# Is Gravity an entropic force after all?

Recently, there was a rapid communication published in Phys.Rev.D (PRD 83, 021502), titled "Gravity is not an entropic force", that claimed that an experiment performed in 2002 with ultra cold neutrons in a gravitational field, disprove Verlinde's entropic approach to Gravity.

The neutrons experiment gave results consistent with the predictions of Newtonian gravity for the lowest energy state.

As I understand it, the author claims that the fact that Verlinde's entropic force comes from a thermodynamic process that is irreversible (or approximately reversible), leads to non-unitarity in the evolution of quantum systems. The non-unitarity then exponentially suppresses the eigenfunctions, predicting results very much different than the Newtonian. Thus, that experiment is in contradictions to what is expected if Verlinde's approach is correct.

My questions are,

1. First of all, is there anything else essential that I am missing?
2. Is there any response to that argument?
3. Is that a fatal problem with Verlinde's entropic approach?
4. Is that a fatal problem for any entropic approach?

1. There is also this recent comment arxiv.org/abs/1104.4650
2. Once more: gravity is not an entropic force arxiv.org/abs/1108.4161
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I don't really understand what Eric Verlinde is after, and why he "re-discovered" Newtonian gravity as an emergent theory, while he seems to ignore Ted Jacobson's work published 15 years before his, see arxiv.com/abs/gr-qc/9504004. If you'd like to discuss gravity as an entropic force, the discussion should revolve around Ted Jacobson's work, not Verlinde's, IMHO. –  Tim van Beek Jan 31 '11 at 14:33
I am only mentioning Verlinde's approach because that paper is written as a comment to his work. But the question is general and extends to all entropic approaches. –  Vagelford Jan 31 '11 at 14:45
@Tim - or for that matter why he ignored Padmanabhan's work arxiv.org/abs/0912.3165 published a month before his own. –  user346 Feb 1 '11 at 6:14
@Tim van Beek --Lubos discusses Ted's work on <motls.blogspot.com/search?q=verlinde+entropic+gravity>; –  Gordon Feb 2 '11 at 7:41
Why do you attribute this argument to the authors of the rapid communication? It's Lubos Motl's argument. –  Ron Maimon Sep 2 '11 at 0:18

This question has introduced me to the whole "entropic force" area which has several papers during 2010. I see that there are references to "entropic force" explanations for Coulomb's law and other areas too. Here is a link to a simple introduction to these ideas.

The Verlinde paper and others however are deriving Newtons Law, Einstein's GR etc as classical theories. The underlying formulation of course being a stochastic behaviour of unknown microstates. Despite the presence of $\hbar$ and the motivation from the Black Hole area formulae the Verlinde paper does not introduce an explicit link with quantum mechanics. Thus there is no derivation of Schrodinger's equation and no introduction of $\Psi$.

The Kobakhidze paper says "One starts with a "holographic screen” S which contains macroscopically large number of microscopic states which we denote as $\left|i(z)\right\rangle$, $i(z) = 1, 2, ...,N(E(z), z).$ The screen is then described by the mixed state

$$\rho(z)=\Sigma p_{i(z)}\left|i(z)\rangle \langle i(z)\right|$$

However Verlinde does not explicitly introduce microstates as quantum states, with density matrices etc, although this is a tempting extension.

Now it might be that this is the only sensible quantum development of the stochastic basis of the "entropic idea", but Verlinde has not taken it. So what is disproved is a theory that Verlinde has not written down.

Having said this, there is a resemblance between "entropy" and the idea of introducing "stochastics" into quantum theory. One such attempt is known as "Stochastic Electrodynamics" (link to Wikipedia). As you will see from the summary this has had successes with e.g. the Unruh effect, but problems modelling genuine quantum phenomena.

I dont know whether anyone has considered combining the two areas directly.

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This is an interesting way of looking at it. GR orbital dynamics do not consider a noisy interaction with the gravitational source. The screen here is a bound on the entropy and the Bohr orbit of a particle may interact with it, say an electron in orbit around a tiny black hole. –  Lawrence B. Crowell Feb 9 '11 at 14:17

Dear Vagelford, you're totally right. Gravity cannot be an entropic force because

1. its phenomena would be irreversible
2. the degeneracy of the states coming from the entropy would destroy the interference patterns that have been measured e.g. by neutron interferometry.

More than a year ago, this was also explained on my weblog

and Erik Verlinde much like some of his junior Dutch colleagues tried to react but as far as I can say, none of their reactions has ever made any sense.

