# LOG#041. Muons and relativity.  QUESTION: Is the time dilation real or is it an artifact of our current theories?

There are solid arguments why time dilation is not an apparent effect but a macroscopic measurable effect. Today, we are going to discuss the “reality” of time dilation with a well known result:

Muon detection experiments!

Muons are enigmatic elementary particles from the second generation of the Standard Model with the following properties:

1st. They are created in upper atmosphere at altitudes of about 9000 m, when cosmic rays hit the Earth and they are a common secondary product in the showers created by those mysterious yet cosmic rays.
2nd. The average life span is $2\times 10^{-6}s\approx 2ms$
3rd. Typical speed is 0.998c or very close to the speed of light.
So we would expect that they could only travel at most $d=0.998c\times 2 \times 10^{-6}\approx 600m$
However, surprisingly at first sight, they can be observed at ground level! SR provides a beautiful explanation of this fact. In the rest frame S of the Earth, the lifespan of a traveling muon experiences time dilation. Let us define

A) t= half-life of muon with respect to Earth.

B) t’=half-life of muon of the moving muon (in his rest frame S’ in motion with respect to Earth).

C) According to SR, the time dilation means that $t=\gamma t'$, since the S’ frame is moving with respect to the ground, so its ticks are “longer” than those on Earth.

A typical dilation factor $\gamma$ for the muon is about 15-100, although the value it is quite variable from the observed muons. For instance, if the muon has $v=0.998c$ then $\gamma \approx 15$. Thus, in the Earth’s reference frame, a typical muon lives about 2×15=30ms, and it travels respect to Earth a distance $d'=0.998c\times 30ms\approx 9000m$.

If the gamma factor is bigger, the distance d’ grows and so, we can detect muons on the ground, as we do observe indeed!

Remark:  In the traveling muon’s reference frame, it is at rest and the Earth is rushing up to meet it at 0.998c. The distance between it and the Earth thus is shorter than 9000m by length contraction. With respect to the muon, this distance is therefore 9000m/15 = 600m.

An alternative calculation, with approximate numbers:

Suppose muons decay into other particles with half-life of about 0.000001sec. Cosmic ray muons have speed now about v = 0.99995 c.
Without special relativity, muon would travel $d= 0.99995 \times 300000 km/s\times 0.00000156s=0.47 km$ only!

Few would reach earth’s surface in that case. It we use special relativity, then plugging the corresponding gamma for $v=0.99995c$, i.e., $\gamma =100$, then muons’ “tics” run slower and muons live 100 times longer. Then, the traveled distance becomes $d'=100\times 0.9995\times 300000000 m/s\times 0.000001s= 30000m$

Conclusion: a lot of muons reach the earth’s surface. And we can detect them! For instance, with the detectors on colliders, the cosmic rays detectors, and some other simpler tools.

# LOG#029. Interstellar trips in SR. My final article dedicated to the memory of Neil Armstrong. The idea is to study quantitatively the relativistic rocket motion with numbers, after all we have deduced the important formulae, and we will explain what is happening in the two frames: S’-frame (in motion), S-frame (in rest on Earth). There are many “variations” of this problem, also called “Langevin’s paradox” or “the Langevin’s interstellar trip” problem by some authors. Here, we will follow the approach suggested in the book Gravitation, by Misner, Wheeler and Thorne, and we will study the interstellar trip (in the frame of special relativity) with the following conditions:

1st. Spacetime is locally minkovskian (i.e., spacetime is flat). The solution we will expose would not be valid in the case of an interstellar trip to a very far away quasar, or a very very long distance (about thousand millions of lightyears) where we should take into account the effect of the expansion of the Universe, i.e., that on large scales, spacetime is curved (in particular, accordingly to the current data, it is pseudoriemannian). Thus, we can use special relativity in order to calculate distances, velocities and accelerations from the purely kinematical sense. After all, it is logical, since General Relativity says that locally, in small enough regions of spacetime, spacetime is described by a minkovskian metric.

2nd. We select a 4 stage accelerated motion with our rocket. We will assume that our rocket is 100% efficient in the sense it uses photons as propellant particles. The four stages are: acceleration from rest to g (acceleration step 1), decceleration to rest with -g until we approach the destination (that would be the one-way trip), acceleration with -g (seen from the S-frame) and decceleration with +g ( with respect to the S-frame) to reach Earth again in rest at the end of the round-way trip. Schematically, we can draw a spacetime sketch of this 4 stage journey: Please, note the symmetry of the procedure and the different travel steps.

3rd. We set g=9.8m/s², or, as we saw in one of our previous posts, so we use g=1.03lyr/yr² (in units where c=1).

4th. We can not refuel during the travel.

5th. In the case we return to Earth, we proceed to come back after we stop at the destination inmediately. In this case, there is no refuel of the starship or rocket in any point of the trip.

