Unraveling the Mysteries of Muon Experiments

  • Thread starter Thread starter psmitty
  • Start date Start date
  • Tags Tags
    Experiments Muon
psmitty
Messages
16
Reaction score
0
Hi guys, just wanted to ask a question related to muon
experiments (and all other which can be presented this
way) to get some things clearer.

Ok, here goes:

We have lots of muons traveling towards Earth. Their
mean lifetime, measured in lab conditions (at rest) is
2.2µs. Their concentrations (flux) have been measured
at different altitudes, and their speed (0.99c) has
been measured near the Earth surface. By comparing
their concentrations at an altitude of 15km and at sea
level, it has been shown that many more survive than
expected, considering their speed and their mean
lifetime.

SR calculation follows:

Note: I took delta_x (change of height) to be negative,
because it's decreasing, but this is a matter of choice.

Speed of light is c=299792458m/s
Speed of muon is -0.99c = -296794533,4 m/s
Lorentz factor is then: gamma = 7.08881205

EARTH's frame
delta_x: -15km = -15000m (height decreased by 15km)
delta_t: delta_x/v = 50.54µs
It takes 50µs for the muon to travel 15km. v=0.99c.

MUON's frame
delta_t': gamma*(delta_t-(v*delta_x)/(c*c)) = 7.13µs
delta_x': gamma*(delta_x-v*delta_t) = 0m (in this frame, muon is stationary)
distance to Earth at t'=0: delta_x/gamma = -2116m
It takes 7µs for the muon to travel 2km. v=0.99c.

Ok, so far everyhing is as Relativity predicts.

Now the strange part.

What if we started with the fact that it takes 7µs for
the muon to travel 2km at that speed and want to find
out delta_t in Earth's frame? Let's say that muon is
stationary and Earth is traveling towards the muon.

MUON's frame
delta_x: -2.116km = -2116m
delta_t: delta_x/v = 7.13µs

Now we are in muon's frame, and want to find out
the time and distance Earth needs to travel in Earth's
frame. We should get 50µs, distance of 0m, but
we should be able to calculate muon's distance also.

Using exactly the same reasoning as when we started,
we get:

EARTH's frame
delta_t': gamma*(delta_t-(v*delta_x)/(c*c)) = 1.01µs
delta_x': gamma*(delta_x-v*delta_t) = 0m (in this frame, Earth is stationary)
distance to muon at t'=0: delta_x/gamma = -298.5m

Shouldn't we be able to get our starting results (50µs, 0m, -15km)?
 
Physics news on Phys.org
Ok, nevermind, I found the answer elsewhere. I made a mistake in my calculations:

EARTH's frame
delta_x: -15km = -15000m (height decreased by 15km)
delta_t: delta_x/v = 50.54µs
It takes 50µs for the muon to travel 15km. v=0.99c.

MUON's frame
delta_t': gamma*(delta_t-(v*delta_x)/(c*c)) = 7.13µs
delta_x': gamma*(delta_x-v*delta_t) = 0m (in this frame, muon is stationary)
distance to Earth at t'=0: delta_x/gamma = -2116m
It takes 7µs for the muon to travel 2km. v=0.99c.

But distance to Earth at t'=0 (2km) is not the total distance between Earth and muon, because (which gets clearer from the Minkowski diagram) Earth in muon's frame starts to travel before t'=0.

To get the distance to Earth in muon's frame, I should have used delta_x=0 (because Earth is not moving in its frame), and then delta_x' would be 106km (which is the total distance to Earth in muons frame).

2km is the distance at t'=0, which Earth travels for the last 7µs of its trip.

Thanks anyway! Cheers!
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
Thread 'Relativity of simultaneity in actuality'
I’m attaching two figures from the book, Basic concepts in relativity and QT, by Resnick and Halliday. They are describing the relativity of simultaneity from a theoretical pov, which I understand. Basically, the lightning strikes at AA’ and BB’ can be deemed simultaneous either in frame S, in which case they will not be simultaneous in frame S’, and vice versa. Only in one of the frames are the two events simultaneous, but not in both, and this claim of simultaneity can be done by either of...
Back
Top