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where v == (Kl /7r) 2 Using cylindrical coordinates and changing the variable K =
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Jk; + k~ to Kl/7r and kz to w == kzl/7r, we can rewrite the second term inside the braces on the right-hand side of (2.203) as
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2 7r 1ic cos () = [ 4 io
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where again v == (Kl/7r)2. Thus, introducing the cutoff, we can rewrite (2.203) as Pcas(x,y) =
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Using the method of Section 2.5.1, we can easily show (Problem 2.13) that
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7r 2 1ic Pcas(x, y) = -z240l 4 '
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which confirms the result of Section 2.5.1, showing that the pressure on the plate at z = l is directed toward the other plate (Le., it is attractive) and has a magnitude of 7r 2 1ic/240l 4
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2.5.3 The vacuum catastrophe: Vacuum zero-point energy and gravity
As in most of physics only energy differences are important, one frequently eliminates the zero-point energy by shifting the origin of energy [i.e., by adding a constant (but infinite) term to the Hamiltonian of the electromagnetic field]. In general relativity, however, one cannot shift the origin of energy, as according to Einstein, energy is equivalent to mass and must generate a gravitational field. So not only the electromagnetic zero-point energy, but the zero-point energies of all the fundamental boson fields of nature (including that of the graviton if it really exists) should give rise to a gravitational field: Interestingly, fermion fields would tend to cancel this effect, as their zero-point energy is negative. Some versions of supersymnietry that predict a fermion partner for every boson would then lead to no gravitational effect [5]. But despite intense experimental effort, no supersymmetry partners of the known elementary particles have been observed yet. If they do exist at large energy scales than are now accessible, the effect would be the same as that of a cutoff to the zero-point spectrum at these energies. In empty space, the electromagnetic vacuum zero-point energy is uniform. The same can be assumed for the zero-point energies of the other fields in nature. What kind of observable gravitational effect would such uniform energy distribution produce As a simple example, we can consider the effect of this uniform density on the motion of the outer planets of our solar system [5]. Unlike the inner planets, which are significantly affected by the other planets (and in the case of our Earth, by the Moon), the motion of the outer planets is dominated by the Sun. Now a uniform density outside a sphere does not produce any gravitational force inside. So for a planet undergoing a circular orbit of radius r centered around the Sun, only the zero-point energy inside a sphere of radius r will contribute to the total gravitational force pulling it toward the Sun. Moreover, the resulting gravitational force will be the same as if the mass equivalent to this energy would be all located at the Sun. Assuming that astronomical observations agree with Newton's theory of gravity, any gravitational force due to the vacuum zero-point energy must be very small in comparison with that due to the Sun. An estimate using the average orbital radius of Pluto yields [5] an upper bound for the total zero-point energy density of the vacuum of about 10 18 GeV/m 3. Although the simple estimation above give us a pedagogical example of an observable effect of a uniform mass density filling the whole of space, there are several problems with it. First, the orbit of Pluto is actually very far from being circular. At its closest approach, Pluto is less than 30 astronomical units43 from the Sun, whereas half an orbit later, Pluto is almost 50 astronomical units from the Sun. Moreover, it is well known in the literature that there are unexplained irregularities in the orbit of Pluto. In fact, we can get a lower44 bound to the cosmological constant this way (Le.,
43Theastronomicai unit is the average distance between the Sun and Earth (i.e., about 149.597.870 km). 44But this lower bound is higher than the upper bound that can be estimated from other observations.