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vectors

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Poirot

Banned
Feb 15, 2012
250
We have y''(t)=B x y'(t), where y:R->R^3, and B is a constant vector in R^3

1)If y represents a trajectory of a particle, show the speed of the particle is constant.

2)Show that the component of the particle's velocity in the direction of the vector B is also constant.

My thoughts: For 1) I think I may need to show that y''(t) is normal to y(t).
For 2) If y'(t) is in the direction of B, then is y'(t) x B=0.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
1) There's probably something "clever" you can do with this part. Perhaps you could do this:

\begin{align*}
\ddot{\mathbf{y}}&=\mathbf{B}\times \dot{\mathbf{y}}\\
\ddot{\mathbf{y}}\cdot \dot{\mathbf{y}}&=(\mathbf{B}\times \dot{\mathbf{y}})\cdot \dot{\mathbf{y}}.
\end{align*}
What is the RHS now? Could you then do something to this equation that would give you $\dot{\mathbf{y}}\cdot \dot{\mathbf{y}}=(\dot{y})^{2}$ on the LHS?

2) What you've written is correct, but I'm not sure it'll help you much, though I could be wrong. You're interested in $\dot{\mathbf{y}}_{\mathbf{B}}$, my notation for the component of $\dot{\mathbf{y}}$ in the direction of $\mathbf{B}$. Note that if you take your original equation and dot it with $\dot{\mathbf{y}}_{\mathbf{B}}$, a similar sort of thing will happen to what happened in the first part. You might be able to work with that. Also note that
$$\dot{\mathbf{y}}_{\mathbf{B}}=\frac{\mathbf{B} \cdot \dot{ \mathbf{y}}}{\mathbf{B} \cdot \mathbf{B}}\,\mathbf{B},$$
which is the projection of the vector $\dot{\mathbf{y}}$ onto the vector $\mathbf{B}$. Basically, this expression answers the question, "How much of $\dot{\mathbf{y}}$ is in the direction of $\mathbf{B}$?"
 
Last edited:

CaptainBlack

Well-known member
Jan 26, 2012
890
We have y''(t)=B x y'(t), where y:R->R^3, and B is a constant vector in R^3

1)If y represents a trajectory of a particle, show the speed of the particle is constant.

2)Show that the component of the particle's velocity in the direction of the vector B is also constant.

My thoughts: For 1) I think I may need to show that y''(t) is normal to y(t).
For 2) If y'(t) is in the direction of B, then is y'(t) x B=0.
Trivially y'' is normal to y' since the direction of a cross product is normal to the plane defined by the two vectors being crossed.

You may without loss of generality assume \({\bf{B}}=(b,0,0)\), that is use a coordinate system with the x-axis pointing along the direction of the magnetic field. Then I would assume a trial solution of the form \( {\bf{y}}=[\dot{y}_1(0) t, \rho \cos(\omega t+\phi), \rho \sin(\omega t+\phi)]\).

If that works it will immediately answer part 2) as well.

You may ask why we would assume something like that, well it is because we know that an electron spirals around the direction of a magnetic field.

CB
 
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Poirot

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Feb 15, 2012
250
How do I get |y'(t)|^2 from y''(t).y'(t) and what are trying to achieve by this?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
How do I get |y'(t)|^2 from y''(t).y'(t) and what are trying to achieve by this?
The speed is equal to the magnitude of the velocity. One expression to obtain the speed is
$$s=\sqrt{v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}=\sqrt{v_{x}\cdot v_{x}+v_{y}\cdot v_{y}+v_{z}\cdot v_{z}}=\sqrt{\mathbf{v}\cdot\mathbf{v}}.$$
Squaring both sides yields
$$s^{2}=\mathbf{v}\cdot\mathbf{v}.$$
In your case, $\mathbf{v}=\dot{\mathbf{y}}.$ So if you can obtain the expression $\dot{\mathbf{y}}\cdot\dot{\mathbf{y}}$, you take the square root, and you're done.

So, how can you get $\dot{\mathbf{y}}\cdot\dot{\mathbf{y}}$ from $\ddot{\mathbf{y}}\cdot\dot{\mathbf{y}}$? Well, supposing we write it out:
$$\ddot{y}_{1}\dot{y}_{1}+\ddot{y}_{2}\dot{y}_{2}+\ddot{y}_{3}\dot{y}_{3}.$$
Here I've switched to numbers for the indices to avoid confusion arising from the fact that we're using a non-standard notation for the trajectory. What we'd like to get at is
$$\dot{y}_{1} \dot{y}_{1}+\dot{y}_{2} \dot{y}_{2}+\dot{y}_{3} \dot{y}_{3}=
\dot{\mathbf{y}} \cdot \dot{\mathbf{y}}.$$

Now let's look at just one of those previous terms: $\ddot{y}_{1}\dot{y}_{1}$. Remember that each of those is a function of time. I have a function $\dot{y}_{1}$ times its derivative $\ddot{y}_{1}$, and I'd like an expression involving just the function $\dot{y}_{1}$. How could you get that?