The doppler radar trap paradox on the path to gravity.

In summary, the police man fires a low wattage continuous wave radar gun at a craft passing him and measures the crafts speed with a beat frequency that he measures with the incoming and outgoing waves. The craft is made of a superconductor and acts as a perfect mirror, so the police man can measure the crafts speed accurately. The police man's original calculation of the crafts speed was incorrect because the photons have energy and the craft's distance from the reflector was 1.7329 meters.
  • #71
DaleSpam said:
Your conclusion that there is any missing energy is incorrect.

I didn't conclude that. I asked where it went. A question is NOT a conclusion.
I notice you didn't quote me -- because you can't.
 
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  • #72
DaleSpam said:
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?!
 
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  • #73
andrewr said:
I didn't conclude that. I asked 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:
andrewr said:
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:
andrewr said:
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?
 
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  • #74
andrewr said:
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.
andrewr said:
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.
 
  • #75
DaleSpam said:
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.
 
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  • #76
DaleSpam said:
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.
 
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  • #77
andrewr said:
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?
 
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  • #78
DaleSpam said:
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.
 
  • #79
OK, so you agree that energy is conserved in all frames with both the "photon" and the Maxwell approach.
andrewr said:
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?
 
  • #80
DaleSpam said:
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?
 
  • #81
DaleSpam said:
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.
 
  • #82
andrewr said:
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.
 
  • #83
DaleSpam said:
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 )

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 )
E' = γ( E - | μ0*E/377 |*v )

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
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...
 
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  • #84
andrewr said:
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.

andrewr said:
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.
 
  • #85
As I said to you at the start; In addition to telling you that I NEVER BELIEVED that energy was non-conserved...

andrewr said:
Ηi Dale!

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

The 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.
 
  • #86
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.
 
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  • #87
DaleSpam said:
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.
 
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  • #88
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.
 
  • #89
Mentz114 said:
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.
 
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  • #90
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?
 
  • #91
andrewr said:
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.
This is what I was asking earlier. In what specific way do they not agree? Why do you think both answers can not be correct?

andrewr said:
In the domain of SRT acceleration is not accounted for.
This is a pretty common misconception. SR has no problem with acceleration, it just has trouble with non-inertial reference frames. In other words, you can describe accelerating objects from an inertial reference frame just fine with SR, particularly using four-vectors.

andrewr said:
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.
The frame you are describing here is non-inertial, so none of the normal rules of physics apply in it unless you use GR. E.g. there are time-varying fictitious forces etc. Also, by definition, the change in speed of the mirror is 0 in this frame.

andrewr said:
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?
...
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.
Yes, for a perfect conductor I believe the number of incident photons equals the number of reflected photons in all frames. Otherwise the energy balance would be incorrect and different frames would disagree on the number of ejected electrons via the photoelectric effect.
 
  • #92
andrewr,
BUT one of the external observers -- is correct -- one of the clocks is running slower.
The elapsed time on clocks measured from event 1 to event 2 depends only on the proper length of the clock's worldline between those events.

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.
What observers see is obviously dependent on their frame at the time. Given the full details of each others journeys they can both predict the outcome.

Using the non-paradoxical 'twin paradox' in the context of this thread is just sowing confusion and is irrelevant to your problem, which merely requires stating in a frame-independent way.
 
  • #93
Mentz114 said:
andrewr,

The elapsed time on clocks measured from event 1 to event 2 depends only on the proper length of the clock's worldline between those events.

In the original document from Einsten (pre relativity, but showing the development of his thought) -- the one entitled "On the Electrodynamics of Moving bodies"; world-lines were not discussed. In his later work, which I read some years back -- Although compatible with Relativity, world lines do not represent the theory itself but merely are a suggested framework to keep the ideas straight which he had developed in the preceding chapters. In my book it was more or less an appendix to the principle which he views as consequences of the constancy of the speed of light, and effects deduced because of this constancy.

The proper length of time in either frame of reference is equal to the number of ticks occurring in that frame of reference on the clock (belong to that same frame) -- between measurements. It has nothing to do with an outside observer.
There is a correction that one can make for items which are spaced far enough apart to cause a significant delay between when one measures an event, and when it happened.

When speaking about the same set of events from another frame which is moving relativistically, the said correction is always required. This is not a source of confusion to me:

With respect to measuring each other's velocity -- (and not a third velocity):
So long as distances and times are referred to the same frame of reference; the result will be the same -- Eg: a slower running clock will elapse less time during an event -- and since, for a clock to run slower, any "scale" in the same frame of reference must be shorter -- The Lorentz contraction in length of a measuring stick is canceled by the reduction in the rate of the clock. The other frame being, of course, the variable causing the event being measured.

What observers see is obviously dependent on their frame at the time. Given the full details of each others journeys they can both predict the outcome.

I never said otherwise: But -- in preparation for a future discussion -- many real problems do not HAVE the full details of each other's journey available. In fact; in many real experiments there is no intelligent observer on the other item to have a perception. In such situations; one can't say with certainty what is a doppler shift (only) and what is real about the clock in the "other" frame running slower.
This is a question very similar to that of the EPR / Bell inequality. (I tend to side with Einstein -- but it is purely philosophical at this point.)

It is not always possible to get the other frame of reference to accelerate back toward's "us" to allow the symmetry to be broken. This may not apply to the experiment at hand -- because I can define who accelerated (eg: although I have not done so, previously, I will specify that it is the Mirror -- and now no important questions about the proper way to do the equations can confuse the issue further.)

