# Are charged batteries heavier?

• rootone
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In summary: Interesting link.According to Wikipedia the international standard kilogrammes gains around about 1 micro gram a month by absorbing contaminants from the air despite being under two nested bell jars. That works out at about 33 Pico grams a day.An exception would be the lithium-air battery, which uses oxygen from the atmosphere to oxidise lithium in the discharge reaction:2Li + O2 → Li2O2As the oxygen comes from the atmosphere on discharge and is released to it on charge, it is not part of the battery, which therefore weighs more in the discharged state than the charged state.I did wonder is some sort of balance could be created using an ink jet cartridge...The battery as
rootone
Since you can add energy to a battery then extract it later, considering mass energy equivalence, it should be more massive, no?
Somehow that doesn't make sense though.

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It is more massive. Let's calculate how much. The battery on my laptop stores about 50 W-h = 180,000 Joules. Δm = ΔE/c^2 ~= 2 picograms. Hard to measure!

PumpkinCougar95, WWGD, Vectronix and 8 others
Thanks and yes it would be hard to measure but not impossible.
I expect somebody will be able to confirm that this has been checked out.

Of course mass-energy equivalence has been checked in many different ways. Here's one for example. But I don't think there is any scale accurate enough to measure the mass difference between a charged and uncharged battery.

vanhees71
I heard also about the planes flying in different directions as well, although I think that is GR rather than SR.
Back to batteries though, if they become more massive when charged, even by an immeasurably small amount, where is the extra mass?
Do electrons get more massive?

rootone said:
I heard also about the planes flying in different directions as well, although I think that is GR rather than SR.
Are you thinking of Hafele-Keating? https://en.wikipedia.org/wiki/Hafele–Keating_experiment
Back to batteries though, if they become more massive when charged, even by an immeasurably small amount, where is the extra mass?
Do electrons get more massive?
The battery as a whole is more massive; you can't assign the extra mass to anyone part of it. The mass of the charged battery is greater than the sum of the masses of its constituent parts.

1 Yes that is the experiment I heard of.

2 Aggregated mass of constituents is something I need to think about,.

rootone said:
where is the extra mass?
Do electrons get more massive?
The molecules in a charged battery are different than the molecules in an uncharged battery. The ones in the charged battery are more massive than the ones in he uncharged battery. You cannot assign the mass to any part of the molecules, just the molecules as a whole.

Dale said:
The molecules get more massive.
How?

rootone said:
How?
By being in a configuration with more potential energy.

OK I get the idea.
But what exactly is configuration?
Something to do with Higgs field?

I am puzzled, but hey that's why I am here,

rootone said:
But what exactly is configuration?
Something to with Higgs field?
No. For a chemical reaction it is just electromagnetic. The Higgs is not relevant.

Configuration means the various distances between nuclei and electron orbitals etc. That is what gives chemicals their energy.

Bystander
rootone said:
Since you can add energy to a battery then extract it later, considering mass energy equivalence, it should be more massive, no?
Slightly off-topic comment: this is something that irks me about many sci-fi stories. Some devices, such as weapons, have an incredibly big energy supply, but are still lightweight. Not possible, ##E=mc^2## rules!

Not sure if it's possible to weigh something like a smartphone battery to within 2 pico grams? I know you can weigh much smaller things more accurately but i don't think the method used for that can be applied to something big like a battery.

According to Wikipedia the international standard kilogrammes gains around about 1 micro gram a month by absorbing contaminants from the air despite being under two nested bell jars. That works out at about about 33 Pico grams a day.

An exception would be the lithium-air battery, which uses oxygen from the atmosphere to oxidise lithium in the discharge reaction:
2Li + O2 → Li2O2
As the oxygen comes from the atmosphere on discharge and is released to it on charge, it is not part of the battery, which therefore weighs more in the discharged state than the charged state.

Derek P, dRic2, Ygggdrasil and 3 others
I did wonder is some sort of balance could be created using an ink jet cartridge...

Ink jet cartridges create pico litre droplets of ink so I was thinking that perhaps you could use one to squirt ink onto a counter weight until it was in balance with a discharged battery. Then charge the battery using a wireless charger and count how many additional drops are required to bring it back into balance again.

Then I realized that the tiny drops produced by an ink jet printer would be three orders of magnitude too big! (pico Liters >> pico grams)

phyzguy said:
Of course mass-energy equivalence has been checked in many different ways. Here's one for example. But I don't think there is any scale accurate enough to measure the mass difference between a charged and uncharged battery.

Minor quibble (since I'm an experimentalist): I think you mean 'precise', not 'accurate'. This measurement, for example, requires about 14.5 significant figures (to reliably measure 1 pg changes in a 100g battery). I don't know any existing balance that can do this (maybe the Watt balance). The standard kilgoram calibration campaign quotes about 11 digits (1 ug/kg) using one of these:

https://www.mt.com/dam/P5/labtec/08...tion/03_Datasheet/DS_Vacuum_M_one_M_10_EN.pdf

Andy Resnick said:
Minor quibble...
For absolute mass measurement.
Using a Cavendish torsion balance, the period of oscillation should be comparable to the square root of the difference in mass, or rather 10-6 seconds, and that 'should be' fairly easy to measure with a current time clock.
Just wondering.

Check out https://en.wikipedia.org/wiki/Binding_energy

The binding energy can sometimes be interpreted as a reduction of field energy. For example, an electron and a proton are bound as a hydrogen atom. The electric field has energy ##\frac{1}{2}\epsilon E^2##. When the electron and proton are close together, the E fields overlap and mostly cancel out, reducing the field energy. By rearranging the atoms in matter, you can change the energy even though the number of atoms is equal.

