The Batmobile “fake paradox” helps us to understand Special Relativity a little bit. This problem consists in the next experiment:

There are two observers. Alfred, the external observer, and Batman moving with his Batmobile.

Now, we will suppose that the Batmobile is moving at a very fast constant speed with respect to the garage. Let us suppose that $v=0.866c=\dfrac{\sqrt{3}}{2}c$. Then, we have the following situation from the external observer:

However, with respect to the Batmobile reference frame, we have:

The question is. Who is right? Alfred or Batman? The surprinsig answer from Special Relativity is that Both are correct. Alfred and Batman are right! Let’s see why it is true. For Alfred, there is a time during which the Batmobile is completely inside the garage with both doors closed:

By the other hand, for Batman, the front and rear doors are not closed simultaneously! So there is never a time during which the Batmobile is completely inside the garage with both doors closed.

So, there is no paradox at all, if you are aware about the notion of simultaneity and its relativity!

# 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 $t<0$.