z shift in 'rigid' linearly accelerating reference frames.
Chalky
In the past, I have usually assumed that the z shift distance
relationship in a linearly accelerating reference frame could be
obtained by naive application of the gravitational Doppler shift thus:
1 + z = 1 / (1 - g.d/c^2)
However, since this was generating significant inconsistencies
elsewhere, I decided to analyse this situation from first principles,
and arrived at a slightly different conclusion.
I found that, in the inertial frame instantaneously co-moving with the
accelerating observer, other objects at fixed distances in that
accelerating frame have approach/recession velocities
in the inertial frame which are directly proportional to distance,
thus indicating a z shift in the accelerating frame of
1 + z = beta/(1 - v/c)= beta/(1 - g.d/c^2) (which does, indeed,
resolve the aforementioned inconsistencies).
Consequently I would appreciate clarification of what the 'official
line' is in such relativistic situations.
Shubee
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 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 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 =96L] 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) =961]
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
Chalky
Yes, I have come across this conclusion before.
> This force is given by the equation g(x')=-(c^2)/x'.
It would be helpful if you could define where you believe x' = 0,
here, for the accelerating observer.
If the rocket observer defines x' as distance from him, this equation
tells him he must always experience infinite acceleration.
If he defines x' as his distance from where he started accelerating,
it tells him he needs infinite acceleration to start on his journey.
Both conclusions would be nonsensical.
Shubee
> 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=92s book: Relativity:
Special, General, and Cosmological
http://books.google.com/books?id=fUj_LW51GfQC&pg=PA71&lpg=PA71&dq=%=
22proper+acceleration%22+equation+relativity&source=web&ots=g1SCT8RzqT&=
sig=PW1jNFsw93Nl7Y12PkrF7BInXvo&hl=en&sa=X&oi=book_result&resnum==
9&ct=result
Shubee
http://www.everythingimportant.org/relativity/directory.htm
Chalky
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.
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.
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.
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.
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.
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.
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.
Chalky
> On Jun 25, 10:43 pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > 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=92s book: Relativity:
> Special, General, and Cosmological
>
> http://books.google.com/books?id=fUj_LW51GfQC&pg=PA71&lpg=PA71&dq=%=
> 22proper+acceleration%22+equation+relativity&source=web&ots=g1SCT8RzqT&=
> sig=PW1jNFsw93Nl7Y12PkrF7BInXvo&hl=en&sa=X&oi=book_result&resnum==
> 9&ct=result
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.
Chalky
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 becoming 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')
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 if 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.
> 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
> 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.
Shubert's computations are incorrect, I pointed that out to him
several times, he even dowloaded the correct computations.
You can download the correct formulas here:
Chalky
Thanks. Your derivation is certainly more confidence inspiring, not
only because it does not involve subtractions of infinitie quantities
but also because resultant eqn. 3.5 confirms that the factor 1 - g.L/
c^2 is merely the first term of a binomial expansion, which verifies
my own conclusions, via a different method.
However, your exact solution does leave me a bit 'in the air' since it
leaves me to substitute a complicated function (5.2) into another less
complicated function (5.3) to obtain the result. As someone who knows
from experience that silly blunders can often result, I would
appreciate clarification of whether the resultant function simplifies,
and, if so, to what.
Chalky
> 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'.
After due reflection, this claim disproves itself.
ANY uniformly accelerating body remains at an invariant distance from
an origin located at a distance of c^2/g behind it. By defining g as
inversely proportional to x' you thus define the length of the rocket
to be zero at this origin. That is infinite compression.
Q.E.D.
Chalky
> Shubert's computations are incorrect, I pointed that out to him
> several times
Perhaps what is needed is a clear identification of where we think he
went wrong.
AFAICT he starts with a pair of inertial/accelerating frame transforms
where there is a specific implied relationship between origins. He
then switches to a completely different origin for the accelerating
frame, (to simplify the maths), ignores the possibility that in this
new frame the original accelerating frame origin could be both moving
and accelerating, and proceeds to apply the original transforms
regardless.
I don't know where he got his varying gravitational field equation
from, but it seems to me it could have been derived as follows: Since
all points on the rocket must be at invariant distances from each
other, this can be achieved by making them all invariant distances
from a common origin.
If so, this logic is flawed. For a rigid accelerating rocket, we do
not want the top and the bottom to start accelerating from the same
point in space, but from the same point in time.
If my above analysis is correct, it is a tribute to the mathematical
ingenuity of Schubee that from such an initial comedy of errors he
can still produce an 'exact' verification of the traditional factor 1
- g.d/c^2.
