Wednesday, October 23, 2013

Description of Reality - The EPR Paper Revisited



In a complete theory there is an element corresponding to each element of reality. A sufficient condition for the reality of a physical quantity is the possibility of predicting it with certainty, without disturbing the system. In quantum mechanics in the case of two physical quantities described by non-commuting operators, the knowledge of one precludes the knowledge of the other. Then either (1) the description of reality given by the wave function in quantum mechanics is not complete or (2) these two quantities cannot have simultaneous reality. Consideration of the problem of making predictions concerning a system on the basis of measurements made on another system that had previously interacted with it leads to the result that if (1) is false then (2) is also false. One is thus led to conclude that the description of reality as given by a wave function is not complete.

If you are not familiar with this quotation, it is taken directly from the original EPR paper (See: Can Quantum-Mechanical Description of Physical Reality Be Considered Complete' ? A.EINSTEIN, B.PODOLSKY AND N.ROSEN, Institute for Advanced Study, Princeton, New Jersey (Received March 25, 1935))

There are two distinct errors in the logic to the pre-amble of one of the most famous scientific papers ever written. In the statement, If P then Q, one generally accepts that the premise P implies the conclusion Q. We are making the allusion to the following sentences:

In quantum mechanics in the case of two physical quantities described by non-commuting operators, the knowledge of one precludes the knowledge of the other. Then either (1) the description of reality given by the wave function in quantum mechanics is not complete or (2) these two quantities cannot have simultaneous reality.

(1) Knowledge is one thing, reality is something else. Even Einstein famously quoted, “Do you really believe that the moon isn't there when nobody looks?” Not knowing the position ( or momentum) of a particle does not deny its existence.

(2) Not being able to measure two things simultaneously does not deny the existence of either thing.

Yet it is this erroneous thinking that leads the EPR to conclude that QM is “incomplete”.

The beauty of the Heisenberg Uncertainty Principle (HUP) is that it conveys quite accurately the wave/particle duality. In classical physics, all things were placed into two different bins: (1) particles - those things that bounce off each other; (2) waves - those things that went through each other, amplitudes interfering constructively/destructively, but then went on their way as if nothing had happened. In the subatomic, reluctantly we found objects that don't fit into this framework - hence the need of a different framework that went under the name of Quantum Mechanics (QM).

But the HUP is at the core of this alternative framework.

Δx Δp ≈ ℏ

If we measure a particle, Δx → 0 then Δp →∞, we don’t know where the particle is going.

If we measure with certainty its momentum Δp → 0 then Δx →∞, which is what a wave is.

The best picture we get from QM of these objects is a wave packet for which there is some uncertainty in both the position and momentum.



Quantum Field Theory (QFT)

This is borne out by QFT, in which the theory gives predominance to fields, and particles are regarded as excited states of the fields. To make the theory work, particles are conceived as things being surrounded by a cloud of particles/anti-particles, sometimes interacting with themselves, sometimes with other fields or particles, and sometimes even with the quantum fluctuations of the vacuum. Indeed, these are complicated objects.



History

Historically speaking, the word “incomplete” as used in the EPR paper was technically correct, but it was interpreted incorrectly. If QFT is the house, then QM is the basement. It is the foundation on which QFT was built. However, there were those persuaded by the EPR paper who were drawn to conclude that QM gave the wrong description of reality and an alternative, new theory had to be developed. These attempts may have been brave but led to nowhere. In the meantime, QFT was being built (1930-1970’s), which would complete the house. The irony is that this was accomplished by incorporating into QFT, Special Relativity – Einstein’s own theory that had stunned the world in 1905.

Sunday, July 07, 2013

The Unruh Effect



In this blog, ℏ=c=kb = 1 where ℏ is the reduced Planck constant, c is the speed of light and kb is Boltzmann’s constant.

Preliminary:

In the early 1970’s there were several applications of QFT to GR, the most famous one was done by Hawking in which he derived that a Black Hole has entropy proportional to its area. A less famous, but equally important, is the Unruh Effect. What’s interesting is that this has certain parallel with Einstein derivation of E = mc2. In both cases, there is the use of two observers; in Einstein case, one is in motion relative to the other; in the Unruh case, one is accelerating with respect to the other. Secondly, Einstein calculated the total and kinetic energy for each of the observer, and concluded the only thing that made sense was that a decaying particle must lose mass to the equivalent of E = mc2. (See: Einstein's Derivation of the Famous Equation, E=mc2) On the other hand, Unruh calculated the vacuum energy of each observer, and concluded that the accelerating vacuum energy would radiate as a black body, and its temperature would be proportional to its acceleration.

