E=mc^2

Cecil's Columns/Staff Reports
1

I just read the May 2000 column about Einstein’s equation E=mc^2. The answer was informative but, to my disappointment, didn’t address the question. :mad: The easy answer is this (using the metric system):

E is in joules (it takes about 10 joules to raise a 1 Kg object 1 meter off the floor, and 100 Joules to power a 100 Watt light bulb for 1 second)
m is in Kg (about 2.2 lbs on earth)
c is in meters per second (speed of light = 3x10^8 m/s)

Therefore, 1 joule = 1 Kg.m^2/s^2

So to get an idea of how much energy is contained in 1 Kg of matter:

E=(1 Kg)(3x10^8 m/s)^2 = 9 x10^16 Joules (that’s 9 followed by 16 zeros)

That’s enough energy to keep a 100 Watt light bulb lit for over 28,000 years.

Reference:
http://www.btinternet.com/~j.doyle/SR/Emc2/Equation.htm

(sorry about the 12 year delay in responding. It had something to do with the curvature of time and space) :smiley:

2

May 10, 2000 column

Link: http://www.straightdope.com/columns/read/1775

3

Since this is dredged up, can we get all the superscripts in that column superscripted? It’s a bit jarring to read things like

length2? What is that, and how is it different from length1? Time2 vs time1?

I count 2 places where the digit “2” is not supposed to be a superscript.

Pretty please?

4

No, the question was “In E=mc[sup]2[/sup], what units of measurement was Einstein using?” and the answer correctly says that Einstein wasn’t using any particular system of units. The equation is true regardless of your choice of units.

Now, if the question were “What units are the SI units for energy, mass, and speed?”, your answer would be correct.

5

That’s not literally true; the units must be consistent:

joules, kilograms, meters-per-second
ergs, grams, centimeters-per-second
foot-pounds, slugs, feet-per-second
foot-poundals, pound-masses, feet-per-second

but not:

kilowatt-hours, ounces, miles-per-minute

6

I suppose I could have headed off this criticism by saying “regardless of your choice of a system of units”.

However, since we’re already being picky, I’ll say I don’t really agree. If we want to treat E, m, and c as dimensionless (i.e., unitless) quantities, then sure, the units have to cancel out. E.g., we can get away with this if we have Joules on the left side of the equation and kg times m[sup]2[/sup]/s[sup]2[/sup] on the right, because a Joule is a kg m[sup]2[/sup]/s[sup]2[/sup].

But really, E, m and c are not dimensionless quantities. (Notwithstanding the particle physics convention of setting c=1)

It’s still true that E = m c[sup]2[/sup] even if m is 1 kg, c is 299,792,458 m/s, and E is 6.62887801 x 10[sup]16[/sup] foot-pounds.

6.62887801 x 10[sup]16[/sup] foot-pounds = 1 kg (299,792,458 m/s)[sup]2[/sup]
even though
6.62887801 x 10[sup]16[/sup] does not equal 1 * 299,792,458[sup]2[/sup]

In other words, E = m c[sup]2[/sup] is a statement about physical quantities, not about numbers.