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

[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?

 

[learningphysics]

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.

That's true. But we can define an absolute potential by fixing the lower limit at a specific value.

 

[learningphysics]

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

Gokul never used absolute potential. He calculated the difference between the potential at the center and the potential at infinity.

So he never calculated potential at R--> infinity.

 

[Cyrus]

Ok. So if your saying that he calculated the difference in potential, don't we automatically know that the phi_infinity term, the second term being subtracted is zero. So the only term we have to consider is the potential at the center of the earth. And I thought that its value of potential energy is,: -GmM/R, no matter where you are inside the earth, so the associated potential would be -GM/R, factoring out the mass. But he got a different anwser.

 

[learningphysics]

Ok. So if your saying that he calculated the difference in potential, don't we automatically know that the phi_infinity term, the second term being subtracted is zero.

Right. It doesn't have to be zero. But we can certainly define it as being zero. So if we do define it as zero, the potential at the center is the expression on the right hand side of the equal sign.

So the only term we have to consider is the potential at the center of the earth. And I thought that its value of potential energy is,: -GmM/R, no matter where you are inside the earth, so the associated potential would be -GM/R, factoring out the mass. But he got a different anwser.

When you are inside the earth, you have to be careful. The gravitational field at 0<r<R is only due to the mass inside the radius r. You have to calculate the mass inside a sphere of radius r... and see what the gravitational field due to that mass is. So the mass that's creating the gravitational field at 0<r<R is a function of r... it is the density of the Earth times the volume.
\rho=\frac{M}{\frac{4}{3}\pi R^3}.
m=\frac{4}{3}\pi r^3 \rho

So mass comes out to: \frac{Mr^3}{R^3}

Gravitational field comes out to: -\frac{GMr}{R^3}

Integrate this from r=R to r=0, and we get: -\frac{GM}{2R}

Add this to the integral of the gravitational field from r=infinity to r=R, and we get: -\frac{GM}{2R} + -\frac{GM}{R}

so we get Gokul's answer: -\frac{3GM}{2R}

So if we put a mass m at the center of the earth... and if we take the potential at infinity=0, then the gravitational potential energy would be:
-\frac{3GMm}{2R}

 

[Cyrus]

Thanks for your help. I must say, that certainly clears things up. :-)

 

[learningphysics]

Thanks for your help. I must say, that certainly clears things up. :-)

Cool. Glad to help.

Sleepy time now! :zzz:

 

[michael879]

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.)

oops, I forgot the -

 

[Twukwuw]

the conclusion...

After reading all the posts I now draw some conclusions:

1)When the total energy of an object is changed, its RELATIVISTIC mass must also change.

This is because E = MC^2 for an object, where E is the TOTAL relativistic energy of the object (as observed by an observer), M is the relativistic mass (M = rest mass x gamma), C is the light speed.
(Note that I emphasize that all the terms must be relativistic, because object A may have different amount of energy in reference frames of object B and object C, due to the different relative velocities.)
Since the relativistic mass is a function of m (rest mass) and v (relative velocity), it turns out that the energy level/total energy of an object is also a function of m and v, i.e.
E = f (m,v). Hence, when M is increased, m MAY BE INCREASED but also MAY BE DECREASED, provided there is a greater increase in v when M increases but m decreases!

2)When energy is created, there must be an amount of RELATIVISTIC mass being destroyed, or lost.

For example, during chemical reactions the energy released is related to the lost of RELATIVISTIC mass by ΔE = ΔMC^2. The lost of M can be interpreted this way:
Case 1: The velocity of each chemical particle decreases( averagely ), and M therefore decreses because the relativistic mass of each particle has decreased.
Case 2: The rest masses of chemical particles decrease SOMEHOW!
Case 3: Both the case 1 and case 2 are involved!
With this in mind, we realize that when one does work (which in fact the energy comes from complicated chemical reactions inside his bodies), his RELATIVISTIC mass (with respect to any reference frame) must decrease!
Other than that, when we gain heat energy from a heat reservoir, the relativistic mass of the heat reservoir will also decrease! Why? Because when we extract energy from the heat reservoir, its temperature ( a measure of the average kinetic energy, or average velocity of each particle.) must decrease, hence the relativistic mass must decrease! (What I want to point out here is, when the temperature of an object is changed, its relativistic mass will be also changed.)

3)The potential energy is not a “real” one, it is just created for mathematic convenience.

