# LOG#037. Relativity: Examples(I)

**Posted:**2012/10/07

**Filed under:**Physmatics, Relativity |

**Tags:**boost, events, exercises, frame, length contraction, Physmatics, Relativity, simultaneity, time dilation Leave a comment

**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

$late \beta ct=x$ 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 .