| Topic: |
Science > Physics |
| User: |
"Edward Green" |
| Date: |
09 Oct 2004 03:14:49 PM |
| Object: |
relativistic centrifuge |
I wondered if gravitational time dilation might just be same as the
kinematic time dilation of a centrifuge, expressed in terms of the
acceleration, a. It seems unlikely such a simple possibility would
have escaped notice, but the Nobel prize awaited me, so...
In SR, the rate of a moving clock is slowed by sqrt[1 - v^2/c^2], or
sqrt[1 - aR/c^2]
expressed in terms of acceleration and radius for a centrifuge.
In GR, the rate of a clock on the surface of a planet is slowed by
sqrt[1 - 2aR/c^2]
where a == g, the acceleration of gravity, R the planetary radius.
Damn! Current understanding, borne out by experiment, is that these
are distinct effects. But what an uncanny similarity of form!
What does this mean?
.
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| User: "Old Man" |
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| Title: Re: relativistic centrifuge |
10 Oct 2004 01:34:18 AM |
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"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0410091214.2ff05332@posting.google.com...
I wondered if gravitational time dilation might just be same as the
kinematic time dilation of a centrifuge, expressed in terms of the
acceleration, a. It seems unlikely such a simple possibility would
have escaped notice, but the Nobel prize awaited me, so...
In SR, the rate of a moving clock is slowed by sqrt[1 - v^2/c^2], or
sqrt[1 - aR/c^2]
expressed in terms of acceleration and radius for a centrifuge.
In GR, the rate of a clock on the surface of a planet is slowed by
sqrt[1 - 2aR/c^2]
where a == g, the acceleration of gravity, R the planetary radius.
Damn! Current understanding, borne out by experiment, is that these
are distinct effects. But what an uncanny similarity of form!
What does this mean?
Old Man wonders why, for a centrifuge of radius, R, and
acceleration, a, the ratio of proper time, t_p, to that at the
center of the centrifuge, t_0, isn't given by
t_p / t_0 = sqrt{[ 1 - ( 2 a R / c^2 ) ] [ 1 - ( a R / c^2 ) ]}
[Old Man]
.
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| User: "Ken S. Tucker" |
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| Title: Re: relativistic centrifuge |
10 Oct 2004 05:14:32 AM |
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"Old Man" <nomail@nomail.net> wrote in message news:<a9ydnZyRCdFxSvXcRVn-iQ@prairiewave.com>...
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0410091214.2ff05332@posting.google.com...
I wondered if gravitational time dilation might just be same as the
kinematic time dilation of a centrifuge, expressed in terms of the
acceleration, a. It seems unlikely such a simple possibility would
have escaped notice, but the Nobel prize awaited me, so...
In SR, the rate of a moving clock is slowed by sqrt[1 - v^2/c^2], or
sqrt[1 - aR/c^2]
expressed in terms of acceleration and radius for a centrifuge.
In GR, the rate of a clock on the surface of a planet is slowed by
sqrt[1 - 2aR/c^2]
where a == g, the acceleration of gravity, R the planetary radius.
Damn! Current understanding, borne out by experiment, is that these
are distinct effects. But what an uncanny similarity of form!
What does this mean?
Old Man wonders why, for a centrifuge of radius, R, and
acceleration, a, the ratio of proper time, t_p, to that at the
center of the centrifuge, t_0, isn't given by
t_p / t_0 = sqrt{[ 1 - ( 2 a R / c^2 ) ] [ 1 - ( a R / c^2 ) ]}
[Old Man]
Well the point in the center of the disc is at rest relative
to the point on the rim. Recall they are NOT relatively displacing,
and SR was constructed when x and x' are shifting, i.e. K and K'
CS origins were in relative motion defined by a changing distance
between them. That doesn't happen in the disc experiment.
I'm ok with 2aR == v^2 in the gamma, hence,
1/gamma^2 = 1-v^2 = 1-2aR = 1-(2GM/r^2)r = 1-2GM/r =~ g_00
(SR) (SR+a) (gravity) (field) (GR)
It's a short cut from SR to GR.
Ken
therfore SR does NOT apply.
.
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| User: "Edward Green" |
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| Title: Re: relativistic centrifuge |
10 Oct 2004 01:21:27 PM |
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"Old Man" <nomail@nomail.net> wrote in message news:<a9ydnZyRCdFxSvXcRVn-iQ@prairiewave.com>...
Old Man wonders why, for a centrifuge of radius, R, and
acceleration, a, the ratio of proper time, t_p, to that at the
center of the centrifuge, t_0, isn't given by
t_p / t_0 = sqrt{[ 1 - ( 2 a R / c^2 ) ] [ 1 - ( a R / c^2 ) ]}*
I don't know: just how far is Old Man prepared to discuss the basic
comceptual problems of GR and SR with me? I ask this not from the
stand point of a know-it-all, but as someone who both is conscious of
periodic confusions in his own mind, and does not feel confident all
such features have been purged from the minds of those not considered
unqualified to comment on these things -- aside from obvious
crackpots.
In http://scienceworld.wolfram.com/physics/GeneralRelativity.html, we
meet the following claim:
"The fundamental principle of general relativity asserts that
accelerated reference frames and reference frames in gravitation
fields are equivalent. General relativity states that clocks run
slower in strong gravitational fields (or highly accelerated
frames)..."
I was on the point of criticizing this as follows: in the absence of
gravitational fields GR reduces to SR. And SR includes no component
which predicts time dilation based on acceleration per se. Therefore
neither does GR, and the claim is false.
However, I see a small fly in my ointment. In arguments supporting
the equivalence principle, it is common to imagine an accelerating
spaceship, and
calculate the rate of receipt at the bow of pulses periodically
emitted at the stern, and conclude that the rate of receipt is smaller
than that of emission as controlled by a clock at the stern. So here
is an effect "in an accelerated frame" predicted purely from SR.
However, it would be misleading to offer this effect as, by
implication, something peculiar to GR.
I _think_ it is true that the rate of an accelerating clock is handled
correctly in SR, and that there are no factors beyond (1) the relative
velocity of the clock, and (2) any direct effects of acceleration on
the mechanism of the clock.
This latter takes us perilously close to the Scylla of so-called
Lorentz Aether Theory -- which is a point on which I appear somewhat
cranky, since I insist it is an approach not wrong in spirit nor
alternative to the meat of SR, but an alternative and equivalent way
of understanding what is going on: an antithesis which a full
synthesis must include. If we tell the tale of the accelerating
spaceship within a single atom, we may well have a kind of Lorentzian
Aether Theory for the case of acceleration, which may leave us with a
semantic issue whether this is a fundamental effect of acceleration
which must be added by hand into SR -- a harbinger of GR -- or a
simple consequence of the physics of an accelerated body, analyzed in
SR.
Anyway, moving on, consider this alternative account:
http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html
"It is a common misconception that Special Relativity cannot handle
accelerating objects or accelerating reference frames. It is claimed
that general relativity is required because special relativity only
applies to inertial frames. This is not true. Special relativity
treats accelerating frames differently from inertial frames but can
still deal with them. Accelerating objects can be dealt with without
even calling upon accelerating frames."
Ok. So far, so good.
"The difference between general and special relativity is that in the
general theory all frames of reference including spinning and
accelerating frames are treated on an equal footing. In special
relativity accelerating frames are different from inertial frames.
Velocities are relative but acceleration is treated as absolute. In
general relativity all motion is relative."
Whoa-oh. Not so good!
The alleged distinction refers merely to an accidental formal
difference between the presentations, and not to new physics. GR is
expressed in a way which handles all frames equivalently, but the same
mechanism could be applied to a purely flat world, ruled by SR.
Furthermore acceleration is absolute even in GR: physics continues to
be measureably different on a bucket rotating relative to a first
bucket, whereas physics is not different on a bucket translating
relative to a first bucket. Local experiments can in general
distinguish the first two cases but not the second -- our now unified
formalism notwithstanding.
* (You mean of course, aside from a symbolic factor of "^-1").
.
