Manuscript Em02-A@6 · Em02-A-v6.tex · proposition · cb2f0a9a

Proposition: Non-tip scalar reductions

Route B: the exact non-tip integrals · lines 799–891 · proof included or linked

Em02-A@6#prop:nontip, Em02-A@6#eq:47-integral, Em02-A@6#eq:5a-integral, Em02-A@6#eq:5z-integral, Em02-A@6#eq:response-density

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Status: Proved for displayed mathematical data.

On compact supports within the indicated coordinate domains, the two-line differentiated integrals are the following exact scalar integrals.

For $47$ , put $υ_{47} = - ∣ ω ∣^{2} / [4 (β - γ)]$ . Then

\int \frac{β}{8 π ^{2} ( β - γ )} (Θ_{j, δ} H_{47, δ}) (τ; γ, υ_{47}, ω, 0, v, ξ) d τ d γ d^{2} ω d v d^{2} ξ .

For $5 a$ on $γ < 0$ , put $υ_{5 a} = ∣ ω ∣^{2} / (4 γ)$ . Then

\int \frac{β}{8 π ^{2} ∣ γ ∣} (Θ_{j, δ} H_{5 a, δ}) (τ; γ, υ_{5 a}, ω, 0, v, ξ) d τ d γ d^{2} ω d v d^{2} ξ .

On $υ < 0$ , the alternative formula eliminates $γ = ∣ ω ∣^{2} / (4 υ)$ and uses density $β / (8 π^{2} ∣ υ ∣)$ .

For $5 z$ on $0 < γ < β$ , put

υ_{ω} = \frac{∣ ω ∣ ^{2}}{4 γ}, u_{ω} = \frac{β ∣ ω ∣ ^{2}}{4 γ ( β - γ )} .

Then the integral is

\int \frac{( β - u _{ω} ) ^{2}}{8 π ^{2} γ ( β - γ )} (Θ_{j, δ} H_{5 z, δ}) (τ; γ, υ_{ω}, ω, u_{ω}, v, ξ) d τ d γ d^{2} ω d v d^{2} ξ .

On $υ > 0$ , an alternative eliminates

γ_{ω} = \frac{∣ ω ∣ ^{2}}{4 υ}, u_{ω} = \frac{β υ}{β - γ _{ω}}

and has density

\frac{( β - u _{ω} ) ^{2}}{8 π ^{2} υ ( β - γ _{ω} )}

in $d τ d υ d^{2} ω d v d^{2} ξ$ .

The response measure has density

\frac{β}{2 π} d τ d v d^{2} ξ at r = 0, u = 0.

Its amplitude is the already computed response adjoint, not a newly chosen tangential jet.

Proof. Away from a cone vertex, $R (x) = (2 π)^{- 1} Θ (x^{0}) δ (x^{2})$ . The full null-frame density is

8 (β - u)^{2} d τ d γ d υ d^{2} ω d u d v d^{2} ξ .

For $47$ , the constraint Jacobian in $(u, υ)$ is $16 β (β - γ)$ . For $5 a$ , it is $16 β ∣ γ ∣$ , or $16 β ∣ υ ∣$ in the alternative chart. For $5 z$ , the two Jacobians are respectively $16 γ (β - γ)$ and $16 υ (β - γ)$ . Combining these with $(2 π)^{- 2}$ proves the formulas. The directed support conditions select the indicated domains.

For the response seed, integrate $δ^{(4)} (r)$ in the original $r$ density, and then $δ (4 β u) / (2 π)$ . The result is $β / (2 π)$ .

Displayed formulas in this passage

Em02-A@6#eq:47-integral · line 808
$\int \frac{β}{8 π ^{2} ( β - γ )} (Θ_{j, δ} H_{47, δ}) (τ; γ, υ_{47}, ω, 0, v, ξ) d τ d γ d^{2} ω d v d^{2} ξ .$
Em02-A@6#eq:5a-integral · line 819
$\int \frac{β}{8 π ^{2} ∣ γ ∣} (Θ_{j, δ} H_{5 a, δ}) (τ; γ, υ_{5 a}, ω, 0, v, ξ) d τ d γ d^{2} ω d v d^{2} ξ .$
Em02-A@6#eq:5z-integral · line 838
$\int \frac{( β - u _{ω} ) ^{2}}{8 π ^{2} γ ( β - γ )} (Θ_{j, δ} H_{5 z, δ}) (τ; γ, υ_{ω}, ω, u_{ω}, v, ξ) d τ d γ d^{2} ω d v d^{2} ξ .$
Em02-A@6#eq:response-density · line 860
$\frac{β}{2 π} d τ d v d^{2} ξ at r = 0, u = 0.$
unlabeled · line 832
$υ_{ω} = \frac{∣ ω ∣ ^{2}}{4 γ}, u_{ω} = \frac{β ∣ ω ∣ ^{2}}{4 γ ( β - γ )} .$
unlabeled · line 848
$γ_{ω} = \frac{∣ ω ∣ ^{2}}{4 υ}, u_{ω} = \frac{β υ}{β - γ _{ω}}$
unlabeled · line 853
$\frac{( β - u _{ω} ) ^{2}}{8 π ^{2} υ ( β - γ _{ω} )}$
unlabeled · line 873
$8 (β - u)^{2} d τ d γ d υ d^{2} ω d u d v d^{2} ξ .$

Proposition: Non-tip scalar reductions

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