Is Rest Mass Conserved by the Conservation of Energy and Momentum?

[superg33k]

From my understanding E=mc2 shows that relativistic mass is equivilent to energy. And hence the conservation of relativistic mass is a restatement of the conservation of energy.

Also from my understanding it also indicates that rest mass can be converted into photons. Also I think this happens all the time in particle colliders. Thus rest mass is not conserved.

However I have also been told (but not the most reliable of sources) that the conservation of rest mass is true, and is implied by the combination of conservation of energy and conservation of momentum.

Is rest mass conserved?

 

[Dickfore]

As you pointed out, it is not.

 

[Drakkith]

I think this is a slightly more complicated issue than it seems. Mass can be converted into energy. Energy (arguably) has mass. The photons produced from the annihilation of a particle have momentum (energy) equal to the rest mass of the particles plus the kinetic energy. I would say that both energy and mass are conserved here.

 

[tom.stoer]

E=mc² can be interpreted differently.

If m is the rest mass, then E is a very special energy measured in the rest frame.
If E=E(v) is the energy (as the 0-component of a 4-vector) in an arbitrary reference frame, then m=m(v) is the relativistic mass, not the rest mass.

What is conserved in all processes is the invariant mass defined by (mc²)² = E² - (cp)². This definition guarantuees that the value of m is invariant, i.e. is the same for all reference frames. If we are talking about the special case of one single particle then this m is its rest mass.

 

[superg33k]

E=mc² can be interpreted differently.

If m is the rest mass, then E is a very special energy measured in the rest frame.
If E=E(v) is the energy (as the 0-component of a 4-vector) in an arbitrary reference frame, then m=m(v) is the relativistic mass, not the rest mass.

What is conserved in all processes is the invariant mass defined by (mc²)² = E² - (cp)². This definition guarantuees that the value of m is invariant, i.e. is the same for all reference frames. If we are talking about the special case of one single particle then this m is its rest mass.

And p is relativistic momentum: $$p=\gamma m v$$

So only moving mass can be (partially) converted into photons; and the original rest mass will always be conserved.

If matter is rest mass, and radiation is massless paricles (photons the only one I know about so far). Then I can think of moving matter as matter that has absorbed radiation. Thus the amount of matter and the amount of radiation in the univserse will never change!

I wonder if thinking of moving matter as matter that has absorbed radiation will have any implications to my ideas of space?

But regarding the conservation of mass, thanks for your help guys. I think it all makes sense.

 

[Drakkith]

And p is relativistic momentum: $$p=\gamma m v$$

So only moving mass can be (partially) converted into photons; and the original rest mass will always be conserved.

If matter is rest mass, and radiation is massless paricles (photons the only one I know about so far). Then I can think of moving matter as matter that has absorbed radiation. Thus the amount of matter and the amount of radiation in the univserse will never change!

I wonder if thinking of moving matter as matter that has absorbed radiation will have any implications to my ideas of space?

But regarding the conservation of mass, thanks for your help guys. I think it all makes sense.

Whoa! Hold on a sec. Let's make sure you have this straight. The full mass of a particle, its rest mass and invariant mass and everything else, can and is turned into the momentum of the photons. Also, moving matter didn't have to absorb radiation. I can throw a ball which requires no radiation to happen. Matter and radiation are constantly changing by small amounts in both directions in the universe due to stars, high energy particle collisions, and such.

 

[superg33k]

Whoa! Hold on a sec. Let's make sure you have this straight. The full mass of a particle, its rest mass and invariant mass and everything else, can and is turned into the momentum of the photons. Also, moving matter didn't have to absorb radiation. I can throw a ball which requires no radiation to happen. Matter and radiation are constantly changing by small amounts in both directions in the universe due to stars, high energy particle collisions, and such.

I thought, from (mc²)² = E² - (cp)², we can see that the invariant mass (i.e. the rest mass) is always conserved (since both energy and momentum are conserved (and c is conserved too!)). This would mean that the full (relativistic) mass (i.e. directly propportional to its energy) of a particle can't be changed into photons (unless it had zero rest mass).

