Wednesday, December 21, 2011

Entropy, information and gravity

Recall that the overall entropy of the universe is always increasing. You could have an "isolated system", like the earth for instance, in which the entropy could decrease, but when you enlarge that system, earth + sun, the total entropy is increasing. So looking at the earth by itself is really not a good idea in defining what is considered to be an isolated system -- a common mistake people make.

Now, if we take a system in which gravity plays no role, then generally speaking, an increase in disorder corresponds to an increase in entropy. For instance, an ice cube left on the table will melt to water and then evaporate as water vapour.

Ice (high order or low disorder, low temperature) → water vapor ( high disorder, high temperature).

We see from that equation, that going from low temperature to high temperature, entropy increases.

However, when gravity is factored in, the reverse is true. For instance, at the Big Bang, matter was at a very high temperature. You would think that like the water vapour, the universe had high disorder. But it didn't. We will show below that such a system has low entropy (read low disorder) and as it expands, its temperature decreases and its disorder increases.

Universe at Big Bang (high order or low disorder, high temperature) → Universe at present times (high disorder, low temperature).

Here, we go from high temperature to low temperature, and entropy increases.

Black holes -- as well as any star, our sun included -- are like the universe. They are in a state of low disorder. With Hawking radiation -- our sun radiates mainly electromagnetic radiation -- they are moving from low disorder to high disorder, that is, entropy is increasing.

Modern theory looks at entropy as the lack of information. So you can substitute in the above equation, the following:

low disorder (high order) → high information

high disorder (low order) → low information.

As the universe is expanding, our knowledge of the universe is decreasing -- entropy (or lack of information) is increasing, that is, we are becoming more and more ignorant. At first sight that might sound paradoxical, but think of it this way, as the universe grows bigger and bigger, it is more difficult for us to keep track of every particle in the universe. In fact many galaxies are so far way, that the light from them will never reach us. And this situation gets only worse as time goes by.

Gravity

When an object is orbiting around a larger one (say a satellite around the earth, for instance) and changes its orbit, it will do that by emitting away some energy. The result is that it will move to a smaller orbit, but its speed will increase. Hence its kinetic energy has increased, even though it radiated some energy away. The reason being the case is that the gravitational potential energy is negative.











E = K.E. + P.E.

E = ½ mv2 – GMm/r

So while the kinetic energy increases, the radius being smaller, the potential has become more negative. In absolute value, the change in potential energy is twice the change in kinetic energy. So whatever gain in kinetic energy, there is twice the loss of potential energy.

ΔE = Δ K.E. + ΔP.E.

ΔE = Δ K.E. – 2 Δ K.E.

= – Δ K.E.

But K.E. = NT, where T is the temperature, and N is any positive constant.

So we must have

ΔT = -ΔE /N

But from the definition of specific heat (c),

ΔE = cmΔT, where m is the mass.

This means that gravitational bound system would have a negative specific heat!!!

To rescue this situation, we must conclude that entropy increases when the system decreases in temperature. The less energy the system has, the higher the temperature. This feature of gravitationally bound systems makes them tricky. Only systems with positive specific heat can be in thermal equilibrium with their environment. So gravitationally bound systems can never be in thermal equilibrium with their environment! They always want to keep shrinking, thus losing energy and, by the 2nd of law of thermodynamics, they increase the entropy. In the satellite-earth system, the satellite will emit energy as its orbit degrades into a smaller one.

NOTE: One might think, is the earth orbit shrinking and will it crash into the sun? The earth's orbit shrinks by 3.5×10−13 m per year. However, the effect on the size of the Earth's orbit is negligible over the age of the universe. Unfortunately, in the case for the many satellites that humans have launched and are orbiting the earth, as time goes by, most of them will have their orbit decaying in a much shorter time, and will meet their fatal fate by crashing into the earth.

