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From what I have seen so far, there seem to be two fundamental criteria for considering something to have mass:

  • First is that it must have inertia and therefore momentum,
  • Second is that it interacts gravitationally i.e. it curves spacetime.

Even when we measure the mass of something, we can't do it directly, we calculate it through measuring its inertia or through its weight i.e. gravitational interaction.
But then this seems to create a contradiction in considering photons to be massless, as photons fulfill both1 these2 criteria. So then the definition of mass has to be different, and if so what is the actual definition of mass? If the definition is as I said, then why are photons considered massless?

For more reference to the questions that this question was built on:

If photons have no mass, how can they have momentum?
How is light affected by gravity?

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    Possible duplicates: https://physics.stackexchange.com/q/2229/2451 , https://physics.stackexchange.com/q/34352/2451 and links therein. – Qmechanic Jan 04 '21 at 06:59

2 Answers2

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The mass $m$ of an object is defined in terms of its energy $E$ and momentum $\mathbf p$ by the equation

$$(mc^2)^2=E^2-(\mathbf pc)^2$$

or

$$m^2=E^2-\mathbf{p}^2$$

in units where $c=1$.

This equation has a geometrical interpretation in terms of Minkowski spacetime: the mass is the Lorentz-invariant “length” of the energy-momentum four-vector $(E, \mathbf p)$.

Energy and momentum are important quantities because they are conserved. But observers in different inertial reference frames disagree on their numerical values: energy and momentum are frame-dependent. By contrast, inertial observers all agree on the mass; it is a frame-independent quantity and thus an intrinsic property of the object.

This should not be surprising. For a three-vector in Euclidean space, observers in various frames rotated with respect to one another disagree on the components of the vector but agree on its length.

Photons have both energy and momentum. But these are related by $E=|\mathbf p|c$, so $m=0$ for photons.

An object with nonzero mass moving with velocity $\mathbf v$ has energy

$$E=\frac{mc^2}{\sqrt{1-\frac{\mathbf{v}^2}{c^2}}}$$

and momentum

$$\mathbf p=\frac{m\mathbf v}{\sqrt{1-\frac{\mathbf{v}^2}{c^2}}}.$$

$E$ and $|\mathbf p|$ both become infinite as $v\to c$, which explains why a massive object cannot move at the speed of light.

These formulas are not useful for photons, since they give the indeterminate ratio 0/0. However, you can argue that if $m=0$ then $v$ must equal $c$, otherwise $E$ and $|\mathbf p|$ would both be zero.

G. Smith
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  • I see so qualitatively the definition would be "the total energy of an object minus its momentum". But since momentum of an object is generally determined by $p=mv$ why can't the value for p translate into a value for m? –  Jan 04 '21 at 02:10
  • Isn't it one of the postulates of SR that the laws of physics are always the same for every object in all reference frames? If $p=mv$ is a law of physics and it doesn't apply here, this seems to contradict that postulate –  Jan 04 '21 at 02:17
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    Momentum is not $mv$, it's $mv/\sqrt{1-v^2}$, which diverges for photons. – JEB Jan 04 '21 at 02:21
  • right, the relativistic equation for momentum is different, but in this case if v =1, the momentum would be infinite if mass>0 and undefined if mass=0, which also doesn't make sense –  Jan 04 '21 at 02:30
  • also if we assume p to be a finite value, and m = 0, v would be infinite –  Jan 04 '21 at 03:01
  • if we assume p to be finite and v =1, then m would result in 0 in the final step, but it has to go through an intermediate step where a finite value = an infinite or undefined value –  Jan 04 '21 at 03:41
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    @Neelim you are discussing in comments again – Dale Jan 04 '21 at 03:51
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    the total energy of an object minus its momentum That’s not what the formula says. – G. Smith Jan 04 '21 at 04:35
  • Apologies for the multiple comments, could have fit them into one. But I believe they were relevant to the question. Coming back to the main point I realise that I forgot to include the squares in the definition. Is that what you meant? If not what would be the qualitative definition? –  Jan 04 '21 at 16:49
  • @Neelim Yes, I meant that the mass is the square root of the difference of the squares of the energy and momentum, in units where $c=1$. – G. Smith Jan 04 '21 at 17:51
  • @Neelim of course they are relevant to the question. That is assumed. The issue isn’t relevance, it is purpose. The purpose of comments is to improve the answer, not to have a discussion (regardless of how relevant) and especially not to ask follow-up questions – Dale Jan 05 '21 at 03:50
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The distinction between inertial and gravitational mass makes little sense. Mass is (strictly: was) defined in terms of the standard $1$ kg platinum-iridium block kept in Paris .

It is not defined as "inertial mass". Nor as "gravitational mass". But simply as "mass".

The fundamental "MKS" mechanical units are mass (kg), length (m) and time (s) . Force not being one of these, it has to be defined in terms of them, for instance via Newton's second law. If a force applied to the standard $1$ kg mass in Paris results in an acceleration of $1 \frac{\text m}{\text s ^2}$, then the value of that force is by definition $1$ N.

This allows the masses of other objects to be determined. If a force applied to a body gives an acceleration a; and the same force applied to the standard $1$ kg mass in Paris produces an acceleration $a_1$; then the mass $M$ of that body is by definition $M=a_1/a$.

In possession of operational procedures for measuring quantitatively force and mass, Newton's gravitational constant $G$ can be determined experimentally. And that's it. No separate inertial and gravitational masses. Simply mass.

AlmostClueless
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