# Is there a difference between rest mass and invariant mass?

1. is there a difference between 'rest mass' and 'invariant mass'?
I thought there wasn't...

To put it another way (or maybe this next question is a different question):
2. Is there a difference between the rest mass of a positron/electron pair, and the rest massa of a system containing two fotons flying in opposite directions (the fotons are created when the positron/elektron annihalate.)

Paul

ShayanJ
Gold Member
1) They're just two names given to the m in the equations $E=\gamma m c^2$ and $p=\gamma m v$.
2) The difference is for photons, rest mass is not defined, because they're never at rest. Its just the energy of the photons becomes the rest energy of a positron\electron pair+the kinetic energy in each particle.

jtbell
Mentor
They're just two names given to the m in the equations E=γmc2 E=\gamma m c^2 and p=γmv p=\gamma m v.

What about the equation ##E^2 = (pc)^2 + (mc^2)^2##?

The difference is for photons, rest mass is not defined, because they're never at rest.

Nevertheless, the quantity m in my equation is well-defined for photons, and has the value 0. What you call it is up to you. Real Physicists call it simply "mass." :D

Is there a difference between the rest mass of a positron/electron pair, and the rest massa of a system containing two fotons flying in opposite directions (the fotons are created when the positron/elektron annihalate.)

By "rest mass of a positron/electron pair" I assume you mean the quantity defined by $$m_\textrm{pair}c^2 = \sqrt{(E_\textrm{electron} + E_\textrm{positron})^2 - (|\vec p_\textrm{electron} + \vec p_\textrm{positron}|c)^2}$$ and similarly for the system of two photons: $$m_\textrm{sys}c^2 = \sqrt{(E_\textrm{photon 1} + E_\textrm{photon 2})^2 - (|\vec p_\textrm{photon 1} + \vec p_\textrm{photon 2}|c)^2}$$ ##m_\textrm{pair} = m_\textrm{sys}## because total energy and total momentum are conserved. However, neither the electron/positron pair nor the two photons form a single "object" (bound system), so some people argue that this quantity should not be given the name "mass". Experimental particle physicists do use such quantities in analyzing their data, and call them "invariant mass".

vanhees71
ShayanJ
Gold Member
$p^2 c^2+(mc^2)^2=\gamma^2 m^2 v^2 c^2+m^2 c^4=m^2 c^2(\gamma^2 v^2+c^2)=\\ m^2 c^2(\frac{c^2 v^2}{c^2-v^2}+c^2)=m^2 c^2 \frac{c^2 v^2+c^4-c^2 v^2}{c^2-v^2}=m^2 c^2 \frac{c^4}{c^2-v^2}=\\ m^2 c^2 \frac{c^2}{1-\frac{v^2}{c^2}}=\gamma^2 m^2 c^4=(\gamma m c^2)^2=E^2$

So that equation is a consequence of the equations I wrote, not an independent equation.

Also rest energy of a particle is defined as the energy of the particle in its own inertial frame. But can we assign an inertial frame to photons so that we can say what is their energy in that frame? You can define rest energy for photons only if you have an alternative definition of rest energy that works for photons, i.e. doesn't use the concept of photon's inertial frame. Of course its not that I don't accept the claim "photons have zero rest mass". I somehow think this is an example of the phenomenon Samalkhaiat is referring to in his signature "Physics is an ill-defined mathematical structure." I'm not sure how to think about it though, but the objection I mentioned seems serious to me. I think its just that we know something with rest mass can't go at the speed of light, so we prefer to say photons have zero rest mass, because they're going at the speed of light. Although I think the main reason comes from gauge theory when we see no mass term appears for the abelian gauge field and even Higgs mechanism doesn't give rise to a mass term for it.

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However, neither the electron/positron pair nor the two photons form a single "object" (bound system), so some people argue that this quantity should not be given the name "mass". Experimental particle physicists do use such quantities in analyzing their data, and call them "invariant mass".

1. You say experimental particle physicists call such quantities 'invariant mass'. Would they not use the term 'rest mass', also? Do they make a distinction? And do other physicists sometimes make a distinction between rest mass/invariant mass?
2. If the electron/positron-pair does form a bound system, and also the two fotons do form a bound system (for instance if everything is locked up in some sort of system (like quarks are locked up in a neutron, and are exchanging gluons)), does the 'mass-name' ('rest-mass', 'invariant mass'?) of the system change when inside the system the electron/positron annihalate into the two photons?

Paul

jtbell
Mentor
First, I need to make a disclaimer. The following statements come from my memories of when I was a graduate student 30+ years ago, working on my Ph.D. in experimental particle physics. I have not worked actively in research since then, only in teaching. Others are welcome to confirm or correct these statements as they apply to experimental particle physicists today. I would not be surprised if there were occasional exceptions to these statements back in those days, or today; nevertheless, these are my general impressions. Feel free to mentally insert weasel-words like "generally" or "usually" where appropriate.

