# How can we really state that mass (or energy ) is conserved?

1. Sep 1, 2016

### Ahan Sha

If we consider a box( in our quantum world ) with three balls having total mass 'M= m1 + m2 +m3' . After changing their positions, now, calculating the mass will give another value which is not equal to 'M' but 'M*=m*1 +m*2 + m*3)', which implies our mass is not conserved. then how could energy be conserved (E= mc2) ??!

2. Sep 1, 2016

### Lucas SV

The equation $E=mc^2$ is only correct for a particle at rest. For a moving particle of momentum $p$, instead you should use $E^2=m^2c^4+p^2c^2$.

3. Sep 1, 2016

### Ibix

Why do you think the particles will have a different mass if they are in different positions?

4. Sep 1, 2016

### vanhees71

In relativistic physics the mass (and I mean invariant mass as usual) is not a conserved quantity in reactions (e.g., take annihilation $e^+ + e^- \rightarrow 2 \gamma$, where on the left-hand side the particles have $m_e+m_e=2m_e$ total mass, and on the right-hand side you have $2m_{\gamma}=0$. Of course, energy is conserved in all processes.

5. Sep 1, 2016

### tionis

In general relativity, energy is not conserved .

6. Sep 1, 2016

### pervect

Staff Emeritus
I'm not sure what the context of the question is - other than the reference to QM, which raises the possibility that the question belongs in the quantum forum. Because the question is being asked in the relativity forum, though, I'd assume that the interest was in either special relativity or general relativity (which are both classical theories). Unfortunately a detailed answer would depend on whether you were asking the question in the context of special relativity or in the contecxt of general relativity.

7. Sep 1, 2016

### Ibix

It was originally posted in General Discussion. I reported it for being in the wrong place, suggesting either relativity (based on the $E=mc^2$ reference) or quantum (based on the obvious). The Mentors apparently went for relativity.

8. Sep 1, 2016

### DrStupid

1. If energy and momentum are conserved than mass is conserved too.

9. Sep 1, 2016

### Fervent Freyja

What? $E=mc^2$ applies to any system in physics or chemistry, not just a particle at rest!

For example, this still holds true for a molecule consisting of three atoms. Since the atoms are overlapping (sharing electrons), there is a slight loss of mass, around 1 billionth of 1% less than the total mass of the three individual atoms, I think. However, the difference in mass is equivalent to the kinetic energy that keeps the atoms bound together as a molecule (called binding energy). As long as energy is still conserved in this case, then so is mass.

10. Sep 1, 2016

### Staff: Mentor

That depends on whether you interpret the $m$ in $E=mc^2$ as the rest mass or the relativistic mass. Lucas SV is of course referring to the rest mass, as is the normal practice when it's not specified.

Rest mass is frame-independent but not conserved.
Total energy is conserved but frame-dependent.
The rest mass of a composite system is not necessarily equal to the sum of the rest masses of its individual components.

11. Sep 2, 2016

### Staff: Mentor

It might be more accurate to say that in in general relativity we have to be more careful about how we define conservation of energy.

If a region of space is large enough for general relativistic effects to matter, there is no way of defining the total amount of energy that is contained in that region at any given moment - it will depend on the simultaneity convention we choose. And of course it can't be conserved if we can't even define it. However, there's another way of stating energy conservation, one that does work with GR: Within any arbitrarily small volume the change in the energy contained within that volume at any moment will be the net flow of energy across the boundary of that volume.

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html is a good overview.

12. Sep 2, 2016

### PAllen

I agree with Dr. Stupid's point here: if E and P are conserved, invariant mass = rest mass is conserved. Both E and P are frame dependent, while invariant mass is frame independent and conserved. What makes this true is that (as you state) you cannot add component invariant masses to get invariant mass of a system. Non conservation of invariant mass of a system would imply that there is non conservation of energy in the center of momentum frame, which is absurd (in SR)

13. Sep 2, 2016

### vanhees71

Here one must be more careful. It's often also a matter of context, whether you call something an"energy" (often defined in a specific frame) or a "mass".

In my example case of a collision, usually you use $(p_1+p_2)^2=s$ (Mandelstam variable $s$), where $p_{1,2}$ are the four-momenta of the incoming particles to characterize the incoming state. It's meaning becomes clear, if you go to the center-momentum frame, in which $\vec{p}_1+\vec{p}_2=0$. Then it's clear that $\sqrt{s}$ is the total energy in the center-mass frame, i.e., in this case you characterize this quantitiy as an energy (in a specific frame, in this case the center-momentum frame).

On the other hand, you may have a composite system, say an atomic nucleus. Then you define its mass as the energy in the center-momentum frame, which also includes the binding energy, and indeed here the mass is smaller than the sum of the masses of its constituents. Another extreme example are the usual hadrons like the proton, where most of its mass is understood to be via the dynamics of the strong interaction. The current quark masses of the light quarks (u and d) are in the MeV range, while the nucleon mass is 940 MeV, i.e., most of its mass is dynamically generated.