# LOG#052. Chewbacca’s exam.

I found this fun (Spanish) exam about Special Relativity at a Spanish website:

Solutions:

1) $v=25/29 c$

2) $1.836 \times 10^{12} m = 12 A.U.$

3) t=13.6 months = 13 months and 18 days.

Calculations:

1) We use the relativistic addition of velocities rule. That is,

$V=(u-v)/(1-(uv/c^2))$

where u=Millenium Falcon velocity, v=imperial cruiser velocity= c/5, y V=relative speed=4c/5.

Using units with c=1:

4/5=(v-1/5)/(1-v/5)

4/5(1-v/5)=v-1/5

4/5-4/25v=v-1/5

29/25 v=1

v=25/29

Then, $v=25/29 c$ reinserting units.

2) This part is solved with the length contraction formula and the velocity calculated in the previous part (1). Moreover, we obtain:

$\Delta x'=\Delta x/\gamma$

Using the result we got from (1), and plugging that velocity v and the fact that $\Delta t'$ is equal to one hour, then es

$\Delta x'=v\Delta t'=\Delta x/\gamma$ , and from this

$\Delta x=\gamma v\Delta t'$

Substituting the numerical values, we obtain the given solution easily.

$\Delta x =1.97 ( 25/29 c )1hour =1.7 hc=1.836 \times 10^{12} =12A.U.$

3) Simple application of time dilation formula provides:

$\Delta t'=\gamma \Delta t$

Inserting, in this case, our given velocity, we obtain the solution we wrote above:

$\Delta t' = 1.97 ( 9 months) = 13.6 months = 13 months 18 days$.

# 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#037. Relativity: Examples(I)

Problem 1. In the S-frame, 2 events are happening simultaneously at 3 lyrs of distance. In the S’-frame those events happen at 3.5 lyrs. Answer to the following questions: i) What is the relative speed between frames? ii) What is the temporal distance of events in the S’-frame?

Solution. i) $x'=\gamma (x-\beta c t)$

$x'_2-x'_1=\gamma ((x_2-x_1)-\beta c (t_2-t_1))$

And by simultaneity, $t_2=t_1$

Then $\gamma=\dfrac{x'_2-x'_1}{x_2-x_1}=\dfrac{7}{6}$

$\beta=\sqrt{1-\gamma^{-2}}\approx 0.5$

ii) $ct'=\gamma (ct-\beta x)$

$c(t'_2-t'_1)=-\gamma \beta (x_2-x_1)$

since we have simultaneity implies $t_2-t_1=0$. Then,

$c\Delta t'\approx -1.8 lyrs$

Problem 2. In S-frame 2 events occur at the same point separated by a temporal distance of 3yrs. In the S’-frame, $D'=3.5yrs$ is their spatial separation. Answer the next questions: i) What is the relative velocity between the two frames? ii) What is the spatial separation of events in the S’-frame?

Solution. i) $ct'=\gamma (ct-\beta x)$ with $x_1=x_2$

As the events occur in the same point $x_2=x_1$

$c(t'_2-t'_1)=\gamma c (t_2-t_1)$

$\gamma=\dfrac{t'_2-t'_1}{t_2-t_1}=\dfrac{7}{6}$

$\beta=\sqrt{1-\gamma^{-1}}\approx 0.5$

ii) $x'=\gamma (x-\beta c t)$

$x_1=x_2$ implies $x'_2-x'_1=-\gamma \beta c (t_2-t_1)\approx -1.8 lyrs$

Therefore, the second event happens 1.8 lyrs to the “left” of the first event. It’s logical: the S’-frame is moving with relative speed $v\approx c/2$ for $3.5 yrs$.

Problem 3. Two events in the S-frame have the following coordinates in spacetime: $P_1(x_0=ct_1,x_1=x_0)$, i.e., $E_1(ct_1=x_0,x_1=x_0)$ and $P_2(ct_2=0.5x_0, x_2=2x_0)$, i.e., $E_2(ct_2=x_0/2,x_2=2x_0)$. The S’-frame moves with velocity v respect to the S-frame. a) What is the magnitude of v if we want that the events $E_1,E_2$ were simultaneous? b) At what tmes t’ do these events occur in the S’-frame?

Solution. a) $ct'=\gamma (ct-\beta x)$

$t'_2-t'_1=0$ and then $0=\gamma (c(t_2-t_1)-\beta (x_2-x_1))$

$\beta =\dfrac{c(t_2-t_1)}{(x_2-x_1)}=-\dfrac{0.5x_0}{x_0}=-0.5$

b) $t'=\gamma ( 1-\beta x/c)=\gamma ( 1-\beta x/c)$

$t'_1=\dfrac{1}{\sqrt{1-(-0.5)^2}} \left(x_0/c+0.5x_0 /c)\right)\approx 1.7x_0/c$

$t'_2=\dfrac{1}{\sqrt{1-(-0.5)^2}} \left(x_0/c+0.5\cdot 2\cdot x_0/c)\right)\approx 1.7x_0/c$

Problem 4. A spaceship is leaving Earth with $\beta =0.8$. When it is $x_0=6.66\cdot 10^{11}m$ away from our planet, Earth transmits a radio signal towards the spaceship. a) How long does the electromagnetic wave travel in the Earth-frame? b) How long does the electromagnetic wave travel in the space-ship frame?

