In Special Relativity Theory the basis of "Twin Paradox" is that "moving clocks run slow". What is not clear is that, when a 'clock runs slow' whether its frequency continuously decreases or not. Can anyone clarify this point?
In sci.physics V.Gopal <vgopa...@rediffmail.com> wrote:
> In Special Relativity Theory the basis of "Twin Paradox" is that > "moving clocks run slow". What is not clear is that, when a 'clock > runs slow' whether its frequency continuously decreases or not.
On the contrary, that is perfectly clear. The "speed" of the clock is directly related to its velocity. If the velocity is constant then so is the clockrate.
> In Special Relativity Theory the basis of "Twin Paradox" is that > "moving clocks run slow". What is not clear is that, when a 'clock > runs slow' whether its frequency continuously decreases or not. Can > anyone clarify this point?
Nothing runs "slow" in any refernce frame. When otherwise identical clocks - one of which sustained a velocity relative to the other - are brought together to compare, one is seen to have less elapsed time than its twin due to its hyperbolic rotation through 4-space. The skewing factor is sqrt[1-(v^2/c^2)]. No clock need be accelerated for the Twin Paradox to occur.
Given any achievable velocities V1 and V2 and any finite lightspeed, the sum of the velocities as viewed by any inertial observer cannot exceed
(V1 + V2)/[1 + (V1)(V2)/c^2]
This is transformation of velocities parallel to the direction of motion. For velocities at an arbitrary angle theta,
Makes no difference who does the looking. Any inertial observer has a valid reference frame. Disparities are only noted upon comparison of reference frames. Spacetime is four-dimensional; travel through spacetime is a hyperbolic rotation of all four coordinates. However, the units of time are seconds not meters. ct is a nice length, but the this fundamental unit of length is a very long one.
At nominal velocities you don't travel much along ct compared to x,y,x. Galilean transforms and Newtonian physics are good enough approximations. At relativistic speeds you propagate along ct comparable to a material dimension. You need Lorentzian transforms and relativity to explain initially synchronized clocks separated and then brought together for comparison thereafter.
Acceleration of one observer has nothing to do with it, BTW.
1) Acceleration is an absolute measurement. There is no doubt who is accelerated. 2) Acceleration is irrelevant.
We have three identical clocks that are off (a state of not running) and zeroed. Each clock has a very short toggle jiggger switch sticking out. We load them in individual spaceships and set up the experiment.
CLOCK 1: That's our clock. It sits stationary in our inertial reference frame with a little jigger sticking out. Touch the jigger and the "off" state becomes "on" or the "on" state becomes "off." Clock 1 is "off."
CLOCK 2: In a spaceship traveling at 0.999c relative to our inertial frame of reference. Clock 2 is "off." It skims past Clock 1, the jiggers touch, both Clocks 1 and 2 are now "on" and synchronized.
CLOCK 3: In a spaceship traveling at 0.999c relative to our inertial frame of reference, but 180 degrees counter in direction to Clock 2. Clock 3 is "off." Some arbitrary time after Clocks 1 and 2 synchronize and turn "on" by touching, Clocks 2 and 3 brush past each other, touching jiggers. Clock 2 is now "off," Clock 3 is now "on."
CLOCK 1: That's our clock. It sits stationary in our inertial reference frame with a little jigger sticking out. Clock 3 rushes past, jiggers touch. Clocks 3 and 1 are now off. All clocks are off. No clock has accelerated while "on."
BOTTOM LINE: Get all three clocks together and compare elapsed times. Elapsed time #2=#3, but elapsed time #2+#3 does not equal #1, the local stationary reference frame summation. The sum of #2+#3 elasped time is only about 4.5% that than of #1's accumulated elapsed time. You now have the Twin Paradox without any running clock having been accelerated.
-- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) "Quis custodiet ipsos custodes?" The Net!
>>In Special Relativity Theory the basis of "Twin Paradox" is that >>"moving clocks run slow". What is not clear is that, when a 'clock >>runs slow' whether its frequency continuously decreases or not. Can >>anyone clarify this point?
> Nothing runs "slow" in any refernce frame. When otherwise identical > clocks - one of which sustained a velocity relative to the other - are > brought together to compare, one is seen to have less elapsed time > than its twin due to its hyperbolic rotation through 4-space. The > skewing factor is sqrt[1-(v^2/c^2)]. > No clock need be accelerated for the Twin Paradox to occur.
> Given any achievable velocities V1 and V2 and any finite lightspeed, > the sum of the velocities as viewed by any inertial observer cannot > exceed
> (V1 + V2)/[1 + (V1)(V2)/c^2]
> This is transformation of velocities parallel to the direction of > motion. For velocities at an arbitrary angle theta,
> Makes no difference who does the looking. Any inertial observer has a > valid reference frame. Disparities are only noted upon comparison of > reference frames. Spacetime is four-dimensional; travel through > spacetime is a hyperbolic rotation of all four coordinates. However, > the units of time are seconds not meters. ct is a nice length, but > the this fundamental unit of length is a very long one.
> At nominal velocities you don't travel much along ct compared to > x,y,x. Galilean transforms and Newtonian physics are good enough > approximations. At relativistic speeds you propagate along ct > comparable to a material dimension. You need Lorentzian transforms > and relativity to explain initially synchronized clocks separated and > then brought together for comparison thereafter.
> Acceleration of one observer has nothing to do with it, BTW.
> 1) Acceleration is an absolute measurement. There is no doubt who > is accelerated. > 2) Acceleration is irrelevant.
> We have three identical clocks that are off (a state of not running) > and zeroed. Each clock has a very short toggle jiggger switch
An infinitesimal length, if there is to be no acceleration of any part of the clocks, unless acceleration of parts of the clocks does not matter. Since, to move a jigger, one must accelerate it and possibly the whole clock, does that not count as an acceleration of a clock? Larry
> sticking out. We load them in individual spaceships and set up the > experiment.
> CLOCK 1: That's our clock. It sits stationary in our inertial > reference frame with a little jigger sticking out. Touch the jigger > and the "off" state becomes "on" or the "on" state becomes "off." > Clock 1 is "off."
> CLOCK 2: In a spaceship traveling at 0.999c relative to our > inertial frame of reference. Clock 2 is "off." It skims past Clock > 1, the jiggers touch, both Clocks 1 and 2 are now "on" and > synchronized.
> CLOCK 3: In a spaceship traveling at 0.999c relative to our > inertial frame of reference, but 180 degrees counter in direction to > Clock 2. Clock 3 is "off." Some arbitrary time after Clocks 1 and 2 > synchronize and turn "on" by touching, Clocks 2 and 3 brush past each > other, touching jiggers. Clock 2 is now "off," Clock 3 is now "on."
> CLOCK 1: That's our clock. It sits stationary in our inertial > reference frame with a little jigger sticking out. Clock 3 rushes > past, jiggers touch. Clocks 3 and 1 are now off. All clocks are > off. No clock has accelerated while "on."
