Einstein, Pythagorean, E=MC Squared, and the String Theory of Everything
A Simple Little Challenge
I THOUGHT I would take a break from my normal topics and begin a hub in another area that has always held great fascination for me ... science. As I have mentioned in my profile and other places, Science aka Natural Philosophy, plays a major role in my overall philosophical beliefs. For example, I think science holds the key to understanding Free Will, but, that is not the purpose of this hub. What I would like to do in a few short sections is 1) introduce why Pythagorean's Theorem works the way it does (you remember this one don't you; Hypotenuses, sum of squares, and all that? If not. patience) and 2)
derive, in layman's terms, Albert Einstein's famous equation, E = MC2. Shouldn't be too hard, don't you think?
How did this project come about? On a road trip from Hot Springs, AR back to my home in Florida. When I take these trips I entertain myself by listening to lectures on various subjects of interest; for me, this is often music to my ears and since I drive by myself, nobody else has to suffer my strange affliction. Anyway, on this trip I played a lecture title "Superstring Theory: The DNA of Reality " by Professor S. James Gates, Jr., University of Maryland at College Park. In the course of this lecture, Professor Gates uses the Pythagorean Theorem in many of his descriptions on String Theory, so, he laid out the foundation behind the theorem in a way I have never seen before and in doing so made something that was basically opaque to me, clear. At the same time he stated you can use the principals of this ancient theorem to derive Einstein's famous equation that relates energy and matter, E = MC2
Pythagorean Theorem: Simplest Form in 2-Dimensions
WHAT I am about to show is probably well known to many but was brand new me; this shows you how much I paid attention in college and I was a math major to boot, lol; rote is a wonderful thing. OK, for those who don't recognize Pythagorean's Theorem yet, it is theorem that says:
"The square of the length of the hypotenuse (c) of a right triangle is equal to the sum of the square of each leg (a, b) of the triangle " or C2 = A2 + B2
As an aside, it is interesting to note that Pythagoras of Samos (570 BC - 495 BC) is not who actually derived this equation, it was well known by the Babylonians 1,000 years earlier.
I suspect my high school instructors tried to teach me why this equation worked but, if they did, it never sunk in. All I ever knew was the formula, when and how to apply it. Well, to understand how we get from C2 = A2 + B2 to E = MC2 we need to actually know why Pythagorean's Theorem really works; so, here goes.
If you look at Chart 1, you will see I drew two squares of equal size; in this case all sides are 5. That means, of course, that the Area of each square must be 25. Now, as you can also see that I stacked the two squares on top of each other so that they have one side in common; that side is the base of one square and the top of the other. From this it is easy to see that the Areas of the two squares are and must be the same.
Now, what is a right triangle? It is simply a triangle which has the property that one of its angles is exactly 90 degrees; nothing more, nothing less. Since a triangle, by definition, is made of three sides and three angles, we can label these sides A, B, and C; and angles < a, < b, < c, respectively. By convention, the hypotenuse, the side opposite the 90 degree angle is labeled C.
In our first example, Chart 1, something is missing, side 'B'; it is shown with length zero. Even though this picture looks like two squares stacked on top of each other, it really is a Right Triangle. How, you ask? Simple, I say. One of the three angles is zero degrees leading to the side opposite (B) being length zero.
Since this is really a right triangle, Pythagorean's Theorem applies. Consequently, you should be able to see what the equation is actually saying is that the area of the square attached to the hypotenuse (C) is equal to the sum of the area of the squares attached to the lines opposite the other two angles of the triangle. In this first case, since one of the angles is zero, the side that would be opposite of that angle is non-existent and we are left with the stacked squares.
In Chart 2, you see we raised one corner of the Green square a bit while maintaining the length of side 'C' so that the area of the square does not change. Well, when we do this, two things happen: side 'A' of the Red square gets shorter and we create side 'B' of a new square, the Blue square; remember, we are dealing with a right triangle here. What is happening here? We are maintaining equality, that is what.
Because we are dealing with a closed system, the Green and Red squares comprise the total system and they must be equal in all dimensions because they are squares and share a common side, the initial equality must be maintained. Just because we change the position of one of the squares, so long as we retain the integrity of the right triangle, we don't invalidate the relationship.
So, as we lift the Green square we create a recognizable right triangle, but, in doing so we shrunk the Red square, in our example for 5 units to 4 units. Given side 'A' is now 4, that means the area of Red square is 16 which is now less than the Green square. This means, of course, that we need to bring the total area of the non-Green squares back up to 25. This is accomplished with the creation of the new leg 'B' and the Blue square. As you can see, the Blue square requires an area of 9 so that with the Red square we still have a total area of 25.
No matter how little or how much you raise the Green square, this must be true. In order to maintain the equality within this closed system, you will have to add enough area to the Blue square such that, when combined with the Red square, it equals the area of the Green square.
To bring us back from the areas of the squares to length of the legs of a right triangle all you need to note is that the area of any one of those squares is exactly one of its sides multiplied by itself or, said another way, one of its sides squared.
Pythagorean Theorem in 3-Dimensions
Expanding Our View
Pathagorean's Theorem, as we normally understand it, works in two dimensions; some paired combination of length, width, or height where any two of these dimensions correspond to the 'A' and 'B' legs of the right triangle. Without going into any proof, let me state the obvious, Pathagorean's Theorem also works in three dimensions, length (L), width (W), and height (H). There is nothing tricky about the new formula, it is simply adding one more term to the old formula. For reasons which will become apparent shortly, I am going to replace the 'A' and 'B' in the equation with either 'L', 'W'. or 'H' while leaving the hypotenuse the same, 'C'.
