| Topic: |
Science > Physics |
| User: |
"Ben Rudiak-Gould" |
| Date: |
04 Feb 2005 04:08:37 PM |
| Object: |
An interesting SR puzzle |
Here's a two-part SR puzzle:
1. I'm standing still (i.e. moving inertially) while whirling a
clock around my head at a constant speed. (If you're worried
about conservation of momentum, imagine that I'm whirling two
clocks: it doesn't make any difference). I'm wearing a
wristwatch. Which of these--the clock or the wristwatch--will
appear to me to tick faster, and by how much?
2. Now I'm a different person, sitting on the clock as it whirls
around, looking at the clock and at the wristwatch of the
person doing the whirling. Which (if either) will appear to
me to tick faster, and by how much?
I already know what the answer to this puzzle is, and why. But I find it
very interesting from a pedagogical perspective. Despite doing well in my
undergrad SR course, and despite the puzzle's simplicity (no calculation
required), I doubt I could have solved part 2 had it been on an exam. I
would have become hopelessly mired in trying to figure out the reference
frame of the person sitting on the clock, because SR, as it was taught to
me, was about relating the reference frames of different observers. I didn't
understand at the time that the coordinate system you choose to solve a
problem needn't be the rest frame of the measurement device, and that there
isn't even a well-defined notion of /the/ rest frame of an object in
general. I conjecture that this kind of confusion is quite common.
This puzzle seems to highlight this point more clearly than any I've seen
before. More generally, it's interesting as an example of a problem
involving relative motion for which the Lorentz transformation is really no
help at all. And its solution might help to dispel the common misconception
that SR can't deal with acceleration.
One can also add parts 1b and 2b which ask whether the clock and wristwatch
are redshifted or blueshifted, and by what factor. Not until years later did
I understand the relationship between Doppler shift and the apparent rate of
clocks.
In short, I like this puzzle a lot, and I wish it were taught alongside the
twin paradox (or instead of it!) in undergraduate courses and textbooks. I'm
curious to know if anyone here has taught it, or thought about doing so, or
decided against it, or whatever.
-- Ben
.
|
|
| User: "Eli Botkin" |
|
| Title: Re: An interesting SR puzzle |
04 Feb 2005 08:38:02 PM |
|
|
"Ben Rudiak-Gould" <br276deleteme@cam.ac.uk> wrote in message
news:cu0rp6$r7a$1@gemini.csx.cam.ac.uk...
Here's a two-part SR puzzle:
1. I'm standing still (i.e. moving inertially) while whirling a
clock around my head at a constant speed. (If you're worried
about conservation of momentum, imagine that I'm whirling two
clocks: it doesn't make any difference). I'm wearing a
wristwatch. Which of these--the clock or the wristwatch--will
appear to me to tick faster, and by how much?
2. Now I'm a different person, sitting on the clock as it whirls
around, looking at the clock and at the wristwatch of the
person doing the whirling. Which (if either) will appear to
me to tick faster, and by how much?
I already know what the answer to this puzzle is, and why. But I find it
very interesting from a pedagogical perspective. Despite doing well in my
undergrad SR course, and despite the puzzle's simplicity (no calculation
required), I doubt I could have solved part 2 had it been on an exam. I
would have become hopelessly mired in trying to figure out the reference
frame of the person sitting on the clock, because SR, as it was taught to
me, was about relating the reference frames of different observers. I
didn't understand at the time that the coordinate system you choose to
solve a problem needn't be the rest frame of the measurement device, and
that there isn't even a well-defined notion of /the/ rest frame of an
object in general. I conjecture that this kind of confusion is quite
common.
This puzzle seems to highlight this point more clearly than any I've seen
before. More generally, it's interesting as an example of a problem
involving relative motion for which the Lorentz transformation is really
no help at all. And its solution might help to dispel the common
misconception that SR can't deal with acceleration.
One can also add parts 1b and 2b which ask whether the clock and
wristwatch are redshifted or blueshifted, and by what factor. Not until
years later did I understand the relationship between Doppler shift and
the apparent rate of clocks.
In short, I like this puzzle a lot, and I wish it were taught alongside
the twin paradox (or instead of it!) in undergraduate courses and
textbooks. I'm curious to know if anyone here has taught it, or thought
about doing so, or decided against it, or whatever.
-- Ben
Hi Ben:
I haven't heard this one before. My response is:
Each person will claim that the other clock is running slower than his own.
Also, if they are viewing each others clocks, the images will appear to
advance in synchrony with their personal clocks (or wristwatchs).
If you don't agree, I would very much like your take on the correct
solution.
Eli
.
|
|
|
| User: "Androcles Androcles@ MyPlace.org" |
|
| Title: Re: An interesting SR puzzle |
04 Feb 2005 09:14:39 PM |
|
|
"Eli Botkin" <elibotkin@optonline.net> wrote in message
news:1107571138.ff6c55df2a7871da8f6d9a58c3280785@teranews...
Hi Ben:
I haven't heard this one before. My response is:
Each person will claim that the other clock is running slower than his
own.
Why should anyone do that?
Also, if they are viewing each others clocks, the images will appear
to advance in synchrony with their personal clocks (or wristwatchs).
If you don't agree, I would very much like your take on the correct
solution.
Well, I don't agree, and you can see the take at
http://www.androc1es.pwp.blueyonder.co.uk/KoksDoppler.htm
Androcles
Eli
.
|
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|
|
| User: "Paul B. Andersen" |
|
| Title: Re: An interesting SR puzzle |
06 Feb 2005 03:32:24 PM |
|
|
Eli Botkin wrote:
"Ben Rudiak-Gould" <br276deleteme@cam.ac.uk> wrote in message
news:cu0rp6$r7a$1@gemini.csx.cam.ac.uk...
Here's a two-part SR puzzle:
1. I'm standing still (i.e. moving inertially) while whirling a
clock around my head at a constant speed. (If you're worried
about conservation of momentum, imagine that I'm whirling two
clocks: it doesn't make any difference). I'm wearing a
wristwatch. Which of these--the clock or the wristwatch--will
appear to me to tick faster, and by how much?
2. Now I'm a different person, sitting on the clock as it whirls
around, looking at the clock and at the wristwatch of the
person doing the whirling. Which (if either) will appear to
me to tick faster, and by how much?
I already know what the answer to this puzzle is, and why. But I find it
very interesting from a pedagogical perspective. Despite doing well in my
undergrad SR course, and despite the puzzle's simplicity (no calculation
required), I doubt I could have solved part 2 had it been on an exam. I
would have become hopelessly mired in trying to figure out the reference
frame of the person sitting on the clock, because SR, as it was taught to
me, was about relating the reference frames of different observers. I
didn't understand at the time that the coordinate system you choose to
solve a problem needn't be the rest frame of the measurement device, and
that there isn't even a well-defined notion of /the/ rest frame of an
object in general. I conjecture that this kind of confusion is quite
common.
This puzzle seems to highlight this point more clearly than any I've seen
before. More generally, it's interesting as an example of a problem
involving relative motion for which the Lorentz transformation is really
no help at all. And its solution might help to dispel the common
misconception that SR can't deal with acceleration.
One can also add parts 1b and 2b which ask whether the clock and
wristwatch are redshifted or blueshifted, and by what factor. Not until
years later did I understand the relationship between Doppler shift and
the apparent rate of clocks.
In short, I like this puzzle a lot, and I wish it were taught alongside
the twin paradox (or instead of it!) in undergraduate courses and
textbooks. I'm curious to know if anyone here has taught it, or thought
about doing so, or decided against it, or whatever.
-- Ben
Hi Ben:
I haven't heard this one before. My response is:
Each person will claim that the other clock is running slower than his own.
Also, if they are viewing each others clocks, the images will appear to
advance in synchrony with their personal clocks (or wristwatchs).
If you don't agree, I would very much like your take on the correct
solution.
Eli
The "whirling clock" is accelerated.
That makes all the difference.
The correct answer is they will both agree
that the wrist watch runs faster.
Paul
.
|
|
|
| User: "glbrad01" |
|
| Title: Re: An interesting SR puzzle |
06 Feb 2005 11:00:59 PM |
|
|
"Paul B. Andersen" <paul.b.andersen@deletethishia.no> wrote in message
news:cu62db$oiq$1@dolly.uninett.no...
Eli Botkin wrote:
"Ben Rudiak-Gould" <br276deleteme@cam.ac.uk> wrote in message
news:cu0rp6$r7a$1@gemini.csx.cam.ac.uk...