256bits said:
Using a Cavendish torsion balance, the period of oscillation should be comparable to the square root of the difference in mass, or rather 10-6 seconds, and that 'should be' fairly easy to measure with a current time clock.
Just wondering.

Why go through the effort? The uncertainty/statistical spread of a discharged battery's mass is likely many times larger than a pg.

Edit- I just realized, how did you get 10-6 seconds fractional difference for the oscillation period?

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Andy Resnick said:
Why go through the effort? The uncertainty/statistical spread of a discharged battery's mass is likely many times larger than a pg.
Why is that a problem? Wouldn't you would measure the frequency of oscillation of _a_ battery while it was being discharged looking for the change in mass.

CWatters said:
Why is that a problem? Wouldn't you would measure the frequency of oscillation of _a_ battery while it was being discharged looking for the change in mass.

How would you discharge it while in the device? How much of an effect would you expect to see?

Dale said:
No. For a chemical reaction it is just electromagnetic. The Higgs is not relevant.

Configuration means the various distances between nuclei and electron orbitals etc. That is what gives chemicals their energy.
So, to bring this back to the OP's question:
the inflow of electricity, in the form of electrons, causes molecules to change shape, and some of the electrons' orbital energy is converted back to mass ... by way of muons?

Or am I word salading here?

See a battery on table, put a table mat underneath the battery and wa la, battery weighs more. Not a lot more, I'll give you that!

CWatters said:
Why is that a problem? Wouldn't you would measure the frequency of oscillation of _a_ battery while it was being discharged looking for the change in mass.

If you think this measurement is feasible, why don't you go ahead and give it a try? I concur with Andy Resnick. No measurement technique exists that can measure a change in mass of 1 part in 10^14.

rootone said:
Something to do with Higgs field?

A common misconception, the Higgs field is only responsible for about 1% of the mass of an atom,

Explained here by Frank Wilczek...

Full video here http://techtv.mit.edu/videos/15942-the-origin-of-mass-and-the-feebleness-of-gravity

rootone said:
Somehow that doesn't make sense though.

Why stop there, if you think about it nothing makes sense. For example, once upon a time there was a huge cloud of molecular hydrogen and not much else, then by a convoluted series of condensations a conscious being emerged wondering why things don't make sense.

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rootone
DaveC426913 said:
by way of muons?
No, why would there be muons? There are no nuclear reactions or cosmic rays in a battery.

jerromyjon
Dale said:
No, why would there be muons? There are no nuclear reactions or cosmic rays in a battery.

I was thinking that the change of an electron between orbitals would involve a particle energy-mass conversion.

bland said:
once upon a time there was a huge cloud of molecular hydrogen and not much else, then by a convoluted series of condensations a conscious being emerged wondering why things don't make sense.
Thanks for those.
Are you related to Douglas Adams by any chance?

jerromyjon
DaveC426913 said:
So, to bring this back to the OP's question:
the inflow of electricity, in the form of electrons, causes molecules to change shape, and some of the electrons' orbital energy is converted back to mass ... by way of muons?
No, it's much more straightforward than that. The mass of a multi-particle system is generally not equal to the sum of the masses of the particles of which it is composed.,

Vectronix and jerromyjon
phyzguy said:
If you think this measurement is feasible, why don't you go ahead and give it a try? I concur with Andy Resnick. No measurement technique exists that can measure a change in mass of 1 part in 10^14.
Oh I don't think it's feasible I just didn't understand why differences between batteries would be the problem if only one is used.

Nugatory said:
The mass of a multi-particle system is generally not equal to the sum of the masses of the particles of which it is composed.,

Which is fundamentally why atoms and molecules form, because like water nature generally wants to live on the ground floor.

bland said:
Which is fundamentally why atoms and molecules form, because like water nature generally wants to live on the ground floor.
That's not correct. Energy is not the relevant quantity, its entropy!

DaveC426913 said:
I was thinking that the change of an electron between orbitals would involve a particle energy-mass conversion.
It does, but the particle in question is the whole atom or molecule, not the electron. An excited atom is more massive than an atom in the ground state. In the transition the energy is lost to photons and/or KE.

Δ√QUOTE="Andy Resnick, post: 5951029, member: 20368"]The uncertainty/statistical spread of a discharged battery's mass is likely many times larger than a pg[/QUOTE]
Looking at a harmonic oscillator, torsional:
Originally, I had thought that since the period is proportional to the square root of the mass, ... could be faulty thinking.

Background:
Time period of oscillation of twisting wire,

or frequency,
ω = √[ (k/I )
I - moment of inertia of the mass(s)

https://en.wikipedia.org/wiki/Cavendish_experiment

For the Cavendish balance,
This is the regular formula, for two masses separated horizontal distance 2L, and hanging from a wire.

T -period
m - mass
L - distance to wire
k - torsion coefficient of the wire

The mass hanging from the wire could be a single battery, or two on extended arms similar to the Cavendish setup.

We should have, with the battery uncharged
T1 = C√( m )
where C is a constant composed of 2, π, √L2/2k

Similarly, with the battery charged,
T2 = C√( m + Δm )

Squaring each equation, and subtracting
T12 - T22 = C2 Δm

or
Δm = T12 - T22 / C2

Like you said could be impractical due to errors in measurement for C.

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