# 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)  And by simultaneity, Then  ii)  since we have simultaneity implies . Then, Problem 2. In S-frame 2 events occur at the same point separated by a temporal distance of 3yrs. In the S’-frame, 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) with As the events occur in the same point    ii)  implies 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 for .

Problem 3. Two events in the S-frame have the following coordinates in spacetime: , i.e., and , i.e., . 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 were simultaneous? b) At what tmes t’ do these events occur in the S’-frame?

Solution. a)  and then  b)   Problem 4. A spaceship is leaving Earth with . When it is 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, and for the signal . From these equations, we get and , and it yields and thus for the intersection point. But, and . Putting this value in the intersection point, we deduce that the intersection point happens at . Moreover, b) We have to perform a Lorentz transformation from to , with . and . Then . And thus, we obtain that . The Lorentz transformation for the two events read Remarks: a) Note that and differ by 3 instead of . 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 . They are separated by a distance d. What is the electric force between them?  In the S-frame, we obtain the Lorentz force: 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: with when the temporal origin coincides, i.e., at the time .  Suppose now two points that for the rest observer provide: and . For the electric field we get: and  Then, implies that      There are two special cases from the physical viewpoint in the observed electric fields:

a) When P is directly above the charge q. Then b) When P is directly in front of ( or behind) q. Then, for a=0, Note that we have if .

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