Here's a two-part SR puzzle:
1. I'm standing still (i.e. moving inertially) while whirling a
clock around my head at a constant speed. (If you're worried
about conservation of momentum, imagine that I'm whirling two
clocks: it doesn't make any difference). I'm wearing a
wristwatch. Which of these--the clock or the wristwatch--will
appear to me to tick faster, and by how much?
2. Now I'm a different person, sitting on the clock as it whirls
around, looking at the clock and at the wristwatch of the
person doing the whirling. Which (if either) will appear to
me to tick faster, and by how much?
I already know what the answer to this puzzle is, and why. But I find it
very interesting from a pedagogical perspective. Despite doing well in my
undergrad SR course, and despite the puzzle's simplicity (no calculation
required), I doubt I could have solved part 2 had it been on an exam. I
would have become hopelessly mired in trying to figure out the reference
frame of the person sitting on the clock, because SR, as it was taught to
me, was about relating the reference frames of different observers. I
didn't understand at the time that the coordinate system you choose to
solve a problem needn't be the rest frame of the measurement device, and
that there isn't even a well-defined notion of /the/ rest frame of an
object in general. I conjecture that this kind of confusion is quite
common.
This puzzle seems to highlight this point more clearly than any I've seen
before. More generally, it's interesting as an example of a problem
involving relative motion for which the Lorentz transformation is really
no help at all. And its solution might help to dispel the common
misconception that SR can't deal with acceleration.
One can also add parts 1b and 2b which ask whether the clock and
wristwatch are redshifted or blueshifted, and by what factor. Not until
years later did I understand the relationship between Doppler shift and
the apparent rate of clocks.
In short, I like this puzzle a lot, and I wish it were taught alongside
the twin paradox (or instead of it!) in undergraduate courses and
textbooks. I'm curious to know if anyone here has taught it, or thought
about doing so, or decided against it, or whatever.
-- Ben
Hi Ben:
I haven't heard this one before. My response is:
Each person will claim that the other clock is running slower than his
own.
Also, if they are viewing each others clocks, the images will appear to
advance in synchrony with their personal clocks (or wristwatchs).
If you don't agree, I would very much like your take on the correct
solution.
Eli
The "whirling clock" is accelerated.
That makes all the difference.
The correct answer is they will both agree
that the wrist watch runs faster.
Paul
That is not the correct answer.
Brad
.
|
|
|
| User: "RP" |
|
| Title: Re: An interesting SR puzzle |
07 Feb 2005 01:08:43 AM |
|
|
glbrad01 wrote:
"Paul B. Andersen" <paul.b.andersen@deletethishia.no> wrote in message
news:cu62db$oiq$1@dolly.uninett.no...
Eli Botkin wrote:
"Ben Rudiak-Gould" <br276deleteme@cam.ac.uk> wrote in message
news:cu0rp6$r7a$1@gemini.csx.cam.ac.uk...
Here's a two-part SR puzzle:
1. I'm standing still (i.e. moving inertially) while whirling a
clock around my head at a constant speed. (If you're worried
about conservation of momentum, imagine that I'm whirling two
clocks: it doesn't make any difference). I'm wearing a
wristwatch. Which of these--the clock or the wristwatch--will
appear to me to tick faster, and by how much?
2. Now I'm a different person, sitting on the clock as it whirls
around, looking at the clock and at the wristwatch of the
person doing the whirling. Which (if either) will appear to
me to tick faster, and by how much?
I already know what the answer to this puzzle is, and why. But I find it
very interesting from a pedagogical perspective. Despite doing well in my
undergrad SR course, and despite the puzzle's simplicity (no calculation
required), I doubt I could have solved part 2 had it been on an exam. I
would have become hopelessly mired in trying to figure out the reference
frame of the person sitting on the clock, because SR, as it was taught to
me, was about relating the reference frames of different observers. I
didn't understand at the time that the coordinate system you choose to
solve a problem needn't be the rest frame of the measurement device, and
that there isn't even a well-defined notion of /the/ rest frame of an
object in general. I conjecture that this kind of confusion is quite
common.
This puzzle seems to highlight this point more clearly than any I've seen
before. More generally, it's interesting as an example of a problem
involving relative motion for which the Lorentz transformation is really
no help at all. And its solution might help to dispel the common
misconception that SR can't deal with acceleration.
One can also add parts 1b and 2b which ask whether the clock and
wristwatch are redshifted or blueshifted, and by what factor. Not until
years later did I understand the relationship between Doppler shift and
the apparent rate of clocks.
In short, I like this puzzle a lot, and I wish it were taught alongside
the twin paradox (or instead of it!) in undergraduate courses and
textbooks. I'm curious to know if anyone here has taught it, or thought
about doing so, or decided against it, or whatever.
-- Ben
Hi Ben:
I haven't heard this one before. My response is:
Each person will claim that the other clock is running slower than his
own.
Also, if they are viewing each others clocks, the images will appear to
advance in synchrony with their personal clocks (or wristwatchs).
If you don't agree, I would very much like your take on the correct
solution.
Eli
The "whirling clock" is accelerated.
That makes all the difference.
The correct answer is they will both agree
that the wrist watch runs faster.
Paul
That is not the correct answer.
Brad
I agree.
The correct answer is that in this system GR must be applied. The
whirling clock sees a gravitational field, and it is deeper in the
well than is the wristwatch. Though we can correctly derive the
relative ticking rates from the FoR of the wristwatch using SR, we
cannot reciprocate the observations, because only the wristwatch is
inertial in this system. That the SR solution in this case gives the
correct ticking rate offset is a direct result of the fact that GR was
specifically derived to this end. This is just Einstein's rotating
disk argument. SR is valid only for inertial frames. This is not to
say that it cannot account for the acceleration of objects wrt an
inertial frame, only that it cannot account for the observations made
by accelerated observers. This is the very reason that GR was derived.
Richard Perry
.
|
|
|
|
| User: "RP" |
|
| Title: Re: An interesting SR puzzle |
07 Feb 2005 01:06:14 AM |
|
|
glbrad01 wrote:
"Paul B. Andersen" <paul.b.andersen@deletethishia.no> wrote in message
news:cu62db$oiq$1@dolly.uninett.no...
Eli Botkin wrote:
"Ben Rudiak-Gould" <br276deleteme@cam.ac.uk> wrote in message
news:cu0rp6$r7a$1@gemini.csx.cam.ac.uk...
Here's a two-part SR puzzle:
1. I'm standing still (i.e. moving inertially) while whirling a
clock around my head at a constant speed. (If you're worried
about conservation of momentum, imagine that I'm whirling two
clocks: it doesn't make any difference). I'm wearing a
wristwatch. Which of these--the clock or the wristwatch--will
appear to me to tick faster, and by how much?
2. Now I'm a different person, sitting on the clock as it whirls
around, looking at the clock and at the wristwatch of the
person doing the whirling. Which (if either) will appear to
me to tick faster, and by how much?
I already know what the answer to this puzzle is, and why. But I find it
very interesting from a pedagogical perspective. Despite doing well in my
undergrad SR course, and despite the puzzle's simplicity (no calculation
required), I doubt I could have solved part 2 had it been on an exam. I
would have become hopelessly mired in trying to figure out the reference
frame of the person sitting on the clock, because SR, as it was taught to
me, was about relating the reference frames of different observers. I
didn't understand at the time that the coordinate system you choose to
solve a problem needn't be the rest frame of the measurement device, and
that there isn't even a well-defined notion of /the/ rest frame of an
object in general. I conjecture that this kind of confusion is quite
common.
This puzzle seems to highlight this point more clearly than any I've seen
before. More generally, it's interesting as an example of a problem
involving relative motion for which the Lorentz transformation is really
no help at all. And its solution might help to dispel the common
misconception that SR can't deal with acceleration.
One can also add parts 1b and 2b which ask whether the clock and
wristwatch are redshifted or blueshifted, and by what factor. Not until
years later did I understand the relationship between Doppler shift and
the apparent rate of clocks.
In short, I like this puzzle a lot, and I wish it were taught alongside
the twin paradox (or instead of it!) in undergraduate courses and
textbooks. I'm curious to know if anyone here has taught it, or thought
about doing so, or decided against it, or whatever.
-- Ben
Hi Ben:
I haven't heard this one before. My response is:
Each person will claim that the other clock is running slower than his
own.
Also, if they are viewing each others clocks, the images will appear to
advance in synchrony with their personal clocks (or wristwatchs).
If you don't agree, I would very much like your take on the correct
solution.
Eli
The "whirling clock" is accelerated.
That makes all the difference.
The correct answer is they will both agree
that the wrist watch runs faster.
