The doppler radar trap paradox on the path to gravity.

[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})
<br /> \math{B}&#039;=(\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[/color] 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[/color] my "conclusion" please.

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

1 a: a reasoned judgment[/color] : inference b: the necessary consequence[/color] 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[/color] -- 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[/color] or certain <expect to be forgiven> <expect that things will improve> b: to consider reasonable[/color], 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.

 

[andrewr]

Your conclusion that there is any missing energy is incorrect.

I didn't conclude that. I asked[/color] where it went. A question is NOT a conclusion.
I notice you didn't quote me -- because you can't.

 

[andrewr]

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.

How is it, that if they agree to the same absolute VELOCITY. that they do not agree on a delta velocity?
Before and after the impact, the frames are inertial. Therefore, before and after -- they must agree to the same velocity. That is, there will be two velocities -- the one before impact, and the one after. Each individual (separation) velocity, by the Lorentz transform must be agreed to in both frames of reference. The reversal of sign is a convention.

Are you saying that they will measure different speeds after impact!? That flat contradicts the idea that there is any definite separation velocity between the two objects.

Gedanken:

The policeman sees the impact, using whatever formula you like -- he says, the acceleration caused a change in velocity of Δv. Therefore, the final velocity is 0.5c + Δv. The mirror says, I was moving at -0.5C, impact caused a velocity change of δv. Therefore the separation velocity is now -0.5c - δv.

Are you saying that these two results which agreed before impact to |v| = 0.5c will no longer agree?
eg: | 0.5c + Δv | != | -0.5c - δv |
And yet the Lorentz tranform guarantees that the actual separation velocity between the two is the same?!

 

[Dale]

I didn't conclude that. I asked[/color] where it went. A question is NOT a conclusion.
I notice you didn't quote me -- because you can't.

Sigh, I didn't bother to quote since it was a running theme of the whole thread. I didn't expect you to pretend that it wasn't.

Not only did you specifically conclude that there was missing energy:

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.

But your question assumes that there is some missing energy:

Where did this energy go?

We have been over this multiple times. All of the energy is accounted for in all frames using any analysis technique. Any change in the photon's energy is completely accounted for by the work done on or by the mirror.

I don't know why you are making such an enormously big deal about my use of the word "conclusion". From the very first post it was clear that you believed that there was some missing energy that was not conserved in this situation and that the different standard descriptions needed to be fixed to account for the missing energy. If you no longer believe that there is any missing energy or if you believe that all of the standard methods agree that there is no missing energy then please be explicit. Otherwise, what are you complaining about?

 

[Dale]

How is it, that if they agree to the same absolute VELOCITY. that they do not agree on a delta velocity

My apologies, I thought that you already understand this point. I thought that is what you meant in your "BUT:" comment and I was just confirming and emphasizing a point you already understood.

BUT: *This would not be true if a third object were included in the measurement which is why this gedanken avoids that situation.

Take, for example, the case where we have an enormous photon that accelerates the mirror to .1 c in the inertial frame where it is initially at rest. Now, consider the reference frame where the mirror was initially moving at .99 c. Obviously, they cannot agree on the delta velocity or in the moving frame the mirror's final velocity would be 1.09 c! The delta velocity is defined by the standard velocity addition formula.

 

[andrewr]

Sigh, I didn't bother to quote since it was a running theme of the whole thread. I didn't expect you to pretend that it wasn't.

Not only did you specifically conclude that there was missing energy:

Dale, do you deny that a photon of 100MHz has less energy than one of 300MHz?
The energy is not missing, it was never there to begin with. The question is "how many photons are reflected" and how do we know?

But your question assumes that there is some missing energy:

The fact that it is a question means it is not a conclusion.
The fact that you insist at every step saying "your wrong" and having fantasies about what you think I said means you have a very arrogant approach.

We have been over this multiple times.