Using the non-paradoxical 'twin paradox' in the context of this thread is just sowing confusion and is irrelevant to your problem, which merely requires stating in a frame-independent way.

Are you trying to be funny or just stupid? In the last two posts you have made unqualified remarks which indicate that Einstein's choices are bad. If you are trying to encourage belief in Einstein's credibility; why is it that you make statements which tear him down in order to get at me?
Einstein himself gave the name "paradox" to the twin situation. Einstein himself used purely imaginary clocks (by definition: a Gedanken Experiment). If using a non-paradox as a paradox is "sowing confusion" -- Einstein is more guilty of it than I am. I am merely following in the master's footsteps.

Your objection, and many of Dale's, to be blunt -- have nothing to do with the thread at all. They are purely aesthetic questions. I will remind you, that I am the one who started the thread to explore a question -- I have clarified ad nausium the use of "paradox" -- and I have had to deal with criticism of the word "Paradox" which is NOT a source of confusion to me in the slightest. I would like to have you two stop beating a dead horse; at least Dale appears to be partially focusing (now) on items of "real" value -- like the photon issue regarding the photo-electric effect. I am not interested in doing "world" lines; Minkowsky vectors; or other items which --- also to be blunt --are not NECESSESARY to solve the problem I have which deals with Einsten's theory as he put it forth.

I can understand you have a preference for these later mathematics as "clearer" in your mind. You are more than welcome to use them *yourself* in the privacy of your own home or with others who are interested. Stop ripping on me about them, OK?
 
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  • #94
DaleSpam said:
This is what I was asking earlier. In what specific way do they not agree? Why do you think both answers can not be correct?

To give a specific example will take a bit of time -- I was too sick today to think clearly -- but if I didn't mess up; I have a VERY clear example by inequality. So, assuming I will be able to post it some time tomorrow or Tuesday:
Let me start be restating things which I think we agree on -- and include something which contains (non-obviously) the crux of the issue; but which I want to make certain we agree on before I use it to describe a very specific case.

1. The problem consists of two inertial frames.
a, The fact that we have accelerations caused by photons does not impact 1 very much.
(Einstein ignored acceleration by photon for many of his Gedanken.)
b. I received a perceptive e-mail about (a) -- which correctly asks why I am ignoring the acceleration caused by the radar gun shooting out the photon in the first place. (recoil).
The reason I ignore it, is because the (adjusted for free mirror) problem statement takes place with the photon already in flight and the pre-flight conditions (which are different) are assumed to be exactly canceled by ejection, eg: ejection creates the proper conditions at time of flight. (I apologize to anyone else this may have confused, No one else asked.)

You make a further comment on this issue which I'll address in a separate section below.

2. In Post #27 I derived for a rest frame mirror, with an identical rest frame radar gun (eg: relative velocity=0)
v = 1/sqrt( c**-2 + (m0*c/(Eph*2))**2 )
You checked it and see no problem.

3. Equation #2 says that as mass increases, the amount of energy transferred by a single photon increases. That is; the lower the frequency of a photon or the greater the mass of the mirror -- the less percentage energy the photon is capable of transferring to the mirror.

4. Derivation #2, according to your interpretation, can not be used in the experiment with the separation velocity initially at 1/2C (eg: not just any frame, but see assertion #6) -- although I see the magnitude of the change in percentage as inexplicable. This is something I need to show clearly -- so please don't jump the gun and try to answer my objection until I do so. In order for me to accept your interpretation, I will have to understand how it is that a lower energy photon all the sudden becomes more absorbed due to velocity of the target -- causing a reversal of the trend expressed in equation 2.

5. I agree to your assertion that the Doppler equation -- as we have discussed it: eg: when expressed in non-instantaneous integral/differential form which is too difficult for this thread, but would circumvent the issue -- must insert two different velocities to predict the final color of the photon after reflection. I understand your assertion to be: the initial velocity of separation requires one relativistic Doppler shift to predict the color at arrival. The color of the return photon must include the final velocity of the mirror *after* the entire momentum transfer has taken place -- that is 2x the momentum of the incident photon when taken in the appropriate frame.

6. here's where we might disagree, but I think a consensus can be reached.
In the case of the problem given, with a free mirror which is moving at time of flight of the photon, with a velocity of 1/2C -- an acceleration will occur during reflection. The proper way to do the problem (in my mind) is to use the straight forward Δm * c**2 = ΔKE. Either frame can be taken to be at rest so long as the proper values are used. Eg: Similar (in fashion) to the equation I gave before concerning riding along with a ball which hits a wall -- one can arrive at a relativistic equation where the difference of the gamma's of the before and after velocity are computed -- From the new gamma, a new velocity between the police and the mirror can be computed correctly. -- That computation was not explicitly shown, and rather than fight over its' derivation -- I will instead focus on an ERRONEOUS computation which I warned about in the post where I meant the above equation -- for a moment.

If one mistakenly introduces a third reference frame at rest with respect to mirror before impact; and then computes a velocity change with respect to that extra frame -- eg: by using equation in assertion #2 which is valid for the velocity with respect to that third frame only --
one will necessarily compute a delta velocity which is too big (and to be anal -- so long as the velocity is nonzero) with respect to the velocity that the police man measures..

