Potential Energy Explained: Does It Contradict Conservation of Mass/Energy?

[PBRMEASAP]

So potential energy really is the energy stored in the field acting between the parts,...?

That's right.

 

[El Hombre Invisible]

PBRMEASAP... you're my kind of acronym.

 

[michael879]

Your question is a good one. I'll attempt an answer for you.

If you take a distant mass that has 0 gravitational potential energy and 0 kinetic energy with respect to the Earth or sun, its energy is its relativistic mass x c^2 relative to the Earth frame (at this point its potential energy is 0). So:

actually, the farther something away from us is the greater the Potential Energy. PE = mgh

 

[michael879]

Bang on again. It's always good when someone cleverer than myself tells me something that I tell myself I always knew really. So potential energy really is the energy stored in the field acting between the parts, or in QT the bosons knocking around at the time... energy the parts WILL get, just not yet..?

thats exactly what I said in my first post! PE isn't real energy, its just the promise of future energy. This means its basically a BS concept that people use to prove conservation of energy. It doesn't actually exist.

 

[Gokul43201]

No michael, that's just not right. Potential energy is more than just a promise...it is very real.

It is a "feeling" that comes from having a force acting on you. Everywhere, things have potential energies from various kinds of interactions. A dipole in an electric field has a potential energy of -p.E; a magnetic moment, mu, in a magnetic field has a potential energy -mu.H; a mass in a gravitational potential, $\phi$, has a potential energy $m\phi$.

From binding energies of subatomic particles to the energy that is gained when a spring is compressed to the thing that causes stars and planets to form is nothing but one form or another of potential energy.

 

[learningphysics]

El Hombre:

You know more than I do, but I thought the idea of "relativistic mass" was not a particularly useful concept. In any case, I realize that as far as the relativity aspect went, I was abrupt with the OP and should have explained things better, but I'm shaky with the stuff myself. I'd best be keeping my mouth shut. In any case, I do know that we have rest energy ("mass energy" as you called it) given by:

$$E_0 = m_0 c^2$$

and as I understand it, rest energy + kinetic energy (for an object in motion) is given by:

$$E = \frac{m_0 c^2}{\sqrt{1 - \frac{v^2}{c^2}}}$$

What I'm not clear on are the nuances/interpretations of this result. For instance, I understand that this formula for "relativistic energy" clearly shows that there is an upper limit to the velocity of an object having mass, that it can never attain c. If nothing else, infinite energy would be required. But is it really correct to say that the object's "mass" in some sense, increases as it speeds up? As you pointed out, you have to clear on what you mean by the term "mass". Rest mass does not change

Hi cepheid. I believe that in relativity, potential energy is rest mass. Einstein's $$E=mc^2$$ paper gives a nice demonstration using the case of an elastic collision between two balls. When the two particles collide the rest mass (of the system?) must increase. In other words with the $$E_0 = m_0 c^2$$, $$m_0$$ must increase.

I'm not sure about if the increase in rest mass goes into the particles themselves or into the fields between the two particles...

But if the two particles collided together... compressed and locked together in the compressed state... then if you measured the total rest mass (using a super sensitive weighing machine)... it would weigh more than the sum of the rest masses of the two separate particles.

Please correct me if I'm wrong.

 

[learningphysics]

I never really thought about this until the other day when someone asked me why something accelerating towards a planet doesn't make the planet lose mass. It makes sense according to conservation of mass/energy right? the object gains energy (speed) therefore the planet must lose energy or mass.

But the system does lose "rest mass" according to relativity.

Potential energy exists in the form of "rest mass".

 

[Cyrus]

Please help me out here, but I was under the impression that potential energy is tied up in the rest mass term. Because E=K+mc^2, but nowhere is the potential in the equation, which implies it has to be included in mc^2. As an example: Let's say I have two blocks of ice of equal mass and frozen liquid. They are in a PERFECT insulating box, on EXACT scales, ok maybe not exact due to Heisenburg, but there good scales
:-). Now let's say I melt the ice of one block on one scale, and compare its weight of liquid, to the weight of the still froze ice block on the scale next to it. Would we find that the mass of the now defrosted ice block is MORE than the frozen block, OR would the energy go into increasing the kinetic energy term and NOT the mc^2 term. My instinct would be to say that the kinetic energy is the one that changes, and not the mass.