Then why don't you go away and let someone with better manners attempt a discussion?
In my first post I said I am a slow starter.
You apparently are not trained to work with people who have a handicap, I perhaps need a special ed teacher -- hey!
There's no shame in that.

All of the energy is accounted for in all frames using any analysis technique. Any change in the photon's energy is completely accounted for by the work done on or by the mirror.

Then, what you are saying is that there is only one photon reflected -- period. That is the answer you did not give previously.

I don't know why you are making such an enormously big deal about my use of the word "conclusion". From the very first post it was clear that you believed that there was some missing energy that was not conserved in this situation...

That is a lie. You are making a straw man heretic argument.

and that the different standard descriptions needed to be fixed to account for the missing energy. If you no longer believe that there is any missing energy or if you believe that all of the standard methods agree that there is no missing energy then please be explicit. Otherwise, what are you complaining about?

I am complaining about you putting words in my mouth that are not there.

 

[andrewr]

My apologies, I thought that you already understand this point. I thought that is what you meant in your "BUT:" comment and I was just confirming and emphasizing a point you already understood.Take, for example, the case where we have an enormous photon that accelerates the mirror to .1 c in the inertial frame where it is initially at rest. Now, consider the reference frame where the mirror was initially moving at .99 c. Obviously, they cannot agree on the delta velocity or in the moving frame the mirror's final velocity would be 1.09 c! The delta velocity is defined by the standard velocity addition formula.

I didn't set the problem up the way your are doing it. You are adding things to what I said.
Show the example in terms of the Gedanken I actually gave in post #72.
And I warn you, I did NOT compute the change of velocity with respect to anything except the police man.
Don't go inventing a "third" reference frame.
I never stated that one ought to add the rest calculated value to the police man's value. etc.
I said the final velocity will be agreed upon by both observers.

I know my statements are somewhat vague -- that is because I am allowing for more than one interpretation.
To be explicit, I only need to show the monotonic nature of the transform to justify what I have said.
Qualitatively, LESS acceleration is what I predicted as a certainty.
The hypothesis was concerning the amount of energy adsorbed by the mirror.
The question was concerning the number of photons reflected and how this is known.

The next step will be to ask for experiments verifying the choice of answer given by whomever gives it.

 

[Dale]

I am complaining about you putting words in my mouth that are not there.

OK, so as not to put any words in your mouth: Do you now believe that energy is conserved in all reference frames and that all of the various analysis techniques agree?

 

[andrewr]

OK, so as not to put any words in your mouth: Do you now believe that energy is conserved in all reference frames and that all of the various analysis techniques agree?

:)

I have always believed they will agree. yes. I do believe energy is conserved, yes.
I also believe is that the physics I learned allows the energy to be balanced in more than one way.
Eg: more than one reflected photon; or not.

I am not convinced that every formula which "conserves" energy is necessarily correct. Eg. When two competing formulas predict different results -- but both conserve energy -- nature must have some way of determining which actually happened.

For the case at hand: One may balance energy by saying every photon reflected gives up all the excess energy to the mirror.
Alternately, one may balance energy by saying that for every n photons reflected, one more is created.

Again, Maxwell's equations are an average; they are an expectation value. Tunnenling and other instantaneous violations are permitted.

My interest, as my physics prof challenged the class before every exam. -- If you were on a desert Island knowing only the physics you now know, could you prove X.

In order to prove X, the shortcut was not allowed on the exam. Those who passed the class understood the material.

 

[Dale]

OK, so you agree that energy is conserved in all frames with both the "photon" and the Maxwell approach.

When two competing formulas predict different results -- but both conserve energy -- nature must have some way of determining which actually happened.

Again, not wanting to put words in your mouth, does this mean that you think the Maxwell and the photon approach predict some other different result besides the energy? If so what do they predict differently?

 

[andrewr]

Hi andrewr,
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.

Yes, that one caught me. I assumed based on the symmetry of the equations that the sign would carry through -- but it didn't. Taking the shortcut was bad.

