Gravitational potential energy -- Why is it always negative?

[Hamiltonian]

the gravitational potential energy of a body at any point is defined to be negative of the work done by the conservative force(gravity in this case) from bringing it to that point from a given reference point. if the reference point is taken to be at infinity and the potential energy at this point is taken to be zero, shouldn't the gravitational potential energy at a point closer to the source be positive because the gravitational force is attractive and on moving a body away from infinity(toward the source i.e. the distance between the body and source is decreasing) the conservative force(gravity) will do +ve work on it and not -ve and hence the potential energy at this point should hence be +vealso, does potential energy on its own have any significance or its actually change in potential energy that has some physical meaning?

 

[Ibix]

As you wrote, the gravitational force does positive work on the body, giving it kinetic energy. For energy to be conserved, something must lose energy. That something is the gravitational potential energy of the body, so it must decrease.

 

[etotheipi]

You correctly identified the gravitational potential energy of the system as the negative of the work done by the gravitational forces internal to that system to establish the system from infinite separation to the final state. If you hold one body still and bring the other body toward the other body from infinity, you are also correct that the work done by the gravitational force will be positive,$$W = \int_{\infty}^{r} -\frac{Gm_1 m_2}{r'^2} dr' = \left[ \frac{Gm_1 m_2}{r'} \right]_{\infty}^{r} = \frac{Gm_1 m_2}{r}$$Just take the negative of this to find the gravitational potential energy of the system, since $\vec{F}_1= -\nabla_{\vec{r}_1} U_{12}$ $$U = -\frac{Gm_1 m_2}{r}$$

 

[etotheipi]

also, does potential energy on its own have any significance or its actually change in potential energy that has some physical meaning?

As far as I am aware, at least in non-relativistic physics, no, it's only meaningful up to an additive constant. I have no idea if that's different in SR or GR - perhaps someone who knows what they're talking about can weigh in?

(I remember John Baez talked about this using fancier lingo, that energy differences lie in the real numbers but energies lie in an "R-torsor")

 

[Hamiltonian]

You correctly identified the gravitational potential energy of the system as the negative of the work done by the gravitational forces internal to that system to establish the system from infinite separation to the final state. If you hold one body still and bring the other body toward the other body from infinity, you are also correct that the work done by the gravitational force will be positive,$$W = \int_{\infty}^{r} -\frac{Gm_1 m_2}{r'^2} dr' = \left[ \frac{Gm_1 m_2}{r'} \right]_{\infty}^{r} = \frac{Gm_1 m_2}{r}$$Just take the negative of this to find the gravitational potential energy of the system, since $\vec{F}_1= -\nabla_{\vec{r}_1} U_{12}$ $$U = -\frac{Gm_1 m_2}{r}$$

if the body moves away from the source the work done by gravity would be negative and hence would the gravitational potential energy be +ve as the gravitational potential energy is negative of the work done by gravity.

 

[Ibix]

As far as I am aware, at least in non-relativistic physics, no, it's only meaningful up to an additive constant. I have no idea if that's different in SR or GR - perhaps someone who knows what they're talking about can weigh in?

The concept is the same although the functional form is slightly different. But differences in potential are still the only physically meaningful thing.

The major difference is that gravitational potential is only definable in spacetimes that have a timelike symmetry ("don't change with time", roughly speaking), so the concept isn't as generally useful as it is in Newtonian physics.

 

[etotheipi]

if the body moves away from the source the work done by gravity would be negative and hence would the gravitational potential energy be +ve as the gravitational potential energy is negative of the work done by gravity.

Yeah, that's right - the gravitational potential energy of the system would increase.

I don't understand $\vec{F}_1= -\nabla_{\vec{r}_1} U_{12}$

Whoops yes, sorry, I should have clarified. A conservative force is derivable from a potential energy, e.g. in one dimension you might have seen $F_x = -\frac{dU}{dx}$. Generalising that to the full vector expression means that the derivative becomes a gradient, $\vec{F} = -\nabla_{\vec{r}_1} U$, where the $U$ is the potential energy of the system of the two particles, and it's evaluated at the position of the particle on which you want to know the force, which in this case I've just called $\vec{r}_1$.

