Electron in a magnetic field problem

[Saitama]

Homework Statement

Homework Equations

The Attempt at a Solution

I was completely clueless about the hint so I tried the following:

The force acting on the electron is $\vec{F}=e(\vec{v_e}\times \vec{B})$. The speed of electron in the magnetic always stays constant, only the direction changes.

Since $\vec{F}=m_e\vec{\ddot{s_e}}=e(\vec{v_e}\times \vec{B})$. For proton $m_p\vec{\ddot{s_p}}=e(\vec{v_p}\times \vec{B})$

As per the question, the proton follows the same trajectory, hence, $\vec{\ddot{s_e}}=\vec{\ddot{s_p}}=\vec{\ddot{s}}$. This implies that $v_e/m_e=v_p/m_p$. But this gives the wrong answer. Where is the mistake in this approach?

Okay, getting back to the hint, I see that it says to compute the derivative of unit vector along the velocity with respect to s but I am not sure how to use this.

Also, $\vec{ds}=\vec{v_e(t)}dt=|\vec{v_e}|\hat{v_e(t)}dt$ but still I do not see how to proceed.

Any help is appreciated. Thanks!

 

[voko]

The equality of accelerations is not sufficient. For example, at the sea level, all free falling bodies experience equal accelerations. Yet you can have very different motions depending on the initial velocity. You get identical trajectories only when the initial velocities are identical. But if the velocities in your case are identical, then the proportion you got cannot hold. Contradiction.

But the thing is, accelerations do not have to be equal in identical trajectories. Trajectories are not motions, different motions can still have identical trajectories. For example, you can walk across your room slowly or swiftly: different motions, identical trajectories.

 

[Saitama]

The equality of accelerations is not sufficient. For example, at the sea level, all free falling bodies experience equal accelerations. Yet you can have very different motions depending on the initial velocity. You get identical trajectories only when the initial velocities are identical. But if the velocities in your case are identical, then the proportion you got cannot hold. Contradiction.

But the thing is, accelerations do not have to be equal in identical trajectories. Trajectories are not motions, different motions can still have identical trajectories. For example, you can walk across your room slowly or swiftly: different motions, identical trajectories.

Hi voko! :)

Thanks for the explanation, I understand the mistake now but I am still clueless about approaching the problem. Can I please have a few hints?

 

[voko]

The hint is that you should convert the equation of motion, which is time-dependent, into an equation of trajectory, which is time-independent. The problem already supplies a very strong hint, which is that you use natural parametrization of the curve. All I can add is "use the chain rule".

 

[Saitama]

The hint is that you should convert the equation of motion, which is time-dependent, into an equation of trajectory, which is time-independent. The problem already supplies a very strong hint, which is that you use natural parametrization of the curve. All I can add is "use the chain rule".

I honestly cannot proceed.

Do you ask the following, pardon if this looks foolish.

$$\frac{d}{ds}\frac{\vec{v}}{v} =\frac{d\hat{v}}{dt}\frac{1}{ds/dt}=\frac{d\hat{v}}{dt}\frac{1}{\vec{v}}$$

 

[voko]

Start from something simpler. What is $\frac {ds} {dt}$? What is $\vec v = \frac {d \vec r} {dt}$ in terms of s? It is also useful to keep in mind your observation that $v$ is a constant in this problem.

 

[Saitama]

Start from something simpler. What is $\frac {ds} {dt}$? What is $\vec v = \frac {d \vec r} {dt}$ in terms of s? It is also useful to keep in mind your observation that $v$ is a constant in this problem.

$\frac{ds}{dt}=v$, correct? What is r?

 

[voko]

$\vec r$ is the position of the particle, standard notation if you ask me :)

 

[Saitama]

$\vec r$ is the position of the particle, standard notation if you ask me :)

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k} \Rightarrow \vec{v}=(dx/dt)\hat{i}+(dy/dt)\hat{j}+(dz/dt)\hat{k}$

But I don't know how I can express $\vec{s}$ in terms of x,y and z. :(

 

[voko]

There is no $\vec s$. $s$ is a scalar defined as shown in #1. It is the natural parameter of the curve (trajectory). You need to convert the diff. eq. of motion (time-dependent) to the diff. eq. of the trajectory (s-dependent). Then find out the condition for equal trajectories.

 

[Saitama]

There is no $\vec s$. $s$ is a scalar defined as shown in #1. It is the natural parameter of the curve (trajectory). You need to convert the diff. eq. of motion (time-dependent) to the diff. eq. of the trajectory (s-dependent). Then find out the condition for equal trajectories.

Since I am still clueless, I am referring a book. Here is the paragraph I am looking at:

Do I have to somehow make use of what is shown in the image?

