| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
21 Apr 2006 11:58:30 AM |
| Object: |
Muon variant |
This is not a trick question---no paradox problem for me if correct.
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.
Is it correct that B will detect more particles than A?
Thanks
-tg
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| User: "Ben Rudiak-Gould" |
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| Title: Re: Muon variant |
21 Apr 2006 02:30:45 PM |
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wrote:
B and P reach Xa at the same time WRT A, and the particle group is wide
enough to impact both films.
The "WRT A" is meaningless here. Reaching a location at the same time is
frame-independent.
Is it correct that B will detect more particles than A?
No. They will detect the same number of particles.
-- Ben
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| User: "" |
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| Title: Re: Muon variant |
21 Apr 2006 02:48:28 PM |
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Ben Rudiak-Gould wrote:
tgdenning@earthlink.net wrote:
B and P reach Xa at the same time WRT A, and the particle group is wide
enough to impact both films.
The "WRT A" is meaningless here. Reaching a location at the same time is
frame-independent.
Is it correct that B will detect more particles than A?
No. They will detect the same number of particles.
-- Ben
Can you explain? Particles have a higher relative speed WRT B.
-tg
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| User: "Randy Poe" |
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| Title: Re: Muon variant |
21 Apr 2006 03:20:18 PM |
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wrote:
Ben Rudiak-Gould wrote:
wrote:
B and P reach Xa at the same time WRT A, and the particle group is wide
enough to impact both films.
The "WRT A" is meaningless here. Reaching a location at the same time is
frame-independent.
Is it correct that B will detect more particles than A?
No. They will detect the same number of particles.
Can you explain? Particles have a higher relative speed WRT B.
How would that affect the number of detection events?
There is one event for every particle, in every frame.
- Randy
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| User: "" |
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| Title: Re: Muon variant |
21 Apr 2006 04:18:57 PM |
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Randy Poe wrote:
tgdenning@earthlink.net wrote:
Ben Rudiak-Gould wrote:
tgdenning@earthlink.net wrote:
B and P reach Xa at the same time WRT A, and the particle group is wide
enough to impact both films.
The "WRT A" is meaningless here. Reaching a location at the same time is
frame-independent.
Is it correct that B will detect more particles than A?
No. They will detect the same number of particles.
Can you explain? Particles have a higher relative speed WRT B.
How would that affect the number of detection events?
There is one event for every particle, in every frame.
I'm thinking about particle decay, as in the muon experiment on time
dilation.
The travel time is the same but the speed is different, so less time
passes for those particles approaching B, and fewer should decay.
-tg
- Randy
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| User: "Ben Rudiak-Gould" |
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| Title: Re: Muon variant |
24 Apr 2006 12:59:16 PM |
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wrote:
I'm thinking about particle decay, as in the muon experiment on time
dilation.
The travel time is the same but the speed is different, so less time
passes for those particles approaching B, and fewer should decay.
I think what makes your thought experiment difficult to understand is that
the muons aren't produced anywhere. It doesn't really make sense to speak of
a beam of muons arriving from infinity, since they would all have decayed
infinitely long ago. Try introducing a specific time and place at which the
pulse of muons was produced, and then use the Lorentz transform to figure
out where and when they were produced with respect to the other frame. Then
you'll find that a faster speed (hence slower decay rate) is balanced by a
longer total travel time, such that the ultimate particle count is the same.
-- Ben
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| User: "" |
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| Title: Re: Muon variant |
25 Apr 2006 07:12:40 AM |
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Ben Rudiak-Gould wrote:
tgdenning@earthlink.net wrote:
I'm thinking about particle decay, as in the muon experiment on time
dilation.
The travel time is the same but the speed is different, so less time
passes for those particles approaching B, and fewer should decay.
I think what makes your thought experiment difficult to understand is that
the muons aren't produced anywhere. It doesn't really make sense to speak of
a beam of muons arriving from infinity, since they would all have decayed
infinitely long ago. Try introducing a specific time and place at which the
pulse of muons was produced, and then use the Lorentz transform to figure
out where and when they were produced with respect to the other frame. Then
you'll find that a faster speed (hence slower decay rate) is balanced by a
longer total travel time, such that the ultimate particle count is the same.