This is due to the nonlinear and monotonic nature of the lorentz transform when accelerations (and therefore incremental velocity) are monotonic: eg: the mirror is ONLY accelerated in one direction -- that which increases the speed of the mirror.

Qualitatively: My only assertion previous to this post, is that if one increases the velocity of the mirror in the same direction -- a larger increase in velocity before the transform is made, must equate to a larger transformed velocity.

That is: given base velocity v, and delta velocities computed with the "third" frame, say deltas a and b where a > b , though they all (v,a & b) share the same sign; The transforms of these velocities through relativistic addition must be a' > b'. When these include the base velocity in the sum, and a' contains delta a, etc. This is a very narrow point and ought to flow naturally from the relativistic addition of velocities which you have mentioned.

If any of these points are confusing for some reason, please ask clarifying questions -- it will perhaps make the conversation remain civil.

;--------------------------- To the unanswered part (and separate ideas) in the rest of your post ----------------------

concerning acceleration:
This is a pretty common misconception. SR has no problem with acceleration, it just has trouble with non-inertial reference frames. In other words, you can describe accelerating objects from an inertial reference frame just fine with SR, particularly using four-vectors.

I don't think I misconceived anything ( or perhaps I missed what you said, was it subtle? )--
NON inertial are accelerated frames. The distinction you make is artificial in that you are assigning a history to the frame from which the other is measured.

In the physics forums stickys -- such accelerations (which I use in this thread) -- are justified for SRT. I was aware of theat before I made my statements. But the problem I am highlighting, is (in fact) the very reason Einstein wrote about the so called "twin paradox" after the theory of relativity was already published; a bit of an academic no-no for technical purists (I am told). For, what that does is indicate the theory is non-rigorous, and one can edit by omission what the theory means.
Because of this "loose" definiing, one can speak of accelerations being or not being part of SRT.
What is certain, is that effects of acceleration are NOT EXPLICITLY AND COMPREHENSIVELY SPELLED OUT as to how they should be dealt with in SRT as originally published.

Case in point; I nowhere see in relativity where Einstein limits the energy transform of elastic impact to the one equation (and example) he works in the 1905 paper already mentioned in this thread. If Einstein does not explicitly deny a possibility, he may allow it. His theory, at best, is ill defined in that Einstein had (at least once) to make a clarifying statement about what is and is not part of his theory -- at most, one will end up arguing that in most of his thought experiments, the lab frame can't possibly be inertial because photons accelerate and decelerate the frames -- if but (MICROSCOPICALLY).

I sometimes wonder if this is why he never got a nobel prize for it, although he did get one for the photoelectric effect...And then (as I have anxiety problems anyway, and have to reject conspiracy theory actively) I sometimes wonder if the reason Einstein did not make certain clarifications may have had to do with world war II -- and top secret information which would have helped the Nazis(later russians...) make the bomb if he published it. etc.
Or most likely, perhaps he just hated lawyers of trivia -- which is why minute details of quantum theory bit him so much...

As to the conclusion of your statement and four vectors -- it is irrelevant to whether or not my computations are correct. I have (in the previous post) made a statement which considers the implications of the twin paradox for my Gedanken.

I don't want an argument of mere words to continue. The horse is dead.

And lest someone try to resurrect another dead horse -- My statement concerning Max Plank and the RJ (ultraviolet catastrophe) is not a misconception either. I know full well that Plank had no idea he was solving a problem formulated by someone else -- it is irrelevant, he still solved the problem. etc.

The frame you are describing here is non-inertial, so none of the normal rules of physics apply in it unless you use GR. E.g. there are time-varying fictitious forces etc. Also, by definition, the change in speed of the mirror is 0 in this frame.

Yes it is zero -- and zero is a perfectly valid velocity. If you go back to my classical physics problem/analogy -- it too has a zero velocity. I omitted (set to zero) the momentum in the rest frame.
This leads to an (apparent) violation of the conservation of momentum -- but not a real contradiction.
The normal rules of physics still apply, with respect paid to the changes of one's frame during the first and second half of the computations -- not unlike your four-vectors, but without all the extra hoopla.
I have explained the nature of energy (classical) which contains a *square* of velocity -- hiding the sign of momentum -- such that if energy is truly conserved, momentum is already conserved -- the switch of sign notwithstanding.

As a side note: one could also derive a "law" of conservation of velocity for classical elastic collisions. The velocity, abs(v), of the objects coming together must be equal to the velocity of the objects separating. This law is valid for any two perfectly elastic objects wherein a single velocity magnitude is definable... but hey, most physicists don't bother to stop and smell the mathematical roses... (That's off the top of my head from 16 years ago...I wonder if my memory is still accurate...)

Yes, for a perfect conductor I believe the number of incident photons equals the number of reflected photons in all frames. Otherwise the energy balance would be incorrect and different frames would disagree on the number of ejected electrons via the photoelectric effect.

I am not sure I see how such a disagreement would arise -- although your thought here is exactly the kind of argument which might be used to decide the question I have posed. I have come across experiments on Compton effect (which is/is similar to photoelectric effect, but with momentum calculated and perhaps a disparity in the valid target material) that allow *multiple* electron ejections.

I will note: Even in the case where only one photon is allowed -- the energy changes depending on reference frame. In order to predict electron ejection, one has to take into account the velocity of the target as well as the color of the photon hitting it. Such effects occur regardless of the number of photons involved. A photon, incapable of ejecting an electron -- could eject it if the target happened to be moving toward the photon in the frame that the color is known.

