# B Negative potential energy and negative mass

1. Jun 13, 2017

### 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)

2. Jun 13, 2017

### CWatters

Why do you say gravitational potential energy is normally negative?

3. Jun 13, 2017

### Vicara

Because Ep = -G•M•m/r

4. Jun 13, 2017

### 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.

5. Jun 13, 2017

### Staff: Mentor

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.

6. Jun 13, 2017

### 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

7. Jun 13, 2017

### Vicara

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

8. Jun 13, 2017

### Staff: Mentor

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

9. Jun 13, 2017

### 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?

10. Jun 13, 2017

### PAllen

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.

Last edited: Jun 13, 2017
11. Jun 13, 2017

### DrStupid

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.

12. Jun 13, 2017

### Staff: Mentor

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.

13. Jun 13, 2017

### DrStupid

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

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.

14. Jun 13, 2017

### Vicara

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...

15. Jun 13, 2017

### PAllen

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.

16. Jun 13, 2017

### Staff: Mentor

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/

17. Jun 14, 2017

### Vicara

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?

18. Jun 14, 2017

### Matter_Matters

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.

19. Jun 14, 2017

### Ibix

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.

20. Jun 14, 2017

### Vicara

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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted