Thus, e is the initial letter of "energy" and m of "mass."
As for c, that is the speed of light in a vacuum, and if you ask why c, the answer is that it is the initial letter of celeri tas, the Latin word meaning "speed."
This is not all, however. For any equation to have mean ing in physics, there must be an understanding as to the units being used. It is meaningless to speak of a mass of 2.3, for instance. It is necessary to say 2.3 grams or 2.3 pounds or 2.3 tons; 2.3 alone is worthless.
Theoretically, one can choose whatever units are most convenient, but as a matter of convention, one system used in physics is to start with "grams" for mass, "centimeters" for distance, and "seconds" for time; and to build up, as far as possible, other units out of appropriate combinations of these three fundamental ones.
Therefore, the m in Einstein's equation is expressed in grams, abbreviated gm. The c represents a speed-that is, a distance traveled in a certain time. Using the fundamental units, this means the number of centimeters traveled in a certain number of seconds. The units of c are therefore centimeters per second, or cm/sec.
(Notice that the word "per" is represented by a fraction line. The reason for this is that to get a speed represented in lowest terms, that is, the number of centimeters traveled in one second, you must divide the number of centimeters traveled by the number of seconds of traveling. _If you travel 24 centimeters in 8 seconds, your speed is 24 centi meters - 8 seconds, or 3 cm/sec.)
But, to get back to our subject, c occurs as its square in the equation. If you multiply c by c, you get C2. It is, how ever, insufficient to multiply the numerical value of c by it self. You must also multiply the unit of c by itself.
A common example of this is in connection with meas urements of area. If you have a tract of land that is 60 feet by 60 feet, the area is not 60 x 60, or 3600 feet. It is 60 feet x 60 feet, or 3600 square feet.
Similarly, in dealing with C2, you must multiply cm/sec 'by cm/sec and end with the units CM2 /seC2 (which can be read as centimeters squared per seconds squared).
The next question is: What is the unit to be used for e?
Einstein's equation itself will tell us, if we remember to treat units as we treat any other algebraic symbols. Since e = mc 2, that means the unit of e can be obtained by mul tiplying the unit of m by the unit Of C2. Since the unit of m is gm and that of c2 is CM2 /seC2, the unit of e is gm x CM2/seC2. In algebra we represent a x b as ab; conse quently, we can run the multiplication sign out of the unit of e and make it simply gm CM2/SCC2 (which is read "gram centimeter squared per second squared).
As it happens, this is fine, because long before Einstein worked out his equation it had been decided that the unit of energy on the gram-centimeter-second basis had to be gm CM2 /seC2. I'll explain why this should be.
The unit of speed is, as I have said, cm/sec, but what happens when an object changes speed? Suppose that at a given instant, an object is traveling at 1 cm/sec, while a second later it is travelling at 2 cm/sec; and another second later it is traveling at 3 cm/sec. It is, in other words, "ac celeratin " (also from the Latin word celeritas).
In the case I've just cited, the acceleration is 1 centi meter per secondevery second, since each successive sec ond it is going I centimeter per second faster. You might say that the acceleration is I emlsec per second. Since we are letting the word "per" be represented by a fraction mark, this may be represented as 1 cm/sec/sec.
As I said before, we can treat the units by the same manipulations used for algebraic symbols. An expression like alblb is equivalent to alb b, which is in turn equiva lent to alb x Ilb, which is in turn equivalent to alb2. By the same reasoning, I cm/sec/sec is equivalent to 1 cm/ seC2 and it is CM/SCC2 that is therefore the unit of accelera tion.
A "force" is defined, in Newtonian physics, as some thing that will bring about an acceleration. By Newton's First Law of Motion any object in motion, left to itself, will travel at constant speed in a constant direction forever.
A speed in a particular direction is referred to as a t'veloc ity," so we might, more simply, say that an object in mo tion, left to itself, will travel at constant velocity forever.
This velocity may well be zero, so that Newton's First'Law also says that an object at rest, left to itself, will remain at rest forever.
As soon as a force, which may be gravitational, electro magnetic, mechanical, or anything, is applied, however, the velocity is changed. This means that its speed of travel or its direction of travel or both is changed.
The quantity of force applied to an object is measured by the amount of acceleration induced, and also by the mass of the object, since the force applied to a massive ob ject produces less acceleration than the same force applied to a light object. (If you want to check this for yourself, kick a beach ball with all your might and watch it accel erate from rest to a good speed in a very short time. Next kick a cannon ball with all your might and observe-while hopping in agony-what an unimpressive acceleration you have imparted to it.)
to assure yourself, first, of a supply of nine hundred quin tiflion ergs.
This sounds impressive. Nine hundred quintillion ergs, wow!
But then, if you are cautious, you might stop and think:
An erg is an unfamiliar unit. How large is it anyway?
After all, in Al Capp's Lower Slobbovia, the sum of a billion slobniks sounds like a lot-until you find that the rate of exchange is ten billion slobniks to the dollar.
So-How large is an erg?
Well, it isn't large. As a matter of fact, it is quite a small unit. It is forced on physicists by the lo 'c of the gram-cen 91 timeter-second system of units, but it ends in being so small a unit as to be scarcely useful. For instance, consider the task of lifting a pound weight one foot against gravity.
That's not difficult and not much energy is expended. You could probably lift a hundred pounds one foot without completely incapacitating yourself. A professional strong man could do the same for a thousand pounds.
Nevertheless, the energy expended in lifting one pound one foot is equal to 13,558,200 ergs. Obviously, if any trifling bit of work is going to involve ergs in the tens of millions, we need other and larger units to keep the nu merical values conveniently low.
For instance, there is an energy unit called a joule, which is equal to 10,000,000 ergs.
This unit is derived from the name of the British physi cist James Prescott Joule, who inherited wealth and a brew ery but spent his time in research. From 1840 to 18 9 e ran a series of meticulous experiments which demonstrated conclusively the quantitative interconversion of heat and work and brought physics an understanding of the law of conservation of energy. However, it was the erman sci entist Hermann Ludwig Ferdinand von Helmholtz who first put the law into actual words in a paper presented in 1847, so that he consequently gets formal credit for -,the discov ery.
(The word "joule," by the way, is most commonly pro nounced "jowl," although Joule himself probably pro 167 nounced his name "jool." In any case, I have heard over precise people pronounce the word "zhool" under the im pression that it is a French word, which it isn't. These are the same people who pronounce "centigrade" and "centri fuge" with a strong nasal twang as "sontigrade" and "son trifugp,," under the impression that these, too, are French words. Actually, they are from the Latin and no pseudo French pronunciation is required. There is some justifica tion for pronouncing "centimeter" as "sontimeter," since that'is a French word to begin with, but in that case one should either stick to English or go French all the way and pronounce it "sontimettre," with a light accent on the third syllable.)
Anyway, notice the usefulness of the joule in everyday affairs. Lifting a pound mass a distance of one foot against gravity requires energy to the amount, roughly, of 1.36 joules-a nice, convenient figure.