So you’ve got a theory of consciousness/gravity/space-time/everything, what do you do? This is something a lot of ‘non-mainstream’ people on forums want to know, either explicitly or they would find useful.

My answer is that you should submit your work to a reputable journal for review. Posting it on a forum will not get it the attention you want. People on forums are either not in the mainstream research community or if they are they won’t have a very good view of someone ‘publishing’ their work on a forum. Posting it on a forum, deliberately avoiding a journal, reflects poorly on you because if you’re truly wanting to get your work to the attention of the mainstream journals are the best way.

This might sound like a lot of work but Rome wasn’t built in a day. If you really have knocked over relativity you’re going to have to walk people through your work in extreme detail, you have to convince the nay-sayers. This isn’t anything actual researchers aren’t expected to do. Coming up with a result might only take a few seconds of inspiration, writing it up can take months.

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 $v = 0.9894c$.

Conclusions

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.

This is for Sylwester Kornowski, after the issue of measuring $\alpha_{S}$, the strong coupling constant, came up in yet another thread of his over on SciForums. This not being the first time he and I have been down this particular road I’m writing it here for future convenience.

Sylwester makes, among his repetitive ramblings, two regular claims, that the Standard Model (SM) is a farce which will very soon be over turned, its proponents shamed and that he is able to predict the value of $\alpha_{S}$ with incredible accuracy and predict it’s running better than the SM. Let’s consider what that means for a moment, namely what precisely $\alpha_{S}$ is. It is the strong force’s version of the fine structure constant $\alpha$ in electromagnetism.

In quantum electrodynamics (QED) the photon gauge field $A_{\mu}$ couples to a charged field, say the electron $\psi$ , by a term of the form $e\bar{\psi}\gamma^{\mu}A_{\mu}\psi$. If $e=0$ there is no coupling and the electron is not charged under electromagnetism. The charge is not dimensionless but its possible to construct a dimensionless quantity which depends on it, namely the fine structure constant $\alpha = \frac{e^{2}}{4\pi \epsilon_{0}}$. When a quantum field theorist wants to answer a question like “What’s the scattering behaviour of bouncing electrons off one another?” they will do plenty of calculations involving Feynman diagrams and obtain an answer which is a function of $\alpha$. This means he (or she) has a formula which will relate physically measurable things like momentum and angles and particle counts to things which are not directly measurable, namely the value of $\alpha$. It’s by this method which $\alpha$ can be calculated. For obvious reasons all electromagnetic processes have a dependency on $\alpha$, so by doing experiments involving charged particles we can obtain QED based predictions for the value of $\alpha$ in many different ways. If the values from different experiments didn’t agree we’d know there’s something wrong with QED.

Now the important point here is that the value we obtain for $\alpha$ is dependent on the model we used. Suppose we thought there wasn’t just an electron, muon and tau but a fourth charged lepton more massive than the tau. In some experiments it would contribute to the behaviour of the particles and when we work through the predictions we’d end up getting a different value for $\alpha$. It’s precisely things like this which allowed us to discover things like the W/Z bosons and the top/bottom quarks in the first place, as they contribute to the processes dependent on all of the coupling forces, $\alpha$, $\alpha_{S}$ and $G_{F}$ (the electroweak Fermi constant). So processes which involve quarks, gluons, anything which is charged under the strong force will allow us to compute a prediction for $\alpha_{S}$ but it dependent on the model we use. Altering the SM would alter the predicted $\alpha_{S}$ values.

Now recall Sylwester doesn’t like the SM, he thinks its rubbish and should be binned without delay. Yet at the same time he likes to tell people how his ‘theory’ outputs the value of $\alpha_{S}$ so accurately. So he’s simultaneously denouncing a model while promoting the fact his work is consistent with it?

Any model attempting to replace the SM would have to demonstrate that when it applied to raw experimental data it is consistent. It’s a serious job, a great many physicists spend their careers converting raw data into SM predictions and then comparing them. For example, here is a paper published by an experimental collaboration showing the model dependent processing which must be done, followed by a statistical analysis of the results and a comparison with other methods. Here is a similar collaboration paper which looks at slghtly different processes but does much the same analysis. If widely different answers were obtained the model would be called into question.

This processing of data goes on all the time, CERN had to set up dedicated high speed links to other research facilities around the world in order to transmit the many petabytes of data the LHC produces. It takes a long time to boil that down to values for the parameters in the SM. Sylwester seems completely unaware this is even done, let alone necessary and that’s after I’ve explained it to him several times over the course of years. Until Sylwester does this he is failing to justify his claims about correctly modelling experiments.

Last month a paper (which I use in its vaguest sense) was put on viXra, the website for cranks and nuts who can’t can’t meet arXiv’s basic requirements of sanity and science, pertaining to special relativity. Of course this isn’t a new thing, SR is a favourite topic of people who slept through science class and never made it to university level physics. However, it’s on a specific thing all too often brought up on forums (in my particular experience that being Sciforums), light spheres.

As such I spent a bit of yesterday evening knocking up a refutation of the article, based on posts I made on Sciforums here. The pdf can be found here. The fatal mistake the author makes is that he doesn’t transform into the second frame properly. He doesn’t explicitly compute the Lorentz images of the two points in question, nor the Lorentz image of the light sphere. If he had done this then it would become clear why the Lorentz image of the second point is hit first, despite being hit second in the initial frame.

More formally, we denote $p_{1},p_{2}$ as the points in frame S such that $(p_{1})^{x} < (p_{2})^{x}$. Since $(p_{1})^{y} = (p_{2})^{y}$ and the Lorentz boost $\Lambda$ is by speed v in the x direction we need only consider the $x'$ components of $q_{i} = \Lambda(p_{i})$. While it is true $(q_{1})^{x'} < (q_{2})^{x'}$ for sufficiently large values of v the boost is such that $\vert (q_{1})^{x'}\vert > \vert(q_{2})^{x'}\vert$ and thus $(q_{1})^{t'} > (q_{2})^{t'}$ as both frames see the light sphere as centred on the origin and emitted at time 0.