h:

My question though: is it possible to measure time dilation effects in
a lab, at say, sea level only, without repeating the experiment at a
high altitude. After all the muons will be stoping in the scintillator,
so the lifetime you will be measuring will be the lifetime of still
muons. This lifetime can then be compared to the lifetime of free
muons, assumign you know it.

Since the muons will be stopped, all you will be measuring is the
lifetime of the muon, not any time dilation.

Is this a valid way to measure time dilation effects? Reading The
paper by Frisch + Smith, I can't understand if they tried this also
or not.

Yes, but you'll need at least one more detector and a coincidence
box.

The only complication i can think of with this method is the fact that
both negative and positive muons are comming in, and the negative muons
are captured by the scintillator material which will lead to a shorter
average lifetime and perhaps cancel out any time dilation effetcs.

I'm not sure if this is right at all, and would like comments.

To see the time dilation, what you need is the equivalent of a
change in altitude. To get this without carting your equipment
up a mountain, build a detector telescope, i.e., use two detectors
which are arranged:

\muon
|
v With this arrangement, you can take a coincidence
--- between the muons which traverse both detectors.
That gives you a path through the atmosphere with
a length you can look up.

---


Now, you want to make the entire assembly such that it will pivot
about the center, so that you can rotate the telescope like this:



\
\
\
\


The muons will now have to traverse a longer distance, before passing
through your telescope. If the distance to the top of the atmosphere is d,
and the radius of the earth is r_e, then the path will become longer by an
amount, d' = (d + r_e \cos(A)), where A is the angle of your detector. As
you rotate the telescope, you should find that the number of muons you
count in a given time interval will decrease as a function of angle. Note
that a thorough analysis of the data is going to involve more than
plugging numbers into a formula. There will be systematic effects to deal
with, a couple of which a


(1) Does the actual muon flux depend on the angle of the telescope?

(2) As you turn the telescope, the distance through the earth will
decrease and the number of muons going through the detectors
from the bottom up will increase.

(I'm sure there are additional ones, but since I just made this
experiment up, I haven't thought much about it).

The above are easy to check in two ways. First, you'll want to
measure the muon flux throughout the day to see if there is
a variation as the earth rotates (e.g., the source of primary
cosmics might not be uniform). Second, you can measure the
flux at 0 and 90 degrees without regard as to which order
the detectors are triggered. In the 0 degree position, most (almost
all) of your muons will be from the top down. Very few will pass
all the way through the earth. At 90 degrees, things will be
different. Muons will traverse (mainly) the atmosphere either
direction, so the number of muons passing through the detector
could be _larger_ than the number at vertical, even though more
have decayed. You'll have to calculate what you think you'd expect
and see if it agrees.

Now, you'll also have to obtain (or cut and polish your own)
pieces of scintillator with an appropriate geometry. The
scintillator should be relatively thin, or at least the top
one (call it delta E) should. The bottom one could be thicker,
or even thick enough to stop (most) of the muons, but it doesn't
have to be.

To have a large number of steps, you should make one dimension of
the scintillator small. How small depends upon how far apart you
space the detectors. The other dimension should be larger.
Something like a rectangle that's 2 cm x 30 cm x .1 cm thick
would be ok. You will want to check this, as I just guestimated.
You could most likely get away with somewhat thicker scintillator
of the top detector.


Your electronics will look sort of like:

top det

|pmt 1|---[cfd]--[delay]--+
|
-------
| coinc | --- single channel scalar
---+---
|
|pmt 2|----[cfd]----------+

bottom det

cfd = constant fractions discriminator
you will want the coincidence to only register if the top detector
fires first in some cases and allow either to fire first in
others. It depends upon what you find when you get started.

To make the experiment more sophisticated, you could perform a time
of flight measurement between the scintillators by starting and
stopping the TAC. You could also use different thicknesses of
rangeout material between the top and bottom detector in order to
determine the average energy at various angles.

