Accelerating frame red/blue shift.
Chalky
This is a variation on an earlier posting at sci.physics.research, as
the original version has not yet received a response there.
Problem is, I derived this result pregeometrically, and have been
dabbling in this obscure (dark?) science for so long that my knowledge
of standard relativistic approaches has atrophied to the extent that I
am beginning to wonder if senile dementia is kicking in.
The 'naive' approach referred to below is clearly inaccurate since, in
such a frame, g varies with distance, so the integral of g wrt d is
not that simple.
However, I would appreciate a clear indication of whether the revised
result conforms to standard approaches, in view of the abovementioned
feared approaching senile dementia.
Revision of original posting follows:
In the past, I have tended to assume that the z shift distance
relationship in a 'rigid' linearly accelerating reference frame could
be
obtained to a good approximation 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 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).
Consequently I would appreciate clarification of whether conventional
relativistic approaches give the same result in such situations.
[PS to Charles Francis (assuming he is current moderator). Spammers
have got more clever of late and have cracked my crude antispamming
device so I have had to close this email account entirely, some time
ago. Consequently, if you want to contact me by email you will have to
use John's slightly more sophisticated antispamming email ruse, and he
can then forward it to me at my better protected address.]
Oh No
>
>
>
>Revision of original posting follows:
>
>
>In the past, I have tended to assume that the z shift distance
>relationship in a 'rigid' linearly accelerating reference frame could
>be
>obtained to a good approximation 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 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).
>
>Consequently I would appreciate clarification of whether conventional
>relativistic approaches give the same result in such situations.
It is not clear to me that we can use any form of naive approach. I
think we actually have to think carefully about what geometry means, and
then use the correct solution
http://www.teleconnection.info/rqg/LargeScaleStructure
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Chalky
> Thus spake Chalky <chalkys...@bleachboys.co.uk>
>
>
>
>
>
> >Revision of original posting follows:
>
> >In the past, I have tended to assume that the z shift distance
> >relationship in a 'rigid' linearly accelerating reference frame could
> >be
> >obtained to a good approximation 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 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).
>
> >Consequently I would appreciate clarification of whether conventional
> >relativistic approaches give the same result in such situations.
>
> It is not clear to me that we can use any form of naive approach. I
> think we actually have to think carefully about what geometry means, and
>
> Regards
>
> --
> Charles Francis
> moderator sci.physics.foundations.
> charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
> braces)
>
> http://www.teleconnection.info/rqg/MainIndex
I think we must agree to differ here. I totally agree with you in your
comments in the above link, except to the extent that my own
(laborious) pregeometric research does appear to provide a perfectly
rational relativistic physical cause for what approximates in
Einstein's theory to the 'cosmoliogical constant' fudge factor.
However, as that work has not yet been published, I am not even
allowed to say so at e.g. sci.astro.research.
Part of the reason for not publishing yet is that I want to make
absolutely sure that I have every aspect of the theoretical background
rock solid first.
However, you seem to have 'jumped the gun' here. This specific
question has nothing to do with cosmology per se. It is a cosmology
independent question whose answer should be directly derivable from
SR.
Regards
C
Oh No
If that is so, I have indeed misunderstood the question :-). If one is
to think of the relationship of gravity with accelerating observers, a
good place to start (as did Einstein) is with the equivalence principle.
http://www.teleconnection.info/rqg/TheEquivalencePrinciple
The twin paradox is a sort of link which is treatable in sr, and set him
on the path toward gr. Of course, this only works as simply for a
uniform gravitational field. But that is where the geometrical
interpretation of gravity begins.
Chalky
> Thus spake Chalky <chalkys...@bleachboys.co.uk>
>
> >question has nothing to do with cosmology per se. It is a cosmology
> >independent question whose answer should be directly derivable from
> >SR.
>
> If that is so, I have indeed misunderstood the question :-).
Indeed.
So let me put it differently. At the present point in time I am
interested specifically in the redshift/distance relationship for a
rigid linearly accelerating reference frame.
This is a sufficiently well known theoretical relativistic construct
that there must be some formal conclusion somewhere, and I would like
to know what it is.
> If one is
> to think of the relationship of gravity with accelerating observers, a
> good place to start (as did Einstein) is with the equivalence principle.