Shubee
> On Jun 30, 6:53 am, Dono <sa...@comcast.net> wrote:
>
> > Shubert's computations are incorrect, I pointed that out to him
> > several times
>
> Perhaps what is needed is a clear identification of where we think he
> went wrong.
I think it would be more entertaining if the moderator that approved
the posting of http://groups.google.com/group/sci.physics.research/msg/c1c0b5807d6f9ad3
would chime in and explain why the recommended link http://www.savefile.com/files/1624470
isn't painfully clumsy and amateurish.
I'm not implying that Dono got a wrong answer. Yet I am amused that
both Dono and Chalky protested the simplicity of my final equation and
failed to see that the solution for time t in Dono's equation 5.1 goes
to infinity as (1-gh/c^2) goes to zero. These two obviously failed to
grasp the elegance of my approach.
http://groups.google.com/group/sci.physics.foundations/msg/ead3339930e7557f
http://www.everythingimportant.org/SDA/viewtopic.php?f=14&t=969
> AFAICT he starts with a pair of inertial/accelerating frame transforms
> where there is a specific implied relationship between origins. He
> then switches to a completely different origin for the accelerating
> frame, (to simplify the maths), ignores the possibility that in this
> new frame the original accelerating frame origin could be both moving
> and accelerating, and proceeds to apply the original transforms
> regardless.
The equation for the accelerating frame origin obeys the same Lorentz
invariant equation as every other point x' > 0.
(x')^2 =x^2-(ct)^2
Here x and t refer to the ordinary coordinates in an inertial frame of
reference and x' > 0 is a fixed point in the accelerated coordinate
system.
> I don't know where he got his varying gravitational field equation
> from, ...
It's a standard result in special relativity. I first studied that
equation in a course taught at UTD by Wolfgang Rindler 20 years ago
but I especially remember it from Rindler's book, Relativity: Special,
General, and Cosmological pp. 71-73.
Here is a straightforward method for deriving that equation.
Let x' = D represent a fixed point in the accelerating coordinate
system. 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.
Recall my stipulation that D > 0.
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' continues to be true all along the
path of the motion of the point x(t) = D sqrt(1+ (ct)^2/D^2).
> If so, this logic is flawed. For a rigid accelerating rocket, we do
> not want the top and the bottom to start accelerating from the same
> point in space, but from the same point in time.
It is easy to see from the transformation equations, which express
events (x',t') in the accelerated coordinates to the inertial
coordinate system:
x=x'cosh(ct'/x')
t=(x'/c)sinh(ct'/x') for all x'>0.
that x = x' at t = 0.
I said nothing to suggest that multiple points x' > 0 in an
accelerating coordinate system issue from a single point at t = 0.
> If my above analysis is correct, it is a tribute to the mathematical
> ingenuity of Schubee that from such an initial comedy of errors he
> can still produce an 'exact' verification of the traditional factor 1
> - g.d/c^2.
Shubee
http://www.everythingimportant.org/relativity/directory.htm
Chalky
> On Jun 28, 1:10 pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
> > My question is simple enough. What is your answer.
>
> I apologize for the original link becoming 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.
Well, I did have enough sense to string the corrupted link back
together again, but still no joy.
Even attempting to scroll through to these pages at that site still
gives "pages 65 to 76 are not part of this book review.
Chalky
> On Jul 1, 9:01 pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
> I'm not implying that Dono got a wrong answer. Yet I am amused that
> both Dono and Chalky protested the simplicity of my final equation and
> failed to see that the solution for time t in Dono's equation 5.1 goes
> to infinity as (1-gh/c^2) goes to zero.
Well, duh! As I have repeatedly pointed out, once you have defined
coincidence of origins, that is an inevitable consequence of inertial
frame physics. The accelerating frame event horizon of that origin
occurs only when the accelerating observer has achieved light speed in
that inertial frame. Can't you understand that?
> > AFAICT he starts with a pair of inertial/accelerating frame transforms
> > where there is a specific implied relationship between origins. He
> > then switches to a completely different origin for the accelerating
> > frame, (to simplify the maths), ignores the possibility that in this
> > new frame the original accelerating frame origin could be both moving
> > and accelerating, and proceeds to apply the original transforms
> > regardless.
>
> The equation for the accelerating frame origin obeys the same Lorentz
> invariant equation as every other point x' > 0.
Since you keep changing your mind about what that origin is, it is
obviously not invariant, Lorentz or otherwise.
