"relativistic mass" still a no-no?

[nomadreid]

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?

 

[Dr. Courtney]

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?

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

 

[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).

 

[Mister T]

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".

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.

 

[PeterDonis]

The mass $m$ is the same mass that's used in Newtonian physics.

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.

 

[Mister T]

No, it isn't, and that's a big part of the problem.

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

In the modern language of relativity theory there is only one mass, the Newtonian mass m, which does not vary with velocity; hence the famous formula $E=mc^2$ has to be taken with a large grain of salt.

 

[PeterDonis]

From Lev Okun's June 1989 article in Physics Today

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.

 

[Mister T]

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.

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.

 

[PeterDonis]

are you saying he's wrong?

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.

 

[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 :-).

 

[pervect]

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.

 

[PeterDonis]

In Newtonian as in SRT mechanics $m$ is independent of the velocity of the particle.

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.)

 

[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).

 

[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.

 

[Mister T]

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.

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.

 

[Mister T]

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.

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".

 

[DrStupid]

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).

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

 

[Mister T]

And it is identical with Newton's quantity of matter.

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.

 

[DrStupid]

No, the relativistic mass isn't, and neither is the mass.

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

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

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).

 

[PeterDonis]

In Newtonian physics there's one parameter $m$

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).

 

[PeterDonis]

The termal energy doesn't change the amount of substance but it increases the quantity of matter as defined by Newton

What definition of "quantity of matter" from Newton or Newtonian physics are you using? Can you give a reference?

 

[Mister T]

The thermal 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).

If you define "quantity of matter" to be the mass, then what you say is of course true. Because it's a tautology.

But I would argue that it doesn't match what most people think of when they think of the quantity of matter. When you raise the temperature of a gold brick you increase its mass, but you don't increase the amount of gold it contains. The notion that we can measure its mass to determine how much gold it contains is part of the Newtonian approximation and ignores the mass-energy equivalence. If the increase in mass with temperature were large enough to be significant (and it isn't by a long shot, which is what makes the Newtonian approximation safe to use) then merchants would be forced to use some method other than measuring its mass to determine its value. And its value is based on the amount of gold it contains, not on the energies of any of its constituents.

 

[vanhees71]

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.

That's precisely NOT why I declare. That's true in Newtonian physics for bodies of constant mass, but in relativistic physics, I prefer manifestly covariant formulations. Force is anyway a murky concept, but if you want a force, then I prefer the use of the Minkowski force. With the definition of the four-momentum
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} \, \Rightarrow \; p_{\mu} p^{\mu}=m^2c^2=\text{const}$$
the equation of motion reads
$$\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=K^{\mu}(x,p),$$
where due to the on-shell condition
$$p_{\mu} K^{\mu}=0$$
is a constraint on the Minkowski force.

 

[vanhees71]

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).

There's nothing in Newtonian physics making me define $m_2$. Unfortunately in high school they tell you this again and again, but this doesn't make it better. Newton's definition is the logical concept, and the logic of Leges I and II is
$$\vec{p}=m \vec{v}, \quad \dot{\vec{p}}=\vec{F}.$$

The logic behind $m_3$ I've explained in my previous posting. It's a fundamental empirical fact that $m_3$ is identical with $m_1$ (modulo a unit convention). Within Newtonian physics it cannot be derived, but it's a consequence of the Newtonian limit of GR, where the underlying symmetry principles provide a very convincing derivation of the equivalence between inertial and gravitational mass in the sense of the Newtonian limit. It also shows that not mass is the source of the gravitational field but energy, momentum, and stress, with a necessarily universal coupling of the gravitational field to the energy-momentu-stress tensor of matter (and radiation).

 

[DrStupid]

If you define "quantity of matter" to be the mass, then what you say is of course true.

No and no. I don't define "quantity of matter" to be the mass and that wouldn't be true.

When you raise the temperature of a gold brick you increase its mass, but you don't increase the amount of gold it contains.

Don't confuse quantity of matter with amount of substance.

 

[DrStupid]

What definition of "quantity of matter" from Newton or Newtonian physics are you using?

As Newtons definition 1 ($m := V \cdot \rho$) is not useful anymore because we use it the other way around to define density, I refer to the inplicite definition by his definition 2 ($p := m \cdot v$) including the explanatory note ($p_{total} = \sum {p_i }$), Lex 2 ($F = \dot p$), Lex 3 ($F_1 = - F_2$), Isotropy and the transformation (Galilean in classical mechanics or Lorentz in relativity). If you don't like Lex 2 and 3 you can use conservation of momentum instead.

Can you give a reference?

I. Newton. Philosophiae Naturalis Principia Mathematica. 1687
A. Einstein. Zur Elektrodynamik bewegter Körper. 1905

 

[harrylin]

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". [..]

