If ever an equation has come into its own it is Ein stein's e = mc 2. Everyone can rattle it off now, from the highest to the lowest; from the rarefied intellectual height of the science-fiction reader, through nuclear physicists, college students, newspapers reporters, housewives, busboys, all the way down to congressmen.

Rattling it off is not, of course, the same as understand ing it; any more than a quick paternoster (from which, in cidentally, the word "patter" is derived) is necessarily evi dence of deep religious devotion.

So let's take a look at the equation. Each letter is the initial of a word representing the concept it stands for.

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.

Meanwhile, physicists who were studying heat had in vented a unit that would be convenient for their purposes.

This was the "calorie" (from the Latin word color meaning "heat"). It can be abbreviated as cal. A calorie is the amount of heat required to raise the temperature of I gram of water from 14.5' C. to 15.5' C. (The amount of heat necessary to raise a gram of water one Celsius degree varies slightly for different temperatures, which is why one must carefully specify the 14.5 to 15.5 business.)

Once it was demonstrated that all other forms of energy and all forms of work can be quantitatively converted to heat, it could be seen that any unit that was suitable for heat would be suitable for any other kind of energy or work.

By actual measurement it was found (by Joule) that 4.185 joules of energy or work could be converted into pre cisely I calorie of heat. Therefore, we can say that I cal equals 4.185 joules equals 41,850,000 ergs.

Althouo the calorie, as defined above, is suitable for physicists, it is a little too small for chemists. Chemical re actions usually release or absorb heat in quantities that, under the conventions used for chemical calculations, re sult in numbers that are too large..For instance, I gram of ,carbohydrate burned to carbon dioxide and water (either in a furnace or the human body, it doesn't matter) liberates roughly 4000 calories. A gram of fat would, on burning, liberate roughly 9000 calories. Then again, a human being, doing the kind of work I do, would use up about 2,500,000 calories per day.

The figures would be more convenient if a larger unit were used, and for that purpose a larger calorie was in vented, one that would represent the amount of heat re quired to raise the temperature of 1000 grams (1 kilo gram) of water from 14.50 C. to 15.5' C. You see, I sup pose, that this larger calorie is a thousand times as great as the smaller one. However, because both units'are called calorie," no end of confusion has resulted.

Sometimes the two have been distinguished as "small calorie" and "large calorie"; or "gram-calorie" and "kilo gram-calorie"; or even "calorie" and "Calorie." (The last alternative is a particularly stupid one, since' in speech and scientists must occasionally speak-there is no way of distinguishing a C and a c by pronunciation alone.)

My idea of the most sensible way of handling the matter is this: In the metric system, a kilogram equals 1000 grams; a kilometer equals 1000 meters, and so on. Let's call the large calorie a kilocalorie (abbreviated kcal) and set it equal to 1000 calories.

In summary, then, we can say that 1 kcal equals 1000 cal or 4185 joules or 41,850,000,000 ergs.

Another type of energy unit arose in a roundabout way, via the concept of "power." Power is the rate at which work is done. A machine might lift a ton of mass one foot against gravity in one minute or in one hour. In each case the energy consumed in the process is the same, but it takes a more powerful heave to lift that ton in one minute than in one hour.

To raise one pound of mass one foot against gravity takes one foot-pound (abbreviated I ft-lb) of energy. To expand that energy in one second is to deliver 1 foot pound per second (1 ft-lb/sec) and the ft-lb/sec is there fore a permissible unit of power.

The first man to make a serious effort to measure power accuratel was James Watt (1736-1819). He compared y the power of the steam engine he had devised with the power delivered by a horse, thus measuring his machine's rate of delivering energy in horsepower (or hp). In doing so, he first measured the power of a horse in ft-lb/sec and decided that I hp equals 550 ft-lb/sec, a conversion figure which is now standard and official.

The use of foot-pounds per second and horsepower is perfectly legitimate and, in fact, automobile and airplane engines have their power rated in horsepower. The trouble with these units, however, is that they don't tie in easily with the gram-centimeter-second system. A foot-pound is 1.355282 joules and a horsepower is 10.688 kilocalories per minute. These are inconvenient numbers to deal with.

The ideal am centimeter-second unit of power would be ergs per [email protected] (erg/sec). However, since the erg is such a small unit, it is more convenient to deal with joules per second (joule/sec). And since I joule is equal to 10, 000,000 ergs, 1 joule/sec equals 10,000,000 erg/sec, or 10,000,000 gM CM2/sec3.

Now we need a monosyllable to express the unit joule/ see, and what better monosyllable than the monosyllabic name of the gentleman who first tried to measure power.

