Rate of change of angle between a particle and a field

In summary, the conversation is about finding the time rate of change of the angle between a particle's velocity vector and a static field vector. The provided expression for this is d(theta)/dt = |F x (v . ∇)F| / |F|^2. The request is for a derivation or proof of this expression.
  • #1
James R
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I am wondering if anybody here can help me with a proof.

A particle is moving with velocity [itex]\vec{v}[/itex] in a field [itex]\vec{F}[/itex] (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 [itex]\theta[/itex], then I need a vector expression for [itex]d\theta / dt[/itex].

Here's an expression I think is correct:

[tex] \frac{d\theta}{dt} = \frac{|\vec{F} \times (\vec{v} \cdot \nabla)\vec{F}|}{|\vec{F}|^2}[/tex]

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

Can anybody help?
 
Last edited:
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  • #2
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?
 
Last edited:
  • #3


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
 

1. What is the rate of change of angle between a particle and a field?

The rate of change of angle between a particle and a field refers to how quickly the angle between the two changes over time. It is a measure of how the particle is affected by the field and how it moves in response to it.

2. How is the rate of change of angle between a particle and a field calculated?

The rate of change of angle between a particle and a field is calculated by taking the derivative of the angle with respect to time. This can be represented by the equation dθ/dt, where θ is the angle and t is time.

3. What factors can affect the rate of change of angle between a particle and a field?

The rate of change of angle between a particle and a field can be affected by factors such as the strength of the field, the mass of the particle, and the properties of the medium the particle is moving through.

4. How does the rate of change of angle between a particle and a field relate to the particle's motion?

The rate of change of angle between a particle and a field is directly related to the particle's motion. A larger rate of change of angle indicates a faster change in the particle's direction, while a smaller rate of change of angle indicates a slower change in direction.

5. Can the rate of change of angle between a particle and a field be negative?

Yes, the rate of change of angle between a particle and a field can be negative. This would indicate that the angle between the two is decreasing over time, meaning the particle is moving closer to the field or changing direction in a way that decreases the angle.

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