-- Ben
Afterthought:
Perhaps my misunderstanding has to do with the travel time. As I
understand the experiment, the travel time is simply what A and B
measure in their own frames. Slower particles take the same time to
cover shorter distance as fast to cover longer. What does one frame
have to do with the other?
-tg
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| User: "" |
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| Title: Re: Muon variant |
25 Apr 2006 06:21:33 AM |
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Ben Rudiak-Gould wrote:
tgdenning@earthlink.net wrote:
I'm thinking about particle decay, as in the muon experiment on time
dilation.
The travel time is the same but the speed is different, so less time
passes for those particles approaching B, and fewer should decay.
I think what makes your thought experiment difficult to understand is that
the muons aren't produced anywhere. It doesn't really make sense to speak of
a beam of muons arriving from infinity, since they would all have decayed
infinitely long ago. Try introducing a specific time and place at which the
pulse of muons was produced, and then use the Lorentz transform to figure
out where and when they were produced with respect to the other frame. Then
you'll find that a faster speed (hence slower decay rate) is balanced by a
longer total travel time, such that the ultimate particle count is the same.
-- Ben
Thanks for the follow-up. But I have considered a "starting point" as
follows:
At some point Xc in the path of the particles, there are two films of
half-width in line with A and B, so that we sample the planar density
of particles on their way to Xa. We end up with 4 pieces of film.
Either all the films show the same density, or not. If not, I can
explain it by time dilation. I don't find this outcome paradoxical, but
others apparently do.
(I think this is a fair representation of how the actual experiment was
done, albeit with my slow decaying thought experiment particles. A
measurement is taken at high altitude and then at low, and the numbers
are compared.)
-tg
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| User: "Randy Poe" |
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| Title: Re: Muon variant |
25 Apr 2006 07:45:30 AM |
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wrote:
Ben Rudiak-Gould wrote:
wrote:
I'm thinking about particle decay, as in the muon experiment on time
dilation.
The travel time is the same but the speed is different, so less time
passes for those particles approaching B, and fewer should decay.
I think what makes your thought experiment difficult to understand is that
the muons aren't produced anywhere. It doesn't really make sense to speak of
a beam of muons arriving from infinity, since they would all have decayed
infinitely long ago. Try introducing a specific time and place at which the
pulse of muons was produced, and then use the Lorentz transform to figure
out where and when they were produced with respect to the other frame. Then
you'll find that a faster speed (hence slower decay rate) is balanced by a
longer total travel time, such that the ultimate particle count is the same.
-- Ben
Thanks for the follow-up. But I have considered a "starting point" as
follows:
At some point Xc in the path of the particles, there are two films of
half-width in line with A and B, so that we sample the planar density
of particles on their way to Xa. We end up with 4 pieces of film.
Either all the films show the same density, or not.
They show the same.
If not, I can
explain it by time dilation. I don't find this outcome paradoxical, but
others apparently do.
You're talking about events that only exist in some
frames. That's not explainable by time dilation. If an event
occurs, it occurs with some spacetime coordinates in all
frames.
- Randy
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| User: "" |
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| Title: Re: Muon variant |
25 Apr 2006 09:26:47 AM |
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Randy Poe wrote:
tgdenning@earthlink.net wrote:
Ben Rudiak-Gould wrote:
tgdenning@earthlink.net wrote:
I'm thinking about particle decay, as in the muon experiment on time
dilation.
The travel time is the same but the speed is different, so less time
passes for those particles approaching B, and fewer should decay.
I think what makes your thought experiment difficult to understand is that
the muons aren't produced anywhere. It doesn't really make sense to speak of
a beam of muons arriving from infinity, since they would all have decayed
infinitely long ago. Try introducing a specific time and place at which the
pulse of muons was produced, and then use the Lorentz transform to figure
out where and when they were produced with respect to the other frame. Then
you'll find that a faster speed (hence slower decay rate) is balanced by a
longer total travel time, such that the ultimate particle count is the same.
-- Ben
Thanks for the follow-up. But I have considered a "starting point" as
follows:
At some point Xc in the path of the particles, there are two films of
half-width in line with A and B, so that we sample the planar density
of particles on their way to Xa. We end up with 4 pieces of film.
