If you have any kind of electrical device higher than the wall socket, say a PC on a desk, and the electricity has to move UP the cable to get to it, is it fighting gravity in even the most undetectable (to us currently (punny?)) sense?
My younger brother, when he was a lad, would crimp the phone cord while I was talking. He was old enough to know that I knew he was kidding. 
Well, an electron has a mass of 9.10938188 × 10-31 kilograms, so there would be a force, but it would be on the order of about 10^-30 Newtons. This would be dominated by the force due to voltage difference, so it is left out to make things easier.
IIRC, I don’t think that quantum gravity is well-understood enough to give a definite answer.
Yes. Let’s try an extreme example.
Say we have a big power station, which develops 3,000 MW, at sea level. It supplies a city at the top of a 5,000 m mountain.
In one second, the power station produces 3 GJ. The mass equivalent of 3 GJ is:
m = E / c[sup]2[/sup]
= 3x10[sup]9[/sup] / (3x10[sup]8[/sup])[sup]2[/sup]
= 3.3x10[sup]-8[/sup] kg.
The energy required to raise that mass 5,000 m is:
E = m g h
= 3.3x10[sup]-8[/sup] x 9.8 x 5,000
= 1.6x10[sup]-3[/sup] J.
So you lose 1.6x10[sup]-3[/sup] / 3x10[sup]9[/sup] = 5.4x10[sup]-11[/sup] percent of your energy by having to push it up 5,000 m.
Similarly, your TV wastes 1.1x10[sup]-14[/sup] percent of its energy if it’s 1 m above the outlet.
That compares pretty favorably with the 5-10% that gets wasted in heating losses on the way to the outlet.
I think Desmostylus has the right approach, all that moves up the cable is energy. The net movement of electrons is zero since the supply is AC.
D0pBOt, make sure you switch off your empty power outlets so the electricity doesn’t leak out.
Er…the outlet isn’t ‘pumping’ electrons one-way into the device, is it?
It’s a circuit, and if there is a force required to move the electrons up the wire, they’ll gain gravitational potential energy and will use it on the way back down into the socket. So it’ll even out. If there is quantum gravity or whatever.
I was talking about DC…I think. Which we dont have in power outlets, but with that argument and Small Clanger’s (better) argument about there being no net movement, we can say that if there is a force of gravity, however small, it’s not going to effect efficiency and whatnot.
I believe you are quoting the rest mass of an electron. For all practical purposes, aren’t electrons taken as being massless for they move at the speed of light?
- Jinx
Um, no.
Er, no. They can move close to the speed of light, if they have enough energy, but they do not move at the speed of light.
Neither are they massless.
If the room was really really cold (brrr), would that slow it down?
:eek:
DC still has the electrons coming back down.
Only one nitpick about Desmostylus’s calculation. You can’t equate the mass flow of electrons with the energy delivery, even in DC.
That’s because while the electricity moves quicly through the lines, the electrons’ drift speed is incredibly slow, usually on the order of thousandths of a meter per second. So it might take months for and individual electron to flow up that mountain, even if it were direct current. And of course, since it’s AC, it’s not going anywhere.
Electrons have mass?!? I didn’t know they were Catholic ! :smack:
He isn’t. His calculations are based entirely on the e=mc[sup]2[/sup] mass-energy equivalence of the energy consumed by the hypothetical circuit he describes. The mass of the electron never enters the picture in any way, shape or form.
I tend to think of electrons in a wire like water in a pipe. It is a pretty good first cut analogy.
If the electrons were running up the wire and being sprayed into the air, like water coming out of a hose, then there would be a tiny effect of gravity. (Analogy, your water pressure is lower when you are on your roof cleaning the gutters than when you use the same hose to wash your car on the ground.)
But what we have is a circuit. For every electron going “uphill”, there is an electron on the other side going “downhill”. So the gravitational effects cancel out. (Analogy, hold the end of the hose right near the spigot. The water pressure is the same even if you lift the middle part of the hose up off the ground. For all the water molecules running uphill, there are water molecules running downhill.)
So the only force we have to overcome in a circuit is the one that counters any motion. In a hose, that would be friction with the side of the hose, and in a wire it is the resistance of wire.
As you said, counting the mass of electrons is a first-cut approximation, which has already been shown to have a net zero effect. What Demostylus is figuring is beyond that, a second-cut approximation, figuring the mass of the energy that’s going up the wire but not returning.
You do have a two wire circuit, i.e. two hot wires, and possibly a third green ground wire?
What goes up must also go down. With A.C. the electrons are just doing a little dance back and forth, up and down.
In any event there is NO net gravitational force on the cable, however miniscule.
keeper0
CurtC
You beat me to it. :smack: