The Physics
Opus in profectus


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momentum and energy separately

These ideas are completely disorganized. Keep that in mind when reading this.

Relativity has a different equation for (almost) everything. It's like classical physics just isn't good enough. There's a different one for time (time dilation) and a different one for space (length contraction) and now there's a different one for momentum (relativistic momentum) and another different one for energy (relativistic energy).

The equation for relativistic momentum looks like this…

p =  mv
√(1 − v2/c2)

When v is small (as it is for the kinds of speeds we deal with in everyday life) the denominator is approximately equal to one and the equation reduces to its classical version…

v ≪ c  ⇒  p ≈ mv

which is as it should be. Relativity doesn't replace classical physics, it supplements it. All equations in special relativity should reduce to classical equations at low velocities. This is known as the correspondence principle.

The equation for relativistic energy looks like this…

E =  mc2
√(1 − v2/c2)

Applying the correspondence principal to give us the classical equations is not so easy here. Once again, at low speeds the denominator is one, but the numerator we're left with is something new. Something with no classical counterpart. Something famous.

v ≪ c  ⇒  E ≈ mc2

This equation says that an object at rest has energy, which is why it is sometimes called the rest energy equation. It also says that the reason an object at rest has any energy at all is because it has mass, which is why this equation is also known as the mass-energy equivalence.

Let's try a more sophisticated approach and see where it takes us. The binomial expansion is an equation for transforming a binomial raised to a power into a sum of terms. In its most general form, it looks like this…

(a + b)n = 
k = 0

 an − kbk

Some readers might recognize this as the equation used to generate the terms in Pascal's triangle. Each row of the triangle contains the expansion coeficients for a non-negative integer power (n = 0, 1, 2, 3, …). Although the expansion generates an infinite number of terms, only the first n + 1 of them are non-zero.


Comparing (a + b)n to the relativistic gamma…

γ =  1  = (1 − v2/c2)−½
√(1 − v2/c2)

gives the following parameters for the binomial expansion…

a =  1
b =  v2/c2
n =  −½

When n is a fraction, the expansion is truly infinite. Here's what the first six terms of the relativistic energy equation look like. Only the first two are interesting (n = 0 and n = 1).

E = mc2
1 +  1   v2  +  3   v4  +  5   v6  +  35   v8  +  63   v10  +…
2 c2 8 c4 16 c6 128 c8 256 c10

Distribute mc2 across all the terms The zeroeth term is the rest energy.

E0 = mc2

The first term is the classical equation for kinetic energy.

E1 =  1  mv2

The remaining terms are higher order corrections that become more and more significant as an object's speed approaches the speed of light. I don't know of any practical use for these terms. They do look fancy, however.

E2 =  3   mv4
8 c2
E3 =  5   mv6
16 c4
E4 =  35   mv8
128 c6
E5 =  63   mv10
256 c8

The energy added to an object to take it from an initial speed of zero to a final speed of something is its kinetic energy.

K = E − E0


K =  mc2  −  mc2
√(1 − v2/c2) √(1 − v02/c2)

Let initial velocity equal zero.

K =  mc2  −  mc2
√(1 − v2/c2) 1

Factor out like terms and in the end we get an equation for relativistic kinetic energy that looks like this in expanded notation…

K = 
1 − 1
√(1 − v2/c2)

…and like this in gamma notation.

K = (γ − 1)mc2

momentum and energy together

In relativistic mechanics, the momentum equation…

p =  mv
√(1 − v2/c2)

and the energy equation…

E =  mc2
√(1 − v2/c2)

have a common feature — the Lorentz factor, a.k.a. the relativistic gamma…

γ =  1
√(1 − v2/c2)

which means they can be written in a more compact form like this…

p = γmv E = γmc2

For no immediately apparent reason, start with this expression…

E2 − p2c2

Replace energy and momentum with their gamma versions like this…

γ2m2c4 − γ2m2v2c2

The identity rule allows us to multiply the second term by 1 in the form of c2/c2.

γ2m2c4 − γ2m2v2c2(c2/c2)

Using the commutative and associative properties of multiplication, move things around in the second term.

γ2m2c4 − γ2m2(v2/c2)(c2c2)

Simplify a little…

γ2m2c4 − γ2m2(v2/c2)c4

and pull out like terms.

γ2m2c4(1 − v2/c2)

Notice that the stuff in parentheses is the reciprocal of γ2, which means the stuff on the left cancels the stuff on the right and the stuff in the middle stays put.


This means that…

E2 − p2c2 = m2c4


E2 = p2c2 + m2c4

This is the relativistic energy-momentum relation. For massed particles at rest we get the famous mass-energy relationship or the rest energy equation…

v = 0  ⇒  E = mc2

For massless particles we get the much less famous energy-momentum relationship…

m = 0  ⇒  E = pc

If the equations of relativity are to be believed, then nothing with mass can travel at the speed of light. If it did, it would have either undefined energy (the mathematicians answer) or infinite energy (the physicists answer). If v = c, then √(1 − v2/c2) = 0 and as everyone knows you can't divide by zero. This is the mathematicians argument. There's a breakdown of logic. As v approaches c, 1/√(1 − v2/c2) approaches infinity and finite things with infinite characteristics are seen as completely unrealistic. This is the physicists argument. There's a disconnect from what we can observe. Interestingly, the ∞ symbol means both undefined and infinite.

  E =  mc2
√(1 − v2/c2)
  E =  mc2
√(1 − c2/c2)
  E =  mc2
√(1 − 1)
  E =  mc2
  E = 

But what if an object with zero mass was traveling at the speed of light? Now the relativistic energy equation would have zero in the numerator and zero in the denominator. What does everyone have to say about that?

  E =  mc2
√(1 − v2/c2)
  E =  0c2
√(1 − c2/c2)
  E =  0
  E = ?

Well the mathematicians are still unhappy. Division by zero is just not allowed under any circumstances. But the physicists have a different opinion. They often see extreme values as limits on the behavior of numbers, not as logical statements. What is the physical "reality" of dividing mc2 = 0 by √(1 − v2/c2) = 0? Mathematicians have found ways to deal with the limit of zero divided by zero and physicists often think of extremes as limits rather than actual values. One of the interesting things about physics is that it appears that measurable reality can be described mathematically. One of the other interesting things about reality is that the only thing that's real about it are the measurements — and they don't give a crap about you and your mathematics. In fact, sometimes I think the universe is daring us to try and figure out how it works. (Yo humans. Check this out. High temperature superconductivity. What do your big monkey brains have to say about that?)

Since a fraction whose value can be expressed as 0/0 may have a finite limit under certain circumstances, there's no logical reason that physical entities with zero mass traveling at the speed of light cannot exist. Light appears to be made of particles with zero mass that travel at the speed of light in a vacuum. There is nothing in relativistic mechanics (or the associated mathematics) that disagrees with this statement.

Let me say it right now. Light is composed of particles (called photons) that have no mass and travel at the speed of light in a vacuum. They have no mass, but somehow they still transfer energy (kinetic energy, to be specific) and momentum. That's the way it appears to be so I'm going to say, with only the most microscopic amount of doubt, that that's the way it is. When you observe something different you let me know.