But for something as simple as raising a mass in a gravity field, you are increasing the potential energy, and since E=K+mc^2, then potential has to be inside mc^2, so you should detect a mass change no?

I think one person said that you neglect the mc^2 term because its so small, but that's not right, don't you neglect it because its so BIG, but occurs on both sides of the equality, so it in effect cancels out on both sides?

Sorry If I am asking the same question over again. Feel free to cane me.

 

[Andrew Mason]

actually, the farther something away from us is the greater the Potential Energy. PE = mgh

That is what I said. It is 0 at infinity. It decreases from that as you get closer. Gravitational PE is never positive.

AM

 

[Andrew Mason]

Hi cepheid. I believe that in relativity, potential energy is rest mass. Einstein's $$E=mc^2$$ paper gives a nice demonstration using the case of an elastic collision between two balls. When the two particles collide the rest mass (of the system?) must increase. In other words with the $$E_0 = m_0 c^2$$, $$m_0$$ must increase.

I'm not sure about if the increase in rest mass goes into the particles themselves or into the fields between the two particles...

But if the two particles collided together... compressed and locked together in the compressed state... then if you measured the total rest mass (using a super sensitive weighing machine)... it would weigh more than the sum of the rest masses of the two separate particles.

Please correct me if I'm wrong.

Actually in his E=mc^2 paper Einstein used the example of a body emitting two photons at the same time, each in opposite directions. He demonstrated that if a body gives off energy E (in the form of radiation) its mass diminishes by E/c^2. When the radiation is absorbed by another body, that body increases its mass by E/c^2. Thus a transfer of energy transfers inertia between bodies.

AM

 

[learningphysics]

Actually in his E=mc^2 paper Einstein used the example of a body emitting two photons at the same time, each in opposite directions. He demonstrated that if a body gives off energy E (in the form of radiation) its mass diminishes by E/c^2. When the radiation is absorbed by another body, that body increases its mass by E/c^2. Thus a transfer of energy transfers inertia between bodies.

AM

Hi Andrew. Sorry, I was referring to the wrong paper. This is the one that I meant:
http://www.ams.org/bull/2000-37-01/S0273-0979-99-00805-8/S0273-0979-99-00805-8.pdf

It was written much later... in 1935.

Einstein demonstrates that an inelastic collision results in a change in rest mass.

 

[El Hombre Invisible]

By the time the two bodies collide surely all potential energy that CAN be converted to kinetic energy HAS been converted to kinetic energy - i.e. by being at its minimum it is effectively 0 (unconventionally measuring potential energy as a positive quantity). If the momenta of the bodies are equal and opposite, then to conserve momentum the two bodies would stop. What you'd be counting as rest mass here is kinetic energy surely, not potential energy. Potential energy kind of implies distance, since it is the energy stored/passed between two bodies. Is that right?

 

[Andrew Mason]

By the time the two bodies collide surely all potential energy that CAN be converted to kinetic energy HAS been converted to kinetic energy - i.e. by being at its minimum it is effectively 0 (unconventionally measuring potential energy as a positive quantity). If the momenta of the bodies are equal and opposite, then to conserve momentum the two bodies would stop. What you'd be counting as rest mass here is kinetic energy surely, not potential energy. Potential energy kind of implies distance, since it is the energy stored/passed between two bodies. Is that right?

There seem to be several concepts here that are being confused. Why would a body have to have converted all of its potential energy (with respect to what, exactly?) before colliding with another body?

Two colliding bodies do not have to stop in order to conserve momentum. Elastic collisions result in the two bodies separating at the same speed as they approached each other before the collision.

AM

 

[El Hombre Invisible]

You're over-complicating things. Ball, planet. Ball is held above ground, is released. Potential energy is converted to kinetic energy until the ball hits the ground. At this point that the two are in contact, i.e. at the ball's lowest height, the potential energy is at its minimum and kinetic energy is at its maximum. The question was whether or not the change in potential energy alters the rest mass of the object. I was just saying that it is the change in KINETIC energy that alters the relative mass, and nothing alters the rest mass.

