The Cost of Speed

To accelerate a mass to a speed requires energy and energy costs money to buy.

To accelerate a mass m, to a speed v requires adding kinetic energy, E

E = mc^2(1/(1-v^2/c^2)^0.5 -1)

where c is the speed of light.

The term 1/(1-v^2/c^2)^0.5 occurs so often in relativity that it is given the name gamma, g.

E = mc^2 (g - 1)

The Proof

Particle accelerators all over the world verify this equation every day llook at their electric bill. e.g. Stanford Linear Accelerator Center.

A set of three 250 KV power ines feeds power to SLAC.

Explore The graph

The cost goes up asymptotically as the speed of light is approached.

The Math

For example, at a speed of v = 0.866, g = 2.

Sample calculation: How many Joules does it take to accelerate 1 kilogram to 0.866 c ?

E = mc^2(g -1) = 1 * (3*10^8)^2 * 1 = 10^17 joules.

The cost of energy can be estimated from the cost of electricity as about \$0.3 per kilowatt-hour where a kilowatt-hour is about 3.6 * 10^6 joules.

So the cost is about \$ 10^-7 per joule. ( a tenth of a microdollar per joule)

So the cost for 10^17 joules will be Cost = 10^17 J * 10^-7 \$/J = 10^10 \$

About ten billion dollars to accelerate a kilogram to 0,866 times the speed of light.

(It actually costs twice this much for a trip because you have to slow down at the far end.)

The real cost will be greater than this since I assumed perfect conversion of electric energy to kinetic energy of motion.

Notice that the relativistic equation for kinetic energy is not the same as the Newtonian version E = 1/2 mv^2

However, the relativistic equation becomes the Newtonian version at low speeds.

 Scientific Explorations with Paul Doherty © 2005 8 May 2005