Weight difference between a charge and discharged battery

[leothorn]

Assuming an a suitably accurate weighing device ,

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

 

[soothsayer]

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

 

[leothorn]

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

 

[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.

 

[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.

 

[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?

 

[jbriggs444]

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?

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

 

[soothsayer]

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.

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...

 

[soothsayer]

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

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.

 

[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.

 

[Nugatory]

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.

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.

 

[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.

 

[soothsayer]

Assuming an 12v auto battery good for about 10⁶ Joules to total discharge, we might expect to lose about 10 nanograms.

These are 10 nanograms of relativistic mass?

 

[Nugatory]

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.

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.

 

[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.

 

[res3210]

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

 

[soothsayer]

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.

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.

 

[BruceW]

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.

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.

 

[soothsayer]

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.

Exactly my thoughts...

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

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...

 

[soothsayer]

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.

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.

 

[Nugatory]

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.

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...

 

[Nugatory]

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.

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]

 

[Nugatory]

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.

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.

 

[Drakkith]

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.

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

 

[BruceW]

However, it makes no difference whether nuclear forces are involved or not; if a system radiates energy away, its total invariant mass is reduced.

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.

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.

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).

 

[soothsayer]

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.

Wait, but doesn't that equation you have for E² neglect potential energy? That's just a statement of total energy = kinetic energy + rest energy. Obviously p = 0, so you just have $E = mc^2$ the whole time. But this doesn't make any sense, because you're saying that the energy of the battery is ONLY related to the mass of the battery, but obviously the chemical makeup of the battery has something to do with it too, doesn't it?

Consider we have a box at the top of a slope that has friction, and it starts at rest, p=0. We release the box from the top of the slope, and it accelerates down to the bottom until it hits a flat point, and then briefly slides before coming to a halt, having been stopped by friction. The momentum of the box is zero at both the beginning of the end of the system, but the box clearly had initial potential energy, which it lost, so we can look at the change in potential, or the change in heat from friction and use $E = mc^2$ to determine how much rest mass the box lost??

 

[soothsayer]

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

How? What about the potential has a rest mass? The potential is just created by an electric field between the anode and cathode. The electric field is just composed of gauge bosons--photons, so the field is massless, yes? Where does the mass come from?

 

[Drakkith]

How? What about the potential has a rest mass? The potential is just created by an electric field between the anode and cathode. The electric field is just composed of gauge bosons--photons, so the field is massless, yes? Where does the mass come from?

Let's put it this way. If you have two identical boxes, and one is filled with light while the other is a vacuum, the first box has more mass than the second due to the energy of the light. The same concept applies to the battery. While charged it has more energy than when discharged. This energy, in whatever form, has mass.

This is exactly the same thing as how the sum of the masses of an electron and proton are greater when they are separated than when they are bound together in a hydrogen atom. When the electron binds with the proton energy is released in the form of light. Yes, this light is massless. Yet it removes energy from the SYSTEM, which makes the system less massive.

During discharge, energy is removed from the battery. This causes the battery to lose mass. As has been said, you can calculate the mass loss by using Einstein's famous equation.

 

[soothsayer]

Let's put it this way. If you have two identical boxes, and one is filled with light while the other is a vacuum, the first box has more mass than the second due to the energy of the light.

So if we have two identical boxes, and we fill one of them with light, even though the light has no rest mass, the energy from the light will contribute to the rest mass of the box? I mean I get that it would increase the weight, like, on a scale...

How does the box know it is filled with light, and how does that change its inherent properties? Is it because the light is being absorbed by the box? What if we have two identical boxes and poke a hole in one to let light in?

I guess I missed the lesson on this one

This is exactly the same thing as how the sum of the masses of an electron and proton are greater when they are separated than when they are bound together in a hydrogen atom. When the electron binds with the proton energy is released in the form of light. Yes, this light is massless. Yet it removes energy from the SYSTEM, which makes the system less massive.

Ah, right. I forgot about this. I was also just thinking about how the rest mass of mesons/hadrons are greater than the sum of the masses of the quarks they are made of due to interactions between the quarks. This still seems a little different than your light in a box example above, though.