Or put simpler: an object with rest mass cannot be fully converted into photons. (This is my current thinking).

Basially I'm getting that 2 conservation laws are needed and the third will follow from: (mc²)² = E² - (cp)². If instead of conservation of momentum and energy we go for conservation of mass and energy, then conservation of rest mass implied conservation of matter (since I defined matter as rest mass) and since photons have zero rest mass they will be completely defined by energy, which is also conserved!

 

[superg33k]

I realize I have the explination skills of a goat, if you would like me to try again and be clearer please let me know.

 

[Sam Gralla]

I thought, from (mc²)² = E² - (cp)², we can see that the invariant mass (i.e. the rest mass) is always conserved (since both energy and momentum are conserved (and c is conserved too!)).

I'm guessing the context in which you're talking about conservation is particle collisions. Let's do 2->1 scattering for definiteness. In this case it's true that E_final = E_1 + E_2 (and same for momentum), but it is not true that m_final = m_1 + m_2, because m is quadratic in E and p. So rest mass is not conserved. You could only make the argument you made if rest mass were linear in E and p.

 

[superg33k]

I'm guessing the context in which you're talking about conservation is particle collisions. Let's do 2->1 scattering for definiteness. In this case it's true that E_final = E_1 + E_2 (and same for momentum), but it is not true that m_final = m_1 + m_2, because m is quadratic in E and p. So rest mass is not conserved. You could only make the argument you made if rest mass were linear in E and p.

I don't know what 2->1 scattering is, or why m is quadratic in E and p. However, if you are saying that rest mass isn't conserved, that would mean that one of: (mc²)² = E² - (cp)², conservation of energy or conservation of (relativistic) momentum isn't true. I'm interested to know which law breaks down?

I also thought these were laws of special relativity. Does special relativity and particle scattering not mix?

 

[ZealScience]

In all nuclear reactors, fuels are losing masses and emitting photons, also in stars, poles of black holes. So who told you that rest mass is conserved? Gluons in baryons have more mass than quarks themselves, which can easily be converted into radiations (here I mean the residual of the color force between baryons). More common case is that electrons emit X-ray during jumping to free electrons from orbital electrons and lose mass. Obviously it is not conserved.

 

[superg33k]

In all nuclear reactors, fuels are losing masses and emitting photons, also in stars, poles of black holes. So who told you that rest mass is conserved? Gluons in baryons have more mass than quarks themselves, which can easily be converted into radiations (here I mean the residual of the color force between baryons). More common case is that electrons emit X-ray during jumping to free electrons from orbital electrons and lose mass. Obviously it is not conserved.

But it takes energy to put uranium together (from protons and neutrons) in the first place. I was thinking that's where it came from. If we were looking at objects at that level the 'rest mass' of uranium would be the sum of the individual masses of protons and electrons and neutrons, which would be conserved. This is the same with the electron jump.

The only reason I have to believe that rest mass is conserved is because my current understanding is that 1- energy is conserved. 2- relativistic momentum is conserved, and 3- (mc²)² = E² - (cp)² is always true, m is rest mass, E energy, and p is relativistic momentum.

 

[tom.stoer]

Please read carefully: What is conserved in all processes is the invariant mass defined by (mc²)² = E² - (cp)². This definition guarantuees that the value of m is invariant, i.e. is the same for all reference frames. If we are talking about the special case of one single particle then this m is its rest mass.

Now we come to the scattering of one photon with a charged particle initually at rest where the charged particles receives some small amount of momentum and where the photon is converted into two other particles (e.g. electron + positron).