As to the multiplicity of photons ejected -- again: I see in the literature where Compton scattering can eject more than one electron from a single photon. etc. (I am aware of objections to these experiments which also might nullify their defense on my behalf -- but they are inconclusive at this time). The probability of an electron being ejected from metal in the photoelectric effect is somewhere in the range of 1 electron for 1000 photons (special silicon or alloy detectors are required to increase the ejection efficiency, but even then it is 1 in 3 photons).

What I am getting at -- is that even in standard experiments such an effect as we are discussing would not readily lend itself to easy detection. Considering that the multiplication of photons could be reversed depending on the target velocity -- I see no specific evidence in the the thought you are providing at this point (I am open to further inquiry).

I will propose a thought experiment, hopefully tomorrow, which will perhaps bring the paradoxical nature of the question into clear relief which can be decided for or against the proposition I have made. I do this, as no experiments clearly proving the point are yet forthcoming.
 
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  • #95
andrewr said:
To give a specific example will take a bit of time -- I was too sick today to think clearly -- but if I didn't mess up; I have a VERY clear example by inequality.
OK, I look forward to the example.

andrewr said:
1. The problem consists of two inertial frames.
I disagree with this although I have not made a big deal of it. Any problem can be expressed in an infinite number of inertial reference frames. There is no requirement that a reference frame be tied to an object. For example, one reference frame which is particularly convenient for this kind of elastic collision problem is the "center of momentum frame", and in this frame there is no single object at rest.

However, the reason that I have not made a big deal of this is that I don't mind restricting the analysis to any specific inertial frame(s) you wish. Just carefully specify which two inertial frames you wish to consider and I will gladly restrict my analysis to those also. As long as you realize that there is no reason that the analysis could not just as well be done from any other inertial frame.

andrewr said:
2. In Post #27 I derived for a rest frame mirror, with an identical rest frame radar gun (eg: relative velocity=0)
v = 1/sqrt( c**-2 + (m0*c/(Eph*2))**2 )
You checked it and see no problem.
As you mention here, this is a specific case, not a general equation.

andrewr said:
3. Equation #2 says that as mass increases, the amount of energy transferred by a single photon increases. That is; the lower the frequency of a photon or the greater the mass of the mirror -- the less percentage energy the photon is capable of transferring to the mirror.
I believe you meant "decreases" where I highlighted in red if so then: Yes, in the inertial frame where the mirror is initially at rest.

andrewr said:
4. Derivation #2, according to your interpretation, can not be used in the experiment with the separation velocity initially at 1/2C (eg: not just any frame, but see assertion #6) -- although I see the magnitude of the change in percentage as inexplicable. This is something I need to show clearly -- so please don't jump the gun and try to answer my objection until I do so. In order for me to accept your interpretation, I will have to understand how it is that a lower energy photon all the sudden becomes more absorbed due to velocity of the target -- causing a reversal of the trend expressed in equation 2.
Correct. And, though it almost causes me physical pain, I will not "jump the gun". :smile:

andrewr said:
5. I agree to your assertion that the Doppler equation -- as we have discussed it: eg: when expressed in non-instantaneous integral/differential form which is too difficult for this thread, but would circumvent the issue -- must insert two different velocities to predict the final color of the photon after reflection. I understand your assertion to be: the initial velocity of separation requires one relativistic Doppler shift to predict the color at arrival. The color of the return photon must include the final velocity of the mirror *after* the entire momentum transfer has taken place -- that is 2x the momentum of the incident photon when taken in the appropriate frame.
Excellent, I am glad to see that is a point of agreement now.

andrewr said:
6. here's where we might disagree, but I think a consensus can be reached.
In the case of the problem given, with a free mirror which is moving at time of flight of the photon, with a velocity of 1/2C -- an acceleration will occur during reflection. The proper way to do the problem (in my mind) is to use the straight forward Δm * c**2 = ΔKE. Either frame can be taken to be at rest so long as the proper values are used.
I believe you understand this already, but just for clarity the acceleration occurs in all inertial frames, not just the one where the mirror is initially moving at 0.5 c.

However, we do have a different substantive disagreement here. The proper way to do the problem is most definitely not to simply apply a formula derived at rest. When you derive an equation at rest then all of the terms with a velocity drop out. That does not mean that they do not exist in general, simply that they are 0 for this specific case. Also, because the equation was derived at rest it is not clear if the mass in the equation is the rest mass or the relativistic mass. I suspect it is the invariant rest mass which does not increase with increasing speed.

Earlier you gently chastized me for not immediately identifying my first Doppler equation as being a special case for when the mirror was initially at rest. So, now in a similar vein I gently chastize you for using the special case formula in a general situation where it simply does not apply. You must either derive the equation for a general velocity (in which case it should reduce to your specific case in the limit v0->0) or do a proper transformation to a moving frame (in which case you should get the same answer as if you had derived it for a general velocity).
 
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  • #96
andrewr said:
What is certain, is that effects of acceleration are NOT EXPLICITLY AND COMPREHENSIVELY SPELLED OUT as to how they should be dealt with in SRT as originally published.

This is a given, Acceleration is not fully covered by SR, this is why it is called Special Relativity, it assumes the special case of flat spacetime.

Acceleration violates the principle stating that observers in uniform relative motion can not distinguish if the other is at rest, I'm not sure why this is such a hang up, but I see it brought up time and time again.

