Wednesday, May 29, 2013

Is Newtonian Gravitational Fields Valid Within General Relativity?

We will demonstrate that the Einstein Field Equations reduce to Newton's Law of Gravity in the case of a weak field and slow-motion of a particle (v less than the speed of light, c).

As we have already seen, Newton's Gravitation Law can be written as

(1) ∇2φ = 4πGρ, Equation (16) in Newton's Law of Gravity.

In free-fall, a particle satisfies,

(2) d2x/dt2 = F/m= –∇φ, Equations (4),(15) in Newton's Law of Gravity.

In tensor notation, this is written as,

(3) φ,ii ≡ 4πGρ

(4) d2x/dt2 ≡ –φ,i

Corresponding to (1) in General Relativity is the Einstein Field Equations, which can be written in the trace-reverse form,

(5) Rμν = (8πG/c4)[Tμν – ½Tgμν]

And corresponding to (2), is the geodesic equation,

(6) d2xα/dτ2 = –Γαβγ(dxβ/dτ)(dxγ/dτ)

Our task is to show that (5) will reduce to (1) in low gravity, low velocity.

The first approximation we make is that the particle is moving at velocity near zero,

(7)(dxβ/dτ)≈ (dt/dτ,0,0,0)

The only non-zero term for the Γ's in (6) will be α =i, β = γ =0. The equation becomes,

(8) d2xα/dτ2 = –Γi00(dx0/dτ)(dx0/dτ)


(9) d2xα/dτ2 = –Γi00(dt/dτ)(dt/dτ), (x0 =t)

But by the chain rule of calculus,

(10) d2xα/dτ2 = (dt/dτ)(dt/dτ)d2xα/dt2


(11) d2xα/dt2 = –Γi00

Equation (4) and (11) yields,

(12) φ,i = Γi00

Using the Christoffel symbols of the second kind ( Torsion =0 in GR),

(13) Γαβγ = ½ gδα(gδβ,γ + gδγ,β – gβγ,δ)

Again setting α =i, β = γ =0, equation (12) becomes,

(14) φ,i = ½ gδi(gδ0,0 + gδ0,0 – g00,δ)

Since the time derivative of the metric is zero (Einstein Equivalence Principle), the only surviving term is δ = i. Then,

(15) φ,i = ½ gii(– g00,i)= (–½g00,i)

A solution is,

(16) g00 = – c2 – 2φ

Turning to the Einstein Field Equations (5), and from (16), we only need the time-time equation, μ = ν = 0

(17) R00 = (8πG/c4)[T00 – ½Tg00]


(18) T00 = ρc4 and T = g00T00 = (–1/c2)ρc4= –ρc2

Substituting (18) into (17), we get,

(19) R00 = 4πGρ

Now to get to Newton's Law, equation (1), we need to work on the left-hand side of equation (19).

For that we use the definition of the Ricci tensor,

(20) Rμν = ∂Γγμν/∂xγ – ∂Γγμγ/∂xν + ΓγμνΓδγδ – ΓγμδΓδνγ

From (19), μ = ν = 0,

(21) R00 = ∂Γγ00/∂xγ – ∂Γγ/∂x0 + Γγ00Γδγδ – ΓγΓδ

In low gravity, the square of the Γ's is zero, and its time derivative is also zero. The only surviving term is,

(22) R00 = ∂Γi00/∂xi ≡ Γi00,i

But from (12), repeating below,

(23) Γi00 = φ,i

Take the derivative, and then use (23)

(24) Γi00,i = φ,ii = R00

Using (19),

(25) φ,ii ≡ ∇2φ = 4πGρ

And that is Newton's Law of Gravity (1).

Tuesday, May 28, 2013

Newton's Law of Gravity

Newton’s law of gravity

Fig 1

(1) F = –(GmM)/R2

In vector form, this reads as:

(2) F = – [(GmM)/R2]er

Consider a system of particles. We label the particles 1, 2, 3... i...j...

