In this blog, ℏ=c=k

_{b}= 1 where ℏ is the reduced Planck constant, c is the speed of light and k

_{b}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 = mc

^{2}. 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 = mc

^{2}. (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) ds

^{2}= -dt

^{2}+ dx

^{2}

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) ds

^{2}= –dt

^{2}+ dx

^{2}= 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 = dx

^{μ}/dτ = x'

^{μ}, and the acceleration, a

^{μ}= x''

^{μ}

Consider the square of the acceleration,

(5) a

^{2}= a

^{μ}a

_{μ}= – (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

^{aτ}

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

(8) her time coordinate is ξ

^{0}= τ, and her spatial one is ξ

^{1}= 0.

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

(9) U = ξ

^{1}– ξ

^{0}= – τ, and V = ξ

^{1}+ ξ

^{0}= τ

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

(10) u = (1/a)e

^{aU}, v =(1/a)e

^{aV}

Note that,

(11) U = a

^{-1}ln(ua) and V = a

^{-1}ln(va)

Taking the derivative of equations (10),

(12) du = e

^{aU}dU, dv = e

^{aV}dV

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

(13) ds

^{2}= dudv = e

^{a(U+V)}dUdV

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

(14) For Bob, s

_{Bob}= ∫ ∂

_{u}φ∂

_{v}φdudv;

for Alice, s

_{Alice}= ∫ ∂

_{U}φ∂

_{V}φdUdV

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

(15) For Bob, ∂

_{u}∂

_{v}φ = 0; for Alice, ∂

_{U}∂

_{V}φ = 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)

^{-½}(a

^{−L}

_{ω}e

^{−iωu}+ a

^{+L}

_{ω}e

^{iωu}) + ∫ dω(2ω)

^{-½}(a

^{−R}

_{ω}e

^{−iωv}+ a

^{+R}

_{ω}e

^{iωv})

(18) For Alice, φ

_{Alice}= ∫ dΩ(2Ω)

^{-½}(b

^{−L}

_{Ω}e

^{−iΩU}+ b

^{+L}

_{Ω}e

^{iΩU}) + ∫dΩ(2Ω)

^{-½}(b

^{−R}

_{Ω}e

^{−iΩV}+ b

^{+R}

_{Ω}e

^{iΩV})

• Where (2ω)

^{-½}and (2Ω)

^{-½}are normalizing factors.

• a

^{−L}

_{ω}and a

^{−R}

_{ω}are Bob's annihilation operators; b

^{−L}

_{Ω}and b

^{−R}

_{Ω}are Alice's annihilation operators.

• a

^{+L}

_{ω}and a

^{+R}

_{ω}are Bob's creation operators; b

^{+L}

_{Ω}and b

^{+R}

_{Ω}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

^{−L}

_{ω}→ a

^{−}

_{ω}, b

^{−L}

_{Ω}→ b

^{−}

_{Ω}.

Both observers will define their vacuum state as such,

(20) a

^{−}

_{ω}|0>

_{Bob}= 0; b

^{−}

_{Ω}|0>

_{Alice}= 0

The question is: are these two vacua the same?

Let’s assume we can write,

(21)b

^{−}

_{Ω}= ∫ dω (α

_{Ωω}a

^{−}

_{ω}+ β

_{Ωω}a

^{+}

_{ω}), and

b

^{+}

_{Ω}= ∫ dω (α*

_{Ωω}a

^{+}ω − β*

_{Ωω}a

^{−}

_{ω})

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

^{−}

_{ω}coefficient which is,

(22) ∫dΩ(2Ω)

^{-½}∫dω(α

_{Ωω}e

^{−iΩU}− β*

_{Ωω}e

^{iΩU})

For Bob, from equation (17), the coefficient for a

^{−}

_{ω}is

(23) ∫dω (2ω)

^{-½}e

^{−iωu}

For these two to match, we need,

(24) (2ω)

^{-½}e

^{−iωu}= ∫dΩ' (2Ω')