When we say an object has an extra energy potential energy, we are saying that IN THE FUTURE it will acquire an amount of energy which is the same as the potential difference. And, the energy comes from other resource; here it is the MASS who set up its gravity field, but NOT the gravity field (or the curvature of space-time)!
To give a clearer picture, here I give an “analogy…”
Suppose there is a box in a zero-gravity-field region in the outer space. Inside the box there are 2 identical stone and MOST importantly a person.
It is a rectangular box. Now, stone A is at a higher level, while stone B and the person are at the “bottom” of the box. Both the stones are STATIONARY. First we ask, with respect to the person, are the energies of the 2 stones different? Not! Why? Because they have the SAME REST MASS and the SAME RELATIVE VELOCITY (quickly linked to the situation in the Earth in which 2 identical stones are at different height levels!)
Now, the person say: I will do a work of magnitude ΔW on the stone B such that it will reach stone A and at the same time having a velocity V. When we are informed about the person’s decision, we now say:
BESIDE ITS REST ENERGY, the stone B now also have POTENTIAL ENERGY of ΔW!
Why? Because the person PROMISES that he will give stone B an extra energy in the FUTURE!
So, that is all! That is what we mean by potential energy! It is nothing but merely a term INFORMING us that in the FUTURE an extra energy will be given to stone B!
Now, let’s again have a question, are the masses of the 2 stones different because stone B have a potential energy? NO! They are the same! (Again, quickly linked to the situation in which 2 identical stones are at different height levels, do they have different relativistic masses? You know it! )

Well, these are my opinions and viewpoints. However, I understand that I may be mistaken. Please feel free to write back to me if you find me wrong!
Thank you for your reading!

 

[Janus]

.

I'll just address a couple of quick points here.

This is because E = MC^2 for an object, where E is the TOTAL relativistic energy of the object (as observed by an observer), M is the relativistic mass (M = rest mass x gamma), C is the light speed.
(Note that I emphasize that all the terms must be relativistic, because object A may have different amount of energy in reference frames of object B and object C, due to the different relative velocities.)

No, in the formula
E = mc^2

m represents the rest mass of the object and E the energy equivalence of that rest mass.

The equation that gives the total energy of the object (and the equation from which E=mc² is derived) is

E_t =\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}

The equation that would give the kinetic energy for the object would be

KE =mc^2 \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1 \right)

2)When energy is created, there must be an amount of RELATIVISTIC mass being destroyed, or lost.

'Relativistic mass' and 'energy' are just different terms for the same thing. In fact, the term relativistic mass has fallen into relative disuse among scientists and they use the term energy almost exclusively instead.

 

[learningphysics]

Now, let’s again have a question, are the masses of the 2 stones different because stone B have a potential energy? NO! They are the same! (Again, quickly linked to the situation in which 2 identical stones are at different height levels, do they have different relativistic masses? You know it! )

Well, these are my opinions and viewpoints. However, I understand that I may be mistaken. Please feel free to write back to me if you find me wrong!
Thank you for your reading!

The potential energy is in the person's body. When he does work on the stone, he loses rest mass. The person's body will weigh less than before he did work on the stone.

Note that for a person at rest... relativistic mass and rest mass are the same.

Potential energy in the body is equivalent to rest energy.

 

[catch.yossarian]

What if you look at it from another way...

Saying that something accelerating, and thus speeding it up, towards Earth is increasing its [Kinetic] energy (and thus losing Potential Energy) , and thus something must lose mass to compensate.

Isn't that like saying: "okay, since I'm adding a ton of mass to this object, it will suddenly speed up and fly away from here." ? This doesn't happen, though...

Or maybe I'm not even interpreting this thread. :\

 

[El Hombre Invisible]

Say you had two identical blocks of radioactive uranium and you put one at the bottom of the ocean and one at the top of mount everest. Now as rest mass is converted into energy, would you expect the one with the greater rest mass - the one at the top of the mountain - to yield more energy than the one at the bottom of the ocean?

 

[whozum]

Yaeh I've got to agree with yossarian, I really don't see where this thread is going. In the beginning we were talking about the translation of potential energy to kinetic energy, and now were bringing about relativity and rest masses, but I don't really see their connection to the initial question.

A body's energy due tohaving mass is independent of its energy due to position in a field..

 

[learningphysics]

Yaeh I've got to agree with yossarian, I really don't see where this thread is going. In the beginning we were talking about the translation of potential energy to kinetic energy, and now were bringing about relativity and rest masses, but I don't really see their connection to the initial question.

A body's energy due tohaving mass is independent of its energy due to position in a field..

