[Mentor's note: this is a continuation of a thread in the Quantum Physics forum: https://www.physicsforums.com/showthread.php?t=144296 which turned into a discussion of the various kinds of mass in relativity.] Originally Posted by Los Bobos "It is actually more consistent to say, that from SR we know that mass does not increase with velocity. Here you would have to think what is momentum in SR. And of course the biggest problem of elementary QM, the difference in the handling of spatial and temporal coordinates." Well, no. I haven't studied either of QM or SR in real depth.I'm still in high school.:( How mass does not incrase with velocity? I will be very thankful if you clearly explain it in bit detail. Thanks
The increase of mass with velocity is just an evil pedagogical plan and a historical misunderstanding (this was included in the Lorentz-Fitzgerald theory (if I remember correclty, transverse masses and so on)). This was a good text about the subject http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html. But nevertheless we have only one mass which is invariant and anything else makes very little physical sence.
When people talk about "mass" in SR, they are usually referring to one of two different kinds of "mass": invariant mass (also often called rest mass) which does not change with velocity, and relativistic mass which does change with velocity. Relativistic equations involving mass can be written using either kind of mass. Of course the equations look different depending on which kind of mass is being used, but if you use them properly they predict the same results for physical experiments. It's ultimately a matter of convention or preference which kind of mass people prefer to talk about, and there have been repeated arguments about this here. Most popular-level books and articles about relativity talk about relativistic mass. In serious introductory textbooks (college/university level) there has been a trend in the last couple of decades away from relativistic mass and towards invariant mass, which some people disagee with. In elementary particle physics, people generally use invariant mass. (I worked in that field as a grad student, and I personally do not remember anyone ever using relativistic mass, among the people that I worked with or among the papers that I read.) Further discussion of this issue really belongs in the relativity forum.
That's fine. Thanks Los Bobos and jtbell.This means that invariant mass and relativistic mass are the two forms of one single entity. What is that "entity"? hope my question is not silly:) But one question popping out. What about the gravitational mass? How is it related to any of these two ?
The term "physical mass" is rather vague. I usually stress the fact that independence of the frame of reference is a highly desirable quality, and it is this independence of frame that invariant mass has (at least for isolated systems), and that relativistic mass lacks.
That "entity" is that particle's momentum-4vector [tex]p^a[/tex]. Up to conventional signs and factors of c, Invariant mass is the norm of that 4vector: [tex]\sqrt{p_ap^a}[/tex]. Relativistic mass is the temporal-component of that 4vector for some observer with unit 4velocity [tex]u^a[/tex]: [tex]p_au^a[/tex]. Note that both are scalars, which are invariant...the same for all observers [with unit 4velocity [tex]v^a[/tex]]. The relativistic mass always requires specification of an associated observer with unit 4velocity [tex]u^a[/tex].
Originally Posted by"Robphy" "That "entity" is that particle's momentum-4vector . Up to conventional signs and factors of c, Invariant mass is the norm of that 4vector: . Relativistic mass is the temporal-component of that 4vector for some observer with unit 4velocity : . Note that both are scalars, which are invariant...the same for all observers [with unit 4velocity ]. The relativistic mass always requires specification of an associated observer with unit 4velocity ." By that I suppose that mass is a scalar qty and hence invariant. But the relativistic mass is just due to relative motion between two observers. But then how mass is defined in SR and QM and GR ? Is the definition of mass taken the same in all these fields? Please explain. Principle of equivalence is applied in GR while not in SR and QM why? Thanks.
mass? That is a very lucide explanation. When we start deriving the bazic equations of relativistic dynamics we start with the invariant m0. Volens-nolens we arive at a given point of the derivation to gammam0, Physicists, as well trained name givers call it relativistic mass. Others multiply it with c^2 and call it energy. Is there more to say for a beginner? Regards
Please have a critical look at Physics, abstract physics/0605203 Relativistic dynamics without conservation laws We show that relativistic dynamics can be approached without using conservation laws (conservation of momentum, of energy and of the centre of mass). Our approach avoids collisions that are not easy to teach without mnemonic aids. The derivations are based on the principle of relativity and on its direct consequence, the addition law of relativistic velocities. Regards
But my question reamains unanswered! As in post #8 "By that I suppose that mass is a scalar qty and hence invariant. But the relativistic mass is just due to relative motion between two observers. But then how mass is defined in SR and QM and GR ? Is the definition of mass taken the same in all these fields? Please explain. Principle of equivalence is applied in GR while not in SR and QM why?" At least please answer my questions.
Mass is the norm of the momentum 4-vector. It is the same and the one and the only everywhere. Your second question I don't understand.
Please read posting #3 again. There is no single quantity that everyone agrees on calling simply "mass" in relativity. There is no single quantity in relativity that has all the properties of classical mass and serves all the purposes of classical mass.
jtbell, you are completely right in the statistical sense , but for me there is only one mass. Relativistic mass causes some logical inconsistensies rising from incorrect use of simultaneity (at least it seems so to me). Relativity is not classical and therefore it makes no sense to me in explaining it in classical terms. Is "in explaining it in" english? :).
That's why a mathematical formulation (as I gave #7) helps clarify what is going on. Of course, one may still have to interpret what is going on... but at least we have a mathematical structure and rules of mathematics to guide us. Certainly, if the mathematics leads us somewhere that is inconsistent with the physics, we need a better mathematical model.
Yes, but for me there is no logic in calling the temporal part of the 4-momentum "mass" as it is not mass which is the norm of this 4-vector :). I would like to see for example the lagrangian of QED written with this weird mass. "The road to emitter theory is paved with relativistic masses" ;).
I'm not necessarily advocating its use. I'm advocating a less-ambiguous mathematical language of what has already been defined by others... so as to avoid "loose talk" using overused (overloaded in computer-speak) terms inspired by classical thinking. Once given a mathematical formulation, its physical meaning [especially what it depends on, etc] is clarified. Then one can judge how useful or not it is in whatever context you want to apply it in.
Ok then. what should be the correct answer to my question posted - "Can we mix SR and uncertainty principle? for ex. from uncertainty principle we have [tex]\Delta x \Delta p \geq h/4\pi [/tex]Books then write [tex]m \Delta x \Delta v \geq h/4\pi [/tex] with an assumption that m can be measured accurately. However form SR we know that mass depends on velocity; with mass increasing with velocity. Now if we can't measure v exactly how can we measure m exactly?" I am bit confused with all these discussions. Please help me out.
Maybe it will help in this case to think of the momentum in terms of the invariant (rest) mass instead of the relativistic mass: [tex]p = \frac{m_0 v} {\sqrt{1 - v^2 / c^2}}[/tex] Then we have [tex]\Delta x \Delta p = m_0 \Delta x \Delta \left( \frac{v} {\sqrt{1 - v^2 / c^2}} \right)[/tex] and we can express that complicated [itex]\Delta[/itex] on the right in terms of [itex]\Delta v[/itex]. I'm in a bit of a hurry so I won't do that last step right now.