| Topic: |
Science > Physics |
| User: |
"kenseto" |
| Date: |
30 Apr 2006 01:21:43 PM |
| Object: |
The SRians are making contradictory claims |
The SRians (specifically PD) are making the following contradictory
claims:
Claim #1
PD said: When two observers A and B are in relative motion the passage
of a clock second on A's clock does not coincide with the passage of a
clock second in B's clock.
Claim #2
PD (SR) also said that in the twin paradox situation the elapsed clock
seconds in B's clock (the traveling clock) can be compared directly
with the elapsed clock seconds in A's clock when B return after a
journey.
It seems that the SRians will go to the extend of making contradictory
claims to explain their theory. Go figure.
.
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| User: "Tom Roberts" |
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| Title: Re: Energy slows down clocks. |
08 May 2006 06:56:44 PM |
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Daryl McCullough wrote:
Well, if you include gravity, then we can approximately
give the proper time tau on a clock as a function of its
velocity and height above the Earth as follows:
tau = T (1 -(KE - PE)/mc^2)
where KE = time-averaged kinetic energy (1/2 mv^2),
PE = time-averaged Newtonian potential energy (-GMm/r),
T = the time as measured by a clock far from the earth
at rest relative to the earth.
So adding kinetic energy to the clock slows it down,
while adding potential energy to the clock speeds it
up.
Yes, but you forgot to mention the caveats: this is valid only for
systems in which gravity is "small" and speeds are "slow" (<< c), and
one must use inertial coordinates appropriate to Newtonian mechanics on
earth.
Your writing it this way helped me recognize that (KE - PE) as the
Lagrangian of the clock treated as a particle. I don't know what
significance that has....
Tom Roberts
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| User: "" |
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| Title: Re: Energy slows down clocks. |
09 May 2006 12:31:11 AM |
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In article <e3p6dr02gi1@drn.newsguy.com>, (Daryl McCullough) writes:
Tom Roberts says...
Daryl McCullough wrote:
Well, if you include gravity, then we can approximately
give the proper time tau on a clock as a function of its
velocity and height above the Earth as follows:
tau = T (1 -(KE - PE)/mc^2)
Yes, but you forgot to mention the caveats: this is valid only for
systems in which gravity is "small" and speeds are "slow" (<< c), and
one must use inertial coordinates appropriate to Newtonian mechanics on
earth.
Your writing it this way helped me recognize that (KE - PE) as the
Lagrangian of the clock treated as a particle. I don't know what
significance that has....
Ah! That's a question I know the answer to!
In Newtonian physics, a freefall trajectory minimizes the action
A = integral of L_classical dt.
In GR, a freefall trajectory minimizes the proper time. That means
that it minimizes
Tau = Integral of square-root(g_uv dx^u dx^v)
To see the relationship between these two, we can view x^u as a function
of time t to write the integral as
Tau = Integral of square-root(g_uv dx^u/dt dx^v/dt) dt
which means that we are minimizing an action with an effective
lagrangian
L_GR = square-root(g_uv dx^u/dt dx^v/dt)
In the limit of small velocities and small curvatures, then
these two lagrangians, L_classical and L_GR, must give the same
trajectories. That doesn't mean that they are *equal* in this
limit, but it turns out that they are simply related:
L_GR ==> 1 - L_classical/mc^2
Now, that was a useful post. Thank you.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Daryl McCullough" |
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| Title: Re: Energy slows down clocks. |
09 May 2006 05:48:32 AM |
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says...
In the limit of small velocities and small curvatures, then
these two lagrangians, L_classical and L_GR, must give the same
trajectories. That doesn't mean that they are *equal* in this
limit, but it turns out that they are simply related:
L_GR ==> 1 - L_classical/mc^2
Now, that was a useful post. Thank you.
You're welcome!
--
Daryl McCullough
Ithaca, NY
--
NewsGuy.Com 30Gb $9.95 Carry Forward and On Demand Bandwidth
.
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| User: "Tom Roberts" |
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| Title: Re: Energy slows down clocks. |
09 May 2006 09:00:29 PM |
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wrote:
In article <e3p6dr02gi1@drn.newsguy.com>, (Daryl McCullough) writes:
Tom Roberts says...
Daryl McCullough wrote:
Well, if you include gravity, then we can approximately
give the proper time tau on a clock as a function of its
velocity and height above the Earth as follows:
tau = T (1 -(KE - PE)/mc^2)
Yes, but you forgot to mention the caveats: this is valid only for
systems in which gravity is "small" and speeds are "slow" (<< c), and
one must use inertial coordinates appropriate to Newtonian mechanics on
earth.