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| User: "Old Man" |
|
| Title: Re: relativistic centrifuge |
10 Oct 2004 09:02:18 PM |
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"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0410101021.2ee424e@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:<a9ydnZyRCdFxSvXcRVn-iQ@prairiewave.com>...
Old Man wonders why, for a centrifuge of radius, R, and
acceleration, a, the ratio of proper time, t_p, to that at the
center of the centrifuge, t_0, isn't given by
t_p / t_0 = sqrt{[ 1 - ( 2 a R / c^2 ) ] [ 1 - ( a R / c^2 ) ]}*
I don't know: just how far is Old Man prepared to discuss the basic
comceptual problems of GR and SR with me? I ask this not from the
stand point of a know-it-all, but as someone who both is conscious of
periodic confusions in his own mind, and does not feel confident all
such features have been purged from the minds of those not considered
unqualified to comment on these things -- aside from obvious
crackpots.
In http://scienceworld.wolfram.com/physics/GeneralRelativity.html, we
meet the following claim:
"The fundamental principle of general relativity asserts that
accelerated reference frames and reference frames in gravitation
fields are equivalent. General relativity states that clocks run
slower in strong gravitational fields (or highly accelerated
frames)..."
I was on the point of criticizing this as follows: in the absence of
gravitational fields GR reduces to SR. And SR includes no component
which predicts time dilation based on acceleration per se. Therefore
neither does GR, and the claim is false.
For global observations, like time dilation, SR and GR
aren't required to agree. The EP of GR concerns local
observations and has naught to with global observations.
Other than that, Old Man is even less certain than Ed.
Here's why:
WRT a centrifuge:
When applying a = v^2 / R, for gravitational time dilation,
we have a factor of 2 in the term
GR: (2 G M) / (R c^2) => (2 a R) / (c^2),
whereas for relativistic time dilation, we have,
SR: (v / c)^2 => (a R) / (c^2)
wherein there is no factor of 2. Therefore, GR and SR
don't predict the same global result, but locally, there's no
difficulty. Still, Old Man wonders if we haven't used an
erroneous radius, R, because, for centrifugal force ( |v| is
constant ),
F = dp / dt = m (dv / dt) (gamma)
thus, for a = dv / dt = v^2 / R_p, we get ( R_p is the
proper radius and, R_0, the coordinate radius)
R_0 / R_p = (gamma)
This is correct in the case of a uniform magnetic field,
whereof F = q (v x B). See "Gravity" by J. B. Hartle,
ISBN 0-8053-8662-9
So, In the substations,
(2 G M / R) => (2 a R) and (v^2) => (a R),
is R to be taken as the coordinate radius in both cases ?
The same question applies to the acceleration, a.
However, this correction isn't enough to make-up that
factor of 2. After all this rigmarole, no matter what, it
seems that the global predictions of GR and SR, WRT
time dilation for a centrifuge, remain divergent. That's
OK with Old Man.
However, I see a small fly in my ointment. In arguments supporting
the equivalence principle, it is common to imagine an accelerating
spaceship, and
calculate the rate of receipt at the bow of pulses periodically
emitted at the stern, and conclude that the rate of receipt is smaller
than that of emission as controlled by a clock at the stern. So here
is an effect "in an accelerated frame" predicted purely from SR.
However, it would be misleading to offer this effect as, by
implication, something peculiar to GR.
I _think_ it is true that the rate of an accelerating clock is handled
correctly in SR, and that there are no factors beyond (1) the relative
velocity of the clock, and (2) any direct effects of acceleration on
the mechanism of the clock.
This latter takes us perilously close to the Scylla of so-called
Lorentz Aether Theory -- which is a point on which I appear somewhat
cranky, since I insist it is an approach not wrong in spirit nor
alternative to the meat of SR, but an alternative and equivalent way
of understanding what is going on: an antithesis which a full
synthesis must include. If we tell the tale of the accelerating
spaceship within a single atom, we may well have a kind of Lorentzian
Aether Theory for the case of acceleration, which may leave us with a
semantic issue whether this is a fundamental effect of acceleration
which must be added by hand into SR -- a harbinger of GR -- or a
simple consequence of the physics of an accelerated body, analyzed in
SR.
Anyway, moving on, consider this alternative account:
http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html
"It is a common misconception that Special Relativity cannot handle
accelerating objects or accelerating reference frames. It is claimed
that general relativity is required because special relativity only
applies to inertial frames. This is not true. Special relativity
treats accelerating frames differently from inertial frames but can
still deal with them. Accelerating objects can be dealt with without
even calling upon accelerating frames."
Ok. So far, so good.
"The difference between general and special relativity is that in the
general theory all frames of reference including spinning and
accelerating frames are treated on an equal footing. In special
relativity accelerating frames are different from inertial frames.
Velocities are relative but acceleration is treated as absolute. In
general relativity all motion is relative."
Whoa-oh. Not so good!
The alleged distinction refers merely to an accidental formal
difference between the presentations, and not to new physics. GR is
expressed in a way which handles all frames equivalently, but the same
mechanism could be applied to a purely flat world, ruled by SR.
Furthermore acceleration is absolute even in GR: physics continues to
be measureably different on a bucket rotating relative to a first
bucket, whereas physics is not different on a bucket translating
relative to a first bucket. Local experiments can in general
distinguish the first two cases but not the second -- our now unified
formalism notwithstanding.
* (You mean of course, aside from a symbolic factor of "^-1").
That acceleration and gravitation are guaranteed to be locally
indistinguishable, doesn't imply that the globally observed time
dilation for acceleration is equal to that of gravitation.
[Old Man]
.
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| User: "Ken S. Tucker" |
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| Title: Re: relativistic centrifuge |
11 Oct 2004 11:18:02 AM |
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"Old Man" <nomail@nomail.net> wrote in message news:<uLqdnQ-o-aYxdPTcRVn-gQ@prairiewave.com>...
....
WRT a centrifuge:
When applying a = v^2 / R, for gravitational time dilation,
we have a factor of 2 in the term
GR: (2 G M) / (R c^2) => (2 a R) / (c^2),
whereas for relativistic time dilation, we have,
SR: (v / c)^2 => (a R) / (c^2)
wherein there is no factor of 2.
I'm not sure if this is helpful, but what I use is,
v = at , s = (1/2)at^2 then 2as = v^2
then using Newtons a=GM/r^2, s=r , 2GM/r = 2as = v^2.
Ken
....
[Old Man]
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| User: "Bjoern Feuerbacher" |
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| Title: Re: relativistic centrifuge |
11 Oct 2004 11:58:56 AM |
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Ken S. Tucker wrote:
"Old Man" <nomail@nomail.net> wrote in message news:<uLqdnQ-o-aYxdPTcRVn-gQ@prairiewave.com>...
...
WRT a centrifuge:
When applying a = v^2 / R, for gravitational time dilation,
we have a factor of 2 in the term
GR: (2 G M) / (R c^2) => (2 a R) / (c^2),
whereas for relativistic time dilation, we have,
SR: (v / c)^2 => (a R) / (c^2)
wherein there is no factor of 2.
I'm not sure if this is helpful, but what I use is,
v = at , s = (1/2)at^2 then 2as = v^2
Err, if you did not notice, they are talking about *centripetal*
acceleration here, for which aR = v^2 holds. *Not* 2aR = v^2.
[snip]
Bye,
Bjoern
.
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| User: "Ken S. Tucker" |
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| Title: Re: relativistic centrifuge |
12 Oct 2004 12:40:40 AM |
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Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<ckee4g$1fp$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
"Old Man" <nomail@nomail.net> wrote in message news:<uLqdnQ-o-aYxdPTcRVn-gQ@prairiewave.com>...
...
WRT a centrifuge:
When applying a = v^2 / R, for gravitational time dilation,
we have a factor of 2 in the term
GR: (2 G M) / (R c^2) => (2 a R) / (c^2),
whereas for relativistic time dilation, we have,
SR: (v / c)^2 => (a R) / (c^2)
wherein there is no factor of 2.
