# Relatively Energetic

We all know what the formula for kinetic energy is. Take the mass, multiply by the velocity squared, divide by two. In 1905, Einstein raised his hand and said that we’re not quite right. In fact the actual expression for kinetic energy is

where c is the speed of light and

That doesn’t look much like our old formula:

Not a c in sight. Obviously this is going to make a big difference where v is close to c. When that happens, gamma gets huge. Arbitrarily huge – that’s one of the things that makes it impossible to accelerate to the speed of light. You’d have to add infinite kinetic energy to do so. And on the other end of things, if v = 0 then γ = 1. That results in zero kinetic energy, as we expect. But what about small v, just like we see in everyday life? If the relativistic expression gives us something crazy, it can’t be right.

In our old expression, a 1 kg mass moving at 1 m/s has a kinetic energy of 0.5 joules. Plug in those same numbers to our new relativistic expression and you get that the kinetic energy is 0.5000000000000000042 joules. You can see why we might not have noticed the inaccuracy of our old formula. At normal speeds it simply is so close to right that we thought it was exactly right.

We might like to prove that this is true. The way to do so would be to expand our new kinetic energy equaiton in a power series. I won’t do it in detail because it’s one of those things where if you don’t know how a power series works this example won’t help, and if you do you’ll be bored. But it turns out that the first term in the power series of the relativistic expression is the old classical kinetic energy expression. The next few terms are

The terms other than the first one are very small for v << c because c to some power is absolutely tremendous in magnitude. So there you have it. You don't have to use Einstein’s bright idea at low speeds, but if you do it won’t hurt!

## Comments

1. #1 Yoo
August 26, 2008

It’s also cool to see the formula for relativistic kinetic energy being derived from the classical definition of work, as long as you define a force as the derivative of momentum against time (which is the same definition as in classical mechanics, but isn’t equivalent to F=ma in a relativistic regime).

At least I recall deriving it that way and finding it fascinating …

2. #2 CCPhysicist
August 26, 2008

If your focus is on the transition from low v to relativistic v, that equation is a bit cleaner if you keep the expansion terms separate:

(1/2)mv^2 + {(3/4)(v/c)^2}*(1/2)mv^2 + …

That way you see more clearly that v = 0.1c leads to an error of 0.75% while v = 0.3c will result in an error of 6.75% if you don’t include relativistic effects.

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