Ramifications of electrons not having gravitational mass (split)

[John Cuthber]

It's actually an interesting question.

Imagine, for example, that the electron has inertial mass, but not gravitational mass.

How would you know?

All determinations of the gravitational constant G are made using macroscopic lumps of uncharged matter.

So every electron is accompanied by several thousand times its mass of nucleons.

 

It's not clear how we could tell if all the "G" we measure is due to the heavy bits.

 

We can measure the inertial mass of electrons (etc) with a mass spectrometer- but that's got nothing to do with the determination of their tendency to "fall" under gravity.

[Sensei]

I didn't read whole thread, so sorry if it's been already said,

but if f.e. electrons would not be accounted in gravity then all calcs are going to be invalid, including mass of Earth..

 

Take for example mass of Iron-56

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

It has mass 55.9349375 u = 9.28821407766474E-026 kg

In electric neutral state, it has 26 electrons, each with mass-energy around 510998.928 eV/c^2 (ignoring energy needed to release free electron)

26 * 510998.928 * 1.602176565e-19 / 299792458^2 = 2.36843955843932E-029 kg

2.36843955843932E-029 kg / 9.28821407766474E-026 kg = 0.0002549941

0.0255% of mass of Iron are it's electrons. In Hydrogen it's 0.05446%

For single atom, it's not huge difference, but for mass of Earth it's really huge difference.

As it would be 1.5e+21 kg difference..

 

Average m^3 of Earth is 5505 kg or so, 0.0255% from it, is 1.4 kg.

Edited March 13, 2016 by Sensei

[John Cuthber]

I didn't read whole thread, so sorry if it's been already said,

but if f.e. electrons would not be accounted in gravity then all calcs are going to be invalid, including mass of Earth..

 

Take for example mass of Iron-56

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

It has mass 55.9349375 u = 9.28821407766474E-026 kg

In electric neutral state, it has 26 electrons, each with mass-energy around 510998.928 eV/c^2 (ignoring energy needed to release free electron)

26 * 510998.928 * 1.602176565e-19 / 299792458^2 = 2.36843955843932E-029 kg

2.36843955843932E-029 kg / 9.28821407766474E-026 kg = 0.0002549941

0.0255% of mass of Iron are it's electrons. In Hydrogen it's 0.05446%

For single atom, it's not huge difference, but for mass of Earth it's really huge difference.

As it would be 1.5e+21 kg difference..

 

Average m^3 of Earth is 5505 kg or so, 0.0255% from it, is 1.4 kg.

And we know the mass of the earth from how much it attracts thing- we gate a value for the gravitational constant G by experiments like Cavendish's.

But if the electron doesn't have a gravitational mass then our measurement of G will be exactly wrong enough to calcel ot the "error" you say we would see.

[Robittybob1]

Comparative mass: Proton is 1837 times heavier than an electron.

Neutron = 1
Proton = 0.99862349
Electron = 0.00054386734

 

So for every electron you have a proton and a neutron so the ratio is roughly 1:2*1837 or approx. 0.000272183 Is that the ratio you are using?

[Sensei]

And we know the mass of the earth from how much it attracts thing- we gate a value for the gravitational constant G by experiments like Cavendish's.

But if the electron doesn't have a gravitational mass then our measurement of G will be exactly wrong enough to calcel ot the "error" you say we would see.

Not exactly.

Oil drop experiment,

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

ionize oil drop, rip off electrons from oil drop, get rid of them.

If you have oil drop prior ripping off its electrons, having the same gravitational mass, as after ripping off, it would be revealed in this experiment.

 

Why do you mention G being different.. ?

If gravitational mass is changed, it must be changed for either object failing down, and mass of Earth, at the same time.

G constant is not affected.

F=G*m*M/r^2

after subtracting some mass from either m, and M,

G could (should) remain the same.

Edited March 13, 2016 by Sensei

[John Cuthber]

there are essentially two ways we measure mass. either we hang it on some sort of spring- which we calibrate against the gravitational attraction of 9in the end) a lump of metal kept in France.

And the other way we do it is to measure charge to mass ratios in various sorts of accelerators.

But you can only do the "hang it on a balance" type of measurement on something which is uncharged, and you can only od the "mass spectrometer" thing on something which is charged.

So you can never strictly reconcile the two measurements.

 

 

In the case of the oil drop experiment, the assumption is made (quite reasonably) that the change in mass of the oil drop when it picks up an electron (or more likely an ionised atom or molecule of air) is very small.

The mass of the electron isn't included in that calculation because it's entirely negligible compared to that of even the smallest oil drops.