The neutron interferometry experiments are pretty impressive. They not only show that the interference pattern survives the action by the force of gravity. But it is exactly as shifted as the equivalence principle implies.

And in fact, the changes of the phases have been measured so accurately that the experimenters may deduce not only the zeroth order gravitational acceleration but also the higher-order corrections to it such as the tidal forces. All of these effects preserve the interference pattern - which wouldn't be possible if there were many states representing a macroscopic configuration - and this pattern exactly moves and behaves according to general relativity.

For these two and other reasons, gravity cannot be entropic. We also know it from the explicit models in the AdS/CFT correspondence and elsewhere: only event horizons may produce a large entropy of this order. A cold binary star doesn't carry any entropy associated with the gravitational attraction, certainly not an entropy comparable to the black hole entropy which is what Erik Verlinde claims.

But a multi-million euro grant has already paid by some European politicians to endorse this "research" so it may be unreasonable to expect that too many people aside from the two of us will say these obvious things too comprehensibly and loudly. After all, many people can be bought very easily and inexpensively.

One additional disclaimer: If you originally encountered the proofs that gravity can't be entropic on my blog, you shouldn't treat this answer as an independent confirmation of my previous claims. ;-)

All the best, LM

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The change in entropy has to do with the displacement of a holographic screen. A black hole, which is at the maximum entropy for the “gravity force,” increases its entropy if the black hole absorbs mass-energy. In that case the holographic screen has been displaced. The entropy associated with the force is just a measure of the upper limit on entropy (Bekenstein bound etc) which acts as the gravity source. It is curious that people think somehow this should result in decoherence or entropy with interferometer experiments. This is a misinterpretation of the hypothesis. –  Lawrence B. Crowell Jan 31 '11 at 17:30
"+" Dang it, I've run out of votes, but I'll be back tomorrow. Just what you said. –  Carl Brannen Feb 10 '11 at 4:43
No Lubos, you're wrong. There is no reason at all why entropic forces should be irreversible. Entropic forces are observed most clearly in fully teversible molecular dynamics simulations. A toy model leading to reversible entropic gravity is discussed in: science20.com/hammock_physicist/… –  Johannes Feb 19 '11 at 3:16
@Johannes: observing entropic forces in classical MD is irrelevant, the point is not "irreversibility" in the sense of "flip all the simulation v's and watch it go backwards", the point is that the action of entropic forces can't produce quantum coherent behavior. Quantum coherence requires no entanglement between the particle and the environment, and the statistical fluctuations of a traditional entropic force inevitably entangle the particle with the source of the force. –  Ron Maimon Sep 2 '11 at 0:47
@Ron - the key point here is: the particle ('it') is defined solely in terms of the environment (the 'bits'). (See the above quoted Mikado model.) The bit representation of the particle cannot decohere the particle itself... –  Johannes Sep 2 '11 at 22:50

I think there has been some confusion over this matter. Of course if makes little sense to think a trajectory around a black hole will exhibit an entropy increase. Verlinde proposed an entropy force of gravity from which Newton's law of gravity may be derived. This is a thermodynamic principle for the entropy variation of a holographic screen $$\Delta S~=~2\pi k_B(mc/\hbar)\Delta x$$ where $\Delta x$ is the distance between the holographic screen and a test particle of mass $m$. The entropy is then in increments of $2\pi k_B$ according to displacements of the screen equal to the Compton wavelength $\lambda~=~\hbar/mc$. The standard entropy formula $S~=~k_BA/4L_p^2$ indicates a proportionality with respect to area. For the radius of the screen adjusted $S_0~\rightarrow~S~=$ $S_0~+~\Delta S$ by the radial change in the screen $r~=~r_0~+~\Delta r$ then $$ΔS~=~k_B/4L_p^2(A~–~A_0)~=~(2\pi k_B/L_p^2)r\Delta r~=~(2\pi k_Bc^3/G\hbar)r\Delta r,$$ which is linear in the radial displacement. By equating $\hbar/mc~=~G\hbar/rc^3$ gives a radius $r~=~Gm/c^2$, which is appropriate for the Newtonian result, but is half the Schwarzschild result.

The entropy for an orbit of a test mass is constant, and this entropy is a measure of the holographic screen. So if you place a Gaussian surface around a gravitating radially symmetric body that cloaks the configuration of the body, the entropy of the screen is the maximum entropy of the system. For a particle orbiting the body the entropy is constant, or $\Delta S~=~0$, for the screen remains constant.