Wednesday, September 28, 2011

Why FTL violates causality

Suppose we have two observers, one in relative motion with respect to the other. We can represent this in fig 1 with one of the observer at rest -- frame of reference in black -- while the second one in motion -- frame of reference in blue. The line cutting in the middle represents the speed of light, c = 1. In particular, from the point of view of a stationary observer, an observer moving at constant velocity has a coordinate frame whose space and time axes are “tilted” towards the light cone (blue).














Two events a and b are silmultaneous in the moving frame -- they both lie on the x' axis, that is, t' = 0. But for the stationary observer, these are read off at ta ≠ tb. Therefore these two events are not simultaneous. Welcome to the world of Einstein's theory of Special Relativity.

But now suppose that we have invented a device that can send signal at speed greater than c, and to make it easy for illustration purpose, we take that speed to be infinite, that is, sending a signal at P arrives instantaneously as Q. This is indicated by the red line in fig 2.











In fig 3, events P and Q are simultaneous in the stationary frame ( X-T frame), while events Q and R are simultaneous in the moving frame ( X' - T' frame). So one could send a signal to himself, which would arrive in his past!

















Or to put in another way, I would receive signals from my future self. This would violate causality.

Friday, September 23, 2011

Why massive particles can't travel faster than the speed of light

In the news, there is a team at CERN stating that they have measured the speed of neutrinos to be greater than the speed of light. What's wrong with that?

One of the fundamental concept in physics is Lorentz invariance. What this means is that if I apply the Lorentz transformation laws to an equation that I think is valid in one frame of reference, this equation would not changed in another frame of reference.

Now, in the Lorentz transformation laws, we get this factor:
γ =(1 - (v/c)2).

Should the speed of a particle be greater than the speed of light (v > c), then the γ factor becomes an imaginary number! This brings humongus headache to the theory. One way out is to postulate that the particle has an imaginary mass (tachyons) since this factor often multiplies the mass of the particle, and the product of two imaginary number is a real number. However in the real world, masses are real quantity, not imaginary, so particles with a mass must travel at a speed less than the speed of light, and only massless particles ( m = 0) can travel at the speed c.

In the case of the recent findings about the neutrinos traveling at a speed greater than light, that would contradict this theory which has been around for over 100 years. That's why the findings, if confirmed, would be disturbing, to say the least.

Gauge Theory and Higgs Mechanism

HERE'S A GENERAL RUNDOWN OF THE THEORY:

(1) in QM: x → operator
But to satisfy Relativity, in which time is on an equal footing with space, in QFT: x → parameter, and Φ(x) → operator. Now Φ(x), a function of x, is called the “field”.

(2) L = T – V. The Lagrangian plays an important role. From Noether’s theorem, we know that if the Lagragian is invariant under a symmetry, this symmetry points to a conservation law.

Corresponding to L there is a Hamiltonian, H = T + V. The Hamiltonian is known to measure the energy of a system.

(3)In classical mechanics, let v = dx/dt, then L = ½ mv2 – V(x). The corresponding Hamiltonian is, H = ½ mv2 + V(x). Quantizing this, (ℏ =1),we get the Schroedinger equation:

i∂Ψ(x)/∂t =( -½m∆2 + V(x))Ψ(x).

(4) In Relativity, the energy equation is:

E2= p2c2 + m2c4.

Quantizing this, (c =1) yields the K-G equation:

½(∂μΦ)(∂μΦ) + ½mΦ2 = 0.

From this, the Lagrangian can be deduced as:

L = ½ (∂μΦ) 2 – ½mΦ2.


(5) In QFT, the general Lagrangian is:

L = ½ (∂μΦ) 2 – V(Φ).

(6) Comparing (5) and (4), if V(Φ) contains any terms with Φ 2, its coefficient is taken to be the mass of the field quanta (particles).

Gauge theory:

From electromagnetism, it was known that Maxwell’s equations were gauge invariant. In QM, gauge invariance of the Lagragian involves three important steps:

(7) the wave function is transformed as Φ → eiqXΦ
(8) the operator ∂μ → ∂μ + iqAμ
(9) the electromagnetic field Aμ → Aμ - ∂μX

(10) In QED, in equation (5), V(Φ) → - ¼ Fμν Fμν,
where Fμν = ∂μAν - ∂νAμ

If you apply, 7,8,9,10 to equation (5), you get the invariance of the Lagrangian under gauge transformation, in which the photon mediates the electromagnetic force. Note that the photon has no mass.