> BOTTOM LINE: Get all three clocks together and compare elapsed > times. Elapsed time #2=#3, but elapsed time #2+#3 does not equal #1, > the local stationary reference frame summation. The sum of #2+#3 > elasped time is only about 4.5% that than of #1's accumulated elapsed > time. You now have the Twin Paradox without any running clock having > been accelerated.
> >>In Special Relativity Theory the basis of "Twin Paradox" is that > >>"moving clocks run slow". What is not clear is that, when a 'clock > >>runs slow' whether its frequency continuously decreases or not. Can > >>anyone clarify this point?
> > Nothing runs "slow" in any refernce frame. When otherwise identical > > clocks - one of which sustained a velocity relative to the other - are > > brought together to compare, one is seen to have less elapsed time > > than its twin due to its hyperbolic rotation through 4-space. The > > skewing factor is sqrt[1-(v^2/c^2)]. > > No clock need be accelerated for the Twin Paradox to occur.
> > Given any achievable velocities V1 and V2 and any finite lightspeed, > > the sum of the velocities as viewed by any inertial observer cannot > > exceed
> > (V1 + V2)/[1 + (V1)(V2)/c^2]
> > This is transformation of velocities parallel to the direction of > > motion. For velocities at an arbitrary angle theta,
> > Makes no difference who does the looking. Any inertial observer has a > > valid reference frame. Disparities are only noted upon comparison of > > reference frames. Spacetime is four-dimensional; travel through > > spacetime is a hyperbolic rotation of all four coordinates. However, > > the units of time are seconds not meters. ct is a nice length, but > > the this fundamental unit of length is a very long one.
> > At nominal velocities you don't travel much along ct compared to > > x,y,x. Galilean transforms and Newtonian physics are good enough > > approximations. At relativistic speeds you propagate along ct > > comparable to a material dimension. You need Lorentzian transforms > > and relativity to explain initially synchronized clocks separated and > > then brought together for comparison thereafter.
> > Acceleration of one observer has nothing to do with it, BTW.
> > 1) Acceleration is an absolute measurement. There is no doubt who > > is accelerated. > > 2) Acceleration is irrelevant.
> > We have three identical clocks that are off (a state of not running) > > and zeroed. Each clock has a very short toggle jiggger switch
> An infinitesimal length, if there is to be no acceleration of any part > of the clocks, unless acceleration of parts of the clocks does not > matter. Since, to move a jigger, one must accelerate it and possibly > the whole clock, does that not count as an acceleration of a clock? > Larry
[snip]
You are so are so ignorantly hopelessly fill of shit. You have no idea what is going on. Replace the jiggers with complimentary fiberoptic emitters and pinhole photodiode detectors. Have the ships skim within 1 millimeter of each other to transfer synchronization. Run your spew to bull bore. Show how:
"Elapsed time #2=#3, but elapsed time #2+#3 does not equal #1, the local stationary reference frame summation. The sum of #2+#3 elasped time is only about 4.5% that than of #1's accumulated elapsed time"
does not obtain. Now assume the experiment is 3 years elapsed time in the stationary frame and show it again. Show us how three teeny dinks make up for over 1045 days missing between the sum of the moving clocks' elapsed time and the stationary clock's elasped time. Now do it for 10 years, 100 years. How good is your spew now?
-- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) "Quis custodiet ipsos custodes?" The Net!
This begs the question of how you are supposed to be able to measure what a clock says "now" when it is somewhere else.
: When otherwise identical : clocks - one of which sustained a velocity relative to the other - are : brought together to compare, one is seen to have less elapsed time : than its twin due to its hyperbolic rotation through 4-space. The : skewing factor is sqrt[1-(v^2/c^2)].
You can't assert this symmetrically. It cannot be the case BOTH that clock a has less time elapsed than clock b AND that clock b has less time elapsed than clock a.
: No clock need be accelerated for the Twin Paradox to occur.
Please. The clocks have to be acclerated just to get back to the same place, in order for the comparison to occur.
: Makes no difference who does the looking. Any inertial observer has a : valid reference frame. Disparities are only noted upon comparison of : reference frames.
What does it even mean to compare two whole frames, as opposed to just two clocks?
: Acceleration of one observer has nothing to do with it, BTW. : : 1) Acceleration is an absolute measurement. There is no doubt who : is accelerated. : 2) Acceleration is irrelevant.
It's not irrelevant to your thought-experiment below.
: We have three identical clocks that are off (a state of not running) : and zeroed. Each clock has a very short toggle jiggger switch : sticking out. We load them in individual spaceships and set up the : experiment. : : CLOCK 1: That's our clock. It sits stationary in our inertial : reference frame with a little jigger sticking out. Touch the jigger : and the "off" state becomes "on" or the "on" state becomes "off." : Clock 1 is "off." : : CLOCK 2: In a spaceship traveling at 0.999c relative to our : inertial frame of reference. Clock 2 is "off." It skims past Clock : 1, the jiggers touch, both Clocks 1 and 2 are now "on" and : synchronized. : : CLOCK 3: In a spaceship traveling at 0.999c relative to our : inertial frame of reference, but 180 degrees counter in direction to : Clock 2. Clock 3 is "off."
No acceleration so far.
: Some arbitrary time after Clocks 1 and 2 synchronize and turn : "on" by touching, Clocks 2 and 3 brush past each other, : touching jiggers. Clock 2 is now "off," Clock 3 is now "on."
: CLOCK 1: That's our clock. It sits stationary in our inertial : reference frame with a little jigger sticking out. Clock 3 rushes : past, jiggers touch. Clocks 3 and 1 are now off. All clocks are : off. No clock has accelerated while "on."
NOR while off, EITHER, yet, in this scenario.
: BOTTOM LINE: Get all three clocks together
But THAT will REQUIRE acceleration. If their velocities don't change then the clocks will NEVER "get together".
: and compare elapsed times. Elapsed time #2=#3, but elapsed time : #2+#3 does not equal #1, the local stationary reference frame : summation. The sum of #2+#3 elasped time is only about 4.5% : that than of #1's accumulated elapsed time. You now have the : Twin Paradox without any running clock having been accelerated.
But you also have the presumption that you know what a clock reads EVEN though it is not HERE for you to be able to actually OBSERVE, now. Getting it here would require acceleration, EVEN if it is off. Somehow I think the fact that acceleration must still occur must still be relevant, EVEN if the clock is off.
-- --- "It's difficult ... you need to be united to have any strength, but internal issues have to be addressed." --- E. Ray Lewis, on liberalism in America
> Uncle Al <Uncle...@hate.spam.net> writes: > : Nothing runs "slow" in any reference frame.
> This begs the question of how you are supposed to > be able to measure what a clock says "now" when it > is somewhere else.
Another ignorant comment by a sci.math-er.
There are truly classical prescriptions developed by Einstein to enable synchronization of clock times between co-moving frames. The one fundamentally important fact is that the speed of light is a constant in any frame and that exceeds any speed attainable by material objects such as clocks and observers.
> Uncle Al <Uncle...@hate.spam.net> writes: > : Nothing runs "slow" in any reference frame.
> This begs the question of how you are supposed to > be able to measure what a clock says "now" when it > is somewhere else.
Did you read it? Did you really? You have the clock off. You have the clock on. You have the clock off. When the clock is off the accumulated time doesn't change. Take the battery out of your analog watch. The hands don't move. Put the watch in a deep gravitational well, send it out to Alpha Centuri and back, put it in a Egyptian pyramid for 5000 years. Has the indicated time changed? Write a number on a piece of paper. Put the inscribed paper in a Beckman Ultra-Max centrifuge at ONE MILLION GEES for year. Retrieve the paper. HAS THE NUMBER CHANGED?
There is nothing so perfect that monstrous stupidty cannot ruin it. Do you have any mind at all? Do you really? "Thought experiment" is meaningless if you do not think.
> : When otherwise identical > : clocks - one of which sustained a velocity relative to the other - are > : brought together to compare, one is seen to have less elapsed time > : than its twin due to its hyperbolic rotation through 4-space. The > : skewing factor is sqrt[1-(v^2/c^2)].
> You can't assert this symmetrically. > It cannot be the case BOTH that clock a has less time elapsed > than clock b AND that clock b has less time elapsed than clock a.
It's called "The Twin Paradox." This is an especially fine rendering of it. THAT IS WHY IT IS A PARADOX. Acceleration has been removed and velocity is relative, yet the clocks do not agree. Go ahead, resolve it.
> : No clock need be accelerated for the Twin Paradox to occur.
> Please. The clocks have to be acclerated just to get back > to the same place, in order for the comparison to occur.
You stupid bastard. The crew WRITES DOWN the elapsed time. You may now burn the clock or go at it with sledgehammers for all I care. The figures for elapsed time were never accumulated including an acceleration.
> : Makes no difference who does the looking. Any inertial observer has a > : valid reference frame. Disparities are only noted upon comparison of > : reference frames.
> What does it even mean to compare two whole frames, as opposed > to just two clocks?
You stupid bastard. Words fail me. THE CLOCKS ARE THE FRAMES.
> : Acceleration of one observer has nothing to do with it, BTW. > : > : 1) Acceleration is an absolute measurement. There is no doubt who > : is accelerated. > : 2) Acceleration is irrelevant.
> It's not irrelevant to your thought-experiment below.
You stupid bastard. Read the foregoing. I'm not going to repost it.
> : We have three identical clocks that are off (a state of not running) > : and zeroed. Each clock has a very short toggle jiggger switch > : sticking out. We load them in individual spaceships and set up the > : experiment. > : > : CLOCK 1: That's our clock. It sits stationary in our inertial > : reference frame with a little jigger sticking out. Touch the jigger > : and the "off" state becomes "on" or the "on" state becomes "off." > : Clock 1 is "off." > : > : CLOCK 2: In a spaceship traveling at 0.999c relative to our > : inertial frame of reference. Clock 2 is "off." It skims past Clock > : 1, the jiggers touch, both Clocks 1 and 2 are now "on" and > : synchronized. > : > : CLOCK 3: In a spaceship traveling at 0.999c relative to our > : inertial frame of reference, but 180 degrees counter in direction to > : Clock 2. Clock 3 is "off."
> No acceleration so far.
No clock ever accelerates while it is on. Acceleration while it is off is irrelevant. The crew brings aboard parts, accelerates to 0.999c, then builds the clock from scratch, zeroes it, and installs the jigger. Jigger, jigger. The crew writes down the elapsed time then throws the clock overboard. The piece of paper bearing the number is then decelerated for comparison with the other two elapsed times. How can you pee without wetting yoru pants?
The crew brings aboard ore, a smelter, and a machine shop. It accelerates to 0.999c, then acceleration goes to zero. The crew then refines the ore, machines the metal, and builds the clock. Jigger jigger. The elapsed time number is etched an inch deep into a big block of Corning Ultra Low Expansion Glass. The clock is then blown out the plasma warp engines. The inscribed glass slab is decelerated and compard with the other two inscribed glass slabs.
The crew brings aboard a fission reactor and sodium chloride. It accelerates to 0.999c, then acceleration goes to zero. The NaCl is put into the reactor to neutron activate to radioactive Na-24 (15 hr), Cl-36 (3x10^5 yr), and Cl-38 (37.2 min) for a triple calibration. The clock is counting the clicks and noting the decrease in intensity for each of the three channels in the interval between the jigger touches. The number is written down. When they get back they shove the radiologically hot salt up your ass.
> : Some arbitrary time after Clocks 1 and 2 synchronize and turn > : "on" by touching, Clocks 2 and 3 brush past each other, > : touching jiggers. Clock 2 is now "off," Clock 3 is now "on."
> : CLOCK 1: That's our clock. It sits stationary in our inertial > : reference frame with a little jigger sticking out. Clock 3 rushes > : past, jiggers touch. Clocks 3 and 1 are now off. All clocks are > : off. No clock has accelerated while "on."
> NOR while off, EITHER, yet, in this scenario.
> : BOTTOM LINE: Get all three clocks together
> But THAT will REQUIRE acceleration. If their velocities > don't change then the clocks will NEVER "get together".
THE CLOCKS ARE OFF, you jackass. THE ELAPSED TIME NUMBERS WERE WRITTEN DOWN AND THE CLOCKS WERE DESTROYED.
> : and compare elapsed times. Elapsed time #2=#3, but elapsed time > : #2+#3 does not equal #1, the local stationary reference frame > : summation. The sum of #2+#3 elasped time is only about 4.5% > : that than of #1's accumulated elapsed time. You now have the > : Twin Paradox without any running clock having been accelerated.
> But you also have the presumption that you know what a clock > reads EVEN though it is not HERE for you to be able to actually > OBSERVE, now. Getting it here would require acceleration, EVEN > if it is off. Somehow I think the fact that acceleration must > still occur must still be relevant, EVEN if the clock is off.
You stupid bastard. THE ELAPSED TIME NUMBERS WERE WRITTEN DOWN. Will the written numbers change no matter how you spew and fart?
-- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) "Quis custodiet ipsos custodes?" The Net!
> Uncle Al <Uncle...@hate.spam.net> writes: > : Nothing runs "slow" in any reference frame.
> This begs the question of how you are supposed to > be able to measure what a clock says "now" when it > is somewhere else.
> : When otherwise identical > : clocks - one of which sustained a velocity relative to the other - are > : brought together to compare, one is seen to have less elapsed time > : than its twin due to its hyperbolic rotation through 4-space. The > : skewing factor is sqrt[1-(v^2/c^2)].
> You can't assert this symmetrically. > It cannot be the case BOTH that clock a has less time elapsed > than clock b AND that clock b has less time elapsed than clock a.
The situation is not symmetrical. Draw a world line for a stationary clock. Draw a world line for a moving clock which starts out and ends up where the stationary clock is.
> : No clock need be accelerated for the Twin Paradox to occur.
> Please. The clocks have to be acclerated just to get back > to the same place, in order for the comparison to occur.
No. Use your loaf. This has been explained countless times in the ng.
> : Makes no difference who does the looking. Any inertial observer has a > : valid reference frame. Disparities are only noted upon comparison of > : reference frames.
> What does it even mean to compare two whole frames, as opposed > to just two clocks?
Oh dear.
> : Acceleration of one observer has nothing to do with it, BTW. > : > : 1) Acceleration is an absolute measurement. There is no doubt who > : is accelerated. > : 2) Acceleration is irrelevant.
> It's not irrelevant to your thought-experiment below.
It is irrelevant. You have not shown anywhere below where the acceleration of anything affected anything.
> : We have three identical clocks that are off (a state of not running) > : and zeroed. Each clock has a very short toggle jiggger switch > : sticking out. We load them in individual spaceships and set up the > : experiment. > : > : CLOCK 1: That's our clock. It sits stationary in our inertial > : reference frame with a little jigger sticking out. Touch the jigger > : and the "off" state becomes "on" or the "on" state becomes "off." > : Clock 1 is "off." > : > : CLOCK 2: In a spaceship traveling at 0.999c relative to our > : inertial frame of reference. Clock 2 is "off." It skims past Clock > : 1, the jiggers touch, both Clocks 1 and 2 are now "on" and > : synchronized. > : > : CLOCK 3: In a spaceship traveling at 0.999c relative to our > : inertial frame of reference, but 180 degrees counter in direction to > : Clock 2. Clock 3 is "off."
> No acceleration so far.
> : Some arbitrary time after Clocks 1 and 2 synchronize and turn > : "on" by touching, Clocks 2 and 3 brush past each other, > : touching jiggers. Clock 2 is now "off," Clock 3 is now "on."
> : CLOCK 1: That's our clock. It sits stationary in our inertial > : reference frame with a little jigger sticking out. Clock 3 rushes > : past, jiggers touch. Clocks 3 and 1 are now off. All clocks are > : off. No clock has accelerated while "on."
> NOR while off, EITHER, yet, in this scenario.
> : BOTTOM LINE: Get all three clocks together
> But THAT will REQUIRE acceleration. If their velocities > don't change then the clocks will NEVER "get together".
> : and compare elapsed times. Elapsed time #2=#3, but elapsed time > : #2+#3 does not equal #1, the local stationary reference frame > : summation. The sum of #2+#3 elasped time is only about 4.5% > : that than of #1's accumulated elapsed time. You now have the > : Twin Paradox without any running clock having been accelerated.
> But you also have the presumption that you know what a clock > reads EVEN though it is not HERE for you to be able to actually > OBSERVE, now. Getting it here would require acceleration, EVEN > if it is off. Somehow I think the fact that acceleration must > still occur must still be relevant, EVEN if the clock is off.
> > In Special Relativity Theory the basis of "Twin Paradox" is that > > "moving clocks run slow". What is not clear is that, when a 'clock > > runs slow' whether its frequency continuously decreases or not. Can > > anyone clarify this point?
> Nothing runs "slow" in any refernce frame. When otherwise identical > clocks - one of which sustained a velocity relative to the other - are > brought together to compare, one is seen to have less elapsed time > than its twin due to its hyperbolic rotation through 4-space. The > skewing factor is sqrt[1-(v^2/c^2)]. > No clock need be accelerated for the Twin Paradox to occur.
> Given any achievable velocities V1 and V2 and any finite lightspeed, > the sum of the velocities as viewed by any inertial observer cannot > exceed
> (V1 + V2)/[1 + (V1)(V2)/c^2]
> This is transformation of velocities parallel to the direction of > motion. For velocities at an arbitrary angle theta,
> Makes no difference who does the looking. Any inertial observer has a > valid reference frame. Disparities are only noted upon comparison of > reference frames. Spacetime is four-dimensional; travel through > spacetime is a hyperbolic rotation of all four coordinates. However, > the units of time are seconds not meters. ct is a nice length, but > the this fundamental unit of length is a very long one.
> At nominal velocities you don't travel much along ct compared to > x,y,x. Galilean transforms and Newtonian physics are good enough > approximations. At relativistic speeds you propagate along ct > comparable to a material dimension. You need Lorentzian transforms > and relativity to explain initially synchronized clocks separated and > then brought together for comparison thereafter.
> Acceleration of one observer has nothing to do with it, BTW.
> 1) Acceleration is an absolute measurement. There is no doubt who > is accelerated. > 2) Acceleration is irrelevant.
> We have three identical clocks that are off (a state of not running) > and zeroed. Each clock has a very short toggle jiggger switch > sticking out. We load them in individual spaceships and set up the > experiment.
> CLOCK 1: That's our clock. It sits stationary in our inertial > reference frame with a little jigger sticking out. Touch the jigger > and the "off" state becomes "on" or the "on" state becomes "off." > Clock 1 is "off."
> CLOCK 2: In a spaceship traveling at 0.999c relative to our > inertial frame of reference. Clock 2 is "off." It skims past Clock > 1, the jiggers touch, both Clocks 1 and 2 are now "on" and > synchronized.
> CLOCK 3: In a spaceship traveling at 0.999c relative to our > inertial frame of reference, but 180 degrees counter in direction to > Clock 2. Clock 3 is "off." Some arbitrary time after Clocks 1 and 2 > synchronize and turn "on" by touching, Clocks 2 and 3 brush past each > other, touching jiggers. Clock 2 is now "off," Clock 3 is now "on."
> CLOCK 1: That's our clock. It sits stationary in our inertial > reference frame with a little jigger sticking out. Clock 3 rushes > past, jiggers touch. Clocks 3 and 1 are now off. All clocks are > off. No clock has accelerated while "on."
> BOTTOM LINE: Get all three clocks together and compare elapsed > times. Elapsed time #2=#3, but elapsed time #2+#3 does not equal #1, > the local stationary reference frame summation. The sum of #2+#3 > elasped time is only about 4.5% that than of #1's accumulated elapsed > time. You now have the Twin Paradox without any running clock having > been accelerated.
When you can define time, then maybe your input will be interesting. Clocks do not measure the passage of time, they 'register' an internal velocity (of some mass or system of mass) times it's displacement, either linear or angular depending upon the type of clock, an internal velocity that must change in order for the registered 'ticks' to get out of sync with another clock. A change in velocity does not a change in time equal. Now suppose I adopt the motion of an elastic ball between to walls as my time-piece: 1) How will this clock differ fundamentally from any other clock? 2) Since when will an acceleration of the elastic ball constitute a change in the rate that I am passing through time, rather than constituting simply an increase in its intrinsic frequency wrt time?