Paul
That is not the correct answer.
Brad
I agree.
The correct answer is that in this system GR must be applied. The
whirling clock sees a gravitational field, and it is deeper in the
well than is the wristwatch. Though we can correctly derive the
relative ticking rates from the FoR of the wristwatch using SR, we
cannot reciprocate the observations, because only the wristwatch is
inertial in this system. That the SR solution in this case gives the
correct ticking rate offset is a direct result of the fact that GR was
specifically derived to this end. This is just Einstein's rotating
disk argument. SR is valid only for inertial frames. This is not to
say that it cannot account for the acceleration of objects wrt an
inertial frame, only that it cannot account for the observations made
by accelerated observers. This is the very reason that GR was derived.
Richard Perry
Richard Perry
.
|
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|
| User: "Jesse Mazer" |
|
| Title: Re: An interesting SR puzzle |
07 Feb 2005 03:35:38 AM |
|
|
RP wrote:
glbrad01 wrote:
"Paul B. Andersen" <paul.b.andersen@deletethishia.no> wrote in
message news:cu62db$oiq$1@dolly.uninett.no...
Eli Botkin wrote:
"Ben Rudiak-Gould" <br276deleteme@cam.ac.uk> wrote in message
news:cu0rp6$r7a$1@gemini.csx.cam.ac.uk...
Here's a two-part SR puzzle:
1. I'm standing still (i.e. moving inertially) while whirling a
clock around my head at a constant speed. (If you're worried
about conservation of momentum, imagine that I'm whirling two
clocks: it doesn't make any difference). I'm wearing a
wristwatch. Which of these--the clock or the wristwatch--will
appear to me to tick faster, and by how much?
2. Now I'm a different person, sitting on the clock as it whirls
around, looking at the clock and at the wristwatch of the
person doing the whirling. Which (if either) will appear to
me to tick faster, and by how much?
I already know what the answer to this puzzle is, and why. But I
find it very interesting from a pedagogical perspective. Despite
doing well in my undergrad SR course, and despite the puzzle's
simplicity (no calculation required), I doubt I could have solved
part 2 had it been on an exam. I would have become hopelessly
mired in trying to figure out the reference frame of the person
sitting on the clock, because SR, as it was taught to me, was
about relating the reference frames of different observers. I
didn't understand at the time that the coordinate system you
choose to solve a problem needn't be the rest frame of the
measurement device, and that there isn't even a well-defined
notion of /the/ rest frame of an object in general. I conjecture
that this kind of confusion is quite common.
This puzzle seems to highlight this point more clearly than any
I've seen before. More generally, it's interesting as an example
of a problem involving relative motion for which the Lorentz
transformation is really no help at all. And its solution might
help to dispel the common misconception that SR can't deal with
acceleration.
One can also add parts 1b and 2b which ask whether the clock and
wristwatch are redshifted or blueshifted, and by what factor. Not
until years later did I understand the relationship between
Doppler shift and the apparent rate of clocks.
In short, I like this puzzle a lot, and I wish it were taught
alongside the twin paradox (or instead of it!) in undergraduate
courses and textbooks. I'm curious to know if anyone here has
taught it, or thought about doing so, or decided against it, or
whatever.
-- Ben
Hi Ben:
I haven't heard this one before. My response is:
Each person will claim that the other clock is running slower than
his own.
Also, if they are viewing each others clocks, the images will
appear to advance in synchrony with their personal clocks (or
wristwatchs).
If you don't agree, I would very much like your take on the correct
solution.
Eli
The "whirling clock" is accelerated.
That makes all the difference.
The correct answer is they will both agree
that the wrist watch runs faster.
Paul
That is not the correct answer.
Brad
I agree.
The correct answer is that in this system GR must be applied.
No, SR can deal with the question of the proper time recorded by an
accelerating clock by considering its path as seen in an inertial frame.
Just integrate [squareroot(1 - v(t)^2/c^2) dt] between two times t1 and
t2 in your frame, with v(t) being the clock's instantaneous velocity at
time t in your frame; this will give the correct answer for the time
elapsed on the accelerating clock during the interval (t1, t2) in your
frame (you also have to take into account the finite speed of light to
figure out how much time you'd actually *see* elapse on the accelerating
clock during that time-interval, but in this problem the answer would be
the same). Then imagine that your own clock is sending signals to the
accelerating clock at the speed of light, and figure out what
time-signal the accelerating clock will be receiving at time t1 and what
time-signal it will be receiving at time t2; this tells you what amount
of time an observer sitting on the accelerating clock would see elapsed
on your clock during the same interval. If the accelerating clock is
maintaining a constant distance from your clock, and a constant velocity
in your frame, it's pretty easy to show that an observer on the
accelerating clock will see your clock sped up by the same amount that
you see the accelarating clock slowed down.
Jesse
.
|
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|
| User: "Dirk Van de moortel" |
|
| Title: Re: An interesting SR puzzle |
07 Feb 2005 08:04:46 AM |
|
|
"Jesse Mazer" <vze2ztqw@mail.verizon.net> wrote in message news:4207605E.4090903@mail.verizon.net...
RP wrote:
glbrad01 wrote:
"Paul B. Andersen" <paul.b.andersen@deletethishia.no> wrote in
message news:cu62db$oiq$1@dolly.uninett.no...
[snip]
The "whirling clock" is accelerated.
That makes all the difference.
The correct answer is they will both agree
that the wrist watch runs faster.
Paul
That is not the correct answer.
Brad
I agree.
The correct answer is that in this system GR must be applied.
No, SR can deal with the question of the proper time recorded by an
accelerating clock by considering its path as seen in an inertial frame.
Just integrate [squareroot(1 - v(t)^2/c^2) dt] between two times t1 and
t2 in your frame, with v(t) being the clock's instantaneous velocity at
time t in your frame; this will give the correct answer for the time
elapsed on the accelerating clock during the interval (t1, t2) in your
frame
You can see this in action on
http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
And SR is even sufficient in a non-inertial frame:
http://www.geocities.com/slithytove5/AccelClocks.htm
http://groups-beta.google.com/group/sci.physics.relativity/msg/806a08d082376014
http://groups.google.co.uk/groups?&threadm=buphbm$m65$1@dolly.uninett.no
In that case a factor (1+a(t) x(t) / c^2 ) must be appended
to the square root.
It's a bit messy, but it works and it's still SR.
Dirk Vdm
(you also have to take into account the finite speed of light to
figure out how much time you'd actually *see* elapse on the accelerating
clock during that time-interval, but in this problem the answer would be
the same). Then imagine that your own clock is sending signals to the
accelerating clock at the speed of light, and figure out what
time-signal the accelerating clock will be receiving at time t1 and what
time-signal it will be receiving at time t2; this tells you what amount
of time an observer sitting on the accelerating clock would see elapsed
on your clock during the same interval. If the accelerating clock is
maintaining a constant distance from your clock, and a constant velocity
in your frame, it's pretty easy to show that an observer on the
accelerating clock will see your clock sped up by the same amount that
you see the accelarating clock slowed down.
Jesse
.
|
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| User: "Mike" |
|
| Title: Re: An interesting SR puzzle |
07 Feb 2005 08:39:06 AM |
|
|
Dirk Van de moortel wrote:
"Jesse Mazer" <vze2ztqw@mail.verizon.net> wrote in message
news:4207605E.4090903@mail.verizon.net...
RP wrote:
glbrad01 wrote:
"Paul B. Andersen" <paul.b.andersen@deletethishia.no> wrote in
message news:cu62db$oiq$1@dolly.uninett.no...
[snip]
The "whirling clock" is accelerated.
That makes all the difference.
The correct answer is they will both agree
that the wrist watch runs faster.
Paul
That is not the correct answer.
Brad
I agree.
The correct answer is that in this system GR must be applied.
No, SR can deal with the question of the proper time recorded by an
accelerating clock by considering its path as seen in an inertial
frame.
Just integrate [squareroot(1 - v(t)^2/c^2) dt] between two times t1
and
t2 in your frame, with v(t) being the clock's instantaneous
velocity at
time t in your frame; this will give the correct answer for the
time
elapsed on the accelerating clock during the interval (t1, t2) in
your
frame
You can see this in action on
http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
[snip]
Good think I forgot to KILLFILE you Donkey. The ecerpt from the above
imbecile's link:
"Observer I can parametrize the worldline of A with the same
proper time T, so he will see infinitesimally consecutive velocities
of A as v(T) and v(T+dT).
