Oct 12, 2018 11:22 PM
Special Relativity follows..
The purpose of this post isn't really to discuss the muon experiment - I'm really trying to clarify the nature of spacetime as a geometry. In the following analysis c appears as a constant in the same manner as there is a constant used to convert litres to gallons. It would have been surprising if the French had randomly chosen the natural unit for measuring liquids (the gallon) and so we have a conversion factor between one and the other. In the same way we have c as a conversion factor between time and distance.
Basically we have muons created in the upper atmosphere heading down at close to the speed of light (about 0.98 of the speed of light). They are counted at two points one 10km above the other. Muons decay very rapidly so the difference between the two counts is a good indication of the elapsed time between the two counting stations.
In the simplest notation - in one dimension of length and one of time, where s is the spacetime interval, c is a constant (found by experiment), x is a distance, t is time and v is the muon velocity in the Earth frame
s²=x²-c²t² .... we can derive this later if there is any interest
Compare this with Pythagoras' Theorum
r²=p²+q² <- the spacetime interval is the same geometryish thing but with distance and time (and a conversion factor).
In the nature of special relativity the time in the Earth frame isn't going to be the same as the time in the muon frame so I'll put a ' after times and distances in the Earth frame. So t' is Earth time and t is muon time.
In the muon frame elapsed time =t we have
s²=-c²t² < note no distance here*, the first and second counts are in the same place as far as the muon is concerned.
In the Earth frame t' and x' (where x' ~10km) we have
s²=x'²-c²t'²
since s² is the same in both frames
c²t²=x'²-c²t'²
we know x' = vt' from Newton and the definition of velocity
so
-c²t²=(vt')²-c²t'²
c²t²=c²t'²-(vt')²
t= t'√(1-v²/c²)
so t=t_muon is less than t'=t_Earthclock
The t= t'√(1-v²/c²) should be the same as wiki or Einstein or anyone else gets.
There may be mistakes and/or typos or I may simply be wrong.
For more details and another analysis see:-
http://hyperphysics.phy-astr.gsu.edu/hba.../muon.html
I'd like to emphasise that I've used the speed of light (as a constant) to predict the result of an experiment that doesn't involve light. Just time, distance, velocity and muons.
The s²=x²-c²t² might seem assplucked - deriving it takes a few more lines of high school algebra.
* - it took me years to work out why.
The purpose of this post isn't really to discuss the muon experiment - I'm really trying to clarify the nature of spacetime as a geometry. In the following analysis c appears as a constant in the same manner as there is a constant used to convert litres to gallons. It would have been surprising if the French had randomly chosen the natural unit for measuring liquids (the gallon) and so we have a conversion factor between one and the other. In the same way we have c as a conversion factor between time and distance.
Basically we have muons created in the upper atmosphere heading down at close to the speed of light (about 0.98 of the speed of light). They are counted at two points one 10km above the other. Muons decay very rapidly so the difference between the two counts is a good indication of the elapsed time between the two counting stations.
In the simplest notation - in one dimension of length and one of time, where s is the spacetime interval, c is a constant (found by experiment), x is a distance, t is time and v is the muon velocity in the Earth frame
s²=x²-c²t² .... we can derive this later if there is any interest
Compare this with Pythagoras' Theorum
r²=p²+q² <- the spacetime interval is the same geometryish thing but with distance and time (and a conversion factor).
In the nature of special relativity the time in the Earth frame isn't going to be the same as the time in the muon frame so I'll put a ' after times and distances in the Earth frame. So t' is Earth time and t is muon time.
In the muon frame elapsed time =t we have
s²=-c²t² < note no distance here*, the first and second counts are in the same place as far as the muon is concerned.
In the Earth frame t' and x' (where x' ~10km) we have
s²=x'²-c²t'²
since s² is the same in both frames
c²t²=x'²-c²t'²
we know x' = vt' from Newton and the definition of velocity
so
-c²t²=(vt')²-c²t'²
c²t²=c²t'²-(vt')²
t= t'√(1-v²/c²)
so t=t_muon is less than t'=t_Earthclock
The t= t'√(1-v²/c²) should be the same as wiki or Einstein or anyone else gets.
There may be mistakes and/or typos or I may simply be wrong.
For more details and another analysis see:-
http://hyperphysics.phy-astr.gsu.edu/hba.../muon.html
I'd like to emphasise that I've used the speed of light (as a constant) to predict the result of an experiment that doesn't involve light. Just time, distance, velocity and muons.
The s²=x²-c²t² might seem assplucked - deriving it takes a few more lines of high school algebra.
* - it took me years to work out why.