I'm not sure if this is helpful, but what I use is,
v = at , s = (1/2)at^2 then 2as = v^2
ken
Err, if you did not notice, they are talking about *centripetal*
acceleration here, for which aR = v^2 holds. *Not* 2aR = v^2.
Bjoern
Right, Vo = circular orbital velocity, Ve = escape velocity.
Vo^2 = GM/r , Ve^2 = 2GM/r
The Ve^2 is what to insert into SR's sqrt(1-v^2) to get a GR
equivalent. I pointed out in a previous post SR does not apply,
and is useless in centifuges, because the relative velocity of
the center of the disc and the rim is zero. In a centrifuge,
the metrics "g_uv" aren't constant either, so GR is required,
for anything except a superficial analysis.
Ken
.
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| User: "Old Man" |
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| Title: Re: relativistic centrifuge |
12 Oct 2004 02:15:31 AM |
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"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:2202379a.0410112140.5af5dfbc@posting.google.com...
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message
news:<ckee4g$1fp$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
"Old Man" <nomail@nomail.net> wrote in message
news:<uLqdnQ-o-aYxdPTcRVn-gQ@prairiewave.com>...
...
WRT a centrifuge:
When applying a = v^2 / R, for gravitational time dilation,
we have a factor of 2 in the term
GR: (2 G M) / (R c^2) => (2 a R) / (c^2),
whereas for relativistic time dilation, we have,
SR: (v / c)^2 => (a R) / (c^2)
wherein there is no factor of 2.
I'm not sure if this is helpful, but what I use is,
v = at , s = (1/2)at^2 then 2as = v^2
ken
Err, if you did not notice, they are talking about *centripetal*
acceleration here, for which aR = v^2 holds. *Not* 2aR = v^2.
Bjoern
Right, Vo = circular orbital velocity, Ve = escape velocity.
Vo^2 = GM/r , Ve^2 = 2GM/r
The Ve^2 is what to insert into SR's sqrt(1-v^2) to get a GR
equivalent. I pointed out in a previous post SR does not apply,
and is useless in centifuges, because the relative velocity of
the center of the disc and the rim is zero. In a centrifuge,
the metrics "g_uv" aren't constant either, so GR is required,
for anything except a superficial analysis.
Ken
Well, ... gravitational time dilation does depend upon Newtonian
gravitational potential,
t_p / t_0 = sqrt[ 1 + 2 U(R) / c^2 ]
where, U(R) = - GM / R,
and not upon field acceleration. So, escape velocity for U => 0
as R => infinity does seem most appropreate for gravitation.
Lets calculate the work required to move to the center of the
centrifuge where U = 0. That's easy. It's just the kinetic energy,
(1 / 2) m v^2. Thus,
U(R) = - (1 / 2) v^2
and gravitational time dilation in a centrifuge is then
gravitational: t_p / t_0 = sqrt[ 1 - (v / c)^2 ]
which is also that of SR. So, SR and GR do agree on globally
observed time dilation. That's fine with Old Man.
[Old Man]
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| User: "Mike" |
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| Title: Re: relativistic centrifuge |
12 Oct 2004 09:44:49 AM |
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"Old Man" <nomail@nomail.net> wrote in message news:<ms2dna-K24QKGfbcRVn-tg@prairiewave.com>...
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:2202379a.0410112140.5af5dfbc@posting.google.com...
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message
news:<ckee4g$1fp$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
"Old Man" <nomail@nomail.net> wrote in message
news:<uLqdnQ-o-aYxdPTcRVn-gQ@prairiewave.com>...
...
WRT a centrifuge:
When applying a = v^2 / R, for gravitational time dilation,
we have a factor of 2 in the term
GR: (2 G M) / (R c^2) => (2 a R) / (c^2),
whereas for relativistic time dilation, we have,
SR: (v / c)^2 => (a R) / (c^2)
wherein there is no factor of 2.
I'm not sure if this is helpful, but what I use is,
v = at , s = (1/2)at^2 then 2as = v^2
ken
Err, if you did not notice, they are talking about *centripetal*
acceleration here, for which aR = v^2 holds. *Not* 2aR = v^2.
Bjoern
Right, Vo = circular orbital velocity, Ve = escape velocity.
Vo^2 = GM/r , Ve^2 = 2GM/r
The Ve^2 is what to insert into SR's sqrt(1-v^2) to get a GR
equivalent. I pointed out in a previous post SR does not apply,
and is useless in centifuges, because the relative velocity of
the center of the disc and the rim is zero. In a centrifuge,
the metrics "g_uv" aren't constant either, so GR is required,
for anything except a superficial analysis.
Ken
Well, ... gravitational time dilation does depend upon Newtonian
gravitational potential,
t_p / t_0 = sqrt[ 1 + 2 U(R) / c^2 ]
where, U(R) = - GM / R,
and not upon field acceleration. So, escape velocity for U => 0
as R => infinity does seem most appropreate for gravitation.
Lets calculate the work required to move to the center of the
centrifuge where U = 0. That's easy. It's just the kinetic energy,
(1 / 2) m v^2. Thus,
U(R) = - (1 / 2) v^2
and gravitational time dilation in a centrifuge is then
gravitational: t_p / t_0 = sqrt[ 1 - (v / c)^2 ]
which is also that of SR. So, SR and GR do agree on globally
observed time dilation. That's fine with Old Man.
[Old Man]
They do not agree locally because centrifugal acceleration seems to be
real. If you add the magnitudes of centripetal and centrifugal for
local observers in rotation you get 2v^2/R.
That says a lot. Although GR transfers out ficticious accelerations by
adding a 4th dimension in its metric, in reality, you need to account
for that ficticious accelerations to match the predictions of GR.
Thus, GR hides out the so called ficticious force using math but in
reality they manifest and are not ficticious at all because they
contribute to time dilation.
Mike
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| User: "Bjoern Feuerbacher" |
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| Title: Re: relativistic centrifuge |
12 Oct 2004 04:15:58 AM |
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Ken S. Tucker wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<ckee4g$1fp$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
"Old Man" <nomail@nomail.net> wrote in message news:<uLqdnQ-o-aYxdPTcRVn-gQ@prairiewave.com>...
...
WRT a centrifuge:
When applying a = v^2 / R, for gravitational time dilation,
we have a factor of 2 in the term
GR: (2 G M) / (R c^2) => (2 a R) / (c^2),
whereas for relativistic time dilation, we have,
SR: (v / c)^2 => (a R) / (c^2)
wherein there is no factor of 2.
I'm not sure if this is helpful, but what I use is,
v = at , s = (1/2)at^2 then 2as = v^2
ken
Err, if you did not notice, they are talking about *centripetal*
acceleration here, for which aR = v^2 holds. *Not* 2aR = v^2.
Bjoern
Right, Vo = circular orbital velocity, Ve = escape velocity.
Vo^2 = GM/r , Ve^2 = 2GM/r
The Ve^2 is what to insert into SR's sqrt(1-v^2) to get a GR
equivalent.
Why?
I pointed out in a previous post SR does not apply,
and is useless in centifuges, because the relative velocity of
the center of the disc and the rim is zero.
Why on earth do you think so??? What is the velocity Vo above, in your
opinion, if not the velocity of the rim?
[snip]
Bye,
Bjoern
.
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| User: "Ken S. Tucker" |
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| Title: Re: relativistic centrifuge |
12 Oct 2004 12:32:06 PM |
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Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<ckg7ce$f04$2@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<ckee4g$1fp$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
"Old Man" <nomail@nomail.net> wrote in message news:<uLqdnQ-o-aYxdPTcRVn-gQ@prairiewave.com>...
...
WRT a centrifuge:
When applying a = v^2 / R, for gravitational time dilation,
we have a factor of 2 in the term
GR: (2 G M) / (R c^2) => (2 a R) / (c^2),
whereas for relativistic time dilation, we have,
SR: (v / c)^2 => (a R) / (c^2)
wherein there is no factor of 2.
I'm not sure if this is helpful, but what I use is,
v = at , s = (1/2)at^2 then 2as = v^2
ken
Err, if you did not notice, they are talking about *centripetal*
acceleration here, for which aR = v^2 holds. *Not* 2aR = v^2.
