Physics Van 3-site Navigational Menu

Physics Van Navigational Menu

Q & A: using E=mc^2

Learn more physics!

Most recent answer: 10/22/2007
Q:
I have heard of Einstien’s theory of relativity (E=mc2) But how would you figure out the energy of something using this formula? What does mass mean in this context? And are you measuring the speed of light in miles per hour, miles per second, lightyears per year etc.? And what is the speed of light? (c) Thanks!
- Tom (age 11)
U.S.
A:

In this context the mass is the inertial mass, the same mass that tells how much momentum (p) something traveling at velocity v has, p=mv. In many cases this is almost entirely the familiar rest mass of a material.

There is no one set of ’right’ units. You have to be careful to use a consistent set of units for mass, velocity, energy, etc. The standard international units measure mass in kilograms, velocity in meters per second, and energy in Joules. Since one Joule is defined as one kilogram*meter^2/second^2, in these units you can use E=mc^2 with no conversion factors.
Another convenient set of units is the ’cgs’ system: mass in grams, velocity in centimeters per second, and energy in ergs, with one erg being one gram*centimeter^2/second^2.

There are also ’fundamental’ unit systems used in high-energy physics and some other areas, in which all velocities are measured in units of ’c’ itself. That is, c is defined to be one. In these units, the relation becomes very simple: E=m. The ’c^2’ factor arises just because we use some historical units.

Units like ’miles per hour’ are pretty useless for doing any physics because they have not been incorporated into any coherent system where the energy units are simply related to the mass and velocity units.

Mike W.

The speed of light is 299,792,458 meters/second. The second is defined in terms of an atomic transition frequency, and the meter is defined so that the speed of light comes out to exactly this number (or you can think of it as the meter is defined as a fixed number of wavelengths of the same atomic transition).

Tom

(published on 10/22/2007)

Follow-up on this answer.