Originally Posted by **Deco** Again I'd argue it's a mass-equivalence. |

But you haven't argued it other than saying "because $\displaystyle p = \frac{h}{\lambda}$" which really isn't an argument. All you did is to suggest that it

*should more appropriately named* "relativistic mass-equivalence

**.**" Making a suggestion is far from putting forward an arguement. What are the

*reasons* for your suggestion. The relationship $\displaystyle p = \frac{h}{\lambda}$ is a quantum mechanical relationship and not something of practical use in classical mechanics since in practice the wavelength can't be measured for macroscopic objects. I recommend that you merely propose a definition for the term

*mass* and the limits in which your definition should hold. It is then a mere matter of calculation to determine whether it depends on speed. What you call other quantities is merely a matter of taste after that. Some physicists refer to relativistic mass using such terms as "apparent mass" etc.

Originally Posted by **Mr. Rogers** Okay. Yeah, that makes sense to me... though it may make some calculations difficult. |

That's the nature of the beast, i.e. special relativity is more complicated than non-relativistic mechanics.

Originally Posted by **Mr. Rogers** All equations would have to incorporate the fact that an object mass would change as it's velocity does... well, at least when the velocity of the object is significant. |

That is done by replacing

*m* with $\displaystyle \gamma m$.

*m* is then referred to as the

*proper mass* of the object.

Originally Posted by **Mr. Rogers** I've spent a good amount of time trying to accurately determine the Kinetic energy of a 10kg mass moving at the speed of light. |

No object which can be at rest in any inertial frame of reference can move at the speed of light. Any answer you've gotten trying to make such a calculation would be wrong. The derivation for the kinetic energy of a body is given in my website at

http://www.geocities.com/physics_world/sr/work_energy.htm. In that page I prove the well known result that the kinetic energy

*K* of a body is given by the relation

$\displaystyle K = (\gamma - 1)mc^2$

As v -> c, $\displaystyle \gamma$ -> infinity.

Originally Posted by **Mr. Rogers** .. my point is that applying this concept is insignificant unless dealing with a significant velocity (nearing the speed of light). |

Agreed. The first thing you learn when you study relativity is that when the speed of light is small compared to the speed of light one can ignore relativistic effects in many cases. But even when the speed is small there are other effects which might not be ignorable. I'm sure you haven't run into them yet, especially since most physicists are unaware of if but it's a fact that stress contributes to inertia. That means that if body is under stress then the stress itself contributes to its inertia. I.e. if you take two bodies which are otherwise identical (i.e. have the same energy as measured in the bodies rest frame) then the momentum will be different. The momentum will also depend on the bodies orientation with respect to its motion.

Originally Posted by **Mr. Rogers** I'm sorry, It appears I've gotten too involved in the concept and have moved way from the question. So in answer to your original question, the most advanced physics class in my high school did not mention this at all, we never even discussed the formula "E=mc^2" in very much detail if at all. So, naturally, we've not discussed the debate regarding this concept. |

Don't appologize Mr. Rogers. I appreciate your thoughts.

Thanks folks