Lorentz Force- Diff Eqn Solution

[Arman777]

Homework Statement

Solve the Lorentz Force Equation (Find $\vec{r}(t)$) under constant $\vec {B}$ and $\vec {E}$.
İnitial Velocity: $\vec{v_i}=<0,0,0>$
İnitial position: $\vec{r_i}=<0,0,0>$
($\vec {B}$ and $\vec {E}$ are given values)

Homework Equations

Lorentz Force:
$\vec {F}=q(\vec {E}+\vec{v} × \vec {B})$

The Attempt at a Solution

So I write the equaiton in a diff eqn form like,
$m\vec{a}=q(\vec {E}+\vec{v} × \vec {B})$
then
$\ddot {\vec{r}}-\frac {q} {m}(\dot {\vec{r}}× \vec {B})=\frac {q\vec {E}} {m}$

Now the right side is constant and left side have also constant coefficients. I am not sure how to proceed after this. I can use auxiliary equation maybe ? but since these equations are about vectors I am not sure how would that work out.

 

[Orodruin]

Do not try to solve the second order differential equation directly. As the position itself does not enter, just solve the first order differential equation for the velocity. If you want the position you can integrate later.

What you have is a first order linear ODE.

 

[Ray Vickson]

Homework Statement

Solve the Lorentz Force Equation (Find $\vec{r}(t)$) under constant $\vec {B}$ and $\vec {E}$.
İnitial Velocity: $\vec{v_i}=<0,0,0>$
İnitial position: $\vec{r_i}=<0,0,0>$
($\vec {B}$ and $\vec {E}$ are given values)

Homework Equations

Lorentz Force:
$\vec {F}=q(\vec {E}+\vec{v} × \vec {B})$

The Attempt at a Solution

So I write the equaiton in a diff eqn form like,
$m\vec{a}=q(\vec {E}+\vec{v} × \vec {B})$
then
$\ddot {\vec{r}}-\frac {q} {m}(\dot {\vec{r}}× \vec {B})=\frac {q\vec {E}} {m}$

Now the right side is constant and left side have also constant coefficients. I am not sure how to proceed after this. I can use auxiliary equation maybe ? but since these equations are about vectors I am not sure how would that work out.

You can write a system of three coupled linear differential equations in each of the three coordinate components separately. Just expand out the vector product on the right into its component form.

 

[Arman777]

Hmm I have

$\dot {\vec{v_x}}-\frac {q} {m}< 0,-{v_x}{B_z},{v_x}{B_y}>=\frac {q\vec {E}} {m}$
$\dot {\vec{v_y}}-\frac {q} {m}< {v_y}{B_z},0,-{v_y}{B_x}>=\frac {q\vec {E}} {m}$
$\dot {\vec{v_z}}-\frac {q} {m}< -{v_z}{B_y},{v_z}{B_x},0>=\frac {q\vec {E}} {m}$

are these right or this is what should I do ?

 

[Orodruin]

Actually, writing it out in components is rather unnecessary and is going to be cumbersome in general. In this case it is a better idea to create a basis based on the electric and magnetic fields to expand the solution in (the case when they are linearly dependent is trivial).

 

[Arman777]

this case it is a better idea to create a basis based on the electric and magnetic fields to expand the solution in (the case when they are linearly dependent is trivial).

Well I don't think I can do that. Because B and E could be anything. We never learned in class the vector diff eqn we just learned diff eqn solutions for some type of equations. Umm that's kind of why I can't move on..Okay let's not write in components then but write in general. Then what should I do..or if there's a site that explains these stuff I can look there too

 

[Orodruin]

Well I don't think I can do that. Because B and E could be anything.

This does not matter. In fact, it is the reason I propose it. Of course you could introduce a coordinate system such that $\vec B = B \vec e_3$ and $\vec E = E_3 \vec e_3 + E_1 \vec e_1$, but there is no reason for doing so. The point is that, as long as $\vec E$ and $\vec B$ are linearly independent, they form a complete basis together with $\vec E\times \vec B$. Of course, this is still writing in components of a basis, just a more cleverly chosen basis.