I would expect that this experiment would take several weeks worth
of counting, but I think it will do what you want. However, if I
suddenly realize I overlooked something, I'll post it.

kenseto wrote:

(formerly)" dlzc1.cox@net wrote in message
news:l0Agb.47226$gv5.34510@fed1read05...
Dear kenseto:

"kenseto" wrote in message
...

(formerly)" dlzc1.cox@net wrote in message
news:afrgb.47150$gv5.18057@fed1read05...
...
atmosphere as 60 feet thick, if I head out in a spaceship and
accelerate
to
close to c, will I now measure my distance to Alpha Centaure to be
less
than
the 4.3 light years it is from the Earth?

This is correct. The 4.3 ly is a "rest frame" measurement for Earth.
As
your speed increases, the distance is decreased... for you. You will
in
no
circumstances get there in less than 4.3 years, of Earth time.

This is because the spaceship's clock second is worth gamma*earth clock
second. There is no decrease in distance. What this mean is that you
can't
compare spaceship's clock reading with the earth clock reading directly.
In
order to do the comparison you must transform the spaceship's clock
reading
to the earth frame.

Ken, this ("There is no decrease in distance") is not true.

It is true.

In the
spaceship's frame, the distance is foreshortened, and the spaceship's
determination of its speed (without dereferencing to your "absolute"
frame)
will be affected as well.

You can say that the distance is foreshortened.but it is more correct to say
that the spaceship has a higher state of absolute motion than the earth
toward the star. Notice that both interpretations give the same end result.
The spaceship clock second has a higher absolute time content and thus it
takes a shorter spaceship clock time to reach the star and you (SR)
interpret this as distance foreshortening.

Only in a single frame is their no change in
distance. When someone is in one frame then changes speed to another
frame, the distance will be seen to change _in_that_frame_. Concur?

No....distance never change. What is changing is that a spaceship clock
second has a higher absolute time content than the earth clock second.

Ken Seto

How about considering that the distance and the time are independent of
speculations about it. Altered measuring instruments do not equate to
changes in time or distance, only to changes in the measure of them.
There is quite a difference between the two. Processes in nature occur
completely independently of our perceptions of them, and thus only those
mathematical laws that are devoid of frame dependent variables are valid
descriptions, IOW, all others are contradictory to the PoR.

Richard Perry

http://www.cswnet.com/~rper
Electromagnetism: First Principles

"Richard" wrote in message
...


kenseto wrote:

(formerly)" dlzc1.cox@net wrote in message
news:l0Agb.47226$gv5.34510@fed1read05...
Dear kenseto:

"kenseto" wrote in message
...

(formerly)" dlzc1.cox@net wrote in message
news:afrgb.47150$gv5.18057@fed1read05...
...
atmosphere as 60 feet thick, if I head out in a spaceship and
accelerate
to
close to c, will I now measure my distance to Alpha Centaure to
be
less
than
the 4.3 light years it is from the Earth?

This is correct. The 4.3 ly is a "rest frame" measurement for
Earth.
As
your speed increases, the distance is decreased... for you. You
will
in
no
circumstances get there in less than 4.3 years, of Earth time.

This is because the spaceship's clock second is worth gamma*earth
clock
second. There is no decrease in distance. What this mean is that you
can't
compare spaceship's clock reading with the earth clock reading
directly.
In
order to do the comparison you must transform the spaceship's clock
reading
to the earth frame.

Ken, this ("There is no decrease in distance") is not true.

It is true.

In the
spaceship's frame, the distance is foreshortened, and the spaceship's
determination of its speed (without dereferencing to your "absolute"
frame)
will be affected as well.

You can say that the distance is foreshortened.but it is more correct to
say
that the spaceship has a higher state of absolute motion than the earth
toward the star. Notice that both interpretations give the same end
result.
The spaceship clock second has a higher absolute time content and thus
it
takes a shorter spaceship clock time to reach the star and you (SR)
interpret this as distance foreshortening.

