Why gravity is incompatible with quantum theory?

Question

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Answer 1

I'm not sure about all that irrelevant math - the basic reason is very, very simple.

The current theory of gravity - general relativity - is a classical theory. That means that it is smooth at all scales. In other words its predictions can be scaled to arbitrarily small scales and arbitrarily small energies smoothly and continuously.

This is absolutely fine most of the time, because the universe is a very gravitationally smooth place. It does not matter even at the normal quantum mechanical distances we encounter - the atomic scale - because the force of gravity is around 24 orders of magnitude (that is 1,000,000,000,000,000,000,000,000) smaller than any of the other forces acting - in other words, it is entirely negligible.

Quantum mechanics, on the other hand, is not a smooth theory. It predicts that at small scales and small energies the universe is not continuous but discrete. Energy comes in packets that can only have certain values, for instance.

This is clearly a problem. If we encounter a situation where both quantum mechanics and general relativity are trying to predict say energy levels that fall into the same range, then the quantum mechanics ones will have gaps between them and the general relativity ones will not. In other words, one theory will predict things that the other theory says cannot happen.

This is exactly the situation we find just after the big bang or perhaps in black holes, where matter is so compacted that the force of gravity is comparable to the other forces.

Answer 2

The intuitive content is simple. If a fluid is flowing in some area, and we wish to know how much fluid flows out of a certain region within that area, then we need to add up the sources inside the region and subtract the sinks. The fluid flow is represented by a vector field, and the vector field's divergence at a given point describes the strength of the source or sink there. So, integrating the field's divergence over the interior of the region should equal the integral of the vector field over the region's boundary. The divergence theorem says that this is true.

The divergence theorem is thus a conservation law which states that the volume total of all sinks and sources, the volume integral of the divergence, is equal to the net flow across the volume's boundary.

Suppose V is a subset of Rn (think of the case n=3) which is compact and has a piecewise smooth boundary. If F is a continuously differentiable vector field defined on a neighborhood of V, then we have

\iiint\limits_V\left(\nabla\cdot\mathbf{F}\right)dV=\iint\limits_{\part V}\mathbf{F}\cdot d\mathbf{S},

where ∂V is the boundary of V oriented by outward-pointing normals, and dS is shorthand for ndS, the outward pointing unit normal of the boundary ∂V. In terms of the intuitive description above, the left-hand side of the equation represents the total of the sources in the volume V, and the right-hand side represents the total flow across the boundary ∂V.

The divergence theorem is frequently applied in these following variants (cf. vector identities):

\iiint\limits_V\mathbf{F}\cdot \left(\nabla g\right) + g \left(\nabla\cdot \mathbf{F}\right)dV=\iint\limits_{\part V}g \mathbf{F}\cdot d\mathbf{S}

(this is the basis for Green's identities, if \mathbf{F}=\nabla f),

\iiint\limits_V \nabla g dV=\iint\limits_{\part V} g d\mathbf{S},

\iiint\limits_V \mathbf{G}\cdot\left(\nabla\times\mathbf{F}\right) - \mathbf{F}\cdot \left( \nabla\times\mathbf{G}\right) dV = \iint\limits_{\part V}\left(\mathbf{F}\times\mathbf{G}\right)\cdot d\mathbf{S},

\iiint\limits_V \nabla\times\mathbf{F} dV = \iint\limits_{\part V}d\mathbf{S} \times\mathbf{F}.

Note that the divergence theorem is a special case of the more general Stokes theorem which generalizes the fundamental theorem of calculus.

Example

Suppose we wish to evaluate \iint\limits_S\mathbf{F}\cdot \mathbf{n}dS, where S is the unit sphere defined by x2 + y2 + z2 = 1 and F is the vector field \mathbf{F} = 2 x\mathbf{i}+y^2\mathbf{j}+z^2\mathbf{k}. The direct computation of this integral is quite difficult, but can be simplified using the divergence theorem:

\iint\limits_S\mathbf{F}\cdot \mathbf{n} dS=\iiint\limits_W\left(\nabla\cdot\mathbf{F}\right)dV=2\iiint\limits_W\left(1+y+z\right)dV
= 2\iiint\limits_W dV + 2\iiint\limits_W y dV + 2\iiint\limits_W z dV

By symmetry,

\iiint\limits_W y dV = \iiint\limits_W z dV = 0

Therefore,

2\iiint\limits_W\left(1+y+z\right)dV = 2\iiint\limits_W dV = \frac{8\pi}{3}

because the unit ball W has volume 4π/3.

] Applications


Used to simplify the conservation of mass, momentum and energy equations.

Electrostatics

Applied to an electrostatic field we get Gauss's law: the divergence is a constant times the volume charge density.

Gravity

Applied to a gravitational field we get that the surface integral is -4πG times the mass inside, regardless of how the mass is distributed, and regardless of any masses outside.

Spherically symmetric mass distribution

In the case of a spherically symmetric mass distribution we can conclude from this that the field strength at a distance r from the center is inward with a magnitude of G/r² times the total mass at a smaller distance, regardless of any masses at a larger distance.

For example, a hollow sphere does not produce any gravity inside. The gravitational field inside is the same as if the hollow sphere were not there (i.e. the field is that of any masses inside and outside the sphere only).

See also the shell theorem.

Cylindrically symmetric mass distribution

In the case of an infinite cylindrically symmetric mass distribution we can conclude that the field strength at a distance r from the center is inward with a magnitude of 2G/r times the total mass per unit length at a smaller distance, regardless of any masses at a larger distance.

For example, an infinite hollow cylinder does not produce any gravity inside.

Bouguer plate

We can conclude that for an infinite, flat plate (Bouguer plate) of thickness H gravity outside the plate is perpendicular to the plate, towards it, with magnitude 2πG times the mass per unit area, independent of the distance to the plate (see also gravity anomalies).

More generally, for a mass distribution with the density depending on one Cartesian coordinate z only, gravity for any z is 2πG times the difference in mass per unit area on either side of this z value.

In particular, a combination of two equal parallel infinite plates does not produce any gravity inside.

Answer 3

No quantized gravity effect has ever been seen.

Answer 4

You'd have to ask the people that believe in "The Secret" :)