The Doppler Shift Equation For An Accelerating Frame of Reference
Shubee
> At the present point in time I am interested specifically
> in the redshift/distance relationship for a rigid linearly
> accelerating reference frame.
The equation you seek is easy to derive. It is a part of my standard
reply for frequently asked questions regarding a constantly
accelerating frame of reference.
The equations for a coordinate system undergoing constant proper
acceleration are:
x=x'cosh(ct'/x')
t=(x'/c)sinh(ct'/x') for all x'>0.
Permit me to illustrate their meaning. Think of an accelerating
rocket. c is the speed of light. x' is a general position location in
the accelerating rocket. (Call x' and t' rocket coordinates if you
like). Believe it or not, in relativity, if a long rocket accelerates
without any part of the ship being compressed or stretched during its
motion, then an astronaut at the bottom of the rocket will feel a
greater acceleration than an astronaut at the tip of the rocket. This
force is given by the equation g(x')=-(c^2)/x'.
Many interesting problems can be solved in special relativity by first
learning how to interpret and use these three elementary equations.
Suppose a rocket ship, which looks like a long rod, begins to
accelerate at t=t'=0. Let’s say that the ship has a rest length of L.
(x=x' at t=0). Suppose that the tip of the rocket ship undergoes a
constant proper acceleration g. For that case, assign the tip of the
ship the fixed point x'_b =(c^2)/g. Then the bottom of the ship
undergoes a proper acceleration of g' =(c^2)/(x'_a) where x'_a = x'_b
- L. Every point on the accelerated ship is carrying a clock. If some
event happens on the ship, it will be at some x' with the clock at
that point reading time t'. According to the coordinates of the
stationary frame, the event will happen at point x at time t. All
clocks are synchronized to read zero time at the instant acceleration
begins.
From these equations, I shall derive the exact redshift/distance
equation for light that is moving from the bottom of the rocket to the
furthermost tip of the rocket. I propose the following strategy.
One helpful insight in physics is the observation that if we had a
constant gravitational field, then the equation for the Doppler shift
would be trivial. I reason as follows:
Let f be the initial frequency of a photon at some initial point in a
constant gravitational field. Suppose that the photon is moving
radially. Since the force felt by the photon is constant along its
path, the Doppler shift would depend only on the distance traveled by
the photon through the constant accelerative force. In this instance,
the Doppler shift equation for a photon traveling a distance x would
be
f' = fH(x).
If the photon would move an additional distance y, then
f'' = f'H(y).
Consequently, f''=fH(x+y) = fH(x)H(y) and we are left with the simple
functional equation
H(x+y) = H(x)H(y)
The only continuous solution for this equation is of the form
H(x) =exp (alpha x) where alpha is some constant.
Comparing this equation with Einstein’s approximation leads us to the
conclusion that alpha = -g/c^2.
Now for the more general problem of a photon moving radially in a
varying gravitational field where the accelerative force -g is a known
function of x, simply recall baby calculus and divide the photon path
into infinitely many infinitesimal layers, at least in the limit,
where the gravitational force is approximately constant at each layer.
The infinite product
f|(final) =f [exp g(x'1)dx'/c^2] [exp g(x'2)dx'/c^2]... [exp g(x'n)dx'/
c^2] would convert to the integral
f|(final) =f exp[ integral g(x')dx'/c^2 ]
where g(x') = -c^2 / x'
The limits of integration would be from x'= x'_a to x'= x'_b.
Remembering to divide g(x') by c^2, the exponential of the integral
that is to be evaluated between the limits of [c^2/g –L] to [c^2/g] is
easily computed to be 1-gL/c^2.
So f' = f (1-gL/c^2).
It is easy to check that this reasoning is correct. I shall now derive
the equation in a completely different way and without exploiting
Einstein's approximation. Consider the instantaneously co-moving
inertial frame of reference at the moment that a photon at the base of
the rocket (x'= x'_a) is emitted toward the tip (x'= x'_b). I will
first determine the time it takes for the photon to arrive, in
inertial coordinates, given that the front of the rocket (x'= x'_b) is
moving at a constant proper acceleration g. I shall then figure out
the final velocity from that.