E'=γ(E - v*|B|)
B'=γ(B - v*|E|)

Thank you. Regarding the equations that followed, converting to H -- the sign error is there as well.
Is the factor of c correctly inserted, or is it superfluous?

 

[andrewr]

OK, so you agree that energy is conserved in all frames with both the "photon" and the Maxwell approach.Again, not wanting to put words in your mouth, does this mean that you think the Maxwell and the photon approach predict some other different result besides the energy? If so what do they predict differently?

I am not sure they "predict" anything. Let me explain:

The photon idea is consistent with AE theory. That is, his theory indicates that energy is proportional to frequency. However, he does not derive the value of h from first principles. That is, the proportionality constant is decoupled from the theory. When we speak of a photon having such and such an energy, we don't know what that looks like (for certain) in Maxwell's equations. This makes the question very difficult to answer.

Secondly-- momentum and energy are aspects of the same law. In special relativity, and in the example I gave, one can convert momentum to energy and work the problem totally in the energy domain.
In the example of classical physics I presented (as perverted as it seemed) at the start of the thread shows that one can get the "right" answer while appearing to violate conservation of momentum. It is a purely fictional violation.

Thirdly -- the idea of relativistic mass -- if it has any analogy to normal mass -- would suggest that one can plug F=ma; where m is replaced by a relativistic mass and get very close to the correct results for small variations. eg: That is to say, if something is twice as massive due to relativity -- it ought to be *around* (not exactly) twice as hard to accelerate. So long as one takes a differential approach, correcting the mass for each differential increment of velocity; the error ought to be quite small.
I could be wrong here, but it is one of the views I came to from my physics lectures -- and part of understanding is putting notions to the test.

Now: to the prediction question.
I don't know.

It depends on what you mean by the photon approach vs. Maxwell.

I tend to think, that if I were able to work out a sinusoidal example of Maxwell's equations with a moving mirror -- and plugged the instantaneous E and H fields in for a pure sine wave truncated after a long period of time; it must give me the average result -- eg: The average experimental result for repeating the experiment many times with sequentially striking photons. A sinusoidal train of energy represents many photons serially hitting the mirror of an exact frequency and photon energy.

Taking the result of such a calculation, and dividing it by the number of photons represented would be satisfactory to approximate a single (average) photon collision. I could then say, how many photons were reflected for every photon which hit. If the answer is '1' then no further inquiry is needed. If the answer is fractional (as in non integerial by an amount unexplained by round off error from the approximation of a pure sine wave that is actually trucated) -- Then more needs to be done for me to understand what Maxwells and the Photon approach mean.

 

[Dale]

Regarding the equations that followed, converting to H -- the sign error is there as well.
Is the factor of c correctly inserted, or is it superfluous?

I didn't check in detail, but from a quick glance it looked right. The way to tell is to check units. If the units are consistent then you must have the factors of c in the right spot.

 

[andrewr]

I didn't check in detail, but from a quick glance it looked right. The way to tell is to check units. If the units are consistent then you must have the factors of c in the right spot.

I apologize in advance for all the edits -- I can't seem to tie my shoes right since yesterday...

Well, based on units alone,

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

In SI units -- which are typical for engineering -- an E field is volts/meter, (capital V/m ) velocity=little v.
A B field is kg / ( A * s**2 ) -- which is why I prefer H fields which are simpler ( A/m ) until making things actually move...
Alternately B = V*s

BUT:
With no adjustment at all -- just SI units -- E' --> (unitless gamma) * ( V/m - kg / (A * s**2 ) * m / s )
power (P) is V*A, and is also kg * m**2 / s**3. (Elementary Mr. watt )
So, that works out to : E' = ( V/m - kg * m / (A * s**3 ) ) = V/m - P/( m * A) = V/m - V*A/(m*A) = V/m - V/m
In other words, no correction was necessary at all. The hint was misleading in the case of E'
E' = γ( E - |B|*v )[/color]

For B ( to be thorough, no shortcutsx ) -- (Made a mistake anyway the first try.)
B' = (unitless)( V*s - V/m * m/s )
V*s - V/s .. nope, and c doesn't make it better. I'll check again later and Edit if I figure it out.