Fig 2

(3)F = – ∑i≠j[(GmiMj)/Rij2]er

Define Gravitational Field:

(4) A(x) = F(x)/mi

Substituting (3),

(5) A = = – ∑i≠j[(GMj)/Rij2]er

Gauss Theorem

(6) ∫v A dV = ∫s A•dS

Substitute (5) into the right hand side of (6),

(7) ∫s A•dS = – ∫s[(GM)/R2]er•dS

However, the magnitude of the infinitesimal area element dA is just the area of the infinitesimal solid angle dΩ, given by,

(8) dS = R2er

Substitute (8) into (7),

(9) ∫s A•dS = – ∫s[(GM)/R2]er•R2er

= – GM ∫serer

= – GM ∫sdΩ ,(er2=1)


(10) ∫s A•dS = –4πGM

But the mass can be written in terms of the density,

(11) M = ∫v ρdV

Therefore equation (10) becomes,

(12) ∫s A•dS = –4πG∫v ρdV

Using Gauss Law , equation (6),

(13) ∫s A•dS = ∫v A dV = –4πG∫v ρdV


(14) ∇•A = – 4πGρ

Define the Gravitational Potential:

(15) A = –∇φ

We get,

(16) ∇2φ = 4πGρ

Note that Gauss theorm allows us to say that a particle is attracted as if all the other matter was concentrated at the center of a sphere.

Monday, May 20, 2013

Mach–Zehnder interferometer
Particle or Wave?

This is a continuation of what we have covered in the two-slit experiment. If you have forgotten the main concepts, kindly review them.

We will try to shed some light on what makes up light: particles or waves.

In papers written about this subject, such as of the Mach-Zehnder interferometer, you will often read: "The explanation for this result is that it appears a single photon travels both paths and engages in interference with itself." Language fails us when we are trying to describe a QM system. However, the language of mathematics can compensate.

So first, a few words on the devise in question we will use for this blog. The Mach-Zehnder interferometer is made of a collimated beam which is split by a half-silvered mirror or a beam splitter. The two resulting beams are each reflected by a mirror. The two beams then pass a second beam splitter and enter a number of detectors - in figure 1(a),(b), there are two detectors A and B; in figure 1(c) there three detectors, C being the additional one.

Fig 1

The enigma is in figure 1(a). So before we tackle this situation, we will look at figure 1(c), reproduced below.

Fig 1(C)

To keep track of the different beams, we use E, for beaming going East, S for south, N for north, etc. The incoming beam is labelled I, the outgoing one, O. (See figure 3, below)

After the incoming wave goes through the split beam, (top left-hand corner) we have.

| I > → |E> + |S>

The East going state will reflect at the mirror and then go South, to undergo another split at the bottom right-hand corner, and then go the detector A and B. We can write this whole sequence as,

|E> → |S> → 2-1/2(a|A> + b|B>)

Here the constants a and b are needed as we must normalize the state vector. This is another reason why we shouldn't take the wavefunction in QM as real. A better language is the state vector using the Dirac notation as we are doing. We normalize it because the state vectors in QM allow us to calculate probabilities, and a simple mathematical rule is that the sum of all probabilities in a given situation adds up to 1. So we need to put another constant for the South going state, which will be reflected by the bottom left-hand mirror, to go Eastward.

|S> → |E> → c|C>

Therefore, | I > = 2-1/2(a|A> + b|B>) + c|C>


< I | I > = a2/2 + b2/2 + c2 = 1

By symmetry, we can choose: a = b

a2 + c2 = 1

Giving us a = b = c = 2-1/2

Putting this altogether:

| I > = 2-1|A> + 2-1|B> + 2-1/2|C>

A little calculation will give:

Probability at A is the square of the amplitude:

P(A) = │< I | A >│2 = 1/4.

Similarly, P(B) = 1/4 and P(C) = 1/2

Particle or Wave

Fig 2

At the heart of the mystery - particle or wave - is what's happening in figure 1(a), reproduced above in figure 2. Keep in mind that the light intensity has been reduced so that only one photon at a time travels through the apparatus. What is seemingly unexplanable are the numbers, which read as: 0% in detector A, and 100% in detector B. Even if the single photon would interfer constructively with itself at the beam-splitter(bottom right-hand corner), after that, each wave should continue along their path, and 50% should go to A, 50% to B. But we don't get that, instead we get 100% at B, none at A. That would mean the wave coming from the southern branch moving from left to right, goes 100% (of 50%)through the beam-splitter(bottom right-hand corner), IOW, doesn't split and continues to B (depicted in fig 1b), while the northern branch, which is now moving downward after reflection with the mirror(top right-hand corner), but now makes an extraordinary 90 degree turn at the beam-splitter (bottom right-hand corner), without splitting to move 100% (of 50%) from left to right to reach B (depicted in fig 1a)!!!