 

[Arman777]

This does not matter. In fact, it is the reason I propose it. Of course you could introduce a coordinate system such that $\vec B = B \vec e_3$ and $\vec E = E_3 \vec e_3 + E_1 \vec e_1$, but there is no reason for doing so. The point is that, as long as $\vec E$ and $\vec B$ are linearly independent, they form a complete basis together with $\vec E\times \vec B$. Of course, this is still writing in components of a basis, just a more cleverly chosen basis.

Hmm I see your point I guess. So what will change then we have 3 bases E B and ExB. So in vxB we are simplfying the result ?

I also searched online and saw that when E and B are othogonal to each other things get more simpler..

 

[Orodruin]

Hmm I see your point I guess. So what will change then we have 3 bases E B and ExB. So in vxB we are simplfying the result ?

Why don't you try it and show us what you get?

I also searched online and saw that when E and B are othogonal to each other things get more simpler..

Indeed, but it really just separates out one component. That component is anyway irrelevant for the evolution of the others (when they are not orthogonal that component also has a contribution that depends on the solution of the other components, but it does not enter into the solution for the others so you can solve for those first and then just integrate).

 

[Arman777]

It will be something like

$\dot {\vec{v}}-\frac {q} {m}< {B_y}{v_z},0,-{B_y}{v_x}>=\frac {q\vec {E}} {m}$

Also we need a base transformation matrix from Standart basis $S=<\vec {i},\vec {j},\vec {k}>$ to this new basis $C=<\vec {E},\vec {B},\vec {E}×\vec {B}>$

so it would be like$T=\begin{pmatrix} \vec {E}~~ \vec {B}~~ \vec {E}×\vec {B} \\ \end{pmatrix}$

or we need transpoze of it actually to change the basis from C to S. Even I write these again I don't know how to solve vector ODE.

 

[Orodruin]

Why do you need a basis transformation? You are not asked to express the solution in any particular set of coordinates. The absolutely easiest thing is to forget about standard bases altogether and just work in terms of the basis introduced by the fields themselves. In other words, any vector $\vec w$ can be written on the form
$$
\vec w = w_e \vec E + w_b \vec B + w_\times \vec E \times \vec B.
$$
Use this to expand the velocity in this basis and identify terms on both sides of the differential equation. You should get a set of three coupled first order ODEs for the components of the velocity.

 

[Arman777]

Why do you need a basis transformation? You are not asked to express the solution in any particular set of coordinates. The absolutely easiest thing is to forget about standard bases altogether and just work in terms of the basis introduced by the fields themselves. In other words, any vector $\vec w$ can be written on the form
$$
\vec w = w_e \vec E + w_b \vec B + w_\times \vec E \times \vec B.
$$
Use this to expand the velocity in this basis and identify terms on both sides of the differential equation. You should get a set of three coupled first order ODEs for the components of the velocity.

I find
$$\vec{F}=(q-v_×)\vec{E}+v_e(\vec{E}×\vec{B})$$
after this we know that
$$m\dot{\vec{v}}=(q-v_×)\vec{E}+v_e(\vec{E}×\vec{B})$$
$$\dot{\vec{v}}= \dot{v_e} \vec E + \dot{v_b} \vec B + \dot{v_×}(\vec E \times \vec B)$$
so
$$m[\dot{v_e} \vec E + \dot{v_b} \vec B + \dot{v_×}(\vec E \times \vec B)]=(q-v_×)\vec{E}+v_e(\vec{E}×\vec{B})$$

so $m\dot{v_e} \vec E=(q-v_×)\vec{E}$
$\dot{v_b} \vec B=0$
and
$\dot{v_×}(\vec E \times \vec B)=v_e(\vec{E}×\vec{B})$

is this makes sense ?

Actually it would be better If I turn them to standart basis..but I can do that at the end I guess

 

[Orodruin]

The general idea is correct, but I think you are missing a lot of the stuff resulting from your basis not being orthonormal (this is of course the drawback). Can you detail how you did the cross product $\vec v \times \vec B$?

 

[Arman777]

The general idea is correct, but I think you are missing a lot of the stuff resulting from your basis not being orthonormal (this is of course the drawback). Can you detail how you did the cross product $\vec v \times \vec B$?

