Talk:Potential energy

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I assume electrical potential energy refers mainly to the energy gained by placing charges closer or farther apart than they would naturally be inclined to go? (You could probably state this more generally in terms of fields.) Does it also refer to electromagnetic radiation, e.g. light, or would that be "actual energy" instead of "potential energy"? --Ryguasu 00:19 Nov 21, 2002 (UTC)

1 Gravitational nitpicking
2 Electrical potential energy = electrostatic potential energy = coulomb energy?
3 Magnetic potential energy
4 Does Mass-Energy Equivalence Apply?
5 Cleanup
6 Gravitational potential integral
- 6.1 Method originally presented
- 6.2 A simpler way
7 Revision of gravitational potential energy section
8 Rest Mass Energy

[edit] Gravitational nitpicking

I just added a paragraph pointing out that gravitational potential energy is also imparted to the Earth.

I also commented out this:

[We may also want to link to an explanation of that second term (the gravitational forces created by hollow spherical shells)]

— PhilHibbs | talk 18:23, 6 September 2005 (UTC)

[edit] Electrical potential energy = electrostatic potential energy = coulomb energy?

Are these terms equivalent? If so, they should be redirected and mentioned in the article. -- Kjkolb 09:24, 25 November 2005 (UTC)

[edit] Magnetic potential energy

Should magnetic potential energy be included here? -- Kjkolb 09:57, 27 November 2005 (UTC)

I believe that is covered by electrical potential energy, since electricity and magnetism are two aspects of the same fundamental force. Perhaps the section on electrical potential energy should be modified to say electromagnetic potential energy. Wbrameld 23:22, 25 January 2006 (UTC)

[edit] Does Mass-Energy Equivalence Apply?

Does potential energy have mass? For example, take any two stars in our galaxy, let's say Polaris and Rigel, to pick two at random. The gravitational attraction and physical distance between Polaris and Rigel embody an enormous amount of potential energy. If you solve E=mc^2 for mass:
m = E/c^2
and if you plug in the gravitational potential energy between Polaris and Rigel as E in that equation, you get a significant mass value. But does this mass make any real-world sense? Does the mass of that potential energy have its own gravitational field, and if so, what shape does it take? (And to get really way out there, has this been considered by people working on the problem of 'dark' matter?) Wbrameld 22:55, 25 January 2006 (UTC)

It does apply, but doesn't solve the dark matter problem (in fact, it was one of the original reasons for conjecturing that there was a large amount of unseen matter). Among other things, this is why the nuclei of atoms weigh less than their constituent protons and neutrons would in isolation (they're stuck way down at the bottom of a Strong-force potential well, and the mass difference is equal to the energy needed to pull them up out of that well). Cosmologically, the flatness problem relates to the difference between the gravitational potential energy of the universe and the mass of all objects within it. See that article for more information on the subject. --Christopher Thomas 21:26, 26 April 2006 (UTC)

[edit] Cleanup

This article has been edited in bits by many people, and has ended up as a random collection of facts and approaches. The overall organization needs to be redone, I think. -- SCZenz 20:38, 26 April 2006 (UTC)

[edit] Gravitational potential integral

The removal of this math was reviewed at Wikipedia talk:WikiProject Physics the first time around. User:Enormousdude thought it was in error, and removed it. User:SCZenz thought it was in error, and agreed with the removal. I doublechecked it, and concluded that it was, indeed, in error.

It turns out we were wrong.

After spending an hour or more trying to figure out what the heck they did, it turns out it's valid, but (almost, but not quite) the most confusing possible way of working that math. As a result, I'm leaving it out of the article. By all means replace it with something else that makes the same point less confusingly. The reasoning, as near as I can tell, is below. I'm including an alternate way of working the problem that's much easier to follow. --Christopher Thomas 06:54, 27 April 2006 (UTC)

Yes, I see now. I'm impressed you went through the effort of clarifying the situation. I don't think any information of significance is gained by the more complex approach, but certainly the easier way should be added to the article. -- SCZenz 07:08, 27 April 2006 (UTC)

[edit] Method originally presented

Original equation:

$U_g = \int_{h_0}^h {GmM \over r^2} dr + \int_0^{h_0} {GmM \over h_0^2} {r \over h_0} dr$

What's happening is that they're trying to integrate force (across distance) on a test particle (with mass $m$ ) moved from the centre of a uniform sphere (with mass $M$ ) outwards. They move the test particle from radius 0 to radius $h 0$ (the sphere's radius), integrating force applied by the mass within radius $r$ , then as a separate integral integrated from radius $h 0$ out to the final radius $h$ .

The first term of the equation above represents the integration outside the sphere. I don't think anyone disputed that.