> When you can define time, then maybe your input will be interesting. > Clocks do not measure the passage of time, they 'register' an internal > velocity (of some mass or system of mass) times it's displacement, > either linear or angular depending upon the type of clock, an internal > velocity that must change in order for the registered 'ticks' to get out > of sync with another clock.
Radioactivity, shithead. You look at the declining rate of decay and that is your clock. Why don't you tell us where the "internal velocity" is in radioactivity?
[snip of more uninspired idiocy]
-- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) "Quis custodiet ipsos custodes?" The Net!
George Greene <gree...@eagle.cs.unc.edu> wrote in message <news:xeslm4niu65.fsf@eagle.cs.unc.edu>... > Uncle Al <Uncle...@hate.spam.net> writes: > : Nothing runs "slow" in any reference frame.
> This begs the question of how you are supposed to > be able to measure what a clock says "now" when it > is somewhere else.
> : When otherwise identical > : clocks - one of which sustained a velocity relative to the other - are > : brought together to compare, one is seen to have less elapsed time > : than its twin due to its hyperbolic rotation through 4-space. The > : skewing factor is sqrt[1-(v^2/c^2)].
> You can't assert this symmetrically. > It cannot be the case BOTH that clock a has less time elapsed > than clock b AND that clock b has less time elapsed than clock a.
> : No clock need be accelerated for the Twin Paradox to occur.
> Please. The clocks have to be acclerated just to get back > to the same place, in order for the comparison to occur.
> : Makes no difference who does the looking. Any inertial observer has a > : valid reference frame. Disparities are only noted upon comparison of > : reference frames.
> What does it even mean to compare two whole frames, as opposed > to just two clocks?
> : Acceleration of one observer has nothing to do with it, BTW. > : > : 1) Acceleration is an absolute measurement. There is no doubt who > : is accelerated. > : 2) Acceleration is irrelevant.
> It's not irrelevant to your thought-experiment below.
> : We have three identical clocks that are off (a state of not running) > : and zeroed. Each clock has a very short toggle jiggger switch > : sticking out. We load them in individual spaceships and set up the > : experiment. > : > : CLOCK 1: That's our clock. It sits stationary in our inertial > : reference frame with a little jigger sticking out. Touch the jigger > : and the "off" state becomes "on" or the "on" state becomes "off." > : Clock 1 is "off." > : > : CLOCK 2: In a spaceship traveling at 0.999c relative to our > : inertial frame of reference. Clock 2 is "off." It skims past Clock > : 1, the jiggers touch, both Clocks 1 and 2 are now "on" and > : synchronized. > : > : CLOCK 3: In a spaceship traveling at 0.999c relative to our > : inertial frame of reference, but 180 degrees counter in direction to > : Clock 2. Clock 3 is "off."
> No acceleration so far.
> : Some arbitrary time after Clocks 1 and 2 synchronize and turn > : "on" by touching, Clocks 2 and 3 brush past each other, > : touching jiggers. Clock 2 is now "off," Clock 3 is now "on."
> : CLOCK 1: That's our clock. It sits stationary in our inertial > : reference frame with a little jigger sticking out. Clock 3 rushes > : past, jiggers touch. Clocks 3 and 1 are now off. All clocks are > : off. No clock has accelerated while "on."
> NOR while off, EITHER, yet, in this scenario.
> : BOTTOM LINE: Get all three clocks together
> But THAT will REQUIRE acceleration. If their velocities > don't change then the clocks will NEVER "get together".
> : and compare elapsed times. Elapsed time #2=#3, but elapsed time > : #2+#3 does not equal #1, the local stationary reference frame > : summation. The sum of #2+#3 elasped time is only about 4.5% > : that than of #1's accumulated elapsed time. You now have the > : Twin Paradox without any running clock having been accelerated.
> But you also have the presumption that you know what a clock > reads EVEN though it is not HERE for you to be able to actually > OBSERVE, now. Getting it here would require acceleration, EVEN > if it is off. Somehow I think the fact that acceleration must > still occur must still be relevant, EVEN if the clock is off.
All calculations in Special Theory of Relativity are based on the law of conservation of momentum and kinetic energy. Acceleration is questionable in Special Relativity Theory because force and accelerational, in general, are not codirectional in special relativity. Moreover, in general action and reaction are not equal and opposite. The distinction between force and power as well as the distinction between momentum and kinetic energy are blurred. Special relativity begins by assuming conventional mechanics as its premise (the starting point) and contradicts its own premise. How can a conclusion contradict its own premise? A clock cannot be said to run slow unless it has angular deceleration. Therefore everybody seems to agree that clocks 'become' slow - they decelerate during linear acceleration of the point mass. It is obvious that if speed of light is the maximum speed attainable then linear acceleration must become zero when speed of light is reached. If we want to make linear acceleration of the mass zero then in, acceleration=(V2-V1)/T, the clock that shows T in the denominator must stop or take infitite time to sweep through an arc equal to the arc that shows one unit of time in a clock at rest. The real problem is how to make the value of linear acceleration zero when a mass attains the speed of light. Neither Newton nor Einstein nor any of us can find the right answer to this problem.
In order for the jiggers of the clocks of spaceships 1 and 2 to click each other on, the two spaceships have to have a certain orientation relative to each other. We'll say that spaceship 2 zooms by on the right of spaceship 1, so spaceship 2's clock jigger is sticking out on the left.
Now in order for the jigger of the clock of spaceship 3 to click off the jigger of spaceship 2, spaceship 3 has to occupy the same position relative to spaceship 2 that spaceship 1 did. In order to hit the clock jigger of spaceship 2, which sticks out on the left, spaceship 3 has to pass spaceship 2 on the left of spaceship 2.
Now in order to follow the same trajectory that spaceship 2 followed, only backwards and translated by one spaceship-width, spaceship 3 will ram into spaceship 1 at 0.999 c, a collision from which we expect to be unable to recover any clock fragments, leaving only one clock with nothing to compare it to.
-Doug Magnoli [Delete the two and the three for email.]
> > In Special Relativity Theory the basis of "Twin Paradox" is that > > "moving clocks run slow". What is not clear is that, when a 'clock > > runs slow' whether its frequency continuously decreases or not. Can > > anyone clarify this point?
> Nothing runs "slow" in any refernce frame. When otherwise identical > clocks - one of which sustained a velocity relative to the other - are > brought together to compare, one is seen to have less elapsed time > than its twin due to its hyperbolic rotation through 4-space. The > skewing factor is sqrt[1-(v^2/c^2)]. > No clock need be accelerated for the Twin Paradox to occur.
