Does an electrical charge have weight?

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steve066 said:
And you think I'm confused.
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I most certainly do. You can't seem to distinguish the difference between a particle in a system and the mass of a system.

You think that you have to add a particle to a system to increase its mass. You are absolutely wrong.

I did not say energy and mass are the same thing, only that mass is a form of energy. Pay attention.

You can define system mass such that it includes the particles and the energy inside a system. It is done all the time. When they calculate the mass of the sun, do you think they are leaving out the photons bouncing around inside?

I will say that a system containing only one photon is a massless system for the same reason that a photon is considered massless: there is no rest frame for the system and thus no rest mass.

If you can find the center of momentum for a system, then you can find the rest mass even if the system has more than just particles inside.

I'm talking about the mass of the capacitor system because we were talking about how much the weight would change if put on a scale of sufficient resolution to measure everything inside the system boundary. That includes all internally stored energies.

We are ultimately talking about energy conversions here. Categorizing the different energy forms differently doesn't mean the weight of the capacitor does not change.

We can debate whether the scale deflects because of a gravitational field or a distortion in space-time but the end result is the same.
 
How do you do that?

How do you do that?

__dan said:
electrons have stable mass. If there are theories where the electron gets lighter, it's not from charging in a battery.

Photons have no mass, adding photons to an electron does not affect mass.

The battery electrolyte has specific gravity and a change in density from charging. Charging the battery causes the electrolyte to becomes less dense, it occupies more space at the same mass (I could easily have that backwards, I know charging affects density, specific gravity). The battery plates (one of them) gets a coating from charging, it looks like a dusting with charcoal dust when fully charged on, I am guessing, the positive plate.

Adding electrons to a cap adds the weight of the additional electrons.
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IMO, electrons are moved from one plate of the cap to the other. It would seem that one plate would weigh more than the other, but overall there would be no change in mass.
 
Steve,

Please step away from the term 'mass' when reading mivey's posts; you and he are using the term differently. You are using 'mass' to mean _particles_, mivey is using 'mass' to mean 'mass-energy'.

Instead, where mivey says 'mass', instead read 'anything which creates or responds to a gravitational field'. While particles certainly both create and respond to gravitational fields, an electric field is itself a thing which will both create and respond to gravity.

This is the key to the discussion: does an electric field itself create or respond to a gravitational field. If I simply re-arrange the same set of particles, without changing the number or types of particles, such that I change the total energy contained by that set of particles, will I change the amount of stuff that creates or responds to a gravitational field.

The example is a simple parallel plate vacuum capacitor. Simply by moving electrons from one plate to the other, without changing the total number of electrons, I can create an electric field between the electrodes. Moving these electrons around requires effort, supplied by an outside source, and stores energy in the capacitor. The total amount of _matter_ in that capacitor is unchanged; simply moved around.

I claim that the electric field itself will both respond to and create a gravitational field. Because I've added something to the capacitor which responds to a gravitational field, I have increased the total 'anything which creates or responds to a gravitational field' contained within the capacitor case. The capacitor gets heavier (presuming it is sitting someplace with gravity.) Mivey would say that the electric field itself has mass. No particles, simply energy, and that energy has mass.

Electromagnetic waves moving in free space are _known_ to be attracted by gravity. The measurement of the deflection of starlight around the sun was one of the earliest tests of relativity. So we _know_ that electromagnetic fields respond to gravitational fields. We've not made the measurement in the opposite direction, but I would be tremendously surprised if EM waves violate 'every action has an equal and opposite reaction'. I claim that light will both respond to and create a gravitational field.

Sound waves move through media by mechanically moving the atoms and molecules around, storing energy by pushing molecules closer together or pulling them apart. The forces that hold atoms and molecules in their various positions are essentially electrical in nature; a sound wave means energy stored in the electrical fields surrounding the molecules. These electrical fields create and respond to gravitation, thus the sound leaving the speakers must create and respond to gravitation.

-Jon
 
winnie said:
...This is the key to the discussion...
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That's it in a nutshell.

The problem comes in when we try to mix the terms mass and energy. You have to define what you are talking about. We could restrict the whole discussion to energy alone and just forget the word mass and use the better(?) term of "rest-energy". Then there is the discussion about what defines "rest-energy", so you haven't avoided much.

Also, if you use "rest-energy" some people start thinking you are trying to throw away the mass. Mass is actually on the other side of the E=mc^2 equation from energy. Sometimes it is more prudent to speak of mass or "rest-mass" because it keeps you from having to explain "rest-energy". Considering "rest-energy" is another way to say that mass is a type of energy.

BTW, the real term for Einstein's "m" is inertia, but again requires more explaining so it becomes easier to use mass. People often confuse this with matter.

