + v1))

Go ahead and test the general one as much as you like, I tested it in Mathematica with low velocities and relativistic velocities and with low energy photons and high energy photons, and then I verified the equality analytically. I would be glad to post the code if you use Mathematica, but there are four-vectors in the derivation.

I don't have Mathematica (at least the graphing part doesn't appear online) -- I'm poor thanks to my wife. Now, I have to use my brain.
The difference in energy was off by a factor of two -- the actual (absolute) frequency was off by far less than 1%.
Even though I worked a specific case, I was careful to post a general formula for energy transfer v. photon energy.

Your solution was not qualified as "ad-hoc" when you posted it; nor did you explain "why" the equation I gave was "wrong" at first.

Secondly, now that the gnat has been strained -- let's look at the elephant:
Please justify using two different velocities on physical grounds.
The receiver and transmitter is the same surface -- that of the mirror.

The division of the process into two velocities simultaneous (for the impact certainly has transmission and reception happening at the same time) is rather without explanation.

If that isn't paradoxical -- like I said when I started the thread -- nothing is.
Do you disagree?

It would seem an assumption is being made which no one has stated. It also would seem that there is the issue of energy transfer vs. mass coming on strong. What, before I speculate, is your explanation for the prediction that photons of higher energy are more adsorbed than those of low energy? and why is it, that motion -- which causes a photon to DROP in energy -- would appear to cause the photon to be more adsorbed?

I don't mind that relativity is used -- but I wan't to see the other side -- eg: I do want to verify the individual steps against Maxwell's equations because the paradox seemingly revolves around differences between these two methods -- and that will allow me to UNDERSTAND (the object of my inquiry). I do not doubt that the solution provided by a relativistic transform can be just as valid as Maxwell -- I am not trying to disprove relativity. But, when I do engineering -- and pass it along to other engineers -- the common language is Maxwell.

 

[andrewr]

Set l=1, m=n=0. These are the components of a unit vector, the direction of the plane wave.

OK. That saves me a lot of time. I tried that solution and thought I must be making a mistake -- but since you seem to agree, then I can remove several possibilities.
There in section 7, where AE breaks the variables out as l' = ... followed in my printout by something that looks like another prime ('), but perhaps that is really a comma? eg: He is separating the list into four elements with commas. It doesn't mean the right hand variables are in the translated frame.

If so, then it means l' = 1 as well. Correct?

In my original equation back in the first handfull of posts, I spoke about the relative length contraction of a wave. In order to get that length in the relative frame -- I used interference of the reflected wave with the transmitted wave to figure out where a null spot would be in the lab frame.
Of course this is with a BSEE text which does not clarify whether or not one can use the method of a short in a moving frame -- and there is some indication that there is a danger in the assumption.
At very least, the amplitude is wrong.

Now, whether moving or not, a neutral superconducting mirror is a neutral superconducting mirror...
It's properties don't change as far as the lab frame is concerned. Eg: there is an infinitely thin layer at the surface which acts as the actual conductor -- and internally all magnetic and electric fields are rejected. (eg: one can float a superconductor on a magnet for that reason).

So, as far a Lorentz contraction is concerned -- there really is no change of the electrical (excluding mass) properties of the mirror. An infinitely thin layer -- is still infinitely thin.

In any event, I figured I was safe -- since the error I was making (if any) had to repeat consistently -- and therefore I was still finding a spot where the E and B fields were the same in any frame that they were at the surface of the mirror -- eg: that the transformation would have to be consistent between frames -- and therefore any error of the analysis of reflection would necessarily cancel. I was rewarded with the predicted length of AE.

But as a side effect, I learned that it is the modulation waveform (an AM modulation) which translates the "length" of a wave when viewing a moving frame from a lab. I don't calculate the actual wavelength expansion through Lorentz transform alone, or an error results in what the other frame would perceive.