> Given any achievable velocities V1 and V2 and any finite lightspeed, > the sum of the velocities as viewed by any inertial observer cannot > exceed
> (V1 + V2)/[1 + (V1)(V2)/c^2]
> This is transformation of velocities parallel to the direction of > motion. For velocities at an arbitrary angle theta,
> Makes no difference who does the looking. Any inertial observer has a > valid reference frame. Disparities are only noted upon comparison of > reference frames. Spacetime is four-dimensional; travel through > spacetime is a hyperbolic rotation of all four coordinates. However, > the units of time are seconds not meters. ct is a nice length, but > the this fundamental unit of length is a very long one.
> At nominal velocities you don't travel much along ct compared to > x,y,x. Galilean transforms and Newtonian physics are good enough > approximations. At relativistic speeds you propagate along ct > comparable to a material dimension. You need Lorentzian transforms > and relativity to explain initially synchronized clocks separated and > then brought together for comparison thereafter.
> Acceleration of one observer has nothing to do with it, BTW.
> 1) Acceleration is an absolute measurement. There is no doubt who > is accelerated. > 2) Acceleration is irrelevant.
> We have three identical clocks that are off (a state of not running) > and zeroed. Each clock has a very short toggle jiggger switch > sticking out. We load them in individual spaceships and set up the > experiment.
> CLOCK 1: That's our clock. It sits stationary in our inertial > reference frame with a little jigger sticking out. Touch the jigger > and the "off" state becomes "on" or the "on" state becomes "off." > Clock 1 is "off."
> CLOCK 2: In a spaceship traveling at 0.999c relative to our > inertial frame of reference. Clock 2 is "off." It skims past Clock > 1, the jiggers touch, both Clocks 1 and 2 are now "on" and > synchronized.
> CLOCK 3: In a spaceship traveling at 0.999c relative to our > inertial frame of reference, but 180 degrees counter in direction to > Clock 2. Clock 3 is "off." Some arbitrary time after Clocks 1 and 2 > synchronize and turn "on" by touching, Clocks 2 and 3 brush past each > other, touching jiggers. Clock 2 is now "off," Clock 3 is now "on."
> CLOCK 1: That's our clock. It sits stationary in our inertial > reference frame with a little jigger sticking out. Clock 3 rushes > past, jiggers touch. Clocks 3 and 1 are now off. All clocks are > off. No clock has accelerated while "on."
> BOTTOM LINE: Get all three clocks together and compare elapsed > times. Elapsed time #2=#3, but elapsed time #2+#3 does not equal #1, > the local stationary reference frame summation. The sum of #2+#3 > elasped time is only about 4.5% that than of #1's accumulated elapsed > time. You now have the Twin Paradox without any running clock having > been accelerated.
> -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > "Quis custodiet ipsos custodes?" The Net!
> Now in order to follow the same trajectory that spaceship 2 followed, > only backwards and translated by one spaceship-width, spaceship 3 will > ram into spaceship 1 at 0.999 c, a collision from which we expect to be > unable to recover any clock fragments, leaving only one clock with > nothing to compare it to.
It's a thought experiment to illustrate how it works.. I hope you're joking when you use this to disprove the experiment.
> > > In Special Relativity Theory the basis of "Twin Paradox" is that > > > "moving clocks run slow". What is not clear is that, when a 'clock > > > runs slow' whether its frequency continuously decreases or not. Can > > > anyone clarify this point?
> > Nothing runs "slow" in any refernce frame. When otherwise identical > > clocks - one of which sustained a velocity relative to the other - are > > brought together to compare, one is seen to have less elapsed time > > than its twin due to its hyperbolic rotation through 4-space. The > > skewing factor is sqrt[1-(v^2/c^2)]. > > No clock need be accelerated for the Twin Paradox to occur.
> > Given any achievable velocities V1 and V2 and any finite lightspeed, > > the sum of the velocities as viewed by any inertial observer cannot > > exceed
> > (V1 + V2)/[1 + (V1)(V2)/c^2]
> > This is transformation of velocities parallel to the direction of > > motion. For velocities at an arbitrary angle theta,
> > Makes no difference who does the looking. Any inertial observer has a > > valid reference frame. Disparities are only noted upon comparison of > > reference frames. Spacetime is four-dimensional; travel through > > spacetime is a hyperbolic rotation of all four coordinates. However, > > the units of time are seconds not meters. ct is a nice length, but > > the this fundamental unit of length is a very long one.
> > At nominal velocities you don't travel much along ct compared to > > x,y,x. Galilean transforms and Newtonian physics are good enough > > approximations. At relativistic speeds you propagate along ct > > comparable to a material dimension. You need Lorentzian transforms > > and relativity to explain initially synchronized clocks separated and > > then brought together for comparison thereafter.
> > Acceleration of one observer has nothing to do with it, BTW.
> > 1) Acceleration is an absolute measurement. There is no doubt who > > is accelerated. > > 2) Acceleration is irrelevant.
> > We have three identical clocks that are off (a state of not running) > > and zeroed. Each clock has a very short toggle jiggger switch > > sticking out. We load them in individual spaceships and set up the > > experiment.
> > CLOCK 1: That's our clock. It sits stationary in our inertial > > reference frame with a little jigger sticking out. Touch the jigger > > and the "off" state becomes "on" or the "on" state becomes "off." > > Clock 1 is "off."
> > CLOCK 2: In a spaceship traveling at 0.999c relative to our > > inertial frame of reference. Clock 2 is "off." It skims past Clock > > 1, the jiggers touch, both Clocks 1 and 2 are now "on" and > > synchronized.
> > CLOCK 3: In a spaceship traveling at 0.999c relative to our > > inertial frame of reference, but 180 degrees counter in direction to > > Clock 2. Clock 3 is "off." Some arbitrary time after Clocks 1 and 2 > > synchronize and turn "on" by touching, Clocks 2 and 3 brush past each > > other, touching jiggers. Clock 2 is now "off," Clock 3 is now "on."
> > CLOCK 1: That's our clock. It sits stationary in our inertial > > reference frame with a little jigger sticking out. Clock 3 rushes > > past, jiggers touch. Clocks 3 and 1 are now off. All clocks are > > off. No clock has accelerated while "on."
> > BOTTOM LINE: Get all three clocks together and compare elapsed > > times. Elapsed time #2=#3, but elapsed time #2+#3 does not equal #1, > > the local stationary reference frame summation. The sum of #2+#3 > > elasped time is only about 4.5% that than of #1's accumulated elapsed > > time. You now have the Twin Paradox without any running clock having > > been accelerated.
> When you can define time, then maybe your input will be interesting. > Clocks do not measure the passage of time, they 'register' an internal > velocity (of some mass or system of mass) times it's displacement, > either linear or angular depending upon the type of clock, an internal > velocity that must change in order for the registered 'ticks' to get out > of sync with another clock. A change in velocity does not a change in > time equal. Now suppose I adopt the motion of an elastic ball between to > walls as my time-piece: > 1) How will this clock differ fundamentally from any other clock? > 2) Since when will an acceleration of the elastic ball constitute a > change in the rate that I am passing through time, rather than > constituting simply an increase in its intrinsic frequency wrt time?
