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  Newtonian gravity from the holographic principle?

+ 7 like - 0 dislike

Can one understand Newton's law of gravitation using the holographic principle (or does such reasoning just amount to dimensional analysis)?

Following an argument similar to one given by Erik Verlinde, consider a mass $M$ inside a spherical volume of space of radius $R$.

The holographic principle says that the mass $M$ must be completely described on the surface of the sphere in terms of the energy in elementary entities or strings.

By the equipartition theorem, the total energy, $E=Mc^2$, on the surface is given by the number of strings, $N$, times the number of degrees of freedom per string, times $1/2\ k_BT$ per degree of freedom:

$$E = N \times d_f \times \frac{1}{2} k_B T.\ \ \ \ \ \ \ \ \ \ (1)$$

The number of strings on the surface is given by the Bekenstein-Hawking formula:

$$N = \frac{A}{4},$$

where $A$ is the sphere surface area in units of the Planck area $G\hbar/c^3$. In terms of the radius $R$ the number $N$ is given by:

$$N = \frac{\pi c^3 R^2}{G \hbar}.$$

By substituting into equation (1) we obtain an expression for temperature $T$:

$$M c^2 = \frac{\pi c^3 R^2}{G \hbar} \times d_f \times \frac{1}{2}k_BT$$

$$T = \frac{4}{d_f}\frac{\hbar}{2\pi ck_B}\frac{GM}{R^2}.$$

If there are $d_f=4$ degrees of freedom per string then the above formula gives the Unruh temperture for an object falling through the surface with acceleration $g$:

$$g = \frac{GM}{R^2}.$$

This of course is the acceleration given by Newton's law of gravity due to the presence of a mass $M$.

P.S. Could $d_f=4$ come from the 4 dimensions of spacetime?

This post imported from StackExchange Physics at 2015-03-23 09:23 (UTC), posted by SE-user John Eastmond
asked Aug 12, 2013 in Theoretical Physics by John Eastmond (55 points) [ no revision ]
Related: arxiv.org/abs/1308.1977 section 2, "Holography in pseudo-Newtonian gravity"

This post imported from StackExchange Physics at 2015-03-23 09:23 (UTC), posted by SE-user Mitchell Porter

2 Answers

+ 1 like - 0 dislike

If we do have the degrees of freedom coming from that of spacetime, then, we first need to define string dynamics in four dimensions. After doing so, we need to then state that, to satisfy the four D problem, the string is such that only translations of the string actually affect it (Wrong!). Then, we could say that quantum Newtonian gravity does not work at the string scales, and this can be foreseen. Of course, you could be referring to the quanta of Newtonian gravity. But, by the above statement, we can say that the quanta of Newton's gravity would have a spin 0, the same as that of the Goldstone bosons, and thus, maybe, your argument talks about the relation between the Goldstone bosons and quantum Newtonian gravity!

This post imported from StackExchange Physics at 2015-03-23 09:23 (UTC), posted by SE-user Sanath Devalapurkar
answered Aug 18, 2013 by Sanath Devalapurkar (25 points) [ no revision ]
+ 1 like - 0 dislike

A lot of things about Verlinde's calculation are in the air. He does $S=A$. Followed by $E=AKT$. $T=a$, therefore,

$$a = E/A=M/r^2$$.(Now add dimensions )

Which really if you think about it is just gauss's law, because he used area and not more. and previously established results like $T=a$ have been used out of context to somehow obtain newtons laws.

This post imported from StackExchange Physics at 2015-03-23 09:23 (UTC), posted by SE-user Prathyush
answered Sep 17, 2013 by Prathyush (705 points) [ no revision ]

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