The doppler radar trap paradox on the path to gravity.

[andrewr]

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?

 

[Max™]

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.

 

[andrewr]

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![/color]
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![/color])
# ------ 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?

 

[Dale]

One very minor error and we have agreement:

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.

 

[andrewr]

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.

 

[andrewr]

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.

Aye. I understood this all along -- I am just a poor at using language unambiguously.
I spend too much time re-reading and editing as it is...

It is just that when keeping track of time on a clock, one needs to know if it was accelerated historically.
Eg: that sudden jerk which goes from 0 to whatever in no time...

Once the jerk took place (dirac) -- the relative non-accelerating velocity will continue to make the clock tick slower relative to a clock which has never been accelerated at any time during the experiment.

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.

Aye. Thanks Max.

 

[Dale]

That (~2) would not even be *exactly* true in an inertial reference frame. ... I don't think it will work out to 4/3rds by gut reaction *Perhaps you mean just shy of 4/3? *

Correct, that is what I meant by the "~". The "~2" means "approximately 2" and the "~4/3" means "approximately 4/3". Sorry if the notation was unclear.

I was taking the 2x momentum as a given from physics texts which specify that -- so, in fact, those texts are wrong .. ?

I wouldn't say "wrong", it is true, but only for the special case of a non-accelerating mirror at rest. If they didn't say that then they were being sloppy. My textbook (Serway) for example is sloppy, but uses it correctly in the context where there is a restoring force on the mirror and so the mirror's acceleration is 0.

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

Don't forget that energy is a scalar quantity and momentum is a vector quantity. So although they are related to each other they are not the same. Also, notice that E²=p²c² has both a positive and a negative root for p (in one dimensional collisions). So the magnitude of the momentum changed by ~2/3 (i.e. from 1 to ~±1/3), but the direction also changed (i.e. from 1 to ~-1/3) resulting in a change in momentum with magnitude >1 (i.e. ~-4/3). This is why it is important to know that both energy (a scalar) and momentum (a vector) are conserved and why the problem cannot be worked entirely using just one or the other.

 

[andrewr]

Correct, that is what I meant by the "~". The "~2" means "approximately 2" and the "~4/3" means "approximately 4/3". Sorry if the notation was unclear.

It was clear -- I was tired -- eg: insomnia. I just missed it.

I wouldn't say "wrong", it is true, but only for the special case of a non-accelerating mirror at rest. If they didn't say that then they were being sloppy. My textbook (Serway) for example is sloppy, but uses it correctly in the context where there is a restoring force on the mirror and so the mirror's acceleration is 0.

My text is the same -- sloppy.

Don't forget that energy is a scalar quantity and momentum is a vector quantity. So although they are related to each other they are not the same.

Sure, although true conservation of energy guarantees conservation of momentum -- so that if one were to assign a direction to energy, all the information can be handled at that level.
Taking the derivative of energy (classic) -- where the energy is balanced, will give momentum.
That was the original incentive in Physics for computing a scalar field from vector ones -- eg: the E field can be integrated to arrive at a potential energy field (a scalar).

Also, notice that E²=p²c² has both a positive and a negative root for p (in one dimensional collisions). So the magnitude of the momentum changed by ~2/3 (i.e. from 1 to ~±1/3), but the direction also changed (i.e. from 1 to ~-1/3) resulting in a change in momentum with magnitude >1 (i.e. ~-4/3). This is why it is important to know that both energy (a scalar) and momentum (a vector) are conserved and why the problem cannot be worked entirely using just one or the other.

I'll have to think about that. But, my math teachers always told us to test ideas at the extremes -- because these are the most likely places for failures to occur, and mistakes to be made visible.
The issue, with respect to energy there -- is that if the mass of the mirror is allowed to go infinite -- the reflected photon has a momentum change -- but the mirror does not.
Only the sign of the momentum changes, though, and not the value.

Since no acceleration occurs on the mirror, even if moving, -- at infinite mass -- then one has the issue that the mathematics OUGHT to reflect there being NO energy transfer. I would expect, mathematically, that the energy transfer drops as the mass increases because of the limiting case.

A second thing that I notice, is that the wave solution to the Maxwell version of the problem, seems to assume that the mirror is immersed in a magnetic field -- which, from careful inspection, is not the case.
The mirror moves OUT of the field as the photon reflects. We can balance energy and momentum by assigning (mathematically) various scenarios -- but that will always leave open the question of experimental verification. I will still need to work the problem various ways -- and I have equipment coming to do some rudimentary tests to satisfy my curiosity.

 

[Dale]

Sure, although true conservation of energy guarantees conservation of momentum -- so that if one were to assign a direction to energy, all the information can be handled at that level.

I agree and will just point out that that is exactly what the four-momentum does.

The issue, with respect to energy there -- is that if the mass of the mirror is allowed to go infinite -- the reflected photon has a momentum change -- but the mirror does not.

No, the mirror does have a momentum change which is opposite to the momentum change of the photon. It just does not have a change in velocity, but that is expected since a finite change in momentum (force) cannot change the velocity of (accelerate) an infinite mass. All of the above works for an infinite mass mirror (or a mirror that does not accelerate due to an external force), simply replace all of the "approximates" with "exactlys".