Since v(T+dT) is the standard SR composition ('addition') of the
velocities v(T) and dV(T), we can write (using c=1):
v(T+dT) = [ v(T) + dV(T) ] / [ 1 + v(T) dV(T) ]
which, using the expression for dV(T), becomes:
v(T+dT) = [ v(T) + a(T) dT ] / [ 1 + v(T) a(T) dT ]
To calculate dv(T)/dT we use
[ v(T+dT) - v(T) ] / dT = a(T) [1-v^2(T)] / [ 1 + v(T) a(T) dT
]
Take the limit dT --> 0
dv(T)/dT = a(T) [ 1 - v^2(T) ]"
Now, you think you can fool people with this stupid derivation of
yours. You ain't going to fool me Dinky-Donkey.
The error in the above derivation is basic. Only a Dinky-Donkey could
err that way.
Are you so stupid Dink-Donk?
Hint: maybe mixing Lorentz and Galilean relativity?
Mike
Dirk Vdm
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| User: "Dirk Van de moortel" |
|
| Title: Re: An interesting SR puzzle |
07 Feb 2005 09:50:41 AM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message news:1107787146.699646.255680@c13g2000cwb.googlegroups.com...
Dirk Van de moortel wrote:
"Jesse Mazer" <vze2ztqw@mail.verizon.net> wrote in message
news:4207605E.4090903@mail.verizon.net...
RP wrote:
glbrad01 wrote:
"Paul B. Andersen" <paul.b.andersen@deletethishia.no> wrote in
message news:cu62db$oiq$1@dolly.uninett.no...
[snip]
The "whirling clock" is accelerated.
That makes all the difference.
The correct answer is they will both agree
that the wrist watch runs faster.
Paul
That is not the correct answer.
Brad
I agree.
The correct answer is that in this system GR must be applied.
No, SR can deal with the question of the proper time recorded by an
accelerating clock by considering its path as seen in an inertial
frame.
Just integrate [squareroot(1 - v(t)^2/c^2) dt] between two times t1
and
t2 in your frame, with v(t) being the clock's instantaneous
velocity at
time t in your frame; this will give the correct answer for the
time
elapsed on the accelerating clock during the interval (t1, t2) in
your
frame
You can see this in action on
http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
[snip]
Good think I forgot to KILLFILE you Donkey. The ecerpt from the above
imbecile's link:
"Observer I can parametrize the worldline of A with the same
proper time T, so he will see infinitesimally consecutive velocities
of A as v(T) and v(T+dT).
Since v(T+dT) is the standard SR composition ('addition') of the
velocities v(T) and dV(T), we can write (using c=1):
v(T+dT) = [ v(T) + dV(T) ] / [ 1 + v(T) dV(T) ]
which, using the expression for dV(T), becomes:
v(T+dT) = [ v(T) + a(T) dT ] / [ 1 + v(T) a(T) dT ]
To calculate dv(T)/dT we use
[ v(T+dT) - v(T) ] / dT = a(T) [1-v^2(T)] / [ 1 + v(T) a(T) dT ]
Take the limit dT --> 0
dv(T)/dT = a(T) [ 1 - v^2(T) ]"
Now, you think you can fool people with this stupid derivation of
yours. You ain't going to fool me Dinky-Donkey.
The error in the above derivation is basic. Only a Dinky-Donkey could
err that way.
Are you so stupid Dink-Donk?
Hint: maybe mixing Lorentz and Galilean relativity?
You want the low speed limit or you want the Galilean
limit?
Okay, either take the limit for v(T) --> 0 of
dv(T)/dT = a(T) [ 1 - v^2(T) ]"
You get:
Limit { v(T) --> 0 ; dv(T)/dT }
= Limit { v(T) --> 0 ; a(T) [ 1 - v^2(T) ] }
= a(T)
or, after re-inserting c, take the limit for c to infinity of
dv(T)/dT = a(T) [ 1 - v^2(T)/c^2 ]"
You get
dv(T)/dT = a(T)
as well.
Does it hurt?
Anything else you have trouble grasping?
Maybe you should have a look at the definition of velocity.
Maybe you should have a look at the definition of derivative.
Maybe you should have a look at the definition of limit.
You want an epsilon delta definition?
Did you have as much trouble with this as you had with
square roots?
Dirk Vdm
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| User: "Mike" |
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| Title: Re: An interesting SR puzzle |
07 Feb 2005 10:14:05 AM |
|
|
Dirk Van de moortel wrote:
[snip]
You want the low speed limit or you want the Galilean
limit?
Okay, either take the limit for v(T) --> 0 of
dv(T)/dT = a(T) [ 1 - v^2(T) ]"
You get:
Limit { v(T) --> 0 ; dv(T)/dT }
= Limit { v(T) --> 0 ; a(T) [ 1 - v^2(T) ] }
= a(T)
or, after re-inserting c, take the limit for c to infinity of
dv(T)/dT = a(T) [ 1 - v^2(T)/c^2 ]"
You get
dv(T)/dT = a(T)
as well.
Does it hurt?
Your derivation is good only in the limit cases. You Dink-Donk cannot
see it's not valid in anything in between. Either rest or action at a
dostance. You just conformed Newton stupid.
Anything else you have trouble grasping?
YOu are the only one.
Maybe you should have a look at the definition of velocity.
You should look at the proper use of infinitesimals.
Maybe you should have a look at the definition of derivative.
That's what I meant.
Maybe you should have a look at the definition of limit.
Yeh, right. Donkey will reach us limits now.
You want an epsilon delta definition?
With open or closed intervals stupid?
Did you have as much trouble with this as you had with
square roots?
You are the one who still debates whether swrt is a function or a
definition.
Dink-Dink, Dinky-Donkey, the bell tolls for thee.
Mike
Dirk Vdm
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| User: "Dirk Van de moortel" |
|
| Title: Re: An interesting SR puzzle |
07 Feb 2005 10:47:00 AM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message news:1107792844.993607.285850@z14g2000cwz.googlegroups.com...
Dirk Van de moortel wrote:
[snip]
You want the low speed limit or you want the Galilean
limit?
Okay, either take the limit for v(T) --> 0 of
dv(T)/dT = a(T) [ 1 - v^2(T) ]"
You get:
Limit { v(T) --> 0 ; dv(T)/dT }
= Limit { v(T) --> 0 ; a(T) [ 1 - v^2(T) ] }
= a(T)
or, after re-inserting c, take the limit for c to infinity of
dv(T)/dT = a(T) [ 1 - v^2(T)/c^2 ]"
You get
dv(T)/dT = a(T)
as well.
Does it hurt?
Your derivation is good only in the limit cases. You Dink-Donk cannot
see it's not valid in anything in between. Either rest or action at a
dostance. You just conformed Newton stupid.
Ha, now I see, you object to the usage of my using the
relativistic composition of velocities in
| "Observer I can parametrize the worldline of A with the same
| proper time T, so he will see infinitesimally consecutive velocities
| of A as v(T) and v(T+dT).
| Since v(T+dT) is the standard SR composition ('addition') of the
| velocities v(T) and dV(T), we can write (using c=1):
| v(T+dT) = [ v(T) + dV(T) ] / [ 1 + v(T) dV(T) ]
You probably think this should be
v(T+dT) = v(T) + dV(T)
Well, I have some bad news for you. You missed the very
essence of the derivation. Look at the title: "SR treatment of..."
You know that SR says that
"When and object P has a relative velocity u w.r.t.
to an inertial frame Q, of which the observer itself
has a relative velocity v w.r.t. an inertial observer R,
then the object P has a relative velocity
w = (u+v)/(1+u v)
w.r.t. the observer R."
Now replace
P => observer A at proper time T+dT
Q => instantaneously comoving inertial frame of observer
at proper time T of observer
R => initial rest frame I
u => dV(T)
v => v(T)
w => v(T+dT)
We must do this because it must be valid for *all* values
of T, and v(T) becomes quite large, so using the Galilean
transformation here would be totally wrong.
Together with the fact that dV(T) can be written as
dV(T) = a(T) dT
because dT is infinitesimal, that is the very essence of
the derivation.
You see, it's not only mathematics that is doing its job
here. It's physics. It's knowing what the symbols mean.
Too bad you just made a fool of yourself by missing
the crucial part.
Dirk Vdm
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| User: "Mike" |
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| Title: Re: An interesting SR puzzle |
07 Feb 2005 04:18:03 PM |
|
|
Dirk Van de moortel wrote:
[snip]
| "Observer I can parametrize the worldline of A with the same
| proper time T, so he will see infinitesimally consecutive
velocities
| of A as v(T) and v(T+dT).
| Since v(T+dT) is the standard SR composition ('addition') of the
| velocities v(T) and dV(T), we can write (using c=1):
| v(T+dT) = [ v(T) + dV(T) ] / [ 1 + v(T) dV(T) ]
You probably think this should be
v(T+dT) = v(T) + dV(T)
[snip]
Obviously not. I did not object to SR velocity addition. By now, you
should be able to figure out your error. It's in your next step.