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The experiment that we are talking about was performed with neutrons bounded between a linear potential and an infinite wall (something like piled on top of a hard surface or more like bouncing balls). So, we are not talking about "free falling" neutrons for which as I understand what you are saying, there is no entropy change and thus the evolution is unitary. Would you like to elaborate on that? –  Vagelford Feb 2 '11 at 16:15
The distinction does not seem relevant. It is a standard problem in QM to find the wavelength of a quantum ball bouncing on a surface. The problem of a bouncing ball, free fall plus a rebounding potential, does not seem to change the result much. After all we can't get Bohr orbits of particles around the Earth very easily. –  Lawrence B. Crowell Feb 9 '11 at 14:13

The entropy of a gravitating particle in Verlinde's theory CANNOT be constant. Consider a screen placed at a distance $r$ from a mass $M$. It has an entropy $S_{screen}(r)$, and this is the maximal entropy the given region surrounded by the screen can have. Now, if you place a particle of mass m at distance $r+\delta r$, the entropy of the screen becomes $S_{screen}(r+\delta r)$. This later entropy is the entropy of the screen placed at $r+\delta r$, where the particle represented by certain microstates. Integrating out this microstates (coarse graining the screen) gives back $S_{sreen}(r)$. Since, the entropy is an additive quantity, the entropy on the screen with a test particle is $S_{screen- without-neutron}(r+\delta r)+S_{neutron}(r+\delta r) = S_{screen}(r)+S_{neutron}(r+\delta r)$. Thus, $S_{neutron}(r+\delta r)=\delta S_{screen}$, that is, entropy of a test particle $m$ depends on the distance from $M$. Is not that trivial?

Second point -- Time evolution of the system seems to be unitary, since the energy eigenvalues obtained are real. There are systems in QM which have such properties, see e.g. enter link description here

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Gravity as the Pressure of ether

Pressure gravitation theory is more than 3 hundred years old and is discarded mainly on the basis of Le Sage theory where gas filled ether. But today we know that space is filled with electromagnetic waves (let’s call it ether), so the first condition for pressure gravity the totally filled space is given. If we presume that there are ether segments we cannot measure and they interact mass, the only way we can detect their effect is through gravity-like forces. So gravity and ether proves each other.

In the next I’d like to present my interpretation of pressure gravity force through the Pioneer and Fly-by anomalies. In pressure gravity theory gravitation is the difference between the ‘attacking force’ of ether ‘AF’ and the weakened ether leaving the mass ‘LF’. So $$g=AF-LF=Gm/r^2$$

I stemmed pressure gravity force equations between masses from the $$g/AF = y/(g/r^2) = q$$ ratios, where $AF=$ pressure of ether; $g=$ surface gravity of parent mass; $r=$ distance from parent mass; $y=$ gravity deviation in other mass (extra acceleration toward parent mass caused by ether weakening in other mass) and $q=$ gravity coefficient. From this: $$AF= \frac{(g/r)^2}{y}$$ Calculating the pressure of ether from Pioneer anomaly where $y= 8.7 \times 10^{-10}$ (extra acceleration force in Pioneer toward Sun at 70 AU) I got ~ $380 000 m\text{ m/s}^2$ for the pressure of ether (at 20 AU it would be ~ $4.646 \text{ million m/s}^2$).

Solving fly-by anomaly $$y=q \times g/r^2$$ Inserting $AF$ to this equation for Earth I found that $q=2.5087 \times 10^{-5}$ and $y=~ 0.21\text{ mm/s}^2$ at ~600 km distance from Earth, which means that Galileo needed some 20sec and NEAR some 60 sec to reach their acceleration anomaly not counting with other factors. So the solution of pressure gravity theory is acceptable, and can be managed to these anomalies.

Modified Newtonian acceleration forces according to Pressure gravity: $$F_1=\left(\frac{Gm_2}{r^2}+y_2\right) \times m_1 = \left(\frac{Gm_1}{r^2}+y_1\right) \times m_2=F_2$$ $$\frac{m_1}{m_2}=\frac{y_1}{y_2}$$ Where $G$ is the gravitational constant and $m_1$ and $m_2$ are different masses.

The spacecraft anomalies show not only that pressure gravity theory works, but that these anomalies may be the consequences of pressure gravitation. While to prove attractive force gravity is highly difficult (particles cannot call each other), to deny pressure force gravity is impossible, as there is no bottom limit for $y$. Smaller $y$ means only higher pressure of ether.

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