Bjoern
Right, Vo = circular orbital velocity, Ve = escape velocity.
Vo^2 = GM/r , Ve^2 = 2GM/r
The Ve^2 is what to insert into SR's sqrt(1-v^2) to get a GR
equivalent.
Why?
Well GR Prof's really don't like the step, but it seems
to work out "superficially". The way Old Man worked it
out using kinetic energy seems ok to me.
I pointed out in a previous post SR does not apply,
and is useless in centifuges, because the relative velocity of
the center of the disc and the rim is zero.
Why on earth do you think so??? What is the velocity Vo above, in your
opinion, if not the velocity of the rim?
Well calculate "v" using the relativity of the center and
the rim, v=dr/dt=0.
In GR rotating CS's are permissable, as well.
[snip]
Bye,
Bjoern
.
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| User: "Gregory L. Hansen" |
|
| Title: Re: relativistic centrifuge |
12 Oct 2004 01:00:32 PM |
|
|
In article <2202379a.0410120932.5f9d7b10@posting.google.com>,
Ken S. Tucker <dynamics@vianet.on.ca> wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message
news:<ckg7ce$f04$2@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in
message news:<ckee4g$1fp$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
"Old Man" <nomail@nomail.net> wrote in message
news:<uLqdnQ-o-aYxdPTcRVn-gQ@prairiewave.com>...
...
WRT a centrifuge:
When applying a = v^2 / R, for gravitational time dilation,
we have a factor of 2 in the term
GR: (2 G M) / (R c^2) => (2 a R) / (c^2),
whereas for relativistic time dilation, we have,
SR: (v / c)^2 => (a R) / (c^2)
wherein there is no factor of 2.
I'm not sure if this is helpful, but what I use is,
v = at , s = (1/2)at^2 then 2as = v^2
ken
Err, if you did not notice, they are talking about *centripetal*
acceleration here, for which aR = v^2 holds. *Not* 2aR = v^2.
Bjoern
Right, Vo = circular orbital velocity, Ve = escape velocity.
Vo^2 = GM/r , Ve^2 = 2GM/r
The Ve^2 is what to insert into SR's sqrt(1-v^2) to get a GR
equivalent.
Why?
Well GR Prof's really don't like the step, but it seems
to work out "superficially". The way Old Man worked it
out using kinetic energy seems ok to me.
I pointed out in a previous post SR does not apply,
and is useless in centifuges, because the relative velocity of
the center of the disc and the rim is zero.
Why on earth do you think so??? What is the velocity Vo above, in your
opinion, if not the velocity of the rim?
Well calculate "v" using the relativity of the center and
the rim, v=dr/dt=0.
\vec{v} = d\vec{r}/dt = (-wR sin(wt), wR cos(wt))
|\vec{v}| = sqrt((-wR sin(wt))^2 + (wR cos(wt))^2)
In GR rotating CS's are permissable, as well.
Rotations are part of the Lorentz group.
--
"Usenet is like a herd of performing elephants with diarrhea -- massive,
difficult to redirect, awe-inspiring, entertaining, and a source of
mind-boggling amounts of excrement when you least expect it. "
-- Gene Spafford, 1992
.
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| User: "Ken S. Tucker" |
|
| Title: Re: relativistic centrifuge |
13 Oct 2004 01:20:18 AM |
|
|
(Gregory L. Hansen) wrote in message news:<ckh640$ts4$1@hood.uits.indiana.edu>...
In article <2202379a.0410120932.5f9d7b10@posting.google.com>,
Ken S. Tucker <dynamics@vianet.on.ca> wrote:
....
Well calculate "v" using the relativity of the center and
the rim, v=dr/dt=0.
\vec{v} = d\vec{r}/dt = (-wR sin(wt), wR cos(wt))
|\vec{v}| = sqrt((-wR sin(wt))^2 + (wR cos(wt))^2)
In GR rotating CS's are permissable, as well.
Rotations are part of the Lorentz group.
I think SR applies to relative displacement. I can rotate the
entire universe by spinning in my chair, and
ds^2 = dt^2 - dr^2 stays the same,
rotations *seem* to make no difference.
Imagine a space ship moving in a straight line speeding
past you. You're an interstellar cop with a Doppler Radar
unit, and you will find a moment where the space ship is
instantaneously at rest relatively to you, as it goes from
approaching to receding
Also at that moment, a pedestrian is standing in line
directly behind the spaceship at rest relative to you,
and you measure at that moment, you, the spaceship, and
pedestrian are all relatively stationary.
At that moment all 3 clocks will have the same rate.
Regards
Ken S. Tucker
.
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| User: "Gregory L. Hansen" |
|
| Title: Re: relativistic centrifuge |
13 Oct 2004 09:19:31 AM |
|
|
In article <2202379a.0410122220.20f2fe40@posting.google.com>,
Ken S. Tucker <dynamics@vianet.on.ca> wrote:
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
news:<ckh640$ts4$1@hood.uits.indiana.edu>...
In article <2202379a.0410120932.5f9d7b10@posting.google.com>,
Ken S. Tucker <dynamics@vianet.on.ca> wrote:
...
Well calculate "v" using the relativity of the center and
the rim, v=dr/dt=0.
\vec{v} = d\vec{r}/dt = (-wR sin(wt), wR cos(wt))
|\vec{v}| = sqrt((-wR sin(wt))^2 + (wR cos(wt))^2)
In GR rotating CS's are permissable, as well.
Rotations are part of the Lorentz group.
I think SR applies to relative displacement. I can rotate the
entire universe by spinning in my chair, and
ds^2 = dt^2 - dr^2 stays the same,
rotations *seem* to make no difference.
That is a metric for straight-line motion. You may as well let dr->dx.
Rotational motion has a d(theta) term, too.
If you're spinning in your chair, you're in an accelerated (rotating)
reference frame. The d(theta) term may be transformed away, but the
simple metric you gave above will pick up inertial terms corresponding to
centrifugal and Coriolis forces, and a "gravitational" time dilation.
Imagine a space ship moving in a straight line speeding
past you. You're an interstellar cop with a Doppler Radar
unit, and you will find a moment where the space ship is
instantaneously at rest relatively to you, as it goes from
approaching to receding
There is no such point, although there will be a point where the Doppler
shifting is zero.
Also at that moment, a pedestrian is standing in line
directly behind the spaceship at rest relative to you,
and you measure at that moment, you, the spaceship, and
pedestrian are all relatively stationary.
At that moment all 3 clocks will have the same rate.
The spaceship will have the clock rate dt'=dt/sqrt(1-v^2/c^2). Doppler
shifting is not clock rate. Work it out from the Lorentz transform.
--
"In any case, don't stress too much--cortisol inhibits muscular
hypertrophy. " -- Eric Dodd
.
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| User: "Bjoern Feuerbacher" |
|
| Title: Re: relativistic centrifuge |
13 Oct 2004 04:51:54 AM |
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|
Ken S. Tucker wrote:
[snip]
Imagine a space ship moving in a straight line speeding
past you. You're an interstellar cop with a Doppler Radar
unit, and you will find a moment where the space ship is
instantaneously at rest relatively to you, as it goes from
approaching to receding
You make no sense at all. The relative velocity of the space ship
relative to me remains the *same* all the time. It is never zero.
Let my position vector be r_c and the position vector of the space ship
be r_s. The velocity of the space ship wrt me is then d/dt (r_c - r_s).
I am at rest in my own CS, hence d/dt r_c = 0. The space ship is moving
with constant velocity, hence d/dt r_s = v, where v is a *constant*
vector. Hence the velocity of the space ship wrt me is *constant*.
The *direction* of this vector is first pointing towards me and then
away from me. But that does in no way mean that at the time when it goes
from approaching to receding, its relative velocity is zero, i.e. it is
at rest!
Also at that moment, a pedestrian is standing in line
directly behind the spaceship at rest relative to you,
and you measure at that moment, you, the spaceship, and
pedestrian are all relatively stationary.
