# Newton's G or Planck's Length?

1. Jul 6, 2006

### arivero

Time ago there was an interesting article on the arxiv title "Trialogue", about h, c, and G as fundamental dimensionful constants. A doubt can be raised additionaly: which is the more interesting mathematical object, $$G_N$$ or [/tex]L_p[/tex].

If Planck's length is the fundamental object, one can recover Gauge QFT in a single limit L-->0, but a double limit is needed to recover Newton's constant: h-->0, L--->0. And the single limit h-->0 is rather strange, because if Planck's length (and c) remains finite then G goes to infinite.

2. Jul 6, 2006

Staff Emeritus
So you regard G as derivitive from c, h, and l_p via the Planck relations? What do you get (you made the claim; you do the work ) when you plug this definition of G into Newton's inverse square relation or Schwartzschild's expression for the horizon in his geometry?

3. Jul 7, 2006

### Chronos

If I had to vote for 'the most fundamental' constant, it would definitely be gravity [g]. I can see how you might derive other fundamental constants using 'g', but, not how you could derive 'g'.

4. Jul 8, 2006

### arivero

$$F= {l_p^2 c^3 \over \hbar} {m_1 m_2 \over r_{12}^2}$$

$$R_M=2 M {c \over \hbar} l_p^2$$

And yes, here is the paradox: F is a classical quantity, so not c neither h should appear; and R is a relativistic unquantised object, so we do not expect h to appear. If we want to be free to remove h in the second expression, lp^2 must go to 0 too. And if we want to remove both h and c in the expression for Newton force, we need a different kind of dependence of lp.

Still, string theory postulates that the fundamental quantity is lp (or its cousin, Planck force) and the same claim seems to emerge from LQG.

Last edited: Jul 8, 2006
5. Jul 8, 2006

### arivero

Hmm, in function of the stress tension of the fundamental string, call it T_P, we should get
$$F\propto = {c^4 \over T_P} {m_1 m_2 \over r_{12}^2}$$
and
$$R_M\propto =2 M {c^2 \over T_P}$$

so at least here we do not have explicit h around, but the non relativistic limit is still to be got when c goes to infinite, then again T_P must go to infinity fast enough to counterweight c in Newton's force.

Last edited: Jul 8, 2006
6. Jul 8, 2006

### rbj

i presume you mean $G$. why is that so fundamental? like $\hbar$ and $c$, it just a number that is a human construct (as a function of the anthropometric units we use). there is no intrinsic strength of gravity in the universe, but using our units like meters, kg, and second, we measure that strength.

7. Jul 8, 2006

### arivero

rbj, I already stressed "dimensionful" in the starting post of the thead. You can argue that dimensionful quantities are human constructs, but still Planck's Mass seems more fundamental than, say, the kilogram; I expect you notice the difference.

8. Jul 8, 2006

### rbj

certainly the Planck Mass is more fundamental (being less anthropocentric in definition) than the kilogram. i just couldn't understand what the other poster meant by "g" other than G.

i tend to think that the Planck units (with a possible substitution of $4 \pi G$ for $G$ and $\epsilon_0$ for $4 \pi \epsilon_0$) simply have to be the natural units of physical reality. it obviates the need to answer the questions of why Nature would bother to take some quantity (say, flux density) and scale it with some number to get some other directly related quantity (in this case, field strength). those conversion factors just go away so we wouldn't need to try to explain why Nature would be fiddling with them. (there are always the dimensionless quantities like $\alpha$ that still need explaining.)

9. Jul 10, 2006

### arivero

Posibly better, $$c^4/G$$ instead of G. Two reasons: it is the string tension , and it is the quantity one needs to match the units of Einstein-Hilbert action.

10. Jul 10, 2006

### robphy

Just a comment: It seems to me that there is an implicit assumption about the dimensionality of space[time] here. For example, if one generalizes the Poisson equation to n-dimensions, the analogue of the Newton constant "G" would depend on n. A similar argument could be applied to the Coulomb constant. One implication is that the resulting "dimensionless constants" would take different forms in different spacetime-dimensions.

11. Jul 10, 2006

### arivero

Hmm, it should if you extract it from the original newton formula. On the very very opposite side, string tension is always a tension. Just now I am not sure which is the position of Einstein-Hilbert action between these two extremes.

12. Jul 10, 2006

### George Jones

Staff Emeritus
Chapter 3 (and problems 3.9, 3.10) of Zwiebach's book has an interesting discussion of electromagnetism, gravity, and the Planck lenghth in higher dimensions.

13. Jul 10, 2006

### arivero

A pity, Zwiebach lecture notes do not include problems 3.9, 3.10.

It is very interesting the notation $$G^{(n)}$$ meaning this dimension-dependent Newton constant.

Still (see http://pancake.uchicago.edu/~carroll/notes/ or the wikipedia), it sounds different when we look at Einstein-Hilbert action:
$$S= {c^4 \over 16 \pi G} \int R \sqrt {-g} dx^4$$
and Zwiebach does not care about it in that chapter.

Last edited by a moderator: Apr 22, 2017
14. Jul 10, 2006

### arivero

OK, they are not so different. I think I get it. The point is that the units of $$\int R \sqrt {-g} dx^n$$ are not L^(n-2) but $$[L]^{n-2} \over [c]$$, and thus the compensation factor gets an extra c, becoming $$c^{4}/ G^{(n)}$$. This happens because the time coordinate differential, dx^(0), carries this factor c around to be homogeneus with space. Is it?

Last edited: Jul 10, 2006