The second term represents integration inside the sphere. This starts off expressed as:

$+ \int_0^{h_0}{\left ( \frac{G m}{r^2} \right ) \left ( \frac{r}{h_0} \right )^3 M} dr$

Cancelling gives:

$+ \int_0^{h_0}{\frac{G m M r}{h_0^3}dr}$

...And then they grouped the terms within this integral strangely.

Integrating gives:

$U_g = G m M \left [ - \left ( \frac{1}{r} \right ) \vert_{h_0}^{h} \right ] + \left ( \frac{G m M}{h_0^3} \right ) \left [ \left ( \frac{1}{2} \right ) r^2 \vert_0^{h_0} \right ]$

$U_g = G m M \left [ - \frac{1}{h} + \frac{1}{h_0} \right ] + \left ( \frac{G m M}{h_0^3} \right ) \left [ \left ( \frac{1}{2} \right ) h_0^2 - 0 \right ]$

$U_g = G m M \left [ \frac{1}{h_0} - \frac{1}{h} \right ] + G m M \left [ \left ( \frac{1}{2} \right ) \frac{1}{h_0} \right ]$

$U_g = G m M \left [ \left ( \frac{3}{2} \right ) \frac{1}{h_0} - \frac{1}{h} \right ]$

...So the equations as-given were correct, just incredibly non-intuitive.

Setting the final distance $h$ as infinity makes $\frac{1}{h}$ vanish, giving:

$U_g(\infty) = G m M \left ( \frac{3}{2} \right ) \frac{1}{h_0}$

Declaring $U_g(\infty)$ to be zero gives:

$U_g(h) = G m M \left [ \left ( \frac{3}{2} \right ) \frac{1}{h_0} - \frac{1}{h} \right ] - G m M \left ( \frac{3}{2} \right ) \frac{1}{h_0}$

$U_g(h) = - G m M \left ( \frac{1}{h} \right )$

As mentioned above, this is pretty close to being the most complicated way of deriving the gravitational potential energy. --Christopher Thomas 06:54, 27 April 2006 (UTC)

[edit] A simpler way

If you want to use a "work to take a particle from point A to point B" approach to get gravitational potential energy, and you know the particle will always be outside the sphere, just integrate outside the sphere, and consider it a point mass with mass $M$ . Here, we're taking a test mass with mass $m$ from radius $r 1$ to radius $r 2$ , as measured from the centre of the sphere:

$U_g(r_2) - U_g(r_1) = \int_{r_1}^{r_2} \frac{G m M}{r^2} dr$

$U_g(r_2) - U_g(r_1) = G m M \left [ - \left ( \frac{1}{r} \right ) \vert_{r_1}^{r_2} \right ]$

$U_g(r_2) - U_g(r_1) = G m M \left ( \frac{1}{r_1} - \frac{1}{r_2} \right )$

We want a common reference point for potential energy. We could set it to the radius of the sphere (making potential energy relative to the surface), but that isn't very good as a common reference point, because our spheres may vary in size and density. Setting it at infinity works, though, as making radius tend towards positive infinity doesn't cause our equations to blow up. We can therefore define gravitational potential energy as GPE relative to $U_g(+\infty)$ :

$U_g(r) \equiv U_g(r) - U_g(+\infty)$

$U_g(r) \equiv G m M \left ( \frac{1}{+\infty} - \frac{1}{r} \right )$

$U_g(r) \equiv - \frac{G m M}{r}$

...This gives the usual convention for defining gravitational potential energy using Newtonian gravitation. --Christopher Thomas 06:54, 27 April 2006 (UTC)

[edit] Revision of gravitational potential energy section

This section was a horrible mess, so although I'm not an expert I've attempted a tidy-up on the basis that I could hardly make it any worse. In an ideal world someone might review it and make any necessary corrections. In particular, the original text contained the statement "Note that the potential energy of both objects is the same, so the potential energy of the whole system is 2× $U g$ (as defined above)." It seems to me that this statement is untrue, so I have rewritten that paragraph based on the asumption that $U g$ , as defined earlier, measures the total potential energy of the system.

Matt 00:19, 28 July 2006 (UTC).

[edit] Rest Mass Energy

By the definition given

Potential energy is the of energy that is by virtue of the relative positions (configurations) of the objects within a physical system

How is rest mass an example potential energy? Note that $\nabla E_0 = 0$ , It doesn't depend whatsoever on the position relative to other object. 131.215.7.198 06:09, 2 December 2006 (UTC)

Ahh.. Never mind I figure that out that consider A -> B + C and if we label state by number of A, B and C, then chemical potential(relativistically correct) will be non-relativistic+mc^2131.215.7.198 11:37, 2 December 2006 (UTC)

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