In the weak force, the bosons involved have mass, and one had to figure out how to include a mass term, keeping the Lagrangian gauge invariant.

There is where number (6) comes into play under the notion of SPONTANEOUS SYMMETRY BREAKING.

Higgs Mechanism:

Basically, I will only look at U(1) symmetry. Electroweak interactions need a U(1) x SU(2) symmetry, but SU(2) requires 2 by 2 matrices, and the software on this forum is inadequate to deal with matrices. But you can get the flavor just by doing U(1) symmetry and how mass is introduced in the Lagrangian of equation (5).

I will rewrite this equation as:

(11) L = ∂μΦμΦ - ¼ Fμν Fμν – V(ΦΦ).

(12) where V(ΦΦ) = (m2)/(2φ2) {ΦΦ - φ2} 2

Three important things to note:

(13) The field Φ is now a complex number, denoted by (Φ1, Φ2) or Φ = Φ1 + iΦ2 ( i being the imaginary number, square root of – 1), and Φ = Φ1 – iΦ2.

(14) the minimum field energy is obtained when ΦΦ = φ2.

(15) The number of possible vacuum states is infinite. We break this symmetry by requiring that Φ is real, we take the vacuum state to be (φ,0), and expand:

Φ = φ + (½ ½)h

Substituting 7,8,9, 12, and 15 into 11, we get

(17) L = {(∂μ - iqAμ)( φ + (½ ½)h)}{( ∂μ + iqAμ)( φ + (½ ½)h} - ¼ Fμν Fμν - (m2)/(2φ2) {2½φh + ½h2}2

After calculating the Lagrangian, we separate it into two parts:

(18) L = Lfree + Lint

where

(19) Lfree = ½∂μh∂μh - m2h2 - ¼ Fμν Fμν + q2φ2AμAμ

All the remaining terms are lumped into Lint, which offer no interest.

So, we can see that by breaking the symmetry, we end up with two massive particles. In equation 19, the second term refers to a scalar particle with mass equal to 2½m, associated with h (the higgs field) and the fourth term, a vector boson with mass 2½qφ, associated with Aμ( the electromagnetic field).

NOTE: in the Weinberg electroweak theory, equation 19 would have three extra terms for the vector boson instead of a single term, each one was identified with the W+, W -, and Z bosons, with 2x2 matrices that would be groups obeying the algebra under SU(2). This prediction, which was confirmed subsequently in the following years, earned Weinberg, Salam and Glashow the Nobel prize.

Monday, September 05, 2011

Twin Paradox



Two twins, one stays at home, while the other travels to the stars at speeds close to the speed of light. Upon her return, the traveling twin will look much younger than her stay-at-home twin, why?
















The twin who leaves and returns has the trajectory depicted in red-blue in fig 1. While for the twin staying on earth, the trajectory is in green. The diagram is deceiving as it gives the appearance that the traveling twin has covered more grounds. But remember that the “distance” on a space-time graph is really t2 – x2, which is the proper time. Moving clocks are slower, and therefore the traveling twin’s clock will register a smaller time than the stay-at-home twin.

We will assume that the traveling twin moves from a1 to a2 at a velocity v, and when she returns back, she moves with the same velocity, but in the opposite direction from a2 to a3. Therefore it takes an equal amount of time T for each leg of the trip. According to the stay-at-home twin, the round trip amounts to 2T. For the traveling twin, on the first leg of the trip, from a1 to a2,

Ta1 to Ta2 = T(1 – v2/c2)½

On the return trip,

Ta2 to Ta3 = T (1 – (–v)2/c2)½ = T(1 – v2/c2)½

Therefore the total trip = 2T(1 – v2/c2)½, and this is less than 2T.

