Answer to Question #109075 in Astronomy | Astrophysics for Brant

Question

I need help with an early step in the derivation of the parker solar wind formula. My question is regarding the sign of dP/dr right here: density • dv/dt = -dP/dr - GM(sun)•density/r^2. Why is the sign of dP/dr negative. This is the first step in developing the radially symmetric outflow. Regarding the forces involved, the left side of the equation represents the force driving the wind and the right side is the overall outward force due to pressure minus the force of gravity. IE. residual outward force, equals total outward force minus inward force. Why isn’t dP/dr positive? Thank you.

Answer 1

The reason for the negative sign is simple. Chapman (1057) took part in the derivation of this equation and considered corona to be in hydrostatic equilibrium:

$$\frac{dP}{dr}=-\rho g.$$

Why is it negative? Because as we go further from the center of the sun, the pressure decreases and the pressure gradient is negative. Roughly speaking,

$$dP=P_2-P_1<0, \text{ while } P_2<P_1.$$

In order to count this and avoid negative pressure, we write minus before the derivative.