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tgdenning@earthlink.net
science forum beginner
Joined: 20 Mar 2006
Posts: 18
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 Posted: Mon Apr 24, 2006 3:00 pm Post subject:
Variant of SR muon experiment
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I tried this on the other physics groups but I got very little
feedback, perhaps because I didn't claim to refute relativity or use
the word "conspiracy". Some replies are 'no' but no explanation.
Let there be a group of particles P moving from right to left. They
are arranged in a single plane perpendicular to the X axis. The
particles are subject to decay.
Located at Xa along the axis of motion is a single-molecule thick
piece of film A which is sensitive to the particles. Also of course
oriented perpendicular to the X axis.
There is a second film B which is moving relative to A from left to
right and displaced so that they don't overlap.
B and P reach Xa at the same time WRT A, and the particle group is
wide enough to impact both films.
Since the speed relative to B is greater, is it correct that B will
detect more particles than A due to time dilation? (Fewer decay.)
If not, what is the mechanism?
Thanks
> -tg |
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Jon Bell
science forum beginner
Joined: 07 May 2005
Posts: 45
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| Posted: Tue Apr 25, 2006 12:58 pm Post subject:
Re: Variant of SR muon experiment
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In article <1145880711.340789.164200@e56g2000cwe.googlegroups.com>,
<tgdenning@earthlink.net> wrote:
| Quote: |
Since the speed relative to B is greater, is it correct that B will
detect more particles than A due to time dilation? (Fewer decay.)
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No. This is easiest to see in the reference frame in which the particles
are at rest. In this frame both detectors A and B are moving, but they
arrive at the particles at the same time. If the particles were all
created at the same time, then equal numbers of them must have decayed as
far as both detectors are concerned, because the time interval between
creation and detection is the same for both detectors in this frame.
To see what happens in each detector's rest frame, here's a specific
numeric example. I'll measure time in seconds (s) and distance in
light-seconds (ls) (the distance light travels in one second), because
this makes c = 1 ls/s which makes the arithmetic simpler.
Let frame S be the one in which detector A is at rest at x = 10 ls. Let
frame S' be the one in which detector B is at rest at x' = 0. Let the
origins of the two frames coincide (x = x' = 0 at t = t' = 0) so we can
use the usual Lorentz transformation equations to relate them. Let the
relative velocity (i.e. the velocity of detector B relative to A) be 0.7
ls/s (0.7c) to the right. The corresponding gamma factor
1/sqrt(1-v^2/c^s) is 1.400. Then in frame S, detector B travels the 10 ls
to detector A in (10 ls)/(0.7 ls/s) = 14.29 s.
Let the particles be created at x = 0 and t = 3 s in frame S. Then, in
order to arrive at detectors A and B when they meet, in frame S they must
travel a distance of 10 ls in 14.29 - 3 = 11.29 s, so their velocity in S
must be (10 ls)/(11.29 s) = 0.886 ls/s to the right. The corresponding
gamma factor is 2.157, so their mean lifetime in S is 2.157 times their
mean lifetime in their own rest frame.
Let the particles have a mean lifetime of 5 s in their own rest frame.
Then their mean lifetime in S is (2.157)(5 s) = 10.785 s. In S, the time
between creation and detection corresponds to 11.29/10.785 = 1.047 mean
lifetimes. This determines how many of them decay in S.
Now let's see what this looks like in frame S'. In frame S, the particles
were created at x = 0 and t = 3 s. According to the Lorentz
transformation, in S' they are created at
x' = gamma * (x - vt)
= 1.400 * (0 - (0.7 ls/s)(3 s))
= -2.94 ls
t' = gamma * (t - vx/c^2)
= 1.400 * ((3 s) - (0.7 ls/s)(0 s)/(1 ls/s)^2)
= 4.2 s
In frame S the particles arrive at the detectors at x = 10 ls and t =
14.29 s. Using the Lorentz transformation again, in S' they arrive at the
detectors at
x' = 1.400 * (10 ls - (0.7 ls/s)(14.29 s))
= 0 ls
t' = 1.400 * (14.29 s - (0.7 ls/s)(10 ls)/(1 ls/s)^2)
= 10.206 s
So, in frame S' the particles travel a distance 0 - (-2.94) = 2.94 ls in
10.206 - 4.2 = 6.006 s, which gives a speed of 0.490 ls/s and a gamma
factor of 1.147. Their mean lifetime is S' is (1.147)(5 s) = 5.735 s. In
S', the time between creation and detection corresponds to 6.006/5.735 =
1.047 mean lifetimes, the same as in frame S. Therefore the same number
of particles decay between creation and detection, in the two frames.
