1

Lecture 9: Shocks - revised

Dr Graham M. HarperSchool of Physics, TCD

PY4A04 Senior Sophister

Physics of the Interstellar and Intergalactic Medium

What a good physicist does best - Simplify

Veil Nebula~8000 yr old

2

What is a 1-D shock?

� Infinite plane-parallel geometry� good approximation if the radius of curvature of the actual shock is

much greater than the thickness of the shock

1x∆2x∆

( ) ( )AreaxAreax 2211 ∆=∆ ρρ

Shock front

In unit time

Use handout

3

Conservation Equations: Mass in the frame of the shock

Conservations of mass: written for a narrow (1-D) shock

In the frame of the shock the flow is steady (time independent)

In passing through the shock mass is conserved

( ) ( )0=

∂∂+

∂∂=⋅∇+

∂∂

x

u

tu

t

ρρρρ

]1[2211 uu ρρ =

( )Constu

x

u

t=⇒=

∂∂+

∂∂ ρρρ

0

Conservation Equations: Momentum in the frame of the shock

Conservations of momentum: in unit time mass ρu1 enters shockwith momentum u1(ρ1u1). The mass leaves the shock with

momentum u2(ρ1u1) the difference must be given by the force per unit area

Using Eq [1] we obtain

]2[2222

2111 uPuP ρρ +=+

12211111 PPuuuu −=− ρρ

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Conservation Equations: Energy in the frame of the shock

Conservations of energy: requires that the rate at which gas pressure does work per unit area (Pu) and the rate of flow of both internal (U) and kinetic energy (1/2ρu2) is constant across the shock

Non radiating shocks: If gas acts as a perfect gas on either side ofthe shock then the internal energy is given by

]3[21

21

2211121112

2222 uPuPUuuUuu −=

+−

+ ρρ

]4[1

1PU

−=

γ

Some detail ...

Substitute Eq [4] into Eq. [3] and reorg.

−++=

−++

11

121

11

121

121112

2222 γ

ργ

ρ PuuPuu

−+=

−+

12

12

1

12111

2

22222 γ

γρ

ργ

γρ

ρ Puu

Puu

]5[1

21

2

1

121

2

222 −

+=−

+γ

γργ

γρ

Pu

Pu

Use Eq. [1] to divide both sides (now symmetric)

5

Three Jump (J-Shock) Conditions

]1[2211 uu ρρ =

]5[1

21

2

1

121

2

222 −

+=−

+γ

γργ

γρ

Pu

Pu

]2[2222

2111 uPuP ρρ +=+

Mach number

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Mach number

1

2112

1

1111

11

P

uM

Pcc

uM

γρ

ργ

=

==

Sound speed in the upstream gas c1

Mach number = ratio of the inflowing gas speed (as seen by the shock) to its sound speed: γ is the ratio of specific heats =CP/CV - adiabatic index, or adiabatic exponent

E.g., M supergiant outflow moving at 15 kms-1

into 60K interstellar cloud

B star wind (600 kms-1) into WIM 8000 K

169.0

151 ≈=M

5012

6001 ≈=M

After some algebra ...

7

Rankine-Hugoniot Relations

][11

12 2

11

2 AMP

P

+−−

+=

γγ

γγ

Relations between gas pressure, density, and velocity on either side of the shock front

][1

12

11

212

1

1

2 BMu

u

++

+−==

γγγ

ρρ

8

Strong Non-Radiating Shocks: M>>1

][1

1

2

1

12

12

1

1

2 BMu

u

++

+−==

γγγ

ρρ

?35

11

2

1

2

1 ≈⇒=≈+−≈

ρργ

γγ

ρρ

forLimiting value for monatomic gas

(translation)

?5

7

2

1 ≈⇒=ρργ diatomic molecule

(translation+rotation)

Can neglect pressure in the incoming flow (P1) for monatomic gas

211

222

2112 4

3uuuP ρρρ =−≈ 2

12 163

uk

mT

m

kTP

µµρ ≈=

Rest Frames: v1,2 velocities fixed frame

� Observationally work in a fixed reference e.g. Supernova, ionizing star. For fast shocks we can often (but not always) neglect v1 Vv

Vv−=−=

22

11

u

u

VvVv-V

VvVv

1

2

4

3

4

12

2

1

2 =⇒≈−−==

u

u

212 4

3 Vρ≈P2

2 16

3 Vk

mT

µ≈

Gas behind the shock follows in the same direction as the shock

100 kms-1 ~150,000 K2V

32

9

2

3

2

2int,2 ==

ρP

e 222,2 32

9

2

1 V== υKEe

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Entropy considerations

� The Rankine-Hugoniot relations do not in themselves forbid a time reversal of the shock, namely expansive shocks where sub-sonic hot gas expanding to become supersonic cool gas – a rarefaction shock

� Entropy decreases in this process as the flow becomes more ordered

� 2nd Law of Thermodynamics forbids rarefactions shocks, but does allows compressive shocks.

