Special Relativity : Relativistic Dynamics

Question 1

Recap the classical mechanics conservation of momentum.

Answer 1

F1 =dp1/dt
F2 =dp2/dt

F1 + F2 = d(p1+p2)/dt
F1 = -F2
p1 + p2 = pt = constant (integrate to see)

Isolated system

Question 2

Elastic collision derivations and facts?

Answer 2

For equal mass elastic scattering, out x going particle must be orthogonal.

p = p1 + p2
p*2/2m = (p1+p2)^2/2m = p1^2/2m + p2^2/2m + 2p1p2/2m

Bubble chamber shows particles appearing from nowhere, but we don’t lose momentum or energy in SR. NO!

Consider orthogonal direction motion in frame S’.
y’=uy’.t’
Assume p = kmv where k is a function of v/c
y = y’
t’ = t/y
u(y)=u’(y)/y

Now consider two particles scattering (A & B)
A moving fast along x’ and slow up y’. A scattering against B moving slow downwards. Collision occurs and directions vertically swap.

In frame S’ A has horizontal velocity and B only vertical, opposite in S frame.
Speed along y = U, U &laquo_space;c, horizontal relativistic speeds.

S’ : u’(B)(y) = U, S : u’(A)(y)

Question 3

Momentum of A & B particles seen in different frames?

Answer 3

In S motion of B is p(B)= 2kmu’(B) = 2kmU/y
In S motion of A is p(A)= 2mU
Conservation shows that k=y
p=ymv
Particle behaves as though mass has increased by y.

Question 4

Relativistic Energy now.

Answer 4

F = d(ymv)/dt
Work done : Fdx=dT
dT = md(yv)dx/dt
dT = m(yvdv + v^2dy)

Now differentiating the Lorentz factor, you find mc^2dy= ymvdv + mv^2dy, so now dT = mc^2dy

Integrating you find T = ymc^2 + A
When T = 0, v = 0 , y = 1 => A = -mc^2
So T = (y-1)mc^2

T = kinetic , mc^2 = rest energy. So ymc^2 = mc^2 + T

Question 5

Treating relativistic energy as classical.

Answer 5

v«c
Taylor expand Lorentz to find it approx. to = 1 + v^2/2c^2
Sub this back in you find that T = (1/2)mv^2
!!!

Question 6

Describe and think about the invariant relativistic energy quantity.

Answer 6

E^2 - (pc)^2 is the invariant quantity, it doesn’t change
Sub in p = ymv, E =ymc^2

E^2 - (pc)^2 =(mc^2)^2
E^2=(pc)^2 + (mc^2)^2

Sanity check? Stationary particle, p = 0, E =mc^2
Remember sanity checking with Lorentz is super powerful

E^2 can be treated like Pythagoras. Another sanity check method.