# Can Energy-Mass really be conserved ?

1. Sep 28, 2015

### Shinyoung Noh

Conservation of mass from chemistry and conservation of energy from physics were combined into conservation of mass and energy as special relativity proved mass-energy equivalence.
In short, if we count all the mass in using E=mc^2, total sum of energy is always conserved even with nuclear fission.

Then, I do not see how that can work as I thought of this question.
here is the question.

1. I lifted a stone from the floor to the table. Assume 100% efficiency I spent 20 J and the stone now has 20 J of potential energy.
2. Suddenly, nuclear fission happened on the other side of the earth. a half of the Earth's mass is now converted into a massive energy, following e=mc^2. Energy is conserved as nuclear fission made the energy accordingly.
3. somehow, my stone on the table was not affected from the nuclear fission, it was not burnt or anything.
But now g is a half, so the potential energy it has is a half. So it will gain only 10J when it is dropped back to the floor.

Then the question is, where is my another 10 J gone ?

2. Sep 28, 2015

### Staff: Mentor

No, $g$ doesn't change. In general relativity, energy as well as mass contributes to gravity. If I have a given mass somewhere I'll find the same gravitational field around it as if I had concentrated the equivalent (by $E=mc^2$) amount of energy in the same place.

(There is quite a bit more to the conservation of mass-energy in relativity than this, but these additional complexities don't matter for this particular question).

3. Sep 28, 2015

### Staff: Mentor

Into the KE of the fission products.

4. Sep 28, 2015

### Shinyoung Noh

I am a bit confused.
Nugatory says that it will still be 20 J as energy created will contribute to the gravity and potential energy does not really change,
While DaleSpan says potential energy of the stone does reduce and nuclear fission will not be e=mc^2 but will produce equivalent additional energy.

Both answer my question. However, now it brings me to another questions.

Nugatory, Energy is not a substance. We do see many forms of energy such as heat, electricity, and even mass, but as far as I know, there is no such 'substance' as energy. Only the number exists. Then would you explain how can 'energy' exert gravitational force to the stone ?

DaleSpan, does that mean E does not equal mc^2 ?

5. Sep 28, 2015

### Staff: Mentor

Well, you didn't specify your scenario clearly. Nugatory is correctly describing the scenario where the energy from the fission remains here. I am describing the scenario where the energy is radiated away. I don't know which one you intended.

6. Sep 28, 2015

### vanhees71

The explanation is very simple: In relativity invariant mass is not conserved. If you have a container of gas, the invariant mass of the container+gas depends on the temperature of the gas (particle number kept fixed). In nuclear physics, the invariant mass of a nucleus is not the sum of the masses of the protons and neutrons it is composed of, but
$$M_{\text{nucleus}}=Z m_{\text{p}}+(A-Z) m_{\text{n}}-\frac{E_{\text{bind}}}{c^2}.$$
$E_{\text{bind}}$ is the binding energy, holding the nucleons together in the nucleus $Z$ is the charge number (i.e., the proton number), and $A$ the nucleon number (and thus $A-Z$ the neutron number).

For hadrons as bound states of light quarks like protons and neutrons it's the opposite: the mass of the constituents is only a tiny fraction of the total mass of the object, i.e., most of the proton mass is due to dynamics of QCD.

Further, in General Relativity, which is the correct relativistic theory of gravity, the total energy and momentum of matter (and radiation) is the source of gravity, not mass density. This is very important, because the energy-momentum tensor is locally conserved, not mass density.

7. Sep 28, 2015

### Staff: Mentor

All the forms of energy gravitate as if they had mass given by $E=mc^2$.

If I have a sealed box containing a positron and an electron, it will have a certain mass: the weight of the box, plus two electron masses from the two particles inside it. When the electron and positron meet, they will annihilate and turn into electromagnetic energy; the amount of energy of course is given by $E=mc^2$ with $m$ equal to twice the electron mass. This electromagnetic energy will eventually be absorbed by the walls of the box, heating them slightly. Throughout all of this, if the box is truly sealed so that none of the energy inside it can leak out.... The weight of the box and the gravitational force it exerts will be unchanged, even though we went from two electrons to an equivalent amount of electromagnetic energy to an equivalent amount of heat energy.

8. Sep 28, 2015

### pervect

Staff Emeritus
We can definitely say that it's wrong to say that "g suddenly and instantaneously drops to half" in this situation. To say more requires a better description of the problem, and there is the additional complication that even if we had an unambiguous and clear statement of the problem, most versions of the problem will be extremely difficult to solve using full General Relativity.

Rather than explore this specific problem, it's probably more helpful to point out that the correct answer to "is mass-energy conserved in General Relativity" is "maybe", rather than "yes". See for instance the sci.physics.faq http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

If you limit the discussion to special relativity (leaving gravity out of the picture) you can say that mass-energy is conserved. You probably won't recognize the math required to fully and correctly express what we mean when we say "mass-energy is conserved in special relativity", unless you have some graduate level math. It looks something like $\nabla_a T^{ab} = 0$.

