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In Special Relativity, there are two kinds of mass one can define for a system, right?:

$$m_{rel} = \frac{E_{tot}}{c^2}$$

and

$$m_0 = \frac{\sqrt{E_{tot}^2 - p_{tot}^2c^2}}{c^2}$$

Both have their pros and cons. The first one is conserved (because $E_{tot}$ is) but not invariant. The second one is invariant (because $E_{tot}^2 - p_{tot}^2c^2$ is) but not conserved.

Usually, the second one is preferred over the first. Why is that? Why is in-variance preferred over conservation?

PhyEnthusiast
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  • the invariant form furnishes the basis for the way relativity is currently taught at the college level. the claim is made by practitioners that this approach facilitates the teaching of more advanced topics in the field of relativity. The bottom line is that while either formulation will always yield the correct answer, the invariant form is "preferred". – niels nielsen Aug 26 '18 at 06:58
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    Well, the first “mass” is just energy up to a constant factor which we usually set to 1. So we’re already considering it. It’s pointless to give the same conserved quantity another name. Might as well use the word “mass” for something else, so it’s actually useful. – knzhou Aug 26 '18 at 07:04
  • Your first equation is false for photons while your second equation is true for photons – John Rennie Aug 26 '18 at 07:09
  • @JohnRennie Couldn't we just define a "relativistic mass for photons" of frequency $\nu$ to be $h\nu/c^2}$ – PhyEnthusiast Aug 26 '18 at 08:17
  • @PhyEnthusiast but you can't feed that mass into Newton's equations and expect it to produce any sensible predictions. – John Rennie Aug 26 '18 at 08:22
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    The quantity $m_\text{rel}$ defined by the OP is the timelike component of the momentum vector. We don't normally give a special name to a particular component of a vector. –  Aug 26 '18 at 12:48
  • Both of your expressions are written for single particles, but you haven't really explored the consequences of various definitions unless you consider their implications for the mass of systems. https://physics.stackexchange.com/q/387971/520 https://physics.stackexchange.com/q/398570/520 https://physics.stackexchange.com/q/289470/520 Of course both points of view lead to the same physics, but I content that one is clearer, more teachable, and simply easier than the other. – dmckee --- ex-moderator kitten Aug 26 '18 at 17:23
  • Also you are wrong to imply that the invariant mass is not conserved (or more likely you are trying to use the inapplicable notion that you find the mass of a system by adding the mass of its parts which simply isn't true). To be clear: the invariant mass of a closed system is a conserved quantity. Why? Because the four-momentum of a closed system is a conserved quantity. So the question is founded on a misapprehension. – dmckee --- ex-moderator kitten Aug 26 '18 at 17:29
  • @dmckee I am not sure I understand. The invariant mass of an electron and a positron together in a closed system is $2m_e$. But after annihilation, the invariant mass of two photons is $0$. So, invariant mass was not conserved, ryt? – PhyEnthusiast Aug 27 '18 at 08:02
  • You find the invariant mass of systems by summing the four-vectors of the parts and then finding the norm of that sum. That means the mass of the system depends on the motion of the particles relative the CoM of the system, but not on the motion of the CoM relative the observer. Consequently the invariant mass of a system consisting of an electron and a positron is not in general $2m_e$. It also means the invariant mass of a closed system is a conserved quantity. – dmckee --- ex-moderator kitten Aug 27 '18 at 17:31
  • @dmckee I think I understand now. So, a system of 2 photons could have a total invariant mass $\neq 0$? Can you write an answer, so that I may accept it? – PhyEnthusiast Aug 28 '18 at 00:56

2 Answers2

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The invariant mass $m_0$ is conserved by definition of its being a constant in time. There is no other equation for $m_0$ but $m_0=\text{const}$.

The mass $M_0$ of a compound system can be calculated (expressed) via the masses of interacting particles.

Vladimir Kalitvianski
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  • Invariant mass is not conserved. For example if two objects with mass $m$ collide and fuse to form a new single object the invariant mass of the new object will be greater than $2m$. This happens because the kinetic energy originally present is converted to mass. – John Rennie Aug 26 '18 at 09:19
  • @JohnRennie: What you are saying is that masses are not additive in general case. – Vladimir Kalitvianski Aug 26 '18 at 10:21
  • I'm saying that total invariant mass is not conserved in the general case. Your question starts by claiming that it is. I assume you mean that only for the special case of elastic collisions, but if so you should say so as right now it's very misleading. – John Rennie Aug 26 '18 at 10:41
  • @JohnRennie: $(m_0)_i$ are constants in the equations of motion for each interacting particle. The rest is calculable from them, so the compound system properties are unambiguous functions of $(m_0)_i$. $M_0$ of a compound system makes only sense if you can write an equation of motion for the compound (interacting) system as a whole which is a rare case. – Vladimir Kalitvianski Aug 26 '18 at 10:59
  • @John, the invariant mass of a closed system is conserved (because 4-momentum is conserved). The sum of the masses of the components of a system is not conserved, but that is not the mass of the system. Nor is the sum of the relativistic masses of a closed system conserved unless you include the fields in the sum. – dmckee --- ex-moderator kitten Aug 26 '18 at 17:17
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To an extent this is a matter of opinion since both invariant and relativistic mass can be used in calculations. However I suspect the opinion of most physicists these days would be that the invariant mass is a more useful concept.

If we start with Newtonian physics then we know that the basic equation of motion is Newton's second law:

$$ F = ma = \frac{dp}{dt} $$

If we now consider relativistic velocities then we find this no longer correctly describes the motion. However, if we use the relativistic mass $m_r$ we find the equations do work. This I suspect was why the concept caught on originally. If you regard Newton's laws as fundamental then it seems natural to redefine the mass to keep Newton's laws working.

But the force, acceleration, momentum, etc used in Newtonian mechanics are three-vectors and we know that special relativity is most naturally formulated using four-vectors. Indeed if you use four vectors you can simply write:

$$ \mathbf F = m\mathbf a = \frac{d\mathbf p}{d\tau} $$

and now $m$ is the invariant mass i.e. the same as the mass of the object in the objects rest frame. So if you regard four-vectors as fundamental then it seems natural to redefine Newton's laws and keep the mass constant.

As to which is better ultimately that is a matter of opinion, though I doubt you will find many physicists today who consider the concept of relativistic mass better. The invariant mass has a natural physical interpretation as the norm of the four-momentum (which immediately explains why it is an invariant since the norm of any four-vector is a scalar invariant).

Using this formalism also gives us the relativistic equation for the total energy:

$$ E^2 = p^2c^2 + m^2c^4 $$

that applies to everything - massless particles as well as massive ones. Unless you subscribe to this view you have a problem explaining why massless particles can have energy without any mass.

John Rennie
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