> (x')^2 =x^2-(ct)^2
>
> Here x and t refer to the ordinary coordinates in an inertial frame of
> reference
The ordinary coordinates of what?
> and x' > 0 is a fixed point in the accelerated coordinate
> system.
Why isn't x' = 0 a fixed point too?
> I said nothing to suggest that multiple points x' > 0 in an
> accelerating coordinate system issue from a single point at t = 0.
No you didn't. Nobody said you did. You said, by implication, that
they issued from a single point in SPACE not in TIME.
Dono
> I think it would be more entertaining if the moderator that approved
> the posting ofhttp://groups.google.com/group/sci.physics.research/msg/c1c0b5807d6f9ad3
> would chime in and explain why the recommended linkhttp://www.savefile.com/files/1624470
> isn't painfully clumsy and amateurish.
>
Because your solution is wrong. I explained that to you but to no
avail.
> I'm not implying that Dono got a wrong answer.
Good. Because, as opposed to you, I got the correct answer. Now, it is
good that you downloaded the file and you are learning how to correct
your errors.
Chalky
>
> I'm not implying that Dono got a wrong answer. Yet I am amused that
> both Dono and Chalky protested the simplicity of my final equation and
> failed to see that the solution for time t in Dono's equation 5.1 goes
> to infinity as (1-gh/c^2) goes to zero. These two obviously failed to
> grasp the elegance of my approach.
Check your dictionary. Elegance is synonymous with simplicity, not
with mathematical complexity and maximum dependency on additional
premises.
The beauty of Dino's approach is in its conceptual simplicity.
If Dino had only used accelerating frame time instead of inertial
frame time, his first derivation could have been a one liner:
v = gt = gd/c, hence red/blue shift from SR Doppler shift formula.
Why does your solution fail to confirm this obvious consequence of
Newtonian dynamics, even for short rockets?
I can think of only one rational explanation, unless you have made a
silly blunder in the subsequent maths:
Your initial premises are logically flawed.
Dono
>
> I'm not implying that Dono got a wrong answer. Yet I am amused that
> both Dono and Chalky protested the simplicity of my final equation and
> failed to see that the solution for time t in Dono's equation 5.1 goes
> to infinity as (1-gh/c^2) goes to zero.
h is the rocket height. Hopefully you know what g is (I am not sure).
So , gh<<c^2.
This is 9-th grade algebra. Better luck disproving me next time.
Chalky
> On Jul 2, 12:11 pm, Shubee <e.Shu...@gmail.com> wrote:
>
> > I think it would be more entertaining if the moderator that approved
> > the posting ofhttp://groups.google.com/group/sci.physics.research/msg/c1c0b5807d6f9ad3
> > would chime in and explain why the recommended linkhttp://www.savefile.com/files/1624470
> > isn't painfully clumsy and amateurish.
>
> Because your solution is wrong. I explained that to you but to no
> avail.
>
> > I'm not implying that Dono got a wrong answer.
>
> Good. Because, as opposed to you, I got the correct answer.
I have just checked the binomial expansions and my answer is the same
as yours to the level of accuracy given in formula 3.5. However, I
note that many of your references relate to experimental tests, and I
would also appreciate clarification of the level of accuracy to which
such experimental tests may have verified theory, if at all possible.
Shubee
> Dono <sa...@comcast.net> wrote:
> > http://www.savefile.com/files/1624470
> However, your exact solution does leave me a bit 'in the air' since it
> leaves me to substitute a complicated function (5.2) into another less
> complicated function (5.3) to obtain the result. As someone who knows
> from experience that silly blunders can often result, I would
> appreciate clarification of whether the resultant function simplifies,
> and, if so, to what.
I just now finished reducing Dono's embarrassingly large, inelegant
algebraic expression to 1-gh/c^2. The effort should only require a
little mathematical courage from you. I rank the simplification as an
easy exercise in high school algebra.
Shubee
http://www.everythingimportant.org/relativity/directory.htm
Chalky
> Chalky wrote:
> > Dono <sa...@comcast.net> wrote:
> > >http://www.savefile.com/files/1624470
> > However, your exact solution does leave me a bit 'in the air' since it
> > leaves me to substitute a complicated function (5.2) into another less
> > complicated function (5.3) to obtain the result. As someone who knows
> > from experience that silly blunders can often result, I would
> > appreciate clarification of whether the resultant function simplifies,
> > and, if so, to what.
>
> I just now finished reducing Dono's embarrassingly large, inelegant
> algebraic expression to 1-gh/c^2.
How?