On May 13, 5:51*am, Albertito wrote:
On May 13, 11:24 am, PD wrote:
[snipped]



Of course you can. You didn't read what I wrote. You time a muon in
flight the same way you could a car on the highway: you time its
crossings on successive, spaced gates. (Note you don't have to follow
the car all the way from the garage to do so.) You make a scintillator
paddle triplet, with each paddle separated by some appreciable
difference (say, 20m), and then the signals from each of the
scintillators to an o'scope or a TDC by an equal length cable. 20 m
would be about 60 ns if the speed were as high as c, trivial to
measure by either of the devices mentioned. The presence of the MIP
signal in all three paddles assures that you are seeing the passage of
the muon. You measure the distance, and you measure the time it takes
the muon to cover that distance. You have a direct measurement of the
velocity. It does not exceed c. *Measured*.

It might do you good to go into the laboratory to see for yourself,
rather than just jacking around with piddly little ideas.

PD

That only proves those experiments are biased by
relativistic formulas.

Uh-huh. Congratulations, you've just fell face-first into Swamp Henri,
where one emerges from the muck with the belief that if simple
experiments show that one's ideas are hooey, then there is obviously
something wrong with the experiments and a cover-up to boot.

Time dilation is not an observable
by definition,

What on Earth gave you that ridiculous idea?

it is just an interpretation to explain an
observable. You can't label single muons to keep track
of them,

Why not? I produce them one at a time and I watch them leave little
energy deposits in plates of PVT I put in their path. It's not rocket
science.

the same way you can't label single electrons.
Perform a double-slit experiment to keep track of a single
muon.

This isn't a double slit. There is no interference. Perhaps you don't
understand the simple word-picture I used to describe the apparatus.
Perhaps you are groping.

Can you clarify the path a muon can follow in that
double-slit experiment? If you can't clarify a muon path
why are so sure a muon has travelled a distance R in time
t'? Didn't you know that intermediate measurements alter
the experimental results.

To order hbar, sure. That isn't a whole lot compared to 60ns over 20m.
Want to check your math again? Or are you happy just clamping your
hands over your ears and hollering, "NANANANA! I can't HEAR you!!!"

PD

On May 13, 3:02*pm, Albertito wrote:
On May 13, 6:59 pm, Igor wrote:





On May 13, 1:52 pm, Albertito wrote:

On May 13, 6:26 pm, Igor wrote:

On May 11, 2:26 pm, Albertito wrote:

On May 11, 7:10 pm, Igor wrote:

On May 9, 7:44 am, Albertito wrote:

Let us consider the simple case of addition of velocities
along a straight line. The incompleteness of SR resides
in the fact that a speed v can't currently be experimentally
distinguished from its rapidity r = arctanh(v/c), for values
of that beta = v/c *below the third-order term approximation.
The power series expansion of r = arctanh(v/c) is

* * * * * r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ...

Provide references of any experimental test of SR, showing
that the rapidity r can be distinguished from its beta v/c, beyond
its second-order approximation. Prove at least that the third-order
term v*3/3c^3 lives outside the error bars. Since, we can't *still
perform such accurate experimental tests, we must conclude
the addition of velocities still remains within the euclidean sum
of vectors

* * * * * * * *w = u + v.

That sum can't still be experimentally distinguished from the sum
of rapidities

* * * * * * *arctanh(w/c) = arctanh(u/c) + arctanh(v/c).

In addition, we must also conclude that the relativistic Doppler

* * * * * * * *f' = Exp(-r) f ,
* * * * * * * *where r = arctanh(v/c),

can't still be experimentally distinguished from this one

* * * * * * * *f' = Exp(-v/c) f

Uhh, you do understand that v/c = tanh r don't you? *I didn't think
so.

Really?, didn't you?
Didn't *you realize that both v/c and its rapidity,
r = arctanh(v/c), currently are experimentally
indistinguishable?

You must understand that if I know one of them, I know both of them.
So what is your point?

The point is that you can't know both of them.
How do you know that v/c is the beta you
have measured for a moving body, but it is
not its rapidity r = arctanh(v/c). Error bars
on (v/c ± d) do not allow you to discriminate
v/c from its rapidity beyond its second-order
term.

Now you're just being either silly, stupid, or both. *I don't need to
measure both of them. *Or maybe you haven't noticed that they are
functions of each other?

It's unnecessary to blurt out curses. I have not
said you need to measure both of them,
I've said that if you measure v/c, its error bars
make its rapidity useless for relativistic corrections,
for example relativistic addition of velocities.

You like to spout gibberish.