# I "relativistic mass" still a no-no?

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1. Nov 11, 2016

The Wiki article https://en.wikipedia.org/wiki/Mass_in_special_relativity seems to advise "don't use the expression 'relativistic mass'; stick to 'relativistic momentum' pγ". So what does one do if Alice were to measure Bob's mass while Bob is traveling at velocity v with respect to Alice, and Alice wants to compare her measurement to his rest mass that she happens to know. Would one say that Alice can't measure Bob's mass directly; that she could only measure his momentum?

2. Nov 11, 2016

### Dr. Courtney

Any time you think of measuring a mass, you need to specify a technique, which invariably requires whether you are measuring the gravitational mass, the inertial mass, or the mass-energy (E = mc^2).

Here's a nice paper showing how gravitational and intertial masses can be measured at a distance between objects in relative motion, but here, the earth-moon system is used, and odds are Bob and Alice are moving in a straight line relative to each other.

Lunar Laser Ranging Tests of the Equivalence Principle

3. Nov 11, 2016

### vanhees71

It's good advise not to use old-fashioned concepts from times where the mathematics of SR hasn't been understood in its whole glory yet. That was a very short period between 1905 and 1907, when Minkowski introduced the four-dimensional tensor formalism of the spacetime manifold, named after him Minkowski space, which is a pseudo-Euclidean affine space with a pseudometric of signature (1,3) or equivalently (3,1), depending on the convention you are used to. I'm used to the high-energy-particle physicists' "west-coast convention" with the (1,3) signature.

This leads to the covariant definition of the fundamental dynamical quantities energy, and momentum. Using the fact that after introducing a Minkowski reference frame (inertial frame) you can describe the trajectory of a massive particle by the world line, $x^{\mu}=x^{\mu}(\tau)$, where $x^{\mu}$ are the (contravariant) components $(x^{\mu})=(x^0,\vec{x})=(ct,\vec{x})$ where $c$ is the speed of light in a vacuum, and $\tau$ is the proper time of the particle, defined by
$$\mathrm{d} \tau=\frac{1}{c} \mathrm{d} s = \frac{1}{c} \sqrt{\eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}}, \qquad (*)$$
where $(\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$ are the Minkowski-metric components.

Then with the invariant (Minkowski scalar) mass of the particle one defines the four-momentum (which is a Minkowski four-vector) by its components
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
It obeys the constraint
$$p_{\mu} p^{\mu} = (p^0)^2-\vec{p}^2=m^2 c^2,$$
which follows from the definition of the proper-time increment.

To interpret the components we check the non-relativistic limit, when $|\vec{v}|=|\mathrm{d} \vec{x}/\mathrm{d} t| \ll c$. To that end we express the four-momentum in terms of derivatives with respect to the coordinate time. The spatial components are
$$\vec{p}=m \frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} = m \frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} \tau} = m \gamma \vec{v},$$
where
$$\gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}} \simeq 1+\frac{1}{2} \frac{\vec{v}^2}{c^2} + \mathcal{O(v^4/c^4)}.$$
Thus for $|\vec{v}| \ll \vec{c}$ we see that
$$\vec{p} \simeq m \vec{v} [1+\mathcal{O}(v^2/c^2)],$$
which identifies the invariant mass, $m$, with the Newtonian limit of the mass, i.e., $m$ is just the mass known from Newtonian mechanics.

For the temporal component
$$p^0=\sqrt{m^2 c^2 + \vec{p}^2} =m c \sqrt{1+\vec{p}^2/(mc)^2} \simeq m c[1+\vec{v}^2/(2 c^2) + \mathcal{O}(v^4/c^4) ] \simeq \left (m c^2 + \frac{m}{2} \vec{v}^2 \right) \frac{1}{c}.$$
Thus we have
$$c p^0= E =c \sqrt{m^2 c^2+\vec{p}^2} \simeq m c^2+\frac{m}{2} \vec{v}^2,$$
i.e., up to an additive constant $E_0=m c^2$ (the socalled "rest energy of the particle") $E$ is the relativistic generalization of kinetic energy of Newtonian mechanics. That the rest energy is included here is crucial to make $(p^{\mu})=(E/c,\vec{p})$ a four-vector, i.e., up to a unit-conversion factor $1/c$ energy and momentum are components of a four-vector.

To describe relativistically moving particles and find corresponding dynamical laws, it's much simpler to use the manifest covariant four-vector formalism than to guess around with the non-covariant (1+3)-dimensional formalism of the early days of relativity! That's why it's not wise to use old-fashioned ideas of "relativistic mass" (sometimes they had to distinguish even between "longitudinal" and "transverse mass", leading to a complete mess compared to the modern treatment in terms of Minkowski's four-dimensional tensor formalism).

