Expressing equation of motion in Cartesian components

[bigevil]

Homework Statement

The general equation of motion of a non-relativistic particle of mass m and charge q when it is placed in a region where there is a magnetic field B and an electric field E is

$$m\bold{\ddot{r}} = q(\bold{E} + \bold{\dot{r}} \times \bold{B})$$

where r is the position of the particle at time t and $$\bold{\dot{r}} = v_o \bold{k}$$

Write the above equation in terms of Cartesian components of the vectors involved.

When $$\bold{B} = B \bold{j}$$ and $$\bold{E} = E \bold{i}$$ and the particle starts from the origin at t=0 and $$\bold{\dot{r}} = v_o \bold{k}$$, prove that the particle continues along its initial path.

There's a further third part which is even more complicated, but I'm trying to wrap my head around the first two.

2. The attempt at a solution

I'm definitely missing something here that I should know but don't.

$$\dot{\bold{r}} \times \bold{B} = -Bqv_o \bold{i}$$
$$\bold{E} = E \bold{i}$$

Given these, shouldn't $$m\ddot{x} = qE - Bv_{o}q$$? And shouldn't the rest be effectively zero because both E and the cross vector only have values in the i direction!

The answer given is this:

$$m\ddot{x} = -\frac{(Bq)^2}{m}x + qE - Bv_{o}q$$
$$\ddot{y} = 0$$

No answer for $$\ddot{z}$$ is given.

I'd appreciate any hints, please don't work everything out for me...

 

[HallsofIvy]

First, that's in 3 dimensions, not two, isn't it?

If $\vec{B}= B\vec{j}$ then
$$\vec{r}'\times\vec{B}= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ \frac{d x}{dt} & \frac{d y}{dt} & \frac{d z}{dt} \\ 0 & B & 0\end{array}\right|$$
[tex]= -B\frac{dz}{dt}\vec{i}+ 0\vec{j}+ B\frac{dx}{dt}\vec{k}[/itex]<br /> <br /> Put those into your component equations.[/tex]

 

[bigevil]

Thanks Halls. It's 3 dimensions ijk/xyz. I already did that. I used $$B_x, B_y, B_z$$.

My own equation for the x-component equation comes up to:

$$m\ddot{x} = qE_x + qB_{z}\dot{y} - qB_{y}\dot{z}$$, of which the middle term equals zero. And I still don't have a function in terms of x, as per the answer. I think I've done the component equation wrongly, but I don't know what or why.

 

[gabbagabbahey]

To extend on what Halls said; you seem to be assuming that $$\mathbf{\dot{r}}(t)=v_0\mathbf{k}$$ for all $$t$$ when you are computing $$\mathbf{\dot{r}}\times\mathbf{B}}$$, but you are only told that $$\mathbf{\dot{r}}(0)=v_0\mathbf{k}$$.

You should get that at a general time $t$, $$\mathbf{\dot{r}}\times\mathbf{B}}=B(-\dot{z}\mathbf{i}+\dot{x}\mathbf{k})$$ (Halls' result contains an error)...And hence you should find that you obtain coupled DE's which you must decouple in order to obtain the given result.

 

[bigevil]

I thought of that too gabba, but my reasoning was that the particle starts from the origin, so integrating $$\bold{\dot{r}}$$, $$\bold{r}(t) = v_{o}t + c$$, where c=0, so the relation holds as I assumed when differentiating back. Is this right? (EDIT: of course not.)

Let me chew on what you said for a while... thanks gabba.

 

[bigevil]

I did get at the second equation you mentioned, gabba, but still I don't understand how to get the coupled DEs. Presumably you mean changing variables or applying chain rule after reaching this second equation, still trying something out, but still lost nonetheless...

I'm just working on the x-component equation now. What I did think of is expanding $$-B \frac{dz}{dt}$$. That would generate two functions if B was a vector. But it isn't. So I'm just stuck with $$\frac{dz}{dt}$$ now.

 

[HallsofIvy]

I just noticed that I had the derivatives of "x" and "z" reversed in my formula for the cross product so I corrected my post. That, I think, is the error gabbagabbahey was referring to.

Since $\vec{E}= E\vec{j}$ and $\vec{r}\times\vec{B}= -\frac{dz}{dt}\vec{i}+ \frac{dx}{dt}\vec{k}$ you equations will be
mx"= -B z', my"= -E, mz"= Bx'

It is easy to solve for y separately. To solve for x and z, either write it as a matrix equation,
$$m\left[\begin{array}{c}x" \\ z"\end{array}\right]= \left[\begin{array}{cc}0 & -B \\ B & 0\end{array}\right]\left[\begin{array}{c} x \\ z\end{array}\right]$$
and find eigenvalues or, simpler in concept but more computation,
differentiate the first equation to get mx"'= -B z"= -B(Bx'/m) or x"'= -(B/m)²x'. That's easy to solve for x' and then integrate to get x. Once you have that us z'= -(m/b)x" to find z' and then integrate to get z.

 

[bigevil]

Thanks a lot Halls and gabba, I've got it. The matrix equation you were talking about doesn't display though.

Please look through my work here, hopefully it's right:

$$\bold{E} + \bold{\dot{r}} \times \bold{B} = (E - \dot{z}B) \bold{i} + \dot{x}B \bold{k}$$
as discussed above.

a) $$m\ddot{x} = qE - \dot{z}Bq$$
b) $$\ddot{z} = \dot{x}\frac{Bq}{m}$$

Integrate b) to get $$\dot{z} = \frac{Bq}{m} x + c$$, but we already know that $$\dot{z}(0) = v_o$$, so $$v_o = c$$ and $$\dot{z} = \frac{Bq}{m}x + v_o$$

Then, $$m\ddot{x} = qE - (\frac{Bq}{m}x + v_o)qB$$ giving the answer required.

 

[gabbagabbahey]

Thanks a lot Halls and gabba, I've got it. The matrix equation you were talking about doesn't display though.

Please look through my work here, hopefully it's right:

$$\bold{E} + \bold{\dot{r}} \times \bold{B} = (E - \dot{z}B) \bold{i} + \dot{x}B \bold{k}$$
as discussed above.

a) $$m\ddot{x} = qE - \dot{z}Bq$$
b) $$\ddot{z} = \dot{x}\frac{Bq}{m}$$

Integrate b) to get $$\dot{z} = \frac{Bq}{m} x + c$$, but we already know that $$\dot{z}(0) = v_o$$, so $$v_o = c$$ and $$\dot{z} = \frac{Bq}{m}x + v_o$$

Then, $$m\ddot{x} = qE - (\frac{Bq}{m}x + v_o)qB$$ giving the answer required.

Looks Good to me!