Mike
Dirk Vdm
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| User: "Dirk Van de moortel" |
|
| Title: Re: An interesting SR puzzle |
08 Feb 2005 04:38:44 AM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message news:1107814683.872500.141820@o13g2000cwo.googlegroups.com...
Dirk Van de moortel wrote:
[snip]
| "Observer I can parametrize the worldline of A with the same
| proper time T, so he will see infinitesimally consecutive
velocities
| of A as v(T) and v(T+dT).
| Since v(T+dT) is the standard SR composition ('addition') of the
| velocities v(T) and dV(T), we can write (using c=1):
| v(T+dT) = [ v(T) + dV(T) ] / [ 1 + v(T) dV(T) ]
You probably think this should be
v(T+dT) = v(T) + dV(T)
[snip]
Obviously not. I did not object to SR velocity addition. By now, you
should be able to figure out your error. It's in your next step.
The derivation is just as airtight as the plastic bag you have
put around your head. You are turning blue again.
Dirk Vdm
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| User: "Mike" |
|
| Title: Re: An interesting SR puzzle |
08 Feb 2005 04:44:10 AM |
|
|
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1107814683.872500.141820@o13g2000cwo.googlegroups.com...
Dirk Van de moortel wrote:
[snip]
| "Observer I can parametrize the worldline of A with the same
| proper time T, so he will see infinitesimally consecutive
velocities
| of A as v(T) and v(T+dT).
| Since v(T+dT) is the standard SR composition ('addition') of
the
| velocities v(T) and dV(T), we can write (using c=1):
| v(T+dT) = [ v(T) + dV(T) ] / [ 1 + v(T) dV(T) ]
You probably think this should be
v(T+dT) = v(T) + dV(T)
[snip]
Obviously not. I did not object to SR velocity addition. By now,
you
should be able to figure out your error. It's in your next step.
The derivation is just as airtight as the plastic bag you have
put around your head. You are turning blue again.
Dirk Vdm
Check it out again and again. Refrain from making psychotic statements
that demonstrate beyond any doubt your ratten character. Focus on the
sh** you keep on vomitting in these ng's.
Mike
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| User: "Dirk Van de moortel" |
|
| Title: Re: An interesting SR puzzle |
08 Feb 2005 05:14:09 AM |
|
|
"Mike" <eleatis@yahoo.gr> wrote in message news:1107859450.024900.182650@z14g2000cwz.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1107814683.872500.141820@o13g2000cwo.googlegroups.com...
Dirk Van de moortel wrote:
[snip]
| "Observer I can parametrize the worldline of A with the same
| proper time T, so he will see infinitesimally consecutive
velocities
| of A as v(T) and v(T+dT).
| Since v(T+dT) is the standard SR composition ('addition') of
the
| velocities v(T) and dV(T), we can write (using c=1):
| v(T+dT) = [ v(T) + dV(T) ] / [ 1 + v(T) dV(T) ]
You probably think this should be
v(T+dT) = v(T) + dV(T)
[snip]
Obviously not. I did not object to SR velocity addition. By now,
you
should be able to figure out your error. It's in your next step.
The derivation is just as airtight as the plastic bag you have
put around your head. You are turning blue again.
Dirk Vdm
Check it out again and again. Refrain from making psychotic statements
that demonstrate beyond any doubt your ratten character. Focus on the
sh** you keep on vomitting in these ng's.
Well, it's obvious that you are choking by now, and it's not
the first time, remember?
Do keep digging.... and squeal, pig, squeal :-)
Dirk Vdm
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|
| User: "Mike" |
|
| Title: Re: An interesting SR puzzle |
08 Feb 2005 06:18:50 AM |
|
|
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1107859450.024900.182650@z14g2000cwz.googlegroups.com...
Dirk Van de moortel wrote:
"Mike" <eleatis@yahoo.gr> wrote in message
news:1107814683.872500.141820@o13g2000cwo.googlegroups.com...
Dirk Van de moortel wrote:
[snip]
| "Observer I can parametrize the worldline of A with the
same
| proper time T, so he will see infinitesimally
consecutive
velocities
| of A as v(T) and v(T+dT).
| Since v(T+dT) is the standard SR composition
('addition') of
the
| velocities v(T) and dV(T), we can write (using c=1):
| v(T+dT) = [ v(T) + dV(T) ] / [ 1 + v(T) dV(T) ]
You probably think this should be
v(T+dT) = v(T) + dV(T)
[snip]
Obviously not. I did not object to SR velocity addition. By
now,
you
should be able to figure out your error. It's in your next
step.
The derivation is just as airtight as the plastic bag you have
put around your head. You are turning blue again.
Dirk Vdm
Check it out again and again. Refrain from making psychotic
statements
that demonstrate beyond any doubt your ratten character. Focus on
the
sh** you keep on vomitting in these ng's.
Well, it's obvious that you are choking by now, and it's not
the first time, remember?
Do keep digging.... and squeal, pig, squeal :-)
Dirk Vdm
You will be exposed when the right time comes Donkey. The bell tolls
for thee...
Mike
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| User: "RP" |
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| Title: Re: An interesting SR puzzle |
07 Feb 2005 12:58:20 PM |
|
|
Jesse Mazer wrote:
RP wrote:
glbrad01 wrote:
"Paul B. Andersen" <paul.b.andersen@deletethishia.no> wrote in
message news:cu62db$oiq$1@dolly.uninett.no...
Eli Botkin wrote:
"Ben Rudiak-Gould" <br276deleteme@cam.ac.uk> wrote in message
news:cu0rp6$r7a$1@gemini.csx.cam.ac.uk...
Here's a two-part SR puzzle:
1. I'm standing still (i.e. moving inertially) while whirling a
clock around my head at a constant speed. (If you're worried
about conservation of momentum, imagine that I'm whirling two
clocks: it doesn't make any difference). I'm wearing a
wristwatch. Which of these--the clock or the wristwatch--will
appear to me to tick faster, and by how much?
2. Now I'm a different person, sitting on the clock as it whirls
around, looking at the clock and at the wristwatch of the
person doing the whirling. Which (if either) will appear to
me to tick faster, and by how much?
I already know what the answer to this puzzle is, and why. But I
find it very interesting from a pedagogical perspective. Despite
doing well in my undergrad SR course, and despite the puzzle's
simplicity (no calculation required), I doubt I could have solved
part 2 had it been on an exam. I would have become hopelessly
mired in trying to figure out the reference frame of the person
sitting on the clock, because SR, as it was taught to me, was
about relating the reference frames of different observers. I
didn't understand at the time that the coordinate system you
choose to solve a problem needn't be the rest frame of the
measurement device, and that there isn't even a well-defined
notion of /the/ rest frame of an object in general. I conjecture
that this kind of confusion is quite common.
This puzzle seems to highlight this point more clearly than any
I've seen before. More generally, it's interesting as an example
of a problem involving relative motion for which the Lorentz
transformation is really no help at all. And its solution might
help to dispel the common misconception that SR can't deal with
acceleration.
One can also add parts 1b and 2b which ask whether the clock and
wristwatch are redshifted or blueshifted, and by what factor. Not
until years later did I understand the relationship between
Doppler shift and the apparent rate of clocks.
In short, I like this puzzle a lot, and I wish it were taught
alongside the twin paradox (or instead of it!) in undergraduate
courses and textbooks. I'm curious to know if anyone here has
taught it, or thought about doing so, or decided against it, or
whatever.
-- Ben
Hi Ben:
I haven't heard this one before. My response is:
Each person will claim that the other clock is running slower than
his own.
Also, if they are viewing each others clocks, the images will
appear to advance in synchrony with their personal clocks (or
wristwatchs).
If you don't agree, I would very much like your take on the correct
solution.
Eli
The "whirling clock" is accelerated.
That makes all the difference.
The correct answer is they will both agree
that the wrist watch runs faster.
Paul
That is not the correct answer.
Brad
I agree.
The correct answer is that in this system GR must be applied.
No, SR can deal with the question of the proper time recorded by an
accelerating clock by considering its path as seen in an inertial frame.
