武汉大学刘觉平电动力学第四次作业

Exercises 2-6-8
Prove that
(1) The interval of the proper times for any two world points with light-like separation is zero;
Solution:
For any two world points with light-like separation we have ()2
0s ∆= ()()()22220s c t l ∆=∆-∆=, so l v c t
∆==∆
For special Lorentz transformation we have
00//τττ==
0τ=So 0τ must be zero.
(2) For three world points 123,,P P P with time-like separation, the Minkowski inequality follows 322131123, for t t t τττ+≤<<
where ji τ is the proper time interval of j P with respect to i P , and the equality is valid if and
only if 123,,P P P lie on the same straight world line.
Solution:
Because 13,P P are in time-like separation, so
()()()222
'c t c t l ∆=∆-∆ Which means we can chose a system of reference, in which 13,P P happened at the same
displace but different time.
So we have:
()
()
(
)313121213232///t t t t t t τττ-=-=-= And
3132210
v v v ==-
So
()3131t t τ=-
(
(21213232t t t t ττ=-=-312132τττ≥+
Only if 210v =, which means the three world points lie on the same straight world line.
(3) For two points 1P and 2P with time-like separation and 12t t <, the proper time interval of a
point particle moving from 1P to 2P through a curve is less than that through a straight line.
Solution:
We separate the curve line into some straight line
121lim n i n i s l →∞==∑
According to the result of the formal question, we have For each two pieces of the straights ij mj im τττ>+,
So ()21121lim n i i n i τττ→∞=>+∑ which means the proper time interval of a
point particle moving from 1P to 2P through a curve is less than that through a straight line.
Exercises 2-6-9
There is a pair of twins, A and B . B flies away from the earth with the velocity v , and when
B has covered a distance of L , he comes back immediately with the velocity -v and finally he meets A again. Calculate the difference of their ages when they meet again. Are the observed results to the difference of their ages by A and B accordant with each other.
Solution:
The total time observed by A is 2/A L v τ=
The time of B observed by A is
()()12
/2//2/B B B A A A
c L c c L c τττγτβγτβγτ=+=-++=
So the difference of their ages is ()(2/1A B L v ττ-=
The difference of their ages is the same, but the observer is always older.
Exercises 2-7-6
Derive the simple relation between tan /2θ and tan /2θ' from Eq. (2.7.10), where θ and θ' are the polar angle of the velocity of a point particle in the inertial frame S and S ', respectively. Assume that the coordinate axes of the two inertial frames S and S ' are parallel to each other, S ' has a constant velocity of V =x V e with respect to S .
Solution:
First we have
sin θ=and ()''cos /cos 1/cos V c V c θθθ+=+ And then 2sin 2sin 1cos 22tan 2sin cos 2sin cos 222θθθ
θθθθθ
-=
=
= So
()
''1/cos cos /tan 2tan V c V c θθθ
+-+==
Exercises 2-7-7
In view of the inertial frame fixed at a star, a rocket starts from the stationary state at 0x =, and moves with an acceleration along x-axis. The acceleration g of the rocket measured by an astronaut in this rocket is a constant. Find the distance the rocket has flied away when its velocity is reached at the magnitude v .
Solution:
According ()33'21/x
x x a a v V c γ'=+
In the view of the inertial frame S ', which is static respect to the rocket at time t ,
We have '0x v =, so ()
32221/a g v c
=- Then
()3222
1/ds ds dv ds v a dt dv dt dv ds g v c dv
====- So
()(
)()
32220
3222202
1/1/21/1v v v s ds v c dv g v c dv g c g --==--==⎰⎰⎰。