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.

Selected Questions -Answers From All Experts Astronomy Forum (Gravity Wells)

Question -

I actually have two questions, I know that the answers may be highly
theoretical because of limited knowledge we have of black holes, but any
answer - even theory - is welcome.

1. If a photon has no mass, how come light is drawn into a black hole?

2. What method of propulsion do black holes use the move themselves
through space? How is their movement in a vacuum achieved?

Thanks for your time mate

Answer -

You are quite correct in assuming that the answers are 'highly
theoretical', and I don't see how these can be made much simpler than
provided below.

First, a physical definition. For a term we call "inertia". This is any
resistance to change in position or motion.

You can think of it as a kind of "foundational" mass. It is the most
basic, fundamental concept we have by which "mass" can be recognized.

That is, when we perform an experiment and find that 'X' (whatever 'X' is)
exhibits a resistance to change its position or motion.

Keeping that thought in mind, we come to the photon. We already know that
light carries energy, usually expressed in a simple form as:

E = hf

where h is the Planck constant, and f is the frequency of the light.

But what many general readers do not know is that light also carries
momentum. For example, whenever light is emitted from some source there is
a tiny but definite "recoil" effect. (Something like you experience from
the stock of a rifle, if you've ever fired on a rifle range).

This recoil betrays the presence of what we call "momentum" to do with the
photon.

This is given by the product of one over its speed (1/c) times the energy,
E. Writing the momentum as p:

p = E x (1/c)  = E/c


We already have a good hint that photons have mass-inertia associated
with them from Einstein's equation:

E =  m c ²     (E = mass x speed of light squared).

Therefore, the mass is:

m =  E /   c ² (Energy divided by speed squared).

This equation (above) can be combined with the earlier one we obtained

 (E   = p c) to give:

m = p / c

Thus, the "mass" associated with a photon is its momentum divided by its
speed, c.

Other experiments validate this, including one in which a (true) radiometer is
bombarded with light, and turns it vanes - spins them around. This is a
result of the pressure of light exerted on the vane surfaces. (You perhaps
have seen this device in action, though most of them are really toys that
operate on a different principle from the real ones).

Basically, light cannot possibly exert any "pressure" unless it also
carries mass, or inertia.

Having said all this, it is important to recognize that in all these cases
I've noted we are looking at photons MOVING!

Once they cease to move, they no longer exhibit mass.

Thus, what you are really thinking of when you say "if a photon has no
mass", is REST MASS. The mass it exhibits when at rest.

And, of course, this is zero.

Hopefully, you can see from the above, that there is no contradiction. The
light that is prevented from escaping from black holes or drawn into them,
is moving - NOT at rest. Hence, it exhibits all the properties of mass or
inertia.


Regarding your second question, black holes like the stars they collapsed
from occupy specific regions of space-time. Hence, they are not
"propelled" like rockets or anything.

Rather, as part of a dynamical system (say the galaxy) they will move in
accord with the Newton's law of gravitation - given their distance from
the mass center of the galaxy, and the nature and masses of nearby bodies.

Since space itself is a vacuum, this motion is no more mysterious than the
planets of our Solar System moving "through a vacuum". Vacua provide no
impediment to motion, and indeed, every body in the cosmos will move under
the influence of the gravitational forces at work in its neighborhood.

The bottom line is that the presence of gravity means that NO body in the
universe can remain "still" or "motionless".