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 $l x_0=6.66\cdot 10^{11}m$ 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

$\beta ct=x$ 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 $l t_1=5t_0$ . 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:

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