| Oregon Magazine | Kick the habit at Serenity Lane |
| Relativity for Dummies
The mass gain problem: (For amateur physics nuts,
only.)
Email #1 E=MC2 Energy and mass (OMED: matter) are the two fundamental kinds of existence. Under the proper conditions, each may become the other. At some future time, each may switch back to the opposite state. Regardless of the form involved, a unit of one contains the potential of the other. Mass contains potential energy. Energy contains potential mass. Or is it something more than just "potential?" The velocity of light is the speed limit
of the universe. .When matter of any kind accelerates, it does
so as a result of applied energy, and gains mass. At a velocity close to
the speed of light, the mass gain is rapid.. Not potential mass gain,
but real mass gain. A dime accelerated to 99.999999% of the velocity
of light does not contain just the potential mass that would be expressed
in collision with another body, but in addition as a result of the acceleration
has actually gained nearly infinite, presently operative mass of its own
– mass identical to that of the matter of a slowly moving neutron star
in the sense that to add velocity to the dime's journey would require a
nearly infinite number of units of real, not potential, energy.
The need for ever increasing amounts of energy to accelerate the dime is the result of increased resistance in the direction of travel (if the dime were passing through a medium like air or water), or is the result of inertial resistance of the sort described in Newtonian physics. That is “real” inertia. Real mass resisting a change in uniform motion. It suggests to me that what we call reaction mass does not push the rocket to the left because it throws mass out to the right. It suggests rather that reaction mass pushes with part of its volume, and transfers part of those of its qualities known as inertial rest mass to the vehicle it is accelerating. The acceleration process adds weight to the object accelerated. Why then does the accelerated object not weigh more when it comes to rest on a surface? Because the added real mass is transformed into energy during the deceleration process? Reaction mass acceleration adds real mass to a ship, while reaction mass deceleration subtracts an equal amount of real mass from the same ship? (The mass is shed instantly in the form of heat and radiation during impact, or shed slowly by the braking effects of rockets or some medium like atmosphere.) In all my reading, I have never seen an explanation of where the mass increase comes from. The above is my best current reasoning on the subject. The problem with it is that if written in one of your treasured formulae, it would state that mass and momentum are equivalent in some degree or other.. M=mv/tf Where M is momentum, m is mass, v is velocity and tf is a transformation constant of some sort like Hubble's or Planck's, representing a heretofore undesignated mass transfer mechanism activated by the adding of velocity to any body in the state we call matter. It is no wonder that you people are crazy. Jacques Clouseau Email #2 M=mv/tf Where M is momentum, m is mass, v is velocity and tf is a transformation constant of some sort like Hubble's or Planck's, representing a heretofore undesignated mass transfer mechanism activated by the adding of velocity to any body in the state we call matter. That may be backwards Maybe it's M for mass gain in any given volume of matter is equal to the velocity times the transformation factor. M=Vtf But, that doesn't do it either. It's closer, but not right. It doesn't work in a geometric progression, if that's what you call it. Increased increases with each unit of acceleration. You probably have some symbol to attach to the tf function that would identify it as a multiplier with a regularly rising exponent, if that's the right term. (I know I used mass interchangeably for matter in that other email.) (Ignore me. You and I both know that I don't know [expletive deleted] about this. It just fascinates me.) L Email #3 M=mv/tf I'm back to the first one. It expresses a mass value for M (momentum) as defined by an exponential transformation mechanism additive. Momentum is equivalent to mass in some way related to velocity expressed as a percentage of C, and as defined by some sort of progressive rheostat with the high gain numbers on the far end.. Being a progressive, you should love that. Now, go away and stop bothering me. If you have any other questions about relativistic physics, just email them to me and I'll get to them when I have space-time. LL Postscript: The astrophysicist sent me a note. He said my formula was correct, except for the definition of "tf," the transformational factor. It isn't, he said, a fundamental mathematical expression that desribes a property of the universe like Planck's Constant, but rather a simple (abstract) computational expression, like a fraction. A tool that must be used to predict the values that will result from a particular physical event. At least, I think that's what he said. © 2002 Oregon Magazine |
| Around
Oregon News Digest | Arts&Lettres
| Business
| Editorial
| Events | Life&Styles
Natural History | Outdoor | SciTech | Sports | Travel | Peg's Bottom Gazette | Contact |