4. Nov 11, 2016

### Mister T

Are you referring to this expression? $$p=mv \gamma$$
The debate over this issue has nothing to do with how you measure the mass, but is instead about what you call the mass. That is, whether you prefer to call $\gamma m$ the mass or you prefer to call $m$ the mass. The mass $m$ is the same mass that's used in newtonian physics. Calling $\gamma m$ the mass is something that was done by authors of books, much more so in the past than now. If you look, for example, at introductory physics textbooks, the ones used in courses to educate freshman and sophomore college and university students, you will see that prior to the 1990's almost all of them called $\gamma m$ the mass, but ever since then they almost all call $m$ the mass.

Physicists who work in high energy physics have always called $m$ the mass when doing their physics. Many of them called $\gamma m$ the mass in books that they wrote, much more so in the past than now.

You might think that giving them different names could resolve the issue. For example calling $m$ the mass and calling $\gamma m$ the relativistic mass. It doesn't because then the debate simply shifts to the usefulness of having more than one kind of mass. A lot has been written about the best way to teach and learn this basic concept. Those of us who prefer to have only one kind of mass seem to be winning, or have already won, the debate.

5. Nov 11, 2016

### Staff: Mentor

No, it isn't, and that's a big part of the problem. In Newtonian physics, the concept of "mass" conflated several different things that, in relativistic physics, turn out to be different. One of those things (roughly, "quantity of matter") turns out to correspond with $m$; another (roughly, "amount of inertia") turns out to correspond with $\gamma m$ (with some caveats, since the relationship between force and acceleration is direction-dependent in SR). And yet a third (gravitational mass) turns out to correspond with neither, since in GR the source of gravity is not "mass" but the stress-energy tensor.

6. Nov 11, 2016

### Mister T

From Lev Okun's June 1989 article in Physics Today:

7. Nov 11, 2016

### Staff: Mentor

Yes, but that just means that Okun picked out one particular characteristic of the mass in Newtonian physics ("does not vary with velocity") and ignored all the others. In other words, "does not vary with velocity" is not a definition of "Newtonian mass"; it's an empirical claim that all of the different properties that Newtonian physics conflates under the term "mass" do not vary with velocity. And we now know this empirical claim to be false.

8. Nov 11, 2016

### Mister T

So, are you saying he's wrong? Because he continues in that article to refer to the $m$ that appears in relations like $p=mv \gamma$ as "the ordinary mass, the same as in Newtonian mechanics"? And I thought that that is what I was also doing.

9. Nov 12, 2016

### Staff: Mentor

I'm saying that I think he is using the term "Newtonian mass" or "mass as used in Newtonian physics" in a more restricted sense than was implied by your earlier post. If he did actually mean "Newtonian mass" to cover all of the ways the term "mass" is used in Newtonian physics, then yes, I think he was wrong. But I think it's more likely that he only meant "Newtonian mass" in a more restricted sense and didn't stop to consider the other senses of the word "mass" in Newtonian physics that do not correspond to $m$ (rest mass) in relativity.

10. Nov 12, 2016

### vanhees71

I'm fully agreeing with Okun. In Newtonian as in SRT mechanics $m$ is independent of the velocity of the particle. The quantity $m \gamma$ is just the relativistic energy of the particle divided by $c^2$ (see my posting above).

From a group-theoretical point of view, however, you are right that the notion of mass is completely different in Newtonian as compared to specila-relativistic physics. In Newtonian physics the mass is a nontrivial central charge of the Lie algebra of the quantum Galileo group, while it is a Casimir operator of the Lie algebra of the proper orthochronous Poincare group in the relativistic case. That explains why the mass superselection rule of non-relativistic QM doesn't hold in nature, which of course is relativistic :-).

11. Nov 12, 2016

### pervect

Staff Emeritus
Arguing about relativistic mass is mostly pointless. I'd recommend using whatever approach gets you the right answer - and being sure to be clear about which mass you are using if there's any question.

12. Nov 12, 2016

### Staff: Mentor

You are ignoring the fact that the $m$ you speak of refers to different things in Newtonian mechanics and SRT. In fact, in Newtonian mechanics the symbol $m$ refers to different things in different equations. So you can't even make the statement quoted above meaningful without specifying which Newtonian equation you are talking about. And depending on which Newtonian equation you choose, the statement quoted above, if you make it meaningful, might be false. (Strictly speaking, you also need to specify that by $m$ in SRT you mean rest mass/invariant mass, not "relativistic mass", but that convention is pretty well established now.)

13. Nov 12, 2016

### vanhees71

I admit that for the sake of clarity one should always write "invariant mass" when arguing within the theory of relativity because of the longevity of the bad notion of a relativistic mass ;-)).

I don't understand what you mean with your statements in reference to Newtonian mechanics. In Newtonian physics there's one parameter $m$, the (inertial) mass of a body. Newtonian mechanics was conjectured by Newton, and there the mass appears in $\vec{p}=m \vec{v}$ and it has a specific meaning in the equation of motion $\dot{\vec{p}}=\vec{F}$, given the force $\vec{F}$ acting on the body. The only other meaning in Newtonian physics is the appearence of $m$ in the Newtonian theory of gravity $F_g=G m M/r^2$, the "gravitational mass". That here the same $m$ as before in the sense of an inertial mass (modulo a convention of units) appears is an empirical fact within Newtonian physics.

