v(r, t) = . (6.70)

4πr 2

Matching this to an outgoing wave gives

q(t)

™ ‚φ 1 r 1 r

= ’ 2f t ’ ’f t’

= v1 (r, t) = . (6.71)

4πr 2 ‚r r c rc c

Close to the origin, in the near ¬eld , the term ∝ f /r2 will dominate, and so

1

’ q(t) = f (t).

™ (6.72)

4π

Further away, in the far ¬eld or radiation ¬eld , only the second term will

survive, and so

‚φ 1 r

≈’ f t’

v1 = . (6.73)

‚r rc c

The far-¬eld velocity-pulse pro¬le v1 is therefore the derivative of the near-

¬eld v1 pulse-pro¬le.

The pressure pulse

ρ0 r

™

P1 = ’ρ0 φ = q t’

¨ (6.74)

4πr c

is also of this form. Thus, a sudden localized expansion of gas produces an

outgoing pressure pulse which is ¬rst positive and then negative.

160 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS

v v or P

x x

Near field Far field

Three-dimensional blast wave.

This phenomenon can be seen in (hopefully old) news footage of bomb blasts

in tropical regions. A spherical vapour condensation wave can been seen

spreading out from the explosion. The condensation cloud is caused by the

air cooling below the dew-point in the low-pressure region which tails the

over-pressure blast.

Now consider what happens if we have a sheet of explosive, the simultane-

ous detonation of every part of which gives us a one-dimensional plane-wave

pulse. We can obtain the plane wave by adding up the individual spherical

waves from each point on the sheet.

r

P

x

s

Sheet-source geometry.

6.2. WAVE EQUATION 161

Using the notation de¬ned in the ¬gure, we have

√

x2 + s2

1

∞

√ f t’

φ(x, t) = 2π sds (6.75)

c

x2 + s2

0

with f (t) = ’q(t)/4π, where now q is the rate at which volume is being

™ ™

intruded per unit area of the sheet. We can write this as

√

√

x2 + s2

∞

d x2 + s2 ,

f t’

2π

c

0

t’x/c

= 2πc f („ ) d„,

’∞

c t’x/c

=’ q(„ ) d„.

™ (6.76)

2 ’∞

√

In the second line we have de¬ned „ = t ’ x2 + s2 /c, which, inter alia,

interchanged the role of the upper and lower limits on the integral.

Thus, v1 = φ (x, t) = 1 q(t ’ x/c). Since the near ¬eld motion produced

™

2

1

by the intruding gas is v1 (r) = 2 q(t), the far-¬eld displacement exactly re-

™

produces the initial motion, suitably delayed of course. (The factor 1/2 is

because half the intruded volume goes towards producing a pulse in the neg-

ative direction.)

In three dimensions, the far-¬eld motion is the ¬rst derivative of the near-

¬eld motion. In one dimension, the far-¬eld motion is exactly the same as

the near-¬eld motion. In two dimensions the far-¬eld motion should there-

fore be the half-derivative of the near-¬eld motion ” but how do you half

di¬erentiate a function?

An answer is suggested by the theory of Laplace transformations as

1

™

d 1 F („ )

t

2

def

√

F (t) = √ d„. (6.77)

dt π t’„

’∞

Exercise: Use the calculus of improper integrals to show that, provided

F (’∞) = 0, we have

™ ¨

d 1 F („ ) 1 F („ )

t t

√ √

√ =√

d„ d„. (6.78)

t’„ t’„

dt π π

’∞ ’∞

This means that 1 1

d d d d

2 2

F (t) = F (t). (6.79)