Only in a single frame is their no change in
distance. When someone is in one frame then changes speed to another
frame, the distance will be seen to change _in_that_frame_. Concur?

No....distance never change. What is changing is that a spaceship clock
second has a higher absolute time content than the earth clock second.

Ken Seto

How about considering that the distance and the time are independent of
speculations about it.

Distance never change.The rate of a clock changes with acceleration.
Acceleration, in turn changes the state of absolute motion of the clock and
the rate of a clock is dependent on the state of absolute motion of the
clock. When you measure distance with a clock the distance is shortened
according to the state of absolute motion of the clock.

Altered measuring instruments do not equate to
changes in time or distance, only to changes in the measure of them.

The rate of a clock is dependent on the state of absolute motion of the
clock. The light path length of a rod is dependent on the state of absolute
motion of the rod. This explains why the speed of light is a constant math
ratio in all frames as follows:
Light path length of rod (299,792,458m)/the absolute time content for a
clock second co-moving with the rod.


Ken Seto

The stronger the gravity the slower the pendulum. The greater the speed
the slower of the pendulum. Want to see what it is like going at"C"
enter a blackhole. Want to have the same force as entering a blackhole
accelerate to light speed. So it is written Bert

On Mon, 6 Oct 2003 17:39:22 -0700, \(formerly\)" dlzc1.cox@net
wrote:

Dear Thomas Jones:

"Thomas Jones" wrote in message
...
"Paul B. Andersen" wrote in message
...

"Richard" skrev i melding




I am in no way arguing for or against any of the points made. But, by
"longer" he meant stayed on the bike for a longer amount of time not for
a
longer distance. So if you equate the time a muon decays to the time the
boy falls off the bike his analogy does make some sense. May not be
correct
but it doesn't have the flaw you just blasted him for.

Richard's argument invokes large spinning gyroscopes for stability, namely
bicycle wheels. The same arguments he makes do not apply to skiers,
surfers, or anything else. In all cases, knowledge of what you are doing
provides the longer ride. He will not agree that solar-generated muons are
smarter or more experienced though, and *that* is why they survive longer.

The muon would not survive in the numbers that they do, even if they just
travelled faster but still less than c. They would have to be superlumenal
(as HenriWilson would believe) in order to make the trip. Unless time
dilation were, in fact, a factor in comparing the two realities. The muon
sees the Earth's atmosphere as some 60 feet thick, and we see its lifetime
dilated. That is spacetime. It is symmetric that way.

David A. Smith


Where you brainwashed unfortunates go wrong is that the increased distance
travelled is measured in the ground frame not that of the moving muon. So is
the supposed muon lifetime.

The muon's own frame doesn't play any part in this experiment.

In the ground frame, the muons are assumed to decay at the same rate as they do
in other circumstances. They are observed to reach the ground although that is
regarded as impossible according to SR.

Obvious answer: many of the muons are initially traveling at c.

This is quite possible since they are produced in collisions rather than in
accelerating fields.

SR is plainly wrong - but some of us knew that anyway.

Henri Wilson.

"Whenever a relativist moves, half the universe shrinks and the other half expands".

See my animations at:
http://www.users.bigpond.com/HeWn/index.htm

On Mon, 6 Oct 2003 21:25:39 -0700, \(formerly\)" dlzc1.cox@net
wrote:

Dear Thomas Jones:

"Thomas Jones" wrote in message

for."

I disagree, but that is neither here nor there.

...
And there is one question that I've always wondered about and your last
sentence makes this a good place to ask it. If the muon sees the Earth's
atmosphere as 60 feet thick, if I head out in a spaceship and accelerate
to
close to c, will I now measure my distance to Alpha Centaure to be less
than
the 4.3 light years it is from the Earth?