This is where ordinary language fails us, and we must concentrate on the mathematical framework on which QM is built.

In terms of our state vectors, we can depict the situation in figure 2, with figure 3, below.

Fig 3

We notice that this is very similar to ordinary vectors, depicted below.

Fig 4

We have (we will not normalize because it is redundant as you will see),

| I > → |E> + |S> , Figure 4a

| E > → |S> , at the top right-hand mirror

| S > → |E> , at the bottom left-hand mirror


| I > → |S> + |E>

And at the second splitter,

|S> + |E> → | O > , Figure 4b


| I > = | O >

All the particles will end up at detector B.

By symmetry, if we run the experiment backward:

| O > → |W> + |N> → |N> + |W> → | I >

This is a testimony of the power of the mathematical framework of Quantum Mechanics. Notice: we've obtained the correct result with not a word on what is at stakes, particles or waves! The only assumptions we've made is that quantum states are vectors (in a Hilbert space) and obey the same mathematical rules as ordinary vectors, and that the probability is equal to the square of the amplitude.

Sunday, May 05, 2013

Spooky Action at a Distance and Bell's Theorem Revisited

You have a theorem based on two assumptions:

1. Assumption A (logic)
2. Assumption B (locality)

You design an experiment which violates the theorem. ( Hint: it's a quantum system)

What can you conclude?

Either A is false, or B is false, or both.

But we know that A is false on account that Bell used a mathematical framework for classical system. In classical physics, we need to know the position and momentum in order to know everything about the particle, and these are points in phase space. So set theory, points in set theory to represent position and momentum, and Boolean algebra is the right mathematical framework for classical physics. Call that mathematical framework classical logic. So from that, we can say that Bell's theorem tells us that what we have is a classical system.

But for a quantum system, we need to represent the state of a particle by a vector in Hilbert space, and observables by operators acting on those states, not points from set theory. This is a totally different framework than a classical system. Call that quantum logic. So Bell's theorem applied to a quantum system is not going to work. Violations of Bell's theorem does not prove non-locality, or what was called spooky action at a distance.

Going back to our two assumptions:if A is false, we cannot conclude that B is false. Therefore we don't have any conclusive proof of non-locality.

Thought Experiment

You have to understand that Bell's theorem applies to classical system.

Now, you devise an experiment. You look at your results and they don't fit with that theorem.

You wonder why. You study that theorem carefully and you find that it is based on two assumptions. You don't know which one is false. Is it A, is it B, is it both?

Then some smart physicist called Susskind comes to you, and say, listen, the first assumption is wrong.

You ask why. He demonstrates. The logic applies to classical physics, and then points out that your experiment is about a quantum system.

Would you deduct from this that assumption B is true or false? I think not.

What I'm saying is if you want to investigate whether or not non-locality is a fundamental feature of the universe, you need to forget about Bell's theorem. You are going to need another yardstick from which you can design an experiment that will allow you to investigate that issue.

One More Argument

In Bell's theorem:

We make two assumptions in the proof. These are:

A. Logic is a valid way to reason.
B. Parameters exist whether they are measured or not. For example, when we collected the terms Number(A, not B, not C) + Number(A, B, not C) to get Number(A, not C), we assumed that either not B or B is true for every member.

Consider any measurements A, B and C.

Classical system:

You use Boolean algebra, based on set theory, you get (Bell's inequality):

(1) Number(A, not B) + Number(B, not C) is greater than or equal to Number(A, not C)

Quantum system:

You use a different mathematical framework, in which states are vectors in a Hilbert Space, and you get (violations of Bell's inequality):

(2) Number(A, not B) + Number(B, not C) is not greater than or equal to Number(A, not C)

What you need to realize is that CLASSICAL logic was used to derive Bell's inequality. A quantum system violating Bell's inequality can only mean that a classical system is different than a quantum system, which is based on a different kind of logic, quantum logic.