# Weight difference between a charge and discharged battery

1. Jul 26, 2013

### leothorn

Assuming an a suitably accurate weighing device ,

Would the weight of a completely discharged battery be different from a completely charged battery.

2. Jul 26, 2013

### soothsayer

It would not. Batteries simply convert their chemical potential into electrical potential, when attached to a circuit. They do not lose any mass.

3. Jul 26, 2013

### leothorn

What about the fact that the chemical reaction leads to change in weight of the constituent parts of the reaction /

4. Jul 26, 2013

### soothsayer

Regardless, any changes in weight should be all self contained. If there is weight change in constituent parts, then some parts are gaining mass from other parts, but it should all be conserved.

5. Jul 26, 2013

### jbriggs444

Assuming a suitably accurate weighing device...

The internal energy of a charged battery contributes to the rest energy of the battery and, accordingly, to its rest mass. The conversion factor is e = mc2. For a chemical cell, this contribution is not normally measureable.

6. Jul 26, 2013

### soothsayer

Look at it this way, if the weight of a battery decreased after being depleted, it would have to mean that the weight of the circuit it was attached to increased. Power this circuit for years with many batteries, and the mass of the circuit would necessarily have to increase arbitrarily. Is this sensical?

7. Jul 26, 2013

### jbriggs444

The circuit radiates thermal energy, thus draining mass into the environment.

8. Jul 26, 2013

### soothsayer

That's untrue, the E in E = mc2 is an internal energy, a mass equivalency energy. The chemical potential in a fresh battery is not equivalent to this kind of energy.

By this logic, an object would gain mass simply by being raised in a gravitational field...

9. Jul 26, 2013

### soothsayer

No, the thermal energy comes from resistance in the circuit, which dissipates the electrical potential, which is generated from a chemical potential. There is no mass-energy conversion in this system.

10. Jul 26, 2013

### BruceW

In relativity, the relativistic mass of the battery would decrease. This is because relativistic mass is synonymous with energy (in most places that I have seen it written). And also, an object does gain relativistic mass by being raised in a gravitational field (in general relativity).

edit: ah wait, other way around, as an object is raised in a gravitational field, its energy decreases, so if you stand on the top floor, then the elevator loses energy as it moves upwards. Also, we need to be careful about what coordinate system we are using, as soon as we start talking about general relativity.

edit again: well, I really confused myself. uh... In Einstein's field equations, the 'gravitational energy' is not included in the stress-energy tensor, so in this sense, the energy of an object does not include any 'gravitational energy' it may have.

Last edited: Jul 26, 2013
11. Jul 26, 2013

### Staff: Mentor

Soothsayer, you're right that there is no mass-energy conversion going on.

However, the battery+circuit system does lose energy when it radiates away some of the heat generated in the circuit by the flowing current. That energy loss means that the total amount of energy in the battery+circuit system decreases, and therefore its weight. Assuming an 12v auto battery good for about 106 Joules to total discharge, we might expect to lose about 10 nanograms.

12. Jul 26, 2013

### soothsayer

I don't believe in relativistic mass. :tongue:

Though, I suppose that if the battery IS losing energy, then according to GR, the gravitational force acting on it is lessened, which means that its weight WOULD technically have decreased? Even though its invariant mass (I hate having to add that qualifier) does not change.

13. Jul 26, 2013

### soothsayer

These are 10 nanograms of relativistic mass?

14. Jul 26, 2013

### Staff: Mentor

This has nothing to do with relativistic mass, and it is the invariant mass that it is changing.

It's basically the same situation as if you were to gather up the products of a uranium fission reaction. You'd find that the sum of the masses of the daughter nuclei and the stray neutrons released in the reaction is less than that of the original uranium nucleus; and the missing mass is the energy released in the explosion.

15. Jul 26, 2013

### BruceW

Invariant mass is conserved (unless there are nuclear forces at work). In a battery, I would guess the forces are mostly electromagnetic, so the invariant mass should be conserved.

edit: in other words, I am saying that most chemical reactions are mostly electromagnetic, not nuclear. I'm pretty sure this is true. But maybe not, maybe there are exceptions.