The Unruh Effect

The standard metric in Special Relativity is (see equation (6) in Relativistic Doppler Effect ),

(1) ds2 = -dt2 + dx2

The light-cone coordinates are defined as,

(2) u = – t + x, v = t + x

Taking the differential,

(3) du = – dt + dx, dv = dt + dx

The metric becomes,

(4) ds2 = –dt2 + dx2 = dudv

Now consider two observers: Alice, who is in an accelerating frame; and Bob, who is at rest or in uniform motion.

We will write the velocity, v = dxμ/dτ = x'μ, and the acceleration, aμ = x''μ

Consider the square of the acceleration,

(5) a2 = aμ aμ = – (t'')2 + (x'')2

A solution to that equation is,

(6) t = (1/a ) sinh(aτ), x = (1/a ) cosh(aτ).



We see that in fig.1 both Alice and Bob are space-like, and Alice lies on a hyperbola.

The light-cone coordinates, equation (2), becomes,

(7) u = (1/a)e-aτ, v =(1/a)e

We want to find a coordinate system in which Alice is going nowhere, that is,

(8) her time coordinate is ξ0= τ, and her spatial one is ξ1= 0.

In this new coordinate system, the light-cone coordinates would be,

(9) U = ξ1 – ξ0 = – τ, and V = ξ1 + ξ0 = τ

Substituting equation (9), (8) into (7), we get,

(10) u = (1/a)eaU, v =(1/a)eaV

Note that,

(11) U = a-1ln(ua) and V = a-1ln(va)

Taking the derivative of equations (10),

(12) du = eaUdU, dv = eaVdV

Now the metric, equation (4), takes on the form,

(13) ds2 = dudv = ea(U+V)dUdV

Both observers will write down their action according to the coordinates they live in.

(14) For Bob, sBob = ∫ ∂uφ∂vφdudv;
for Alice, sAlice = ∫ ∂Uφ∂VφdUdV

The equation of motion is the well-known wave equation,

(15) For Bob, ∂uvφ = 0; for Alice, ∂UVφ = 0

Since both are similar equations, we will write the solution for one, keeping in mind that the other is just a matter of changing, u → U, and v →V. So the solution can be written as,

(16) φ = A(u) + B(v), where A(u) are the left-moving waves and B(v) are the right-moving waves.

In terms of QFT, these solutions are written as,

(17) For Bob, φ Bob = ∫ dω(2ω) (a−Lωe−iωu + a+Lωeiωu) + ∫ dω(2ω)(a−Rωe−iωv + a+Rωeiωv)

(18) For Alice, φ Alice = ∫ dΩ(2Ω)(b−LΩe−iΩU + b+LΩeiΩU) + ∫dΩ(2Ω)(b−RΩe−iΩV + b+RΩeiΩV)

• Where (2ω) and (2Ω) are normalizing factors.

• a−Lω and a−Rω are Bob's annihilation operators; b−LΩ and b−RΩ are Alice's annihilation operators.

• a+Lω and a+Rω are Bob's creation operators; b+LΩ and b+RΩ are Alice's creation operators.

Again, because the solution is symmetric between left-moving waves and right-moving waves, we need to focus on one of them. Thus, we will use for both observers, the left-moving waves and drop the L and R index. That is,

(19) a−Lω → aω , b−LΩ → bΩ.

Both observers will define their vacuum state as such,

(20) aω |0>Bob = 0; bΩ |0>Alice = 0

The question is: are these two vacua the same?

Let’s assume we can write,

(21)bΩ = ∫ dω (αΩωaω + βΩω a+ω), and
b+Ω = ∫ dω (α*Ωω a+ω − β*Ωω aω)

Substitute equation (21) into equation (18), again keeping only the left-moving waves, and the annihilation operator terms, that is, we want the aω coefficient which is,

(22) ∫dΩ(2Ω)∫dω(αΩω e−iΩU − β*ΩωeiΩU)

For Bob, from equation (17), the coefficient for aω is

(23) ∫dω (2ω)e−iωu

For these two to match, we need,

(24) (2ω)e−iωu = ∫dΩ' (2Ω')Ω'ωe−iΩ'U − β*Ω'ωeiΩ'U)

Multiply both sides by eiΩU and then integrate over all U,

(25) ∫(2ω)e−iωu eiΩUdU = ∫dΩ'(2Ω')∫dU(αΩ'ωe−iΩ'UeiΩU − β*Ω'ωeiΩ'UeiΩU)

On the right-hand side, the first term inside the bracket is just the definition of the Dirac delta function,

