# Under which angles does the solar wind leave the Sun and arrive at it's desination?

As the sun emits charged particles outwards, those particles will travel long distances to reach far away points in the solar system. They could be influenced by other bodies, electromagnetic fields, but I guess that their initial angle of ejection from the solar surface is the most important factor.

The (post) ejection angles could be straightened by magnetic fields, particle interaction or something like this, the influence of other bodies in the solar system is I'm guessing minor (except maybe for the local system of bodies with a magnetic field).

My question is: Does the solar wind move in a mostly straight line from the sun to let's say Earth, Mars or a random point in the interplanetary void? Are most arriving particles perpendicular to the projected disk which represents Earth?

Nice to have:

As a follow up, what is the distribution of ejection angles from the sun and the distribution of arriving angles on Earth (or some other random circular reference)?

Second follow up: If Venus was passing through the particles intended for Earth, would there be some lensing or significantly reduced particle bombardment from the influenced particles?

# Background

There is a nice review at http://solarphysics.livingreviews.org/Articles/lrsp-2013-5/articlese2.html that shows several nice figures. The following are excerpts from this review.

The solar wind is composed of an ionized gas called a plasma, which means the particles are subject to the Lorentz force.

The plasma leaves the sun mostly following the magnetic field along nearly radial trajectories. However, becaues the sun rotates and the photospheric magnetic fields are frozen-in to the plasma, the magnetic field appears like it is being "wound up" in a pattern similar to an Archimedean spiral.

Let's assume a constant bulk flow velocity and magnetic flux conservation, then the radial component of the solar magnetic field, $B_{r}$, will decrease with inverse distance squared or $\propto r^{-2}$. In spherical coordinates, we can write: $$B_{r}\left( r, \theta, \phi \right) = B_{ro} \left( \frac{ R_{\odot} }{ r } \right)^{2} \tag{1}$$ where $B_{ro}$ is the field at the point connected to the sun at radius $R_{\odot}$, longitude $\phi_{o}$, and colatitude $\theta$. Since we assumed magnetic flux conservation (i.e., plasma is "frozen-in" to the magnetic field), then a plasma streamline will coincide with the magnetic field. Therefore, in the co-rotating frame of the sun we have: $$\frac{ B_{\phi}\left( r, \theta, \phi \right) }{ B_{r}\left( r, \theta, \phi \right) } = \frac{ V_{\phi} }{ V_{r} } = \frac{ - \boldsymbol{\Omega} \times \mathbf{r} }{ V_{r} } \tag{2}$$ where $V_{r}$ is the assumed constant radial solar wind speed, $V_{\phi}$ is the azimuthal velocity resulting from the co-rotating reference frame, and $\Omega$ is the angular frequency of rotation of this co-rotating reference frame (i.e., the solar rotation rate ~ 24-27 day period, depending on scheme or ~$14.713^{\circ}/day$ at the equator or ~$2.972 \times 10^{-6} \ rad/s$). Thus, we can see that we can rewrite $B_{\phi}\left( r, \theta, \phi \right)$ as: $$B_{\phi}\left( r, \theta, \phi \right) = - B_{ro} \frac{ \Omega \ R_{\odot}^{2} \ \sin{\theta} }{ V_{r} \ r } \tag{3}$$

To summarize, under these assumptions $B_{\phi} \rightarrow 0$ at the poles (i.e., where $\theta \rightarrow 0$ or $\pi$). The angle that the IMF makes with respect to the radial direction we will call the spiral angle. The model described above is loosely referred to as the Parker Spiral after Eugene Parker.

Does the solar wind move in a mostly straight line from the sun to let's say Earth, Mars or a random point in the interplanetary void? Are most arriving particles perpendicular to the projected disk which represents Earth?

If we use $r$ = 1 AU, $\theta = \pi/2$, and $V_{r}$ = 400 km/s, then spiral angle of the IMF is ~$48^{\circ}$. If we change $r$ to 0.7 AU (i.e., roughly Venus' orbit), then this angle drops to ~$38^{\circ}$. If we change $r$ to 1.5 AU (i.e., roughly Mars' orbit), then this angle drops to ~$59^{\circ}$.

As a follow up, what is the distribution of ejection angles from the sun and the distribution of arriving angles on Earth (or some other random circular reference)?

This, unfortunately, cannot be answered because the solar magnetic field is so variable and inhomogeneous that we can only make generalized approximations like the model described above.

If Venus was passing through the particles intended for Earth, would there be some lensing or significantly reduced particle bombardment from the influenced particles?

We are much too far away to notice any significant wake-like effects from Venus. We can, on occasion, detect particles that have come from Jupiter but those are highly energetic and unrelated to typical solar wind processes.

• Thanks for your answer, I deleted my attempt, it was far, far too simplistic and totally ignored Solar rotation. – user108787 Aug 5 '16 at 19:35
• @count_to_10 - Well I am not sure too simplistic is necessarily a bad thing. I often strive to find the simplest way to explain things without loosing the essence and physics of the phenomena... – honeste_vivere Aug 5 '16 at 20:29
• Thank you for your answer, a very nice explanation indeed. Maybe Venus was a bad example, a magnetic disturbance at the L1 (0.01AU) point could be more relevant for this. An object at the exact L1 point wouldn't change much I guess, because of the angles, but an object which is at the distance of 0.01AU in the particle path probably would make a difference though. – WalyKu Aug 8 '16 at 8:20
• @Kurtovic - Look up research on the ion foreshock of the Earth's bow shock. There you will find all sorts of electromagnetic waves and structures that actively and directly affect both the bow shock and the terrestrial magnetosphere. – honeste_vivere Aug 8 '16 at 11:51