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


(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


(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.