Re: The Cosmic Muons Paradox

On Sep 3, 9:32 am, blackhead <larryhar...@xxxxxxxxxxxx> wrote:
On 3 Sep, 13:42, "kens...@xxxxxxxxxx" <kens...@xxxxxxxxxx> wrote:





On Sep 2, 6:47 pm, "Inertial" <relativ...@xxxxxxxx> wrote:

<kens...@xxxxxxxxxx> wrote in message

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On Sep 1, 10:35 pm, blackhead <larryhar...@xxxxxxxxxxxx> wrote:
On 1 Sep, 13:05, "kens...@xxxxxxxxxx" <kens...@xxxxxxxxxx> wrote:

On Aug 31, 2:52 pm, blackhead <larryhar...@xxxxxxxxxxxx> wrote:

On 31 Aug, 15:29, "kens...@xxxxxxxxxx" <kens...@xxxxxxxxxx> wrote:

On Aug 31, 10:13 am, blackhead <larryhar...@xxxxxxxxxxxx> wrote:

On 31 Aug, 14:08, "kens...@xxxxxxxxxx" <kens...@xxxxxxxxxx>
wrote:

On Aug 31, 7:07 am, blackhead <larryhar...@xxxxxxxxxxxx> wrote:

On 29 Aug, 21:55, "kens...@xxxxxxxxxx" <kens...@xxxxxxxxxxx>
wrote:

On Aug 29, 9:47 am, blackhead <larryhar...@xxxxxxxxxxxx>
wrote:

[snipped]

According to your theory, if the the Earth observer
measures the speed
of the Earth to be v, then the muon observer measures the
speed of the
Earth to be -gamma*v, correct?

No.....gamma_muon is derive from v_e (v_e is velocity of
earth as
measured by the muon clock). So v_e cannot be gamma*v.
Gamma_muon is
used to determine the clock rate of the earth. In IRT the
muon
observer gives the following two predictions for the rate
of the earth
clock:
T_e = gamma_muon*T_muon........(1)
OR
T_e = T_muon/gamma_muon........(2)

Obviously equation (1) is the correct prediction.

Ken Seto

According to your theory, what's the relationship between the
velocity
of the Earth measured in the muon frame, and the velocity of
the muon
measured in the Earth
frame?

According to the muon observer the velocity of the earth is:
V_me = Lambda(F_mm - F_me)
V_me = veloocity of the earth as measured by the muon observer.
Lambda = wavelength of a standard light source in the muon
frame.
F_mm = Frequency of a standard light source in the muon frame.
F_me = Tranverse Doppler frequency of an identical standard
light
source in the earth frame as measured by the muon observer.

According to the earth observer the velocity of the muon is:
 V_em = Lambda(F_ee - F_em)
V_em = velocity of the muon as measured by the earth observer.
Lambda = wavelength of a standard light source in the earth
frame.
F_ee = Frequency of a standard light source in the earth frame.
F_em = Transverse Doppler frequency of an identical standard
light
source in the muon frame as measured by the earth observer.

Thanks. So if V_em was measured by the Earth observer and he took
his
clock and ruler with him into the frame of the muon, can't your
theory
predict what V_me he would measure?

I don't understand your question. The earth observer becomes the
muon
observer and he would measure V_me as ahown above.

Would he measure:

V_me = V_em?

No....why? Because 1 second on the muon clock has a longer duration
(absolute time) than 1 second on the earth clock.

V_me = gamma_me* V_em?

No....I guess that:
V_me = V_em/gamma_me

V_me = f(V_me)*V_em?

 > ?????

f(V_me) is an arbitary function of V_me. From above you seem to think
that f(V_me) = 1/gamma_me.

For SR, V_me = -V_em, so I thought you might have a similar
expression
for your theory.

This SR assertion is wrong. It assumes that a clock second used to
measure speed is a universal interval of time....In other words, the
passage of a clock second in the earth frame corresponds to the
passage of a clock second in the muon frame.

If you can't give an exact answer, can you at least
say whether V_em > V_em for all V_em?

?????????????????????
How can V_em>V_em????

Whoops, should be V_em > V_me for all V_em.

In conclusion, you have:

V_em = gamma_me*V_me, gamma_me = 1/sqrt(1 - (v_me/c)^2).

So if the limiting velocity of a muon in the Earth frame is c, then:

V_em = gamma_me*V_me = c ==> V_me < c
i.e the limiting velocity in the
muon frame is less than the limiting velocity in the Earth frame
meaning the speed of light is different in both frames. Do you know
about this, or am I wrong?

You are wrong. In IRT the speed of light is a constant math ratio in
all frames as follows:
Light path length of ruler (299.792,458 m long physically)/the
absolute time content for a clock second co-moving with the ruler.
What this mean is that there is no absolute value for the speed of
light since a clock second use to measure the speed of light is not a
universal interval of time.

So is the measured speed of light as physics defines it (with local rulers
and clocks) the same in all frames of reference?

Wrong....the one-way or two way speed of light never been measured
with local physical ruler.
The speed of light is defined to be a
constant ratio in all inertial frames and this definition is based on
a redefinition of a meter length: 1 meter = 1/299,792,458 light-
second.

The speed of light isn't defined to be a *constant* ratio in all
frames, it's used to define 1 meter.

Yes it is defined as a constant ratio in all frames.....if not then a
clock second must represent a universal interval of time.

You can still take that 1 meter
standard with the time standard to another frame and measure the
velocity of light there, forming a comparison with that in the
previous frame.

ROTFLOL.....using that standard the speed of light is: 1 light-second/
1 second. This is a circular ratio.

According to your theory, these measurements would be
different.

Sigh....how many time I have to tell you that my theory says that the
speed of light is a constant math ratio in all frames as follows:
Light path length of ruler(299,792,458 m long physically)/the absolute
time content for a clock second co-moving with the ruler.

Ken Seto




Ken Seto

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Relevant Pages

  • Re: The Cosmic Muons Paradox
    ... of the Earth to be v, then the muon observer ... Earth to be -gamma*v, correct? ... measured by the muon clock). ... of the Earth measured in the muon frame, ...
    (sci.physics.relativity)
  • Re: The Cosmic Muons Paradox
    ... Because 1 second on the muon clock has a longer duration ... > than 1 second on the earth clock. ... It assumes that a clock second used to ... > passage of a clock second in the muon frame. ...
    (sci.physics.relativity)
  • Re: The Cosmic Muons Paradox
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    (sci.physics.relativity)
  • Re: The Cosmic Muons Paradox
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    (sci.physics.relativity)
  • Re: The Cosmic Muons Paradox
    ... Earth to be -gamma*v, correct? ... measured by the muon clock). ... used to determine the clock rate of the earth. ... of the Earth measured in the muon frame, and the velocity of the muon ...
    (sci.physics.relativity)
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