The first hints that SR wouldn't be able to fully model acceleration were included in the very end of the 1905 paper "On the Electrodynamics of Moving Bodies", section 10. The case of the slowly accelerated electron reveals the difficulties of trying to model acceleration in terms which SR deals with naturally.

You CAN take a timesliced approach where you insert the new values by hand at each new slice, but it is very inelegant and contrived, this was one of the major motivations behind producing GR for Einstein.

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

Einstein said:
10. Dynamics of the Slowly Accelerated Electron

*snip*

With a different definition of force and acceleration we should naturally obtain other values for the masses. This shows us that in comparing different theories of the motion of the electron we must proceed very cautiously.

We remark that these results as to the mass are also valid for ponderable material points, because a ponderable material point can be made into an electron (in our sense of the word) by the addition of an electric charge, no matter how small.

*snip*

andrewr said:
Case in point; I nowhere see in relativity where Einstein limits the energy transform of elastic impact to the one equation (and example) he works in the 1905 paper already mentioned in this thread. If Einstein does not explicitly deny a possibility, he may allow it. His theory, at best, is ill defined in that Einstein had (at least once) to make a clarifying statement about what is and is not part of his theory -- at most, one will end up arguing that in most of his thought experiments, the lab frame can't possibly be inertial because photons accelerate and decelerate the frames -- if but (MICROSCOPICALLY).

Einstein said:
All problems in the optics of moving bodies can be solved by the method here employed. What is essential is, that the electric and magnetic force of the light which is influenced by a moving body, be transformed into a system of co-ordinates at rest relatively to the body. By this means all problems in the optics of moving bodies will be reduced to a series of problems in the optics of stationary bodies.

More specifically, in the opening portion of the paper, and at several points throughout, including the 4th section of the paper: 4. Physical Meaning of the Equations Obtained in Respect to Moving Rigid Bodies and Moving Clocks, he specifies the consideration of rigid bodies. I must be missing the reference to elastic collisions which you keep bringing up, would you be so kind as to point it out to me?

Regarding rest mass and relativistic mass, just an example of their less than obvious relationship comes from the quark model. The masses of the quarks themselves are insufficient to give the observed mass for protons and neutrons, unless they are in motion relative to each other, gaining enough relativistic mass to give the observed values.

"All mass is interaction." as Feynman put it so well.
 
  • #97
Max™ said:
Acceleration is not fully covered by SR, this is why it is called Special Relativity, it assumes the special case of flat spacetime.
You can certainly have acceleration in flat spacetime. SR handles acceleration just fine from an inertial reference frame in flat spactime via the four-acceleration. See: http://www.mth.uct.ac.za/omei/gr/chap2/node4.html or do a quick Google search on "four acceleration" for other links.
 
  • #98
Oh, I know this very well.

I was simply trying to clarify to andrewr that his objection to SR (which apparently is based on it's omission to some degree of considerations of acceleration) is not based on the 1905 paper as he seemed to be stating it was.

It handles acceleration like I handle juggling knives... sure I can do it, I can juggle golf balls for hours, but in general I'd like to leave the knifeplay to an expert.

The expert in the case of acceleration is General Relativity, which produces ways to describe and calculate it without having to stress the case, as you do with SR.
 
  • #99
Max™ said:
It handles acceleration like I handle juggling knives... sure I can do it, I can juggle golf balls for hours, but in general I'd like to leave the knifeplay to an expert.

The expert in the case of acceleration is General Relativity, which produces ways to describe and calculate it without having to stress the case, as you do with SR.
Agreed, but I would identify the required level of experise as four-vectors rather than general relativity which I would reserve for non-inertial coordinate systems or curved spacetime. So maybe GR is juggling knives and acceleration in SR is juggling those stage torches that burn bright, but at a low temperature that doesn't really hurt to touch. Some skill is required, but not as much as the audience thinks.
 
  • #100
True, I'm probably just odd because I began reading up on GR first after hearing about Black Holes, then later learned about SR as a flat spacetime approximation of GR. So it seems counterintuitive to me to work strictly in SR.
 
  • #101
Hi Max,

I expect to get into GR shortly -- I just don't wish to do so for this first part.
I have mulled the word-play over in my mind, and I think I could accept this view:

Einstein's theory of relativity matured to some degree after he wrote his paper in 1905.
In the books that I read, the objections to minor details -- such as photons accelerating the reference frames, etc. Were dealt with because he was writing to the general public in those books.
Eg: The ones co-authored by Leapold, Infield, etc. (spell?)

I have a note in my book -- must be 20 years old -- regarding the twin paradox and acceleration which supports what Dale says in this way: If Einstein hadn't published the twin paradox, the mathematics in this section appear to be able to correctly calculate the results anyway...
I am not going to quote the rest of the note, but essentially to many -- the twin paradox must be something of a letdown from a lawyer's standpoint.

Minowski and four vectors, though, must be a bit anachronistic -- for I don't see them in the 1905 document, and they appear in the earlier editions of the books as an appendix; In the newest edition of the book entitled just "relativity"; any distinction is lost.

I suppose, this is more of a historical question than a science question -- and it is also more of an argument of degrees rather than of definite yes/no. There is nothing in the 1905 paper to stop one from calculating the first half of the twin paradox as if the accelerated clock is really ticking slowly -- and the second half as if the non-accelerated clock is ticking slowly but the accelerated one is not -- just suffering from doppler shift when read externally.