^{-½}(α

_{Ω'ω}e

^{−iΩ'U}− β*

_{Ω'ω}e

^{iΩ'U})

Multiply both sides by e

^{iΩU}and then integrate over all U,

(25) ∫(2ω)

^{-½}e

^{−iωu}e

^{iΩU}dU = ∫dΩ'(2Ω')

^{-½}∫dU(α

_{Ω'ω}e

^{−iΩ'U}e

^{iΩU}− β*

_{Ω'ω}e

^{iΩ'U}e

^{iΩU})

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}e

^{iΩU}= ∫dU e

^{i(Ω − Ω')U}= 2π δ(Ω −Ω')

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

(27) α

_{Ωω}= (2π)

^{-1}(Ω)

^{½}(ω)

^{-½}∫dUe

^{-iωu}e

^{iΩU}

Similarly, we get,

(28) β

_{Ωω}=(−1)(2π)

^{-1}(Ω)

^{½}(ω)

^{-½}∫dUe

^{+iωu}e

^{iΩU}

Using equation (11), which is repeated below,

(29) U = a

^{-1}ln(ua)

(30) α

_{Ωω}= (2π)

^{-1}(Ω)

^{½}(ω)

^{-½}∫(du/u)e

^{-iωu}e

^{i(Ω/a)lnua}

(31) β

_{Ωω}=(−1)(2π)

^{-1}(Ω)

^{½}(ω)

^{-½}∫(du/u)e

^{+iωu}e

^{i(Ω/a)lnua}

Now we don’t need to evaluate these integrals to find the relationship between α

_{Ωω}and β

_{Ωω}. We note that there are very similar, and we get our answer by using the transformation u → -u in equation (31),

(32)β

_{Ωω}=(−1)(2π)

^{-1}(Ω)

^{½}(ω)

^{-½}∫(du/u)e

^{-iωu}e

^{i(Ω/a)ln(-ua)}

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

(33) e

^{i(Ω/a)ln(-ua)}= e

^{i(Ω/a)ln(ua)}e

^{i(Ω/a)ln(-1)}

Recall Euler’s famous equation:

(34) e

^{iπ}= -1 → iπ = ln(-1). Therefore,

(35) e

^{i(Ω/a)ln(-ua)}= e

^{i(Ω/a)ln(ua)}e

^{i(Ω/a)ln(-1)}= e

^{i(Ω/a)ln(ua)}e

^{-(πΩ/a)}

Putting this back into equation (32),

(36)β

_{Ωω}=(−1)(2π)

^{-1}(Ω)

^{½}(ω)

^{-½}∫{(du/u)e

^{-iωu}

*x*e

^{i(Ω/a)ln(ua)}e

^{-(πΩ/a)}}

(37) β

_{Ωω}= − α

_{Ωω}e

^{-(πΩ/a)}

Squaring both sides and dropping the indices,

(38) |β|

^{2}= |α|

^{2}e

^{-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) ρ

_{mn}= <ψ

_{m}|e

^{-H/T}| ψ

_{n}> = |α|

^{2}e

^{-Em/T}, where H is the Hamiltonian, and E is the energy eigenvalue of H.

Also,

(40) ρ

_{nm}= <ψ

_{n}|e

^{-H/T}| ψ

_{m}> = |β|

^{2}e

^{-En/T}

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

(41) ρ

_{mn}= ρ

_{nm}

We get,

(42) |β|

^{2}= |α|

^{2}e

^{-(En - Em)/T}= |α|

^{2}e

^{-E/T}

But from the Einstein equation,

(43) E = ℏΩ = Ω (ℏ = 1)

Therefore,

(44) |β|

^{2}= |α|

^{2}e

^{-Ω/T}

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

(45) T = a/(2π)

Now putting back the constants ℏ,c, and k

_{b}to make the equation dimensionally consistent,

(46) T = aℏ/(2πck

_{b})

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/s

^{2}, (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 ~ 10

^{22}m/s

^{2}, both cases being beyond our present technology.

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