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Muon velocities and timings

October 4, 2011

This post is to collect into one place a response to the teeth grindingly dense posts of Emil over on SciForums here. He seems to be suffering a similar issue to Sylwester in his inability to grasp what experiments actually measure. The link Emil provides gives all the information needed to understand but when did facts ever get in the way of a crank?

The Scenario

An experiment measures muon counts are two different places, the top of a mountain and the bottom of said mountain. The muons are produced in the upper atmosphere in large amounts, which then move down towards the Earth and some of this ‘rain of muons’ is detected by the counters. Unfortunately muons are not stable and they decay so the rain gets thinner and thinner as it gets closer the surface. We know the half life of muons in their rest frame from other experiments, t_{\tfrac{1}{2}} = 2.25 \times 10^{-6} seconds.

The Measurements

So what’s measured? The count at the top, C_{t} = 550, the count at the bottom, C_{b}=422 and the distance between the counters, L=1219 metres. Nothing else. The counts C_{t} and C_{b} are not perfect counts of all the muons. Instead they are a measure of the density of muons in the region around the detectors. The amount of muons counted by the detector is proportional to the density of the muons in that region. The detectors are like rain gauges, they don’t capture everything but they allow you to compare relative amounts of rain, which is ultimately what we are interested in.

Inferring Velocities

So how does this experiment tell us the velocity of the muons? Well, suppose the muons are made at the top of the mountain (really it’s much higher but bear with me). If the muons weren’t moving they’d never leave the top detector’s vicinity and so the count at the bottom would be C_{b}=0. Conversely if they moved infinitely fast none of the muons would have time to decay so there density of the muon ‘rain’ would be unchanged, giving C_{t}=C_{b}. The truth is somewhere in the middle. So what’s important? The time it takes the muon rain to move the distance from the top to the bottom and the rate at which they decay. The faster they decay the less will reach the bottom but the faster they move the more will reach the bottom. So the question is “How fast does a moving muon decay?”.

Decay Rates and Velocities

We have, from other experiments where we trap muons, that muons at rest (so v=0) will decay with a half life of t_{\tfrac{1}{2}} = 2.25 \times 10^{-6} seconds. Rather than assume velocity has no effect let’s make things as general as possible and include velocity dependency (which may turn out to be trivial), so let t_{\tfrac{1}{2}}(v) be the half life of a muon moving at speed v  such that t_{\tfrac{1}{2}}(0) = t_{\tfrac{1}{2}} = 2.25 \times 10^{-6}. For future convenience we define \tau(v) = \frac{1}{\ln 2}t_{\tfrac{1}{2}}.

Unstable sets of particles decay exponentially, with the half life being involved in the exponent. Making sure to keep the velocity dependency manifest this gives

N_{t} = N_{0}\,2^{-\tfrac{t}{t_{\tfrac{1}{2}}(v)}} = N_{0}\,\exp \left( -\frac{t}{\tau(v)} \right)

The reason for defining \tau(v) now becomes clear.

Some Simple Kinematics

The top detector measures the value of N_{0} = C_{t}. At the bottom of the mountain the muon rain is thinner due to additional decay and we get the count C_{b}. The muons at the bottom are presumed to be older than those at the top, by an amount T, as they have moved further from where they were produced. Therefore we have a relationship between the counts at the top and bottom of the mount as a function of the decay rate and the approximate differences in muon ages,

C_{b} = C_{t}\,\exp \left( -\frac{T}{\tau(v)} \right)

Rearranging this we can express T in terms of everything else,

T = -\tau(v)\,\ln \left( \frac{C_{b}}{C_{t}} \right) = \tau(v)\,\ln \left( \frac{C_{t}}{C_{b}} \right)

Given the time the muons take and the distance they travel we can compute their velocity,

v = \frac{L}{T} = \frac{L}{\tau(v)}\ln \left( \frac{C_{t}}{C_{b}} \right)^{-1}

So What’s The Velocity?

If \tau(v) depends on the velocity then both the left and right hand sides of the previous expression have velocity dependence. There’s a number of ways to sort this out.

  1. Assume \tau(v) = \tau is a constant
  2. Assume the left hand side is the speed of light and set \tau(v) = \gamma(v) \tau, ie use time dilation
  3. Solve numerically for a particular case

Case 1

This is how detractors of special relativity work, they say that time dilation doesn’t exist and so there is no velocity dependency. Thus we obtain

v = \frac{L}{\tau}\ln \left( \frac{C_{t}}{C_{b}} \right)^{-1} \equiv \alpha c .

We have defined \alpha for future convenience.

Case 2

This is an approximation for high energy relativistic muons, as their true speed is close to the speed of light but we keep the velocity in \gamma as unknown. The equation then rearranges to

\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} = \frac{1}{c}\left( \frac{L}{\tau}\ln \left( \frac{C_{t}}{C_{b}}\right)^{-1} \right) = \alpha .

This shows why we made our choice of notation change.

Case 3

This is the full way of doing it in general. Fortunately it can be rearranged so that we obtain the same large factor as the previous cases and for convenience we use \beta = \tfrac{v}{c},

\frac{\beta}{\sqrt{1-\beta^{2}}} = \frac{1}{c} \left( \frac{L}{\tau}\ln \left( \frac{C_{t}}{C_{b}} \right)^{-1} \right) = \alpha .

The right hand side we obtain from the previous case again! As a result we can obtain an analytic expression for \beta,

\beta = \frac{\alpha}{\sqrt{1+\alpha^{2}}} .

All that remains now is to put in the numbers, so let’s do that for all three cases!

Putting In The Numbers

  1. Solving directly gives v = 2.045 \times 10^{9} metres per second, which is approximately 6.82c, so \alpha = 6.82.
  2. Since we know the large bracket gives 6.82c we just solve \gamma = \alpha = 6.82, which gives v = 0.9892c.
  3. Putting in the value of \alpha gives \beta = 0.9894 so v = 0.9894c.


The experiment does not uniquely determine the velocity of the muons. This is because the velocity is not directly measured, it is inferred from the other data using a model and different models give different results. Specifically the results depend on how the model things the muon half life behaves as a function of velocity. Other experiments are needed to determine this and fortunately they have been done, as muon  storage rings form a standard component in accelerators. In order to store the muons they must be kept moving very fast else they decay before they can be used. The motion of the muons is directly measurable by the accelerator and the relationship between velocity and half lifes, ie \tau(v), is found to match the predictions of relativity. In addition the conditions which produce muons in the upper atmosphere are easily replicable in accelerators and all muons produced in experiments where their velocities can be directly measured have always moved slower than light.

To claim the experiment just examined is proof muons move faster than light is false, as there are alternative explanations. However the experiment doesn’t exclude such a possibility either but in the light of other experiments such a possibility is excluded.

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