I don't have a problem with the example you worked above, as far as I have gotten. (about half way).
But I am curious if perhaps you would agree with a gut reaction that I have concerning relativity.
(Others are free to answer as well.)

Given the E field and H field of a propagating wave in free space at a single *point* and instant, eg: a single E vector and H vector at rest in a lab frame-- relativity would seem to allow the transformation of the E and H fields at that point for an observer with a different velocity -- without any knowledge of the E and H fields at any other location. (This may very well be wrong, but what do you think?)

If so, then, some of the E vector becomes H -- and vice versa. Is there a compact expression for this transformation?; eg: if qq amount of E becomes H, then a certain amount of H becomes E. (A linear transform of some kind?)

Thanks.

 

[Dale]

Please justify using two different velocities on physical grounds.
The receiver and transmitter is the same surface -- that of the mirror.

The division of the process into two velocities simultaneous (for the impact certainly has transmission and reception happening at the same time) is rather without explanation.

If that isn't paradoxical -- like I said when I started the thread -- nothing is.
Do you disagree?

Hi andrewr, I guess this should not surprise either of us, but yes I disagree. Your stance seems exactly backwards to me. Please justify using either single velocity on physical grounds. Which velocity should the Doppler formula choose and why? Personally, that seems to make much less sense to me, particularly given the quite reasonable description of a mirror as having two Doppler shifts.

IMO, the only other physically reasonable thing besides using the two velocities would be some sort of average of the two velocities. Certainly using either endpoint alone doesn't make physical sense. Just for giggles I solved for what that average velocity would be. It is:
$$v=\frac{c^2+v_0 v_1 \pm \sqrt{\left(c^2-v_0^2\right) \left(c^2-v_1^2\right)}}{v_0+v_1}$$
I have no idea what it means, but if you prefer to use that "average" v in your formula then it should work. Or if that equation has some significance of which I am unaware then I would be glad to know.

I don't mind that relativity is used -- but I wan't to see the other side -- eg: I do want to verify the individual steps against Maxwell's equations because the paradox seemingly revolves around differences between these two methods -- and that will allow me to UNDERSTAND (the object of my inquiry). I do not doubt that the solution provided by a relativistic transform can be just as valid as Maxwell -- I am not trying to disprove relativity. But, when I do engineering -- and pass it along to other engineers -- the common language is Maxwell.

I already did that and you already rejected it. The key issue being that you didn't like the idea of working the problem in the mirror's frame and then doing a standard transformation to another frame because of the different appearance of the boundary condition. I was unable to find anything about the boundary condition for a moving conductor after a pretty thourogh net search. I found some scientific papers that sounded promising from the abstract, but I wasn't willing to spend ~$20 to see.

Personally, I am satisfied by my derivation and I think that if I hadn't made you angry you would be satisfied too. In the mirror's frame everything checks out: Maxwell's equations, energy, momentum, Doppler, and the boundary conditions. In the transformed frame Maxwell's equations are still satisfied, as are energy, momentum, and Doppler. In the absence of any evidence to the contrary, is it really such an unacceptable stretch of reason to assume that the boundary conditions are correct also? I don't think so, and I suspect that you would not think so either if it had come from someone besides me.

 

[Ich]

Given the E field and H field of a propagating wave in free space at a single *point* and instant, eg: a single E vector and H vector at rest in a lab frame-- relativity would seem to allow the transformation of the E and H fields at that point for an observer with a different velocity -- without any knowledge of the E and H fields at any other location.

Of course. That's why we talk about electromagnetism, where E and B are two aspects of the same thing.

If so, then, some of the E vector becomes H -- and vice versa. Is there a compact expression for this transformation?

§6.

 

[jtbell]

If so, then, some of the E vector becomes H -- and vice versa. Is there a compact expression for this transformation?; eg: if qq amount of E becomes H, then a certain amount of H becomes E. (A linear transform of some kind?)