(26) ∫dU e−iΩ'U eiΩU = ∫dU ei(Ω − Ω')U = 2π δ(Ω −Ω')

And the second term vanishes at infinity. The net result is,

(27) αΩω = (2π)-1(Ω)½(ω) ∫dUe-iωueiΩU

Similarly, we get,

(28) βΩω =(−1)(2π)-1(Ω)½(ω) ∫dUe+iωueiΩU

Using equation (11), which is repeated below,

(29) U = a-1ln(ua)

(30) αΩω = (2π)-1(Ω)½(ω) ∫(du/u)e-iωuei(Ω/a)lnua

(31) βΩω =(−1)(2π)-1(Ω)½(ω) ∫(du/u)e+iωuei(Ω/a)lnua

Now we don’t need to evaluate these integrals to find the relationship between αΩω and βΩω. We note that there are very similar, and we get our answer by using the transformation u → -u in equation (31),

(32)βΩω =(−1)(2π)-1(Ω)½(ω) ∫(du/u)e-iωuei(Ω/a)ln(-ua)

We need to look at the last term in the integral,

(33) ei(Ω/a)ln(-ua) = ei(Ω/a)ln(ua)ei(Ω/a)ln(-1)

Recall Euler’s famous equation:

(34) e = -1 → iπ = ln(-1). Therefore,

(35) ei(Ω/a)ln(-ua) = ei(Ω/a)ln(ua)ei(Ω/a)ln(-1) = ei(Ω/a)ln(ua)e-(πΩ/a)

Putting this back into equation (32),

(36)βΩω =(−1)(2π)-1(Ω)½(ω) ∫{(du/u)e-iωu
x ei(Ω/a)ln(ua)e-(πΩ/a)}

Now comparing (36) and (30), we get

(37) βΩω = − αΩωe-(πΩ/a)

Squaring both sides and dropping the indices,

(38) |β|2 = |α|2e-2π(Ω/a)

This is where Unruh noticed that this result is the same as for a black body radiating at a temperature T.

More specifically, the density of states between any two state m and n is,

(39) ρmn = <ψm|e-H/T| ψn> = |α|2 e-Em/T, where H is the Hamiltonian, and E is the energy eigenvalue of H.

Also,

(40) ρnm = <ψn|e-H/T| ψm> = |β|2 e-En/T

The density state is symmetric under the indices. That is,

(41) ρmn = ρnm

We get,

(42) |β|2 = |α|2e-(En - Em)/T = |α|2e-E/T

But from the Einstein equation,

(43) E = ℏΩ = Ω (ℏ = 1)

Therefore,

(44) |β|2 = |α|2e-Ω/T

Comparing the above with equation (38), we get,

(45) T = a/(2π)

Now putting back the constants ℏ,c, and kb to make the equation dimensionally consistent,

(46) T = aℏ/(2πckb)

An accelerating detector would see particles in a thermal bath. This is the Unruh effect.

We know that the action of accelerating electrons gives off electromagnetic radiation, what we need to transmit to TV’s, satellites, phones, etc. As for the Unruh effect, a typical acceleration of 10m/s2, (earth’s acceleration due to gravity), would yield a temperature of the order, T ~ 10-20 K, or to boil water at 373K, we would need to accelerate it at a ~ 1022 m/s2, both cases being beyond our present technology.

Monday, June 10, 2013

Relativistic Doppler Effect

One of the assumption of Special Relativity is that the velocity of light is constant in every inertial frame of reference. Consider two frames: one at rest (t,x,y,z); and a second one (t',x',y',z') is in motion with velocity V with respect to the first one. We've omitted the z- and z'-axis as representation of 4 dimensions on a 2-D surface is not possible.



Now suppose a light was turn on, and our two observers would be looking at this beam of light spreading throughout space as a sphere. (See fig above)

In the rest frame we have:

(1) x2 + y2 + z2 = (ct)2

Similarly in the moving frame:

(2) x'2 + y'2 + z'2 = (ct')2

Rewriting both equations (1) and (2) as:

(3) -(ct)2 + x2 + y2 + z2 = 0

(4) -(ct')2 + x'2 + y'2 + z'2 = 0

Since both expressions are equal to zero, they are equal to each other. But more fundamental, these two expressions are suggesting that they represent the same fundamental quantity in two different frames. That there are equal means something doesn't change from one frame to the other. Let's represent this quantity as:

(5) s2 = -(ct)2 + x2 + y2 + z2

= -(ct')2 + x'2 + y'2 + z'2

To keep in line with convention, we will express these quantities as small differences. We use the symbol d, which is common use in calculus. We also set c =1, as it is a constant and we can always put it back if we need to do a calculation.

