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 Incoming links: Maximum deflection through a strong perfect-gas shock

Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2021 / Section B / 4 / d / Solution 2026-10-18

Rewrite the relation from part (c) as

(γ - 1) sin β cos (β - θ) = (γ + 1) cos β sin (β - θ) .

The angle-addition identities reduce this to

sin (2 β - θ) = γ sin θ .

For

0 < θ < arcsin (1/ γ)

, the right side lies strictly between zero and one. There are two physically admissible strong oblique-shock angle branches:

β_{weak} β_{strong} = \frac{1}{2} [θ + arcsin (γ sin θ)], = \frac{1}{2} [θ + π - arcsin (γ sin θ)] .

Both lie between

θ

and

π /2

. They merge when

γ sin θ = 1

, proving the maximum deflection through a strong perfect-gas shock,

θ_{m a x} = arcsin (1/ γ), β_{*} = \frac{π}{4} + \frac{θ _{m a x}}{2} .

For small positive

θ

, the formal strong-shock-limit asymptotics are

β_{weak} \sim \frac{γ + 1}{2} θ, β_{strong} \sim \frac{π}{2} - \frac{γ - 1}{2} θ .

The weak branch is normally observed for an unconfined supersonic stream turning through a modest compression corner. It has a smaller normal component of upstream velocity, hence a smaller downstream pressure increase; for small turns the downstream flow remains supersonic. The nearly normal strong branch can be selected by suitable downstream pressure conditions, so it is not mathematically forbidden.

The two limits must be ordered honestly. At finite

M

, the weak oblique shock approaches the Mach angle

arcsin (1/ M)

θ \to 0

, rather than zero. The displayed weak-branch asymptotic applies when the large-

M

limit is taken first, or in the overlapping range

1/ M ≪ θ ≪ 1

, where

M sin β ≫ 1

remains valid.

Figure 1.
Calculated weak and strong shock-angle branches in the infinite-Mach-number limit, for a monatomic gas
. Details.

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Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 70 / 1 / d / Solution Created 2026-10-03 Updated 2026-10-14

Let

U = ∣ u_{1} ∣

and

q = (γ - 1) / (γ + 1)

, the strong-shock limit of the normal velocity ratio. In the plane containing the upstream direction and the shock normal, take

u_{1} = U e_{X}

and

n = cos β e_{X} + sin β e_{Y}

. The normal velocity is multiplied by

q

while the tangential velocity is unchanged, hence

u_{2} = u_{1} - (1 - q) (u_{1} \cdot n) n, u_{X 2} = U [1 - (1 - q) cos^{2} β], u_{Y 2} = - U (1 - q) cos β sin β .

Eliminating

β

gives the strong-shock polar for a perfect gas:

u_{Y 2}^{2} = (U - u_{X 2}) (u_{X 2} - q U) .

Equivalently this is the circle

(u_{X 2} - \frac{γ U}{γ + 1})^{2} + u_{Y 2}^{2} = (\frac{U}{γ + 1})^{2} .

Both signs of the perpendicular component are possible by changing shock orientation. The endpoints

u_{X 2} = q U, U

lie on the upstream axis. An original sketch for

γ = 5/3

illustrates the locus and a tangent ray:

Figure 1.
Strong-shock velocity polar for gamma five thirds and the tangent giving maximum deflection
. Details.

The deflection is the angle of the ray from the origin to the downstream velocity, rather than the shock-normal angle

β

. Its maximum occurs where this ray is tangent to the circle. The circle's center is at distance

γ U / (γ + 1)

and its radius is

U / (γ + 1)

, so the tangent right triangle proves

δ_{m a x} = arcsin (1/ γ) .

This is the maximum deflection through a strong perfect-gas shock; for

γ = 5/3

it is about

36.87

degrees.

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