Negative potential energy and negative mass

[Vicara]

My question is:
If gravitational potential energy is normally negative, and E=m•c^2, doesen't that means that negative mass could exist?
(I don't know much about general relativity so please explain as simple as posible)

 

[CWatters]

Why do you say gravitational potential energy is normally negative?

 

[Vicara]

Because Ep = -G•M•m/r

 

[vanhees71]

Usually you choose the potential being negative, i.e., for a spherically symmetric mass $M$ in the origin of a coordinate system the potential of the force on a test mass $m$ is
$$V(\vec{r})=-\frac{G m M}{r},$$
where $G$ is Newton's gravitational constant. Here masses are always positive.

I've, however no clue, what this might have to do with the rest mass-energy equivalence of special relativity. For a relativistic theory of gravitation you neen general relativity.

 

[Nugatory]

doesen't that means that negative mass could exist?

No. What it does mean is that a system consisting of two widely separated objects will have a mass very slightly greater than if the same two objects were close together. The difference in mass (multiplied by $c^2$ according to $E=mc^2$) will be the energy that was needed to move them apart, which is the difference in potential energy between the two configurations.

It's important that we work with the difference between two configurations because potential energy itself is defined that way. It makes no sense to say that the potential is positive or negative until you've specified what you're calling zero. A very common convention is that we call the potential energy zero when the two objects are infinitely distant, and with this convention the potential energy will always be negative.

 

[Vicara]

I think that I'm just mixing concepts that I don't really understand, its just that I learned this formula (before i used Ep = m•g•h ) and the fact that the other formula gives a negative result for energy confused me and I thought that negative mas could be possible by the mass-energy equivalence

 

[Vicara]

No. What it does mean is that a system consisting of two widely separated objects will have a mass very slightly greater than if the same two objects were close together. The difference in mass (multiplied by $c^2$ according to $E=mc^2$) will be the energy that was needed to move them apart, which is the difference in potential energy between the two configurations.

It's important that we work with the difference between two configurations because potential energy itself is defined that way. It makes no sense to say that the potential is positive or negative until you've specified what you're calling zero. A very common convention is that we call the potential energy zero when the two objects are infinitely distant, and with this convention the potential energy will always be negative.

Hmmm, that makes sense...
And in the case of maseless particles like photons?

 

[Nugatory]

And in the case of maseless particles like photons?

Sorry, you'll have to be more specific about what you're asking.

 

[Vicara]

I mean, if a photon gets closer to a mass, the potential energy will decrease, hence, his mass will decrease too. But it don't have mass, so what will hapend?

 

[PAllen]

I mean, if a photon gets closer to a mass, the potential energy will decrease, hence, his mass will decrease too. But it don't have mass, so what will hapend?

E = mc² is a special case (for a massive body at rest) of the general relation:

E² = p²c² + m²c⁴

For light, m is zero, and energy is related solely to momentum

Also, note that as something moves from higher to lower potential with no other forces acting, its kinetic energy increases. Since the energy of light is pure kinetic energy, this means the locally measured energy of light increases as it approaches a massive body, i.e. it blue shifts.

 

[DrStupid]

What it does mean is that a system consisting of two widely separated objects will have a mass very slightly greater than if the same two objects were close together. The difference in mass (multiplied by $c^2$ according to $E=mc^2$) will be the energy that was needed to move them apart, which is the difference in potential energy between the two configurations.

I do not see how this answers the question. Why is the energy that is needed to move the objects apart always below the total energy of the separated object? As the classical gravitational potential would allow a negative total energy for sufficiently small distances, the question can be answered in the scope of general relativity only.

 

[Nugatory]

I do not see how this answers the question. Why is the energy that is needed to move the objects apart always below the total energy of the separated object?

No matter how far apart I move the objects it will always require the addition of more energy to move them farther apart; so the potential energy at infinity is always greater than the energy at any finite distance, no matter how large. Thus, if you're going to make the (arbitrary, but very convenient) choice to take the energy at infinity to be the zero point, then the potential energy at any finite separation distance must be less than zero. However, we could have taken the zero point of potential to be the minimum when the two objects have approached as closely as possible and compressed, and reshaped themselves under the gravitational forces as they merge. Do this, and the energy will be finite and positive everywhere including in the limit as $r\rightarrow{\infty}$ and whether or not we further offset the zero point by an additional constant $m_1c^2+m_2c^2$ contribution from counting the rest masses.