This is correct. The 4.3 ly is a "rest frame" measurement for Earth. As
your speed increases, the distance is decreased... for you. You will in no
circumstances get there in less than 4.3 years, of Earth time.

Crap.

Clock proper rates do not change with velocity. Ask Paul Anderson.
The on-board clock and the Earth clocks remain exactly in synch all the way.

If you keep on accelerating, you can get to Alpha Centaure in any time you
like. It's just a matter of making the right rocket engines.

Time and length contractions are just an illusion. The gamma term is wrong
anyway.




David A. Smith



Henri Wilson.

"Whenever a relativist moves, half the universe shrinks and the other half expands".

See my animations at:
http://www.users.bigpond.com/HeWn/index.htm

On Tue, 7 Oct 2003 17:13:21 -0400, "kenseto" wrote:


"Richard" wrote in message
...


kenseto wrote:


How about considering that the distance and the time are independent of
speculations about it.

Distance never change.The rate of a clock changes with acceleration.
Acceleration, in turn changes the state of absolute motion of the clock and
the rate of a clock is dependent on the state of absolute motion of the
clock. When you measure distance with a clock the distance is shortened
according to the state of absolute motion of the clock.

Altered measuring instruments do not equate to
changes in time or distance, only to changes in the measure of them.

The rate of a clock is dependent on the state of absolute motion of the
clock. The light path length of a rod is dependent on the state of absolute
motion of the rod. This explains why the speed of light is a constant math
ratio in all frames as follows:
Light path length of rod (299,792,458m)/the absolute time content for a
clock second co-moving with the rod.


Ken Seto


In physics, the only thing more amusing than SR is aether theory. - but at
least the latter is 'logically' sound.


Henri Wilson.

"Whenever a relativist moves, half the universe shrinks and the other half expands".

See my animations at:
http://www.users.bigpond.com/HeWn/index.htm

On Tue, 07 Oct 2003 12:13:56 -0500, EjP wrote:

Thomas Jones wrote:
(formerly)" dlzc1.cox@net wrote in message


superlumenal

(as HenriWilson would believe) in order to make the trip. Unless time
dilation were, in fact, a factor in comparing the two realities. The muon
sees the Earth's atmosphere as some 60 feet thick, and we see its lifetime
dilated. That is spacetime. It is symmetric that way.

David A. Smith





"May not be correct but it doesn't have the flaw you just blasted him for."


Re-read my post. I said I'm not arguing for or against either point. Just
making the observation that his analogy did not suffer from the flaw for
which he got blasted, namely that he was saying it moves farther during its
life because its moving faster. That was not his analogy.

And there is one question that I've always wondered about and your last
sentence makes this a good place to ask it. If the muon sees the Earth's
atmosphere as 60 feet thick, if I head out in a spaceship and accelerate to
close to c, will I now measure my distance to Alpha Centaure to be less than
the 4.3 light years it is from the Earth? If not, why does this happen to
the muon and wouldn't happen to me. If it does, how will I be able to
explain this sudden decrease in the distance to a known place? These are
serious questions, it's something I've always wondered.



The length contraction is trivially required by the "relative" part of
special relativity. Taking your spaceship example, relativity says that
in the "rest" frame of the Sun and Alpha Centauri, the ship's clocks are
and so less time will pass on the trip. However, the rest frame of the
ship is also a valid frame of reference. If the ship is standing
still and the solar systems are moving by, why would the ship take less
time? The answer is that in this frame the distance between the star
systems is shorter when measured in this frame.

As a practical matter, if we ever do get around to flying to
Alpha Centauri, the occupants of the ship will most likely say
"The trip took less time on our clocks because we just spent
a jillion dollars building a ship that could travel near the
speed of light and our clocks ticked more slowly", rather
than "look, the distance to Alpha Centauri just got
shorter", even though both view are valid in the
strictly pedentic sense. There's also a little matter
of the ship accelerating, but that's a separate issue.