Thinking about one photon we have rest mass m=0, i.e. E² = (cp)². So how can this zero rest mass be converted into two particles with non-zero rest mass? It's because that the equation (mc²)² = E² - (cp)² is valid for

a) the photon with mass m_a=0, i.e. (m_ac²)² = E_a² - (cp_a)²
b) the other particle with mass m_b, i.e. (m_bc²)² = E_b² - (cp_b)²
c+d) electron and positron with mass m_c=m_d, i.e. (m_c,dc²)² = E_c,d² - (cp_c,d)²

Now these four cases are unrelated and the equation tells us nothing about conversion or conservation of rest mass. But there are two cases e) and f) which we forgot

e) total energy, momentum and invariant mass of the initial state, i.e. (m_a+bc²)² = E_a+b² - (cp_a+b)²
f) total energy, momentum and invariant mass of the final state, i.e. (m_c+dc²)² = E_c+d² - (cp_c+d)²

Here the invariant mass is conserved, namely m_a+b = m_c+d; this follows directly from the conservation of total energy and momentum.

But in these cases e) and f) these two invariant masses need not be rest masses of a particle b/c my last sentence if we are talking about the special case of one single particle then this m is its rest mass is no longer applicable; we are talking about a pair of particles.

Let's analyse the scenario in the c.o.m. frame where the total momentum of the two particles electron + positron vanishes

p_c+d = p_c + p_d = 0.

Then we have according to c)

(m_electronc²)² = E_electron² + (cp_electron)²

but according to f) we get

E_c+d ² = (E_electron + E_positron)² = 4E_electron²

(m_c+dc²)² = 4E_electron = 4[(m_electronc²)² + (cp_electron)²]

So the conserved inertial mass m_a+b = m_c+d for this scattering process cannot be identified with the rest mass of a given particle.

 

[Dale]

The easiest way to express the basic conservation laws is simply to say that the four-momentum of an isolated system is conserved. This one statement correctly encapsulated the conservation of energy, momentum, and mass. The four-momentum of a system is simply the sum of the four-momenta of each of its constituents.

 

[PAllen]

I don't know what 2->1 scattering is, or why m is quadratic in E and p. However, if you are saying that rest mass isn't conserved, that would mean that one of: (mc²)² = E² - (cp)², conservation of energy or conservation of (relativistic) momentum isn't true. I'm interested to know which law breaks down?

I also thought these were laws of special relativity. Does special relativity and particle scattering not mix?

One can say that initial invariant mass of the system of particles is the same as final invariant mass of the system particles after the collision. That is, if you sum the total energy and momenta of the particles initially, and compute E^2-(pc)^2, and do the same after collision, the result will be the same. Further, this value - invariant mass/enery (unlike energy or momentum separately) will be the same for every frame of reference.

 

[lili_njcb]

But it takes energy to put uranium together (from protons and neutrons) in the first place. I was thinking that's where it came from. If we were looking at objects at that level the 'rest mass' of uranium would be the sum of the individual masses of protons and electrons and neutrons, which would be conserved. This is the same with the electron jump.

The only reason I have to believe that rest mass is conserved is because my current understanding is that 1- energy is conserved. 2- relativistic momentum is conserved, and 3- (mc²)² = E² - (cp)² is always true, m is rest mass, E energy, and p is relativistic momentum.

Some time after the Big Bang, particles are at extremely high energy level, condensing into baryons or nucleus release their energy and mass. You can say that they were hydrogen in first place, but 298 hydrogen nuclei have more energy/mass than one Uranium-298 nucleus. i.e. single protons and neutrons have more mass than a nucleus, though it's not very significant.

To your reason, I think that photons also have momentum, but do they have rest mass? That equation is for bosons that are massless. At least they don't have inertial mass unlike fermions.

 

[superg33k]

Some time after the Big Bang, particles are at extremely high energy level, condensing into baryons or nucleus release their energy and mass. You can say that they were hydrogen in first place, but 298 hydrogen nuclei have more energy/mass than one Uranium-298 nucleus. i.e. single protons and neutrons have more mass than a nucleus, though it's not very significant.

I didn't understand the post big bang bit. But regarding the rest of it.