Your writing it this way helped me recognize that (KE - PE) as the
Lagrangian of the clock treated as a particle. I don't know what
significance that has....
Ah! That's a question I know the answer to!
In Newtonian physics, a freefall trajectory minimizes the action
A = integral of L_classical dt.
In GR, a freefall trajectory minimizes the proper time. That means
that it minimizes
Tau = Integral of square-root(g_uv dx^u dx^v)
To see the relationship between these two, we can view x^u as a function
of time t to write the integral as
Tau = Integral of square-root(g_uv dx^u/dt dx^v/dt) dt
which means that we are minimizing an action with an effective
lagrangian
L_GR = square-root(g_uv dx^u/dt dx^v/dt)
In the limit of small velocities and small curvatures, then
these two lagrangians, L_classical and L_GR, must give the same
trajectories. That doesn't mean that they are *equal* in this
limit, but it turns out that they are simply related:
L_GR ==> 1 - L_classical/mc^2
Now, that was a useful post. Thank you.
I agree. I had not thought of it that way. Thanks!
One minor nit that does not affect the result: in GR a geodesic
_maximizes_ the proper time. Now look at that minus sign.
Tom Roberts
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| User: "" |
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| Title: Re: Energy slows down clocks. |
09 May 2006 09:36:32 PM |
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In article <1Pb8g.14527$Lm5.6697@newssvr12.news.prodigy.com>, Tom Roberts <tjroberts137@sbcglobal.net> writes:
mmeron@cars3.uchicago.edu wrote:
In article <e3p6dr02gi1@drn.newsguy.com>, (Daryl McCullough) writes:
Tom Roberts says...
Daryl McCullough wrote:
Well, if you include gravity, then we can approximately
give the proper time tau on a clock as a function of its
velocity and height above the Earth as follows:
tau = T (1 -(KE - PE)/mc^2)
Yes, but you forgot to mention the caveats: this is valid only for
systems in which gravity is "small" and speeds are "slow" (<< c), and
one must use inertial coordinates appropriate to Newtonian mechanics on
earth.
Your writing it this way helped me recognize that (KE - PE) as the
Lagrangian of the clock treated as a particle. I don't know what
significance that has....
Ah! That's a question I know the answer to!
In Newtonian physics, a freefall trajectory minimizes the action
A = integral of L_classical dt.
In GR, a freefall trajectory minimizes the proper time. That means
that it minimizes
Tau = Integral of square-root(g_uv dx^u dx^v)
To see the relationship between these two, we can view x^u as a function
of time t to write the integral as
Tau = Integral of square-root(g_uv dx^u/dt dx^v/dt) dt
which means that we are minimizing an action with an effective
lagrangian
L_GR = square-root(g_uv dx^u/dt dx^v/dt)
In the limit of small velocities and small curvatures, then
these two lagrangians, L_classical and L_GR, must give the same
trajectories. That doesn't mean that they are *equal* in this
limit, but it turns out that they are simply related:
L_GR ==> 1 - L_classical/mc^2
Now, that was a useful post. Thank you.
I agree. I had not thought of it that way. Thanks!
One minor nit that does not affect the result: in GR a geodesic
_maximizes_ the proper time. Now look at that minus sign.
Yes, right. Didn't notice this detail.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "John C. Polasek" |
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| Title: Re: Energy slows down clocks. |
08 May 2006 10:04:13 PM |
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On Mon, 08 May 2006 23:56:44 GMT, Tom Roberts
<tjroberts137@sbcglobal.net> wrote:
Daryl McCullough wrote:
Well, if you include gravity, then we can approximately
give the proper time tau on a clock as a function of its
velocity and height above the Earth as follows:
tau = T (1 -(KE - PE)/mc^2)
where KE = time-averaged kinetic energy (1/2 mv^2),
PE = time-averaged Newtonian potential energy (-GMm/r),
T = the time as measured by a clock far from the earth
at rest relative to the earth.
So adding kinetic energy to the clock slows it down,
while adding potential energy to the clock speeds it
up.
Snip Excuse snip i lost the original message.
Daryl:
The problem of the speed of a clock moved from earth's surface to a
mountain top, say 1000 meters, is fairly simple work for Dual Space
relativity. Gravity is vastly more important.
We know (or I do) that a clock moved from infinity to the earth will
lose 60 us/day due to gravity, while a clock on a revolving earth
would lose only 0.1 us/day due to the Lorentz effect.
I calculated that raising the clock 1000 m would speed it up
fractionally by 1.09e-13 and the velocity omegaR would reduce it by
3.751e-16.
The ratio is 291 times more for gravity than velocity.
