Thursday, December 29, 2011

The explanation of special relativity & its results (part3 )

Relativistic linear momentum and the Relativistic form of Newton’s laws
   First, recall that the law of conservation of

linear momentum states that when two isolated

objects collide, their combined total momentum

remains constant.
    Suppose that the collision is described in

a reference frame S in which linear momentum is

conserved. If we calculate the velocities in a

second reference frame S` using the Lorentz

velocity transformation equation and the classical

definition of linear momentum,P = mu (where u is the

velocity of either object), we find that
linear momentum is not conserved in S` . However,

because the laws of physics are the same in all

inertial frames, linear momentum must be conserved

in all frames.
In view of this condition and assuming that the

Lorentz velocity transformation equation is

correct, we must modify the definition of linear

momentum to satisfy the following conditions:
1.    Linear momentum p must be conserved in all

2.    The relativistic value calculated for p

must approach the classical value mu as u

approaches zero.
For any particle, the correct relativistic equation

for linear momentum that satisfies these conditions

        p = mu /√1 - u2/c 2 = γ    mu
where u is the velocity of the particle and m is

the mass of the particle. When u is much less than  

c, γ = /√1 - u2/c 2 approaches unity and p

approaches mu.
Therefore, the relativistic equation for p does

indeed reduce to the classical expression when u is

much smaller than c.
The relativistic force F acting on a particle whose

linear momentum is p is defined as
F = dp/dt
This expression, which is the relativistic form
of Newton’s second law, is reasonable because it

preserves classical mechanics in the limit of low

velocities and requires conservation of linear

momentum for an isolated system both

relativistically and classically.
    We have seen that the definition of linear
momentum and the laws of motion require
generalization to make them compatible with the
principle of relativity. This implies that the
definition of kinetic energy must also be modified.
To derive the relativistic form of the work–kinetic
energy theorem, let us first use the definition of
relativistic force.  To determine the work done on
a particle by a force F
W =⌠x1x2 F dx = ⌠x1x2 dp/dt dx
for force and motion both directed along the x
axis. In order to perform this integration and find
the work done on the particle and the relativistic
kinetic energy as a function of u – we can easily
proof that
 w = mc2 /  √1 - u2/c 2   - mc2 = γ mc2 - mc2
Remember that that the work done by a force acting
on a particle equals the change in kinetic energy
of the particle. Because of our assumption that the
initial speed of the particle is zero, we know that
the initial kinetic energy is zero. We
therefore conclude that the work W is equivalent to
the relativistic kinetic energy K:
K = γ mc2 - mc2 .
    This equation is routinely confirmed by
experiments using high-energy particle
    The constant term mc2, which is independent
of the speed
of the particle, is called the rest energy ER of
the particle. The
term     mc2, which does depend on the particle
speed, is therefore the sum of the
kinetic and rest energies. We define γ mc2 to be the
total energy E:
Total energy = kinetic energy +  rest energy
E    =     γ mc2     = K + mc2, so
E = mc2 / √1 - u2/c 2  This is Einstein’s famous
equation about mass–energy equivalence.
The relationship E = K + mc2  shows that mass is a
form of energy, where c2 in the rest energy term is
just a constant conversion factor. This expression
also shows that a small mass corresponds to an
enormous amount of energy, a concept fundamental to
nuclear and elementary-particle physics.
Coming up :