(And do not fall into the trap of thinking that because $F=-Gm_1m_2/r^2$ blows up as $r\rightarrow{0}$ that the minimum of energy is not a finite number that we can take as the zero point. The infinity here is just telling us that the ideal point particle approximation upon which that formula is based breaks down when $r$ becomes small enough).

This is a digression from the original poster's question, so followup should happen in a new thread in the Classical Physics subforum.

 

[DrStupid]

However, we could have taken the zero point of potential to be the minimum when the two objects have approached as closely as possible and compressed, and reshaped themselves under the gravitational forces as they merge.

The question remains: Why is there a closest possible approach with a positive total energy?

This is a digression from the original poster's question, so followup should happen in a new thread in the Classical Physics subforum.

The original question can't be answered in classical physics because there is no mass-energy equvalence and no lower limit for the gravitational potential.

 

[Vicara]

E = mc² is a special case (for a massive body at rest) of the general relation:

E² = p²c² + m²c⁴

I´ve seen this formula before, but I don´t understand how light could have momentum or kinetic energy, as both of them include mass in their expressions...

 

[PAllen]

I´ve seen this formula before, but I don´t understand how light could have momentum or kinetic energy, as both of them include mass in their expressions...

Well, for starters, there is empirical fact: how would a light sail work if light did not have momentum?

For light, momentum is E/c, not anything involving mass.

In SR, kinetic energy is simply total energy minus rest energy. Since light has no rest energy, all of its energy is kinetic. Note that mass, per se, is not involved in this definition.

Summary: for light p = E/c, and KE = E, with E being the total energy of the light.

 

[Nugatory]

?..as both of them include mass in their expressions...

You are thinking of the classical $p=mv$ and $E_k=mv^2$ or the relativistic $p=\gamma{m_0}v$ and $E_k=(\gamma-1)m_0c^2$ which do indeed include the mass. However, these formulas only apply in the special case of particles with non-zero mass and traveling at less than the speed of light, and clearly that doesn't include photons. Do note, however, that the formula @PAllen quoted is equivalent to the two relativistic ones in the $m_0=0$ case.

We also have a FAQ: https://www.physicsforums.com/threads/how-can-light-have-momentum-if-it-has-zero-mass.512541/

 

[Vicara]

Well, for starters, there is empirical fact: how would a light sail work if light did not have momentum?

For light, momentum is E/c, not anything involving mass.

In SR, kinetic energy is simply total energy minus rest energy. Since light has no rest energy, all of its energy is kinetic. Note that mass, per se, is not involved in this definition.

Summary: for light p = E/c, and KE = E, with E being the total energy of the light.

So, what i thought that light could have negative energy (or mas, by the mass-energy equivalence) is wrong because light don't have potential energy?

 

[Matter_Matters]

Because Ep = -G•M•m/r

You should note that the choice of sign for the potential energy is really down to convention. You could easily define it as the following:
$$V(r) = \frac{GM}{r},$$
such that the force associated with the above potential may be given by the usual
$$F = \nabla V(r),$$
which gives the usual gravitational force produced by some body of mass $M$
$$F = -\frac{GM}{r^2}.$$
In relation to it's connection the rest energy in relativity, I think you are confusing the two theories and want to consult the general theory of relativity for a relativistic description of gravity as the other comments suggest.

 

[Ibix]

So, what i thought that light could have negative energy (or mas, by the mass-energy equivalence) is wrong because light don't have potential energy?

It doesn't really matter whether light can have potential energy or not. You get to choose what zero potential is, so whether something has positive potential or negative is up to you. Something you get to choose can't have physical consequences.

 

[Vicara]

It doesn't really matter whether light can have potential energy or not. You get to choose what zero potential is, so whether something has positive potential or negative is up to you. Something you get to choose can't have physical consequences.

Then the energy (and the mass) depends of the frame of reference?