The gravity one is easiest with my gravity law:
c dc/dr = MG/r^2 so
dc/c = MGH/R^2c^2 = gH/c^2 = 1.09e-13
John Polasek
http://www.dualspace.net
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| User: "John C. Polasek" |
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| Title: Re: Energy slows down clocks. |
08 May 2006 10:08:46 PM |
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On Mon, 08 May 2006 23:56:44 GMT, Tom Roberts
<tjroberts137@sbcglobal.net> wrote:
Daryl McCullough wrote:
Well, if you include gravity, then we can approximately
give the proper time tau on a clock as a function of its
velocity and height above the Earth as follows:
tau = T (1 -(KE - PE)/mc^2)
where KE = time-averaged kinetic energy (1/2 mv^2),
PE = time-averaged Newtonian potential energy (-GMm/r),
T = the time as measured by a clock far from the earth
at rest relative to the earth.
So adding kinetic energy to the clock slows it down,
while adding potential energy to the clock speeds it
up.
Snip Excuse snip i lost the original message.
Daryl:
The problem of the speed of a clock moved from earth's surface to a
mountain top, say 1000 meters, is fairly simple work for Dual Space
relativity. Gravity is vastly more important.
We know (or I do) that a clock moved from infinity to the earth will
lose 60 us/day due to gravity, while a clock on a revolving earth
would lose only 0.1 us/day due to the Lorentz effect.
I calculated that raising the clock 1000 m would speed it up
fractionally by 1.09e-13 and the velocity omegaR would reduce it by
3.751e-16.
The ratio is 291 times more for gravity than velocity.
The gravity one is easiest with my gravity law:
c dc/dr = MG/r^2 so
dc/c = MGH/R^2c^2 = gH/c^2 = 1.09e-13
John Polasek
http://www.dualspace.net
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| User: "Daryl McCullough" |
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| Title: Re: Energy slows down clocks. |
08 May 2006 11:39:23 PM |
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Tom Roberts says...
Daryl McCullough wrote:
Well, if you include gravity, then we can approximately
give the proper time tau on a clock as a function of its
velocity and height above the Earth as follows:
tau = T (1 -(KE - PE)/mc^2)
Yes, but you forgot to mention the caveats: this is valid only for
systems in which gravity is "small" and speeds are "slow" (<< c), and
one must use inertial coordinates appropriate to Newtonian mechanics on
earth.
Your writing it this way helped me recognize that (KE - PE) as the
Lagrangian of the clock treated as a particle. I don't know what
significance that has....
Ah! That's a question I know the answer to!
In Newtonian physics, a freefall trajectory minimizes the action
A = integral of L_classical dt.
In GR, a freefall trajectory minimizes the proper time. That means
that it minimizes
Tau = Integral of square-root(g_uv dx^u dx^v)
To see the relationship between these two, we can view x^u as a function
of time t to write the integral as
Tau = Integral of square-root(g_uv dx^u/dt dx^v/dt) dt
which means that we are minimizing an action with an effective
lagrangian
L_GR = square-root(g_uv dx^u/dt dx^v/dt)
In the limit of small velocities and small curvatures, then
these two lagrangians, L_classical and L_GR, must give the same
trajectories. That doesn't mean that they are *equal* in this
limit, but it turns out that they are simply related:
L_GR ==> 1 - L_classical/mc^2
--
Daryl McCullough
Ithaca, NY
--
NewsGuy.Com 30Gb $9.95 Carry Forward and On Demand Bandwidth
.
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| User: "PD" |
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| Title: Re: Energy slows down clocks. |
10 May 2006 07:28:15 AM |
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Daryl McCullough wrote:
Tom Roberts says...
Daryl McCullough wrote:
Well, if you include gravity, then we can approximately
give the proper time tau on a clock as a function of its
velocity and height above the Earth as follows:
tau = T (1 -(KE - PE)/mc^2)
Yes, but you forgot to mention the caveats: this is valid only for
systems in which gravity is "small" and speeds are "slow" (<< c), and
one must use inertial coordinates appropriate to Newtonian mechanics on
earth.
Your writing it this way helped me recognize that (KE - PE) as the
Lagrangian of the clock treated as a particle. I don't know what
significance that has....
Ah! That's a question I know the answer to!
In Newtonian physics, a freefall trajectory minimizes the action
A = integral of L_classical dt.
In GR, a freefall trajectory minimizes the proper time. That means
that it minimizes
Tau = Integral of square-root(g_uv dx^u dx^v)
To see the relationship between these two, we can view x^u as a function
of time t to write the integral as
Tau = Integral of square-root(g_uv dx^u/dt dx^v/dt) dt
which means that we are minimizing an action with an effective
lagrangian
L_GR = square-root(g_uv dx^u/dt dx^v/dt)
In the limit of small velocities and small curvatures, then
these two lagrangians, L_classical and L_GR, must give the same
trajectories. That doesn't mean that they are *equal* in this
limit, but it turns out that they are simply related:
L_GR ==> 1 - L_classical/mc^2
Aside from the minus sign.
I think there is a minor error somewhere in this excellent, terse
account. I believe a geodesic is a path of *maximal* time-like
interval, not a path of *minimal* time-like interval. It is this
difference from Euclidean geometry that is the underlying explanation
of even SR phenomena like the Twin Puzzle.
PD
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| User: "Daryl McCullough" |
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| Title: Re: Energy slows down clocks. |
10 May 2006 10:52:52 AM |
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PD says...
I believe a geodesic is a path of *maximal* time-like
interval, not a path of *minimal* time-like interval. It is this
difference from Euclidean geometry that is the underlying explanation
of even SR phenomena like the Twin Puzzle.
Yes, you're right. The correct word, to cover both cases, is
"extremizing", but that's kind of an awkward word.
--
Daryl McCullough
Ithaca, NY
--
NewsGuy.Com 30Gb $9.95 Carry Forward and On Demand Bandwidth
.
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| User: "Dirk Van de moortel" |
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| Title: Re: Energy slows down clocks. |
10 May 2006 11:30:36 AM |
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"Daryl McCullough" <stevendaryl3016@yahoo.com> wrote in message news:e3t28k02ien@drn.newsguy.com...
PD says...
I believe a geodesic is a path of *maximal* time-like
interval, not a path of *minimal* time-like interval. It is this
difference from Euclidean geometry that is the underlying explanation
of even SR phenomena like the Twin Puzzle.
Yes, you're right. The correct word, to cover both cases, is
"extremizing", but that's kind of an awkward word.
Yes - sounds frightening :-)
"Keeping stationary" sounds better.
Dirk Vdm
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| User: "Hexenmeister" |
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| Title: Re: Energy slows down clocks. |
10 May 2006 01:03:03 PM |
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"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote
in message news:Myo8g.417603$AN3.11491357@phobos.telenet-ops.be...
|
| "Daryl McCullough" <stevendaryl3016@yahoo.com> wrote in message
news:e3t28k02ien@drn.newsguy.com...
| > PD says...
| >
| > >I believe a geodesic is a path of *maximal* time-like
| > >interval, not a path of *minimal* time-like interval. It is this
| > >difference from Euclidean geometry that is the underlying explanation
| > >of even SR phenomena like the Twin Puzzle.
| >
| > Yes, you're right. The correct word, to cover both cases, is
| > "extremizing", but that's kind of an awkward word.
|
| Yes - sounds frightening :-)
| "Keeping stationary" sounds better.
These people keep stationary,
http://www.eurwholesale.co.uk/default.aspx?cam=ggl ,
fucking stupid local village idiot.
Androcles.
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| User: "PD" |
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| Title: Re: Energy slows down clocks. |
10 May 2006 03:33:15 PM |
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Hexenmeister wrote:
"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote
in message news:Myo8g.417603$AN3.11491357@phobos.telenet-ops.be...
|
| "Daryl McCullough" <stevendaryl3016@yahoo.com> wrote in message
news:e3t28k02ien@drn.newsguy.com...
| > PD says...
| >
| > >I believe a geodesic is a path of *maximal* time-like
| > >interval, not a path of *minimal* time-like interval. It is this
| > >difference from Euclidean geometry that is the underlying explanation
| > >of even SR phenomena like the Twin Puzzle.
| >
| > Yes, you're right. The correct word, to cover both cases, is
| > "extremizing", but that's kind of an awkward word.
|
| Yes - sounds frightening :-)
| "Keeping stationary" sounds better.
These people keep stationary,
http://www.eurwholesale.co.uk/default.aspx?cam=ggl ,
fucking stupid local village idiot.
That would be "stationery", you FSLVI you.
Perhaps you would be looking for
http://en.wikipedia.org/wiki/Stationary_point ?