Fast mesons live longer than stationary muons. What goes tick tock in a meson?
> > In Special Relativity Theory the basis of "Twin Paradox" is that > > "moving clocks run slow". What is not clear is that, when a 'clock > > runs slow' whether its frequency continuously decreases or not. Can > > anyone clarify this point?
That would be a second order "running slow", wouldn't it -- an acceleration of slowness.
> Nothing runs "slow" in any refernce frame. When otherwise identical > clocks - one of which sustained a velocity relative to the other - are > brought together to compare, one is seen to have less elapsed time > than its twin due to its hyperbolic rotation through 4-space. The > skewing factor is sqrt[1-(v^2/c^2)]. > No clock need be accelerated for the Twin Paradox to occur.
I doubt this. I've seen you make this claim before. Let's take a closer look ...
> Given any achievable velocities V1 and V2 and any finite lightspeed, > the sum of the velocities as viewed by any inertial observer cannot > exceed
> (V1 + V2)/[1 + (V1)(V2)/c^2]
What? Given any "achievable velocities" v1 and v2, the sum is bounded by 2c. Since the frame in which the sum is 2c - eps is an inertial frame, that is your bound: v1 + v2 < 2c.
Alright Al -- I think you've become a relativity crank in your old age; you go through a lot of manipulation of symbols, and yet your conclusion is clearly wrong from the start (ooh ... I _love_ writing like this) -- acceleration is indeed necessary somewhere for there to be a twin paradox.
Argument: take two clocks in relative motion, and remember, we are forbidden to acceleration either one. Therefore the clocks are in constant inertial relative motion, and are related in their rest frames by a Lorentz transform. Well, we all know about Lorentz transforms, don't we -- there is a reciprocal slowing of time, a feat made possible by the interweaving of time and space. It all works out Ok in the end, but neither clock can unambiguously be said to have lost time wrt the other.
As a special case, we can consider two co-moving clocks. In this case, in the clocks' rest frame, they remain in synchronization up to a constant forever.
Further argument: Normally, although I will not absolutely insist upon it, a "twin paradox" requires two clocks to have been coincident at an event A and at a second event B. The requirement of coincidence is to remove any ambiguity about "synchronization". I agree completely that the clocks do not have to be at mutual rest ... coincident for some interval of proper time ... it is sufficient that they pass each other at a single event, at any relative velocity. A snapshot of the relative times shown at that event is an observer invariant.
Ok. Now, applying what we have learned above, how can two clocks be coincident twice? Well, either they remain coincident at all times, or else they depart from coincidence and then momentarily regain that holy condition. In order for this to happen, at least one clock must be accelerated: otherwise they will cross paths exactly once.
Since this seems crystal clear, any further calculation seems superfluous. What part of my analysis do you disagree with?
<read, read, read>
Oh ... you cheated! You used _three_ clocks. Well, this seems like a pretty Rube Goldberg "twin paradox", but I guess. The out and back clock could be replaced with two clocks running hot, straight and normal, which pass off a token in a kind of relay race. I agree then that the elapsed proper time on this clock pair would be less than on a single inertial clock which they crossed at event A and event B, although it could be argued that the two clocks doing the relay race were the moral equivalent of one clock doing a sharp acceleration.
Still, your prevarication has the valuable property of showing that it is not any intrinsic effect of acceleration on the clock which results in the "paradox", but only on the trajectory.
<Sententious conclusion mode>: Indeed, how could it be otherwise? SR knows nothing of acceleration, contains no terms in acceleration, and hence any effects of acceleration in SR must merely be an integrated compounding of many small increments of constant velocity. To deal with the effects of acceleration per se on matter we need GR. And therefore, we would predict in any particular instance of the two clock twin paradox that the SR result is in fact modified by GR, since at least one of the clocks must undergo acceleration. So you have present the SR-clean version in which no clocks undergo acceleration. Kudos, Al, Kudos (though I choke on them).
Amazing how many iteration the understanding goes through. From the false claim sometimes seen "We need GR to handle acceleration", meaning apparently that we must invoke the GR deity whenever the sacred word "acceleration" appears, to the antithesis that we can in fact handle a lot about acceleration in a pure SR environment, in effect by integration over many contiguous inertial frames, yadda, yadda, to the final triumphant (?) synthesis: we can handle a lot about acceleration in SR but for a _complete_ account of a process involving the acceleration of material bodies even in flat space, I think we must after all invoke GR.
Actually, I'm not 100% sure of the last claim: Is there any "purely GR-ish" effect related to the acceleration of material bodies in flat space, or are all physical effects obtained by combining the predictions of SR with any direct effects of the forces involved on the body?
Some question just never go away. Damn.
I guess the last question could be summed up by experiment: put any kind of clock we like in an ultracentrifuge. Spin liberally. Once we have accounted for (1) the relativistic effects of |u| (magnitude of speed at the end of the arm) (2) any direct effects of the stress field on the clock, is there (3) some residue left over for GR, an "intrinsic effect of acceleration".
No clock need be accelerated for the Twin Paradox to occur.
<snip>
P.S. Calling your three-clock version the "Twin Paradox" is at least misleading, like a magician who calls the audiences attention away from what he is really doing. You should call it the "triplet paradox", maybe. Only half facetious.
> BOTTOM LINE: Get all three clocks together and compare elapsed > times.
???
> Elapsed time #2=#3, but elapsed time #2+#3 does not equal #1, > the local stationary reference frame summation. The sum of #2+#3 > elasped time is only about 4.5% that than of #1's accumulated elapsed > time. You now have the Twin Paradox without any running clock having > been accelerated.
Didn't notice this clause before. How do you propose to "get all three clocks together" without accelerating some of them!!?? Oh... I see ... yet more Rube Goldbergism. You are using stopwatches ... any _running_ clock having been accelerated. Smart aleck.
Franz Heymann wrote: > Fast mesons live longer than stationary muons. What goes tick tock in a > meson?
<Spaceman> "Fast mesons" are a mass in motion and you don't even see it. Motion = ticker. Clock worship god sinking. Blub blub. Research all clocks. -2*-2 = -4 Pressure of uncharged electrons as big as basketballs. LOL. Yeeha! </Spaceman>
> George Greene <gree...@eagle.cs.unc.edu> wrote in message <news:xeslm4niu65.fsf@eagle.cs.unc.edu>... > > Uncle Al <Uncle...@hate.spam.net> writes: > > : Nothing runs "slow" in any reference frame. > All calculations in Special Theory of Relativity are based on the law > of conservation of momentum and kinetic energy.
Special Relativity is a self-conistent hyperbolic geometry. Physics is inserted later. Conservation laws are absolutely *absent* from SR's derivation.
Instead of shitting in our parlour, why don't you learn something about that which you spew?