Its later -- and I don't see the mistake. However, there is another way to get to the solution I am after anyhow.
Free space has 377Ohms impedance (η), So that a freely traveling wave will have E and H in phase and related by a fixed ratio. (Plane wave solutions that I am interested in).
I can simply take the E field and scale it to get the H field - as η must be the same in all frames of reference.

I can also plug that directly into the E' equation to get:
E' = γ( E - |B|*v )
E' = γ( E - | μ0*H |*v )
E' = γ( E - | μ0*E/377 |*v )
E' = γ( E - | μ0*E/377 |*v )

I will stand by this; in SI units -- and assume the authors of the original eqns. were correct.
E' = γ( E - |B|*v )[/color]
E' = γ( E - | μ0*E/377 |*v )[/color]

And the boundary condition I am interested in for the moving conductor falls out perfectly:
E' = γ( E - | μ0*E/377 |*v ) = 0
( E - | μ0*E/377 |*v ) = 0
This is cleaner knowing that η = sqrt( μ0/ε0 ) A.K.A
γ( E - | E |* (v/c) ) = 0[/color]
I was tempted to go one step further, and just do E/γ = 0; but I am not so sure that's a good idea at this time.

The purpose of all the cross products in the originals is to shave off the components of the field which are not orthogonal to the vector v. Those components, apparently, do not transform. Eg. as length contraction happens only in the direction of travel -- E and H "contraction" happens only orthogonal to travel. Hmmm...

Post note: (smelling the roses...editing the grammar / lack therof ... )
Even though Compton scattering is a different effect than the bulk one we are speaking of in the superconductor --
I find it terribly disturbing that in the literature, the wavelength change of a photon impacting on an electron directly -- that the formula reads:
γ2 - γ1 = 0.24*( 1 - cos φ ) ( From google searches. )
These articles indicate that the angle specified is 0 when the impact is head on -- complete reflection in the opposite direction. These, it is claimed, are where the electron is "ejected". The last time I checked, cos(0) = 1, and therefore the above equation would say NO wavelength change when the electron is ejected...
Talk about conservation of energy problems.

Walter and Miller's textbook of radiotherapy: radiation physics, ... - Google Books Result
pp. 43 (see diagram.)

This is typical of the industry...

 

[Dale]

Now: to the prediction question.
I don't know.

OK, so then I think it is safe to say that any "paradox" that might have existed earlier has been resolved.

It depends on what you mean by the photon approach vs. Maxwell.

Well, neither of us has been using any real quantum mechanics in this thread. So I have been considering the "photon" approach to simply be a relativistic analysis of a collision between the mirror with a massless particle of energy E=hf.

 

[andrewr]

As I said to you at the start; In addition to telling you that I NEVER BELIEVED that energy was non-conserved...

Ηi Dale!

A paradox is an apparent (even if illusionary) contradiction.[/color]
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[/color] 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 twin paradox, is, and always will be a paradox. It can be resolved.
Depending on how one views a photon collision with a mirror -- I think it safe to say -- a paradox exists as well. It is possible to look at it -- illusionarily. Zeno's paradox is the same thing,
As I look at the problem different ways, it does resolve into an intelligible answer.
The "answer" to a specific problem -- however -- is not what I am after.
It is how to determine the answer using various tools that I understand.

I have presented what I am after -- eg: the boundary condition -- but instead of helping me find it in term of what I do understand, and then allowing me to compare it to SR -- The thread is littered with an argument that never needed to be. I am stubbornly going to pursue the answer till I understand it well.
I learn best from someone who knows just a *little* more than I do.
Too much; too fast -- and I am simply not going to be able to handle it.

When I fully understand; the paradox will be resolved -- not when someone gives me an answer I don't understand. It has been 15 years since I took Physics -- I'm rusty (to say the least). It has been two years since E&M.