Utter nonsense.
I don't know if you were unable to understand the teachings of Prof.
Sawyer, or if you forget all the stuff you learned by him in the
meantime. But you show here *clearly* that you don't understand quite
basic physics.
[snip]
Bye,
Bjoern
.
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| User: "Ken S. Tucker" |
|
| Title: Re: relativistic centrifuge |
13 Oct 2004 01:45:07 PM |
|
|
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<ckitrq$7dt$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
[snip]
Imagine a space ship moving in a straight line speeding
past you. You're an interstellar cop with a Doppler Radar
unit, and you will find a moment where the space ship is
instantaneously at rest relatively to you, as it goes from
approaching to receding
You make no sense at all. The relative velocity of the space ship
relative to me remains the *same* all the time. It is never zero.
Let my position vector be r_c and the position vector of the space ship
be r_s. The velocity of the space ship wrt me is then d/dt (r_c - r_s).
I am at rest in my own CS, hence d/dt r_c = 0. The space ship is moving
with constant velocity, hence d/dt r_s = v, where v is a *constant*
vector. Hence the velocity of the space ship wrt me is *constant*.
How do you account for the Doppler Effect, even using sound
say as a train approaches and passes me by?
Bye,
Bjoern
.
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| User: "Old Man" |
|
| Title: Re: relativistic centrifuge |
13 Oct 2004 07:39:40 PM |
|
|
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:2202379a.0410131045.4e751816@posting.google.com...
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message
news:<ckitrq$7dt$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
[snip]
Imagine a space ship moving in a straight line speeding
past you. You're an interstellar cop with a Doppler Radar
unit, and you will find a moment where the space ship is
instantaneously at rest relatively to you, as it goes from
approaching to receding
You make no sense at all. The relative velocity of the space ship
relative to me remains the *same* all the time. It is never zero.
Let my position vector be r_c and the position vector of the space ship
be r_s. The velocity of the space ship wrt me is then d/dt (r_c - r_s).
I am at rest in my own CS, hence d/dt r_c = 0. The space ship is moving
with constant velocity, hence d/dt r_s = v, where v is a *constant*
vector. Hence the velocity of the space ship wrt me is *constant*.
How do you account for the Doppler Effect, even using sound
say as a train approaches and passes me by?
radial velocity: v_r = v*sin(A),
tangential velocity: v_t = v*cos(A),
velocity: v^2 = (v_r)^2 + (v_t)^2 = constant.
At A = 0 (closest approach), v_r = 0, and, v_t = v, whereat
only the transverse Doppler effect (time dilation) is effective.
For a centrifuge, v_r = 0, v_t = v, for all angles, A.
[Old Man]
Bjoern
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| User: "Ken S. Tucker" |
|
| Title: Re: relativistic centrifuge |
14 Oct 2004 08:05:49 AM |
|
|
"Old Man" <nomail@nomail.net> wrote in message news:<H_2dnSnUNJ5SV_DcRVn-gg@prairiewave.com>...
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:2202379a.0410131045.4e751816@posting.google.com...
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message
news:<ckitrq$7dt$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
[snip]
Imagine a space ship moving in a straight line speeding
past you. You're an interstellar cop with a Doppler Radar
unit, and you will find a moment where the space ship is
instantaneously at rest relatively to you, as it goes from
approaching to receding
You make no sense at all. The relative velocity of the space ship
relative to me remains the *same* all the time. It is never zero.
Let my position vector be r_c and the position vector of the space ship
be r_s. The velocity of the space ship wrt me is then d/dt (r_c - r_s).
I am at rest in my own CS, hence d/dt r_c = 0. The space ship is moving
with constant velocity, hence d/dt r_s = v, where v is a *constant*
vector. Hence the velocity of the space ship wrt me is *constant*.
How do you account for the Doppler Effect, even using sound
say as a train approaches and passes me by?
radial velocity: v_r = v*sin(A),
tangential velocity: v_t = v*cos(A),
velocity: v^2 = (v_r)^2 + (v_t)^2 = constant.
At A = 0 (closest approach), v_r = 0, and, v_t = v, whereat
only the transverse Doppler effect (time dilation) is effective.
For a centrifuge, v_r = 0, v_t = v, for all angles, A.
[Old Man]
I'll apply first to inertial tangential motion.
The equations of SR do not apply to tangential velocity,
have a glance at SR and read the transformation equations
require motion along X and X' with those parallel and
coincident.
In GR the tangential velocity is transformed away by using
a rotating CS, so the general equation,
ds^2 = g_uv dx^u dx^v becomes
ds^2 = dt^2 - dr^2
Thinking now about a centrifuge in GR..
A centrifuge is complicated by the fact that the geodesic
is constantly perturbed by an Electromagnetic force, such
as the forces that hold the material centrifuge together,
like the inter-molecular forces in the string on Newton's
bucket.
Two approaches are used to address that problem in GR,
the easy way (superficial) is to let the string exert
an absolute force and ergo acceleration on the bucket,
(I can provide refs for that).
The second is to maintain GR's requirement that all
acceleration is relative. That's tougher but you can
learn more about how GR and EMF relate. I've never
read a good treatment yet, so I need to explain it.
Regards
Ken S. Tucker
.
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|
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| User: "Bjoern Feuerbacher" |
|
| Title: Re: relativistic centrifuge |
14 Oct 2004 11:14:32 AM |
|
|
Ken S. Tucker wrote:
"Old Man" <nomail@nomail.net> wrote in message news:<H_2dnSnUNJ5SV_DcRVn-gg@prairiewave.com>...
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:2202379a.0410131045.4e751816@posting.google.com...
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message
news:<ckitrq$7dt$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
[snip]
Imagine a space ship moving in a straight line speeding
past you. You're an interstellar cop with a Doppler Radar
unit, and you will find a moment where the space ship is
instantaneously at rest relatively to you, as it goes from
approaching to receding
You make no sense at all. The relative velocity of the space ship
relative to me remains the *same* all the time. It is never zero.
Let my position vector be r_c and the position vector of the space ship
be r_s. The velocity of the space ship wrt me is then d/dt (r_c - r_s).
I am at rest in my own CS, hence d/dt r_c = 0. The space ship is moving
with constant velocity, hence d/dt r_s = v, where v is a *constant*
vector. Hence the velocity of the space ship wrt me is *constant*.
How do you account for the Doppler Effect, even using sound
say as a train approaches and passes me by?
radial velocity: v_r = v*sin(A),
tangential velocity: v_t = v*cos(A),
velocity: v^2 = (v_r)^2 + (v_t)^2 = constant.
At A = 0 (closest approach), v_r = 0, and, v_t = v, whereat
only the transverse Doppler effect (time dilation) is effective.
For a centrifuge, v_r = 0, v_t = v, for all angles, A.
[Old Man]
I'll apply first to inertial tangential motion.
The equations of SR do not apply to tangential velocity,
Err, then why can one derive the transversal Doppler shift from the
equations of SR?
have a glance at SR and read the transformation equations
require motion along X and X' with those parallel and
coincident.
That is the most simply case of Lorentz transformations. It is not
hard to generalize them.
In GR the tangential velocity is transformed away by using
a rotating CS, so the general equation,
ds^2 = g_uv dx^u dx^v becomes
ds^2 = dt^2 - dr^2
Thinking now about a centrifuge in GR..
Why? The original topic of this thread was how one could explain
a centrifuge *without* using GR.
[snip]
Two approaches are used to address that problem in GR,
the easy way (superficial) is to let the string exert
an absolute force and ergo acceleration on the bucket,
(I can provide refs for that).
First I would like you to explain what you mean by an
"absolute force".
[snip]
Bye,
Bjoern
.
|
|
|
| User: "Ken S. Tucker" |
|
| Title: Re: relativistic centrifuge |
14 Oct 2004 06:05:14 PM |
|
|
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<ckm8l8$695$4@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
"Old Man" <nomail@nomail.net> wrote in message news:<H_2dnSnUNJ5SV_DcRVn-gg@prairiewave.com>...
