Rate of change of angle between a particle and a field

[James R]

I am wondering if anybody here can help me with a proof.

A particle is moving with velocity $\vec{v}$ in a field $\vec{F}$ (which could be an electric or magnetic field, for example). The velocity can change with time as the particle moves, but the field is static in space. The particle sees the field change as it moves from one position to another in space.

Now, I am interested in the time rate of change of the angle between the velocity vector and the field direction as the particle moves.

In other words, if at some point along the particle's trajectory the angle between F and v is $\theta$, then I need a vector expression for $d\theta / dt$.

Here's an expression I think is correct:

$$\frac{d\theta}{dt} = \frac{|\vec{F} \times (\vec{v} \cdot \nabla)\vec{F}|}{|\vec{F}|^2}$$

What I need is a derivation of this expression, or any kind of proof that it is in fact correct.

Can anybody help?

 

[James R]

LaTeX is not working at the moment, so here's a version without LaTeX formatting:

I am wondering if anybody here can help me with a proof.

A particle is moving with vector velocity v in a vector field F (which could be an electric or magnetic field, for example). The velocity can change with time as the particle moves, but the field is static in space. The particle sees the field change as it moves from one position to another in space.

Now, I am interested in the time rate of change of the angle between the velocity vector and the field direction as the particle moves.

In other words, if at some point along the particle's trajectory the angle between F and v is theta, then I need a vector expression for d(theta)/dt.

Here's an expression I think is correct:

d(theta)/dt = |F (cross product) (v.(gradient operator))F| / |F|^2

What I need is a derivation of this expression, or any kind of proof that it is in fact correct.

Can anybody help?

 

[sir-pinski]

There are a few different approaches to proving this expression for the rate of change of angle between a particle and a field. One way is to use the chain rule and the definition of the cross product.

First, let's define some variables. Let \vec{r} be the position vector of the particle, \vec{v} be its velocity, and \vec{F} be the field. We can also define the unit vector \hat{F} as the direction of the field at any given point. Then, the angle between \vec{v} and \vec{F} can be expressed as:

\theta = \cos^{-1}(\hat{F} \cdot \hat{v})

where \hat{v} is the unit vector in the direction of the velocity. Now, taking the derivative of both sides with respect to time, we get:

\frac{d\theta}{dt} = -\frac{\hat{F} \cdot \frac{d\hat{v}}{dt}}{\sqrt{1 - (\hat{F} \cdot \hat{v})^2}}

Using the chain rule, we can express \frac{d\hat{v}}{dt} as:

\frac{d\hat{v}}{dt} = \frac{d\vec{v}}{dt} \cdot \frac{1}{|\vec{v}|} \vec{v} - \frac{\vec{v}}{|\vec{v}|^3} \frac{d|\vec{v}|}{dt}

Substituting this into our previous equation and simplifying, we get:

\frac{d\theta}{dt} = \frac{\hat{F} \cdot \frac{d\vec{v}}{dt} \times \vec{v} - (\hat{F} \cdot \vec{v}) \frac{\vec{v} \times \frac{d\vec{v}}{dt}}{|\vec{v}|^2}}{\sqrt{1 - (\hat{F} \cdot \hat{v})^2}}

Now, using the definition of the cross product, we can rewrite this as:

\frac{d\theta}{dt} = \frac{\hat{F} \cdot (\vec{v} \times \frac{d\vec{v}}{dt})}{|\vec{v