3
$\begingroup$

In An Introduction to Tensor Calculus and Relativity, Lawden provides the definition of mass by considering a collision between two massive particles $p$ and $q$ with masses $m_p$ and $m_q$, respectively. The particles' are given velocities $\mathbf u_p,\mathbf u_q$, respectively, prior to collision, and $\mathbf v_p,\mathbf v_q$, after. It is has been shown experimentally that, in general, $m_p\mathbf u_p+m_q\mathbf u_q=m_p\mathbf v_p+m_q\mathbf v_q$ and from this it follows, or so Lawden claims, that

$$\frac{m_p}{m_q}=\frac{\lVert\mathbf v_q-\mathbf u_q\rVert}{\lVert\mathbf u_p-\mathbf v_p\rVert}.$$

If $q$ is then chosen to have unit mass, we define the mass of any particle by observing a collision between it and $q$.

However, this equation holds only in the case that $\frac{m_p}{m_q}$ is positive, in general we should only have

$$\left\lVert\frac{m_p}{m_q}(\mathbf u_p-\mathbf v_p)\right\rVert=\lVert\mathbf v_q-\mathbf u_q\rVert,$$

which gives us

$$\left\lvert\frac{m_p}{m_q}\right\rvert=\frac{\lVert\mathbf v_q-\mathbf u_q\rVert}{\lVert\mathbf u_p-\mathbf v_p\rVert}.$$

So if we want to define mass this way, we ought to have

$$\frac{m_p}{m_q}=\operatorname{sgn}\left(\frac{m_q}{m_p}\right)\frac{\lVert\mathbf v_q-\mathbf u_q\rVert}{\lVert\mathbf u_p-\mathbf v_p\rVert},$$

which is positive if $m_p,m_q$ have the same sign, and negative otherwise. Since the masses aren't defined a priori, the sign should be a free parameter unless there's some conserved relation between the velocities which necessitates that mass is always positive.

Edit/Remark

It is readily apparent that classical laws of motion are invariant under simultaneous inversion of mass and parity:

$$\mathbf F=m\mathbf a=(-1)^2m\mathbf a=(-m)(-\mathbf a)=-m\frac{\partial^2}{\partial t^2}(-\mathbf x)$$

$$\mathbf p=m\mathbf v=(-1)^2m\mathbf v=(-m)(-\mathbf v)=-m\frac{\partial}{\partial t}(-\mathbf x)$$

$$E^2=p^2c^2+m_0^2c^4=p^2c^2+(-m_0)^2c^2$$

$$\mathbf F=G\frac{m_1m_2}{\lVert\mathbf r_1-\mathbf r_2\rVert^3}\lVert\mathbf r_1-\mathbf r_2\rVert=G\frac{(-1)^2m_1m_2}{\lVert-(\mathbf r_1-\mathbf r_2)\rVert^3}\lVert-(\mathbf r_1-\mathbf r_2)\rVert=G\frac{(-m_1)(-m_2)}{\lVert(-\mathbf r_1)-(-\mathbf r_2))\rVert^3}\lVert-((-\mathbf r_1)-(-\mathbf r_2))\rVert$$

I haven't checked the laws for electromagnetism yet, but I would guess that the same holds with, at most, the inversion of charge. In this sense, the sign of mass is a matter of convention, one which reflects the sign convention for position. However, this is not what I am interested in. What I want to know is whether or not it is necessary that all masses have the same sign as a consequence of some conserved relation which would not apply otherwise.

$\endgroup$
5
  • $\begingroup$ the concept of mass comes from direct observations, and then mathematical tools have modeled it. Have a look at this en.wikipedia.org/wiki/Mass#Phenomena $\endgroup$
    – anna v
    Commented Oct 8, 2023 at 17:46
  • 2
    $\begingroup$ @annav But mass can't be observed directly, it can only be measured by its relation to e.g. acceleration, momentum. I believe that this is the reason Lawden thought it necessary to provide a definition of mass and show that it is invariant with respect to inertial frames (unlike velocity). $\endgroup$
    – R. Burton
    Commented Oct 8, 2023 at 18:07
  • $\begingroup$ proofs belong to mathematical models, measurements are numbers derived in an experiment. acceleration etc are all numbers measurable in experiments, and then can be mathematically modeled . you are talking of the mathematical models. $\endgroup$
    – anna v
    Commented Oct 8, 2023 at 19:08
  • 2
    $\begingroup$ @annav I understand, and again: we do not directly measure mass, we measure time, position, and related quantities and infer mass, energy, etc. from them. There is, to my knowledge, no "massometer" that just measures mass by itself. $\endgroup$
    – R. Burton
    Commented Oct 9, 2023 at 0:04
  • $\begingroup$ true, but the numbers for mass never come out negative, and are fixed for a fixed quantity of matter. $\endgroup$
    – anna v
    Commented Oct 9, 2023 at 3:32

1 Answer 1

0
$\begingroup$

If mass is a numeric property of a particle, independent of its speed, position, etc, then it must be positive. This does not follow strictly from mechanics though, as you have already guessed. Superficially, if one flips the signs of all masses in the world, the collisions between particles will indeed proceed the same way.

The problem will be once you invoke thermodynamics. If the mass is negative then your kinetic energy is negative-determinate, which is clearly a problem as the total energy of a system comprised of such negative-massed particles will be unbounded from below.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.