**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

collisions.

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

is

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.

RELATIVISTIC ENERGY

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

accelerators.

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 :

EQUIVALENCE OF MASS AND ENERGY

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

collisions.

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

is

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.

RELATIVISTIC ENERGY

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

accelerators.

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 :

EQUIVALENCE OF MASS AND ENERGY