From the perspective of the initial co-moving inertial reference frame
ct= L + c^2/g[ sqrt (1+(gt/c)^2) –1]
or equivalently,
1-gL/c^2 = sqrt (1+(gt/c)^2) - gt/c
At this point, I insert the well-known formula:
gt/c = sinh gt'/c where t' is the elapsed proper time for any clock at
x'_b.
Thus 1-gL/c^2 = exp (-gt'/c)
Now recall that v/c = tanh (gt'/c) is the very well-known formula for
the final velocity of a clock undergoing constant proper acceleration
g and recall that
arc tanh x = 1/2 ln [(1+x)/(1-x)] is always applicable for |v/c| < 1.
Consequently,
1-gL/c^2 = exp (-gt'/c) = sqrt [(1-v/c)/(1+v/c)], which again is the
correct factor in the Doppler shift equation.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
John C. Polasek
wrote:
Nifty! God thinks of everything.
rot...@gmail.com
> Let f be the initial frequency of a photon at some initial point in a
> constant gravitational field.
Just to be sure, f wrt (measured wrt) which frame?
>Suppose that the photon is moving
> radially. Since the force felt by the photon is constant along its
> path,
ok...
>the Doppler shift would depend only on the distance traveled by
> the photon through the constant accelerative force.
Why? And, x is measures wrt which frame?
> In this instance, the Doppler shift equation for a
> photon traveling a distance x would be
>
> f' = fH(x).
Why is it proportinal to f? Why not some other relation? (That will
lead to other solutions which satisfy the functional relation)
Shubee
> A few Q's.
>
> > Let f be the initial frequency of a photon at some initial
> > point in a constant gravitational field.
>
> Just to be sure, f wrt (measured wrt) which frame?
The instantaneous, free-falling, local frame of reference seems to be
an acceptable frame to define this measurement in a hypothetical
constant gravitational field. In the real world however, it suffices
to pick that frame that is at rest with respect to the presumed source
of the real gravitational field.
> >Suppose that the photon is moving radially. Since the
> > force felt by the photon is constant along its path,
>
> ok...
>
> > the Doppler shift would depend only on the distance traveled
> > by the photon through the constant accelerative force.
>
> Why? And, x is measures wrt which frame?
>
> > In this instance, the Doppler shift equation for a
> > photon traveling a distance x would be
>
> > f' = fH(x).
>
> Why is it proportinal to f? Why not some other relation? (That will
> lead to other solutions which satisfy the functional relation)
I believe that the simplicity of this assumption comes from Einstein’s
principle of equivalence.
http://www.einstein-online.info/en/spotlights/equivalence_principle/index.html
Shubee
Dono
> On Jun 19, 1:19 pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
I must commend you for trying to calculate something instead of
pushing (again) your crackpot paper.
Your computations are nevertheless incomplete and they produce the
incorrect results. Here are the complete calculations I did some time
ago:
http://www.savefile.com/projects/1046240
DvM gave me the clue for the first part (the one that uses the
Newtonian approximation of accelerated motion). I did the part using
hyperbolic motion.
Dono
>
> 1-gL/c^2 = exp (-gt'/c) = sqrt [(1-v/c)/(1+v/c)], which again is the
> correct factor in the Doppler shift equation.
>
The correct formula is not 1-gL/c^2 but 1-gL/c^2+0.5(gL/c^2)^2 , if
you use the Newtonian approximation.
If you use the eqs of hyperbolic motion, the formula is even more
complicated, see my writeup.
BURT
The Doppler shift for acceleration through space is energy effected by
changing Gamma; multiplied when toward light and divided downward when
away.
Mitch Raemsch
Dono
Dono
mitch.nico...@gmail.com
>
> - Show quoted text -
Gamma for matter in motion through space effects light's energy.