� Complete description is then given by the two Rankine-Hugoniotrelations and the 2nd Law of Thermodynamics

� Rarefaction waves do exist (bicycle pump)

Radiating Shocks

� Behind the shock the kinetic energy is converted to thermal energy� V > 50 kms-1 will ionize hydrogen. Plenty of electrons to excite energy

levels that can radiate energy away – the heated shock starts to cool� Typical ISM shock speeds are 80-2500 km s-1 - X-ray emitting plasma

Thermally unstable

( ) ss TT 1∝Λ

4VV ∝∝ cc tL

323 V∝∝ sc Tt

( )s

sc Tn

nT

ratelossenergy

contentenergyt

Λ∝∝

2

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Isothermal limit of a radiating shock

� If the cooling length is not too long then we can consider the region immediately behind the adiabatic shock and the subsequence cooling phase as a transition zone and set γ=1 in the Rankine-Hugoniotrelations

� Now the density can continue to increase behind the shock without limit� Using the same shock-to-fixed reference frame relations

21

212

1

1

2 111

211

MMu

u ≈+

++−==

γγγ

ρρ

( ) VVvVv-V

VvVv

1

2 ≈=⇒≈−−== 2

122

211

2 111

M-Mu

u

212

21

1

2 Vρ=⇒= PMP

P

agnetic ields

� onsider a galactic magnetic field running parallel to the shock front

� It exerts a pressure on partially ionized gas

� he new shock ump condition for momentum conser ation becomes

π82BPmag =

πρ

πρ

88

222

222

212

111

BuP

BuP ++=++

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Magnetic field in a 1-D shock

� The magnetic field is frozen into the partially ionized gas and thenumber of field lines is conserved across the shock. Since thecompression is in 1-D the conservation of mass leads to

1x∆2x∆

( ) ( )2211

2211

xBxB

AreaxAreax

∆=∆∆=∆ ρρ

Shock front

In unit time2

2

1

1

ρρBB =

Effects of Magnetic Fields

� Typical galactic field of 3x10-6 G consequences ...� Strong non-radiating shock, magnetic pressure increases by 16 giving a

pressure that < 1/5 of the gas pressure = not important� However, in isothermal shocks it may be important

� Strong radiating shocks the compression of gas leads to the increase in B which acts against the shock – limiting the shock gas density

� The maximum density occurs when the magnetic pressure balances the dynamic (ram) pressure

211

2

1

max2

122

88u

BB ρρ

ρππ

=

=

1

1231max B

uρρ ∝

12

Real universe – a tad more complicated

ISM Probes – bow-shocks

High proper motions

Infrared Imaging T. Ueta

If we know the stellar wind properties, we can learn about the ISM

LL Ori

R Hya = R Hydrae

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Betelgeuse α Ori = αOrionis

JAXA/Akari T. UetaThis one looks too circular?

( ) ( ) 2WW

2WWW uuuu RRPP ISMISMISMISMISM ρρρρ ≈⇒+=+ 22

Steady shock (not evolving)In frame of star VS=0

Momentum balance across theShock, where the star-ISM relative velocity is uISM

Dynamic ram pressures exceed local gas pressures

Betelgeuse α Ori = αOrionis

( )

ConstRMdt

dMt

===

=⋅∇+∂∂

WWu

υ

ρπ

ρρ

24

0

&

2ISM

ISM

W

u4u

πρM

RS

&

=

Conservation of mass for stellar “wind” mass-loss

Solve for Rs� Mdot = 3x10-6 solar masses per year� UISM~ 25 km s-1

� UW=17 km s-1

� Rs = distance x angle�distance=200 pc, angle=7 arcmin

� nISM=2 cm-3 10x too big