Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Assume a central body of mass $M$, and call $a$ the acceleration of a test body at a distance $r$ due to any interaction whatsoever with the central body. Is is correct to say that the ratio $a r^2/ M$, for any interaction, is always larger than or equal to the gravitational constant $G$?

share|improve this question

2 Answers 2

No, this is false. Two extremal charged black holes at distance r stand still forever, and are described by this metric (isotropic form, exterior only):

$$ ds^2 = {-dt^2 \over (1 + \phi)^2 } + (1+ \phi)^2(dx^2 + dy^2 + dz^2) $$

And an electric field

$$ F_{0i} = - F_{i0} = \partial_i \phi $$

For $i=x,y,z$ where

$$ \phi = {Q\over |r-r_0|} + {q\over |r-r_1|} $$

Where $r_0$ and $r_1$ are the centers of the black holes (actually this is the entire position of their event horizon, which is collapsed to a point in these cruddy exterior-only coordinates).

If you make q small and Q large, you have a test object with no acceleration due to the cancellation of the electromagnetic and gravitational force. This is a counterexample to your claim. The brane stacks in string theory are exactly analogous constructions with form-fields and gravity in higher dimensions.

You can understand the cancellation as follows: if you make a Kaluza Klein theory, and boost the objects in the 5th circular dimension so that they are going at the speed of light, the charge is equal to the mass, so the black hole is extremal charged. But if you unwrap the KK circle to it's double cover, the attraction is as between two objects highly boosted in the same direction. In this case, special relativistic time dilation makes the force go to zero. This is an explanation for the miraculous class of exact solutions one case of which is given above.

There is another sense in which the gravitational force is a lower bound on other forces (but not on the total force--- there can be cancellation as above!). When black holes are always allowed to decay, as is suspected to be required in consistent quantum gravity, the decay products must be extremal or super-extremal, so that they can always be emitted by an extremal black hole to reduce it's charge and mass keeping it within the allowed realm of black holes. This requires that black holes be always allowed to fall apart no matter what their charge, so that they have a consistent S-matrix.

This restriction tells you that the lightest charged objects must repel more than they attract gravitationally. This bound is suspected to hold in all string theory solutions, but it is not proved yet.

share|improve this answer
    
Very interesting! Don't they deform? Do they remain exactly spherical all the time? In other words: is the solution stable? –  Frank Sheldon Sep 4 '12 at 18:19
    
Some questions arise: Does the solution (mass values do not appear in it, for some strange reasons) not contradict the second principle of thermodynamics? What happens when spin is taken into account? What happens when quantum theory is taken into account? Does the counterexample depend on string theory? And: do extremal black holes exist? Is this really a counterexample of the claim? –  Frank Sheldon Sep 8 '12 at 5:49
    
@FrankSheldon: For an extremal black hole the mass is equal to the charge, so I used one letter "Q" to denote either. Where is the contradiction to any law of thermodynamics? Extremal black holes exist like a zero temperature solid exists, it's an idealization which is only achievable at absolute zero. –  Ron Maimon Sep 8 '12 at 7:27
    
A zero temperature solid contradicts the third law of thermodynamics. Does an extremal black hole as well? –  Frank Sheldon Sep 8 '12 at 13:46
    
@FrankSheldon: Yes, in the same way--- it's physical, you can imagine it, but you have to isolate it and cool it down, and it takes an infinite amount of effort for a macroscopic hole. But you can get close with no problem. –  Ron Maimon Sep 8 '12 at 15:18

Yes, gravity is the weakest possible interaction: it curves space-time in the weakest possible way. Other interactions always ADD to the curvature. It is however, not simple to make this statement observer-independent.

@Ron: The example you give is neglects quantum effects - spin, black body radiation, charge quantization, and black hole entropy. Take them into account, and you will find that the bodies will move with respect to each other; the curvature will be larger than that of pure gravity in several places.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.