But the gist is what you summarized: Is there a change in the gravitational or inertial response?
 
In this same line of thought.

Mivey:

If I have an ideal transformer (no losses) not connected to anything, and it weights "X" pounds. Then I connect it to the utility at full load, will it be heavier at any moment??

If it is yes? then why?
 
Mayimbe said:
In this same line of thought.

Mivey:

If I have an ideal transformer (no losses) not connected to anything, and it weights "X" pounds. Then I connect it to the utility at full load, will it be heavier at any moment??

If it is yes? then why?
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An ideal transformer is just a mathematical ratio so is immune to the effects of the physical world.
 
Charging

Charging

rattus said:
IMO, electrons are moved from one plate of the cap to the other. It would seem that one plate would weigh more than the other, but overall there would be no change in mass.
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The difference between capacitance, charge accumulation Q=CV, which can execute in single wire circuits, and using a capacitor element in a two wire circuit.

One previous poster gave the example, rub a ballon in your hair and it picks up an excess static charge, additional electrons above ground state, and the added weight of more electrons.

There is current flow in single wire tuned resonating circuits, the cap would have one wire on the circuit and the other in the air.

I know this would be easier to present with a diagram. Also, I present this not as an assertion but as an easily testable arrangement.

Take a battery and a capacitor in parallel with isolation switches on both sides of the cap and an earth ground at one side of the battery. Ground the 'pos' side of the battery so the neg, high side of the battery, has the excess of negatively charged electrons to supply.

With both isolating switches open, close the neg switch. The open circuit voltage appears at the other open isolating switch. The cap has picked up some electrons, a negligible open circuit charge, and gained mass. The delta V is across the isolating switch. The delta V across the cap is zero, the dielectric is not being worked. Both cap plates are neg delta V to ground and equal V to each other.

Current has flowed in one wire with the circuit being completed internally in the chemical battery reaction.

Same circuit, close both isolating switches. The battery charges and the delta V is across the dielectric. But the Pos side has been earthed an remains at zero delta V to earth. It has no work to do to go to earth or received any work to change its V to earth.

With the cap charged, open both isolating switches and remove the cap to the air. Its charge remains delta V across and working the dielectic. One side is zero volts to ground so by definition the cap in sum has an excess of electrons, a charge and mass gain.

The cap charged and isolated from a complete circuit is maintained at a higher potential energy by having more electrons relative to the earth. Close the circuit and charge tries to equalize between the plates, current flows.

Equalization of charge between the plates is due to changing the external circuit arrangement, not due to capacitance. The external circuit can be arranged to load an excess of electrons onto the cap.

As an easily testable hypothesis:

Same circuit, battery and cap in parallel with isolating switches, earth the pos battery side. Place two equal resistors in series with the battery supply terminals and metering equipment across the resistors to record current flow.

Close the isolating switch that is earthed. The cap has no work to do, it begins at ground potential, no current flows.

Close the high side isolating switch and the battery sees the ground V across the cap and charges it until the cap delta V matches the battery's. I suspect the current charging waveforms will not match equally. The dielectric is an insulator and for the battery the internal chemical reaction completes the circuit internally until the battery can again maintain an open circuit voltage.

Yes, in AC circuits the cap will have a heavy neutral current.
 
mivey said:
An ideal transformer is just a mathematical ratio so is immune to the effects of the physical world.
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Let's take a closer look. Define system S to be the source up to the ideal transformer. Define system T to be the transformer. There is a mass trapped in system T that weighs "X". Define system L to be the load side of the transformer. Let the systems remain at rest so we don't have to track momentums. Let the source and load systems exchange potential energy.

In an ideal transformation, the input power equals the output power. So the input energy equals the output energy. An ideal transformer is a mathematical ratio and is just a transfer function from system S to system L and it transfers the energy at the speed of light so will never have a time when it has a net increase in energy (or weight).

But let's assume it is not quite ideal. Let's assume the mass is the mechanism for transfer and it transfers at less than the speed of light. Also assume that any energy absorbed by T at the S-T interface must eventually exit at the L-T interface (lossless).

Also, if the transformer has weight, it is being measured relative to something (let's say the earth). We will call the earth system E.

When power and load is applied, there will be a slight propagation delay as the energy moves through the mass in T. This propagation delay will mean there is a net quantity of energy propagating through the mass in T. In this state, the transformer will be heavier. The weight of the S-T-L system remains the same.

So what happened? Before power was applied, there was potential energy from S-E and from T-E and from L-E. Call them SE1, TE1, and LE1 and SE1+TE1+LE1 = PE1

The potential energy of S-E is dropping because it is losing energy. The potential energy of L-E is increasing because it is absorbing energy. When the transfer is taking place, the potential energy of T-E is higher.

So during transfer, at any state i:
SEi<SE1, TEi>TE1, LEi>LE1, PEi=PE1

After power is cut, the energy propagating through T reaches L after a delay and we have, for the final state j:
SEj<SE1, TEj=TE1, LEj>LE1, PEj=PE1
 
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Whats the fallacy with my reasoning?

Whats the fallacy with my reasoning?

E = mc ^2 c= the speed of light


E = 1/2 [ (Q^2)/C ] C= capacitance


mc ^2 = 1/2 [ (Q^2)/C ]

weight gain in charged capacitor
m = {1/2 [ (Q^2)/C ]} / {c^2}
speed of light = 299 792 458 m /s
 
Snowjob said:
E = mc ^2 c= the speed of light


E = 1/2 [ (Q^2)/C ] C= capacitance


mc ^2 = 1/2 [ (Q^2)/C ]

weight gain in charged capacitor
m = {1/2 [ (Q^2)/C ]} / {c^2}
speed of light = 299 792 458 m /s
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That is the relativistic mass gain for the ideal charge scenario, not the weight gain. Multiply by the acceleration of gravity and you will have the weight.
 
mivey said:
The potential energy of S-E is dropping because it is losing energy. The potential energy of L-E is increasing because it is absorbing energy. When the transfer is taking place, the potential energy of T-E is higher.
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alright.

mivey said:
So during transfer, at any state i:
SEi<SE1, TEi>TE1, LEi>LE1, PEi=PE1
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agree.

mivey said:
After power is cut, the energy propagating through T reaches L after a delay and we have, for the final state j:
SEj<SE1, TEj=TE1, LEj>LE1, PEj=PE1
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I have a doubt on this part, specially on the TEj=TE1 expression.
Even in an ideal transformer, you will need energy to magnetize the core right? assuming that when you cut the power, consequently you will cut the load, since the potential energy has dropped to zero on system S. Then with no load to feed, the T system is on a stationary state again, but different to TE1. Since it has been magnetize previously.

Then, from your explanation I understand that the potential energy (rest-mass?) of the system T at state j and state 1 are the same. Right?? But I find it difficult to see. Or is it that at the time of state j, the tranformer has been demagnetize properly by some other system (E)?? If it so, then that system has gained some energy right?? what happened there?? what system gained the energy of the magnetization of the transformer??
 
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Mayimbe said:
I have a doubt on this part, specially on the TEj=TE1 expression.
Even in an ideal transformer, you will need energy to magnetize the core right?
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Not an ideal transformer.
Mayimbe said:
assuming that when you cut the power, consequently you will cut the load, since the potential energy has dropped to zero on system S. Then with no load to feed, the T system is on a stationary state again, but different to TE1. Since it has been magnetize previously.
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Remember, one premise was that all energy entering the S-T interface would have to eventually leave through the L-T interface. But, if you simultaneously cut the load & source, you will trap the energy due to propagation delay in T and system T will have a net energy increase.
Mayimbe said:
Then, from your explanation I understand that the potential energy (rest-mass?) of the system T at state j and state 1 are the same. Right?? But I find it difficult to see. Or is it that at the time of state j, the tranformer has been demagnetize properly by some other system (E)?? If it so, then that system has gained some energy right?? what happened there?? what system gained the energy of the magnetization of the transformer??
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The last one connected gets the excess T energy (I assumed the load).

Consider the lossless transformer. Its natural equilibrium point is TE1. We know this because we said it was lossless. There is no new equilibrium point available.

If a load is connected, the excess T energy will be absorbed by the load because T wants to reach equilibrium. The load is an energy sink and has no real equilibrium point and will take whatever energy it gets.

If the load is gone, then T gets caught in a non-equilibrium state, much like a charged capacitor, and the magnetization energy stays in system T.

These are ideal scenarios. In a real scenario, most of it would probably be converted to heat and radiate away.
 
__dan said:
One previous poster gave the example, rub a ballon in your hair and it picks up an excess static charge, additional electrons above ground state, and the added weight of more electrons.
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But, since the electrons came from your hair, and you're holding the balloon, doesn't your overall weight remain the same?
 
mivey said:
Not an ideal transformer.
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I have always thought that an ideal transformer was the same that a real tranformer, but without loses. But you are right.

mivey said:
If the load is gone, then T gets caught in a non-equilibrium state, much like a charged capacitor, and the magnetization energy stays in system T.
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Does that mean that system T will gain mass in this scenario?

mivey said:
These are ideal scenarios. In a real scenario, most of it would probably be converted to heat and radiate away.
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Isnt the hysteresis phenomenom involved??? I dont believe that is converted to heat.
 
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