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:2202379a.0410131045.4e751816@posting.google.com...
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message
news:<ckitrq$7dt$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
[snip]
Imagine a space ship moving in a straight line speeding
past you. You're an interstellar cop with a Doppler Radar
unit, and you will find a moment where the space ship is
instantaneously at rest relatively to you, as it goes from
approaching to receding
You make no sense at all. The relative velocity of the space ship
relative to me remains the *same* all the time. It is never zero.
Let my position vector be r_c and the position vector of the space ship
be r_s. The velocity of the space ship wrt me is then d/dt (r_c - r_s).
I am at rest in my own CS, hence d/dt r_c = 0. The space ship is moving
with constant velocity, hence d/dt r_s = v, where v is a *constant*
vector. Hence the velocity of the space ship wrt me is *constant*.
How do you account for the Doppler Effect, even using sound
say as a train approaches and passes me by?
radial velocity: v_r = v*sin(A),
tangential velocity: v_t = v*cos(A),
velocity: v^2 = (v_r)^2 + (v_t)^2 = constant.
At A = 0 (closest approach), v_r = 0, and, v_t = v, whereat
only the transverse Doppler effect (time dilation) is effective.
For a centrifuge, v_r = 0, v_t = v, for all angles, A.
[Old Man]
I'll apply first to inertial tangential motion.
The equations of SR do not apply to tangential velocity,
Err, then why can one derive the transversal Doppler shift from the
equations of SR?
Ok, this is important. Set identical atomic clocks on
each, "particle", one is approaching purely tangential
motion (where dr=0) and is *inertial* and have that
particle emit radiation at a frequency set by those clocks.
Will the rest clock find the moving clock to - at any point
- be synchronized?
have a glance at SR and read the transformation equations
require motion along X and X' with those parallel and
coincident.
That is the most simply case of Lorentz transformations. It is not
hard to generalize them.
Prove that to us.
In GR the tangential velocity is transformed away by using
a rotating CS, so the general equation,
ds^2 = g_uv dx^u dx^v becomes
ds^2 = dt^2 - dr^2
Thinking now about a centrifuge in GR..
....
Two approaches are used to address that problem in GR,
the easy way (superficial) is to let the string exert
an absolute force and ergo acceleration on the bucket,
(I can provide refs for that).
First I would like you to explain what you mean by an
"absolute force".
f_u >0 look up Lorentz force in GR.
Ken
.
|
|
|
| User: "Bjoern Feuerbacher" |
|
| Title: Re: relativistic centrifuge |
15 Oct 2004 04:13:12 AM |
|
|
Ken S. Tucker wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<ckm8l8$695$4@news.urz.uni-heidelberg.de>...
[snip]
I'll apply first to inertial tangential motion.
The equations of SR do not apply to tangential velocity,
Err, then why can one derive the transversal Doppler shift from the
equations of SR?
I notice you did not bother to answer that.
Ok, this is important. Set identical atomic clocks on
each, "particle", one is approaching purely tangential
motion (where dr=0)
If it has dr=0, then it is not approaching.
and is *inertial* and have that
particle emit radiation at a frequency set by those clocks.
Will the rest clock find the moving clock to - at any point
- be synchronized?
No, it won't.
have a glance at SR and read the transformation equations
require motion along X and X' with those parallel and
coincident.
That is the most simply case of Lorentz transformations. It is not
hard to generalize them.
Prove that to us.
A Lorentz transformation is defined most generally as a transformation
which leaves the Minkowski metric invariant. Translations, rotations and
Lorentz boosts obey this requirement. So if you want to have a
transformation so that the X and X' are not coincident, combine a
translation and a Lorentz boost. If you want to have a transformation so
that the X and X' axes are not parallel, combine a rotation and a
Lorentz boost.
Hey, this is all quite basic stuff!
[snip]
Two approaches are used to address that problem in GR,
the easy way (superficial) is to let the string exert
an absolute force and ergo acceleration on the bucket,
(I can provide refs for that).
First I would like you to explain what you mean by an
"absolute force".
f_u >0
A covariant vector. Obtained by which metric from the contravariant
vector? And why do you call this "absolute force"?
look up Lorentz force in GR.
A Google search for "Lorentz force" and "General Relativity" yields
2160 hits. A bit too many...
OTOH, adding "absolute force" to the search terms brings no hits at all.
So, can you now please answer my question why you call the above
"absolute force"?
Bye,
Bjoern
.
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| User: "Ken S. Tucker" |
|
| Title: Re: relativistic centrifuge |
15 Oct 2004 08:00:51 PM |
|
|
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cko4b8$l9l$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
....
Ok, this is important. Set identical atomic clocks on
each, "particle", one is approaching purely tangential
motion (where dr=0)
If it has dr=0, then it is not approaching.
and is *inertial* and have that
particle emit radiation at a frequency set by those clocks.
Will the rest clock find the moving clock to - at any point
- be synchronized?
No, it won't.
How can the emission go from relatively blue shifted
to relatively red shifted, without being equal at some
point?
have a glance at SR and read the transformation equations
require motion along X and X' with those parallel and
coincident.
That is the most simply case of Lorentz transformations. It is not
hard to generalize them.
Prove that to us.
A Lorentz transformation is defined most generally as a transformation
which leaves the Minkowski metric invariant. Translations, rotations and
Lorentz boosts obey this requirement. So if you want to have a
transformation so that the X and X' are not coincident, combine a
translation and a Lorentz boost. If you want to have a transformation so
that the X and X' axes are not parallel, combine a rotation and a
Lorentz boost.
Hey, this is all quite basic stuff!
You can't do a translation in GR, but rotations are ok.
Two approaches are used to address that problem in GR,
the easy way (superficial) is to let the string exert
an absolute force and ergo acceleration on the bucket,
(I can provide refs for that).
First I would like you to explain what you mean by an
"absolute force".
f_u >0
A covariant vector. Obtained by which metric from the contravariant
vector? And why do you call this "absolute force"?
look up Lorentz force in GR.
A Google search for "Lorentz force" and "General Relativity" yields
2160 hits. A bit too many...
OTOH, adding "absolute force" to the search terms brings no hits at all.
So, can you now please answer my question why you call the above
"absolute force"?
A few days ago I posted to Old Man a simplified GR
solution and a difficult GR solution to the centrifuge
problem.
The simplified version is found in R.C. Tolman's "Relativity,
Thermo..." Eq.(103.1), and in a more modern treatment in
Weinberg's, "Grav and Cosmo", Eq.(5.2.8)
The difficult solution is in Einstein's 1916 "Foundation
of GR" Eq.(65), where the Lorentz forces,
kappa,sigma = 0
vanish as explained in his following text.
That's to satisfy the GRist's where the problem
comes from.
According to the Principle of General Relativity
absolute acceleration does not exist. The general
equation defining that is the geodesic expressed
by the absolute derivative,
DU^i = 0 = U^i;u U^u ,
{i=1,2,3} {u=0,i}, U^u is 4-velocity.
I think that can be generalized to
DU^u = 0 or DU_u = 0
by association when
g_uv;w =0
is recognized.
The Lorentz force is given by
f_u = qF_uv U^v ,
and if f_u is non-zero in one system then it is
non-zero in all CS's.
Defining that force by
f_u == m*DU_u where m is mass,
then requires the "absolute acceleration" tensor
DU_u =/=0 in all FoR's,
in violation of the PoGR.
That is why Einstein in the above ref. set the
Lorentz force to vanish, and set,
f_u = 0 ,
eliminating absolute force and acceleration from
electromagnetic physics.
(I'm with Einstein)
To anyone familiar with Quantum Theory and in particular
the motion of charges in the presence of EM fields, you
may notice that the constraint f_u =0 is nothing less
than a "quantum geodesic", in it's description of the
relation of EM fields and electron motion, including
quantum events like the absorption of photons, in the
nuclear neighbourhood.
That is how GR subsumes QT, by replacing the ad hoc
basis for QT - as Planck introduced - by the PoGR Einstein
did in 1916.