[swansont]

 

For single atom, it's not huge difference, but for mass of Earth it's really huge difference.

As it would be 1.5e+21 kg difference..

 

But the mass of the earth is 6 x 10^24 kg.

Not exactly.

Oil drop experiment,

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

ionize oil drop, rip off electrons from oil drop, get rid of them.

If you have oil drop prior ripping off its electrons, having the same gravitational mass, as after ripping off, it would be revealed in this experiment.

 

No it wouldn't. The difference in mass from losing or adding an electron or two, as JC points out, is tiny.

[Sensei]

 

Not exactly.

Oil drop experiment,

https://en.wikipedia...drop_experiment

ionize oil drop, rip off electrons from oil drop, get rid of them.

If you have oil drop prior ripping off its electrons, having the same gravitational mass, as after ripping off, it would be revealed in this experiment.

No it wouldn't. The difference in mass from losing or adding an electron or two, as JC points out, is tiny.

 

Where did you get this "one or two electrons"?

 

On one side we have F=qE

on other side we have F=mg

For q=e=1.602176565*10^-19 C (one mentioned by you electron)

mg=eE / e

E=mg/e

for m=10 mg (I am guessing.. what mass might have single oil drop?)

so

E=10^-5 kg * 9.81 m/s^2 / 1.602176565*10^-19 C = 6.1229*10^14 V/m

Don't you think so it's quite too much?

The more ionized atoms (the larger q in equation), by x-ray, the smaller electric field needed to cancel out gravitational force.

While performing experiment, it can't ionize all medium between electrodes. It would ruin experiment, discharge electrodes.

Reasonable workable voltage is probably around a few thousands volts between plates. Adjust q for it.

 

 

Going back to the main subject.

How about this one:

if electron mass would not contribute to gravitation,

then unstable isotopes decaying by beta decay plus, double beta decay plus, beta decay minus, double beta decay minus,

prior decay mass of atom contribute to overall mass of Earth, and failing object,

and after decay would not, making Earth, or object less massive than prior decay?! IMHO absurd.

Either prior decay and after decay mass of Earth, or other object remain the same, unless electron (or other particle) escape it (like (anti)neutrinos).

 

[swansont]

Where did you get this "one or two electrons"?

 

On one side we have F=qE

on other side we have F=mg

For q=e=1.602176565*10^-19 C (one mentioned by you electron)

mg=eE / e

E=mg/e

for m=10 mg (I am guessing.. what mass might have single oil drop?)

so

E=10^-5 kg * 9.81 m/s^2 / 1.602176565*10^-19 C = 6.1229*10^14 V/m

Don't you think so it's quite too much?

The more ionized atoms (the larger q in equation), by x-ray, the smaller electric field needed to cancel out gravitational force.

While performing experiment, it can't ionize all medium between electrodes. It would ruin experiment, discharge electrodes.

Reasonable workable voltage is probably around a few thousands volts between plates. Adjust q for it.

 

 

No, not at all. I did this experiment in college. As I recall, our data was only from drops having a very small number of electrons.

 

The oil drops were micron-sized, so they would have a mass more like a pg. So nine or ten orders of magnitude smaller than what you have. So now we're talking 10^5 V/m. If the separation of the plates is on the order of a cm, then the potential difference is a kilovolt. Not a big deal.

 

There couldn't have been too many charges on (or removed from) the drop, since that would make it nearly impossible to make the measurement. Distinguishing between 1 or 2 electrons is far easier than 999 and 1000. Or even 99 and 100.

 

But even if it was 1000 electrons, that's 10^-24 grams. 12 orders of magnitude lighter than the oil drop.

[John Cuthber]

Where did you get this "one or two electrons"?

 

 

From Milikan's experiment.

Where are you getting anything else?

He needed a microscope to see the oil drops which suggests they are pretty small- perhaps 1 to 10 microns. (one of the errors in the measurement is due to the drops being small enough that air isn't a fluid, it's a bunch of moving lumps)

With a density of about 1 gram per ml and a volume of something of the order of a thousandth of a thousandth of a thousandth of a ml . your "estimate" of a milligram or so looks to be about 7 orders of magnitude out.

 

But even that's not the only problem.

If the Xray source or whatever is producing enough ions to put a thousand electrons on a bigger oil drop in a reasonable period of time then, over the course of the experiment it's likely to add lots more. The number wouldn't be anything near constant so you wouldn't be able to make measurements..

 

Once again, I see you are choosing to argue that everyone is wrong, rather than actually checking.

It's a habit you should get out of.