It's sometimes better to read the physics FAQ about such topics:
http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

 

[PeterDonis]

There's nothing in Newtonian physics making me define $m_2$.

I would not phrase it that way. I would say that Newtonian physics makes the empirical claim that $m_2 = m_1$, i.e., that defining $p = m v$ and $\dot{p} = F$ is equivalent to defining $F = ma$. But we now know that is not the case. So we now have to make a conceptual distinction between $\dot{p}$ and $F / a$. And, as you say, we then prefer to use $\dot{p}$ as our fundamental definition because it is easily made covariant. But that doesn't change the fact that Newtonian physics conflates two things that we now know are not the same.

It's a fundamental empirical fact that $m_3$ is identical with $m_1$ (modulo a unit convention).

Only in the Newtonian limit. But if we're restricting ourselves to the Newtonian limit, this whole discussion is pointless.

Once we go beyond the Newtonian limit, there is no such thing as $m_3$, because gravity is not a Newtonian force, it's spacetime curvature, and the source of spacetime curvature is not a scalar, it's a tensor.

 

[PeterDonis]

I. Newton. Philosophiae Naturalis Principia Mathematica. 1687
A. Einstein. Zur Elektrodynamik bewegter Körper. 1905

You need to be more specific. Where in these references is the term "quantity of matter" explicitly defined? I'm not looking for your own personal interpretation of what that term means. I'm looking for some explicit definition of it in an acceptable source.

 

[Mister T]

I refer to the inplicite definition by his definition 2 ($p := m \cdot v$) including the explanatory note ($p_{total} = \sum {p_i }$), Lex 2 ($F = \dot p$), Lex 3 ($F_1 = - F_2$), Isotropy and the transformation (Galilean in classical mechanics or Lorentz in relativity).

Are you using that as a definition of quantity of matter, or as a definition of mass? You've already told us that the two are not the same by definition, so to make your point that the two are the same in some other way you need separate definitions of the two terms, along with some justification that the two are the same.

Don't confuse quantity of matter with amount of substance.

The $\mathrm{SI}$ unit of mass is the kilogram. Note that a macroscopic unit is being used to define a macroscopic property. When speaking of quantity of matter it's macroscopic objects, and only macroscopic objects, that concerned Newton and his contemporaries. In practice what merchants do is measure the mass by means of a balance (still the most precise method we have for measuring the mass of a macroscopic object) and use that as a measure of the amount of substance. They can get away with that because the Newtonian approximation holds, that is, the mass-energy equivalence can be ignored because its contribution is negligible (can't be determined with the balance).

It's the amount of substance that is equivalent to the quantity of matter, not the mass.

 

[vanhees71]

The SI unit for the "amount of substance" is not kilogram, which is the unit for mass, but mol.

 

[harrylin]

You need to be more specific. Where in these references is the term "quantity of matter" explicitly defined? I'm not looking for your own personal interpretation of what that term means. I'm looking for some explicit definition of it in an acceptable source.

It's not difficult to find Definition no.1 of the Principia - https://en.wikisource.org/wiki/The_Mathematical_Principles_of_Natural_Philosophy_(1846)/Definitions

 

[DrStupid]

You need to be more specific. Where in these references is the term "quantity of matter" explicitly defined? I'm not looking for your own personal interpretation of what that term means. I'm looking for some explicit definition of it in an acceptable source.

According to http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/25 quantity of matter $q$ (I don’t use the symbol $m$ in the following derivation in order to avoid confusions with mass) is the product of volume and density:

$q = V \cdot \rho$

That answers your question for the explicit definition. However, that doesn’t help you to determine whether q depends on velocity or not and if yes how. Without additional information would be at best an unknown function $q \left( q_0 , v \right)$ of the quantity of matter $q_0$ of a body at rest and its velocity $v$. But fortunately there are additional conditions which allow to derive this function:

According to http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/26 the momentum is the product of quantity of matter and velocity

$p\left( {q_0 ,v} \right) = q\left( {q_0 ,v} \right) \cdot v$

In the explanatory note Newton also defined that the momentum of a body is the sum of the momentums of its parts. That includes that the quantity of matter must be additive at least for the special case that all parts have the same velocity

$q\left( {q_0 ,v} \right) = \sum\limits_i {q\left( {q_{0,i} ,v} \right)} = \sum\limits_j {q\left( {q_{0,j} ,v} \right)}$

That results in

$q\left( {q_0 ,v} \right) = q_0 \cdot f\left( v \right)$

with

$f\left( 0 \right) = 1$

The next condition is isotropy. It requires that quantity of matter must be independent from direction. That includes

$f\left( { - v} \right) = f\left( v \right)$

Now let's continue with the laws of motion. According to the http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/46 force is defined as

$F := \dot p$

With the definition of momentum and the properties of quantity of matter we know so far this means