Nowadays we call it a Lorentz transformation. Ich has pointed you to the relevant section in Einstein's 1905 paper. For a more modern presentation, see for example

http://farside.ph.utexas.edu/teaching/em/lectures/node123.html (and the preceding pages for context)

or an E&M textbook at the intermediate undergraduate level or above, e.g. Griffiths, "Introduction to Electrodynamics".

 

[Mentz114]

andrewr,
This is how E and B transform in vector notation,
$$\mathbf{E}'=(\mathbf{E}\cdot \mathbf{n})\mathbf{n} + \gamma(\mathbf{n}\times(\mathbf{E}\times\mathbf{n})+\mathbf{v}\times\mathbf{B})$$
$$\math{B}'=(\mathbf{B}\cdot \mathbf{n})\mathbf{n}+ \gamma(\mathbf{n}\times(\mathbf{B}\times\mathbf{n})-\math{v}\times\mathbf{E})$$
where $\mathbf{n}$ is the unit vector in the direction of $\mathbf{v}$.

Itzykson and Zuber page 9.

 

[Dale]

Just for giggles I solved for what that average velocity would be. It is:
$$v=\frac{c^2+v_0 v_1 \pm \sqrt{\left(c^2-v_0^2\right) \left(c^2-v_1^2\right)}}{v_0+v_1}$$
I have no idea what it means, but if you prefer to use that "average" v in your formula then it should work. Or if that equation has some significance of which I am unaware then I would be glad to know.

I spent some time playing around with this average velocity to try to figure out its significance since it is a fairly complicated expression that is not immediately obvious. However, it turns out to have a pretty clear meaning. It really is the average velocity taking into account the relativistic velocity addition. In other words, if you use the relativistic velocity addition formula to subtract the average from v0 and v1 then you get two equal and opposite velocities. This corresponds to the center of momentum frame where there is no Doppler shift and no energy transfer.

 

[andrewr]

Nowadays we call it a Lorentz transformation. Ich has pointed you to the relevant section in Einstein's 1905 paper. For a more modern presentation, see for example

http://farside.ph.utexas.edu/teaching/em/lectures/node123.html (and the preceding pages for context)

or an E&M textbook at the intermediate undergraduate level or above, e.g. Griffiths, "Introduction to Electrodynamics".

andrewr,
This is how E and B transform in vector notation,
$$\mathbf{E}'=(\mathbf{E}\cdot \mathbf{n})\mathbf{n} + \gamma(\mathbf{n}\times(\mathbf{E}\times\mathbf{n})+\mathbf{v}\times\mathbf{B})$$
$$\math{B}'=(\mathbf{B}\cdot \mathbf{n})\mathbf{n}+ \gamma(\mathbf{n}\times(\mathbf{B}\times\mathbf{n})-\math{v}\times\mathbf{E})$$
where $\mathbf{n}$ is the unit vector in the direction of $\mathbf{v}$.

Itzykson and Zuber page 9.

Thanks, I will have to look at that link.

Oddly, I got an email about a post that doesn't show up here on the forums. Do posts get dropped often, or do I need to check the e-mail logs to see who is pulling my leg. My meds aren't working too well today -- so it may be a little while before I can really respond.

I was actually after something simpler, the fields are all in point form -- so I was wanting to verify that the Lorentz transform is also in point form. In particular, I was wanting to know how much each transformed into the other, given that one transform was known (qq) decimal fraction of E became H, then how much of the H became E?

The formulas from Zuber are probably enough to work it out -- it will just be a couple of days. It is easier to think and type than it is to work math for me ... that's why I asked.

I have often heard it explained that an H field is really just a time delayed E field -- but I have some contradictions (internal) that I would like to remove -- but I am fascinated with the way I can calculate the correct deflection strength when using relativity. Considering how difficult the paradox (eg: that I am still seeing about photon division are to talk about), I'm not really ready to go there ... I sort of need to prepare for the inevitable cross examination that seems to be the norm here.

