Deciphering the Energy-Momentum relation

The Mass-Energy equivalence equation was highlighted on Taipei-101 during the World Year of Physics, 2005
Source: https://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence

"Energy is the ultimate convertible currency"

-Brian Greene

Our Universe is filled with many beautiful things, not just the tall mountains and deep seas, we are today blessed with advancements in astronomy which helps us look at large distant galaxies. We can now learn how the stars are born, talk about what does a black hole look like etc. At the same time, humans have developed massive instruments which aids us in probing the universe at a quantum scale. Understanding of nature at this domain is also as necessary as knowing about large scale structure as both these fields of physics are interdependent on each other. However, there is very little common in these fields. The only language with which we can relate the world of quantum mechanics with the world of classical mechanics (including Astrophysics) is the language of energy. Many of the equations of energy are in one or the other way common to both the microscopic and macroscopic landscape. You can learn more about the scale of energies for different phenomena in our universe from here, however in this post, we are just going to talk in great detail about the famous equation in physics; the Mass-Energy equivalence relation:

\(E=m_{rel}c^2\)

Before we jump directly into this, it is really important to discuss a little about the Special Theory of Relativity (don't worry, we are not going to spend hours here! Maybe sometime later). According to the Special Theory of Relativity, developed by Albert Einstein in 1905, mass of any body is dependent on what speed it is moving with. In other words, greater the velocity of the body, greater is its mass which is what the following equation describes:

\(m_{rel} = \frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}}\)

where \(m_{rel}\) is the relativistic mass (i.e. the mass of the body when it is in motion), \(m_o\) is the rest mass of the body (i.e. when the body is at rest), \(v\) is the velocity of the object and \(c\) is the velocity of light. 

Now, if 

\(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\)

Then the above equation is written as,

\(m_{rel} = \gamma m_o\)

As one can notice, whenever \(v\) is close to speed of light, only then \(\gamma\) is a non-zero quantity. That is to say, relativistic effects are only observed for a body moving at speedds comparable to the speed of light. 

We know that the linear momentum of any body is given by the product of mass and velocity. However, when a body is moving at speeds closer to speed of light, it is the relativistic mass we take into account rather than the rest mass.

\(p = m_{rel}v = \gamma m_ov\)

where \(p\) is the momentum. No carefully carrying out a little algebra and rearranging the first and last equation, we get:

\(E^2-p^2c^2 = m_o^2c^4\gamma^2(1-\frac{v^2}{c^2}) = m_o^2c^4\)

i.e.

\(E^2 = p^2c^2 + (m_oc^2)^2\)

This is our Energy-Momentum relation in Special Theory of Relativity. Now let's try to understand what this equation really means!

If you look at the Energy-Momentum relation carefully, it is of the form \(c^2 = a^2 + b^2\) and this is nothing but the Pythogoras theorem for a right angled triangle. Considering one length of the triangle as \(p^2c^2\) and the other as \((m_oc^2)^2\), as shown in the figure below, our hypotenues then turns out to be \(E^2\). 

Figure 1: Geometrical representation of Energy-Momentum relation.
(a) The case when rest mass is equal to zero
(b) The case when momentum is zero

Now we will try our hand in some special conditions, that is to say; what wil happen if \(E=pc\)? As you can see in Figure 1(a), this is only possible when the other arm of the triangle has zero length i.e. when \(m_o = 0\)(as \(c\) is the speed of light which is always constant, it is only the \(m_o\) which can have zero value). So this means, let's say if we have a body or a particle with zero rest mass, its energy is given by \(E=pc\). Making use of d'Broglie wave equation,

\(p=\frac{h}{\lambda}\)

we get,

\(E=\frac{hc}{\lambda}=h\nu\)

where \(\lambda\) is the wavelength of the matter wave, \(h\) is the Planck's constant and \(\nu\) is the frequency of the radiation. The only known particle in the observable universe which has a zero rest mass is a photon! So this is basically how we define the energy of a photon.

Likewise, when we assume a particle or body at rest with respect to the frame of observation (Figure 1(b)), \(p=0\). And so we get \(E=m_o c^2\). This is the energy a body as at rest by the virtue of its mass. In other words, mass is a condensed form of energy and if we somehow completely convert this body of mass \(m_o\) into energy, we can harness \(E=m_oc^2\) Joules of energy. A very stunning example of rest mass energy is evident from the experiments in nuclear physics, particularly in the phenomena of Nuclear Fission and Nuclear Fusion.

Consider the following Nuclear fusion reaction of two deuterium combining to form a helium-3 isotope. 

\(_{1}^2H + _{1}^2H \rightarrow _{2}^3He + _{0}^1n + 3.27MeV\)

where the mass of a deuterium is 2.014102 amu, mass of neutron is 1.0086654 amu and that of \(_{2}^3He\) is 3.0160293 amu. Clearly as far as atomic weights are concerned, the LHS is not equal to RHS and this loss of weight is compensated by liberation of energy of 3.27MeV. This verifies that mass and energy are interconvertible quantities.

Finally we can now talk about the total energy of a body moving with a velocity \(v\) with rest mass \(m_o\), which is given by the following equation:

\(Total\) \(Energy = Rest\) \(mass\) \(energy + Kinetic\) \(energy\)

i.e. 

\(m_{rel}c^2 = m_oc^2 + KE \)

or,

\(KE = m_{rel}c^2 - m_oc^2\)

\(KE = m_oc^2 [\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1]\)

The plot of relativistic Kinetic Energy and relativistic mass is shown in the following graph:

Figure 2: The graph on the left depicts the nature of relativistic mass as a function of velocity and the graph on the right shows the behavior of KE (in Joules) as a function of velocity of the body.

Now it is really important to appreciate the physics for the entire scale, i.e. for \(\frac{v^2}{c^2}<<1\) and \(\frac{v^2}{c^2}\approx 1\). The case for \(\frac{v^2}{c^2}\approx 1\) is pretty straightforward; when \(v\) approaches \(c\), the value of \(\gamma\) becomes infinity and so does the Kinetic Energy KE. 

Now the expansion of \((a+x)^n\) is given by:

\((a+x)^n = a^n + na^{n-1}x + \frac{n(n-1)}{2!}a^{n-2}x^2 + ...\)

and hence for the former scenario, when the body is moving at speeds much lower as compared to speed of light, our relativistic KE equation takes the form:

\(KE = \frac{1}{2}m_ov^2 + \frac{3}{8}\frac{m_ov^4}{c^2} + \frac{5}{16}\frac{m_ov^6}{c^4} + ...\)

or,

\(KE\approx \frac{1}{2}m_ov^2\)

that is the classical equation for Kinetic energy.

So what did we learn here is that one can speak about mass in terms of energy. Mass and energy are interconvertible quantities or to sound more cool; mass is the codensed form of energy! We also saw how kinetic energy, momentum and mass behave differently for bodies moving at speeds closer to the speed of light. Since momentum is the building block of classical mechanics and every observable thing around us, plethora of physics changes when we look at things in the 'high velocity' domain. 

The universe is indeed a strange place filled with wonders, it is upto us how we perceive its beauty. So the next time you hear about small particles moving with enormous speeds, don't take them lightly.