For the spaceship, $ct=x/\beta$ and for the signal $ct=x+ct_0$. From these equations, we get

$late \beta ct=x$ and $ct=x+ct_0$, and it yields $\beta ct =ct-ct_0$ and thus $t=\dfrac{t_0}{1-\beta}$ for the intersection point. But, $\beta=0.8=8/10=4/5$ and $1-\beta=1/5$. Putting this value in the intersection point, we deduce that the intersection point happens at $t_1=5t_0$. Moreover,

$t_1-t_0=4t_0=4\dfrac{x_0}{0.8c}\approx 11100 s=3.08h=3h 5min$

b) We have to perform a Lorentz transformation from $(ct_0,0)$ to $(ct_1,x_1)$, with $t'=t=0$.

$t_0=x_0/v=2775s$ and $t_1=5t_0=13875s$. Then $x_1=vt_1=5vt_0=5x_0=5\cdot 6.66\cdot 10^{11}m=3.33\cdot 10^{12}m$. And thus, we obtain that $\gamma=5/3$. The Lorentz transformation for the two events read

$(t'_2-t'_1)=\gamma (t_1-t_0)=\gamma (t_1-t_0)-\beta/c(x_1-x_0)=3700 s\approx 1.03h=1h1m40s$

Remarks: a) Note that $t_1-t_0$ and $t'_2-t'_1$ differ by 3 instead of $5/3$. This is due to the fact we haven’t got a time interval elapsing at a certain location but we face with a time interval between two different and spatially separated events.

b)The use of the complete Lorentz transformation (boost) mixing space and time is inevitable.

Problem 5. Two charged particles A and B, with the same charge q, move parallel with $\mathbf{v}=(v,0,0)$. They are separated by a distance d. What is the electric force between them?

$E'=\left(0,\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{d^2},0\right)$

$B'=\left(0,0,\gamma \dfrac{\gamma v q}{4\pi\epsilon_0d^2c^2}\right)$

In the S-frame, we obtain the Lorentz force:

$\mathbf{F}=q\left(\mathbf{E}+\mathbf{v}\times\mathbf{B}\right)=\left(0,\gamma k_C\dfrac{q^2}{d^2}-\gamma \beta^2\dfrac{q^2}{d^2},0\right)=\left(0,\dfrac{k_Cq^2}{\gamma d^2},0\right)$

The same result can be obtained using the power-force (or forpower) tetravector performing an inverse Lorentz transformation.

Problem 6. Calculate the electric and magnetic field for a point particle passing some concrete point.

The electric field for the static charge is: $E=k_C\dfrac{q}{x'^2+y'^2+z'^2}=k_C\dfrac{q}{r'^2}$

with $\mathbf{v}=(v,0,0)$ when the temporal origin coincides, i.e., at the time $t'=t=0$.  Suppose now two points that for the rest observer provide:

$P=(0,a,0)$ and $P'=(-vt',a,0)$. For the electric field we get:

$E'=k_C\dfrac{q}{r'^3}(x',y',z')$ and $E'(t')=k_C\dfrac{q}{(\sqrt{(x'^2+y'^2+z'^2})^3}(x',y',z')$

$E'_(t')=k_C\dfrac{q}{(v^2t'^2+a^2)^{3/2}}(-vt',a,0)$

Then, $E'_p\rightarrow E_p$ implies that $t'=\gamma t=\gamma (t-\dfrac{vx}{c^2})\vert_{x=0}$

$E'_p(t)=k_C\dfrac{q}{(\gamma^2v^2t^2+a^2)^{3/2}}(-\gamma v t,a,0)$

$B'=B'_p(t')=(0,0,0)=B'_p(t)$

$E_p(t)=(E'_{p_x}(t),\gamma E'_{p_y}(t),0)=k_C\dfrac{q}{\gamma^2 v^2t^2+a^2}(-\gamma v t,\gamma a,0)$

$B_p(t)=(B_{p_x},B_{p_y},B_{p_z})=(0,0,\gamma \dfrac{\gamma v E'_p(t)}{c^2})$

$B_p(t)=\dfrac{q}{\gamma^2v^2t^2+a^2}(0,0,\gamma \dfrac{v}{c^2}a)=(0,0,\dfrac{v}{c^2}E_{p_y}(t))$

There are two special cases from the physical viewpoint in the observed electric fields:

a) When P is directly above the charge q. Then $E_p(t=0)=(0,k_C\gamma \dfrac{q}{a^2},0)$

b) When P is directly in front of ( or behind) q. Then, for a=0, $E_p(t)=(-k_C\dfrac{vt}{\gamma^2(v^2t^2)^{3/2}},0,0)$

Note that we have $\dfrac{vt}{(v^2t^2)^{3/2}}\neq \dfrac{1}{v^2t^2}$ if $t<0$.