Shubee
> On Jun 22, 10:58 am, Shubee <e.Shu...@gmail.com> wrote:
>
> > On Jun 19, 1:19 pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > > At the present point in time I am interested specifically
> > > in the redshift/distance relationship for a rigid linearly
> > > accelerating reference frame.
>
> > The equation you seek is easy to derive. It is a part of my standard
> > reply for frequently asked questions regarding a constantly
> > accelerating frame of reference.
>
> > The equations for a coordinate system undergoing constant proper
> > acceleration are:
>
> > x=x'cosh(ct'/x')
> > t=(x'/c)sinh(ct'/x') for all x'>0.
>
> > Permit me to illustrate their meaning. Think of an accelerating
> > rocket. c is the speed of light. x' is a general position location in
> > the accelerating rocket. (Call x' and t' rocket coordinates if you
> > like). Believe it or not, in relativity, if a long rocket accelerates
> > without any part of the ship being compressed or stretched during its
> > motion, then an astronaut at the bottom of the rocket will feel a
> > greater acceleration than an astronaut at the tip of the rocket.
>
> contradiction when combined with another standard special relativistic
> result relative to an accelerating observer; that he will remain
> (relative to him) at an invariant distance from an origin located at g/
> c^2 behind him.
First of all, your memory is faulty. g/c^2 doesn’t have the dimensions
of distance. But c^2/g does. Also, the result that you barely remember
is that if an observer has a head start on a photon, then the photon
will never catch up to the observer if the observer is accelerating at
a uniform acceleration g and if his initial head start distance is L =
c^2/g.
> Clearly another observer at that (inertial) origin cannot experience
> accelerations of zero and infinity at the same time.
I believe that you’re confusing c^2/g with g/c^2. Those are two
different quantities. The first quantity has the dimensions of length.
The second has the dimensions of 1/length. In the coordinate system
given, no observer can be stationed at x'=0 because that point
represents a point of infinite acceleration.
It is important to note that there is no limit to the magnitude of
proper acceleration and however great a number you select for the
acceleration g, the resulting speed of an object will always be less
than the speed of light. In that sense, the speed of light can be
interpreted as representing infinite acceleration.
> > This force is given by the equation g(x')=-(c^2)/x'.
>
> I am afraid this formula doesn't make any sense to me. If the rocket
> observer defines x' as distance from him, the formula tells him that
> he must always experience infinite acceleration.
And it doesn’t make sense to me that you would want to change the
definition of rocket coordinates in midstream. There are good reasons
to read carefully. The choice for the point x'=0 to represent the
event horizon was completely arbitrary. But it does simplify the
math.
> Thanks for the subsequent advice but with this level of confusion at
> the start, I was unable to follow the rest of your reasoning.
Do you remember the bottom line question that you asked?
“At the present point in time I am interested specifically
in the redshift/distance relationship for a rigid linearly
accelerating reference frame.”
I answered your inquiry directly.
I derived the equation:
f' = f (1-gL/c^2).
Do you understand that equation?
L is the length of the rocket.
c is the ordinary speed of light.
g is the acceleration at the top of the rocket.
f is the frequency of a photon emitted at the bottom of the rocket.
f' is the frequency of the photon when it’s received at the top of the
rocket.
Note that if the length of the rocket L approaches c^2/g, then the
Doppler shift approaches zero. Can you guess what the meaning of that
limit point is? It means that L = c^2/g is that critical head start
distance where light can never reach the accelerating observer at the
top of the rocket.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Dono
>
> I answered your inquiry directly.
> I derived the equation:
> f' = f (1-gL/c^2).
>
> Do you understand that equation?
>
> L is the length of the rocket.
> c is the ordinary speed of light.
> g is the acceleration at the top of the rocket.
> f is the frequency of a photon emitted at the bottom of the rocket.
> f' is the frequency of the photon when it’s received at the top of the
> rocket.
Shitbert,
Any time you start selfpromotion you reagin the Shitbert name.
