Manuscript Em06-A@1 · Em06-A-v1.tex · proposition · ee9923c6

Proposition: Nontrivial plateau coefficient

A nonzero example and one evaluated response · lines 932–1023 · proof included or linked

Em06-A@1#prop:witness, Em06-A@1#eq:witness

Choose

0 < t_{1} < t_{2} < t_{3} < t_{4} < T < \frac{π}{2 m} .

Let

a \in C_{c}^{\infty} ((t_{1}, t_{2})), F \in C_{c}^{\infty} ((t_{3}, t_{4}))

be nonnegative and nonzero, and let $ρ \in C_{c}^{\infty} (R^{3})$ be nonnegative with $\int ρ d^{3} x = 1$ . Set

f (t, x) = F (t) ρ (x), h_{00} = h_{0 i} = 0, h_{ij} = - a (t) δ_{ij} .

Choose $χ$ equal to one on a neighborhood of

[t_{1}, t_{4}] \times {spatial projection of D_{f}} .

Such a compact switch separated from $t = 0$ exists.

Choose compact smooth scalar Cauchy data equal to $(1, 0)$ on a ball containing the initial causal dependence region of $D_{f}$ . Then their amplitude belongs to $S (R^{3})$ , and

O_{f} [h; ϕ_{cl}] = - 3 m \int_{R} F (t) \int_{0}^{t} a (s) cos (m s) sin (m (t - s)) d s d t < 0.

In particular $v_{f} \neq = 0$ .

Proof. Finite propagation and uniqueness imply $ϕ_{cl} (t, x) = cos (m t)$ throughout $D_{f}$ . For the stated metric coefficient,

τ (h) = 3 a (t), C (h)^{0} = - \frac{3}{2} a^{'} (t), C (h)^{i} = 0.

The unswitched coordinate solution is consequently

w [h]^{0} (t) = - \frac{3}{2} \int_{0}^{t} a (s) d s, w [h]^{i} = 0.

On the observation dependence region, the switch is one whenever $h$ is nonzero. The scalar density source on the free shell is

Q_{h} ϕ_{cl} = - C (h)^{0} \partial_{t} ϕ_{cl} = - \frac{3 m}{2} a^{'} (t) sin (m t) .

The retarded solution therefore gives

ϕ_{1, χ} (t, x) = \frac{3}{2} \int_{0}^{t} sin (m (t - s)) a^{'} (s) sin (m s) d s

in the observation region. Subtracting the coordinate term yields

ϕ_{1, χ} - w [h]^{0} \partial_{t} ϕ_{cl} = \frac{3}{2} \int_{0}^{t} sin (m (t - s)) a^{'} (s) sin (m s) d s - \frac{3 m}{2} sin (m t) \int_{0}^{t} a (s) d s .

Integrate the first term by parts. Its endpoint terms vanish. The trigonometric identity

sin (2 m s - m t) - sin (m t) = - 2 cos (m s) sin (m (t - s))

gives

ϕ_{1, χ} - w [h]^{0} \partial_{t} ϕ_{cl} = - 3 m \int_{0}^{t} a (s) cos (m s) sin (m (t - s)) d s .

Integrating against $f$ proves (eq:witness).

For $s \in (t_{1}, t_{2})$ , $t \in (t_{3}, t_{4})$ , both $cos (m s)$ and $sin (m (t - s))$ are strictly positive. The integral is therefore strictly negative. If $v_{f} = 0$ , (eq:classicalpairing) and (eq:representation) would make the same value zero, a contradiction.

Displayed formulas in this passage

Em06-A@1#eq:witness · line 961
$O_{f} [h; ϕ_{cl}] = - 3 m \int_{R} F (t) \int_{0}^{t} a (s) cos (m s) sin (m (t - s)) d s d t < 0.$
unlabeled · line 935
$0 < t_{1} < t_{2} < t_{3} < t_{4} < T < \frac{π}{2 m} .$
unlabeled · line 939
$a \in C_{c}^{\infty} ((t_{1}, t_{2})), F \in C_{c}^{\infty} ((t_{3}, t_{4}))$
unlabeled · line 946
$f (t, x) = F (t) ρ (x), h_{00} = h_{0 i} = 0, h_{ij} = - a (t) δ_{ij} .$
unlabeled · line 951
$[t_{1}, t_{4}] \times {spatial projection of D_{f}} .$
unlabeled · line 978
$τ (h) = 3 a (t), C (h)^{0} = - \frac{3}{2} a^{'} (t), C (h)^{i} = 0.$
unlabeled · line 982
$w [h]^{0} (t) = - \frac{3}{2} \int_{0}^{t} a (s) d s, w [h]^{i} = 0.$
unlabeled · line 987
$Q_{h} ϕ_{cl} = - C (h)^{0} \partial_{t} ϕ_{cl} = - \frac{3 m}{2} a^{'} (t) sin (m t) .$
unlabeled · line 993
$ϕ_{1, χ} (t, x) = \frac{3}{2} \int_{0}^{t} sin (m (t - s)) a^{'} (s) sin (m s) d s$
unlabeled · line 999
$ϕ_{1, χ} - w [h]^{0} \partial_{t} ϕ_{cl} = \frac{3}{2} \int_{0}^{t} sin (m (t - s)) a^{'} (s) sin (m s) d s - \frac{3 m}{2} sin (m t) \int_{0}^{t} a (s) d s .$
unlabeled · line 1007
$sin (2 m s - m t) - sin (m t) = - 2 cos (m s) sin (m (t - s))$
unlabeled · line 1012
$ϕ_{1, χ} - w [h]^{0} \partial_{t} ϕ_{cl} = - 3 m \int_{0}^{t} a (s) cos (m s) sin (m (t - s)) d s .$

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section: A nonzero example and one evaluated response