Special Relativity Revisited (5): The Inertia Of Energy

We now conclude our introduction to Einstein's theory of special relativity with what many regard as its most fundamental conclusion: the inertia of energy, or light. This is embodied in Einstein's famous equation:

E = m c²

which is more accurately posed as:

E = (D m) c²

where D m is the "mass defect" or difference, say in a nuclear reaction, and c is the velocity of light.

    Before looking at examples, it's useful to consider the relativistic mass of a particle, in terms of its rest mass m_o. The rest mass, as the term implies is the mass of the object at rest or:

m_o = m [(1 - v²/c²)^½]

Thus, if v = 0 (particle at rest) then we have:

m_o = m(1)^½ = m

so the mass and rest mass are identical.

Now, the relativistic mass is then:

m = m_o / [(1 - v²/c²)^½]


and again, if v = 0 then m_o = m

But what if v = c? (Object moving at the speed of light?)

Then:

m = m_o / [(1 - c²/c²)^½] = m_o / [1 -1]^½ = m_o /0 = ¥

     In other words, m would be infinite! This is another way of saying that to try to achieve the velocity of light one would have to overcome infinite inertia! In other words, it can't be done...not for a material object.

From this, we can also see the relativistic momentum must be:

p = mu = m= m_o u / [(1 - c²/c²)^½ ]

This approaches the classical value (p = mu) as u ® 0

Newton's 2nd law in the relativistic format is simply:

F = ma = m (du/dt) = d/dt[m_o u / [(1 - v²/c²)^½ ]


The relativistic energy is found by taking the integral of : (dp/du) u du

® ò _o ^u  (dp/ du) u du

from 0 to u and obtaining:

W = mc²/ [(1 - u²/c²)^½ ] - mc²



And by the work -energy theorem:

W = K(f) - K(i)

where K(i) is just the initial rest energy, or m_o c²

Then W = m_o  c²/ [(1 - u²/c²)^½ ] - m_o c² =

 (total energy - rest energy) 

A variation entails finding the work done (W') between two points x1 and x2 with velocity v2 at x2 and time t2, and velocity v1 at x1 and time t1.

Then we may write:

W'  =  ò _x1 ^x2  (dp/ dt) dx  =    ò _t1 ^t2  (dp/ dt) (dx/dt) dt = ò _t1 ^t2  v (dp/ dt) dt


=   ò _v1 ^v2  v  dp   =    ò _v1 ^v2  v  (dp/dv)  dv


Note in the above we used the following facts:

v = dx/ dt   and dp = (dp/dv) dv

Here p is given as a function of v such that:


W'=   ò _v1 ^v2  v  (d/dv) [ m_o  v / [(1 - v²/c²)^½ ]   dv

On integrating the preceding equation by parts we find:

W'  =   m_o  c²/ [(1 - v2 ²/c²)^½ ] - m_o c² / [(1 - v1 ²/c²)^½ ]

This equation immediately shows that the effect of the work done is to produce a change in the quantity:

E' =   m_o  c²/ [(1 - v²/c²)^½ ] 

Note how this is different from the classical kinetic energy equation: 

 E = 1/2  ( m v² )


In particular E' does not become 0 when v = 0 (Instead it reduces to: 

E' = m_o c²

Hence, if we desire a quantity which correspond as closely as possible to classical KE we need to define:

E_k   =   m_o c² / [(1 - v ²/c²)^½ ]    -    m_o c²


If indeed this is a correct relativistic  generalization of kinetic energy it must reduce to approximately  1/2  m v²   when  v << c.

This can easily be shown by expanding the binomial   (1 - v ²/c²)^½   in the last eqn. using the binomial theorem, i.e.


(1 - v ²/c²) ^-1/2    =    _{1  +}    v ²/2 c²   +   (3 v ⁴/ 8c⁴  ) +    ......


Example Problem:

Apply the basic mass-energy equation, E = (D m) c², to the case of nuclear fusion.



Consider:

₁H² + ₁H² ®  ₂He³ + ₂He³ + _o n ¹

which actually occurs in the Sun.

We now compile the masses (in atomic mass units) on each side:

2.015 u + 2.015 u ® 3.017 u + 1.009 u

or:

4.030 u ® 4.026 u


Now, the right side is less than the left by an amount equal to the mass defect or:

D m = 4.030 u - 4. 026 u = 0.004 u

To get the energy E:

E = (0.004 u)(931 MeV/u) = 3.7 MeV

where 931 MeV/u is the conversion factor incorporating c²

To transfer to Joules:

3.7 MeV = 3.7 MeV x (1.6 x 10^-13 J/MeV)= 6.0 x 10^-13 J 


Example Problem (2):

Determine the energy required to accelerate an electron from 0.50c to 0.90c.

By the work -energy theorem:

W = K(f) - K(i)

K(i) = m_o c²/ [(1 - u²/c²)^½ ]

u1 = 0.50 c

K(f) = m_o c²/ [(1 - u²/c²)^½ ]

u2 = 0.90c   (where: m_o = 9.1 x 10^-31 kg )

K(f) - K(i) = m_o c²/ [(1 - (0.90c)²/c²)^½ ]

- m_o c²/ [(1 - (0.50c)²/c²)^½ ]

K(f) - K(i) = m_o c²/[(1 - 0.81]^½ - m_o c²/[(1 - 0.25]^½

K(f) - K(i) = 2.294 m_o c² - 1.155 m_o c² = 1.134 m_o c²

Or:

K(f) - K(i) = 9.32 x 10^-14 J = 0.583 MeV 



Problems for Budding Physicists:

1) A student proposes to compute the kinetic energy of a particle relativistically by using the expression  1/2  m v²  with the 'relativistic mass of the particle.   Would this be correct? Explain why or why not.

2) Determine the energy required to accelerate a proton from 0.25c to 0.50c. 

3) Protons emerge from a particle accelerator with a kinetic energy equal to 0.49 mc².   What is the speed of these particles? Compare the result to that obtained from the non-relativistic relation between mass and energy.

4) What is the speed of a particle whose kinetic energy is equal to its rest energy? What percentage error is made if the non-relativistic kinetic energy expression is used?

5) Show that the relativistic  kinetic energy equation:

E_k   =   m_o c² / [(1 - v ²/c²)^½ ]    -    m_o c²

^{Reduces to approximately:}

1/2  m v²   when  v << c.