From the point of view of relativistic physics gravity is described consistently within General Relativity, and there's no notion of a gravitational mass, but the equivalence principle enforces the energy-momentum-stress tensor of matter (and a cosmological constant) to be the sources of the gravitational field, not mass. That then explains the equivalence between the Newtonian inertial and gravitational mass in the non-relativistic limit of GR, where the dominating components of the energy-momentum-stress tensor of matter is provided by $m c^2$. In this sense in relativistic physics there's only one notion of mass, namely the invariant mass.

The old-fashioned relativistic mass, $m/\sqrt{1-v^2/c^2}$ is nothing else than the energy of a body divided by $c^2$ defined such that $(E/c,\vec{p})$ is a four-vector (see one of my previous postings in this thread). There is no need to introduce a relativitic mass anywhere in the formalism, and in my experience with learning and teaching relativity for quite a while, it only leads to confusion when students stumble over these outdated ideas in textbooks. Unfortunately you find even nowadays new textbooks using that outdated concept, and that's why I try to make it very clear that it is a bad and confusing one, although it's not necessarily wrong if used with the proper understanding. Indeed, there are no mistakes in the works by the founding fathers of SR using the relativistic mass, even with more confusion distinguishing a longitudinal and a transverse mass, although of course it's always much simpler to use the manifestly covariant equations of motion with as simple as possible defined quantities (i.e., in this case invariant mass as a scalar instead of a quantity which, as a temporal component in some inertial frame, has no simple transformation rule under Lorentz transformations).

14. Nov 12, 2016

### Ibix

I think @PeterDonis's point is that the rest mass is the surviving term in a lot of places when you take Newtonian limits. But you can declare that what you mean by mass is just $|\vec F|/|\vec a|$ and all this formal limit stuff can take a long walk off a short plank. In that sense, Newtonian mass is relativistic mass.

Now, I think it would be wrong to do that. It's subordinating the more accurate theory's concepts to those of the less accurate theory. But that's what some very clever people did (due to not yet completely grasping relativity, of course), and modern students come at it from the same Newtonian viewpoint. So when you say that rest mass is the same as the Newtonian mass (absolutely true in a take-the-limit sense, as Okun says) you need to be a bit careful that your listeners aren't using the F/a definition of Newtonian mass.

15. Nov 12, 2016

### Mister T

That's what I'm asking you about. In other words, in what way did my post imply that I was using the term in any way that's different from the way Okun is using it?

I do agree with the three points you made about the role of newtonian mass in newtonian physics:

1. It's a measure of the quantity of matter.
2. It's a measure of inertia.
3. It's a measure of the agent responsible for the gravitational force.

And that in relativistic physics the newtonian mass doesn't fill any of those same roles.

But what I don't see is how any of that implies that what Okun wrote in that article is wrong, that what I wrote in my post implies anything different from what Okun states, and that therefore how any of what I wrote is wrong.

I do agree that a learner can be left with the erroneous notion that in relativistic physics the relativistic mass can allow one or more of those three points to continue to be valid, and that that is "a big part of the problem" with introducing it to students, and that that is one (among other) good reasons for abandoning it.

16. Nov 12, 2016

### Mister T

Ahhh ... I take your point. Okun makes that point quite clearly in the article. And I didn't in my post.

When I said "same mass as in newtonian physics" I didn't mean "mass used in the same way as in newtonian physics".

17. Nov 12, 2016

### DrStupid

And it is identical with Newton's quantity of matter. The velocity dependence results from the replacement of Galilean transformation by Lorentz transformation.

18. Nov 12, 2016

### Mister T

No, the relativistic mass isn't, and neither is the mass. In relativistic physics the energies of the constituents of a composite body contribute to the mass of that body, as measured in the rest frame of that composite body.

We don't increase the quantity of matter in a block of copper when we raise its temperature. But we do increase its mass.

That is the entire point of the mass-energy equivalence, something that Newton had no way of appreciating two centuries before it was discovered by Einstein.

19. Nov 12, 2016

### DrStupid

Your "No" suggests that I claimed something like that. That is not correct.

The termal energy doesn't change the amount of substance but it increases the quantity of matter as defined by Newton (even though he wasn't aware of this effect).

20. Nov 12, 2016

### Staff: Mentor

Yes, and, as you note, that one parameter $m$ appears in three different equations: $p = mv$, $F = ma$, and $F_g = GmM / r^2$. So Newtonian physics is making an empirical claim: that there is one single parameter $m$ that correctly appears in all three of the phenomena described by these equations. And we now know that this empirical claim is false. In other words, if we define

$$m_1 = \frac{p}{v}$$

$$m_2 = \frac{F}{a}$$

$$m_3 = \frac{F_g r^2}{GM}$$

then Newtonian physics claims that $m_1 = m_2 = m_3$, but we now know that claim is false; roughly speaking, $m_1 = p / \gamma v$, and $m_2$ and $m_3$ can't be correctly captured by scalars at all ($m_2$ is direction dependent, and $m_3$ is really a tensor).