So, assume first we are dealing with length and width, then we have C2 = L2 + W2 for our two-dimensional world. If we want to talk in terms of all three dimensions, we get, C2 = L2 + W2 + H2. As it turns out, this same expansion can be used regardless of the number of dimensions we want to talk about; all you do keep adding squared terms. For our purposes, however, we are only going to add one more which I will call 'T' so that my new "Pathagorean's Theorem" will read C2 = L2 + W2 + H2 + T2.
Pythagorean Theorem in 4-Dimensions with Units of Measure
WHAT IS this 'T' dimension? Well, remember who we are talking about here, Einstein. What is one of the things Einstein is most famous for? Proving to the world that the passage of Time is not constant but can change. In other words, the passage of 10 seconds as seen by me, may be the passage of 20 seconds as seen by you. The upshot of Albert Einstein's science is that
Time is a dimension no different than length, width, and height; time is simply a fourth dimension and is the 'T' in our expanded Pythagorean Theorem.
With the addition of 'T' dimension, some have started calling the resulting hypotenuse of our four-dimensional right triangle the "Einstein Hypotenuse EC".
I will try to stay as far away from mathematics as possible so that there is at least a modicum of a chance I won't lose my non-math oriented readers but nevertheless some will be necessary.
The first complicating factor we must introduce is that of units. So far in the charts I presented, I used simple numbers with no real representation of what they stood for. Most probably, you took them to mean distances of some sort, but I never really said until I changed the lables for 'A' and 'B' to 'L', etc. Now, however, I do mean distances, and, since I am writing to a mostly American audience, although I must tip my hat to the many Canadians that follow me as well, I will use miles as my distance measure, although it really doesn't matter. For time, I will use the normal unit of seconds.
This immediately presents a problem because, as you can see from Chart 3, we are mixing "miles" and "seconds"; mathematically, you can't do that. As a result, we need to start doing "math magic"; it is also, as it turns out, the first step in turning a "sow's ear into a silk purse".
OK, what is the problem? We have "miles" squared equal to three times "miles" squared plus "seconds" squared; we got to do something about those seconds. What we must find is a constant which relates distance with time and, guess what, we have one, provided by none other than Mr. Einstein ... light or rather the Speed of Light, 'c'. According to Einstein, the speed of light is a constant, some 186,282 miles/sec, so it does not fundamentally disturb anything by multiplying the Time dimension by this constant. But, it does simply things for us a bit because the units of 'c' is miles/sec so, when c is multiplied by Time all you have left, in terms of units, is miles or, in our situation, miles squared. As a result, this "time" term is now in the same units as the rest of the equation and the equation is in balance.
Therefore. referring to Chart 3, we have Einstein's Hypotenuse, EC2 = L2 + W2 + H2 + c2T2, where the units are in terms of length. Even the time dimension is in terms of length because we multiplied time by the speed of light, a constant.
(Note: Einstein did one more thing to adapt the Pythagorean Theorem to his Theory of Special Relativity, he changed the signs on the length terms from positive to negative so that the equation actually reads EC2 = c2T2 -L2 - W2 - H2. Why he did this is beyond my understanding at the moment but the fundamentals behind the Pythagorean Theorem do not change. For my purposes, as you will see, the negative signs do not matter so I will leave the equation alone.)
Einstein's Genius: Representing Momentum and Energy in terms of Pythagorean's Theorem
Getting to E = MC Squared
AS YOU have seen, Pythagorean's Theorem is used to talk about distances, inches, feet, miles, etc. Even so, it was Einsteins genius that saw how it could also be used relative to Momentum and Energy. For those who don't know, Momentum is the Mass of an object times its Velocity while Energy, the ability of a system to do work, is a constant times Mass times Velocity2. Notice also that Velocity is a Distance divided by time. Since both Momentum and Energy are, so to speak, a function of Distance, they can, with the proper mathematical manipulations, be thought of as Areas such as we have in our original formulation of Pythagorean's Theorem. These units are noted in Chart 4 and, when you only consider Pathorean's Theorem in terms of momentum, then it is easy to see the area of the hypotenuse squared is (Mass x Distance/Time)2
Mathematics allows you to multiply both sides of an equation by a constant without changing the nature of the equation. So, if we do that here and multiply each side by the speed of light squared, which has the same units as the existing terms, specifically (distance/time)2. Consequently, as you can see in Chart 4 we can express the left-side of the Pythagorean Theorem as mass2 x c2 or m2c2.
Let's add in, now, the 4th dimension of Energy, where the first three dimensions are momentem in the up-down, left-right, and back-forth directions. The problem with Energy is its terms, mass x distance2/time2. This must be corrected and can be done so by dividing by the speed of light 'c' which gives (mass x distance/time)/c.
So, substituting back into E2, we get ((mass x distance/time)/c)2 or mass2 x (distance/time)2/c2.which looks exactly like the left-hand term we previously developed. Chart 5 shows this.
One more assumption is now required, assuming that the system we are talking about is at rest then an interesting thing happens. Objects with zero velocity have zero momentum, therefore, all the Momentum terms in EInsteing's Hypotenuse equation become zero.
From here it is a simple matter to finish our work. From Chart 5, we see that (mass2 x (distance/time)2 is equal to E2 so we have E2/c2. To put it all together and flipping sides, we get E2/c2 = m2c2. Multiplying each side by c2 you get E2 = m2c4. Taking the square root of each side and guess what, one of the most famous equations in the world emerges
- E = mc2. Voila! Who needs the Calculus, algebra is just fine thank you.
(To you real mathematicians out there, be kind in your comments if you would. It has been a decade or so since I delved this deep. which I realize is still just the surface, into the mechanics of algebra and units. Let me know if I made any logical errors in getting from the two knowns, Pythagorean's Theorem and Einstein's equation relating energy and mass - My Esoteric)
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