Either all the films show the same density, or not.
They show the same.
If not, I can
explain it by time dilation. I don't find this outcome paradoxical, but
others apparently do.
You're talking about events that only exist in some
frames. That's not explainable by time dilation. If an event
occurs, it occurs with some spacetime coordinates in all
frames.
I don't get what you are saying here. All I've done (I think) is take
two separate experiments, one with fast particles over a longer
distance, and one with slow particles over a shorter distance, and
place them next to each other. Since I started thinking about this
because of a philosophical question, it might be appropriate to talk
about 'existence' at some point. But right now it just seems that what
happens in each frame is independent of the other. I'm trying to get
the physics straight first, and time dilation explains measuring
different amounts of decay for particles of different speeds over the
same time period. Surely experiments have been done under exactly those
conditions, using accelerators and targets at different distances?
-tg
- Randy
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| User: "Randy Poe" |
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| Title: Re: Muon variant |
25 Apr 2006 07:40:30 AM |
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wrote:
Randy Poe wrote:
wrote:
Ben Rudiak-Gould wrote:
wrote:
B and P reach Xa at the same time WRT A, and the particle group is wide
enough to impact both films.
The "WRT A" is meaningless here. Reaching a location at the same time is
frame-independent.
Is it correct that B will detect more particles than A?
No. They will detect the same number of particles.
Can you explain? Particles have a higher relative speed WRT B.
How would that affect the number of detection events?
There is one event for every particle, in every frame.
I'm thinking about particle decay, as in the muon experiment on time
dilation.
The travel time is the same but the speed is different, so less time
passes for those particles approaching B, and fewer should decay.
Think about the classic atmospheric muon experiment. Consider
any particular muon coming into existence in the upper
atmosphere.
First consider a muon that decays before hitting a
detector on the ground. A ground observer and an
observer traveling with the muons both agree that
the muon decayed. The traveling observer says the
lifetime was on the order of 2.2 usec, the ground observer
says it was something longer.
Now consider a muon that hits the ground detector.
Again, both observers agree that the detection occurred,
and they disagree on the age of the muon at the time
of detection.
Does that help?
- Randy
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| User: "" |
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| Title: Re: Muon variant |
25 Apr 2006 08:38:21 AM |
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Randy Poe wrote:
tgdenning@earthlink.net wrote:
Randy Poe wrote:
tgdenning@earthlink.net wrote:
Ben Rudiak-Gould wrote:
tgdenning@earthlink.net wrote:
B and P reach Xa at the same time WRT A, and the particle group is wide
enough to impact both films.
The "WRT A" is meaningless here. Reaching a location at the same time is
frame-independent.
Is it correct that B will detect more particles than A?
No. They will detect the same number of particles.
Can you explain? Particles have a higher relative speed WRT B.
How would that affect the number of detection events?
There is one event for every particle, in every frame.
I'm thinking about particle decay, as in the muon experiment on time
dilation.
The travel time is the same but the speed is different, so less time
passes for those particles approaching B, and fewer should decay.
Think about the classic atmospheric muon experiment. Consider
any particular muon coming into existence in the upper
atmosphere.
First consider a muon that decays before hitting a
detector on the ground. A ground observer and an
observer traveling with the muons both agree that
the muon decayed. The traveling observer says the
lifetime was on the order of 2.2 usec, the ground observer
says it was something longer.
Now consider a muon that hits the ground detector.
Again, both observers agree that the detection occurred,
and they disagree on the age of the muon at the time
of detection.
Does that help?
This may well be a case of thinking about something too long and being
stuck in a brain loop. But what would help is an explanation of the
experiment as I constructed it. The classic experiment didn't involve
observers traveling with the muons, but actual measurements at altitude
and on the ground.
As I said in reply to Ben, I understand the travel times to be related
only to the measurement frame. The effect is that fewer particles decay
than expected for a given travel time, depending on the speed relative
to c. If this is correct, then two experiments using different particle
speeds should yield two different results, with the slower showing less
time dilation, with more particles decaying. Is some of this
wrong---does distance come in to it?
-tg
- Randy
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