Electron in a magnetic field problem

[Saitama]

So what do you get all in all?

$$\frac{d^2\vec r}{ds^2}=\frac{e}{mv}\left(\frac{d\vec r}{ds}\times \vec B\right)$$

I don't see how to interpret the equation I have written above. :(

 

[voko]

The equation is correct. It is a differential equation of a naturally parametrized curve. You need to find out the condition that the curves corresponding to an electron and a proton are equal.

 

[Saitama]

The equation is correct. It is a differential equation of a naturally parametrized curve. You need to find out the condition that the curves corresponding to an electron and a proton are equal.

Looking at the equation I have obtained, I would say the mv part should be same for both electron and proton, correct?

voko, sorry to ask foolish question but what does "naturally parametrized curve" mean? Can you please pass a link? Many thanks!

 

[voko]

Looking at the equation I have obtained, I would say the mv part should be same for both electron and proton, correct?

Yes.

voko, sorry to ask foolish question but what does "naturally parametrized curve" mean? Can you please pass a link? Many thanks!

$s$ is a parameter of the curve. If you take any function $s' = f(s)$, $s'$ is also a parameter of the curve. But there is unique (up to an additive constant) way to parametrize the curve so that $s$ is equal to the length of the curve. This is natural parametrization. The condition on $s$ you have in #1 ensures that. Can you see why?

Also, do you see that $\frac {d\vec r} {ds}$ is the unit velocity vector, a. k. a. the unit tangent vector?

Do you see that $\frac {d^2\vec r} {ds^2}$ is the normal acceleration vector?

 

[Saitama]

Yes.

So that gives $U_p=-5.44 \times 10^{-7}$. Thanks voko! :)

$s$ is a parameter of the curve. If you take any function $s' = f(s)$, $s'$ is also a parameter of the curve. But there is unique (up to an additive constant) way to parametrize the curve so that $s$ is equal to the length of the curve. This is natural parametrization. The condition on $s$ you have in #1 ensures that. Can you see why?

Which condition?

Also, do you see that $\frac {d\vec r} {ds}$ is the unit velocity vector, a. k. a. the unit tangent vector?

Do you see that $\frac {d^2\vec r} {ds^2}$ is the normal acceleration vector?

I don't see any of this. :(

Can you please suggest a book or a link where this is discussed?

 

[voko]

Which condition?

The one in the end: $s = \int v(t)dt$.

I don't see any of this. :(

Hmm. You had $\hat v$ for the velocity unit vector. Then you found that it was equal to $\frac {d\vec r} {ds}$.

Now, because it is unit vector, its derivative must be orthogonal to it, so it is normal to the curve.

The total velocity vector is $$ \frac {d\vec r} {dt} = v \frac {d \vec r} {ds} $$ so the total acceleration vector is $$ \frac {d^2\vec r} {dt^2} = \frac {dv} {dt} \frac {d \vec r} {ds} + v^2 \frac {d^2 \vec r} {ds^2} $$ The first term here is the tangential acceleration, and the second term is the normal acceleration.

The magnitude of $\frac {d^2 \vec r} {ds^2}$ is the curvature, and its inverse is the radius of curvature.

Can you please suggest a book or a link where this is discussed?

http://en.wikipedia.org/wiki/Frenet–Serret_formulas

 

[Saitama]

Hmm. You had $\hat v$ for the velocity unit vector. Then you found that it was equal to $\frac {d\vec r} {ds}$.

Now, because it is unit vector, its derivative must be orthogonal to it, so it is normal to the curve.

The total velocity vector is $$ \frac {d\vec r} {dt} = v \frac {d \vec r} {ds} $$ so the total acceleration vector is $$ \frac {d^2\vec r} {dt^2} = \frac {dv} {dt} \frac {d \vec r} {ds} + v^2 \frac {d^2 \vec r} {ds^2} $$ The first term here is the tangential acceleration, and the second term is the normal acceleration.

The magnitude of $\frac {d^2 \vec r} {ds^2}$ is the curvature, and its inverse is the radius of curvature.

Thank you very much voko for the explanations! :)

I am not sure if I fully understand this discussion so I will be looking at the link you have posted and try to learn more about this. Thanks again.

 

[CAF123]

Hi Pranav-Arora,

The beginning of the book by Kleppner and Kolenkow has some treatment on this topic.