If you can parameterise the positions of both particles, e.g. $\vec{r}_1 = \vec{r}_1(s)$ and $\vec{r}_2 = \vec{r}_2(t)$, then you can use the total derivative:$$dU_{12} = \nabla_{\vec{r}_1}U_{12} \cdot \frac{d\vec{r}_1}{ds} ds + \nabla_{\vec{r}_2}U_{12} \cdot \frac{d\vec{r}_2}{dt} dt$$ $$\Delta U_{12} = \int dU_{12} = -\int \vec{F}_{12} \cdot \frac{d\vec{r}_1}{ds} ds - \int \vec{F}_{21} \cdot \frac{d\vec{r}_2}{dt} dt = -(W_{12} + W_{21})$$ i.e. the change in gravitational potential energy is the negative of work done by gravitational forces internal to the system

 

[PeroK]

Summary:: why should gravitational potential energy always be negative if the reference point is taken to be at infinity and the gravitational potential energy at this point is taken to be zero?

The potential energy is at a maximum at infinity. If zero is your maximum, then at all other points the value is negative.

If something had positive potential energy, then it could lose potential energy as it moves further away (towards infinity) and hence gain KE and speed up. Objects cannot speed up as they move away if the force is attractive.

 

[kuruman]

Summary:: why should gravitational potential energy always be negative if the reference point is taken to be at infinity and the gravitational potential energy at this point is taken to be zero?

We have already picked the gravitational potential energy to be zero at infinity from the force center. At infinity the gravitational force is also zero. Now consider a bound system such as the Earth and a satellite in some orbit. If you want this satellite to just escape the Earth and reach infinity with zero kinetic energy, you must add kinetic energy to it. If you must add energy to get the total mechanical energy to become zero at infinity, what sign must the potential energy have when the satellite is closer to the force center than infinity?

 

[vanhees71]

As far as I am aware, at least in non-relativistic physics, no, it's only meaningful up to an additive constant. I have no idea if that's different in SR or GR - perhaps someone who knows what they're talking about can weigh in?

(I remember John Baez talked about this using fancier lingo, that energy differences lie in the real numbers but energies lie in an "R-torsor")

In Newtonian physics and SR absolute energy values have no physical meaning, i.e., you can add constants to the total energy of a system without changing any physics.

This changes in GR, where the energy-momentum-stress tensor is the source of gravity. This implies that the absolute value of the energy density (the time-time component of the energy-momentum tensor) matters, and that implies one of the biggest puzzles of contemporary physics: Why is the energy density so small? According to the cosmological standard model the energy density of the universe is just the critical one, implying a rather small (though dominant) contribution from the "cosmological constant" (if interpreted as being part of the energy-momentum-stress tensor of the right-hand side of the Einstein equations).

From a naive approach to the contribution of the Higgs-mass running in standard model QFT you'd expect a value which is by several orders of magnitude larger (if you take the extrapolation to the Planck scale, it's by a factor of $10^{120}$ larger. On the other hand the total vacuum energy has to be renormalized anyway and you could just argue that one has to subtract the value to get the right one according to the cosmological observations. Some physicists find this ugly, because it means that we have to fine-tune the corresponding parameters of the Standard Model at an accuracy of $10^{120}$, which sounds pretty ad hoc. One way out of this dilemma could be some supersymmetric extension of the Standard model, where the quadratic divergence of the Higgs-mass renormalization constribution gets logarithmic, but at the moment supersymmetry is not so much in favor anymore given that the LHC hasn't provided any hint so far for the existence of any supersymmetric partners of the known elementary particles.

There's a famous review paper by Weinberg on this. Though it's pretty old, I don't think that we know too much more about this puzzle today:

S. Weinberg, The cosmological constant problem, Rev. Mod.
Phys. 61, 1 (1989),
https://link.aps.org/abstract/RMP/V61/P1

 

[Vanadium 50]

B-level everyone.