(View Primary Article: Birth of the Universe)
Some of my friends complained that my article ‘Birth of the Universe’ has some unexplained mathematics. When I read the article myself, I agree that I have used mathematics to explain the phenomenon, however some of the mathematics, I have used need to be explained before the article could be properly understood. Hence I am clarifying the basis of some of the mathematical formula.
1. Gravity
Before I clarify gravity, let me explain that Newtonian theory of gravity has been proved wrong by Albert Einstein’s ‘Theory of general relativity’ which was physically verified by Sir Eddington, the famous astronomer of England. Theory of General Relativity stated that gravity is not due to force of attraction between two masses. In fact gravity is the curvature of space – time due to big mass distorting the elasticity of the space. However the mathematics of the geometry of curvature is so complicated (Even Albert Einstein had to take some lessons on the geometry of the curvature from his class fellow and great mathematician Marcel Grossmann) that we are allowed to use Newtonian formula for gravity and gravitational force. Scientists believe that the ultimate value remains close to the correct one.
Now talking of gravity of the earth, let us consider Mr. X having a mass of m, standing on the face of the earth.
Mass of Earth = M Kg
Mass of Mr. X = m Kg
Radius of earth = r m
Now using the Newtonian formula for force between two masses
Force F = F1 = F2 = G x M x m/r2 Newton (where G is the gravitational constant equal to
6.67 x 10-11 m3/Kg/Sec-2), M and m are the mass of the two objects and r is the distance between them.
Also F = mass x acceleration. Here acceleration is the gravity. Hence
m x g = G x M x m/r2 Newton
Cancelling m from either side of the equation, we have
g = G x M x m/r2 Newton (which is independent of the mass of Mr. X)
Now applying the parameters of earth in this formula;
Mass of Earth = 6 x 1024 Kg (rounded figure)
Diameter of Earth = 12800 KM = 12.8 x 106 m (rounded figure)
Radius of the earth = 6.4 x 106 m (rounded figure)
gearth = 6.67 x 10-11 * 6 x 1024 / (6.4 x 106)2 = 0.977 x 10 = 9.77 m/sec/sec
If we apply the actual figure for the mass and radius of the earth, instead of rounded one, the result will be 9.81 m/sec/sec
Similarly we can calculate the gravity of the Sun, applying mass and radius of the sun.
2. Scape Velocity
Now let us calculate the scape velocity of the earth. This is the velocity with which if an object is thrown away from the earth, the object never returns.
Let us consider Mr. X standing on the face of the earth is throwing away a ball with a speed that the ball never returns back to the earth.
Again using the Newtonian formula for gravity, the Potential Energy working on the ball which tends to pull the ball back to the earth due to gravity is;
P.E. = G x M x m/ r
Where G is the gravitational constant = 6.67 x 10-11 m3/Kg/Sec2
M is the mass of the earth = 6 x 1024 Kg
m is the mass of the ball
r is the distance for the ball to reach so as never to return
There is another energy working on the ball which tends the ball to move away from the earth. It is the Kinetic energy due to the speed of the ball = ½ m V2 . This energy varies throughout the journey of the ball depending upon the speed.
In order to achieve a speed that the ball never returns back to the earth, the Potential Energy (PE) of the ball must be zero. However in the formula of PE, none of the values in the numerator viz G, M or m can be zero. Hence the ball must travel to infinity in order to reduce the PE to zero.
Also when the ball stops at infinity, the Kinetic Energy (KE) which is a factor of speed will also become zero.
That means PE equals KE. Since energy is neither created nor destroyed, PE and KE will always be equal throughout the journey of the ball, working in opposite directions.
Hence in order to achieve a speed for the projectile never returns back to earth, the condition is that KE should be equal to PE at each stage of the travel. Since we are interested in the speed with which the ball left the earth, we consider the values at the earth in order to compute the scape velocity of the earth.
Hence ½ m V2 = G x M x m/ r
Cancelling m from both sides of the equation, we have
½ V2 = G x M / r
Applying the parameters of earth
This is how we calculate the scape velocity of any large mass. In case we want to calculate as to how much a star of certain mass should be squeezed (due to gravitational collapse) to become a black hole, all we need to know is the mass of the star. The scape velocity is ‘C’ (speed of light) for a black hole. C stands for celeritas, which means swiftness in Greek. The scape velocity for a black hole is the speed of light i.e. 3 x 108 m/sec
3. Electromagnetic radiation
Above is the full spectrum of the electromagnetic waves starting from the highest frequency on the left (Gamma rays) to the lowest frequency on the right which is radio waves.
It is quite cumbersome to refer to the electromagnetic waves in terms of their frequency. Instead they are referred to with respect to their wave lengths. The relation is as follows
f = C/ʎ where
f is the frequency of the EM wave in Hz or cycles per second
C is the speed of light 3 x 108 m/s
ʎ is the wavelength in m
Hence the properties of a wave can be referred in terms of f (frequency) or ʎ (wavelength). They are normally referred in terms of wavelength.
The spectrum is as follows.
Gamma rays 1 fm – 10 pm f stands for femto which is 10-15 , p stands for pico which is 10-12
X rays 1 pm – 10 nm n stands for nano which is 10-9
Ultraviolet 10 nm – 380 nm
Visible lights 380 nm – 740 nm
Infrared 740 nm – 100 µm
Microwave 100 µm – 10 cm
Radio waves 10 cm – 100 Km
Energy of the photons of each wave is directly proportional to its frequency. The relation is
E = ħ*f where
E is the energy of the photon
ħ is the Planck constant and the value is 6.626 x 10-36 Joules second. However for photons the energy is so low that they are expressed in terms of eV (electron volt). 1 eV = 1.6 x 10-19 Joules
ʎ is the wavelength
Hence Energy can be expressed in terms of wavelength
E = ħ x C/ ʎ = 6.626 x 10-36x 3 x 108/ ʎ ≈ 20 x 10-28/ ʎ Joules
Equals = 12.5 x 10-9/ ʎ eV
Now it could be understood that the gamma rays photons when they emanate from the core of a star and rush towards the edge of the star, they collide with hydrogen and helium atoms, get sucked in and release themselves. In this process by the time they reach the edge of the star, they have lost most of their energy and subsequently their frequency is lowered.. That means they move towards the right hand of the spectrum. Ultimately what emanates from the edge of the star is not gamma rays but photons of lesser energy (reduced frequency and increased wavelengths) and what we receive from the sun and the stars is , ultraviolet, visible lights and infra-red.