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)


\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?


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')


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)


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.


Leave a Reply

Please log in using one of these methods to post your comment:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s