 

[Mentz114]

Regarding my earlier post of the E, B transformations, there is an implicit factor of c omitted. I&Z like to work with c=1.

 

[andrewr]

Regarding my earlier post of the E, B transformations, there is an implicit factor of c omitted. I&Z like to work with c=1.

Thanks, that would definitely have tripped me up.
I get, with c=1, that the magnitude of E and B are interchangeable -- otherwise, one needs a factor of c * B. I can live with that.
E**2 is comparable to (c*B)**2I am just going to list out a couple of thoughts here, and see how these equations you have shown me work out.

In free space, for freely traveling radiation (non standing wave), E and H are in phase though they may rotate around the axis of travel. eg: circular & elliptical polarization -- which I will ignore for the simple case.
(The more complicated cases I think I can treat by superposition of linearly polarized waves out of phase -- and I find that simpler)
The direction of light propagation, and the mirror, are the same -- so direction of n and v is perpendicular to E and B (or H).

In a plane polarized wave, E is perpendicular to the direction of travel, therefore the dot product of E and n is 0.
That leaves the Lorentz transform scaled portion of E' = γ( n x (E x n) + v x B)

Since E is perpendicular to n, the cross product E x n is the same as |E| in the direction perpendicular to E and n -- which is therefore B -- and then crossed with n a second time, reverting back to the direction of E. So, for the plane wave the n x (E x n) reverts back to E.E' = γ( E + v x B )

The sign of v x B will depend on the direction of v and B.
If the wave travels in the same direction as v, then E x B must be in the direction of v.
Then, (ExB) x B must be in the same direction as the final v x B.
So, that is the opposite direction of the original E, magnitude |v|*|B|
If I take + values of v to be in the direction of the light propagation, and - to be opposing it:

E' = γ( E - v*|B| )
B' = γ( B + v*|E| )

O.K. those are simple enough for me to think about.
I know the relation B = μ0*H, and you indicated the implicit factor of c; by inspection, I would guess:

E' = γ( E - v*|B|*c )
B' = γ( B + v*|E|/c )

and

E' = γ( E - v*|H|*c/μ0 )
H' = γ( H + v*|E|*μ0/c )

An E field causes a force on a charge; I am not sure the metric requires a factor of c.
Hmmm...

If I made a mistake (or was misled) could someone let me know where and what the final expressions ought to be.
These are simple enough for me to work with / familiar territory that I would like to have them for reference.
Also, please note, I will be working on this thread slowly -- so I don't expect fast replies -- but the thread won't be abandoned without my giving a notice of some kind.

Thanks.
Andrew.

 

[andrewr]

Letting the boundary conditions for reflection off a moving mirror remain a mystery (mild) for a while;
I want to shift attention back to the original problems I posed -- but still allowing the free floating mirror (acceleratable) as that seems to be the problem most people are willing to solve...

Given a 1Kg rest mass mirror, and a single 300MHz photon bouncing off of the mirror... which is now moving at 1/2c...

1) The mass of the mirror will increase in the lab frame.
2) I would expect *LESS* energy to be adsorbed by the mirror which is moving -- on the basis of the calculations done at rest. Eg:

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.

The energy adsorbed by the at rest mirror was:

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

That means:
0.000000000000000000000000000000000000000442% of the photons energy went to the mirror.

As the mirror becomes more massive; that value drops.
There is only one distance to measure, and one velocity in the problem. We need not concern ourselves with relativistic addition of velocities, etc.

Proposition a): The velocity that an observer moving with the mirror sees the policeman receding will have the same value as the velocity that the police man sees the mirror receding. This is a result of the length contraction and time dilation both being by the lorentz γ factor.
These cancel when length/time computes velocity. BUT: *This would not be true if a third object were included in the measurement which is why this gedanken avoids that situation.
( I am having trouble verifying this assertion myself; so if you have a disproof -- that would be fine.)Proposition b) The mirror knows its rest mass, and knows a photon of less than 300MHz is striking the mirror (f=300MHz * sqrt( (c-v)/(c+v) ) ). Therefore, the acceleration (and final velocity) of the mirror in the frame of reference of the mirror is calculated as if the mirror were at rest. It must be less than that of a 300MHz photon. The final velocity will be agreed to by both the mirror frame and the police (lab) frame because of proposition a.