I have already shown that your derivation is wrong and so is your
result:
http://groups.google.com/group/sci.physics.relativity/msg/25453b760481476e?
Shubee
> Well, I did ask for clarification of the 'official line' in this
> situation so I guess, in the absence of third party comments to the
> contrary, I must accept that your response represents that 'official
> line', at least as far as SPR contributors are concerned.
> > > It would be helpful if you could define where you believe x' = 0,
> > > here, for the accelerating observer.
>
> > All the basics are explained in Wolfgang Rindler's book: Relativity:
> > Special, General, and Cosmological
>
> > http://books.google.com/books?id=fUj_LW51GfQC
>
> This link does not answer my question. I asked you where _you_ place
> the origin of _your_ accelerating coordinate system
> I did not ask you to give me a link to a 428 page book on relativity
> theory.
>
> My question is simple enough. What is your answer.
I apologize for the original link getting corrupted and thus not
taking you directly to pages 71-73 of Rindler's book. You wanted the
official line. Those 3 pages are as official as you can get. They are
an excellent explanation of the properties of the point x'=0 in _my_
accelerating coordinate system.
> I failed to follow the whole of your derivation, not least because
> there are a couple of instances where non ascii characters have been
> translated into gibberish within your equations, but I did follow
> enough to challenge your claim that your derivation is exact.
I regret the corruption of data. Here are two links where the same
derivation was posted successfully.
http://groups.google.com/group/sci.physics.foundations/msg/ead3339930e7557f
http://www.everythingimportant.org/SDA/viewtopic.php?f=14&t=969
> 1) Your first 2 equations establish that you are defining both frame
> origins as occupying the same point in space at some unspecified point
> in time when clocks are synchronised and set to zero.
It was stated that the rocket ship begins to accelerate at t=t'=0.
When t=0, then x=x' for all x'>0 follows trivially from the equations:
x=x'cosh(ct'/x')
t=(x'/c)sinh(ct'/x')
Note that I specified the restriction that x'>0 earlier. The x' are
called rocket coordinates.
> Your third equation, on the other hand, appears to indicate that you
> are treating the base and the tip of the rocket as starting from the
> same origin, with the tip accelerating first, and the base
> accelerating later. This clearly bears no relation to reality. We do
> not see the tips of rockets fly skyward before the fuel is ignited.
You are misunderstanding the physics. Professor Rindler is clearly
teaching on page 71-73 of his book that the gravitational field in a
uniformly accelerating rocket varies as c^2/X.
> 2 ) But wait, it gets worse. Your final equation indicates that the
> tip of the rocket has a velocity of c relative to the inertial origin.
Not exactly. The equation f' = f(1-gL/c^2) only implies that the tip
of the rocket will eventually attain light speed if that furthermost
front-end of the rocket can maintain a proper acceleration of g
forever, which is how long you'll have to wait for a photon to arrive
from the bottom of the rocket. That assumes that the length of the
rocket L is L = c^2/g.
> Clearly, therefore, this places the point of clock synchronisation an
> infinite time in the past, for the inertial observer. Consequently
> you are assuming that SR can be applied with perfect accuracy on epic
> scales relative to which even the universe shrinks to point size. This
> strikes me as overly ambitious.
Actually, I don't really believe that the photon will ever arrive at
the front-end of the rocket, even if you wait forever. It will always
be at the point x'=0 in the accelerated coordinate system for all
time.
x'^2 =x^2-(ct)^2
> 3) Finally, you arrive at the remarkable conclusion that the
> relativistic Doppler shift is 1 + z = 1 / (1 + g.L/c^2) both when g is
> independent of L and when g varies substantially with changing L. This
> alone should set the alarm bells ringing.
Please write my equation correctly.
My equation is f' = f(1-gL/c^2).
L is the length of the rocket.
c is the speed of light.
g is the constant acceleration at the top of the rocket.
f is the frequency of a photon emitted at the bottom of the rocket.
f' is the frequency of the photon when it's received at the top of the
rocket.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Dono
> Please write my equation correctly.