 

[voko]

$s$ here is a kind of $l$, except it has a special property. Here is another little hint for you: $$ \vec v = \frac {d\vec r} {dt} = \frac {d \vec r} {ds} \frac {ds} {dt} = ? $$

 

[Saitama]

$s$ here is a kind of $l$, except it has a special property. Here is another little hint for you: $$ \vec v = \frac {d\vec r} {dt} = \frac {d \vec r} {ds} \frac {ds} {dt} = ? $$

I still don't see what $d\vec{r}/ds$ represents, I have never seen anything like this.

 

[voko]

$s$ is the parameter of the curve. The equation of the curve (trajectory) is given by $$ \vec r = \vec r(s) $$ The motion along the curve is specified by fixing $s = s(t)$. So $$ \vec r = \vec r(s(t)) $$ Then you just apply the chain rule.

 

[Saitama]

So $$ \vec r = \vec r(s(t)) $$ Then you just apply the chain rule.

Don't we end up with the same thing if the quoted equation is differentiated wrt time?

$$\frac{d\vec{r}}{dt}=\frac{d\vec{r}}{ds}\frac{ds}{dt}$$

What do I have to substitute for $d\vec{r}/ds$?

 

[voko]

You keep $d\vec r / ds$. You need the terms independent of $t$ because you want the equation of the curve.

 

[Saitama]

You keep $d\vec r / ds$. You need the terms independent of $t$ because you want the equation of the curve.

And how do I find the equation of curve?

You stated that the equation of curve is $\vec{r}=\vec{r}(s)$ but I don't see how to figure out the RHS. I am completely blank on this. :(

We have
$$\vec{v}=\frac{d\vec{r}}{ds}v$$

The hint asks to compute $d/ds(\vec{v}/v)$

$$\Rightarrow \frac{d}{ds}\frac{\vec{v}}{v}=\frac{d}{ds}\frac{d\vec{r}}{ds}=\frac{d^2\vec{r}}{ds^2}$$

I will leave for now and go to sleep. Thank you very much voko for the help you have provided so far. :)

 

[voko]

You start with the equation of motion. Which is $$

m \frac {d\vec v} {dt} = \vec F

$$ Try to express as much as you can in terms of $s$ rather than $t$.

 

[Saitama]

You start with the equation of motion. Which is $$

m \frac {d\vec v} {dt} = \vec F

$$ Try to express as much as you can in terms of $s$ rather than $t$.

$$\frac{d\vec{v}}{dt}=v\frac{d}{dt} \frac {d\vec{r}}{ds}=v\frac{d}{ds}\frac{d\vec{r}}{ds}\frac{ds}{dt}=v^2\frac{d^2\vec{r}}{ds^2}$$

Is this what you ask me? What should I do next?

 

[voko]

What about the RHS?

 

[Saitama]

What about the RHS?

$$\vec{F}=mv\frac{d\vec{v}}{ds}$$

 

[voko]

In #1, you had $\vec F = e \left( \vec v \times \vec B \right)$.

 

[Saitama]

In #1, you had $\vec F = e \left( \vec v \times \vec B \right)$.

Sorry.

$$\Rightarrow \frac{d^2\vec r}{ds^2}=\frac{e}{mv^2}(\vec{v}\times \vec{B})=\frac{e}{mv}(\hat{v}\times B)$$

So for the same trajectory, we need to have the above expression same for both the proton and electron, right?

 

[voko]

What is $\hat{v}$?

 

[Saitama]

What is $\hat{v}$?

The speed in the magnetic field is constant so I wrote $\vec v$ as $|\vec v|\hat{v}$ where $\hat v$ is the unit vector along the direction of the velocity vector.

 

[voko]

Again, you want the equation to contain only $\vec r$ and its derivatives with regard to $s$. Does $\hat v$ fit the bill?

 

[Saitama]

Again, you want the equation to contain only $\vec r$ and its derivatives with regard to $s$. Does $\hat v$ fit the bill?

Can you please explain some more? I don't get what you ask me here.

 

[voko]

$\vec v = \frac {d\vec r} {dt}$. It is the derivative of $\vec r$ with respect to $t$. The equation of the curve may only contain things that depend on the parameter of the curve, $s$ in this case. If you want to use the unit vector of velocity, you need to demonstrate that it can be represented via $\vec r$ and its $s$-derivatives.

 

[Saitama]

$\vec v = \frac {d\vec r} {dt}$. It is the derivative of $\vec r$ with respect to $t$. The equation of the curve may only contain things that depend on the parameter of the curve, $s$ in this case. If you want to use the unit vector of velocity, you need to demonstrate that it can be represented via $\vec r$ and its $s$-derivatives.

I had,
$$\vec v=v\frac{d\vec r}{ds}$$

I can replace $\vec v$ with $|\vec v|\hat{v}$ which gives

$$\hat v=\frac{d\vec r}{ds}$$
Is this what you ask me to do?

 

[voko]

So what do you get all in all?

 

[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.