My understanding is the rest mass of hydrogen is the sum of the rest mass of its particles. And that electrons / protrons / neutrons have a well defined rest mass, that does not change.

From these properties, since a atom made of 298 hydrogen atoms has exactly the same number of electrons / protons / neutrons as the 298 individual hydrogen atoms then it will have exactly the same rest mass.

However if they have less energy bound in the 298 atom than free then energy will be released, this must be massless, i.e. photons. The rest mass is the same but the energy is less. Both rest mass and energy are conserved.

Finally to get it back into its free state as it was before, it requires energy (i.e. photons). You could bombard it with very fast massive particles, but to get massive particles to move very fast you need massless energy somewhere down the line.

To your reason, I think that photons also have momentum, but do they have rest mass? That equation is for bosons that are massless. At least they don't have inertial mass unlike fermions.

Correct, photons have momentum but have zero rest mass, i.e. they are massless particles.

 

[tom.stoer]

Could you please show us how rest mass is conserved in the example I discussed above? You have an initial state (photon, heavy nucleus) and a final state (electron, positron) with the masses (0,M) and (m,m) respectively.

I showed explicitly that there is a conserved mass (invariant mass) which in this case cannot be identified with rest mass.

So in which sense is rest mass conserved?

 

[Dickfore]

The easiest way to express the basic conservation laws is simply to say that the four-momentum of an isolated system is conserved. This one statement correctly encapsulated the conservation of energy, momentum, and mass. The four-momentum of a system is simply the sum of the four-momenta of each of its constituents.

What mass are you talking about here? And, if rest mass, care to show us how what you said is sufficient to derive its (rest mass) law of conservation.

 

[PAllen]

What mass are you talking about here? And, if rest mass, care to show us how what you said is sufficient to derive its (rest mass) law of conservation.

Conservarion of 4 momentum implies:

conservartion of total energy (timelike component of 4 momentum)
conservation of momentum (spacelike components of 4 momentum)
conservation (and invariance) of invariant mass(sometimes called 'system' rest mass) = norm of 4 momenum

 

[superg33k]

Could you please show us how rest mass is conserved in the example I discussed above? You have an initial state (photon, heavy nucleus) and a final state (electron, positron) with the masses (0,M) and (m,m) respectively.

I showed explicitly that there is a conserved mass (invariant mass) which in this case cannot be identified with rest mass.

So in which sense is rest mass conserved?

Rest mass and invariant mass are the same thing. Rest mass is the mass of a particle in its own rest frame. Since the definition is in its own rest frame all reference frames will agree with it, i.e. rest mass is invariant. Since its invariant for one particle its invaraint for a sum of particles. Also going through your example I couldn't find anything that contradicted this.

 

[PAllen]

Rest mass and invariant mass are the same thing. Rest mass is the mass of a particle in its own rest frame. Since the definition is in its own rest frame all reference frames will agree with it, i.e. rest mass is invariant. Since its invariant for one particle its invaraint for a sum of particles. Also going through your example I couldn't find anything that contradicted this.

An electron and a positron collide, annihilate, and produce two high energy photons. Sum of particle rest mass before is 2*(electron rest mass). Sum of particle rest mass after is 2*0. In *any given* frame of reference, total energy and total momentum are conserved. Further the invariant mass of the system (which is *not* the sum of particle masses) is not only conserved but is numerically the same in all frames at all times. This is all explained in detail in Tom's post.

 

[Dale]

What mass are you talking about here? And, if rest mass, care to show us how what you said is sufficient to derive its (rest mass) law of conservation.

I am talking about the invariant mass (aka rest mass) of the isolated system. The invariant mass is the norm of the four momentum, so conservation of the four momentum implies conservation of any function of the four momentum.

 

[superg33k]

An electron and a positron collide, annihilate, and produce two high energy photons. Sum of particle rest mass before is 2*(electron rest mass). Sum of particle rest mass after is 2*0. In *any given* frame of reference, total energy and total momentum are conserved. Further the invariant mass of the system (which is *not* the sum of particle masses) is not only conserved but is numerically the same in all frames at all times. This is all explained in detail in Tom's post.