If 1 kg of ice at 0C melts into water at 0C... the water gains mass.

The molecules in the ice or the fields between the molecules (I'm not sure which) weigh more in the water state. But total rest mass increases and this increase is due to the increase in potential energy.

 

[learningphysics]

Say you had two identical blocks of radioactive uranium and you put one at the bottom of the ocean and one at the top of mount everest. Now as rest mass is converted into energy, would you expect the one with the greater rest mass - the one at the top of the mountain - to yield more energy than the one at the bottom of the ocean?

Where gravity is concerned I'm not sure...I don't know anything about general relativity. I'm not sure whether the one at the top of the mountain has greater rest mass.

If we were in an analogous situation with an electrostatic field rather than a gravitational field... I'd say yes, the one at the further distance yields more energy.

 

[learningphysics]

Perhaps this link will be helpful to explain the connection between rest mass and potential energy:

http://musr.physics.ubc.ca/~jess/p200/emc2/node6.html

Note the example with regards to the ship leaving the earth.

 

[pmb_phy]

Perhaps this link will be helpful to explain the connection between rest mass and potential energy:

http://musr.physics.ubc.ca/~jess/p200/emc2/node6.html

Note the example with regards to the ship leaving the earth.

Here is an example from Nuclear Physics that I worked out as a specific example

http://www.geocities.com/physics_world/sr/nuclear_energy.htm

The mass of a particle depends on the position of the particle in a gravitational field (e.g. is a function of the gravitational potential). This works for gravity but does not work for any other field. Here by "mass" I mean m = p/v , not proper mass (aka "rest mass"). For details you can find the derivation in Einstein's text The Meaning of Relativity. Page 100 as I recall. This is Mach's Principle as Einstein saw in this text.

Pete

 

[pmb_phy]

No, in the formula
E = mc^2

m represents the rest mass of the object and E the energy equivalence of that rest mass.

That depends on how one defines mass and the symbol. Twukwuw quite clearly and explicitly defined E/c^2 as "relativistic mass." When the object is a particle or an isolated body then that is quite true.

'Relativistic mass' and 'energy' are just different terms for the same thing.

That is quite incorrect. They most definitely are not the same thing in different units. In fact they are not even proportional in general. If you want to read more on this see Rindler's intro sr text.

In fact, the term relativistic mass has fallen into relative disuse among scientists and they use the term energy almost exclusively instead.

That's not quite accurate either. For a counter example please see

http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

Seems to me that you're thinking mostly of particle physicists and not physicists in general. A search of the internet and a look at some modern relativity texts confirms this. Even particle physicists are being lazy when they do this. Its simply easier to do. They do the same thing with proper time. For example; if you look up the "lifetime" of a neutron in the particle physics literature you'll see that its about 15 minutes. However this is understood to be the "proper lifetime." But constanly using the adverb "proper" is a pain in the butt when everyone understands what it is. Since particle physicists study only the intrinsic properties of particlces its also a pain in the butt. Particle physicists don't study relativity in all its glory. In general the two cannot be interchanged. Especially for extended non-isolated bodies. In such cases the relation E/c^2 = p/v its incorrect and therefore E is not proportional to rel-mass and therefore it cannot be said that they're the "same thing."

I mentioned all this to Dr. David Morin at Harvard. He's the prof there who wrote that great set of lecture notes. He seems to agree on this and he will be going over it during the summer.

In the past I held that it was simply a matter of definition. However I've changed my mind. Long story and I want a cup of tea right at the moment!

Pete

 

[El Hombre Invisible]

learningphysics, I now see where our paths were crossed. I believe the original question of whether potential energy effected rest mass was about one part of the system within which the interactions were occurring. For instance, when a charge A moves away from an opposite charge B, the rest mass of A remains unchanged, likewise for B. However the mass of the system as a whole, A+B, increases due to an increase in potential energy. This, it seems, is what your link refers to. The way I understand it, the rest mass of a system can be deemed to be the relativistic masses of all of its components in a frame of reference such that the system as a whole is at rest. As relativistic momentum includes those of force-carrying particles, the potential energy of all interactions within the system are included in the rest mass of the system as a whole.

 

[El Hombre Invisible]

And then I read to the bottom: "As a result, we might expect the rest mass of a spaceship to be slightly larger after it leaves the Earth than it was on Earth, simply because it has left the "gravity well'' of the Earth. This is the case! However, the mass change is imperceptibly small in this case. " D'oh! Now I go back to confusion again. Woe is me!