Relativity-Physics, Dynamics, Forces and Momentum-Assignment Solution, Exercises of Physics

This course is introduction to Physics. Its includes: acceleration, angular momentum, ballistic motion, center of mass, circular of orbits, Newton laws, drag force, velocity, conservation law of energy, superposition, circular motion, time dilation, work and energy. This assignment includes: Relavite, Velocity, Differential, Form, Lorentz, Transformation, Frame, Moving, Particle, Energy, Collision, Incident

Typology: Exercises

2011/2012

Uploaded on 08/12/2012

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HW Solutions # 14 - 8.01 MIT - Prof. Kowalski
Relativity.
1) 37.20 Relativistic Relative Velocity
Write differential form of Lorentz transformation:
x = γ(x ut)
t = γ(t ux/c2)
1
γ = 1 u2/c2
which is:
dx = γ(dx udt) (1)
dt = γ(dt udx/c2) (2)
Divide (2) by (1) you’ll get (dx/dt = vx):
v= vx u
x 1 uvx/c2 (3)
Where u is the velocity of the moving frame. In this problem you
want to know the relative velocity so it means that your moving
frame has the velocity of one of them and in this frame you want to
know velocity of the other:
vx = 0.9520c
u = 0.9520c
Replace this into equation (3) you’ll get:
v=0.9520c (0.9520c) = 0.9988c
x 1 (0.9520c)(0.9520c)/c2
v .9988c
rel = 0
1
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Relativity.

1 ) 37.20 Relativistic Relative Velocity

Write differential form of Lorentz transformation:

x�^ = γ(x − ut)

t � = γ(t − ux/c 2 )

γ = � 1 − u^2 /c^2

which is:

dx�^ = γ(dx − udt) (1)

dt � = γ(dt − udx/c 2 ) (2)

Divide (2) by (1) you’ll get (dx/dt = vx):

v � =

vx − u x 1 − uvx/c^2

Where u is the velocity of the moving frame. In this problem you

want to know the relative velocity so it means that your moving

frame has the velocity of one of them and in this frame you want to

know velocity of the other:

vx = 0. 9520 c

u = − 0. 9520 c

Replace this into equation (3) you’ll get:

v � =

  1. 9520 c − (− 0. 9520 c) x =^0.^9988 c 1 − (0. 9520 c)(− 0. 9520 c)/c^2

v (^) rel = 0. 9988 c

1

� �

2 ) 37.44 Creating a Particle

a) The total energy of a particle is:

E = mγc 2

Where m is the rest mass and

1 γ = (4) (^2) /c 2

1 − u

Before collision - incident two protons -:

� E = EP + EP = 2 mP γP c 2

After collision - two protons + new particle which are at rest -:

� E � = EP + EP + Eη � 0 = (2mP + mη^0 )c 2

From the energy conservation:

� E =

E

we get:

2 mP + mη 0 γ (^) P = 2 mP

With the numbers given:

γ (^) P= 1. 29

From (4):

� 1 u/c = 1 − γ^2

2

3 ) 37.71 The Pole and Barn Paradox

The reason there is an apparent paradox is that the notion of fitting

the moving pole into the barn contains an implicit assumption that

the length of the pole (or the barn) is not changed by their motion.

Actually the length of a moving object contains the idea of simul

taneity - it is the distance between the ends at the same instant of

time. Simultaneity is destroyed by special relativity - two people

who have a relative velocity will not agree on what is simultaneous!

Hence they can (and do) disagree about the relative size of pole and

barn.

This paradox is often sharpened by imagining that the doors of the

barn are suddenly and simultaneously closed by the farmer. Seeing

the door closing, the runner stops. Then does his pole fit? This

introduces an extraneous element into the problem: when the runner

stops, only the middle of the pole where he is holding it stops - so

when do the ends stop? In reality the pole will break. But if the

ends magically stopped simultaneously in the frame of the runner,

the farmer would see the rear end of the pole stop while the front end

kept going. When the pole, which to the farmer appeared shorter

than the barn while moving, stops in this manner, it will be longer

than the barn. This confirms the original opinion of the runner (who

thought the barn much shorter than his pole), but only because the

ends stopped simultaneously in the frame of the runner.

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4 ) 37.72 Fizeau Experiment

From the relativistic addition of velocities discussed in the book and

derived here (problem 1) we can apply them here. Assume light has

velocity c/n in the direction of water’s velocity, V :

c

  • V c n +^ V^ V v = 1 +

(c/n)V =^

n = (c/n+V )[1− +O(V 2 /c 2 )] = c/n+V −V /n 2 +O(V 2 /c 2 ) 1 + V^ nc c^2 nc

Therefore:

= (c/n)^ +^ (1^ −^1 /n

2 v ∼ )V

Where we threw out O(V 2 /c^2 ) assuming it’s very small. This is the

same as the given formulae with

k = 1 − n^2

With the n=1.333 you’ll get:

k = 0. 4372

which is incredibly close to the experimental value.

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