Thanks Tom and Allen, everyone. I thought I must have been wrong in my original post about massive particles turning into photons. I haven't studied particle physics.

So I know the answer to the question "Is rest mass conserved?". But now I'm more confused about what is conserved! I'll have to read about this invariant mass. Cheers for all the posts.

 

[Drakkith]

Thanks Tom and Allen, everyone. I thought I must have been wrong in my original post about massive particles turning into photons. I haven't studied particle physics.

So I know the answer to the question "Is rest mass conserved?". But now I'm more confused about what is conserved! I'll have to read about this invariant mass. Cheers for all the posts.

Mass is a confusing topic to discuss most of the time thanks to various definitions and meanings. If we stick with the invariant mass you will find that it is always conserved.

 

[Sam Gralla]

So I know the answer to the question "Is rest mass conserved?". But now I'm more confused about what is conserved! I'll have to read about this invariant mass. Cheers for all the posts.

If you work a few problems I'm sure you'll see what people are saying about this "invariant mass" ("system rest mass" would be a much better name).

 

[superg33k]

I'm thinking the answer to this question is stupidly obvious. But that's probably when I make the stupidest assumptions and biggest mistakes. So if we ignore the world of particle physics (I haven't studied any of it yet but readings Toms post makes me want to!). And electrons are always electrons, nuclei are nuclei, and photons are absorbed and make atoms go to energy levels. Then rest mass is conserved?

I think this is a stupidly obvious question because I think the answer is "If the numbers and types of every particle is conserved, then of course the sum of their rest masses is conserved" Or is there something I'm missing?

 

[tom.stoer]

I think you can safely say that rest mass is conserved if the perticles do not change due to interaction, i.e. if the initial and the final state are (up to individual energy and momentum) identical regrdaing the particle species. But then this statement is almost trivial.

As soon as particles do interact and different particles are in initial and final state then the conservation of rest mass becomes meaningless.

 

[DrGreg]

...and photons are absorbed and make atoms go to energy levels. Then rest mass is conserved?

Rest mass of what? The rest mass of each electron, proton and neutron within the atom will not change, but the "system mass" of the whole atom will increase whenever a photon is absorbed (or decrease when one is emitted).

The system mass is the total energy divided by c², measured in a frame where the total momentum of the system is zero.

 

[mquirce]

But what if mass is not eternal and is created by another particle of mater. Why don't go in other hypothesis that elementary particles are not elementary , they have structure? In this case the mass and massles particles will have a common patern, a common base: the particle of mater which never created and never annihilated? So the energy and the mass have the same mather but with some different "flavors" or "colors ", (in fact only different electric and gravity charges). This is a hypothesis by a layman.

 

[Dickfore]

I am talking about the invariant mass (aka rest mass) of the isolated system. The invariant mass is the norm of the four momentum, so conservation of the four momentum implies conservation of any function of the four momentum.

So, is there any physical meaning of this rest mass of the system? For example, what is the rest mass of a bound system of a proton and an electron?

 

[tom.stoer]

It depends on the system.

If in some sense the system is a bound state, e.g. your bound system of a proton and an electron is a hydrogen atom, the invariant mass is the rest mass of this bound state.

If the system is not a bound state, e.g. at HERA you let electrons and protons collide, the invariant mass is something different; you cannot interpret it as "rest mass" of anything.

 

[Dickfore]

So, how can a system of the same constituents have different rest masses? It does not seem a property of the system, yet alone a conserved quantity?

 

[tom.stoer]

The only mass you can define unambiguously is the invariant mass, and as we have seen this is NOT always the rest mass.

 

[PAllen]

So, how can a system of the same constituents have different rest masses? It does not seem a property of the system, yet alone a conserved quantity?

See Dr. Greg post #29. System (rest) mass = invariant masss; conserved and frame independent. Very clear definition given in #29.