(6) ds2 = -dt2 + dx2 + dy2 + dz2

We define the metric as the coefficient of each of the terms in the above:

(7) η00 = -1, η11 = 1,η22 = 1,η33 = 1,and ηij = 0 for i≠j

We also define the proper time τ as,

(8) dτ2 = - ds2

= dt2 - (dx2 + dy2 + dz2)

= dt2 ( 1 - (dx2/dt2 + dy2/dt2 + dz2/dt2))

= dt2 ( 1 - v2)

Where we have use the definition of velocity,
v = (dx/dt,dy/dt,dz/dt).

Taking the square root on both sides,

(9) dτ = dt/γ

(10) where γ = (1 - v2)

Now, in Newtonian physics, it was assumed that time was different from space. But in Relativity, this separation cannot be upheld any longer. We can see in the definition of the proper time, that both time and space form one manifold. So we need to redefine our quantities with this new perspective.

Convention: we use latin indices for 3-dimensional objects. For instance, the velocity, vi, where i = 1,2,3. So a velocity's components which were written as v = (vx,vy,vz), now will be written as v = (v1,v2,v3).

We use Greek letters (α,β,γ,δ...) to denote objects with 4 components. They will take values 0,1,2,3 for t,x,y,z. So, for example, uβ =(u0,v1,v2,v3). Sometimes, we want to break up the components as temporal and spatial. So we write,uβ =(u0,vi), where it is understood, i = 1,2,3, or uβ =(u0,v)

We will measure the velocity with respect to the proper time τ, not the ordinary time t.

(11) uβ =dxβ/dτ

Using the chain rule and equation (10),

(12) uβ = (dt/dτ)dxβ/dt = γ(dt/dt,dx1/dt,dx2/dt,dx3/dt) = γ(1,v1,v2,v3) = γ(1,v)= (γ,γv) .

The dot product between two vectors is now defined with the metric (see equation (7)),

(13) u2 = u•u = ηαβuαuβ

Note that in u2, the 2 means squaring, u2 = u squared, not the 2nd component of u. When there's confusion, we will point out what is meant by an upper index.

Expanding the above into the temporal and spatial components, and using the metric in (7),

(14) u•u = η00u0u0 + ηijuiuj
= (-1)γ2 + γ2v2 = (-1)γ2 ( 1 - v2)

But from (10),

(15) γ2 = (1 - v2)-1

Therefore,

(16) u2 = u•u = -1

Energy and Momentum

Momentum is defined as mass x velocity,

(17) pβ = muβ

Similarly, we define a 4-vector momentum as,

(18) pβ =(p0,pi) = (p0,p)

An important result is to calculate p2, where the 2 means squaring, not the component 2. First using equation (17)

(19)p2 = muβmuβ = m2u2 = - m2

Then using equation (18) and the metric in (7),

(20) p2 = p•p = η00p0p0 + ηijpipj
= (-1)(p0)2 + (p)2

We define p0 = E/c = E , (using the convention, c=1). Putting this altogether, we get,

(21) p2 = - m2 = (-1)E2 + (p)2

Or,

(22) E2 = m2 + (p)2.

Putting c into the equation,

(23) E2 = m2c4 + p2c2.

Notice when the particle is at rest, p = 0, and we get, E = mc2.

Relativistic Doppler Effect

In the last blog, Einstein's Derivation of the Famous Equation, E=mc2 , we were given the energy of the photon as,

(24) γB+ = ½ E(1 + (V/c)cosΦ)(1 – V2/c2) for the incoming photon

γB- = ½ E(1 - (V/c)cosΦ)(1 – V2/c2) for the outgoing photon



The energy is,

(25) Etotal = - pβ uβ = - (η00p0u0 + ηijpiuj)

= - (-1)Eγ - (pcosθ, psinθ, 0)(-γV,0,0)

= Eγ + (pcosθ)γV = Eγ(1 + (V/c)cosθ),

Where the last step , we use p = E/c.

Each photon released will carry half the energy in opposite direction. For the incoming photon,

(26) γB+ = ½ Etotal = ½ E γ(1 + (V/c)cosΦ)
= ½ E(1 + (V/c)cosΦ)(1 – V2/c2)

Where we use equation (10) in the last step.

For the outgoing photon, we get an extra minus sign in the momentum,



(27) pi = (-pcosθ, -psinθ, 0)

Giving,

(28)γB- = ½ E(1 - (V/c)cosΦ)(1 – V2/c2)

Notice that for the energy of the incoming photon is greater than the energy of the outgoing photon. This is the Doppler Effect, as a light coming towards you will appear blueshifted, while a photon moving away from you will be redshifted.