 

[Vicara]

In relation to it's connection the rest energy in relativity, I think you are confusing the two theories and want to consult the general theory of relativity for a relativistic description of gravity as the other comments suggest.

Well, I think my math and physics knowledge isn't enough to understand such advanced theories hahahaha

 

[vanhees71]

I´ve seen this formula before, but I don´t understand how light could have momentum or kinetic energy, as both of them include mass in their expressions...

No, they don't. This comes from an utterly flawed picture of photons. There's no other way to adequately describe what a photon is than relativistic quantum field theory, particularly QED in this case.

It's way better to think in terms of classical electrodynamics and electromagnetic fields first, because without some acquaintance with the classical theory, there's no chance to understand QFT. For the classical electromagnetic field (in Heaviside-Lorentz units) with electric and magnetic components $(\vec{E},\vec{B})$ the energy density is given by
$$\epsilon=\frac{1}{2}(\vec{E}^2+\vec{B}^2)$$
and the three-momentum density is
$$\vec{\Pi}=\frac{1}{c} \vec{E} \times \vec{B}.$$

 

[Ibix]

Then the energy (and the mass) depends of the frame of reference?

Energy does. Whether mass does or not depends on what you mean by mass. The modern relativistic definition is the modulus of the energy-momentum vector, also known as the rest mass, and this is invariant. Some people do talk about relativistic mass, which is frame dependent, but the concept is falling out of fashion as it causes lots of confusion.

 

[Vicara]

No, they don't. This comes from an utterly flawed picture of photons. There's no other way to adequately describe what a photon is than relativistic quantum field theory, particularly QED in this case.

It's way better to think in terms of classical electrodynamics and electromagnetic fields first, because without some acquaintance with the classical theory, there's no chance to understand QFT. For the classical electromagnetic field (in Heaviside-Lorentz units) with electric and magnetic components $(\vec{E},\vec{B})$ the energy density is given by
$$\epsilon=\frac{1}{2}(\vec{E}^2+\vec{B}^2)$$
and the three-momentum density is
$$\vec{\Pi}=\frac{1}{c} \vec{E} \times \vec{B}.$$

I need to study more physics to understand those beautiful formulas hahahaha

 

[Vicara]

Okey, I will put an example of what I first asked to understand the problem better:
We have two objects, M and m. M have a mass of 10^30 kg and m 10^-6 kg. The distance is 66 m.

The potential energi of m is:
Ep = -G•(10^30)•(10^-6)/66
=-10^14 J

Then, the total energy of m is (I gess):
E = m•c^2+Ep (as kinetic energy is 0)
E = (10^-6)•c^2-10^14 = - 9•10^14 J

 

[Nugatory]

Then, the total energy of m is (I gess):
E = m•c^2+Ep (as kinetic energy is 0)
E = (10^-6)•c^2-10^14 = - 9•10^14 J

The potential energy is a property of the entire system consisting of both masses, so it doesn't make sense to add it to the rest energy of one mass and call that sum the "total energy" of anything. It does make sense to talk about the total energy of the entire system, and it does make sense to call the sum of the rest and kinetic energies of an object its total energy. However, you cannot get the total energy of the system just by adding the total energies of each mass that makes up the system.

 

[Vicara]

The potential energy is a property of the entire system consisting of both masses, so it doesn't make sense to add it to the rest energy of one mass and call that sum the "total energy" of anything. It does make sense to talk about the total energy of the entire system, and it does make sense to call the sum of the rest and kinetic energies of an object its total energy. However, you cannot get the total energy of the system just by adding the total energies of each mass that makes up the system.

Okey, now I see why it doesn't makes sense. I need to study harder hahahaha
Thanks to all the people that helped me :D

 

[DrStupid]

Then the energy (and the mass) depends of the frame of reference?

Energy is frame-dependent but rest mass (that's how the term "mass" is used today) is not.

 

[DrStupid]

We have two objects, M and m. M have a mass of 10^30 kg and m 10^-6 kg. The distance is 66 m.