 

[learningphysics]

You're over-complicating things. Ball, planet. Ball is held above ground, is released. Potential energy is converted to kinetic energy until the ball hits the ground. At this point that the two are in contact, i.e. at the ball's lowest height, the potential energy is at its minimum and kinetic energy is at its maximum. The question was whether or not the change in potential energy alters the rest mass of the object. I was just saying that it is the change in KINETIC energy that alters the relative mass, and nothing alters the rest mass.

Kinetic energy is a maximum just before contact... then the ball compresses. Before it bounces back up it's velocity drops to zero. Therefore kinetic energy becomes 0 at this point. So potential energy is a maximum.

You're only considering GPE... there are other potential energies involved also.

Rest mass is altered. As cyrusabdollahi said... the change in potential energy corresponds to a change in rest mass. There is only rest energy and kinetic energy. If kinetic energy drops, rest energy increases (for conservation of energy)

 

[zoobyshoe]

I never really thought about this until the other day when someone asked me why something accelerating towards a planet doesn't make the planet lose mass.

From the little physics I know, I've gotten the impression that gravity is a two way street. The planet is "pulling" the ball, but the ball is also "pulling" the planet. This strikes me as the reason the planet loses nothing in the bargain.

 

[El Hombre Invisible]

There is only rest energy and kinetic energy. If kinetic energy drops, rest energy increases (for conservation of energy)

Okay, well this contradicts an earlier post by someone else so you can understand my confusion. I was under the impression that rest mass, kinetic energy and potential energy were three different things, with potential being in the space in which the force interactions were occurring between the bodies taking part in them. As potential energy drops, kinetic energy increases and vice versa (also for CoE) but with rest mass staying the same. That picture made a lot more sense to me and seemed more consistent with everything else I know. CHEMICAL energy altering rest mass, sure. Potential? That's the first I've heard about it.

 

[michael879]

That is what I said. It is 0 at infinity. It decreases from that as you get closer. Gravitational PE is never positive.

AM

no that's not right, PE is 0 at the center of the Earth (PE due to Earth)

 

[whozum]

PE is 0 at infinity and at the center.

 

[Janus]

no that's not right, PE is 0 at the center of the Earth (PE due to Earth)

No, Andrew is right. GPE is zero at infinity and goes negative as you approach the Earth. That is because it is expressed as
$$GPE =- \frac{GMm}{r}$$
( Note: G = the gravitational constant and is not g which is the acceleration due to gravity.)

 

[learningphysics]

Clarify potential energy - rest mass relation

Perhaps someone can give a short explanation of how potential energy and rest mass energy are connected.

I was under the impression that the energy stored in the fields due to any of the 4 fundamental forces results in a change in rest mass. I'm no expert... Perhaps gravity is excluded?

Can someone clarify? Thanks.

 

[Gokul43201]

PE is 0 at infinity and at the center.

Spot the mistake :

$$\Phi_{center} - \Phi_{\infty} = \int_{\infty}^0 \vec{E} \cdot \vec{dr} = \int_{\infty}^R \vec{E} \cdot \vec{dr} + \int_R^0 \vec{E} \cdot \vec{dr} = \frac{-GM}{R} + \int_R^0 \frac{G\rho}{r^2}~\frac{4}{3}\pi r^3~dr = \frac{-GM}{R} - \frac{GM}{2R} =\frac{-3GM}{2R} \not{=} 0$$

 

[Andrew Mason]

Spot the mistake :

$$\Phi_{center} - \Phi_{\infty} = \int_{\infty}^0 \vec{E} \cdot \vec{dr} = \int_{\infty}^R \vec{E} \cdot \vec{dr} + \int_R^0 \vec{E} \cdot \vec{dr} = \frac{-GM}{R} + \int_R^0 \frac{G\rho}{r^2}~\frac{4}{3}\pi r^3~dr = \frac{-GM}{R} - \frac{GM}{2R} =\frac{-3GM}{2R} \not{=} 0$$

Well, you are assuming constant density, which is not really correct. But other than that, the answer appears to be correct. It is certainly not 0.