Therefore -- for non-contradiction to hold -- the policeman must also come to the conclusion that the photon accelerated the mirror less. Effectively, the mirror is more massive; and therefore, I would expect, it would adsorb *LESS* energy than 1%.

Now, at 1/2 C -- the returning photon is 100MHz -- eg. 66% of the photon's energy is gone.
There was only one photon -- and photons are supposedly "indivisible".

Where did this energy go? (I assume into multiple reflection photons, but if so -- how many, and how does one know for sure?)

 

[Dale]

Hi andrewr,

Excellent work. I am glad to see that you are stepping out of your comfort zone a little and trying new things. It is good to learn these transformations. Your derivation was good, and just had one very small sign error:

The sign of v x B will depend on the direction of v and B.
If the wave travels in the same direction as v, then E x B must be in the direction of v.
Then, (ExB) x B must be in the same direction as the final v x B.
So, that is the opposite direction of the original E, magnitude |v|*|B|
If I take + values of v to be in the direction of the light propagation, and - to be opposing it:

E' = γ( E - v*|B| )

All of the above is good.

B' = γ( B + v*|E| )

But the correct expression here should be B' = γ(B - v*|E|). Since (ExB)xB is in the same direction and since E is perpendicular to B then (ExB)xE is in the opposite direction.

You can check this simply by setting E = (0,Ey,0) and B = (0,0,Bz) which yields a vector of propagation ExB -> (c,0,0) and therefore if v is in the direction of propagation v = (vx,0,0). Plugging them into Mentz114's equations you get:
E' = (0,γ(Ey - v Bz),0)
B' = (0,0,γ(Bz - v Ey))

You can also check this by referring to the last set of equations in http://www.fourmilab.ch/etexts/einstein/specrel/www/#SECTION21" of Einstein's paper and using the same E and B vectors specified above (although Einstein uses E = (X,Y,Z) and B = (L,M,N), so just set X=Z=0 and L=M=0)

 

[Dale]

1) The mass of the mirror will increase in the lab frame.
2) I would expect *LESS* energy to be adsorbed by the mirror which is moving -- on the basis of the calculations done at rest.
...
Where did this energy go?

We know from first principles that http://farside.ph.utexas.edu/teaching/em/lectures/node89.html" and the temporal and spatial symmetry of the laws. So the details of the scenario are completely irrelevant. As long as Maxwell and Lorentz are obeyed energy and momentum are conserved. No exceptions. Any scenario you construct that finds otherwise is simply a math error somewhere.

In this case the source of the error is pretty obvious. You derived a relationship for a specific scenario, the mirror at rest, and mistakenly over-generalized it. You simply assumed that your equation held without modification in the moving frame and further assumed that the mass in your equation was the relativistic mass rather than the invariant rest mass. Since your conclusion is provably wrong then at least one of your assumptions must be wrong.

To get the correct general formula you can either derive it for the general case of a moving mirror or you can take the rest-frame case and Lorentz transform it to the moving frame. Either way you should get the same formula, and either way you must get conservation of momentum and energy.

 

[andrewr]

In this case the source of the error is pretty obvious. You derived a relationship for a specific scenario, the mirror at rest, and mistakenly over-generalized it. You simply assumed that your equation held without modification in the moving frame and further assumed that the mass in your equation was the relativistic mass rather than the invariant rest mass. Since your conclusion is provably wrong then at least one of your assumptions must be wrong.

Dale,
As usual, I don't know what you are talking about. Be specific.
Quote my "conclusion" please.

http://www.merriam-webster.com/dictionary/conclusion

1 a: a reasoned judgment : inference b: the necessary consequence of two or more propositions taken as premises ; especially : the inferred proposition of a syllogism2: the last part of something: as a: result, outcome bplural : trial of strength or skill —used in the phrase try conclusions c: a final summation d: the final decision in a law case e: the final part of a pleading in law 3: an act or instance of concluding

The only necessary consequence I had is that the light would accelerate the mass of the moving mirror less.
Is that incorrect?

Everything else is conditional; I had a hypothesis -- there would be less than 1% energy expended in accelerating the mirror. Is that what you are talking about?

http://www.merriam-webster.com/dictionary/expect
4 a: to consider probable or certain <expect to be forgiven> <expect that things will improve> b: to consider reasonable, due, or necessary <expected hard work from the students> c: to consider bound in duty or obligated <they expect you to pay your bills>

An expectation is not something which is guaranteed; it could be -- but doesn't have to be. As the examples given in the defintion show. The inclusive nature of the "or" denies a strict interpretation of an expectation as a conclusion.

The law according to Einstein is that in each inertial frame of reference, the laws of physics are identical.
By jumping to the frame of reference of the mirror, I verified that the final velocity will be less than in the case of a 300MHz photon. Eg: I USED the result of a specific experiment in the EXACT same environment.
The mirror is at rest in the frame of reference of the mirror. It is accelerated to non-relativistic speeds by a photon.
etc.

As further guarantee of my integrity -- I asked a question at the end of the post. That, clearly, is not a conclusion as you havenot answered it (as usual).

 

[Dale]

As usual, I don't know what you are talking about. Be specific.

Your conclusion that there is any missing energy is incorrect.

 

[andrewr]

Hi andrewr,

Excellent work. I am glad to see that you are stepping out of your comfort zone a little and trying new things. It is good to learn these transformations. Your derivation was good, and just had one very small sign error:
All of the above is good.But the correct expression here should be B' = γ(B - v*|E|). Since (ExB)xB is in the same direction and since E is perpendicular to B then (ExB)xE is in the opposite direction.

You can check this simply by setting E = (0,Ey,0) and B = (0,0,Bz) which yields a vector of propagation ExB -> (c,0,0) and therefore if v is in the direction of propagation v = (vx,0,0). Plugging them into Mentz114's equations you get:
E' = (0,γ(Ey - v Bz),0)
B' = (0,0,γ(Bz - v Ey))

You can also check this by referring to the last set of equations in http://www.fourmilab.ch/etexts/einstein/specrel/www/#SECTION21" of Einstein's paper and using the same E and B vectors specified above (although Einstein uses E = (X,Y,Z) and B = (L,M,N), so just set X=Z=0 and L=M=0)

Hey... Thanks for actually working the problem at my level for once instead of shoving something over
my head at me.

I'll have to look at this more... I want to get it right.

 

[Dale]

Proposition a): The velocity that an observer moving with the mirror sees the policeman receding will have the same value as the velocity that the police man sees the mirror receding. This is a result of the length contraction and time dilation both being by the lorentz γ factor.
These cancel when length/time computes velocity. BUT: *This would not be true if a third object were included in the measurement which is why this gedanken avoids that situation.
( I am having trouble verifying this assertion myself; so if you have a disproof -- that would be fine.)

This is correct, but only for inertial frames. The inverse of the Lorentz transform with velocity v is also a Lorentz transform but with velocity -v. The easiest way to show it is to take the normal http://en.wikipedia.org/wiki/Lorent...rmation_for_frames_in_standard_configuration" and solve for the unprimed variables.

But it only applies for inertial frames wrt each other, and not to non-inertial frames nor to inertial frames wrt other velocities. So, for example, the inertial reference frame of the cop and the inertial reference frame where the mirror is initially at rest will agree on the initial separation speed, but not on the change in velocity.