Just integrate [squareroot(1 - v(t)^2/c^2) dt] between two times t1 and
t2 in your frame, with v(t) being the clock's instantaneous velocity at
time t in your frame; this will give the correct answer for the time
elapsed on the accelerating clock during the interval (t1, t2) in your
frame (you also have to take into account the finite speed of light to
figure out how much time you'd actually *see* elapse on the accelerating
clock during that time-interval, but in this problem the answer would be
the same).
I as much said the same. There is a transverse doppler shift of the
"ticks" of the whirling clock wrt the wristwatch, which is nothing
more than the ticking rate offset being observed. This is calculated
in normal SR fashion. Nothing fancy needed, just plug in the
instantaneous velocity of the whirling clock. But how do you account
for the non-reciprocity of the effect? After all, wrt the whirling
clock your clock should be ticking slow, i.e. he should likewise
observer a transverse doppler red shift. This is directly where the
paradox has its substance. Unlike in the linear acceleration arguments
such as the twins paradox, there is no need for the other twin to
return and thus incur a reduced ticking rate on his trip to the other.
No, in this case there is never a change in scalar displacement and
ticking rate offsets will accrue over time to large differences. The
reunion never involves an excess distance to cover as in the twins
paradox, thus no opportunity to balance the books.
Wrt the accelerated clock, the other ticks fast, and SR cannot provide
this result. As noted, the lorentz transform requires the other clock
to tick slower.
Then imagine that your own clock is sending signals to the
accelerating clock at the speed of light, and figure out what
time-signal the accelerating clock will be receiving at time t1 and what
time-signal it will be receiving at time t2; this tells you what amount
of time an observer sitting on the accelerating clock would see elapsed
on your clock during the same interval.
Yes, same as above, but doesn't tell me what the accelerated clock
sees, only what the inertial clock says that it should see. That they
are the same is beside the point. The paradox boils down to the
question "Why doesn't the orbiting clock see the stationary clock
ticking slower, i.e. why isn't there a reciprocity in this case?".
If the accelerating clock is
maintaining a constant distance from your clock, and a constant velocity
in your frame, it's pretty easy to show that an observer on the
accelerating clock will see your clock sped up by the same amount that
you see the accelarating clock slowed down.
Sure, its easy to show, just plug in the gravitational potentials in
the GR ticking rate offset equation.
I'll keep an open mind that you and/or Dirk can deliver on this within
the context of SR, but it's going to involve some trickery, such as
ignoring the acceleration, using MCIFs instead, then deriving what the
inertial clock predicts and attributing that observation to the
accelerated clock, and burying that switch in frames within mounds of
differential calculus.
If SR can handle accelerated frames, then what need was there of GR?
Tell me, what is my instantaneous downward velocity as I sit here,
that I can derive my ticking rate relative to a clock on the ISS using
SR alone? I'm just accelerating right? Surely SR can be used to derive
our relative ticking rates.
Maybe I'm wrong for reasons as yet unknown to me, but OTOH I've not
seen anyone provide that math, as many times as they've stated that
"it can be done". Again, we're not talking here about acceleration of
of clocks and rulers wrt an inertial frame, but rather about motions
and ticking rates of clocks from within the perspective of an
accelerated frame.
Richard Perry
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| User: "Dirk Van de moortel" |
|
| Title: Re: An interesting SR puzzle |
07 Feb 2005 01:24:28 PM |
|
|
"RP" <no_mail_no_spam@yahoo.com> wrote in message news:36prroF55v1bfU1@individual.net...
[snip]
Maybe I'm wrong for reasons as yet unknown to me, but OTOH I've not
seen anyone provide that math, as many times as they've stated that
"it can be done". Again, we're not talking here about acceleration of
of clocks and rulers wrt an inertial frame, but rather about motions
and ticking rates of clocks from within the perspective of an
accelerated frame.
http://www.geocities.com/slithytove5/AccelClocks.htm
http://groups-beta.google.com/group/sci.physics.relativity/msg/806a08d082376014
http://groups.google.co.uk/groups?&threadm=buphbm$m65$1@dolly.uninett.no
Dirk Vdm
.
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|
| User: "" |
|
| Title: Re: An interesting SR puzzle |
07 Feb 2005 02:08:54 PM |
|
|
[...]Two travelers, Al and Bert, have independent accelerated motions
along a straight line. Their motions may be plotted on a spacetime
diagram FROM THE POINT IF VIEW OF A SINGLE INERTIAL FRAME, that
we’ll call the “earth” reference frame[...]
Just as I suspected Dirk. Maybe I'm not communicating the argument
precisely enough :)
The second link is an argument involving linear accelleartion and
accrued displacement over time, which I also addressed. and don't have
an issue with. The third was a dead link.
Thanks for the links though.
Richard Perry
Richard Perry
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| User: "Dirk Van de moortel" |
|
| Title: Re: An interesting SR puzzle |
07 Feb 2005 02:25:38 PM |
|
|
<no_mail_no_spam@yahoo.com> wrote in message news:1107806934.371652.165750@l41g2000cwc.googlegroups.com...
[...]Two travelers, Al and Bert, have independent accelerated motions
along a straight line. Their motions may be plotted on a spacetime
diagram FROM THE POINT IF VIEW OF A SINGLE INERTIAL FRAME, that
we’ll call the “earth” reference frame[...]
Just as I suspected Dirk. Maybe I'm not communicating the argument
precisely enough :)
The second link is an argument involving linear accelleartion and
accrued displacement over time, which I also addressed. and don't have
an issue with. The third was a dead link.
The third works perfectly for me.
Here's the link with the (bad) beta groups:
http://groups-beta.google.com/group/sci.physics.relativity/msg/22b466002b6ff4ed
In the introduction the second link points to the first link
http://www.geocities.com/slithytove5/AccelClocks.htm
which has a general accelerated frame.
Some inertial frame is used to make the calculations easier.
They could just as well have used Al's comoving
inertial frame at event "a".
The essentail thing is the restult, that describes the
"the rate of an accelerating clock relative to an accelerating observer"
(Tom's words) which amounts to the "motions and ticking rates of
clocks from within the perspective of an accelerated frame" (your
words).
That should satisfy your question:
| Maybe I'm wrong for reasons as yet unknown to me, but OTOH I've not
| seen anyone provide that math, as many times as they've stated that
| "it can be done". Again, we're not talking here about acceleration of
| of clocks and rulers wrt an inertial frame, but rather about motions
| and ticking rates of clocks from within the perspective of an
| accelerated frame."
So you see, it is messy, but "it can be done".
Requiring anything beyond what you get in Tom's article,
would be like requiring that someone derives and describe
the path of a canon ball from the perspective of a jojo in
action on Titan.
Thanks for the links though.
No problem.
Dirk Vdm
.
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|
|
| User: "RP" |
|
| Title: Re: An interesting SR puzzle |
07 Feb 2005 05:46:27 PM |
|
|
Dirk Van de moortel wrote:
<no_mail_no_spam@yahoo.com> wrote in message news:1107806934.371652.165750@l41g2000cwc.googlegroups.com...
[...]Two travelers, Al and Bert, have independent accelerated motions
along a straight line. Their motions may be plotted on a spacetime
diagram FROM THE POINT IF VIEW OF A SINGLE INERTIAL FRAME, that
we’ll call the “earth” reference frame[...]
Just as I suspected Dirk. Maybe I'm not communicating the argument
precisely enough :)
The second link is an argument involving linear accelleartion and
accrued displacement over time, which I also addressed. and don't have
an issue with. The third was a dead link.
The third works perfectly for me.
Here's the link with the (bad) beta groups:
http://groups-beta.google.com/group/sci.physics.relativity/msg/22b466002b6ff4ed
In the introduction the second link points to the first link
http://www.geocities.com/slithytove5/AccelClocks.htm
which has a general accelerated frame.
Some inertial frame is used to make the calculations easier.
They could just as well have used Al's comoving
inertial frame at event "a".
The essentail thing is the restult, that describes the
"the rate of an accelerating clock relative to an accelerating observer"
(Tom's words) which amounts to the "motions and ticking rates of
clocks from within the perspective of an accelerated frame" (your
words).
That should satisfy your question:
| Maybe I'm wrong for reasons as yet unknown to me, but OTOH I've not
| seen anyone provide that math, as many times as they've stated that
| "it can be done". Again, we're not talking here about acceleration of
| of clocks and rulers wrt an inertial frame, but rather about motions
| and ticking rates of clocks from within the perspective of an
| accelerated frame."
So you see, it is messy, but "it can be done".
Requiring anything beyond what you get in Tom's article,
would be like requiring that someone derives and describe
the path of a canon ball from the perspective of a jojo in
action on Titan.
Thanks for the links though.
No problem.