Regards
Ken S. Tucker
Bye,
Bjoern
.
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| User: "Bjoern Feuerbacher" |
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| Title: Re: relativistic centrifuge |
18 Oct 2004 06:40:25 AM |
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Ken S. Tucker wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cko4b8$l9l$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
I notice that you still did not answer my question why one can derive
the transverse Doppler effect from SR if your assertion that SR can not
deal with tangential velocities is right. This time, you simply snipped
this argument...
Ok, this is important. Set identical atomic clocks on
each, "particle", one is approaching purely tangential
motion (where dr=0)
If it has dr=0, then it is not approaching.
I notice you did not bother to address this.
and is *inertial* and have that
particle emit radiation at a frequency set by those clocks.
Will the rest clock find the moving clock to - at any point
- be synchronized?
No, it won't.
How can the emission go from relatively blue shifted
to relatively red shifted, without being equal at some
point?
Err, try to understand the difference between Doppler shift and time
dilation.
have a glance at SR and read the transformation equations
require motion along X and X' with those parallel and
coincident.
That is the most simply case of Lorentz transformations. It is not
hard to generalize them.
Prove that to us.
A Lorentz transformation is defined most generally as a transformation
which leaves the Minkowski metric invariant. Translations, rotations and
Lorentz boosts obey this requirement. So if you want to have a
transformation so that the X and X' are not coincident, combine a
translation and a Lorentz boost. If you want to have a transformation so
that the X and X' axes are not parallel, combine a rotation and a
Lorentz boost.
Hey, this is all quite basic stuff!
You can't do a translation in GR, but rotations are ok.
1) We were talking about SR here, not about GR.
2) Support that assertion, please.
Two approaches are used to address that problem in GR,
the easy way (superficial) is to let the string exert
an absolute force and ergo acceleration on the bucket,
(I can provide refs for that).
First I would like you to explain what you mean by an
"absolute force".
f_u >0
A covariant vector. Obtained by which metric from the contravariant
vector? And why do you call this "absolute force"?
Care to tell me?
look up Lorentz force in GR.
A Google search for "Lorentz force" and "General Relativity" yields
2160 hits. A bit too many...
OTOH, adding "absolute force" to the search terms brings no hits at all.
So, can you now please answer my question why you call the above
"absolute force"?
You did not answer that question below.
A few days ago I posted to Old Man a simplified GR
solution and a difficult GR solution to the centrifuge
problem.
The simplified version is found in R.C. Tolman's "Relativity,
Thermo..." Eq.(103.1), and in a more modern treatment in
Weinberg's, "Grav and Cosmo", Eq.(5.2.8)
The difficult solution is in Einstein's 1916 "Foundation
of GR" Eq.(65), where the Lorentz forces,
kappa,sigma = 0
vanish as explained in his following text.
That's to satisfy the GRist's where the problem
comes from.
According to the Principle of General Relativity
absolute acceleration does not exist. The general
equation defining that is the geodesic expressed
by the absolute derivative,
DU^i = 0 = U^i;u U^u ,
{i=1,2,3} {u=0,i}, U^u is 4-velocity.
I think that can be generalized to
DU^u = 0 or DU_u = 0
by association when
g_uv;w =0
is recognized.
The Lorentz force is given by
f_u = qF_uv U^v ,
and if f_u is non-zero in one system then it is
non-zero in all CS's.
Defining that force by
f_u == m*DU_u where m is mass,
then requires the "absolute acceleration" tensor
DU_u =/=0 in all FoR's,
Why do you call this the "absolute acceleration tensor"?
in violation of the PoGR.
That is why Einstein in the above ref. set the
Lorentz force to vanish, and set,
f_u = 0 ,
eliminating absolute force and acceleration from
electromagnetic physics.
You *still* have not told me why you call f_u "absolute force".
(I'm with Einstein)
To anyone familiar with Quantum Theory and in particular
the motion of charges in the presence of EM fields, you
may notice that the constraint f_u =0 is nothing less
than a "quantum geodesic",
What do you mean with the term "quantum geodesic"?
And what has f_0 = 0 to do with it?
in it's description of the
relation of EM fields and electron motion, including
quantum events like the absorption of photons, in the
nuclear neighbourhood.
What has f_0 = 0 to do with including quantum events?
That is how GR subsumes QT, by replacing the ad hoc
basis for QT - as Planck introduced - by the PoGR Einstein
did in 1916.
I may guess: all that stuff in the last paragraph is your own invention,
right?
Bye,
Bjoern
.
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| User: "Ken S. Tucker" |
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| Title: Re: relativistic centrifuge |
18 Oct 2004 03:08:28 PM |
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|
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cl0a39$ho0$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cko4b8$l9l$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
I notice that you still did not answer my question why one can derive
the transverse Doppler effect from SR if your assertion that SR can not
deal with tangential velocities is right. This time, you simply snipped
this argument...
I explained several times, tangential = rotational,
think about magnetism.
....
How can the emission go from relatively blue shifted
to relatively red shifted, without being equal at some
point?
Err, try to understand the difference between Doppler shift and time
dilation.
Why should I?
....
A few days ago I posted to Old Man a simplified GR
solution and a difficult GR solution to the centrifuge
problem.
The simplified version is found in R.C. Tolman's "Relativity,
Thermo..." Eq.(103.1), and in a more modern treatment in
Weinberg's, "Grav and Cosmo", Eq.(5.2.8)
The difficult solution is in Einstein's 1916 "Foundation
of GR" Eq.(65), where the Lorentz forces,
kappa,sigma = 0
vanish as explained in his following text.
That's to satisfy the GRist's where the problem
comes from.
According to the Principle of General Relativity
absolute acceleration does not exist. The general
equation defining that is the geodesic expressed
by the absolute derivative,
DU^i = 0 = U^i;u U^u ,
{i=1,2,3} {u=0,i}, U^u is 4-velocity.
I think that can be generalized to
DU^u = 0 or DU_u = 0
by association when
g_uv;w =0
is recognized.
The Lorentz force is given by
f_u = qF_uv U^v ,
and if f_u is non-zero in one system then it is
non-zero in all CS's.
Defining that force by
f_u == m*DU_u where m is mass,
then requires the "absolute acceleration" tensor
DU_u =/=0 in all FoR's,
Why do you call this the "absolute acceleration tensor"?
Non zero in all CS's, didn't vanish in any.
in violation of the PoGR.
That is why Einstein in the above ref. set the
Lorentz force to vanish, and set,
f_u = 0 ,
eliminating absolute force and acceleration from
electromagnetic physics.
(I'm with Einstein)
To anyone familiar with Quantum Theory and in particular
the motion of charges in the presence of EM fields, you
may notice that the constraint f_u =0 is nothing less
than a "quantum geodesic",
What do you mean with the term "quantum geodesic"?
And what has f_0 = 0 to do with it?
f_0 = qF_0i U^i = 0
in vectors that means the scalar product,
-> ->
q*E(x).V(x) = 0
and means the test charge "q" can only move prependicular
to E(x), but NOT parallel. A parallel motion would invite
a spiral motion relative to E(x), and hence continuous
variation of system energy, violating quantum theory.
The above is an "equation of motion" consistent with QT,
based on GR.
in it's description of the
relation of EM fields and electron motion, including
quantum events like the absorption of photons, in the
nuclear neighbourhood.
What has f_0 = 0 to do with including quantum events?
What seems to happen, is the photon's EM field changes
the nuclear E(x) direction, (I mentioned above), by the
addition of the photons E-field.
That is how GR subsumes QT, by replacing the ad hoc
basis for QT - as Planck introduced - by the PoGR Einstein
did in 1916.
I may guess: all that stuff in the last paragraph is your own invention,
right?
Ha, maybe in a previous life. Look you should understand
the facts. Einstein won the Nobel prize for his "photo-electrical
effect", where he used Plancks' obscure quantum theory back
in 1905. In the subsequent period from then to his publication
of GR in 1916, he had 11 years to decide whether or not Lorentz's
f_u =0
or not. He chose f_u = 0 to be included in finding
the "energy density" in GR, and comply with GPoR.
