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![/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?