fiveworlds Posted October 14, 2014 Posted October 14, 2014 (edited) Just as a little aside, nobody uses the concept of relativistic mass anymore. It's not very useful and tends to confuse people. Or just because the equation is wrong. I take a number of samples of several different elements all the same mass of course they are each going to release a different amount of energy because different elements aren't the same. As you can see in the graph. Edited October 14, 2014 by fiveworlds -1
ajb Posted October 14, 2014 Posted October 14, 2014 Or just because the equation is wrong. The equation for relativistic mass is perfectly okay. It is just the interpretation and use of the relativistic mass that is not so clear. The modern thinking is just to forget the notion of relativistic mass and stick to mass as the invariant mass. 1
Strange Posted October 14, 2014 Posted October 14, 2014 I take a number of samples of several different elements all the same mass of course they are each going to release a different amount of energy because different elements aren't the same. As you can see in the graph. That has absolutely nothing to do with the topic. 1
elfmotat Posted October 14, 2014 Posted October 14, 2014 Or just because the equation is wrong. I take a number of samples of several different elements all the same mass of course they are each going to release a different amount of energy because different elements aren't the same. As you can see in the graph. [iMAGE] I really don't know why you insist on doing this. I know this is irrelevant nonsense. You (probably) know this is irrelevant nonsense. Just about everybody knows it's irrelevant nonsense, yet you post it anyway. What satisfaction do you get out of constantly reaffirming your ignorance to the rest of the community? 1
fiveworlds Posted October 14, 2014 Author Posted October 14, 2014 (edited) stick to mass as the invariant mass. Very well calculate invariant mass(independent of velocity) for 1 particle. I am curious to see if you can. Edited October 14, 2014 by fiveworlds
Strange Posted October 14, 2014 Posted October 14, 2014 Very well calculate invariant mass(independent of velocity) for 1 particle. Mass is measured, not calculated. (At least, I can't think of an example where the mass of a particle is calculated. Theory will sometimes give an expected range.) 1
fiveworlds Posted October 14, 2014 Author Posted October 14, 2014 (edited) Mass is measured, not calculated. (At least, I can't think of an example where the mass of a particle is calculated. Theory will sometimes give an expected range.) Not an answer. What is the invariant mass of any one particle? I only want one and I don't care which one. This is really difficult because the planet has a velocity. Edited October 14, 2014 by fiveworlds 1
ajb Posted October 14, 2014 Posted October 14, 2014 What is the invariant mass of any one particle? I only want one and I don't care which one. The rest mass of the electron has been measured as about 9.11×10^−31 kilograms. 1
Strange Posted October 14, 2014 Posted October 14, 2014 . This is really difficult because the planet has a velocity. You do know that velocity is relative? 1
fiveworlds Posted October 14, 2014 Author Posted October 14, 2014 The rest mass of the electron has been measured as about 9.11×10^−31 kilograms. That's fine but I am talking about invariant mass independent of velocity. You do know that velocity is relative? I also know that everything has velocity
ajb Posted October 14, 2014 Posted October 14, 2014 (edited) That's fine but I am talking about invariant mass independent of velocity. That mass is the mass as measured in a frame co-moving with the electron. That is all most the definition of the rest mass. There is no dependence on the velocity of anything here, or really the relative velocity is zero. I also know that everything has velocity With respect to? Edited October 14, 2014 by ajb
fiveworlds Posted October 14, 2014 Author Posted October 14, 2014 With respect to? everything. Secondly taking your particle 9.11×10^−31 kilograms. In order to turn this into energy I apply pressure. When this particle turns into energy the amount of energy released is proportional to the difference in pressure between the particle and the outside environment. -1
swansont Posted October 14, 2014 Posted October 14, 2014 Or just because the equation is wrong. I take a number of samples of several different elements all the same mass of course they are each going to release a different amount of energy because different elements aren't the same. As you can see in the graph. ! Moderator Note Nothing to do with the OP, which is a question about a derivation.
Recommended Posts