$F = q_0 \cdot f \cdot a + q_0 \cdot \dot f \cdot v = q_0 \cdot f \cdot a + q_0 \cdot \left( {f' \cdot a} \right) \cdot v$

In order to keep it simple I will limit the following calculations to the one-dimensional case. With

$K\left( v \right) = f\left( v \right) + f'\left( v \right) \cdot v$

the equation for force can be simplified to

$F\left( v \right) = K\left( v \right) \cdot q_0 \cdot a$

with

$K\left( 0 \right) = 0$
$K\left( { - v} \right) = K\left( v \right)$

According to the http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/49 the forces between two interacting bodies with the quantities of matter $q_1$ and $q_2$ (in their own rest frames) and the velocities v1 and v2 add to zero:

$F_1 + F_2 = K\left( {v_1 } \right) \cdot q_1 \cdot a_1 + K\left( {v_2 } \right) \cdot q_2 \cdot a_2 = 0$

The principle of relativity requires that everything mentioned above (including the function f) must be identical in all frames of reference. That means

$F'_1 + F'_2 = K\left( {v'_1 } \right) \cdot q_1 \cdot a'_1 + K\left( {v'_2 } \right) \cdot q_2 \cdot a'_2 = 0$

and therefore

$- \frac{{q_1 }}{{q_2 }} = \frac{{K\left( {v_2 } \right) \cdot a_2 }}{{K\left( {v_1 } \right) \cdot a_1 }} = \frac{{K\left( {v'_2 } \right) \cdot a'_2 }}{{K\left( {v'_1 } \right) \cdot a'_1 }}$

This is where the transformation comes into play. Let me do the calculation for classical mechanics first:

Galilean transformation results in

$v' = v - u$
$a' = a$

and therefore

$\frac{{K\left( {v_2 } \right)}}{{K\left( {v_1 } \right)}} = \frac{{K\left( {v_2 - u} \right)}}{{K\left( {v_1 - u} \right)}}$

This applies to all cases including

$v_1 = 0$
$v_2 = u = v$

That gives

$K\left( v \right) = 1$

and therefore

$q\left( {q_0 ,v} \right) = q_0$
$p = q_0 \cdot v$

That's the expected result for classical mechanics.Now let's go to special relativity:

With c=1 (to make the formulas less ugly) Lorentz transformation results in

$v' = \frac{{v - u}}{{1 - u \cdot v}}$
$a' = a \cdot \left( {\frac{{\sqrt {1 - u^2 } }}{{1 - u \cdot v}}} \right)^3$

and therefore

$\frac{{K\left( {\frac{{v_1 - u}}{{1 - u \cdot v_1 }}} \right)}}{{K\left( {v_1 } \right) \cdot \left( {1 - u \cdot v_1 } \right)^3 }} = \frac{{K\left( {\frac{{v_2 - u}}{{1 - u \cdot v_2 }}} \right)}}{{K\left( {v_2 } \right) \cdot \left( {1 - u \cdot v_2 } \right)^3 }}$

With the special case

$v_1 = 0$
$v_2 = u = v$

this turns into

$K\left( v \right)^2 = \left( {1 - v^2 } \right)^{ - 3}$

The resulting differential equation

$f' = \frac{1}{v}\left( {\sqrt {1 - v^2 } ^{ - 3} - f} \right)$

has only one physical solution:

$f = \sqrt {1 - v^2 } ^{ - 1}$

That means

$q\left( {q_0 ,v} \right) = \frac{{q_0 }}{{\sqrt {1 - v^2 } }}$
$p = \frac{{q_0 \cdot v}}{{\sqrt {1 - v^2 } }}$

This is the expected result for special relativity.

 

[DrStupid]

Are you using that as a definition of quantity of matter, or as a definition of mass?

I use it as a definition of quantity of matter (see above).

The $\mathrm{SI}$ unit of mass is the kilogram.

And so is the unit of quantity of matter.

It's the amount of substance that is equivalent to the quantity of matter, not the mass.

No, it isn’t. You can easily see that in case of your example with the heated gold brick. If the it moves with constant velocity during heating, its momentum and therefore its quantity of matter will be increased (see definition 2) whereas the amount of substance remains unchanged. Therefore amount of substance cannot be equivalent to quantity of matter. However, quantity of matter can also not be equivalent to mass because changing the speed of the brick at constant mass also changes its quantity of matter (see my calculation above).

 

[jbriggs444]

No, it isn’t. You can easily see that in case of your example with the heated gold brick. If the it moves with constant velocity during heating, its momentum and therefore its quantity of matter will be increased

So, for you, "quantity of matter" is the the m in $p=m \gamma v$. And indeed, if you heat up a gold bar (carefully, using flames that move at the same velocity as the gold bar and leave no soot), it will gain momentum.