AM

 

[Cyrus]

Spot the mistake :

$$\Phi_{center} - \Phi_{\infty} = \int_{\infty}^0 \vec{E} \cdot \vec{dr} = \int_{\infty}^R \vec{E} \cdot \vec{dr} + \int_R^0 \vec{E} \cdot \vec{dr} = \frac{-GM}{R} + \int_R^0 \frac{G\rho}{r^2}~\frac{4}{3}\pi r^3~dr = \frac{-GM}{R} - \frac{GM}{2R} =\frac{-3GM}{2R} \not{=} 0$$

Im sorry, but F dot dr is equal to the WORK done by a gravitational field. The gravitational potential energy is given by:

$$U= - \frac{Gm_em}{r}$$

Thus as R--> infinity, U goes to zero.

P.S. I don't understand why you have E*dr, which is typically used for electric flux, and mixed it in with the gravitational formula.

 

[learningphysics]

Im sorry, but F dot dr is equal to the WORK done by a gravitational field. The gravitational potential energy is given by:

$$U= - \frac{Gm_em}{r}$$

Thus as R--> infinity, U goes to zero.

P.S. I don't understand why you have E*dr, which is typically used for electric flux, and mixed it in with the gravitational formula.

Hi cyrusabdollahi. Gokul was using -E for the gravitational field... it is analogous to the electric field.

$$\Phi_{b} - \Phi_{a} = -\int_a^b \vec{E} \cdot \vec{dr}$$

Where $$E=\frac{-GM}{r^2}$$

Integrating force gives the work... Integrating the gravitational field gives the potential.

You can get the gravitational potential energy that a mass has by multiplying the gravitational potential at the point times the mass. Just as you can get the gravitational force acting on a mass by multiplying the gravitational field at that point times the mass.

Just like with electrostatics.

 

[Cyrus]

I do not follow you learningphysics, the integral for gravity also gives you work. It gives you U_2-U1= Work total. So solving the integral with the limits gives you the change in potential energy, which is the work done. If you want the potential at infinity, then its just going to be the value of U_2 alone, not U_2 - U_1.

Edit: I searched the hyperphysics website, I see what your saying now, but then that would imply that the potential at infinity is the integral from a to b, or infinity to infinity, which is zero by definition, because it is defined to be zero at infinity. so the limits a and b would both be infinty.

 

[learningphysics]

I do not follow you learningphysics, the integral for gravity also gives you work. It gives you U_2-U1= Work total. So solving the integral with the limits gives you the change in potential energy, which is the work done. If you want the potential at infinity, then its just going to be the value of U_2 alone, not U_2 - U_1.

Hi cyrusabdollahi. Potential and potential energy are different:

$$\phi=\frac{-GM}{r}$$

gives potential due to mass M(which you get from integrating the field not the force).

$$U=\frac{-GMm}{r}$$ gives potential energy (which you get by integrating the force).

 

[Cyrus]

Yeah, I mixed the two up, they really shouldent name them so similary, arg! Someone ought to fix that nomenclature.

 

[whozum]

Isnt potential defined to be 0 at infinity?

if $$GPE = \frac{-Gmm}{r}$$

Then as r tends to infinity GPE tends to zero.. isn't it also defined this way?

 

[learningphysics]

Edit: I searched the hyperphysics website, I see what your saying now, but then that would imply that the potential at infinity is the integral from a to b, or infinity to infinity, which is zero by definition, because it is defined to be zero at infinity. so the limits a and b would both be infinty.

Hmmm.. not sure exactly what you're saying here.

The potential is defined as 0 at infinity. Or another way to say it is that... the potential at any point b is the given integral with a=infinity.

 

[Cyrus]

Right, so the question is the potential at infinity. So for the limits of integration, as you stated, would automatically make a= infinity. And since we are questioning the potential at infinity, the limit b would be infinity as well. Which would make the limits of integration therefore the same. Although, I am probably not doing something right.

 

[whozum]

The integral gives you a change in potential, not the absolute potential at a point. The change in potential from infinity to infinity is zero.

 

[learningphysics]

Right, so the question is the potential at infinity. So for the limits of integration, as you stated, would automatically make a= infinity. And since we are questioning the potential at infinity, the limit b would be infinity as well. Which would make the limits of integration therefore the same. Although, I am probably not doing something right.

Yes, that's right. Potential at b = infinity is 0 (since the limits are the same).

 

[Cyrus]

Well, if you state that this is the potential,$$\phi=\frac{-GM}{r}$$
, then it still implies that at R--> infinity, the potential also goes to zero as well.

 

[Cyrus]

So how come that Gokul got a nonzero value of potential at R--> infinity?