<p>Conservation of mass?</p>

<p>In a steady-state wind, the mass loss rate through each shell is the same.
$$ \frac{dM}{dt} = - 4\pi r^2 \rho(r) v(r),$$
where $\rho$ and $v$ would be the density and velocity of a spherically symmetric wind, with a fixed mass-loss rate $\dot{M}$.</p>

<p>Thus for a fixed wind speed you expect the density to fall as the square of the radius in spherically symmetric wind - and it looks to me that a little outside the closed field loops then spherical symmetry is approximately recovered.</p>

<p>This alone, would not account for the sharp drop off in density seen, but the other factor is that a radiation driven wind accelerates with radius when close to the star, which exacerbates the decrease in density. Typically, $v$ would increase by a factor of a few between say 1.5 and 5 times the radius. Multiplying this by a $r^{-2}$ factor would give a factor $\sim 100$ decrease in density overall.</p>

<p>The behaviour of the temperature is much more complex, governed by heating and cooling processes and the expansion of the gas.</p>

CC BY-SA 3.0
2016-11-06T16:56:08.453Z
I feel a little bit silly for not taking the velocity into account; thank you.
HDE 226868
CC BY-SA 3.0
2016-11-05T09:31:39.100Z
2016-11-05T09:31:39.100Z
ProfRob
<p>The first answer gives a good summary of the overall density falloff you see in all directions, but since you are saying the temperature is fairly constant it sounds like you might be asking about the density in the plane where the winds are colliding.  Is that what you mean?  If so, I think the problem is that none of those figures show the velocity field.  That is not a "beam" you are seeing, it is a cross section of an axially symmetric disk.  The wind is channeled down from the polar regions by the field lines, but the disk circumference increases in proportion to r, and there can be some centrifugal acceleration because the star is rotating.  So those could be the cause of the density drop with radius.  There's also that high density layer that appears very dark, and some if not all of that could end up falling back on the star, so it's not really a beam.</p>

CC BY-SA 3.0
2016-11-06T04:09:51.760Z
2016-11-06T04:09:51.760Z
Ken G
<p>I recently went to a talk on X-ray emissions from stellar winds of O-type stars, and a paper was explored that in turn referenced <a href="http://mnras.oxfordjournals.org/content/441/4/3600" rel="nofollow noreferrer">ud-Doula et al. (2014)</a>.</p>

<p>Essentially, O-type stars may show strong x-ray emission due to radial shocks in their stellar winds. In general, the winds are isotropic, and so emission is assumed to be spherically symmetric neglecting rotational effects. However, certain O-type stars (e.g. NGC 1624-2) have strong magnetic fields; this means that outflows are channeled into rather narrow beams within a few stellar radii, which eventually spread out.</p>

<p>X-ray emission is characterized by a parameter $X_{T_x}$
$$X_{T_x}\equiv\rho^2\exp(-T_x/T)$$
for density $\rho$ and temperature $T$, with a threshold temperature of $T_x$. Simulations place most of the emissions in a rather narrow spot near the star, because while there are indeed high temperatures experienced all along a beam extending outward from the star, the density quickly falls off, as shown in Figure 4:</p>

<p><a href="https://i.sstatic.net/H5HQa.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/H5HQa.png" alt="enter image description here"></a>
<br>
<sup>From left to right: Logarithmic density, logarithmic temperature, and x-ray emission.</sup></p>

<p>Why does the density drop off so quickly? I would expect that the channeling of more of the wind into a small beam would lead to a rather steady radial density profile along it, as is the case with the temperature graph.</p>

<hr>

<p>This is unrelated to the question, but I'll add that the temperature isn't really constant. 1D and 2D simulations of spherically symmetric show varying behavior due to repeated shocks inside the wind, and I would expect the same to be true here.</p>

<stellar-astrophysics><density><solar-wind><x-ray>
Why is there a density dropoff in the stellar winds of magnetic O-type stars?
HDE 226868
# Why is there a density dropoff in the stellar winds of magnetic O-type stars?

(Asked by: username_0 on 2016-11-04)

I recently went to a talk on X-ray emissions from stellar winds of O-type stars, and a paper was explored that in turn referenced [ud-Doula et al. (2014)](http://mnras.oxfordjournals.org/content/441/4/3600).

Essentially, O-type stars may show strong x-ray emission due to radial shocks in their stellar winds. In general, the winds are isotropic, and so emission is assumed to be spherically symmetric neglecting rotational effects. However, certain O-type stars (e.g. NGC 1624-2) have strong magnetic fields; this means that outflows are channeled into rather narrow beams within a few stellar radii, which eventually spread out.

X-ray emission is characterized by a parameter $X\_{T\_x}$
$$X\_{T\_x}\equiv\rho^2\exp(-T\_x/T)$$
for density $\rho$ and temperature $T$, with a threshold temperature of $T\_x$. Simulations place most of the emissions in a rather narrow spot near the star, because while there are indeed high temperatures experienced all along a beam extending outward from the star, the density quickly falls off, as shown in Figure 4:

[![enter image description here](https://i.sstatic.net/H5HQa.png)](https://i.sstatic.net/H5HQa.png)
  
From left to right: Logarithmic density, logarithmic temperature, and x-ray emission.

Why does the density drop off so quickly? I would expect that the channeling of more of the wind into a small beam would lead to a rather steady radial density profile along it, as is the case with the temperature graph.

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This is unrelated to the question, but I'll add that the temperature isn't really constant. 1D and 2D simulations of spherically symmetric show varying behavior due to repeated shocks inside the wind, and I would expect the same to be true here.

**Tags:** stellar-astrophysics, density, solar-wind, x-ray

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## 2 answers

### Answer by username_1 on 2016-11-05. Score: 3 | Accepted answer

Conservation of mass?

In a steady-state wind, the mass loss rate through each shell is the same.
$$ \frac{dM}{dt} = - 4\pi r^2 \rho(r) v(r),$$
where $\rho$ and $v$ would be the density and velocity of a spherically symmetric wind, with a fixed mass-loss rate $\dot{M}$.

Thus for a fixed wind speed you expect the density to fall as the square of the radius in spherically symmetric wind - and it looks to me that a little outside the closed field loops then spherical symmetry is approximately recovered.

This alone, would not account for the sharp drop off in density seen, but the other factor is that a radiation driven wind accelerates with radius when close to the star, which exacerbates the decrease in density. Typically, $v$ would increase by a factor of a few between say 1.5 and 5 times the radius. Multiplying this by a $r^{-2}$ factor would give a factor $\sim 100$ decrease in density overall.

The behaviour of the temperature is much more complex, governed by heating and cooling processes and the expansion of the gas.

> **Comments:**
> **username_0**: I feel a little bit silly for not taking the velocity into account; thank you.  

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### Answer by username_2 on 2016-11-06. Score: 0

The first answer gives a good summary of the overall density falloff you see in all directions, but since you are saying the temperature is fairly constant it sounds like you might be asking about the density in the plane where the winds are colliding. Is that what you mean? If so, I think the problem is that none of those figures show the velocity field. That is not a "beam" you are seeing, it is a cross section of an axially symmetric disk. The wind is channeled down from the polar regions by the field lines, but the disk circumference increases in proportion to r, and there can be some centrifugal acceleration because the star is rotating. So those could be the cause of the density drop with radius. There's also that high density layer that appears very dark, and some if not all of that could end up falling back on the star, so it's not really a beam.


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