## 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.