Find the group velocity and phase velocity of a particle moving with velocity 0.9c
Answers
g
=0.9c=2.7×10
8
m.s
−1
is the group velocity
v_f=3.33\times 10^8\ m.s^{-1}v
f
=3.33×10
8
m.s
−1
is the phase velocity
Explanation:
Let the mass of particle be, m_0m
0
velocity of particle is its group velocity, v_g=0.9c=2.7\times 10^8\ m.s^{-1}v
g
=0.9c=2.7×10
8
m.s
−1
Since the velocity of the particle is comparable to the velocity of light, we need its relativistic mass:
m=\frac{m_0}{\sqrt{1-(\frac{v_g}{c})^2 } }m=
1−(
c
v
g
)
2
m
0
Now according to Einstein's mass energy equivalence:
E=m.c^2E=m.c
2
.....................................(1)
and the momentum of this relativistic mass:
p=m.v_gp=m.v
g
.......................................(2)
Now the phase velocity is given as:
v_f=\frac{E}{p}v
f
=
p
E
v_f=\frac{m.c^2}{m.v_g}v
f
=
m.v
g
m.c
2
v_f=\frac{c^2}{v_g}v
f
=
v
g
c
2
putting respective values in above eq.
v_f=\frac{9\times 10^{16}}{2.7\times 10^8}v
f
=
2.7×10
8
9×10
16
v_f=3.33\times 10^8\ m.s^{-1}v
f
=3.33×10
8
m.s
−1
TOPIC: theory of relativity
.