PD
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| User: "Hexenmeister" |
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| Title: Re: Energy slows down clocks. |
10 May 2006 07:03:16 PM |
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"PD" <TheDraperFamily@gmail.com> wrote in message
news:1147293194.964532.150490@u72g2000cwu.googlegroups.com...
|
| Hexenmeister wrote:
| > "Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com>
wrote
| > in message news:Myo8g.417603$AN3.11491357@phobos.telenet-ops.be...
| > |
| > | "Daryl McCullough" <stevendaryl3016@yahoo.com> wrote in message
| > news:e3t28k02ien@drn.newsguy.com...
| > | > PD says...
| > | >
| > | > >I believe a geodesic is a path of *maximal* time-like
| > | > >interval, not a path of *minimal* time-like interval. It is this
| > | > >difference from Euclidean geometry that is the underlying
explanation
| > | > >of even SR phenomena like the Twin Puzzle.
| > | >
| > | > Yes, you're right. The correct word, to cover both cases, is
| > | > "extremizing", but that's kind of an awkward word.
| > |
| > | Yes - sounds frightening :-)
| > | "Keeping stationary" sounds better.
| >
| > These people keep stationary,
| > http://www.eurwholesale.co.uk/default.aspx?cam=ggl ,
| > fucking stupid local village idiot.
|
| That would be "stationery", you FSLVI you.
LOL!
So I do have your attention then, *****, and you do look at web pages.
http://tinyurl.com/k2uw6
Androcles.
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| User: "Ken S. Tucker" |
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| Title: Re: Energy slows down clocks. |
10 May 2006 01:12:44 PM |
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Dirk Van de moortel wrote:
"Daryl McCullough" <stevendaryl3016@yahoo.com> wrote in message news:e3t28k02ien@drn.newsguy.com...
PD says...
I believe a geodesic is a path of *maximal* time-like
interval, not a path of *minimal* time-like interval. It is this
difference from Euclidean geometry that is the underlying explanation
of even SR phenomena like the Twin Puzzle.
Recall the geodesic can be written as an absolute
derivative "D" of a 4-velocity (w.r.t the interval) like,
DU^u/ds =0.
That looks like a *minimal*.
Yes, you're right. The correct word, to cover both cases, is
"extremizing", but that's kind of an awkward word.
Yes - sounds frightening :-)
"Keeping stationary" sounds better.
Dirk Vdm
Yeah that's what is used in the translated version
of GR1916 in preamble before Eq.(20),
"$ds is stationary - a geodetic line -"
Ken
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| User: "Hexenmeister" |
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| Title: Re: Energy slows down clocks. |
10 May 2006 03:09:28 PM |
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"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1147284764.024666.119830@v46g2000cwv.googlegroups.com...
|
| Dirk Van de moortel wrote:
| > "Daryl McCullough" <stevendaryl3016@yahoo.com> wrote in message
news:e3t28k02ien@drn.newsguy.com...
| > > PD says...
| > >
| > > >I believe a geodesic is a path of *maximal* time-like
| > > >interval, not a path of *minimal* time-like interval. It is this
| > > >difference from Euclidean geometry that is the underlying explanation
| > > >of even SR phenomena like the Twin Puzzle.
|
| Recall the geodesic can be written as an absolute
| derivative "D" of a 4-velocity (w.r.t the interval) like,
|
| DU^u/ds =0.
|
| That looks like a *minimal*.
|
| > > Yes, you're right. The correct word, to cover both cases, is
| > > "extremizing", but that's kind of an awkward word.
|
| > Yes - sounds frightening :-)
| > "Keeping stationary" sounds better.
| > Dirk Vdm
|
| Yeah that's what is used in the translated version
| of GR1916 in preamble before Eq.(20),
| "$ds is stationary - a geodetic line -"
| Ken
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.PNG
Androcles
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| User: "Tom Roberts" |
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| Title: Re: Energy slows down clocks. |
10 May 2006 03:29:53 PM |
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|
Daryl McCullough wrote:
PD says...
I believe a geodesic is a path of *maximal* time-like interval, not
a path of *minimal* time-like interval.
Yes, you're right. The correct word, to cover both cases, is
"extremizing", but that's kind of an awkward word.
Technically, this is indeed extremizing the Lagrangian. But as I pointed
out earlier, in the approximation Daryl used, minimizing L_classical is
the same as maximizing L_SR, precisely because of the minus sign.
Ken S. Tucker wrote:
DU^u/ds =0.
That looks like a *minimal*.
Nope. Setting an ordinary derivative to zero can select a minimum,
maximum, or inflection point, and you must look at the second derivative
to distinguish them. The same is true of this covariant derivative.
Indeed, for a timelike path this is essentially always a maximum, and
for a spacelike path it is essentially always a minimum.
Exercise for the reader: what is this for a null path?
Exercise for experts: under what conditions can my
"essentially always" be violated? hint: topology.