9. Sep 29, 2015

### vanhees71

What do you mean by "mass-energy". In relativistic physics energy is conserved. The covariant definition of energy (combining it with momentum to an energy-momentum four-vector) includes the rest energies (sometime called "mass-energy") $m c^2$ ($m$: invariant mass) of the particles involved. Mass is not a conserved quantity.

The mathematical reason for this is a deep difference in the spacetime-symmetry groups, Galileo for Newtonian and Poincare for special-relativistic spacetime, which goes far beyond what I'm allowed to post in this thread, which is labeled B.

10. Sep 29, 2015

### jartsa

Mass is not conserved? Ok, how can we make half of Earth's mass disappear? And what is the correct way to measure Earth's mass afterwards, to make sure that half of the mass disappeared?

11. Sep 29, 2015

### vanhees71

What do you mean by that? Of course, you can imagine to create anti matter such as to annihilate half of the Earth's material into photons. Then half of the mass of the Earth disappeared into radiation. Of course the total energy is conserved, i.e., the energy of the matter making up half of the Earth is not irradiated off in form of electromagnetic waves. Fortunately this is science fiction, because I'm pretty sure that it is practically impossible to create such an amount of antimatter. CERN has a hard enough time to create and store a few hydrogen atoms to make experiments with them.

12. Sep 29, 2015

### jartsa

I mean that mass is always conserved. That's what I believe until I'm given a counter example.

So first we have Earth with mass M. And afterwards we have Earth with mass 1/2 M and radiation with mass 0.

And the combined mass of the half-Earth and the radiation is M. Is this wrong? I think that is the correct result of the relativistic addition of the masses.

(There I was considering zero mass radiation moving at speed of light. If the center of mass of the radiation is at rest relative to the Earth, then the radiation has mass 1/2 M )

Last edited: Sep 29, 2015
13. Sep 29, 2015

### Staff: Mentor

There is your example of mass not being conserved; we went from $M$ before to $M/2$ after. Rest mass is not also not conserved in nuclear reactions (some of it is lost) and pair production (it is created).

It's not wrong, but it is more standard to say that what's being conserved here is total energy, or "mass-energy". If we do it your way, we'd be using the word "mass" sometimes to mean a frame-invariant but not conserved quantity and sometimes to mean a different conserved but frame-dependent quantity.

14. Sep 29, 2015

### vanhees71

As I said, the energy is conserved, the mass obviously not!

15. Sep 29, 2015

### vanhees71

Please, define "mass-energy" for me. Precisely in this context you must give clear definitions. I never have seen this word used anywhere!

16. Sep 29, 2015

### Staff: Mentor

I think everybody agrees that the quantity $$\frac{1}{c^2} \sqrt{E_{total}^2 - \left|\vec p_{total}\right|^2c^2}$$ is conserved for an isolated system of particles, whether they are massive or massless (radiation). The question is whether this quantity should always be called "mass." Some people say yes, others prefer to say this only for bound systems. According to the second group, the radiation released in an unconfined annihilation process does not have mass; however, if the process takes place inside a box with impermeable walls, then the system of box + (radiation inside it) does have a mass which is of course equal to the mass of the box + contents before the annihilation process.

Last edited: Sep 29, 2015
17. Sep 29, 2015

### Demystifier

Exactly! The correct relation is
$$E=\sqrt{m^2c^4+p^2c^2}$$

18. Sep 30, 2015

### vanhees71

.
This quantity is NOT conserved. This is the whole point. Take a container of gas. It has a total energy and momentum that depends on temperature. If you enhance the heat energy of the gas by $\Delta Q$ its invariant mass increases [typo corrected] by $\Delta m=\Delta Q/c^2$. So, although you still have the same gas particles each with its invariable invariant mass, the total mass of the composite system is not conserved.

The same argument also holds for other forms of energy. If you have a capacitor it's invariant mass depends on the voltage on it, because there is field energy $E_{\text{em}}=C q U$, where $C$ is the capacitance, $q$ the charge on its plates, and $U$ the potential difference between the plates. It's mass is $m=m+0+C q U/C^2$, where $m_0$ is the mass of the uncharged capacitor, etc.

An extreme example are hadrons. E.g., the proton consists of three valence quarks. The "bare quarks" (which you cannot observe due to confinement) have a mass of a few MeV, the proton has a mass of about 940 MeV. Most of its mass is dynamically generated, i.e., it consists of a complicated many-body (or better quantum field) state.

Last edited: Oct 1, 2015
19. Sep 30, 2015

### Demystifier

You are both right. It is conserved for an isolated system of particles. It is not conserved if you enhance the heat energy, in which case the system is not isolated. It is also not conserved if the number of particles can change (e.g. by radioactive decay), in which case it is not a fixed system of particles.

20. Sep 30, 2015

### vanhees71

Ok, if you understand this under "isolated", I agree. If there is no exchange of energy with "the environment" the invariant mass stays constant, but that's because energy and momentum stay constant. They are the conserved quantities. Even if some particles decay insight the isolated system, the mass stays constant, because energy and momentum are conserved. You must only be sure that the decay products don't leave the system (which then would not be isolated anymore in this strict sense).