The particles have a longer mean lifetime in S than in S', but they also
have to travel a greater distance in S than in S', in exactly equal
proportion.
--
Jon Bell <jtbell@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA |
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tgdenning@earthlink.net
science forum beginner
Joined: 20 Mar 2006
Posts: 18
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| Posted: Thu Apr 27, 2006 9:18 pm Post subject:
Re: Variant of SR muon experiment
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Jon Bell wrote:
| Quote: | In article <1145880711.340789.164200@e56g2000cwe.googlegroups.com>,
tgdenning@earthlink.net> wrote:
Since the speed relative to B is greater, is it correct that B will
detect more particles than A due to time dilation? (Fewer decay.)
No. This is easiest to see in the reference frame in which the particles
are at rest. In this frame both detectors A and B are moving, but they
arrive at the particles at the same time. If the particles were all
created at the same time, then equal numbers of them must have decayed as
far as both detectors are concerned, because the time interval between
creation and detection is the same for both detectors in this frame.
To see what happens in each detector's rest frame, here's a specific
numeric example. I'll measure time in seconds (s) and distance in
light-seconds (ls) (the distance light travels in one second), because
this makes c = 1 ls/s which makes the arithmetic simpler.
Let frame S be the one in which detector A is at rest at x = 10 ls. Let
frame S' be the one in which detector B is at rest at x' = 0. Let the
origins of the two frames coincide (x = x' = 0 at t = t' = 0) so we can
use the usual Lorentz transformation equations to relate them. Let the
relative velocity (i.e. the velocity of detector B relative to A) be 0.7
ls/s (0.7c) to the right. The corresponding gamma factor
1/sqrt(1-v^2/c^s) is 1.400. Then in frame S, detector B travels the 10 ls
to detector A in (10 ls)/(0.7 ls/s) = 14.29 s.
Let the particles be created at x = 0 and t = 3 s in frame S. Then, in
order to arrive at detectors A and B when they meet, in frame S they must
travel a distance of 10 ls in 14.29 - 3 = 11.29 s, so their velocity in S
must be (10 ls)/(11.29 s) = 0.886 ls/s to the right. The corresponding
gamma factor is 2.157, so their mean lifetime in S is 2.157 times their
mean lifetime in their own rest frame.
Let the particles have a mean lifetime of 5 s in their own rest frame.
Then their mean lifetime in S is (2.157)(5 s) = 10.785 s. In S, the time
between creation and detection corresponds to 11.29/10.785 = 1.047 mean
lifetimes. This determines how many of them decay in S.
Now let's see what this looks like in frame S'. In frame S, the particles
were created at x = 0 and t = 3 s. According to the Lorentz
transformation, in S' they are created at
x' = gamma * (x - vt)
= 1.400 * (0 - (0.7 ls/s)(3 s))
= -2.94 ls
t' = gamma * (t - vx/c^2)
= 1.400 * ((3 s) - (0.7 ls/s)(0 s)/(1 ls/s)^2)
= 4.2 s
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Thanks for all the effort. This is the point at which I have had
trouble. I see these two frames as representing two different
experiments. Why does the speed of S' relative to S matter? Moving to
the frame of the particles is very clever and compelling, but (at least
in my apparently confused mind) it suffers from the same conceptual
problem. Maybe I'm making a really elementary mistake, but I thought
that for some set of experiments with speed and distance as parameters
and time as a constant, the amount of time dilation would vary
according to speed.
What I need is a way to break the conceptual connection between two
independent experiments and this. (Or, everyone else could change their
mind... )
-tg
| Quote: | In frame S the particles arrive at the detectors at x = 10 ls and t =
14.29 s. Using the Lorentz transformation again, in S' they arrive at the
detectors at
x' = 1.400 * (10 ls - (0.7 ls/s)(14.29 s))
= 0 ls
t' = 1.400 * (14.29 s - (0.7 ls/s)(10 ls)/(1 ls/s)^2)
= 10.206 s
So, in frame S' the particles travel a distance 0 - (-2.94) = 2.94 ls in
10.206 - 4.2 = 6.006 s, which gives a speed of 0.490 ls/s and a gamma
factor of 1.147. Their mean lifetime is S' is (1.147)(5 s) = 5.735 s. In
S', the time between creation and detection corresponds to 6.006/5.735 =
1.047 mean lifetimes, the same as in frame S. Therefore the same number
of particles decay between creation and detection, in the two frames.
The particles have a longer mean lifetime in S than in S', but they also
have to travel a greater distance in S than in S', in exactly equal
proportion.
--
Jon Bell <jtbell@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA |
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