I do get Dale's point that acceleration itself (so long as I don't extend it to effects on a clock) must needs be calculated properly in SRT.

Perhaps, if Einstein had a better editor ... the twin paradox would have been phrased (and titled) differently; and the present argument could be cleanly decided in favor of the four vectors and grant Einstein a greater victory earlier as to what he did right and when. But history is written.

I do see, looking around at the other "paradoxes" on the website -- that to the average person, the word is quite confusing; I would hesitate to say the word itself causes the confusion -- rather I tend to think that Relativity challenges people's intuition in the first place -- and entices the more black and white / less learned to create stumbling blocks for others. It is a fact of human psychology, that some people feel smarter when they can make others look dumber. That is not what I intended in my thread, and as I noted earlier -- If I started over, I might have renamed it. I more or less invited Dale and others to categorize me in the same way that many other paradoxes on these boards clearly are.

I still believe it bad form to attack a thread on a superficial basis -- wrong is wrong -- but I can understand the frustration which underlies those trying to combat self breeding ignorance.

Have any questions popped up about the #1 - #6 yet, or are they acceptable for the next stage of the question?
 
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  • #102
Actually the 1905 paper wouldn't have had you calculate the clock as if it were accelerating, simply you would have moved the clock along the path from point A to point B at a uniform velocity, and upon arriving next to the rest clock it would have incurred a certain amount of time dilation. Which is trivially unphysical, objects don't just instantly launch off at set velocities without periods of acceleration.

The theory wasn't well suited to describing the full situation, so he strove to restrict it to situations where it was valid until he could produce a more suitable theory for all general situations.

You are right though, they aren't paradoxes upon investigation, merely upon a cursory reading do they appear to be such.

I feel smarter when I help others feel smarter, simply demonstrating knowledge is trivial, imparting knowledge to another shows a true understanding of the subject, teaching is it's own reward... luckily since it doesn't pay well.
 
  • #103
DaleSpam said:
OK, I look forward to the example.

Working on it now.
I believe you meant "decreases" where I highlighted in red if so then: Yes, in the inertial frame where the mirror is initially at rest.

AYE -- a typo. I knew I wasn't thinking too well -- but missed it.

However, we do have a different substantive disagreement here. The proper way to do the problem is most definitely not to simply apply a formula derived at rest.

I never suggested otherwise. There are two ways it could be done with the mirror at rest -- one is to treat the mirror as at rest (although still keeping track of time -- if it becomes important -- such that the mirror is really the one which accelerated) and calculate the energy change such that the difference in gamma with respect to the police man falls out, so that one may compute the new difference in velocity.
Take the difference in velocity, and add it. (The derivation is convoluted, and since we're having difficulty with simple things, ... I can't do your version without deriving it, so I might as well show how to do it.)

The second way, which I suspect you improperly assumed I was doing half asked -- is to calculate the velocity change of the mirror at rest using the formula derived at rest against a third reference frame.
I never said to do this; though! as there is no third item!
Then, the velocity change computed must be added using the relativistic addition formula.

When you derive an equation at rest then all of the terms with a velocity drop out. That does not mean that they do not exist in general, simply that they are 0 for this specific case. Also, because the equation was derived at rest it is not clear if the mass in the equation is the rest mass or the relativistic mass. I suspect it is the invariant rest mass which does not increase with increasing speed.

Again -- I gave a classic example early in the thread where this is worked. The mass at the other end I did not define -- but it is irrelevant. Whatever the mass of the wall -- it is constant in that problem and it works out in general -- that for whatever mass the wall had, energy and momentum are conserved.
It just isn't a plug and chug without using the brain problem any more.

Earlier you gently chastized me for not immediately identifying my first Doppler equation as being a special case for when the mirror was initially at rest. So, now in a similar vein I gently chastize you for using the special case formula in a general situation where it simply does not apply.

Excuse me? you are falsely accusing again. Not only did I not use a general case formula or a specific one -- I never worked the problem !

So where do you get off?

All I did was say a formula existed. Do you deny that it can be worked with relativistic addition?
If so -- even you must admit that it CAN be worked -- and a formula CAN be derived ?
I'm sorry if my words were confusing to you -- but I DID say that if a third reference frame were used, the problem would be wrong. Funny how you half hear what I say -- and rather than ASK what I mean, you sort of fill in the blanks with what you thought I meant (especially when it makes an error that you can accuse me of). Try ASKING...

You must either derive the equation for a general velocity (in which case it should reduce to your specific case in the limit v0->0) or do a proper transformation to a moving frame (in which case you should get the same answer as if you had derived it for a general velocity).

I am not even going to think about that...I am just going to plug in the equations I learned in undergraduate physics which I know work.

I am using bc, a gnu public license program (free), to do the calculations under Linux.
Most likely this functions under windows as well. I will use the # for comments, so that what I do here can be replicated by copy and paste into the calculator.