Umm
$$\vec{v}×\vec{B}=[{v_e} \vec E + {v_b} \vec B + {v_×}(\vec E \times \vec B)]×\vec{B}$$
$$v_e(\vec E \times \vec B)+{v_b} (\vec B×\vec{B})+{v_×}[(\vec E \times \vec B)×\vec{B}]$$
$$\vec{v}×\vec{B}=v_e(\vec E \times \vec B)-{v_×}(\vec E)$$

but I think you are missing a lot of the stuff resulting from your basis not being orthonormal

what kind of stuff

 

[Orodruin]

How did $(\vec E \times \vec B) \times \vec B$ become $- \vec E$? Remember that your basis is generally not orthonormal here, but there might be a vector product rule you can apply.

 

[Orodruin]

Also, do not forget the $q$ appearing in the magnetic part of the Lorentz force.

Could you summarise what you have now?

 

[Arman777]

How did $(\vec E \times \vec B) \times \vec B$ become $- \vec E$? Remember that your basis is generally not orthonormal here, but there might be a vector product rule you can apply.

$$(\vec{E}×\vec{B})×\vec{B}=\vec{B}[\vec{E}⋅\vec{B}]-\vec{E}[\vec{B}⋅\vec{B}]$$
$$=\vec{B}EBcosφ-\vec{E}B^2$$

is this true ? It looks awkward I mean we cannot assume E and B are orthogoanl right ? hence I get this

 

[Arman777]

Also wait ExB is a basis and B is a basis isn't it normal to get -E I mean we are not in standart basis multiplying 3 vector but we are making cross product of 2 basis ?

 

[Orodruin]

Yes, it is correct. Even if the vector is orthogonal to $\vec B$, it has a component in the $\vec B$ direction since $\vec E$ is not generally orthogonal to $\vec B$. For your own peace of mind and to get shorter equations, I suggest you introduce the notations $E_\perp = E \cos\varphi$ and $\omega = qB/m$.

 

[Arman777]

So my post 18 is wrong and I should continue ?

 

[Ray Vickson]

$$(\vec{E}×\vec{B})×\vec{B}=\vec{B}[\vec{E}⋅\vec{B}]-\vec{E}[\vec{B}⋅\vec{B}]$$
$$=\vec{B}EBcosφ-\vec{E}B^2$$

is this true ? It looks awkward I mean we cannot assume E and B are orthogoanl right ? hence I get this

$$(\vec{E} \times \vec{B}) \times \vec{B} = - \alpha \vec{E} + \beta \vec{B}, $$
where $\alpha = \vec{B} \cdot\vec{B}$ and $\beta = \vec{E} \cdot \vec{B}$ are constant scalars. You do not need to compute them unless you have actual fields specified numerically.

 

[Arman777]

$$(\vec{E} \times \vec{B}) \times \vec{B} = - \alpha \vec{E} + \beta \vec{B}, $$
where $\alpha = \vec{B} \cdot\vec{B}$ and $\beta = \vec{E} \cdot \vec{B}$ are constant scalars. You do not need to compute them unless you have actual fields specified numerically.

I need them. Okay I ll continue to do my calculations then

 

[Orodruin]

So my post 18 is wrong and I should continue ?

You are taking a product of basis vectors, yes. But two of the vectors in your basis are not orthogonal (ExB is orthogonal to both E and B by construction but E and B need not be). Only in an orthogonal basis will the cross product of two basis vectors be proportional to the third.

 

[Arman777]

You are taking a product of basis vectors, yes. But two of the vectors in your basis are not orthogonal (ExB is orthogonal to both E and B by construction but E and B need not be). Only in an orthogonal basis will the cross product of two basis vectors be proportional to the third.

I see now okay thanks

 

[Ray Vickson]

I need them. Okay I ll continue to do my calculations then

No, you really do not need them. Just keep them symbolic throughout all the calculations. Then, if you really want to, you can substitute the values of $\alpha$ and $\beta$ in the final answer, AFTER all the hard work has been done. Believe me: it will save you hours of work and eliminate many possible sources of errors if you refrain from calculating them until the very last step. However, don't take my word for it; try it both ways and see which works better. That is the very best way to learn!