Dirk Vdm
I went over the Bert and Al argument again. I missed the section about
the differences in tilts of the simultaneity lines over a small
interval. I can see how this will affect relative ticking rates in a
way not described by the non-generalized transform that I was trying
to apply.
The effects of boost are accounted for in the equations, and these
provide a shift in the time offset between displaced clocks that are
at rest wrt each other, which introduces a variation in relative
ticking rates in itself, in addition to that produced by relative
velocity, i.e. we don't get reciprocity as is encountered with systems
of inertial clocks. Does that sound about right, or am I still
missing the key?
Richard Perry
.
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|
| User: "Dirk Van de moortel" |
|
| Title: Re: An interesting SR puzzle |
08 Feb 2005 04:45:39 AM |
|
|
"RP" <no_mail_no_spam@yahoo.com> wrote in message news:36qcnvF54qb5aU1@individual.net...
Dirk Van de moortel wrote:
<no_mail_no_spam@yahoo.com> wrote in message news:1107806934.371652.165750@l41g2000cwc.googlegroups.com...
[...]Two travelers, Al and Bert, have independent accelerated motions
along a straight line. Their motions may be plotted on a spacetime
diagram FROM THE POINT IF VIEW OF A SINGLE INERTIAL FRAME, that
we’ll call the “earth” reference frame[...]
Just as I suspected Dirk. Maybe I'm not communicating the argument
precisely enough :)
The second link is an argument involving linear accelleartion and
accrued displacement over time, which I also addressed. and don't have
an issue with. The third was a dead link.
The third works perfectly for me.
Here's the link with the (bad) beta groups:
http://groups-beta.google.com/group/sci.physics.relativity/msg/22b466002b6ff4ed
In the introduction the second link points to the first link
http://www.geocities.com/slithytove5/AccelClocks.htm
which has a general accelerated frame.
Some inertial frame is used to make the calculations easier.
They could just as well have used Al's comoving
inertial frame at event "a".
The essentail thing is the restult, that describes the
"the rate of an accelerating clock relative to an accelerating observer"
(Tom's words) which amounts to the "motions and ticking rates of
clocks from within the perspective of an accelerated frame" (your
words).
That should satisfy your question:
| Maybe I'm wrong for reasons as yet unknown to me, but OTOH I've not
| seen anyone provide that math, as many times as they've stated that
| "it can be done". Again, we're not talking here about acceleration of
| of clocks and rulers wrt an inertial frame, but rather about motions
| and ticking rates of clocks from within the perspective of an
| accelerated frame."
So you see, it is messy, but "it can be done".
Requiring anything beyond what you get in Tom's article,
would be like requiring that someone derives and describe
the path of a canon ball from the perspective of a jojo in
action on Titan.
Thanks for the links though.
No problem.
Dirk Vdm
I went over the Bert and Al argument again. I missed the section about
the differences in tilts of the simultaneity lines over a small
interval.
I can't imagine what you mean with "missed the section about
the difference...".
I noticed that the html-file does not preoperly show - at least
not in my Browser. Perhaps you should try to read the
MS-Word version.
http://www.geocities.com/slithytove5/AccelClocks9.doc
I can see how this will affect relative ticking rates in a
way not described by the non-generalized transform that I was trying
to apply.
The effects of boost are accounted for in the equations, and these
provide a shift in the time offset between displaced clocks that are
at rest wrt each other, which introduces a variation in relative
ticking rates in itself, in addition to that produced by relative
velocity, i.e. we don't get reciprocity as is encountered with systems
of inertial clocks. Does that sound about right, or am I still
missing the key?
You know where the house is, but yes, apparently you are
missing the key, the lock and the door. What you are writing
here has nothing to do with the article :-)
Dirk Vdm
.
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| User: "RP" |
|
| Title: Re: An interesting SR puzzle |
08 Feb 2005 07:47:34 AM |
|
|
Dirk Van de moortel wrote:
"RP" <no_mail_no_spam@yahoo.com> wrote in message news:36qcnvF54qb5aU1@individual.net...
Dirk Van de moortel wrote:
<no_mail_no_spam@yahoo.com> wrote in message news:1107806934.371652.165750@l41g2000cwc.googlegroups.com...
[...]Two travelers, Al and Bert, have independent accelerated motions
along a straight line. Their motions may be plotted on a spacetime
diagram FROM THE POINT IF VIEW OF A SINGLE INERTIAL FRAME, that
we’ll call the “earth” reference frame[...]
Just as I suspected Dirk. Maybe I'm not communicating the argument
precisely enough :)
The second link is an argument involving linear accelleartion and
accrued displacement over time, which I also addressed. and don't have
an issue with. The third was a dead link.
The third works perfectly for me.
Here's the link with the (bad) beta groups:
http://groups-beta.google.com/group/sci.physics.relativity/msg/22b466002b6ff4ed
In the introduction the second link points to the first link
http://www.geocities.com/slithytove5/AccelClocks.htm
which has a general accelerated frame.
Some inertial frame is used to make the calculations easier.
They could just as well have used Al's comoving
inertial frame at event "a".
The essentail thing is the restult, that describes the
"the rate of an accelerating clock relative to an accelerating observer"
(Tom's words) which amounts to the "motions and ticking rates of
clocks from within the perspective of an accelerated frame" (your
words).
That should satisfy your question:
| Maybe I'm wrong for reasons as yet unknown to me, but OTOH I've not
| seen anyone provide that math, as many times as they've stated that
| "it can be done". Again, we're not talking here about acceleration of
| of clocks and rulers wrt an inertial frame, but rather about motions
| and ticking rates of clocks from within the perspective of an
| accelerated frame."
So you see, it is messy, but "it can be done".
Requiring anything beyond what you get in Tom's article,
would be like requiring that someone derives and describe
the path of a canon ball from the perspective of a jojo in
action on Titan.
Thanks for the links though.
No problem.
Dirk Vdm
I went over the Bert and Al argument again. I missed the section about
the differences in tilts of the simultaneity lines over a small
interval.
I can't imagine what you mean with "missed the section about
the difference...".
Quote: "Note that the line of simultaneity through a' does not have
the same slope as the line of simultaneity through a due to the
acceleration of Al (which causes his velocity relative to the earth
frame to change in going form a to )."
Methinks that maybe it is you who are looking at the wrong link.
The simultaneity line (in the diagram) relates the clocks to each
other in the momentary inertial frame of one. Obviously the faster
that the latter clock moves wrt the former, the greater the tilt of
that line. Now taking into consideration that none of the synchronized
comoving clocks in the frame of the former will read the same wrt the
latter, then the slight boost of the latter will cause a further tilt
of the line and thus there will be a difference in the difference in
clock readings (not a typo) over an infinitesimally interval, a
difference that is produced not by the relative velocity over that
small interval, but by the change in tilt of the line of simultaneity.
The symmetry is broken.
I'm just providing my interpretation of the diagram, don't sweat it if
it doesn't make sense to you :)
Richard Perry
I noticed that the html-file does not preoperly show - at least
not in my Browser. Perhaps you should try to read the
MS-Word version.
http://www.geocities.com/slithytove5/AccelClocks9.doc
I can see how this will affect relative ticking rates in a
way not described by the non-generalized transform that I was trying
to apply.
The effects of boost are accounted for in the equations, and these
provide a shift in the time offset between displaced clocks that are
at rest wrt each other, which introduces a variation in relative
ticking rates in itself, in addition to that produced by relative
velocity, i.e. we don't get reciprocity as is encountered with systems
of inertial clocks. Does that sound about right, or am I still
missing the key?
You know where the house is, but yes, apparently you are
missing the key, the lock and the door. What you are writing
here has nothing to do with the article :-)
Dirk Vdm
.
|
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| User: "Dirk Van de moortel" |
|
| Title: Re: An interesting SR puzzle |
08 Feb 2005 08:34:13 AM |
|
|
"RP" <no_mail_no_spam@yahoo.com> wrote in message news:36ru11F54hb9gU1@individual.net...
Dirk Van de moortel wrote:
"RP" <no_mail_no_spam@yahoo.com> wrote in message news:36qcnvF54qb5aU1@individual.net...
Dirk Van de moortel wrote:
<no_mail_no_spam@yahoo.com> wrote in message news:1107806934.371652.165750@l41g2000cwc.googlegroups.com...
[...]Two travelers, Al and Bert, have independent accelerated motions
along a straight line. Their motions may be plotted on a spacetime
diagram FROM THE POINT IF VIEW OF A SINGLE INERTIAL FRAME, that
we’ll call the “earth” reference frame[...]