So 1 joule/sec was set equal to 1 watt. The watt may be defined as representing the delivery of I joule of energy per second.

Now if power is multiplied by time, you are back to energy. For instance, if 1 watt is multipled by 1 second, you have I watt-sec. Since 1 watt equals 1 joule/sec, 1 watt-sec equals I joule/sec x see, or I joule sec/sec. The sees can cel as you would expect in the ordinary algebraic manipu lation tG which units can be subjected, and you end with the statement that I watt-sec is equal to 1 joule and is, therefore, a unit of energy.

A larger unit of energy of this sort is the kilowatt-,hour (or kw-hr). A kilowatt is equal to 1000 watts and an hour - is equal to 3600 seconds. Therefore a kw-hr is equal to 1000 x 3600 watt-sec, or to 3,600,000 joules, or to 36, 000,000,000,OGO ergs.

Furthermore, since there are 4185 joules in a kilocalorie (kcal), 1 kw-hr is equal to 860 kcal or to 860,000 cal.

A human being who is living on 2500 kcal/day is de livering (in the form of heat, eventually) about 104 kcal/ hr, which is equal to 0.120 kw hr/hr or 120 watts. Next tirhe you're at a crowded cocktail party (or a crowded sub way train or a crowded theater audience) on a hot evening in August, think of that as each additional person walks in.

Each entrance is equivalent to turning on another one hun dred twenty-watt electric bulb. It will make you feel a lot hotter and help you appreciate the new light of understand ing that science brings.

But back to the subject. Now, you see, we have a variety of units into which we can translate the amount of energy ,resulting from the complete conversion of I gram of mass.

That gram of mass will liberate:

900,000,000,000,000,000,000 ergs, or 90,000,000,000,000 joules, or 21,500,000,000,000 calories, or 21,500,000 000 kilocalories, '600 kilowatt-hours. or 25,000, Which brings us to the conclusion that although the erg is indeed a tiny unit, nine hundred quintillion of them still mount up most impressively. Convert a mere one gram of mass into energy and use it with perfect efficiency and you can keep a thousand-watt electric light bulb running for 25,000,000 hours, which is equivalent to 2850 years, or the time from the days of Homer to, the present.

How's that for solving the fuel problem?

We could work it the other way around, too. We might ask: How much mass need we convert to produce I kilo watt-hour oi energy?

Well, if I gram of mass produces 25,000,000 kilowatt hours of energy, then 1 kilowatt-hour of energy is produced by 1/25,000,000 gram.

You can see that this sort of calculation is going to take us into small mass units indeed. Suppose we choose a unit smaller than the gram, say the microgram. This is equal to a millionth of a gram, i.e., 10-11 gram. We can then say that I kilowatt-hour of energy is produced by the conver sion of 0.04 micrograms of mass.

Even the microgram is an inconveniently large unit of mass if we become interested in units of energy smaller than the kilowatt-hour. We could therefore speak of a micromicrogram (or, as it is now called, a picogram). This is a millionth of a millionth of a gram (10-12 gram) or a trillionth of a gram. Using that as a unit, we can say that:

1 kilowatt-hour is equivalent to 40,000 picograms

I kilocalorie fl, 46.5

1 calorie 0.0465

1 joule 0.0195

I erg 0.00000000195

To give you some idea of what this means, the mass of a typical human cell is about 1000 picograms. If, under conditions of dire emergency, the body possessed the abil ity to convert mass to energy, the conversion of the con tents of 125 selected cells (which the body, with 50,000, 000,000 000 cells or so, could well afford) would supply the boY; with 2500 kilocalories and keep it going for a full day.

The amount of mass which, upon conversion, yields 1 erg of energy (and the erg, after all, is the proper unit of energy in the gram-centimeter-second system) is an incon veniently small fraction even in terms of picograms.

We need units smaller still, so suppose we turn to the picopicogram (10-24 gram), which is a trillionth of a tril lion of a gram, or a septillionth of a gram. Using the pico picogram, we find that it takes the conversion of 1950 picopicograms of mass to produce an erg of energy.

And the significance? Well, a single hydrogen atom has a mass of about 1.66 picopicograms. A uranium-235 atom has a mass of about 400 picopicograms. Consequently, an erg of energy is produced by the total conversion of 1200 hydrogen atoms or by 5 uranium-235 atoms.

In ordinary fission, only 1/1000 of the mass is converted to energy so it takes 5000 fissioning uranium atoms to produce I erg of energy. In hydrogen fusion, 1/100 of the mass is converted to energy, so it takes 120,000 fusing hydrogen atoms to produce 1 erg of energy.

And with that, we can let e mc,2 rest for the nonce.