Tom Roberts
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| User: "Dirk Van de moortel" |
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| Title: Re: Energy slows down clocks. |
10 May 2006 04:13:58 PM |
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"Tom Roberts" <tjroberts137@sbcglobal.net> wrote in message news:53s8g.24821$4L1.3328@newssvr11.news.prodigy.com...
Daryl McCullough wrote:
PD says...
I believe a geodesic is a path of *maximal* time-like interval, not
a path of *minimal* time-like interval.
Yes, you're right. The correct word, to cover both cases, is
"extremizing", but that's kind of an awkward word.
Technically, this is indeed extremizing the Lagrangian. But as I pointed
out earlier, in the approximation Daryl used, minimizing L_classical is
the same as maximizing L_SR, precisely because of the minus sign.
Ken S. Tucker wrote:
DU^u/ds =0.
That looks like a *minimal*.
Nope. Setting an ordinary derivative to zero can select a minimum,
maximum, or inflection point, and you must look at the second derivative
to distinguish them. The same is true of this covariant derivative.
Indeed, for a timelike path this is essentially always a maximum, and
for a spacelike path it is essentially always a minimum.
Exercise for the reader: what is this for a null path?
saddle point
Exercise for experts: under what conditions can my
"essentially always" be violated? hint: topology.
if time could change from going forward to backward?
Dirk Vdm
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| User: "Hexenmeister" |
|
| Title: Re: Energy slows down clocks. |
10 May 2006 06:01:44 PM |
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|
"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote
in message news:qIs8g.418013$YJ1.11230240@phobos.telenet-ops.be...
|
| "Tom Roberts" <tjroberts137@sbcglobal.net> wrote in message
news:53s8g.24821$4L1.3328@newssvr11.news.prodigy.com...
| > Daryl McCullough wrote:
| > > PD says...
| > >> I believe a geodesic is a path of *maximal* time-like interval, not
| > >> a path of *minimal* time-like interval.
| > >
| > > Yes, you're right. The correct word, to cover both cases, is
| > > "extremizing", but that's kind of an awkward word.
| >
| > Technically, this is indeed extremizing the Lagrangian. But as I pointed
| > out earlier, in the approximation Daryl used, minimizing L_classical is
| > the same as maximizing L_SR, precisely because of the minus sign.
| >
| >
| > Ken S. Tucker wrote:
| > > DU^u/ds =0.
| > > That looks like a *minimal*.
| >
| > Nope. Setting an ordinary derivative to zero can select a minimum,
| > maximum, or inflection point, and you must look at the second derivative
| > to distinguish them. The same is true of this covariant derivative.
| >
| > Indeed, for a timelike path this is essentially always a maximum, and
| > for a spacelike path it is essentially always a minimum.
| >
| > Exercise for the reader: what is this for a null path?
|
| saddle point
|
| >
| > Exercise for experts: under what conditions can my
| > "essentially always" be violated? hint: topology.
|
| if time could change from going forward to backward?
|
| Dirk Vdm
ahahahaha... HAHAHAHA.... hahahaha....
Nice one... Truly psychotic.
Desperate since I said time was not a vector, aren't you?
Now begging Humpty Roberts to concur time runs backwards, eh?
He'll probably agree, he's as insane as you.
"Yes, tests of strong fields are few and far between, but there are
some:
the binary pulsars, and observations of accretion disks near black
holes"
`I don't know what you mean by "observations",' Alice said.
Humpty Roberts smiled contemptuously. `Of course you don't -- till I tell
you.
I meant "there's a nice knock-down argument for you!"' <shrug>
`But "observations" doesn't mean "a nice knock-down argument",' Alice
objected.
`When I use a word,' Humpty Roberts said, in rather a scornful tone,
<shrug>,
`it means just what I choose it to mean -- neither more nor less.' <shrug>
`The question is,' said Alice, `whether you can make words mean so many
different things.'
`The question is,' said Humpty Roberts, `which is to be master -- that's
all.' <shrug>
Alice asked "Why *say* observed when it is actually not observed?"
Humpty Roberts let out a great sigh.
" <sigh>", he said.
"The nuances of English. I was discussing the usage of words and
not the concepts they represent."
-- Tom Humpty Roberts
news:ZDmYf.51582$2O6.5573@newssvr12.news.prodigy.com
You are phonier than Humpty Roberts.
http://www.m-w.com/dictionary/phony
Etymology: perhaps alteration of fawney gilded brass ring used in the fawney
rig, a confidence game, from Irish fainne ring, from Old Irish anne -- more
at
*****
ahahaha... HAHAHAHAHA..... hahahaha...
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.PNG
http://www.androcles01.pwp.blueyonder.co.uk/xorimpliesor.PNG
Androcles.
.