>
> http://www.teleconnection.info/rqg/TheEquivalencePrinciple
>
> The twin paradox is a sort of link which is treatable in sr, and set him
> on the path toward gr. Of course, this only works as simply for a
> uniform gravitational field. But that is where the geometrical
> interpretation of gravity begins.
Yes, I know, except that Einstein indicates it was the rotating disk
experiment.
However, with all due respect, your own application of the twin
paradox experiment here is precisely what I meant by 'naive'
application of the general relativistic Doppler shift.
You can, and I have in the past, employed this argument to 'prove'
that the above 'naive' formula is correct.
Problem is, neither this temporal paradox nor its resolution are
physically meaningful in any observable sense.
Jump up and down under the stars on a clear night, and you will not
see them alternating between black & brighter blue.
Calculate the time each twin will actually observe passing for the
other on just the outward and return legs, and you will see the
totals fit SR theory perfectly.
The only reason for this apparent temporal paradox is that Einstein
did not bother to include the pure Doppler shift in his time dilation
formula, and that is by far the most important factor in determining
what we actually see of the passage of time, in this situation.
Oh No
>On Jun 19, 2:20 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> Thus spake Chalky <chalkys...@bleachboys.co.uk>
>>
>> >However, you seem to have 'jumped the gun' here. This specific
>> >question has nothing to do with cosmology per se. It is a cosmology
>> >independent question whose answer should be directly derivable from
>> >SR.
>>
>> If that is so, I have indeed misunderstood the question :-).
>
>Indeed.
>
>So let me put it differently. At the present point in time I am
>interested specifically in the redshift/distance relationship for a
>rigid linearly accelerating reference frame.
>
>This is a sufficiently well known theoretical relativistic construct
>that there must be some formal conclusion somewhere, and I would like
>to know what it is.
>
>> If one is
>> to think of the relationship of gravity with accelerating observers, a
>> good place to start (as did Einstein) is with the equivalence principle.
>>
>> http://www.teleconnection.info/rqg/TheEquivalencePrinciple
>>
>> The twin paradox is a sort of link which is treatable in sr, and set him
>> on the path toward gr. Of course, this only works as simply for a
>> uniform gravitational field. But that is where the geometrical
>> interpretation of gravity begins.
>
>Yes, I know, except that Einstein indicates it was the rotating disk
>experiment.
He did that, but iirc much later. Harry may correct me, but iirc his
initial calculations were done from the twin paradox c1907.
>
>
>However, with all due respect, your own application of the twin
>paradox experiment here is precisely what I meant by 'naive'
>application of the general relativistic Doppler shift.
>You can, and I have in the past, employed this argument to 'prove'
>that the above 'naive' formula is correct.
If your calculation is correct, then your formula is correct.
Unfortunately I don't have a calculation to hand or time to do one, nor
do I have a reference.
>
>Problem is, neither this temporal paradox nor its resolution are
>physically meaningful in any observable sense.
>
>Jump up and down under the stars on a clear night, and you will not
>see them alternating between black & brighter blue.
Actually the effect is there, just too small to observe by eye. It was
used to measure the redshift in the pound rebka experiment.
>
>Calculate the time each twin will actually observe passing for the
>other on just the outward and return legs, and you will see the
>totals fit SR theory perfectly.
As it should.
>
>The only reason for this apparent temporal paradox is that Einstein
>did not bother to include the pure Doppler shift in his time dilation
>formula, and that is by far the most important factor in determining
>what we actually see of the passage of time, in this situation.
>
In fact this was his original argument that there must be a
gravitational shift, as was later measured by Pound & Rebka.
Chalky
Thanks for that. He does not mention this in his 1920 book.
> >However, with all due respect, your own application of the twin
> >paradox experiment here is precisely what I meant by 'naive'
> >application of the general relativistic Doppler shift.
> >You can, and I have in the past, employed this argument to 'prove'
> >that the above 'naive' formula is correct.
>
> If your calculation is correct, then your formula is correct.
> Unfortunately I don't have a calculation to hand or time to do one, nor
> do I have a reference.
>
>
>
> >Problem is, neither this temporal paradox nor its resolution are
> >physically meaningful in any observable sense.
>
> >Jump up and down under the stars on a clear night, and you will not
> >see them alternating between black & brighter blue.
