X-ray Group Meeting

Observations of Hot Star WindsAbsorption Spectroscopy, Radiation (line-) Driving,

and X-rays

9 February 2001

David Cohen

Allison AdelmanDavid Conners

Eric LevyGeneviève de Messières

Kate Penrose

How do we know that hot star winds exist?

We see evidence in ultraviolet spectra due to line scattering in a dense outflow

Aside: The distinction between scattering, on the one hand, and absorption and re-emission, on the other:

Scattering is non-local; it is unaffected by the physical conditions at the place where the scattering occurs. Whereas with absorption followed by reemission, the properties of the radiation you observe are affected by the physical conditions (like temperature) at the location of the reemission.

Radiative excitation followed by spontaneous (radiative) emission

key

previous position of an electron

current position of an electron

movement of an electron

movement of an photon

photon

e external electron

Taken together, these processes constitute line scattering

•A photon is neither created nor destroyed

•Atom/electron ends up in same configuration in which it began

Scattering may not produce or destroy photons, and it may not effect the properties of atoms or electrons

But it can change the direction of photons

This cartoon illustrates the production of P Cygni profiles as scattering destroys blueshifted light from the star and adds emission at nearly all other wavelengths

Observed P Cygni profiles in two hot stars: Pup and Sco

Pup has a much stronger wind (deeper absorption, higher emission)

You can read the maximum wind velocity right off the spectrum -- it’s the maximum blueshift of the absorption

A somewhat more rigorous derivation of the P Cygni profile

Consider spherical shells in the wind. Each annulus leads to emission over a narrow wavelength range (or absorption, if it is in front of the star)

By summing up the contributions of many shells, we can build up the total P Cygni profile

Ultraviolet absorption lines

•Provide evidence of a stellar wind, but also

•Drive the wind (via their mediation of the transfer of momentum from starlight to the matter that makes up the wind)

Light has momentum -- momentum, p, is energy divided by velocity:

p=Ev

Light doesn’t have much momentum per energy, as its velocity is as high as possible

p=Ec

How much of this momentum can get transferred, and at what rate?

Note: Force is the time rate of change of momentum

F =dpdt

This momentum transfer will accelerate the stellar wind

One particle, with an effective cross-sectional area of (cm2)

can “catch” the flux of momentum (momentum cm-2 s-1), f/c, at the rate given by

Frad =σfc

This is the rate at which the particle gains momentum from the photons, which is the radiation force on the particle

Note that another way to see this is that the force, Frad, divided by area over which it is exerted, is a pressure.

In other words, momentum flux and pressure are identical.

Note that the effective cross sectional area of a particle is not really due to a physical size, but instead can be thought of as a propensity, or likelihood, for an atom to interact with a photon -- the more likely the interaction is, the bigger the cross-section

Now, for gas with lots of particles, it makes sense to ask not about the cross section of one particle, but the cross section per unit mass, (cm2/g), which is called the absorption coefficient, or opacity

Substituting for in the last equation gives the radiation force per unit mass, or the radiative acceleration

arad =Frad

m=

κfc

The opacity of the wind material, along with the brightness of the star (i.e. its flux) determines how much momentum gets transferred from the starlight to the wind, and therefore how much the wind is accelerated

arad =Frad

m=

κfc

opacity

flux

Hot star winds have so much opacity, that much of these stars’ flux is missing -- its momentum has been used to accelerate the wind

Note that the opacity of wind material depends on wavelength -- lines are stronger or weaker at different wavelengths

-- also, for a given line, the Doppler shift can affect how an atom absorbs light

Consider the light from the surface of a star--its photosphere--the spectrum of this light is constant over a given frequency range:

Consider a scattering spectral line from a parcel of atoms in the wind

This line’s profile, or frequency-dependent opacity, is shown in green. This is essentially the probability that a photon of a given frequency will be scattered, or absorbed, by the atom

After atoms of this type have absorbed the photospheric light, the remaining light looks like this:

Absorbing the light has accelerated the parcel, however, so the line profile is now blueshifted a bit

The red cross-hatched area shows the light that can be absorbed by this parcel of atoms -- note that the shifting of the profile out of the “Doppler shadow” allows for more momentum to be absorbed, and thus more acceleration, than if the line weren’t blueshifted

The acceleration of a line-driven wind is thus bootstrapped -- blueshifting of a line enables further acceleration -- the physics of a line-driven outflow is self-regulating

But the situation is non-linear...

If the parcel of atoms gets an extra little push, it will “see” more photospheric flux, get a bigger acceleration and thus more blueshift, and therefore receive even more flux, etc.