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| User: "Ken S. Tucker" |
|
| Title: Re: Energy slows down clocks. |
10 May 2006 04:11:24 PM |
|
|
Tom Roberts wrote:
Daryl McCullough wrote:
PD says...
I believe a geodesic is a path of *maximal* time-like interval, not
a path of *minimal* time-like interval.
Yes, you're right. The correct word, to cover both cases, is
"extremizing", but that's kind of an awkward word.
Technically, this is indeed extremizing the Lagrangian. But as I pointed
out earlier, in the approximation Daryl used, minimizing L_classical is
the same as maximizing L_SR, precisely because of the minus sign.
Ken S. Tucker wrote:
DU^u/ds =0.
That looks like a *minimal*.
Nope. Setting an ordinary derivative to zero can select a minimum,
maximum, or inflection point, and you must look at the second derivative
to distinguish them. The same is true of this covariant derivative.
Well Tom when Daryl used the word "minimal",
I researched the literature and found no good
reason to disagree, on the contrary that's fine
for the reasons I detailed. Furthermore, when
Dirk introduced the word "stationary" that too
is also very reasonable, from literature.
Indeed, for a timelike path this is essentially always a maximum, and
for a spacelike path it is essentially always a minimum.
I'll know how to fracture spacetime
into "time" and "space" like paths, like,
R = r-m , T= t+m
Is that what you mean ?
Exercise for the reader: what is this for a null path?
Exercise for experts: under what conditions can my
"essentially always" be violated? hint: topology.
Tom Roberts
Regards
Ken
.
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| User: "Tom Roberts" |
|
| Title: Re: Energy slows down clocks. |
10 May 2006 11:08:42 PM |
|
|
Ken S. Tucker wrote:
Tom Roberts wrote:
Indeed, for a timelike path this is essentially always a maximum, and
for a spacelike path it is essentially always a minimum.
I'll know how to fracture spacetime
into "time" and "space" like paths, like,
R = r-m , T= t+m
Is that what you mean ?
Not at all. You _really_ need to learn the basics.
In relativity, a timelike path is one that follows a timelike direction.
That is, one whose tangent vector is timelike. A spacelike path is
likewise one whose tangent vector is spacelike.
If you don't know how to tell if a 4-vector is timelike or spacelike,
give up playing on the net and go STUDY. In fact, that's clearly a good
idea anyway.
Tom Roberts
.
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| User: "Ken S. Tucker" |
|
| Title: Re: Energy slows down clocks. |
11 May 2006 01:40:00 AM |
|
|
Tom Roberts wrote:
Ken S. Tucker wrote:
Tom Roberts wrote:
Indeed, for a timelike path this is essentially always a maximum, and
for a spacelike path it is essentially always a minimum.
I'll know how to fracture spacetime
into "time" and "space" like paths, like,
R = r-m , T= t+m
Is that what you mean ?
Not at all. You _really_ need to learn the basics.
Tom read your post, ah you're attempting to
fracture motion as if spacetime is newtonian.
*****Your jargon is a fuckin brain fart*****
I politely asked for clarification of your PoV,
and you spewed meaningless BULL ***** below.
Tom ***** or get off the pot!
In relativity, a timelike path is one that follows a timelike direction.
That is, one whose tangent vector is timelike. A spacelike path is
likewise one whose tangent vector is spacelike.
If you don't know how to tell if a 4-vector is timelike or spacelike,
give up playing on the net and go STUDY. In fact, that's clearly a good
idea anyway.
Tom Roberts
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| User: "=?UTF-8?Q?Jeff=E2=80=A6Relf?=" |
|
| Title: Re: Energy slows down clocks. |
08 May 2006 12:14:35 PM |
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|
Hi Daryl_McCullough, You told me:
Well, if you include gravity, then we can approximately
give the proper time tau on a clock as a function of its
velocity and height above the Earth as follows:
tau = T (1 -(KE - PE)/mc^2)
where KE = time-averaged kinetic energy (1/2 mv^2),
PE = time-averaged Newtonian potential energy (-GMm/r),
T = the time as measured by a clock far from the earth
at rest relative to the earth.
So adding kinetic energy to the clock slows it down,
while adding potential energy to the clock speeds it up.
Wow, how simple... I didn't know that... looks about right though.
.
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| User: "tomgee" |
|
| Title: Re: Why the more accelerated twin aged less. |
08 May 2006 01:13:18 AM |
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|
Sam Wormley wrote:
tomgee wrote:
But to say that the mountaintop clock run faster than the
clock on the ground is to say that it moved slower than
the ground clock, but it actually experiences more
acceleration than the ground clock, thus it should run
slower, not faster.