16. Jul 26, 2013

### res3210

If the mass doesn't change, how can the weight?

17. Jul 26, 2013

### soothsayer

I mean, I totally understand how this happens in fission reactions, but I am lost as to how you lose mass in a battery. Maybe it's my understanding of how a battery works that fails me...how does losing thermal energy in the circuit take away from the mass of the battery? The energy present in the battery at the beginning has no invariant mass of which to speak of.

18. Jul 26, 2013

### BruceW

eh? the battery does have invariant mass to begin with. Any everyday object has invariant mass. The thing is that I thought that this won't change in most chemical reactions. But Nugatory seems to think it does. I'm not certain who is right on this one.

19. Jul 26, 2013

### soothsayer

Exactly my thoughts...

In classical physics, it can't, however, in general relativity, the force of gravity on an object is actually related to its momentum, not its mass, and in special relativity, we learn that momentum is not just simply mass x velocity, but you actually need to throw the Lorentz factor in as well to handle v ~ c situations. This is why light can be effected by gravity, even though it has no mass, because it DOES have momentum.

The amount of weight change in a battery would be negligible, and probably impossible to detect. However, with a suitably accurate scale...

20. Jul 26, 2013

### soothsayer

I didn't say the battery didn't have invariant mass, I said that the chemical/electrical potential which drives the whole system doesn't have invariant mass of which to speak. Just like a gravitational potential doesn't have invariant mass.

21. Jul 26, 2013

### Staff: Mentor

Chemical reactions are electromagnetic interactions between electrons and protons - there is zero nuclear force involved.

However, it makes no difference whether nuclear forces are involved or not; if a system radiates energy away, its total invariant mass is reduced. We usually don't consider this effect for chemical reactions but do consider it for nuclear reactions, for two reasons:
1) Chemical reactions are much weaker, so the total amount of energy involved and the total mass defect is usually unmeasurably small. Consider what ten nanograms means to a car battery - nada. By contrast, uranium fission and hydrogen fusion can easily release about 1% of the total mass-energy so we pay attention to it.
2) In many nuclear reactions, we know the mass of the particles going in, and we know the mass of the particles going out, and we want to calculate the energy released in the reaction. Calculating it from the mass change using $E=mc^2$ is a quick and easy way.

You might want to google for "mass defect" or "mass defect in nuclear physics" and the like....

Last edited: Jul 26, 2013
22. Jul 26, 2013

### Staff: Mentor

Well, ten nanograms is pretty insignificant... But the invariant mass of a charged auto battery is about ten nanograms more than the mass of the same battery uncharged (assuming we're keeping it in a sealed container - otherwise we'll lose water vapor, hydrogen, and the like in quantities that overwhelm that small amount).

[Edit: This might be one of the situations where the term "rest mass" is less confusing than "invariant mass". The mass is changing, which makes "invariant" sound a bit silly - but remember, we're also adding and removing something from the system as we charge and discharge the battery]

Last edited: Jul 26, 2013
23. Jul 26, 2013

### Staff: Mentor

It's easy. First, write down the general equation $E^2=(m_0c^2)^2+(pc)^2$. Now the whole shebang is at rest, so $p$ is zero, there's no relativistic mass involved, and the $(pc)^2$ term is zero. Now consider that the total energy of the charged battery is greater than that of the uncharged battery, and solve for $m_0$ in the two cases.

24. Jul 26, 2013

### Drakkith

Staff Emeritus
Yes it does. The potential energy of a charged battery adds to the invariant mass.

25. Jul 26, 2013

### BruceW

Ah, I think I see what you mean now. If we can imagine the battery as a collection of molecules and photons, then if some photons 'escape', the invariant mass of the battery will decrease simply because we are now thinking of the 'later-time' battery as a subsystem of the 'earlier-time' battery.

well, E is the relativistic mass (in natural units) for a general system. So for p=0, the invariant mass is equal to the relativistic mass. But I see your point, it is obvious that I was wrong earlier when I said the invariant mass would not change. (since we assume the total momentum is always zero, and that the energy changes).