# Differential Equation with Vector Product

1. Aug 30, 2008

### debis

If a particle with charge e and mass m is in an arbitrary magnetic field has motion described by:

$$m\frac{d^2\vec{r}}{dt^2}=\frac{e}{c}\frac{d\vec{r}}{dt}\times\vec{H}$$

prove that the speed $$v\equiv\left\vert\frac{d\vec{r}}{dt}\right\vert$$ is constant.

I don't understand how to do this when the field isn't necessarily constant.
Any suggestions would be greatly appreciated. Thanks!

2. Aug 30, 2008

### debis

Vector Product Differential Equation

If a particle with charge e and mass m is in an arbitrary magnetic field has motion described by:

$$m\frac{d^2\vec{r}}{dt^2}=\frac{e}{c}\frac{d\vec{r} }{dt}\times\vec{H}$$

prove that the speed $$v\equiv\left\vert\frac{d\vec{r}}{dt}\right\vert$$ is constant.

I don't understand how to do this when the field isn't necessarily constant.
Any suggestions would be greatly appreciated. Thanks!

i posted this elsewhere here but after browsing the site i thought this was a more appropriate place to put it...

3. Aug 30, 2008

### Staff: Mentor

Re: Vector Product Differential Equation

Does it specify a variable or constant field?

The fact that the speed is constant means that any acceleration is involved in changing the direction rather than tangential velocity.

4. Aug 30, 2008

### debis

Re: Vector Product Differential Equation

it just says arbitrary, so i'm assuming it means it could be variable or constant (it's definitely not reliably constant because the next question is "assuming the field is a constant")

i tried using circular motion with it for the constant speed, but i didn't understand how to apply it using vector products in differential equations.

5. Aug 30, 2008

### Ben Niehoff

Re: Vector Product Differential Equation

I would start by finding

$$\frac {d}{dt} \left| \frac{d \vec r}{dt} \right|$$

Then you should be able to show, from your available equations, that the above expression is zero.

Note that in general, the vector v x H is always perpendicular to v. This should be enough to prove what you want.

6. Aug 31, 2008

### qbert

what happens if you take the dot product with dr/dt?
to the right side? (should get 0).
to the left side? (do you recognize a complete derivative?)

7. Aug 31, 2008

### Defennder

8. Aug 31, 2008

### Defennder

I don't see how that helps, don't you get 0 on the left hand side as well since a vector and its derivative are perpendicular?

9. Aug 31, 2008

### HallsofIvy

Staff Emeritus
A vector and its derivative are NOT necessarily perpendicular. That is true only if the vector has constant length- which is what you are trying to prove.

10. Sep 1, 2008

### debis

i tried to find $$\frac{d}{dt}\left\vert\frac{d\vec{r}}{dt}\right\vert$$ but i don't know how to handle a diff eq with a vector product in it - i understand in theory how proving all the above mentioned things would work, my main question was how to handle a cross product in a differential equation... i looked in all my text books and on all sorts of websites and nothing helped.

11. Sep 1, 2008

### George Jones

Staff Emeritus
For the question in the original post, qbert (post #6) has given, in my opinion, the best hints.

12. Sep 1, 2008

### debis

ah yes, i got it.
i was so caught up in the diff eq part of it i forgot about simple vecor analysis. i always make things more complicated than they are :/
thank you so much everyone for your help, particularly qbert, your hint really helped :)

Last edited: Sep 1, 2008
13. Apr 3, 2009

### atanas1234

Last edited by a moderator: Apr 24, 2017