>
> Actually the effect is there, just too small to observe by eye. It was
> used to measure the redshift in the pound rebka experiment.
We can't be talking about the same effect here. Once you have left the
ground you are accelerating away from the stars at 981 cm/sec^2. If
the formula were correct, everything more than 1 light year away
should be infinitely redshifted.
>From my reading of http://en.wikipedia.org/wiki/Pound-Rebka_experiment
they measured the gravitational redshift of the Earth, not of the
linearly accelerating observer.
> >Calculate the time each twin will actually observe passing for the
> >other on just the outward and return legs, and you will see the
> >totals fit SR theory perfectly.
>
> As it should.
>
> >The only reason for this apparent temporal paradox is that Einstein
> >did not bother to include the pure Doppler shift in his time dilation
> >formula, and that is by far the most important factor in determining
> >what we actually see of the passage of time, in this situation.
>
> In fact this was his original argument that there must be a
> gravitational shift, as was later measured by Pound & Rebka.
True, and I think we can all agree that this can be represented as 1/
(1 - phi) to a first approximation (in natural units), for all
gravitational fields. However, there do appear to be subtle
differences depending on how you produce the gravitational field, and
perform the analysis.
For example, repeating the rotating disk experiment I get sqrt(1 +
2phi), which only equals 1/(1-phi) to a first approximation.
Chalky
>
> they measured the gravitational redshift of the Earth, not of the
> linearly accelerating observer.
>
> > >Calculate the time each twin will actually observe passing for the
> > >other on just the outward and return legs, and you will see the
> > >totals fit SR theory perfectly.
>
> > As it should.
>
> > >The only reason for this apparent temporal paradox is that Einstein
> > >did not bother to include the pure Doppler shift in his time dilation
> > >formula, and that is by far the most important factor in determining
> > >what we actually see of the passage of time, in this situation.
>
> > In fact this was his original argument that there must be a
> > gravitational shift, as was later measured by Pound & Rebka.
>
> True, and I think we can all agree that this can be represented as 1/
> (1 - phi) to a first approximation (in natural units), for all
> gravitational fields. However, there do appear to be subtle
> differences depending on how you produce the gravitational field, and
> perform the analysis.
>
> For example, repeating the rotating disk experiment I get sqrt(1 +
> 2phi), which only equals 1/(1-phi) to a first approximation.
And, reading between the lines of that 1920 book (and MTW), I think we
can also say with reasonable confidence that Einstein only adopted the
figure of 1/(1-phi) for natural gravitational fields, since that
achieves energy conservation for the redshifting of light.
When it comes to man made gravitational fields, all bets are off, on
this subject.
======================================= MODERATOR'S COMMENT:
Moderator: approved, but what is a man-made gravitational field?
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 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 –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
Oh No
>On Jun 19, 11:13 pm, Oh No <N...@charlesfrancis.wanadoo.co.uk> wrote:
>> >Jump up and down under the stars on a clear night, and you will not
>> >see them alternating between black & brighter blue.
>>
>> Actually the effect is there, just too small to observe by eye. It was
>> used to measure the redshift in the pound rebka experiment.
>
>
>We can't be talking about the same effect here. Once you have left the
>ground you are accelerating away from the stars at 981 cm/sec^2. If
>the formula were correct, everything more than 1 light year away
>should be infinitely redshifted.
You would not get that affect unless you had uniform acceleration - not
achieved by jumping up and down.
>
>>From my reading of http://en.wikipedia.org/wiki/Pound-Rebka_experiment
>they measured the gravitational redshift of the Earth, not of the
>linearly accelerating observer.
The equivalence principle says that this is the same for a uniform
gravitational field.
Oh No
>When it comes to man made gravitational fields, all bets are off, on
>this subject.
>
>
>======================================= MODERATOR'S COMMENT:
> Moderator: approved, but what is a man-made gravitational field?
>
moderators have been reminded to use comments only for moderation
issues, as confusion can result from comments on physics which are
better made in posts (in this case confusion arose because I was not the
moderator who made the comment).
Posters should be aware that we cannot redirect rejection notes to
grunged addresses.
I would also remind posters to snip unnecessary quoted text. Our
software does not allow us to do this, and it is also a reason for
rejection.
]
Chalky
> 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.