# For calculation purposes:
scale=1024 # Carry calculations out to 1024 digits after the decimal point.
c=3.0*10^8 # m/s
h=6.62*10^-34 # j/s
v0=c/2.0
f0=300*10^6 # Hz
mmir0=1.0 # kg

define gamma(v,c) {
return 1.0/sqrt( 1.0 - (v/c)^2 )
}

# ------ Step 1: The original photon energy and double momentum in the frame of the police man:
eph=h*f0 # result
# 0.000000000000000000000000198600000000000000000000000 ...
p2ph = 2.0*eph/c
# 0.0000000000000000000000000000000013240000 ...
#
# The doppler formula, applied twice, would give a return frequency of 100MHz,
# But to do this correctly -- first calculate the momentum transferred, then the total momentum,
# to get the velocity after reflection -- then do doppler to the mirror, and a second doppler away from mirror -- using v1=1/2c and v2=computed value.
# This is the way Dale says to do it, and I conceded the point -- as I think he is right on that.
#
# ------ Step 2: Get the momentum of the mirror moving at 1/2c
# KE = m0 * c**2 * ( γ - 1.0 ), TE = m0*c**2*γ
#
emir0 = mmir0 * c^2 * gamma( v0, c )
# 103923048454132637.6116467804903523420165683152286 ... joules.
# E**2 = p**2 * c**2 + m0**2 * c**4
pmir0 = sqrt( (emir0/c)^2 - mmir0^2 * c^2 )
# 173205080.756887729352744634150587236694280 Kg * m/s
#
# ------ Step 3 obtain the delta velocity (which by the way, does not use the relativistic addition formula!)
# ------ we are assuming the police man is @rest for this calculation, but it could be done from the other end after translating all appropriate quantities.
# This is from Physics, Douglas C. Giancoli -- a standard college physics text.
# p = m * v, where m is the relativistic mass.
# So, I derive --
# p = m0 * γ * v
# v* γ = p / m0
# sqrt( v**2/(1-v**2/c**2) ) = p/m0
# 1/v**2 - 1/c**2 = (m0/p)**2
# 1/v = sqrt( (m0/p)**2 + 1/c**2 )
#
dv = 1.0/sqrt( (mmir0/(pmir0 + p2ph))^2 + c^-2 ) - v0
# delta velocity = .000000000000000000000000000000000859963225957947576\
23637710855766563018710034339435398182970816529788474289521921841235\
93889082935627296156077525942014561955826753771071199507312333915490\
21766583502696181329188197570937656089972583469349098515445031611307\
42071447462594770362103995523372663589240037063914572505544232485263\
93912527920055183927652381366247587144511648491621844214689367698438 ... etc.
#
So according to my calculation, this time worked from the police mans side,
vmirror=150000000.0000000000000000000000000000000008599632259579 ...


# Now, I don't know how you would calculate your end speed -- so I would rather just allow you to provide a formula; Otherwise I can't balance energy against the return photon energy to see what is happeneing.


#But I will attempt a consistency check!
#So, according to Dale's algorithm, the color of the returned photon is computed like:
fret = f0 * sqrt( (c-v0)/(c+v0)) * sqrt( (c-v0-dv)/(c+v0+dv) )
# 99999999.999999999999999999999999999999999617794121796467743894943507307 ...
eret = h*fret
# .000000000000000000000000066199999999999999999999999999999999999999...
# ΔE = h*f0 - h*fret = eph - h*fret
de = eph - h*fret
# .000000000000000000000000132400000000000000000000000000000000000000253020291370738353541547398162299843192826856589805482653896357967644 ... Joules.
# This is the amount of Energy eg: approx 2/3rds the original photon energy -- which dale claims went into accelerating the mirror -- and which I deny based on my hypothesis.
# Test:
de/eph*100 # percent.
# 66.66666666666666666666666666666666666666679 ... Check plus. This is what I expected...
#

# Now, I am going to check to see if I can account for all ~2/3rds energy going to acceleration.

# First step, compute the final velocity based on energy transfer alone.
# v_final( Dale ) = Velocity from ( original KE of mirror + ~2/3rds the photon's original energy. )
# KE = m0 * c**2 * ( γ - 1.0 ), TE = m0*c**2*γ
ke0 = mmir0 * c^2 * ( gamma( v0, c ) - 1.0 )
# 13923048454132637.61164678049035234201656831522862283768334841876711598 ... Joules
#
# Add to this the ~2/3rds KE from the photon which is exactly de from above.
# compute the new gamma, and solve for velocity final, eg: Dale's hypothesis.
#
# KE = m0 * c**2 * ( γ - 1.0 )
# γ - 1.0 = KE/(m0 * c**2)
# THEREFORE:
# γ = 1.0 + KE/(m0 * c**2)
# ---- Second derivation, velocity from a gamma value ...
# γ = 1.0/sqrt( 1 - v**2/c**2 ) --> 1 - (v**2/c**2) = γ**-2 --> v**2 = ( 1.0 - γ**-2 ) * c**2
# v = sqrt( 1.0 - γ**-2 ) * c
#
# Quick check -- compute the gamma of some velocity, and see that we get it back.
sqrt( 1.0 - gamma( 1000.0, c )^-2 ) * c
# 999.99999 ... There's roundoff at over 900 digits down, so we're easily good enough.
#
# ------------------ Dale's hypothesis for ~2/3rds the energy of the photon accelerating the mirror ----
#
gammadale = 1.0 + ( ke0 + de )/( mmir0 * c^2 )
# gammadale ~=
#1.15470053837925152901829756100391491129520497365136486314831576407906645571697780250135323885339614326754573203429668916656692846988602...
vdale = sqrt( 1.0 - gammadale^-2 ) * c
# vdale ~= # 150000000.0000000000000000000000000000000005733088173052983841575847 ...
00000000000000000000000000005733088173052983841575847

# comparing that to my result, previous...
vmirror=150000000.0000000000000000000000000000000008599632259579 ...

# Dale's comes out smaller -- which would mean more than ~2/3rd went into acceleration...
# Energy does not balance, hmmm... I'll have to review this and try again later; must have made some kind of typo.
--Andrew.