Appendix

From here on, the speed of light is c.

We will use the following conventions:

(A1) β = v/c

(A2) γ = (1 - v2/c2) = (1 - β2)

For a wave, it is proportional to ei(k∙x - ωt),where,

(A3) k = |k| = 2π/λ, ω = 2πf, and ω = kc

Just like we can define a 4-vector position, xμ = (ct,xi), we can also define a 4-vector wavenumber, kμ = (ω/c,ki). The phase factor, (k∙x - ωt), can now be written as,

(A4) k∙x - ωt = ημνxμkν, where ημν is the Minkowski metric tensor with signature (-1,1,1,1).

The importance of the phase factor, which basically counts the number of peaks and troughs of the wave, must be frame-independent, that is, a Lorentz scalar.

(A5) kμ → k'μ = Lμνkν

Where Lμν is the Lorentz transformation (See diagram below).



In particular, under a Lorentz boost in the +x direction,

(A6) k'x = γ(kx - βω/c)

(A7) ω'x = γ(ω - βckx)
= γ(ω - βckcosθ)


Where θ is the angle between the boost direction in the +x direction and the direction of the wave propagation k.

Using ω = kc (equ. A3), and γ = (1 - β2) (equ. A2) then equation A7 becomes,

(A8) ω' = γ(ω - βckcosθ) = ω(1 - βcosθ)(1 - β2)

Note: if θ = π/2, then ω' = 0. There is no Doppler shift in the transverse direction.
if θ = 0, then ω'/ω = [(1- β)/(1+β)]½
For low velocity, v < c, then

(A9) ω'/ω ≈ (1- β)

Or,

(A10) Δω'/ω ≈ -v/c

Saturday, June 01, 2013

Einstein's Derivation of the Famous Equation, E=mc2

Here’s a quick rundown on one of the most famous equation in physics, E = mc2. Einstein knew from experiments previously done that a particle could decay and release gamma rays. He reasoned that when this happened, the particle would lose kinetic energy, and this could only be accounted by a loss of mass. So how did he come to that conclusion? He analysed the situation both in a rest frame and in a moving frame.



The value of γ was already known, but we can derive it from Relativistic Doppler Effect (to be the topic for another blog A).

By the law of conservation of energy:

The energy before decay = the energy after decay

(1) In the rest frame:

A0 + KA0 = A1 + KA1 + ½ E + ½E = A1 + KA1 + E

(2) In the moving frame:
B0 + KB0 = B1 + KB1 + ½ E(1 – (v/c)cosΦ)(1 – v2/c2)
+ ½ E(1 + (v/c)cosΦ)(1 – v2/c2)

= B1 + KB1 + E(1 – v2 /c2)

Now taking a look at the energy difference of the particle in each of the frame:

(3) In the rest frame:

A1 – A0 = KA0 – E – KA1 = – ΔKA – E

(4) In the moving frame:

B1 – B0 = KB0 – KB1 – E(1 – v2 /c2)
= – ΔKB – E(1 – v2 /c2)

Whether the observer is at rest or moving with respect to the particle, the energy difference should be the same.

(5) – ΔKA – E = – ΔKB – E(1 – v2 /c2)

Calculating the difference in kinetic energy:

((6) ΔK = ΔKA – ΔKB = E(1 – v2 /c2) – E

= E ((1 – v2/c2) – 1)

≈ E ((1 + ½v2/c2) – 1)

= ½ E (v2/c2)

Einstein had reasoned that if the kinetic energy of the particle is smaller by ½ E (v2/c2), the only way this can happen is that the particle must lose mass when emitting radiation.

By definition the kinetic energy is,

(7) K = ½ mv2

If the particle was initially at rest,

(8) ΔK = K = ½ mv2

Equating (6) and (8)

½ mv2 = ½ E (v2/c2)

(9) Therefore E = mc2.

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)

Or

(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

Therefore,

(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]

But

(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)

Therefore,

(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

Therefore,

(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>

Normalization

< 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

Therefore,

| I > → |S> + |E>

And at the second splitter,

|S> + |E> → | O > , Figure 4b

Therefore,

| 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.

Wednesday, March 13, 2013

The Universe Was Never a Singularity

At any given time, the radius of the universe was greater than the Schwarzschild radius. Otherwise, the universe would have collapsed into a black hole. The Schwarzschild radius can only be zero if and only if the universe has no mass and no energy. Since the mass/energy of the universe was never zero, its Schwarzschild radius had to be greater than zero, and the radius of the universe, being greater at all times than its Schwarschild radius, could have never been zero. Therefore, the universe was never a singularity.