Non spinning plant (no motion) the clock experiencing greater
gravitation (bottom of mountain) runs slower--prediction of
general relativity.
That's right, but special relativity says the clock moving faster
runs slower. The mountaintop clock and that of the GPS
fixed in orbit about the earth are moving faster than the clock
on earth's surface.
.
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| User: "=?UTF-8?Q?Jeff=E2=80=A6Relf?=" |
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| Title: Altitude increased the tick rate more than the velocity decreased it. |
08 May 2006 04:16:21 AM |
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|
Hi Tom_Gee, Clocks on GPS birds and on-top mountains tick faster because
the altitude increased the tick rate more than the velocity decreased it.
.
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| User: "tomgee" |
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| Title: Re: Altitude increased the tick rate more than the velocity decreased it. |
08 May 2006 05:45:20 AM |
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|
Jeff...Relf wrote:
Hi Tom_Gee, Clocks on GPS birds and on-top mountains tick faster because
the altitude increased the tick rate more than the velocity decreased it.
Yes, that's correct, and yet, the gravitational pull
decreases with altitude, therefore its effects
should be less the higher up the clock is, and
stronger for the ground clock. At some point
away from the pull of earth's gravity, the time
rates match and then reverse.
.
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| User: "Sam Wormley" |
|
| Title: Re: Why the more accelerated twin aged less. |
08 May 2006 08:27:10 AM |
|
|
tomgee wrote:
Sam Wormley wrote:
tomgee wrote:
But to say that the mountaintop clock run faster than the
clock on the ground is to say that it moved slower than
the ground clock, but it actually experiences more
acceleration than the ground clock, thus it should run
slower, not faster.
Non spinning plant (no motion) the clock experiencing greater
gravitation (bottom of mountain) runs slower--prediction of
general relativity.
That's right, but special relativity says the clock moving faster
runs slower. The mountaintop clock and that of the GPS
fixed in orbit about the earth are moving faster than the clock
on earth's surface.
Calculate the effects models by SR *and* the effects modeled by GTR.
You'll see.
.
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|
| User: "tomgee" |
|
| Title: Re: Why the more accelerated twin aged less. |
09 May 2006 05:11:33 AM |
|
|
Sam Wormley wrote:
tomgee wrote:
Sam Wormley wrote:
tomgee wrote:
But to say that the mountaintop clock run faster than the
clock on the ground is to say that it moved slower than
the ground clock, but it actually experiences more
acceleration than the ground clock, thus it should run
slower, not faster.
Non spinning plant (no motion) the clock experiencing greater
gravitation (bottom of mountain) runs slower--prediction of
general relativity.
That's right, but special relativity says the clock moving faster
runs slower. The mountaintop clock and that of the GPS
fixed in orbit about the earth are moving faster than the clock
on earth's surface.
Calculate the effects models by SR *and* the effects modeled by GTR.
You'll see.
What is it you think I will see then, Worms?
.
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|
| User: "AllYou!" |
|
| Title: Re: Why the more accelerated twin aged less. |
09 May 2006 07:30:37 AM |
|
|
"tomgee" <tyropress@yahoo.com> wrote in message
news:1147169493.306662.9150@e56g2000cwe.googlegroups.com...
Sam Wormley wrote:
tomgee wrote:
Sam Wormley wrote:
tomgee wrote:
But to say that the mountaintop clock run faster than the
clock on the ground is to say that it moved slower than
the ground clock, but it actually experiences more
acceleration than the ground clock, thus it should run
slower, not faster.
Non spinning plant (no motion) the clock experiencing greater
gravitation (bottom of mountain) runs slower--prediction of
general relativity.
That's right, but special relativity says the clock moving faster
runs slower. The mountaintop clock and that of the GPS
fixed in orbit about the earth are moving faster than the clock
on earth's surface.
Calculate the effects models by SR *and* the effects modeled by
GTR.
You'll see.
What is it you think I will see then, Worms?
That you're wrong.
.
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| User: "tomgee" |
|
| Title: Re: Why the more accelerated twin aged less. |
10 May 2006 12:52:12 AM |
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|
AllYou! wrote:
"tomgee" <tyropress@yahoo.com> wrote in message
news:1147169493.306662.9150@e56g2000cwe.googlegroups.com...
Sam Wormley wrote:
tomgee wrote:
Sam Wormley wrote:
SNIP
Calculate the effects models by SR *and* the effects modeled by
GTR.
You'll see.
What is it you think I will see then, Worms?
That you're wrong.
Why, Worms, you never told us there are two of you!
Which one of you two is the clone?
.
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