Peace.

 

[Dale]

I don't get it. You feel that it resolves even as you look at it different ways. So what specifically is still missing? Just the boundary condition? Well, http://www.springerlink.com/content/q70853654x88pv47/" that I found, but since I don't have free access I don't know what level it is written at.

 

[andrewr]

I don't get it. You feel that it resolves even as you look at it different ways. So what specifically is still missing? Just the boundary condition? Well, http://www.springerlink.com/content/q70853654x88pv47/" that I found, but since I don't have free access I don't know what level it is written at.

Perhaps, If I were to re-created the thread, I would call it the radar trap "dilemma".
A lemma is a small proof -- di, indicates half asked / or two ways. (formal definition in Greek may be different...)
There are two solutions -- they resolve into distinct answers which do not agree -- but each conserves energy.

Either proof, taken by itself, solves the problem -- but both answers are not correct.

Consider the "twin paradox". In the domain of SRT acceleration is not accounted for.
That is how the "paradox" arises. As soon as Einstein "solved" (and since he "solved" it why did he bother to print it as a paradox in the first place?) it by breaking the symmetry one solution presented itself.

There is still a dilemma, though, and it is very deadly for people calculating SRT problems.
The twin paradox shows that calculations that do not know the history of acceleration can be in error.

Gedanken illustration #1:
The two people doing the twin paradox experiment have Alzheimers disease. They do not know who "really" accelerated. How do they determine which item to adjust for doppler shift -- and which one to mark as the "real" clock ?

Gedanken illustration #2:
Pooh poohing #1, (the could have written it down) -- let's say the experiment took a long time. The original experimenters made the big bang in the universe. The new generation of experimenters are on a spaceship called earth. How to they correct for Doppler shift?

So, even though Einstein "solved" the problem -- he didn't really resolve it in SRT. He gave a *plausible* answer -- that's all.

One can always imagine for any given twin paradox scenario a history for which the observers are ignorant.

Along the exact same lines of thought: (Anectdotal humor).

#1
I came out of my physics class thinking that Doppler shift was an audio phenomena and did not apply to light because of how my teacher presented the class on the inferometer. Michaelson inferometer was the death knell. (oops).
I was able to work all the problems in class... boy, when my brother got a speeding ticket, was I ever surprised. (Felt stupid too.)

#2
I went to the TA to ask about the fixup of time thing. Eg: in another frame of reference how does one come to the conclusion that the separation speed is the same. I went back to my notes; Guess what the TA told me to do: "Just don't do that".

He indicated that one can work around the problem by codifying that in "my frame of reference" the other person's measurement of their clock is affected by Doppler shift. (DAMN I thought). What difference ought it make, since there is no ether, that my view of their clock is distorted by Doppler shift?
It's THEIR calculation and view which counts.
Are they NOT justified in claiming their clock is as good as mine?

Obviously not (if the TA's explanation was right). Their calculation of my speed depends on my view of them.

As I said at the beginning of the thread -- I am exploring -- climbing mountains (metaphorical) to see what can be seen. I am adamantly refusing to work in a three reference frame system for these first steps in order to grasp what is knowable in the simplest possible system.

I *DO* know, that any answers I get from working the problem in a single frame of reference will be indisputable.

I think what happened a few posts back is indicative of a stumbling block that you are bringing into this thread -- which was meant to be light-hearted, exploritory, and not dogmatic. I am not certain, but I get the feeling that when you read my post -- you thought that the only way to calculate the change in speed for an object at rest -- is with respect to the rest frame. That is, you only know of a formula that pretends a fixed third reference frame that initially coincided with the mirror -- and that you thought I meant for you to calculate the speed change with respect to that -- and that THAT number and add it to the 1/2c. Rather than to take the mirror at rest and the policeman as moving in analogy to the classical wall and ball experiment, where I "rode along" with the elastic collision of the ball.