 

[WannabeNewton]

There is absolutely no need to mention GR (it's not even being used correctly in this thread anyways); this is over-complicating things. Read the following and then reread Nugatory's and Drakkith's posts: http://en.wikipedia.org/wiki/Mass_i...individual_rest_masses_of_parts_of_the_system

 

[soothsayer]

Thanks for the link, Wannabe. I'm actually pretty dismayed that I never learned this in college.

When was GR used in this thread?

 

[WannabeNewton]

GR was mentioned more than once in the thread but never in a correct manner (we don't need it anyways).

 

[Nugatory]

So if we have two identical boxes, and we fill one of them with light, even though the light has no rest mass, the energy from the light will contribute to the rest mass of the box? I mean I get that it would increase the weight, like, on a scale...

Yes, that is exactly what happens. And if you were to open a hole in the box and let out some of the light inside the box, the mass of the box+contents will decrease, just as it would if you opened a hole and let some of the gas inside the box out. Conversely, if more light to were enter through the hole than left, or more gas, the mass of the box would increase.

How does the box know it is filled with light, and how does that change its inherent properties? Is it because the light is being absorbed by the box? What if we have two identical boxes and poke a hole in one to let light in?

Suppose I put a lead shot in the box... How does the box know that it contains a lead shot so has a greater mass?

Here's a thought experiment. The box, when empty, has a rest mass of 1 kilogram. It contains one gram (rest mass) of matter and one gram (rest mass) of antimatter, so its total rest mass is 1.002 kg as long as the matter and antimatter are separated. But eventually the matter and antimatter will come into contact with each other, annihilate, and turn into (an amazing amount of amazingly energetic) photons that have zero rest mass. If the box is strong enough to contain the resulting explosion so that we on the outside stilll see the same closed system, does the mass of the box plus its contents change?

No. The total rest mass is still 1.002 kilograms.

Backing up from here, the easiest way of thinking about this is to say that energy has mass, given by $E=mc^2$. The trick and pitfall is that kinetic energy depends on speed, hence is frame-dependent. Thus, if you're going to use this equation, you can either:
1) Choose a frame in which the speed is zero so there's no kinetic energy and you are dealing with rest mass. This makes the math especially simple, and feels right for this thread where we're talking about a battery sitting on a table in front of us - no movement, no kinetic energy to mess with. The kinetic energy in the system is captured in the $(pc)^2$ term of $E^2=(m_{0}c^2)^2+(pc)^2$, conveniently zero in this frame.
2) Go with the concept of relativistic mass given by $m_r=\frac{m_0}{\sqrt{1-v^2}}$ and use $m_r$ in $E=mc^2$ so that it includes the non-zero kinetic energy, so is not frame-independent.
3) Use $E^2=(m_{0}c^2)^2+(pc)^2$ (and note that if $v$ and hence $p$ is zero this option reduces to #1 above). As in #1, we're only dealing with rest mass and rest energy here; the frame-dependent kinetic energy is captured in the $(pc)^2$ term.

 

[leothorn]

There is absolutely no need to mention GR (it's not even being used correctly in this thread anyways); this is over-complicating things. Read the following and then reread Nugatory's and Drakkith's posts: http://en.wikipedia.org/wiki/Mass_i...individual_rest_masses_of_parts_of_the_system

Do the conclusions of then discussion hold if the concept of mass in special relativity does not hold ? ( A contention held at the end of the article linked above ?


I was thinking on the following lines..

When the battery system is at higher energy that is the constituent particles are at higher state of activation, they hit the confines of the battery at a much faster rate than when the battery is at lower energy. The combined affect of higher average impulse over time should affect the weight of the battery.

Is this thought process wrong. Could this argument be repaired to make it right ( if its incorrect).
Thanks to all those who replied

 

[Nugatory]

When the battery system is at higher energy that is the constituent particles are at higher state of activation, they hit the confines of the battery at a much faster rate than when the battery is at lower energy. The combined affect of higher average impulse over time should affect the weight of the battery.

The energy of the battery isn't stored in faster-moving versus slower-moving particles. It's stored in high-energy chemical bonds, and the energy is released by chemical reactions in which those bonds are replaced by lower-energy bonds. As long as the battery is not overheating (something that battery designers try to avoid), the constituent particles of the battery are bouncing around at the same speed whether the battery is charged or discharged.