## Muons and Relativity
A muon is a type of subatomic particle that is created in particle
collisions. More properly called a “mu-meson”, it has a
charge equal to that of an electron and a mass 206 times
heavier. As such it is often thought of as a heavy electron
[1]. A box containing muons coming down from the sky is
approaching a similar box on the ground. An equal number of muons
were boxed at the same time.
The flashbulb ensures both groups were boxed simultaneously
(this is also an accepted method in SR of
synchronizing two clocks at a distance [3]).
The muons boxed at 2000ft were going at 0.995 - In the top frame, a muon is coming from the left at velocity
**v**and is about to enter the iron. - In the middle frame, the muon is passing through the iron and
experiencing a steady backward force of
**F**. Because of this, its velocity steadily decreases. - In the bottom frame, the muon has come to a halt after travelling
distance
**D**.
If the muon comes to a stop, this means the original kinetic energy of the muon has been fully converted. As you probably know, kinetic energy is given by the equation: E_{(kinetic)} = ½ mass * velocity^{2}By setting these energy equations equal we can determine the stopping distance from a known initial velocity. Alternatively, if we know the stopping distance, we can instead determine the initial velocity. I.e. we can determine how fast the muon was going when it hit the iron. And this is what we are trying to figure out in this situation. The ½ m v^{2} equation however is for classical mechanics.
There is another equation that is used in relativity. It is
of the form:E_{(kinetic)} = E_{relativistic} – E_{rest}= E – E_{0} = m*c^{2} – m_{0}*c^{2}Where m is the rest mass and _{0}m is the relativistic mass. m
is equal to m/sqrt(1-_{0}v/^{2}c),
i.e. the rest mass multiplied by the Lorentz Transform.^{2}What is the difference between these two functions? For low velocities there is basically no difference. But for high velocities, the velocity calculated by the classical formula can allow v>c. Whereas the relativistic formula ensures
v<c.The above is simplified. In reality the force experienced won’t be constant but a complicated function of velocity. The actual function used is called the Bethe Formula. A summary is here: en.wikipedia.org/wiki/Bethe_formulaNotice there are two versions: a non-relativistic one for low velocities (eqn. 2), and a relativistic one for high velocities (eqn. 1). The relativistic one contains the Lorentz Transform. So which equation do you think they use? If you are a believer in relativity you will use the relativistic equation, correct? Otherwise you will get a result that is inconsistent with relativity, and therefore not considered a proper test of relativity. But by using the relativistic energy equation you get different results for v. And you guarantee v<c. Whereas using
the classical equation can give v>c, especially for high-energy
situations. But this possibility is immediately rejected
because everyone “knows” it’s not possible. In fact, it is
usually not even considered.So the experiments for muon decay unknowingly end up with the sqrt(1- v/^{2}c) function built into the outcome. And thus, no
surprise, “confirms” the function to be true!^{2}
## Direct measurement of speed
Muon speeds can also be directly measured by timing how long it takes
them to pass between two locations. Experiments that employ
this method report that the maximum speed recorded is always less than
light. On the face of it this would appear good evidence that
muons, and presumably other particles, are restricted by a light-speed
limit.
## Time dilation or Length contraction?
Returning to the Frisch & Smith video, there is something else of
interest. In the video, time dilation calculations are done to
explain why muons coming down through the atmosphere are decaying slowly
and why they outlive stationary muons at ground level. But as
pointed out earlier, from the view of the muons coming down, they are
motionless and the ground-muons are moving up at exactly the same
speed. So why not do the calculations the other way around?
[1] There are also less-common positive-charged muons of equal mass,
making them “heavy positrons”. |

Copyright © 2017
Bernard Burchell, all rights reserved.