1. For individual particles, or bound systems of particles (e.g. an atom, an atomic nucleus, even a proton or neutron being viewed as a collection of quarks), we always said simply "mass." In this context, "rest mass" and "invariant mass" serve only to distinguish from "relativistic mass", and we never used the concept of "relativistic mass," so there was no need to make the distinction. During at least the last 30-40 years, the concept of "relativistic mass" appears only in writings about relativity for laymen, and in some introductory textbooks, as far as I know.

2. For unbound systems of particles, as I noted in the previous post, we used the term "invariant mass". I think we did not say simply "mass" because there is no single coherent "object" here.

For example, my advisor was studying the production of Λ0 particles in neutrino interactions in a bubble chamber. The signature for this is a V-shaped pair of tracks appearing near a primary interaction, representing the proton and π- produced in the decay of the invisible Λ0. One of my tasks was to calculate the invariant mass of the p and π-, based on measurements of the momenta and energies of the tracks, and compare it to the known mass of the Λ0 (1115.7 MeV/c2) in order to confirm that they did indeed come from a Λ0.

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ChrisVer
Gold Member
1. You say experimental particle physicists call such quantities 'invariant mass'. Would they not use the term 'rest mass', also? Do they make a distinction? And do other physicists sometimes make a distinction between rest mass/invariant mass?

For individual particle it's good to talk for rest mass of the particle [the mass it has at a frame that it's at rest]. So let's say that you have a particle with 4-momentum $p$ then $p^2 \equiv m^2$.
However this cannot be applied for a system of particles that take place in an interacton. The 4-momentum is the sum of the individual 4-momenta, let's say two particles with $p_1, p_2$... The invariant object is then called "invariant mass" $(p_1 + p_2 )^2 = p_1^2 + p_2^2 + 2 p_1 p_2 = m_1^2 + m_2^2 + 2 p_1 p_2 \ne m_1^2 +m_2^2$...
In contrast to the rest mass, the invariant mass can appear as a distribution function to the momenta, with a peak around the rest mass or invariant mass of the initial state....

vanhees71
Gold Member
2021 Award
I can confirm that jtbell's description of the particle physicist's is the same as 30 years ago (with the caution that I'm a theoretical high-energy nuclear/heavy-ion phycisist, but we are pretty familiar with the HEP community's slang too :-)).

ShayanJ
Gold Member
Sorry for posting again, but I feel my last post is incomplete without this!
I think $E=pc$ is not, as usually thought, a special case of $E^2=p^2c^2+(mc^2)^2$ for photons. Because I could prove the latter using the definitions of relativistic momentum and energy but there is no special case of this proof for photons which gives the former equation!

jtbell
Mentor
I can confirm that jtbell's description of the particle physicist's is the same as 30 years ago (with the caution that I'm a theoretical high-energy nuclear/heavy-ion phycisist, but we are pretty familiar with the HEP community's slang too .

However, on this forum, I do use the terms "rest mass" and "invariant mass." Most people asking questions here have seen "rest mass" and "relativistic mass" and I feel I need to make clear which kind of mass I'm talking about. But whenever you talk about "rest mass" here, someone always brings up photons which are never at rest! So I say that I mean "rest mass", but that a better name for it is "invariant mass", but Real Physicists always call it just "mass".

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ChrisVer
Gold Member
Sorry for posting again, but I feel my last post is incomplete without this!
I think $E=pc$ is not, as usually thought, a special case of $E^2=p^2c^2+(mc^2)^2$ for photons. Because I could prove the latter using the definitions of relativistic momentum and energy but there is no special case of this proof for photons which gives the former equation!

The photon obeys both these equations. In fact the photon obeys the same rule as all the other particles do, since you can define a null-4-momentum for it too. $P_\gamma^\mu P_{\gamma \mu} = 0$ - is the photon's case invariant object. You can see this as a "massless" particle ($m^2=0$). As I said that's the photon's case invariant object- doing a Lorentz transformation maybe will change the photon's energy value and spatial momentum, but it will keep their squares equal as this equation tells us [in other words the photon won't be massive in any reference frame hahahahaha if that makes sense to say, since Lorentz transformation doesn't really makes sense for photons]
In fact the four momentum is much more useful than writing the relativistic equations with the $\gamma$ factor.
It's better to say that the 4-momentum is the object with the 0-th component being the energy and the i-th component being the i-th component of the spatial momentum. The last for massive particles gets this factor $\gamma$. The photon doesn't, as a null-object, its energy squared is equal to its spatial momentum squared (killing 1 degree of freedom for the physical photons)

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vanhees71
Dale
Mentor
2021 Award
I prefer the term "invariant mass" over "rest mass". If an object has a rest frame then the "rest mass" in that frame is also equal to the "invariant mass", but the term "invariant mass" also applies to systems or objects without an inertial rest frame and even for non inertial frames.

The other thing that I like about it is that it gives a hint on how it is calculated. You automatically expect to take the square of a timelike quantity (energy) minus the square of a space like quantitum (momentum)

vanhees71