I note that in classical mechanics: E = 0.5 * m * v(t)**2. When one has the integral of something, they already have the derivative accounted for. The change of sign in the classical equation is due to Energy being a square of velocity -- that is, negative signs get *lost*. Taking the derivative of E(t) (above), gets conservation of momentum. The exact same process can be applied to relativistic mechanics. If one carries the sign around for energy (consistently) one does not have to resort to conservation of momentum -- as it is already accounted for.

Now, I graduated with honors before the anxiety sickness set in. Give me a little credit. I mean what I say; when I indicate that a third reference frame is a mistake -- I meant it.

Let me not do to you, what you seem to be doing to me:
Am I understanding you correctly, when I surmised, that you believe all the energy from the returning (reflected or RE-transmitted photon) goes into acceleration of the mirror; and that no extra photons could be produced?

eg: From the literature I note that Compton scattering never speaks of multiple photons, but that there are references in physics literature to a *single* electron emitting multiple photons -- so that the process is not inconceivable. BUT -- In your opinion, the problem at hand (which is neither of those in the literature) only a single photon is ever reflected when a single photon strikes?

Explicity: in the case of a mirror moving 1/2C -- whose mass is 1Kg (or ~1.15Kg relativistically) -- that the 66% of the original photons' energy totally and ONLY goes to accelerating the mirror. It never goes to generating another ~ two photons?

If so, I would like to know -- for I take the opposite view as a hypothesis and recognize that the hypothesis could be decided either way.
As to the boundary condition: I already have solved for it once; See a few posts back.
Now that I have an idea what it looks like, I can use the techniques learned in E&M to arrive at it from totally undergraduate course material.
It will just take time. I have a lot more to cover to get to gravity; and was hoping that of the 1000's reading this, perhaps one would have a hint.
If you all don't want to enter the fray because of the unintentional intensity -- my e-mail is open too. andrew.pub at sophistasis.com
Thanks.

 

[Mentz114]

andrewr,

Gedanken illustration #1:
The two people doing the twin paradox experiment have Alzheimers disease. They do not know who "really" accelerated. How do they determine which item to adjust for doppler shift -- and which one to mark as the "real" clock ?

What is a 'real' clock ? If you're using imaginary ones, it's no wonder your calculations are wrong. There are no corrections to be made. You haven't understood the reason for different elapsed times.

 

[andrewr]

andrewr,What is a 'real' clock ? If you're using imaginary ones, it's no wonder your calculations are wrong. There are no corrections to be made. You haven't understood the reason for different elapsed times.

Not at all. Putiing "real" in quotes does not make an imaginary clock.
Though, Einstein's clocks in the paradox were in fact imaginary -- was Einstein wrong?
The "if" fails.

Each observer has the perfect right to believe his own clock is correct. (it is). However, he does not have the right to judge the other's clock without knowledge of who accelerated. BUT one of the external observers -- is correct -- one of the clocks is running slower. Aircraft experiments with atomic clocks proved that point some time ago. however, one of the observers is viewing an illusion -- Doppler shift gives an appearance of a slower clock even when the clock isn't running slower.

In my last post, where I spoke of the antecedents -- don't judge my belief now based on what someone did to me in the past. If you think you have spotted an actual (tangible & uncorrected ) error -- I'd appreciate your input. If you accused (hypothetically) in error; let us know.

Thanks for the earlier equation, it helped.

 

[andrewr]

Dale, (&all)

I think I see how the calculation of a moving object from Maxwells' equations could be construed to return 1/9th the power -- for 3x the time. I just did the sine wave -- I'm pretty excited here. ( That's not to say I don't see a possible problem, just that I finally understand where the idea comes from and am more open to it now.)

Now, to the general question -- What Experiments have been done to verify or disprove this result?
Does anyone know of a physical experiment which has been done that shows when an object increases in velocity that it becomes more able to adsorb the energy of a photon? ( For that is what the prediction would mean ).

Does anyone know of a QM problem where this result is used?