-E



Clock rates do not change with velocity.

In fact the spaceship never has a 'velocity'. It is always at rest. It has a
velocity wrt the Earth but, so what? It has a different velocity wrt every
other object in space. WRT some, it is moving faster, to others slower.
Obviously its clock cannot have both increased and decreased its physical rate
simultaneously.

The spaceship is, however, in a state of acceleration. In real life, that can
affect its clocks very slightly. Perfect clocks would not be affected by
aceleration.

The spaceship clock and the Earth clock differ only by the integration of this
rate change over time.

It is not relativistic effect at all. Just a mechanical imperfection of the
clock.



Henri Wilson.

"Whenever a relativist moves, half the universe shrinks and the other half expands".

See my animations at:
http://www.users.bigpond.com/HeWn/index.htm

Dear HenriWilson:

"HenriWilson" wrote in message
...
On Mon, 6 Oct 2003 17:39:22 -0700, \(formerly\)"
dlzc1.cox@net
wrote:

Dear Thomas Jones:

"Thomas Jones" wrote in message
...
"Paul B. Andersen" wrote in message
...

"Richard" skrev i melding




I am in no way arguing for or against any of the points made. But,
by
"longer" he meant stayed on the bike for a longer amount of time not
for
a
longer distance. So if you equate the time a muon decays to the time
the
boy falls off the bike his analogy does make some sense. May not be
correct
but it doesn't have the flaw you just blasted him for.

Richard's argument invokes large spinning gyroscopes for stability,
namely
bicycle wheels. The same arguments he makes do not apply to skiers,
surfers, or anything else. In all cases, knowledge of what you are
doing
provides the longer ride. He will not agree that solar-generated muons
are
smarter or more experienced though, and *that* is why they survive
longer.

The muon would not survive in the numbers that they do, even if they
just
travelled faster but still less than c. They would have to be
superlumenal
(as HenriWilson would believe) in order to make the trip. Unless time
dilation were, in fact, a factor in comparing the two realities. The
muon
sees the Earth's atmosphere as some 60 feet thick, and we see its
lifetime
dilated. That is spacetime. It is symmetric that way.

Where you brainwashed unfortunates go wrong is that the increased
distance
travelled is measured in the ground frame not that of the moving muon. So
is
the supposed muon lifetime.

There is no "increased distance travelled", HenriWilson. There is
increased lifetime in the Earth frame, and decreased distance travelled in
the muon frame.

The muon's own frame doesn't play any part in this experiment.

Relativity applies to both frames. But you knew that.

In the ground frame, the muons are assumed to decay at the same rate as
they do
in other circumstances. They are observed to reach the ground although
that is
regarded as impossible according to SR.

Obvious answer: many of the muons are initially traveling at c.

Your opinion is no surprise here. Too bad they are, en masse, travelling
at less than c, no matter what altitude they are measured at.

David A. Smith

Dear HenriWilson:

"HenriWilson" wrote in message
...
On Mon, 6 Oct 2003 21:25:39 -0700, \(formerly\)"
dlzc1.cox@net
wrote:
....
This is correct. The 4.3 ly is a "rest frame" measurement for Earth.
As
your speed increases, the distance is decreased... for you. You will in
no
circumstances get there in less than 4.3 years, of Earth time.

Crap.

Clock proper rates do not change with velocity. Ask Paul Anderson.
The on-board clock and the Earth clocks remain exactly in synch all the
way.

If you keep on accelerating, you can get to Alpha Centaure in any time
you
like. It's just a matter of making the right rocket engines.

Time and length contractions are just an illusion. The gamma term is
wrong
anyway.

Irrespective of your personal beliefs, hydrogen atoms striking your ship at
even 0.1c will create more radiation than you can easily shield (since it
will convert your shielding material over a period of "years").

Your corpse will arrive at Alpha Centauri, and it will do so in more than
4.3 years.

David A. Smith