Just as I suspected Dirk. Maybe I'm not communicating the argument
precisely enough :)
The second link is an argument involving linear accelleartion and
accrued displacement over time, which I also addressed. and don't have
an issue with. The third was a dead link.
The third works perfectly for me.
Here's the link with the (bad) beta groups:
http://groups-beta.google.com/group/sci.physics.relativity/msg/22b466002b6ff4ed
In the introduction the second link points to the first link
http://www.geocities.com/slithytove5/AccelClocks.htm
which has a general accelerated frame.
Some inertial frame is used to make the calculations easier.
They could just as well have used Al's comoving
inertial frame at event "a".
The essentail thing is the restult, that describes the
"the rate of an accelerating clock relative to an accelerating observer"
(Tom's words) which amounts to the "motions and ticking rates of
clocks from within the perspective of an accelerated frame" (your
words).
That should satisfy your question:
| Maybe I'm wrong for reasons as yet unknown to me, but OTOH I've not
| seen anyone provide that math, as many times as they've stated that
| "it can be done". Again, we're not talking here about acceleration of
| of clocks and rulers wrt an inertial frame, but rather about motions
| and ticking rates of clocks from within the perspective of an
| accelerated frame."
So you see, it is messy, but "it can be done".
Requiring anything beyond what you get in Tom's article,
would be like requiring that someone derives and describe
the path of a canon ball from the perspective of a jojo in
action on Titan.
Thanks for the links though.
No problem.
Dirk Vdm
I went over the Bert and Al argument again. I missed the section about
the differences in tilts of the simultaneity lines over a small
interval.
I can't imagine what you mean with "missed the section about
the difference...".
Quote: "Note that the line of simultaneity through a' does not have
the same slope as the line of simultaneity through a due to the
acceleration of Al (which causes his velocity relative to the earth
frame to change in going form a to )."
Methinks that maybe it is you who are looking at the wrong link.
The simultaneity line (in the diagram) relates the clocks to each
other in the momentary inertial frame of one.
I don't understand what you mean with "inertial frame of one".
And the simultaneity line does not "relate the clocks to each other".
The simultaneity line through a is the collection of all events that
are simultaneous with event a, according to Al.
Period.
Obviously the faster
that the latter clock moves wrt the former, the greater the tilt of
that line.
I have no idea whom you are referring to with your words
"former" and "latter".
The faster Al (and his clock) moves wrt to any inertial frame,
the greater the tilt of the line with respect to that same inertial
frame.
Now taking into consideration that none of the synchronized
comoving clocks in the frame of the former will read the same wrt the
latter,
Still no idea about your "former" or "latter".
No one has mentioned "synchronized comoving clocks" in any frame.
then the slight boost of the latter will cause a further tilt
of the line and thus there will be a difference in the difference in
clock readings (not a typo) over an infinitesimally interval, a
difference that is produced not by the relative velocity over that
small interval, but by the change in tilt of the line of simultaneity.
The symmetry is broken.
I'm just providing my interpretation of the diagram, don't sweat it if
it doesn't make sense to you :)
I have no idea what you are trying to say.
I don't even know whether you have a question or an objection
or something in between.
Dirk Vdm
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| User: "Jesse Mazer" |
|
| Title: Re: An interesting SR puzzle |
07 Feb 2005 04:26:54 PM |
|
|
RP wrote:
Jesse Mazer wrote:
RP wrote:
glbrad01 wrote:
"Paul B. Andersen" <paul.b.andersen@deletethishia.no> wrote in
message news:cu62db$oiq$1@dolly.uninett.no...
Eli Botkin wrote:
"Ben Rudiak-Gould" <br276deleteme@cam.ac.uk> wrote in message
news:cu0rp6$r7a$1@gemini.csx.cam.ac.uk...
Here's a two-part SR puzzle:
1. I'm standing still (i.e. moving inertially) while whirling a
clock around my head at a constant speed. (If you're worried
about conservation of momentum, imagine that I'm whirling two
clocks: it doesn't make any difference). I'm wearing a
wristwatch. Which of these--the clock or the wristwatch--will
appear to me to tick faster, and by how much?
2. Now I'm a different person, sitting on the clock as it whirls
around, looking at the clock and at the wristwatch of the
person doing the whirling. Which (if either) will appear to
me to tick faster, and by how much?
I already know what the answer to this puzzle is, and why. But I
find it very interesting from a pedagogical perspective. Despite
doing well in my undergrad SR course, and despite the puzzle's
simplicity (no calculation required), I doubt I could have
solved part 2 had it been on an exam. I would have become
hopelessly mired in trying to figure out the reference frame of
the person sitting on the clock, because SR, as it was taught to
me, was about relating the reference frames of different
observers. I didn't understand at the time that the coordinate
system you choose to solve a problem needn't be the rest frame
of the measurement device, and that there isn't even a
well-defined notion of /the/ rest frame of an object in general.
I conjecture that this kind of confusion is quite common.
This puzzle seems to highlight this point more clearly than any
I've seen before. More generally, it's interesting as an example
of a problem involving relative motion for which the Lorentz
transformation is really no help at all. And its solution might
help to dispel the common misconception that SR can't deal with
acceleration.
One can also add parts 1b and 2b which ask whether the clock and
wristwatch are redshifted or blueshifted, and by what factor.
Not until years later did I understand the relationship between
Doppler shift and the apparent rate of clocks.
In short, I like this puzzle a lot, and I wish it were taught
alongside the twin paradox (or instead of it!) in undergraduate
courses and textbooks. I'm curious to know if anyone here has
taught it, or thought about doing so, or decided against it, or
whatever.
-- Ben
Hi Ben:
I haven't heard this one before. My response is:
Each person will claim that the other clock is running slower
than his own.
Also, if they are viewing each others clocks, the images will
appear to advance in synchrony with their personal clocks (or
wristwatchs).
If you don't agree, I would very much like your take on the
correct solution.
Eli
The "whirling clock" is accelerated.
That makes all the difference.
The correct answer is they will both agree
that the wrist watch runs faster.
Paul
That is not the correct answer.
Brad
I agree.
The correct answer is that in this system GR must be applied.
No, SR can deal with the question of the proper time recorded by an
accelerating clock by considering its path as seen in an inertial
frame. Just integrate [squareroot(1 - v(t)^2/c^2) dt] between two
times t1 and t2 in your frame, with v(t) being the clock's
instantaneous velocity at time t in your frame; this will give the
correct answer for the time elapsed on the accelerating clock during
the interval (t1, t2) in your frame (you also have to take into
account the finite speed of light to figure out how much time you'd
actually *see* elapse on the accelerating clock during that
time-interval, but in this problem the answer would be the same).
I as much said the same. There is a transverse doppler shift of the
"ticks" of the whirling clock wrt the wristwatch, which is nothing
more than the ticking rate offset being observed.
There's no doppler shift in this problem, the rate that I see the
whirling clock tick using light signals is the same as the rate it
actually *is* ticking in my coordinate system. In both cases, it's
slowed down by a factor of squareroot(1 - v^2/c^2) relative to my clock
This is calculated in normal SR fashion. Nothing fancy needed, just
plug in the instantaneous velocity of the whirling clock. But how do
you account for the non-reciprocity of the effect? After all, wrt the
whirling clock your clock should be ticking slow, i.e. he should
likewise observer a transverse doppler red shift.
No he shouldn't. If you actually do the calculation according to the
rules of SR, you will find that the whirling clock sees my clock ticking
faster than his own.
This is directly where the paradox has its substance. Unlike in the
linear acceleration arguments such as the twins paradox, there is no
need for the other twin to return and thus incur a reduced ticking
rate on his trip to the other. No, in this case there is never a
change in scalar displacement and ticking rate offsets will accrue
over time to large differences. The reunion never involves an excess
distance to cover as in the twins paradox, thus no opportunity to
balance the books.
Wrt the accelerated clock, the other ticks fast, and SR cannot provide
this result. As noted, the lorentz transform requires the other clock
to tick slower.
The Lorentz transform only applies to two inertial frames, you can't use
it if you want to figure out the coordinates of events in accelerating
frames (but we don't need to do that here anyway, we just need to figure
out how much time has elapsed on the accelerating clock between the
events of it receiving signals from my clock at two times t1 and t2.
Then imagine that your own clock is sending signals to the
accelerating clock at the speed of light, and figure out what
time-signal the accelerating clock will be receiving at time t1 and
what time-signal it will be receiving at time t2; this tells you what
amount of time an observer sitting on the accelerating clock would
see elapsed on your clock during the same interval.
Yes, same as a | | | | | | | |