PS. My original idea, that I had Sunday/Monday plugged all 2/3'rds into the equation rather than one corrected for the mirror's acceleration. That wouldn't be fair, or produce a usable result. The fact that the present calculation does not work out, although I never got to the comparison stage -- means we're back to square one. Have you double checked your Doppler shift formula idea in any way? eg: the delta velocity I conceded without checking -- I'll recheck my work too -- but if you are right, the above calculation ought to have yielded the same result for the final velocity -- and not a discrepancy.

I could easily be at fault here -- but I just don't see any mistakes.
The momentum is measured from the police frame, so I used the momentum of the photon in the police frame to do the calculation -- would you have done that differently?
 
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  • #104
One very minor error and we have agreement:
andrewr said:
dv = 1.0/sqrt( (mmir0/(pmir0 + p2ph))^2 + c^-2 ) - v0
Remember, a photon's momentum is proportional to its energy. So the photon's momentum only changes by a factor of ~2 in the mirror's inertial rest frame (incident energy approximately equal to reflected energy and in the opposite direction). In the frame where the mirror is moving at 0.5 c the reflected energy is ~1/3 of the incident energy (and in the opposite direction), so the change in momentum is a factor of ~4/3. So, if you use a factor of 4/3 instead of a factor of 2 you will get very close to my result.

dpph = 4/3*eph/c
dv = 1.0/sqrt( (mmir0/(pmir0 + dpph))^2 + c^-2 ) - v0 = 5.73E-34 m/s

I know you don't like the four-vectors but just for your reference:
The four-momentum of a photon traveling in the ±x direction is (hf/c,±hf/c,0,0).
The four-momentum of a mass m traveling in the ±x direction is (γmc,±γmv,0,0).
So, conservation of the four-momentum gives:
(h f0/c, h f0/c, 0, 0) + (γ0 m c, γ0 m v0, 0, 0) = (h f1/c, -h f1/c, 0, 0) + (γ1 m c, γ1 m v1, 0, 0)
Which gives 2 linearly independent scalar equations. Since h, c, m, f0, v0 are given and since γ0=(1-v0²/c²)-1/2 and γ1=(1-v1²/c²)-1/2, the only unknowns are f1 and v1. Solving the two equations in two unknowns we get:

dv = v1-v0 = 5.733E-34 m/s
df = f1-f0 = (-2.000E8 - 2.548E-34) Hz

Which in turn give:
deph = h df = (-1.324E-25 - 1.687E-67) J
demir = (γ1-γ0) m c² = (1.324E-25 + 1.687E-67) J
dpph = h/c df = (-1/3 2.648E-33 + 5.623E-76) Ns
dpmir = γ1 m v1 - γ0 m v0 = (1/3 2.648E-33 - 5.623E-76) Ns

Thus conserving energy and momentum as expected.
 
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  • #105
DaleSpam said:
One very minor error and we have agreement:Remember, a photon's momentum is proportional to its energy.

That's p=E/c, OK...

So the photon's momentum only changes by a factor of ~2 in the mirror's inertial rest frame (incident energy approximately equal to reflected energy and in the opposite direction).

That (~2) would not even be *exactly* true in an inertial reference frame. eg: the problem I worked previously with the mirror at rest with the police man -- suffers the same defect, if I understand you correctly.
Too bad no one mentioned it then...

I was taking the 2x momentum as a given from physics texts which specify that -- so, in fact, those texts are wrong .. ?
Acceleration changes the Energy of the photon, and therefore the reflected photon has less momentum -- this I understood. But that a photon does not transfer 2x (exactly) was something that I didn't expect. This is one of the major reasons I wanted to warm up on a few problems before tackling gravity -- this ought to have been an obvious error in retrospect.

In the frame where the mirror is moving at 0.5 c the reflected energy is ~1/3 of the incident energy (and in the opposite direction), so the change in momentum is a factor of ~4/3. So, if you use a factor of 4/3 instead of a factor of 2 you will get very close to my result.

In the police frame, the light coming back has approximately 1/3 the frequency, so 2/3rds the frequency has changed -- that is directly proportional to energy.
In the mirror frame, the reflected light is approximately the same color -- so very *little* momentum change.

I don't follow you -- It seems you are speaking about the energy in the police frame -- but the change energy is 66.666...% in that frame. Since momentum is related to energy by p = E/c; I would expect momentum to change by 2/3rds as well.

eg: E**2 = p**2 * c**2 + m0**2 * c**4; but photons have 0 rest mass, therefore:
E**2 = p**2 * c**2
or
E = p * c
p = E/c

The root issue is that we have an inconsistent application of the equations from the get-go.
I'll have to think about this for a bit; but I wanted exact equations...
Perhaps doing the integral for Maxwell is important.
Einstein, in his paper, doesn't appear to be correcting for these details either -- though I will need to look again. That really bothers me.

I see an obvious way to iteratively correct for the momentum change in a limiting approach -- and I don't think it will work out to 4/3rds by gut reaction *Perhaps you mean just shy of 4/3? * -- but I will work the problem a little later and see what drops out.
 
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