 

[Arman777]

Okay so I have

$$m\dot{v_e}=q(1-v_xα)$$
$$m\dot{v_b}=qv_xβ$$
$$m\dot{v_×}=qv_e$$

 

[Ray Vickson]

Okay so I have

$$m\dot{v_e}=q(1-v_xα)$$
$$m\dot{v_b}=qv_xβ$$
$$m\dot{v_×}=qv_e$$

Right. Now, if you try a solution of the form $e^{rt}$ in the homogeneous system (without the $+q 1$ in the first right-hand-side) you will get a system of algebraic equations to determine the value or values of $r$ (in terms of $q,m,\alpha, \beta$). That will get you most of the way to solving the non-homogeneous system (with the $+1 q$ in the first right-hand-side).

 

[Arman777]

Right. Now, if you try a solution of the form $e^{rt}$ in the homogeneous system (without the $+q 1$ in the first right-hand-side) you will get a system of algebraic equations to determine the value or values of $r$ (in terms of $q,m,\alpha, \beta$). That will get you most of the way to solving the non-homogeneous system (with the $+1 q$ in the first right-hand-side).

But one of them is $v_e$ and the other one is $v_×$

 

[Arman777]

Wait okay I got it I ll find something

 

[Arman777]

I got something like

$v_x=c_1e^{r_1t}+c_2e^{r_2t}$

where $r_1=(-q^2α+q\sqrt{q^2α^2-4m^2})/2m$
and $r_2=(-q^2α-q\sqrt{q^2α^2-4m^2})/2m$

 

[Ray Vickson]

But one of them is $v_e$ and the other one is $v_×$

Yes, I know that. So, write $v_e = c_e e^{rt}, v_b = c_b e^{rt}$ and $v_x = c_x e^{rt}$ for some constants $c_e,c_b,c_x$.

 

[Orodruin]

I suggest differentiating the expression for $\dot{v}_e$ and inserting the expression for $\dot{v}_\times$ into the result. This should give you a second order ODE that you might recognize...

Once you have solved that the rest is just simple integration and insertion.

 

[Arman777]

Yes, I know that. So, write $v_e = c_e e^{rt}, v_b = c_b e^{rt}$ and $v_x = c_x e^{rt}$ for some constants $c_e,c_b,c_x$.

Well let me look again

 

[Arman777]

so I find $v_e=c_1sin(\sqrt{\frac {qα} {m^2}t})+c_2cos(\sqrt{\frac {qα} {m^2}t})$

for $v_x=c_1sin({\frac {q\sqrt{α}} {m}t})+c_2cos({\frac {q\sqrt{α}} {m}t})$

but for $v_x$ I guess I didnt add the non-homogeneous part which is only $q^2$.

for $v_b=\frac {qβ}{m}∫v_xdt$

 

[Ray Vickson]

so I find $v_e=c_1sin(\sqrt{\frac {qα} {m^2}t})+c_2cos(\sqrt{\frac {qα} {m^2}t})$

for $v_x=c_1sin({\frac {q\sqrt{α}} {m}t})+c_2cos({\frac {q\sqrt{α}} {m}t})$

but for $v_x$ I guess I didnt add the non-homogeneous part which is only $q^2$.

for $v_b=\frac {qβ}{m}∫v_xdt$

One way to do it is to write $v_e(t) = w_e(t) + a_e + c_e t,$, $v_b(t) = w_b(t) + a_b + c_b t,$ and $v_x(t) = w_x(t) + a_x + c_x t,$ where the a's and c's are constants. When you put those forms into the 3 DEs, you will get DEs for the w(t)s that have some constants in them. By choosing the a's and c's appropriately, you can eliminate all the constants and be left with the homogeneous system
$$
\begin{array}{rcr}
m \dot{w_e} &=& -q \alpha w_x \\
m \dot{w_b} &=& q \beta w_x \\
m \dot{w_x} &=& q w_e
\end{array}
$$
in $w_e, w_b, w_x$. The initial conditions on the $v(t)$s translate into initial conditions on the $w(t)$s, which then fixes the solution uniquely.