How can we tell which twin moved and which twin was at rest? In this case, the traveling twin measures both events, departure and arrival, with the same clock on the first leg of the trip. The twin on earth would need two clocks to measure these two events. Ditto for the return trip. Hence the traveling twin's clock measures the proper time, and is designated as the moving clock. Moving clocks slow down.

Thursday, March 31, 2011

Zeta function is ok

This is a sequel to, Is String Theory wrong?.

Some refresher:

The zeta function is given by ζ(s) = Σ 1/ns, where s is any complex number. The earliest calculation of this function was made by Euler with s = 2

ζ(2) = Σ1/n2 = 1/12 + 1/22 + 1/32+ ... = π2/6.

Note: for s = -1

ζ(-1) = 1 + 2 + 3 +... = -1/12

This is what prodded me to investigate this sum, as mentioned in Is String Theory wrong?

In my investigation I came across the functional theorem:

ζ(s)= 2s π s-1sin(πs/2) ζ(1-s)Γ(1-s), where Γ is the well-known gamma function.

So if we let s = -1

ζ(-1)= 2-1 π -2sin(-π/2) ζ(2)Γ(2)
ζ(-1)= (1/2) (1/π 2)(-1) (π 2/6)(1) = -1/12

This yields the dreaded sum:

ζ(-1) = Σ 1/n-1 = Σ n = 1 + 2 + 3 +... = -1/12

There are other ways far too long to get this result, this being the shortest one. So I ask you to forgive my indulgence.

So what can I conclude? After reviewing complex analytic functions, the Cauchy-Riemann equations, complex integration and Cauchy's integral theorem, power series, the residue theorem and analytic continuation, I can only say that the mathematics is consistent.

Wednesday, February 02, 2011

Two-slit experiment

A second look at the two-slit experiment:

In Quantum Mechanics, two states can evolve, and we write this process as,

|A> → |A'>
|B> → |B'>

By the principle of linear superposition and unitarity, the superposition of these two states will also evolve as,

|A> + |B> → |A'> + |B'>

We label the position from which the electrons pass through as such: where they leave is 0; the slits are labelled +1 and –1; and the screen, S. The states representing the electron passing in one slit are |+1> and going through the second slit, |-1>. When the electrons leave position 0, from the symmetry of the setup, we can say that they arrive at position +1 and –1 with equal probability. So we write,

|0> → |+1> + |-1>

When an electron has arrived at +1 or -1, what happens after that when they hit the screen? Experiments show that they can land on any of the points on the screen, so we write,

|+1> → Σ Ψn|n>
|-1> → Σ Φn|n>




Fig. 1







The whole process can be described as,

|0> → |+1> + |-1> → Σ Ψn|n> + Σ Φn|n>

= Σ(Ψn + Φn) |n>

The probability that an electron will arrive at the mth point on the screen is,

Pm = (Ψ*m< m| + Φ*m< m|)(Ψm|m> + Φm|m>)
= Ψ*mΨm< m|m> + Φ*mΦm< m|m> + Ψ*mΦm< m|m> + Φ*mΨm< m|m>

Using < m|m> = 1

Pm = Ψ*mΨm + Φ*mΦm + Ψ*mΦm + Φ*mΨm

The first term Ψ*mΨm represents the probability if only the first slit was open. Similarly, the second term Φ*mΦm represents the probability if only the second slit was open. Classically, we should get the sum of these two terms if both slits were open. But we do not observe that. The interesting aspect of this result from quantum physics is that we get two extra terms, Ψ*mΦm and Φ*mΨm, that correctly explains the interference pattern of the double-slit experiment (see fig. 1). Another major difference between classical physics and quantum physics is that in the first, probabilities add, while in the second, the amplitudes add and then we square the amplitudes to get the probabilities.


The Act of Measuring and Entanglement

Suppose we want to know through which slit the electron has passed. This can be done by inserting a detector at position +1. Furthermore, we prepare the electron at position 0 with a down spin. When it passes through +1, its spin is flipped to an up spin, and when it passes through position –1, nothing happens to the electron (see fig. 2). With this setup, we know when the electron has passed through slit labeled +1.