Line-driving has an inherent instability

We can calculate the strength of a wind and its velocity structure by simply solving Newton’s second law

a=Fm

However, the force on a given parcel depends on how much radiation flux it is receiving…in other words, on its redshift, which is a function of its acceleration

F ∝ a

Note that the force, as I’ve convinced you, is a function of the rate at which a line is blueshifted out of the Doppler shadow, which depends on how the velocity changes with distance, dv/dr, and the acceleration is proportional to this quantity through the chain rule: a=vdv/dr

This is a highly non-linear set of equations to solve!

But we can solve these equations in a time-dependent manner -- here is the velocity as a function of both height above the star and time from one simulation. Note all the structure and variability.

Here is a snapshot at a single time from the same simulation. Note the discontinuities in velocity. These are shock fronts, compressing and heating the wind, possibly producing X-rays.

So we see that line-scattering is the means via which the light from the photosphere of a hot star drives a massive stellar wind.

This line-driving is unstable by its very nature, producing a very chaotic and violent wind, which may be the explanation for the X-rays observed on hot stars.

Furthermore, this same line-scattering also allows us to observe and analyze these winds.

Coda: This process of momentum transfer and line driving is fascinating and powerful. But why isn’t it the explanation for the solar wind?

How much momentum is required?

The total rate of momentum being lost in radiation is L/c, where L is the luminosity of the star in ergs s-1

The total rate of momentum being taken away in the wind is just Mlossvinf, where Mloss is the mass-loss rate in solar masses

per year (Msun yr-1) and vinf is the terminal velocity of the

wind

For the sun,

L =3.9×1033(ergs s-1)

−dpdt

=Lc

=1.3×1023 (g cm s-2)

(cgs units of momentum loss per second)

For the solar wind, the momentum loss rate is:

˙ M ≈10−14

v∞ ≈300

(Msun yr-1)

(km s-1)

dpdt

= ˙ M v∞ ≈2×1019(g cm s-2)

Called this Mloss before

And called this vinf

Thus the sun’s light has more than enough momentum to drive the solar wind…but the light doesn’t drive the wind in the sun’s case

This is because the opacity of the wind is very low and poorly matched to the wavelengths where the sun emits most of its light (optical) -- the solar wind is transparent to optical light, which streams right through it, taking its momentum away rather than transferring it to the wind

recall:

So, the radiation force is proportional to the flux, but also to the opacity,

Note: force is the rate of change of momentum (check the units -- mass times acceleration = mass times velocity per time)

F =dpdt

=σfc

=μκfc

a=Fm

=ρκfc

(g cm s-2)

(cm s-2)

cm2/particle flux (ergs cm2 s-1)

opacity (cm2 g-1)

g/particle

Hot stars do have a lot of radiation at the wavelengths where wind material has its opacity (the ultraviolet)

The momentum in the light of a hot star (which has a huge luminosity) is, for the O star Pup:

This can drive a huge wind, potentially:

It is 1010 times the momentum in the solar wind

L ≈5×1039(ergs s-1) (106 times solar)

−dpdt

=Lc

≈2×1029 (g cm s-2)

Indeed, the wind of zeta Pup has close to this much momentum:

Its wind is very efficient at absorbing light.

To quantify the efficiency at which a given wind makes use of the momentum of light from the star, we can form a ratio of the winds momentum to the momentum available in the radiation:

˙ M ≈5×10−6

˙ M ≈3×1020

v∞ ≈3000

v∞ ≈3×108

dpdt

= ˙ M v∞ ≈1029

(Msun yr-1)

(g s-1)

(km s-1)

(cm s-1)

(g cm s-2)

˙ M v∞cL

(referred to as the “momentum number” or “performance number” of a radiation-driven wind)

Note: in a little over a million years, a star like Pup will have lost an appreciable fraction of its mass! This has very serious implications for stellar evolution models, supernovae rates, etc.

There are stars, called Woft-Rayet stars, which have momentum numbers greater than unity!

How can this be?

Well, once a photon scatters in the wind and gives up its momentum, is it gone? No, it’s just changed its direction. It can transfer its momentum again, and again, and again...

The problem (for theoretical astrophysicists) then becomes: How to provide enough opacity in the wind to keep the photons trapped so that multiple scattering can occur? Note: the atoms that absorb the light’s momentum also absorb some of its energy…there isn’t an infinite supply from a given photon (what do you suppose the ratio of momentum to energy lost in each scattering is?)