Yes, I am aware of this conclusion. However, it appears to introduce a
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.
Clearly another observer at that (inertial) origin cannot experience
accelerations of zero and infinity at the same time.
> 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.
Thanks for the subsequent advice but with this level of confusion at
the start, I was unable to follow the rest of your reasoning.
Shubee
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
Chalky
Yes, of course UR correct. This was a mere slip of the typewriter.
(aka typo)
> 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.
Please don't make disparaging assumptions. I recall that additional
result perfectly well. It is quoted, for example, in MTW.
The result to which I referred, which you appear to be unaware of, is
derived, for example, at
http://physics.nmt.edu/~raymond/classes/ph13xbook/node61.html
This ref was written by a Professor of physics who confirms it is also
a standard text book result.
I suggest you read it very carefully, before commenting again on the
below quoted text.
Chalky
pm, Shubee <e.Shu...@gmail.com> wrote:
> And it doesn’t make sense to me that you would want to change the
> definition of rocket coordinates in midstream.
Ahem. You failed to define the rocket coordinates. The first two
options I considered work equally well for your first two formulae.
>There are good reasons
> to read carefully.
There are good reasons to write what you think you have written.
> The choice for the point x'=0 to represent the
> event horizon was completely arbitrary.
Quite. Your third formula produces nonsense no matter where you locate
that origin.
Chalky
> I answered your inquiry directly.
> I derived the equation:
> f' = f (1-gL/c^2).
>
> Do you understand that equation?
Of course I do. I introduced it in my original posting:
"In the past, I have tended to assume that the z shift distance
relationship in a 'rigid' linearly accelerating reference frame could
be obtained to a good approximation by naive application of the
gravitational Doppler shift thus:
1 + z = 1 / (1 - g.d/c^2) "
Your formula and my above formulae are equivalent.
However, you are now claiming you have proven that this formula is
exactly true.
Your claim is nonsense because your third premise [g(x')=-(c^2)/x'] is
demonstrably nonsense (since it leads to nonsensical conclusions no
matter where you place your spatial origin for x').
Or do you actually believe that if you locate this origin somewhere in
cloud cuckoo land, that third premise will magically become true?
Chalky
> Thus spake Chalky <chalkys...@bleachboys.co.uk>
>
> >> >Jump up and down under the stars on a clear night, and you will not
> >> >see them alternating between black & brighter blue.
>
> >> Actually the effect is there, just too small to observe by eye. It was
> >> used to measure the redshift in the pound rebka experiment.
>
> >We can't be talking about the same effect here. Once you have left the
> >ground you are accelerating away from the stars at 981 cm/sec^2. If
> >the formula were correct, everything more than 1 light year away
> >should be infinitely redshifted.
>
> You would not get that affect unless you had uniform acceleration - not
> achieved by jumping up and down.
The problem with this argument is that the effect was derived from the
twin paradox experiment, using a mere pulse of acceleration.
This derivation produces the same result no matter how small v is, and
the pulse can therefore easily be as little as 1 g for 1 second. This
can be reproduced perfectly well by jumping up and down. When you do,
no such red/blue shift is observed.
This confirms that that method of derivation is insufficiently
rigorous.
Oh No
calculation assumes zero spacetime curvature. Here we have to contend
with the gravitational field of the earth, which we interpret as meaning
that spacetime is not flat.
Shubee
The x' coordinate is defined to be the Lorentz invariant quantity
(x')^2 =x^2-(ct)^2
Please note that this well-understood Lorentz invariant follows
trivially from the equations:
x=x'cosh(ct'/x')
t=(x'/c)sinh(ct'/x')
Shubee
http://www.everythingimportant.org/relativity/special.pdf
Chalky
> On Jun 23, 3:25 pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > On Jun 22, 10:58 am, Shubee <e.Shu...@gmail.com> wrote:
> > > 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.
>
> The x' coordinate is defined to be the Lorentz invariant quantity
>
> (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.
Chalky
Or are you now redefining x and t as the coordinates of the rocket in
the inertial frame?
Perhaps you had better start again by carefully defining both what you
mean by x and t, and what you mean by x' and t'
Shubee
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
Chalky
Well none of this makes any sense either.
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.
Clearly infinity is never 'small' in comparison to a finite quantity.
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.