> My equation is f' = f(1-gL/c^2).
>
<self promotion removed>
Shitbert,
I showed you that your equation is wrong. Why o you persist?
Shubee
> On Jun 29, 4:41 am, Shitbert <e.Shu...@gmail.com> wrote:
>
> > Please write my equation correctly.
> > My equation is f' = f(1-gL/c^2).
>
I persist because there are professional physicists that recognize
clear thinking.
http://groups.google.com/group/sci.physics.research/msg/e02b316cc4a4d152
http://groups.google.com/group/sci.physics.research/msg/2930b5e18f04c73e
http://groups.google.com/group/sci.physics.foundations/msg/ead3339930e7557f
http://groups.google.com/group/sci.physics.foundations/msg/526ab8adb4c7f959
Shubee
Dono
> On Jun 29, 9:05 am, Dono <sa...@comcast.net> wrote:
>
> > On Jun 29, 4:41 am, Shitbert <e.Shu...@gmail.com> wrote:
>
> > > Please write my equation correctly.
> > > My equation is f' = f(1-gL/c^2).
>
> > I showed you that your equation is wrong. Why o you persist?
>
> I persist because there are professional physicists that recognize
> clear thinking.
No, they don't, Shitbert.
As a matter of fact, they are pointing out your errors:
Tom Roberts
I'll answer the question in the subject: no.
In relativity, "rigid" is impossible. But one could construct a "rod"
out of a series of small atoms which are arranged in sequence along its
length, and each atom is given an appropriate proper acceleration so at
all times the proper distance between adjacent atoms is constant. This
is as close as one can come to "rigid". When one does this for
acceleration along the length of the rod, each atom along its length
must be given a different proper acceleration. This is known as Born
rigid motion.
So there is no "the" proper acceleration of a "rigid" rod. There must be
a different proper acceleration at each point along the rod.
Tom Roberts
Dredd
> [...]
>
> I'll answer the question in the subject: no.
>
> In relativity, "rigid" is impossible. But one could construct a "rod"
>
> Tom Roberts
You seems confused Sir
Relativity effects assume 100% rigidity
mitch.nico...@gmail.com
The rod is extended in curved space. It's extension is curved.
Tom Roberts
> Relativity effects assume 100% rigidity
No, they don't. As I said before, rigid objects are inconsistent with
relativity (e.g. the speed of sound would be infinite, and thus > c,
which is not possible).
Tom Roberts
Pentcho Valev
sci.physics.relativity:
Roberts Roberts just a single false axiom (Einstein's 1905 false light
postulate) and an apparently infinite number of idiocies produced!
Albert the Plagiarist did indeed deserve to become Divine Albert. But
you are a somewhat enigmatic teacher Roberts Roberts: you should refer
zombies to sources where the idiocies are analysed in depth:
http://www.people.fas.harvard.edu/~djmorin/book.html
Chapter 11, p. 45, Problem 26: "Break or not break?"
Pentcho Valev
pva...@yahoo.com
Shubee
> On Jun 29, 2:30 pm, Shubee <e.Shu...@gmail.com> wrote:
>
> > On Jun 23, 3:25 pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > > On Jun 22, 10:58 am, Shubee <e.Shu...@gmail.com> wrote:
>
> > > observer defines x' as distance from him, the formula tells him that
> > > he must always experience infinite acceleration.
>
>
> > (x')^2 =x^2-(ct)^2
>
> Well that formula doesn't make any sense either, since it tells us
> that x' is independent of g.
The acceleration g is a function of x' so it doesn't have to appear
explicitly in the definition of x'. Here is a straightforward method
for finding that function.
Let x' = D represent a fixed point in the accelerating coordinate
system. Suppose D > 0. Then
x^2 = D^2 + (ct)^2
Please note that x=x(t). Clearly, x(t) = D sqrt(1+ (ct)^2/D^2)
represents the motion of the point x' = D in terms of time t. Note
that x(0) = D.
