In 1919, the mathematician Theodor Kaluza began to play around with Albert Einstein’s formulas for gravity. Out of curiosity, he reworked the equations to see how they‘d look in five, rather than four, dimensions. The exercise yielded an extra set of equations, which were the same as James Clerk Maxwell’s equations for the electromagnetic field. His idea of a fifth dimension had produced a mathematical unification of gravity and electromagnetism.

Since Kaluza’s calculations yielded an extra set of equations with Einstein’s formulas for gravity, it’s logical to assume that extending the number of known dimensions to five might also extend E=mc^2. This conclusion is supported by the paper “Did Einstein prove E=mc^2” which states, “The mass–energy formula cannot be derived by Einstein's 1905 argument, except as an approximate relation …” (Ohanian, H.C. Studies in History and Philosophy of Modern Physics Part B: Volume 40, Issue 2, May 2009, Pages 167-173) A more precise relation might be obtained in the following way –

The wave-particle duality of quantum mechanics can be described by starting with v = fλ (wave velocity equals frequency times Greek letter lambda which denotes wavelength). Velocity (speed in a constant direction) of a collection of particles like a car equals distance divided by duration. Since distance is a measure that has to do with space while duration is a measure that has to do with time, it equals space divided by time. (Greene, B. in "Speed", part of his "Space, Time and Einstein" course at http://www.worldscienceu.com/courses/1/elements/YhF9pw) Gravitational and electromagnetic wave motion (space-time motion) travels at c, the speed of light ie

(v= fλ) = distance/duration = space/time = c

A particle's velocity, whether the particle be a boson or fermion, is directly dependent on its energy – so it may be said that

E = (v=fλ) = distance/duration = space/time = c

This is not quite right since c represents energy alone, and space-time deals with mass-energy, so it's better to say

E = (v=fλ) = distance/duration = space/time = mc

What about the 2 in E=mc^2? In later papers Einstein repetitively stressed that his mass-energy equation is strictly limited to observers co-moving with the object under study, and co-movement may be represented by the exponent 2.

If Kaluza and the 5th dimension can extend E=mc^2, can they also extend photon-graviton interaction to the two nuclear forces. This would be done by the quantum spin of the photon (one) and the spin of the graviton (two) producing, via Hawking, * 1 +1/2 -1/2=the spin of the weak and strong forces’ bosons (one).

* Professor Stephen Hawking writes, (Hawking, S.W. A Brief History of Time. Bantam Press, pp.66-67) -

"What the spin of a particle really tells us is what the particle looks like from different directions."

Spin 1 is like an arrow-tip pointing, say, up. A photon has to be turned round a full revolution of 360 degrees to look the same.

Spin 2 is like an arrow with 2 tips - 1 pointing up, 1 down. A graviton has to be turned half a revolution (180 degrees) to look the same.

Spin 0 is like a ball of arrows having no spaces. A Higgs boson looks like a dot: the same from every direction.

Spin ½ is logically like a Mobius strip, though Hawking doesn’t specifically say so. This is because a particle of matter has to be turned through two complete revolutions to look the same, and you must travel around a Mobius strip twice to reach the starting point.

📷

Figure 1 – MOBIUS MATRIX (Mobius equals a,b,c,d,e array) (the figure is attached)

(Mobius Matrix and Our Familiar Dimensions) Width a is perpendicular to the length (b or e) which is perpendicular to height c. How can a line be drawn perpendicular to c without retracing b’s path? By positioning it at d, which is then parallel to (or, it could be said, at 180 degrees to) a. d is already at 90 degrees to length b and height c. d has to be at right angles to length, width and height simultaneously if it's going to include the Complex Plane's vertical "imaginary" axis in space-time (the "imaginary" realm is at a right angle to the 4 known dimensions of space-time, which all reside on the horizontal real plane). In other words, d has to also be perpendicular to (not parallel to) a. This is accomplished by a twist, like on the right side of the Mobius strip, existing in the particles of matter composing side a. In other words, a fundamental composition of matter is mathematics' topological Mobius, which can be depicted in space by binary digits creating a computer image. The twist needs to be exaggerated, with the upper right of the Mobius descending parallel to side "a" then turning perpendicular to it at approximately the level of the = sign, then resuming being parallel. Thus, 90+90 (the degrees between b & c added to the degrees between c & d) can equal 180, making a & d parallel. But 90+90 can also equal 90, making a & d perpendicular. (Saying 90+90=90 sounds ridiculous, but it has similarities to the Matrix [of mathematics, not the action-science fiction movie] in which X multiplied by Y does not always equal Y times X. The first 90 plus the second 90 does not always equal the second 90 plus the first 90 because 90+90 can equal either 180 or 90).

90+90=90 is transferred from the quantum world to the macroscopic by 1+1=1 because this second equation is useful in the practical, everyday experience of humans and animals. Referring back to the spin of the weak and strong forces’ bosons, the Mobius Matrix’s 90+90=90 becomes not 1+1=1 but 1 +1/2 -1/2 = 1. The compactification of Kaluza’s 5th dimension by Oskar Klein, as well as the compactification of string theory in which four dimensions of spacetime are macroscopic and the others form a compact manifold too small to be observed, may be represented by the second 1 becoming (+1/2 -1/2).

More Rodney Bartlett's questions See All
Similar questions and discussions