6th. We neglect any external disturbance which can stop us or even destroy us, e.g., micrometeorites, cosmic radiations, comets, and any other body that could alter our route. Indeed, this kind of stuff has to be seriously considered in any realistic travel, but we want to solve an ideal problem consisting in an ideal interstellar trip according to the current knowledge.

The main quantities we have to compute are:

1. In the S’-frame of the rocket $\tau, 2\tau,4\tau$, in years. They are, respectively, the time we are accelerating with +g, the time we are deccelerating with -g, and the total time accelerating supposing we return to earth inmediately from our destination target.

2. Distance in which we are accelerating (x, in lightyears), seen from the S-frame (we will not discuss the problem in the S’-frame, since it involves lenght contraction and it is more subtle in the calculations). According to our previous studies, we have $\boxed{x=\dfrac{c^2}{g}\left(\cosh \left[\dfrac{g\tau}{c}\right]-1\right)}$

3. Maximum depth in space D=2x. This allows us to select our target or destination to set the remaining parameters of the trip. Due to the symmetry of our problem, we have $\boxed{D=2x=\dfrac{2c^2}{g}\left(\cosh \left[\dfrac{g\tau}{c}\right]-1\right)}$

4. Total length travelled by the spaceship/rocket (in the S’frame) in the roundtrip ( to the destination and back). $\boxed{L=4x=\dfrac{4c^2}{g}\left(\cosh \left[\dfrac{g\tau}{c}\right]-1\right)}$

5. Maximum speed to an specific destination, after accelerating in a given proper time. It reads: $\boxed{V=c\tanh \left(\dfrac{g\tau}{c}\right)}$

Indeed, the relativistic gamma factor is $\gamma =\cosh \dfrac{g\tau}{c}$

6. S-frame duration t(years) of the stage 1 (acceleration phase 1). It is $\boxed{t=\dfrac{c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}$

7. One way duration of the trip (according to the S-frame): $\boxed{t'=2t=\dfrac{2c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}$

8. Total duration of the trip in the S-frame ( round way trip): $\boxed{t''=4t=\dfrac{4c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}$

9. Mass ratio of the final mass with the initial mass of the spaceship after 1 stage, 2 stages ( one way trip) and the 4 stages total round way trip: $\boxed{R=\dfrac{M_f}{M_i}=\exp \left(-\dfrac{g\tau}{c}\right)}$ $\boxed{R=\dfrac{M_f}{M_i}=\exp \left(-\dfrac{2g\tau}{c}\right)}$ $\boxed{R=\dfrac{M_f}{M_i}=\exp \left(-\dfrac{4g\tau}{c}\right)}$

10. Fuel mass vs. payload mass ratio (FM/PM) after 1 stage, 2 stages (until the destination) and the total trip: $\boxed{R=\dfrac{M}{m}=\exp \left(\dfrac{g\tau}{c}\right)-1}$ $\boxed{R=\dfrac{M}{m}=\exp \left(\dfrac{2g\tau}{c}\right)-1}$ $\boxed{R=\dfrac{M}{m}=\exp \left(\dfrac{4g\tau}{c}\right)-1}$

We can obtain the following data using these expressions varying the proper time:   Here in the last two entries, data provided to be out of the limits of our calculator ( I used Libre Office to compute them). Now, we can make some final observations:

1st. We can travel virtually everywhere in the observable universe in our timelife using a photon rocket, a priori. However, it shows that the mass-ratio turns it to be theoretically impossible.

2nd. Remember that in the case we select very long distances, these calculations are not valid since we should use General Relativity to take into account the expansion of spacetime.

3rd. Compare $\tau$ with t, $2\tau$ with $2t$, $4\tau$ with $4t$. For instance, for $4\tau=40$ then we get 57700 years, and v=0.99999999773763c, FM/PM=30000 (multiplying por m-kilograms gives the fuel mass).

4th. In SR, the technological problems are associated to the way photon rockets have, the unfavorable mass ratios (of final mass with initial mass) and fuel mass/payload rations neccesary in the voyage (supposing of course, current physics and that we can not refuel during the trip).

5th. You can choose some possible distance destinations (D=2x) and work out the different parameters of the travel with the above equations.

In this way, for instance, for some celebrated known space marks, arriving at complete stop ( you can make other assumptions and see how the answer changes varying g or varying the arrival velocity too), we can easily get:

Example 1: 4.3 ly Nearest star (Proxima Centauri) in $t=2\tau=3.6$ yrs (S’-frame).

Example 2: 27 ly Vega (Contact movie and book) star in $t=2\tau=6.6$ yrs (S’-frame).

Example 3: 30000ly, our galactic center, in $t=2\tau=20$ yrs (S’-frame).

Example 4: 2000000ly , the Andromeda galaxy, in $t=2\tau=28$ yrs(S’-frame).

Example 5: Generally, you can travel n ly anywhere (neglecting curved spacetime) in $t=2\tau=1.94 \cosh^{-1}(n/1.94+1)$ years (S’-frame).