If you wish to gainsay conservation laws, you must discredit Noether's theorem. You don't know shit about Noether's theorem, either. As Noether's theorem is founded in pure mathematics, I'd say there was a pretty good bung driven up you backside - with a sledgehammer.
[snip random idiocy]
-- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) "Quis custodiet ipsos custodes?" The Net!
> In order for the jiggers of the clocks of spaceships 1 and 2 to click > each other on, the two spaceships have to have a certain orientation > relative to each other.
Bullshit. Look at a joystick - 2(pi) steradians of give. Use a fiberoptic source and a pinhole photodiode detector as the ships skim 1 millmeter part -no mechanical contact.
[snip hopeless 2-D confusion]
-- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) "Quis custodiet ipsos custodes?" The Net!
> > > In Special Relativity Theory the basis of "Twin Paradox" is that > > > "moving clocks run slow". What is not clear is that, when a 'clock > > > runs slow' whether its frequency continuously decreases or not. Can > > > anyone clarify this point?
> That would be a second order "running slow", wouldn't it -- an > acceleration of slowness.
> > Nothing runs "slow" in any refernce frame. When otherwise identical > > clocks - one of which sustained a velocity relative to the other - are > > brought together to compare, one is seen to have less elapsed time > > than its twin due to its hyperbolic rotation through 4-space. The > > skewing factor is sqrt[1-(v^2/c^2)]. > > No clock need be accelerated for the Twin Paradox to occur.
> I doubt this. I've seen you make this claim before. Let's take a > closer look ...
> > Given any achievable velocities V1 and V2 and any finite lightspeed, > > the sum of the velocities as viewed by any inertial observer cannot > > exceed
> > (V1 + V2)/[1 + (V1)(V2)/c^2]
> What? Given any "achievable velocities" v1 and v2, the sum is bounded > by 2c. Since the frame in which the sum is 2c - eps is an inertial > frame, that is your bound: v1 + v2 < 2c.
Newtonian physics and Galilean transforms are empirically invalid udner relativistic conditions. I don't know how your universe operates, but in mine and everybody else's the view of lightspeed and directly opposed velocities in any reference frame cannot exceed lightspeed - including lightspeed vs. lightspeed. This is trivially demonstrated in counter-circulating particle accelerators such as CERN or FermiLab.
[snip incredible bulk of ignorant spew invalidated by common observation]
The unverse doesn't care what you think, or if you think at all. Reality is not subject to majority vote. Your crap is religion, and we all know how well that works. Test of faith!
-- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) "Quis custodiet ipsos custodes?" The Net!
> No clock need be accelerated for the Twin Paradox to occur.
> <snip>
> P.S. Calling your three-clock version the "Twin Paradox" is at least > misleading, like a magician who calls the audiences attention away > from what he is really doing. You should call it the "triplet > paradox", maybe. Only half facetious.
> > BOTTOM LINE: Get all three clocks together and compare elapsed > > times.
> ???
> > Elapsed time #2=#3, but elapsed time #2+#3 does not equal #1, > > the local stationary reference frame summation. The sum of #2+#3 > > elasped time is only about 4.5% that than of #1's accumulated elapsed > > time. You now have the Twin Paradox without any running clock having > > been accelerated.
> Didn't notice this clause before. How do you propose to "get all > three clocks together" without accelerating some of them!!?? Oh... I > see ... yet more Rube Goldbergism. You are using stopwatches ... any > _running_ clock having been accelerated. Smart aleck.
Why make it complicated? Each ship has ore, a smelter, and a machine shop. Accelerate to (+/-)0.999c vs. Clock 1, THEN build the ship's clock starting from molten whatever. Run the experiment. Write down the elapsed time number. Blow the clock out the plasma warp drive. Come back home. The clock NEVER accelerates.
-- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) "Quis custodiet ipsos custodes?" The Net!
> In Special Relativity Theory the basis of "Twin Paradox" is that > "moving clocks run slow". What is not clear is that, when a 'clock > runs slow' whether its frequency continuously decreases or not. Can > anyone clarify this point?
If a clock's frequency continuously decreases, it needs winding up.
: George Greene <gree...@eagle.cs.unc.edu> wrote in message : news:xeslm4niu65.fsf@eagle.cs.unc.edu... : > Uncle Al <Uncle...@hate.spam.net> writes: : > : Nothing runs "slow" in any reference frame. : > : > This begs the question of how you are supposed to : > be able to measure what a clock says "now" when it : > is somewhere else. : >
: There are truly classical prescriptions developed by Einstein to enable : synchronization of clock times between co-moving frames.
of course there are, and according to THOSE prescriptions, moving clocks DO run slow.
-- --- "It's difficult ... you need to be united to have any strength, but internal issues have to be addressed." --- E. Ray Lewis, on liberalism in America
>> > Given any achievable velocities V1 and V2 and any finite lightspeed, >> > the sum of the velocities as viewed by any inertial observer cannot >> > exceed
>> > (V1 + V2)/[1 + (V1)(V2)/c^2]
>> What? Given any "achievable velocities" v1 and v2, the sum is bounded >> by 2c. Since the frame in which the sum is 2c - eps is an inertial >> frame, that is your bound: v1 + v2 < 2c.
> Newtonian physics and Galilean transforms are empirically invalid > udner relativistic conditions. I don't know how your universe > operates, but in mine and everybody else's the view of lightspeed and > directly opposed velocities in any reference frame cannot exceed > lightspeed - including lightspeed vs. lightspeed.
Al, read what you wrote, not what you think you wrote. Ed is objecting to the former, not the latter.
The vector sum of two velocities, v1 and v2 in a particular reference frame is itself a vector. It is not, quite, a velocity, but it is expresed in units of velocity. If the individual vectors have values whose magnitudes are strictly less than c then the vector sum will have a magnitude that is strictly less than 2c.
As I said, this sum is not the velocity of any particular object in any particular reference frame. Instead, it is the time rate of change in the (vector) separation between the two objects whose velocities are v1 and v2 as viewed in the reference frame within which those objects have those velocities.
Note that even with your intended interpretation (velocity of object 1 in frame in which object 2 is at rest), the formula you give has a sign error. Two oppositely directed velocities add to give a combined relative velocity whose magnitude is larger than either individual velocity. Your formula would have them subtract to give a relative velocity that is smaller than either. We can forgive you that since the formula is only valid for the one dimensional case and we can assume that you were talking about speeds of colliding particles rather than velocities of overtaking particles.
The more standard setup for the one dimensional velocity composition formula is object 1 moving with velocity v1 (in some rest frame) and object 2 moving with velocity v2 in the frame within which v1 is at rest. The formula you gave then gives the velocity of v2 in the rest frame.
(The above sign error creeps in when you casually switch the roles of object 1 and the rest frame and forget to change v1 to -v1)
> This is trivially > demonstrated in counter-circulating particle accelerators such as CERN > or FermiLab.
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