The potential energi of m is:
Ep = -G•(10^30)•(10^-6)/66
=-10^14 J

Then, the total energy of m is (I gess):
E = m•c^2+Ep (as kinetic energy is 0)
E = (10^-6)•c^2-10^14 = - 9•10^14 J

As Nugatory already explained, Ep it is not the potential energy of m but of the system consisting of m and M. The total energy of this system is

$E = \left( {m + M} \right) \cdot c^2 - G \cdot \frac{{m \cdot M}}{r}$

As it decreases when the objects get closer, your idea sounds reasonable so far. However, to get a negative total energy the objects would need to get so close that the Newtonian gravitational potential is no longer an acceptable approximation. It needs to be replaced by a relativistic potential energy V (which doesn’t need to be limited to gravity). If I understand correctly then you are asking whether

$E = \left( {m + M} \right) \cdot c^2 + V < 0$

is possible in general relativity or not. My intuitive answer is No. I guess that the system will collapse to a black hole (which cannot release energy anymore, exept by Hawking radiatiation) before its total energy drops to zero or even below. But I have no idea how to prove it.

 

[Mister T]

The potential energi of m is:
Ep = -G•(10^30)•(10^-6)/66
=-10^14 J

This is the potential energy of the system, $m$ is the mass of one of the constituents, and $M$ is the mass of the other other constituent.

Then, the total energy of m is (I gess):
E = m•c^2+Ep (as kinetic energy is 0)
E = (10^-6)•c^2-10^14 = - 9•10^14 J

The total energy is the sum of the rest energy and the kinetic energy, so in a frame of reference where the kinetic energy is zero, the total energy is equal to the rest energy. The rest energy of one of the constituents of your system is $mc^2$, the rest energy of the other constituent is $Mc^2$, and the rest energy of the system, as @DrStupid has pointed out, is $$mc^2 + Mc^2 - G \frac{{mM}}{r}.$$

The energy that binds the two constituents together lowers the rest energy of the system. Your mistake was in thinking you have to add the potential energy to the rest energy, but in fact the potential energy is already included in the rest energy term. This is why the total energy is the sum of the kinetic energy and the rest energy.

In another post you mentioned the expression $mgh$. We often hear this cryptically referred to as the potential energy of $m$. Actually, it's the potential energy of a system, and one of the constituents of that system is some object of mass $m$. So, for example, $m$ is the mass of a book and $mgh$ is the potential energy of a system consisting of the book and planet Earth, with $h$ being the height above some arbitrarily chosen height of zero. If you lift the book from the floor to a shelf that's a height $h$ above the floor you've increased the potential energy of the system by $mgh$. This is true whether you choose your zero of height to be the floor or the shelf or the ceiling.

 

[Vicara]

So, for example, $m$ is the mass of a book and $mgh$ is the potential energy of a system consisting of the book and planet Earth, with $h$ being the height above some arbitrarily chosen height of zero. If you lift the book from the floor to a shelf that's a height $h$ above the floor you've increased the potential energy of the system by $mgh$. This is true whether you choose your zero of height to be the floor or the shelf or the ceiling.

So potential energy is negative only because you are asuming that h = ∞ - r so the distance is negative?

 

[Vicara]

As Nugatory already explained, Ep it is not the potential energy of m but of the system consisting of m and M. The total energy of this system is

$E = \left( {m + M} \right) \cdot c^2 - G \cdot \frac{{m \cdot M}}{r}$

As it decreases when the objects get closer, your idea sounds reasonable so far. However, to get a negative total energy the objects would need to get so close that the Newtonian gravitational potential is no longer an acceptable approximation. It needs to be replaced by a relativistic potential energy V (which doesn’t need to be limited to gravity). If I understand correctly then you are asking whether

$E = \left( {m + M} \right) \cdot c^2 + V < 0$

is possible in general relativity or not. My intuitive answer is No. I guess that the system will collapse to a black hole (which cannot release energy anymore, exept by Hawking radiatiation) before its total energy drops to zero or even below. But I have no idea how to prove it.

My original question was referred to a Newtonian system, except the energy-mass equivalence (simply because I haven't studied GR hahaha) but that's a very interesting point of view

 

[Mister T]

So potential energy is negative only because you are asuming that h = ∞ - r so the distance is negative?