In fact, the condition f_u =0, derived from the
generalized GR geodesic places clear constraints on
the motion of charges in fields, as I demonstrated.
The constraint's are in accord with QT, hence I
term that aspect of the GR equation of motion dealing
with QT as a "quantum geodesic".
Regards
Ken S. Tucker
Bjoern
.
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| User: "Bjoern Feuerbacher" |
|
| Title: Re: relativistic centrifuge |
19 Oct 2004 07:29:31 AM |
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Ken S. Tucker wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cl0a39$ho0$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cko4b8$l9l$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
I notice that you still did not answer my question why one can derive
the transverse Doppler effect from SR if your assertion that SR can not
deal with tangential velocities is right. This time, you simply snipped
this argument...
I explained several times,
No, you did not address the transverse Doppler effect so far anywhere,
as far as I can see. If you think you did, could you please provide a link?
tangential = rotational,
Wrong. Tangential does not need to be rotational. When something
approaches me on a straight line which has a distance to me, then that
object has both a tangential and a radial velocity component, although
it is not moving on a circle. Apparently you don't understand polar
coordinates...
think about magnetism.
What on earth has this to do with magnetism?
How can the emission go from relatively blue shifted
to relatively red shifted, without being equal at some
point?
Err, try to understand the difference between Doppler shift and time
dilation.
Why should I?
...
Because the explanation why the emission goes from blue-shifted to
red-shifted comes from the Doppler effect, whereas the stuff about the
clocks being synchronized is addressed by the time dilation effect. You
obviously confuse the two.
[snip]
then requires the "absolute acceleration" tensor
DU_u =/=0 in all FoR's,
Why do you call this the "absolute acceleration tensor"?
Non zero in all CS's, didn't vanish in any.
Let me get this straight. You call something "absolute" if it is
non-zero in all CS's?
[snip]
(I'm with Einstein)
To anyone familiar with Quantum Theory and in particular
the motion of charges in the presence of EM fields, you
may notice that the constraint f_u =0 is nothing less
than a "quantum geodesic",
What do you mean with the term "quantum geodesic"?
And what has f_0 = 0 to do with it?
f_0 = qF_0i U^i = 0
in vectors that means the scalar product,
-> ->
q*E(x).V(x) = 0
and means the test charge "q" can only move prependicular
to E(x), but NOT parallel.
Well, *if* f_0 = 0, then that is right. But why should it be?
Because you think that if f_0 is not zero, it would be an "absolute
force", and that is forbidden???
A parallel motion would invite
a spiral motion relative to E(x),
Huh? Why on earth do you think so?
and hence continuous variation of system energy,
Huh? Why???
violating quantum theory.
What on earth has this to do with quantum theory?????
The above is an "equation of motion" consistent with QT,
based on GR.
You seem to have a weird idea of what QT says.
in it's description of the
relation of EM fields and electron motion, including
quantum events like the absorption of photons, in the
nuclear neighbourhood.
What has f_0 = 0 to do with including quantum events?
What seems to happen, is the photon's EM field changes
the nuclear E(x) direction, (I mentioned above), by the
addition of the photons E-field.
Where on earth do you get this from???
That is how GR subsumes QT, by replacing the ad hoc
basis for QT - as Planck introduced - by the PoGR Einstein
did in 1916.
I may guess: all that stuff in the last paragraph is your own invention,
right?
Ha, maybe in a previous life. Look you should understand
the facts. Einstein won the Nobel prize for his "photo-electrical
effect", where he used Plancks' obscure quantum theory back
in 1905.
Clear. Nothing new so far.
In the subsequent period from then to his publication
of GR in 1916, he had 11 years to decide whether or not Lorentz's
f_u =0
or not. He chose f_u = 0 to be included in finding
the "energy density" in GR, and comply with GPoR.
Where does he say that?
And where does he say that that has anything to do with a "quantum
geodesic"?
In fact, the condition f_u =0, derived from the
generalized GR geodesic
You said "*I* think that can be generalized to..." (emphasize mine),
i.e. this "generalized GR geodesic" is *your* idea, isn't it?
places clear constraints on
the motion of charges in fields, as I demonstrated.
My strong suspicion is that already your starting point, DU^i = 0,
makes no sense.
You seem to get that equation from "absolute acceleration does not
exist"; but what "absolute acceleration does not exist" has to do with
"DU^i = 0" seems to be your private secret. Probably this has again to
do with your strange idea that something is "absolute" if it is non-zero
in all CSs.
[snip]
Bye,
Bjoern
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| User: "Ken S. Tucker" |
|
| Title: Re: relativistic centrifuge |
20 Oct 2004 04:22:08 PM |
|
|
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cl31bb$i8o$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
What do you mean with the term "quantum geodesic"?
And what has f_0 = 0 to do with it?
f_0 = qF_0i U^i = 0
in vectors that means the scalar product,
-> ->
q*E(x).V(x) = 0
and means the test charge "q" can only move prependicular
to E(x), but NOT parallel.
Well, *if* f_0 = 0, then that is right. But why should it be?
Because you think that if f_0 is not zero, it would be an "absolute
force", and that is forbidden???
Yes, that is exactly correct.
....
Bjoern
Ken
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| User: "Bjoern Feuerbacher" |
|
| Title: Re: relativistic centrifuge |
21 Oct 2004 06:06:35 AM |
|
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Ken S. Tucker wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cl31bb$i8o$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
What do you mean with the term "quantum geodesic"?
And what has f_0 = 0 to do with it?
f_0 = qF_0i U^i = 0
in vectors that means the scalar product,
-> ->
q*E(x).V(x) = 0
and means the test charge "q" can only move prependicular
to E(x), but NOT parallel.
Well, *if* f_0 = 0, then that is right. But why should it be?
Because you think that if f_0 is not zero, it would be an "absolute
force", and that is forbidden???
Yes, that is exactly correct.
Please support your assertion that for any quantity X, "X is absolute"
means "X is non-zero is all CSs".
Bye,
Bjoern
.
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| User: "Ken S. Tucker" |
|
| Title: Re: relativistic centrifuge |
21 Oct 2004 03:10:41 PM |
|
|
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cl857r$pjl$3@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cl31bb$i8o$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
What do you mean with the term "quantum geodesic"?
And what has f_0 = 0 to do with it?
f_0 = qF_0i U^i = 0
in vectors that means the scalar product,
-> ->
q*E(x).V(x) = 0
and means the test charge "q" can only move prependicular
to E(x), but NOT parallel.
Well, *if* f_0 = 0, then that is right. But why should it be?
Because you think that if f_0 is not zero, it would be an "absolute
force", and that is forbidden???
Yes, that is exactly correct.
Please support your assertion that for any quantity X, "X is absolute"
means "X is non-zero in all CSs".
^^
Have a look at the energy momentum components "p^u",
and tell me what you think.
Ken
Bye,
Bjoern
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| User: "Bjoern Feuerbacher" |
|
| Title: Re: relativistic centrifuge |
22 Oct 2004 04:49:05 AM |
|
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Ken S. Tucker wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cl857r$pjl$3@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cl31bb$i8o$1@news.urz.uni-heidelberg.de>...
Ken S. Tucker wrote:
What do you mean with the term "quantum geodesic"?
And what has f_0 = 0 to do with it?
f_0 = qF_0i U^i = 0
in vectors that means the scalar product,
-> ->
q*E(x).V(x) = 0
and means the test charge "q" can only move prependicular
to E(x), but NOT parallel.
Well, *if* f_0 = 0, then that is right. But why should it be?
Because you think that if f_0 is not zero, it would be an "absolute
force", and that is forbidden???
Yes, that is exactly correct.
Please support your assertion that for any quantity X, "X is absolute"
means "X is non-zero in all CSs".
^^
Have a look at the energy momentum components "p^u",
and tell me what you think.
Evasion noted. Yet again.
BTW, the energy-momentum vector transforms like all other four-vectors.
p'^u = (del x'^u)/(del x^v) p^v
Bye,
Bjoern
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