Let b = c^2/D^2
Then x(t) = D sqrt(1+ bt^2).
Using baby calculus, we may expand sqrt(1+ bt^2) using the Maclaurin
series. Thus:
sqrt(1+ bt^2) = 1 + b(t^2)/2 - (b^2)(t^4)/8 + ...
So x(t) = D + (c^2/2D)t^2 - (c^4/8D^3)t^4 + ...
If time t is sufficiently small compared to c^2/2D, we may ignore the
fourth and all higher orders of t in the expansion.
Now recall your high school physics. The equation x = (1/2)at^2
represents distance traveled in time t when an object undergoes a
constant acceleration a. Therefore a = c^2/D is a sensible Newtonian
approximation for the initial acceleration of the point x' = D.
Now consider what it means for the equation D^2 = x^2-(ct)^2 to be
invariant under the Lorentz transformation. It means that whatever
speed the point D attains in time t in one frame of reference, we can
simply go to the co-moving frame where acceleration begins afresh from
zero velocity at a new time t'=0 and compute the acceleration from
there by merely repeating the calculation already performed. Therefore
the acceleration a(x') = c^2/x' is well-defined.
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Shubee
Perhaps you have forgotten that smooth mathematical functions can be
approximated by using the Maclaurin series.
http://mathworld.wolfram.com/MaclaurinSeries.html
Is there a chance that you have never studied freshman level
mathematics in any depth and that you only have false memories of
actually taking a course in calculus?
> We have already established from your placing of the accelerating
> system origin at the event horizon of the accelerating observer, that
> this places t=0 an infinite time in the past, in the inertial system.
Who is we? You haven't proven anything. Setting D=0 in the Lorentz
invariant equation D^2 =x^2-(ct)^2 only proves that x=ct or x=-ct.
Those two equations don't connect t=0 with the infinite past in any
significant way.
> Clearly infinity is never 'small' in comparison to a finite quantity.
And it's even clearer that you have never approximated the value of a
function in the neighborhood of a point.
> I have summarised where I think you have gone wrong at SPR. If you
> want to address these issues please do so in a logical manner, not by
> introducing higher and higher levels of mathematical abstraction.
I wouldn't say that freshman level mathematics is attaining a high
level of mathematical abstraction but I will affirm strongly that if
you can't even acknowledge that I agree with professional relativists
about the non-constant gravitational field in a uniformly accelerating
rocket (cf. Wolfgang Rindler, Relativity: Special, General, and
Cosmological pp. 71-73), then you need to ask questions about simpler
problems.
Shubee
http://www.everythingimportant.org/relativity/directory.htm
BURT
> Is there a chance that you have never studied freshman level
> mathematics in any depth and that you only have false memories of
> actually taking a course in calculus?
>
> > We have already established from your placing of the accelerating
> > system origin at the event horizon of the accelerating observer, that
> > this places t=0 an infinite time in the past, in the inertial system.
>
> Who is we? You haven't proven anything. Setting D=0 in the Lorentz
> invariant equation D^2 =x^2-(ct)^2 only proves that x=ct or x=-ct.
> Those two equations don't connect t=0 with the infinite past in any
> significant way.
>
> > Clearly infinity is never 'small' in comparison to a finite quantity.
>
> And it's even clearer that you have never approximated the value of a
> function in the neighborhood of a point.
>
> > I have summarised where I think you have gone wrong at SPR. If you
> > want to address these issues please do so in a logical manner, not by
> > introducing higher and higher levels of mathematical abstraction.
>
> I wouldn't say that freshman level mathematics is attaining a high
> level of mathematical abstraction but I will affirm strongly that if
> you can't even acknowledge that I agree with professional relativists
> about the non-constant gravitational field in a uniformly accelerating
> rocket (cf. Wolfgang Rindler, Relativity: Special, General, and
> Cosmological pp. 71-73), then you need to ask questions about simpler
> problems.
>
>
> - Show quoted text -
Deceleration and rotation also create gravity according to Einstein's
Equivalence Principle.