Fig. 2






We need two labels for the states: one for position, and the other for the spin. We describe the process as,

|0,d> → |+1,u> + |-1,d> → Σ Ψn |n,u> + Σ Φn |n,d>

Due to the presence of the detector, the electrons are entangled through their spins: one is up, the other is down. Note: entanglement means that the system is prepared in such that if we know the spin of one of the particles, we automatically know the spin of the other particle. Again to calculate the probability of finding the electron at the mth position, we square the amplitudes, or multiply the amplitude with their complex conjugate.


Pm = (Ψ*m < m,u| + Φ*m< m,d|)(Ψm|m,u> + Φm|m,d>)
= Ψ*mΨm< m,u|m,u> + Φ*mΦm < m,d|m,d>
+ Ψ*mΦm< m,u|m,d> + Φ*mΨm< m,d|m,u>

Note for the last two terms, because the electrons have opposite spins, the up and down vector states are now orthogonal to each other.

< m,d|m,d> = < m,u|m,u> = 1
< m,u|m,d> = < m,d|m,u> = 0


Therefore,

Pm = Ψ*mΨm + Φ*mΦm

This result is completely different from the previous result. We see now that the very act of detecting the spin of one electron, that is, making some sort of measurement, destroys the interference pattern. This is another markedly difference between a classical system, in which we can always make a measurement without disturbing it, and a quantum system, in which a measurement entails disturbing the system and getting a different result.

Monday, January 03, 2011

Is String Theory wrong?

In String Theory, one of the most crucial calculations involves summing all the numbers from 1 to infinity, which obviously should be infinite. But not in ST, where the Rieman Zeta Function is used, and gives a value of -1/12??? Yes, if you add all the numbers from 1 to a million, you get a big number; if you add all the numbers from 1 to a trillion, you get an even bigger sum. But should you add all of them to infinity, not only do you get a fraction, but a negative one?

Here's a short version of a proof (attributed to Euler):

Euler started with this:

(1) 1 + x + x2 + x3 + · · · =1/(1 − x)

He differentiated both sides:

(2) 1 + 2x + 3x2 + · · · =1/(1 − x)2

He set x = −1 and got this:

(3) 1 − 2 + 3 − 4 + · · · =1/4

Then Euler considered this function (now known as the Riemann Zeta function):

(4) ζ(s) = 1−s + 2−s + 3−s + 4−s + · · ·

He multiplied by 2−s:

(5) 2−s ζ(s) = 2−s + 4−s + 6−s + 8−s + · · ·

Then he subtracted twice the second equation from the first:

(6) (1 − 2·2−s) ζ(s) = 1−s − 2−s + 3−s − 4−s + ···

and setting s = −1, he got:

(7) −3(1 + 2 + 3 + 4 + · · · ) = 1 − 2 + 3 − 4 + · · ·

Since he already knew the right-hand side equals 1/4, he concluded:

(8) 1 + 2 + 3 + 4 + · · · = −1/12

So what is wrong in this deduction? Take equation (3).

(3) 1/4 = 1 − 2 + 3 − 4 + 5 - 6 + 7 - 8 · · ·

We can regroup it as (take the first two, then the next two, etc.):

(3a) 1/4 = − 1 − 1 − 1 − 1 · · ·

Or we can regroup (3) as (take the first and third, second and fourth, etc.):

(3b) 1/4 = 4 − 6 + 8 − 10 + 12 − 14 · · ·

Now regroup (3b) as:

(3c) 1/4 = − 2 − 2 − 2 − 2· · ·

Comparing (3a) and (3c), we get that:

(9) 1/2 = 1/4 ???

We can conclude that Euler's deduction was plain wrong.

In all the textbooks, you will find that equation (1) is true only if |x| < 1 or -1 < x < 1, so by setting s equal to -1 in equation (3), we have made equation (1) divergent.

Now in all String Theory textbooks, we are told that ζ(-1)= -1/12 can be derived vigorously.

Stay tuned.

EDIT: See: Zeta Function is OK