Newton+Maxwell=mc2:UNIZOR.COM - Relativity 4 All - Conservation

Notes to a video lecture on UNIZOR.COM

Newton + Maxwell = m·c²

In this lecture we will use the laws of classic Newtonian mechanics and the properties of light we discussed in the chapter Waves - Electromagnetic Field Waves of this course to derive the famous Einstein's equation E=m·c².

In particular, we will use the results presented in the lecture Momentum of Light of the above mentioned chapter of this course, which we recommend to review prior to studying materials presented in this lecture.

In the Momentum of Light lecture we have derived the relationship between the amount of work W performed by light moving an object with absorbing surface and the momentum p lost by this light during this process of light absorption:
p = W/c (c is the speed of light)
which, basically, means that the energy E contained in certain amount of light and this light's momentum are related as p=E/c.

Now consider the following thought experiment.
Initially, two identical objects, α and β, of mass M each are at rest and positioned on a distance 2L from each other along X-axis with coordinates −L for α and L for β.
We assume that no external forces are acting on these objects.

Since no external forces act on these objects and they are at rest, the total momentum of motion is zero and the center of mass is at X-coordinate x=0.

At some moment an object α sends a short light signal towards object β.
As we know, this light carries some energy E and momentum p=E/c.

For this totally closed isolated system the total momentum must be preserved.
In particular, it means that the center of mass should stay at the original location x=0.

It also necessitates that immediately after issuing a light signal object α must acquire a momentum −p to neutralize the momentum p of the light signal, that is, object α recoils after issuing this light signal.

This momentum of object α means that it has acquired some speed −V (the value V is presumed positive) and moves towards negative direction of the X-axis, while a light signal with speed c moves towards object β in the positive direction of the X-axis.

If light, together with its energy, does not carry some portion of object α's mass, the center of mass would shift with object α's movement, which should not be the case.
So, light signal carries not only some energy E and momentum p=E/c, but also some part of object α's mass m, leaving object α with mass M−m.

But, according to classic Newtonian Mechanics, if the light signal carries mass m moving with speed c, its momentum must be p=m·c (mass times speed of light).
Since, as we stated above, p=E/c, we have a relationship between light signal's mass and energy E it carries:
p=m·c = E/c
from which follows
E = m·c²

The last equation means that energy E lost by an object α by issuing a light signal equals to its lost mass m times a square of the speed of light.

We came to the same Einstein's relationship between mass and energy using a combination of the Maxwell theory of electromagnetic field, which gave us a relationship between light energy E and its momentum p as p=E/c, and Newtonian Mechanics with its definition of a momentum p as mass m times speed that in the case of light equals to c.

Let's continue our analysis of this experiment and find the behavior of all components involved in it.

Obviously, the light signal moves with speed c from α towards β.
It's initial coordinate is x=−L, so its movement can be described as
x_sig(t) = −L + c·t
for all t from 0 to T=2L/c - the moment light hits object β.
Then it's absorbed by object β and disappears as an independent entity (see below the analysis of movement of β object.

Object α is losing mass m and moves to the negative direction of X-axis with speed V_α.
To conserve the momentum, this speed must satisfy the equation
(M−m)·V_α = m·c
Therefore,
V_α = m·c/(M−m)
and the equation of motion for α object is
x_α(t) = −L − m·c·t/(M−m)

Object β stays in place at coordinate x=L up until it gets hit by a light signal at time T=2L/c.
Then β gets hit by a light signal and absorbs it energy, momentum and mass.
From the Law of Conservation of Momentum follows that the momentum of an object that absorbed the light signal, increasing its mass by m, should be equal to a momentum of this signal:
(M+m)·V_β = m·c
Therefore,
V_β = m·c/(M+m)
and the equation of motion for β object, considering it starts by time T later than α, is
x_β(t) = L + m·c·(t−T)/(M+m)

Let's check if the center of mass is not changing the position.
There are two stages of this experiment: before the light signal hit β object and after.

1. For time t less than T we have three objects - α, light signal and β.
Their combined "mass times X-coordinate" is
(M−m)·x_α(t)+m·x_sig(t)+M·L =
= −L·(M−m) − m·c·t −
− L·m + m·c·t + M·L = 0
as it should.

2. For time t equal or greater than T we have only two objects - α and β.
Their combined "mass times X-coordinate" is
(M−m)·x_α(t) + (M+m)·x_β(t) =
= −L·(M−m) − m·c·t +
+ (M+m)·L+m·c·t−m·c·2L/c =0
as it should.

As we see, the center of mass is always at x=0, which is expected from a closed system.