Your equation makes no sense. Potential energy, like anything else, is negative only because it's less than zero. The thing that's confusing you is the situation where the choice of zero potential energy was made in such a way that the potential energy is always less than zero. Make a different choice for where it's zero and then it doesn't have to be always negative.

 

[PAllen]

As Nugatory already explained, Ep it is not the potential energy of m but of the system consisting of m and M. The total energy of this system is

$E = \left( {m + M} \right) \cdot c^2 - G \cdot \frac{{m \cdot M}}{r}$

As it decreases when the objects get closer, your idea sounds reasonable so far. However, to get a negative total energy the objects would need to get so close that the Newtonian gravitational potential is no longer an acceptable approximation. It needs to be replaced by a relativistic potential energy V (which doesn’t need to be limited to gravity). If I understand correctly then you are asking whether

$E = \left( {m + M} \right) \cdot c^2 + V < 0$

is possible in general relativity or not. My intuitive answer is No. I guess that the system will collapse to a black hole (which cannot release energy anymore, exept by Hawking radiatiation) before its total energy drops to zero or even below. But I have no idea how to prove it.

I think a way to make this (nearly) rigorous in GR and get away from any issues of the conventionality of potential energy (as well as the fact that potential energy is not generally definable in GR, certainly not in this case for an initial condition that is dynamically unstable thus cannot be consistently made part of a stationary spacetime) is conceptually as follows. Each step is non-trivial, but within the state of the art:

1) Define a series of asymptotically flat spacetimes evolved from series of initial cauchy surfaces as follows: each cauchy 'melds' 2 SC geometries of chosen mass parameters, such that they are some initially stationery distance apart per some plausible harmonic coordinate system (to use ADM evolution). This is highly non-trivial, but is what is done e.g. when the final inspiral stages of BH are simulated from an arbitrary starting point. The series uses smaller and smaller distances. Note that there is no really consistent past continuation leading to such a state, but the BH inspiral experts work around this (it is easy to reason that any prior state includes gravitational radiation that we are not including in our initial cauchy surface).

2) Compute the ADM mass of each such (incomplete - unspecified before the initial cauchy surface) manifold. For given chosen mass parameters, this should decrease as the initial distance decreases. This ADM mass is physically the total gravitational mass of the system measured at spatial infinity. It will include the effects, per full GR, of the mass discount from binding energy.

I am not aware of anyone actually doing such computations, but agree the likely outcome is that you would be dealing with infinitesimally separated apparent horizons without total ADM mass reaching zero.

Note, that extant BH inspiral calculations that I know of, are not relevant because they simply follow one history from some initial state, and the GW emitted includes only some of the initial potential energy. Part of it becomes kinetic energy and and angualar momentum 'captured' by the final BH.

This would be pretty close to a proof that you can never get negative total energy (when it is definable) by virtue of objects being very close together, in the sense that a mixture of Newtonian potential and SR might seem to imply.

 

[PeterDonis]

If I understand correctly then you are asking whether

$$
E = \left( {m + M} \right) \cdot c^2 + V < 0
$$

is possible in general relativity or not. My intuitive answer is No

A better answer would be that the quantity $E$ is not always well-defined; that is, we can't always define a "total energy" for a system in GR. In particular, whenever there is more than one gravitating mass present, strictly speaking, the spacetime is not stationary, and in a non-stationary spacetime there is no way to define a potential $V$.

In practice, for cases where gravity is weak everywhere and everything is moving slowly compared to the speed of light, the expression you give for $E$ is a reasonable approximation if we define $V = - GmM / r$. But note that for that case, we are no longer free to redefine the "zero point" of $V$, because $E$ is now the total energy of the system as measured from the outside, and that measurement will have a definite result; we can't just arbitrarily adjust it by changing the zero point of $V$. We have to use the zero point of $V$ that correctly predicts what the outside measurement will be, and that zero point is $V \rightarrow 0$ as $r \rightarrow \infty$.

Also, since this approximation is restricted to weak fields and slow motion, it won't work for, e.g., an object falling into a black hole, or two black holes coming together. It is still possible to define a total energy $E$ as measured from the outside for such systems, but it requires more care.