B Helix in intrinsic coordinates?

[etotheipi]

If a particle is in a magnetic field $\vec{B} = B\hat{z}$ with velocity $\vec{v} = v_x \hat{x} + v_y \hat{y} + v_z \hat{z}$, then in Cartesian coordinates we can obtain the pair of differential equations $$\ddot{x} = \frac{qB}{m}\dot{y}$$$$\ddot{y} = -\frac{qB}{m}\dot{x}$$which give the solution$$\vec{r}(t) = \begin{pmatrix}r_L \cos{\omega t}\\r_L \sin{\omega t}\\v_z t\end{pmatrix}$$where $\omega = \frac{qB}{m}$ and $r_L = \frac{mv_{\bot}}{qB}$, i.e. a helix.

Furthermore, we see that the acceleration vector has magnitude $\frac{v_{\bot}^2}{r_L}$ in a direction perpendicular to the velocity (and the magnetic field).

However, I'm having some trouble relating this to the acceleration in intrinsic coordinates. The radius of curvature of the helix given by $\vec{r}(t)$ is $\rho = \frac{r_L^2 + v_z^2}{r_L}$ and the normal component of acceleration in intrinsic coordinates is $\frac{v^2}{\rho} = \frac{v^2 r_L}{r_L^2 + v_z^2}$, with $v^2 = v_z^2 + v_{\bot}^2$.

The normal component of acceleration in intrinsic coordinates should equal $\frac{v_{\bot}^2}{r_L}$, though, since both are the components of acceleration perpendicular to the velocity. I wondered what I've done wrong?

 

[kuruman]

Can you be a little more detailed about your assertions of the expressions for the radius of curvature and the normal component of the acceleration? How did you get them? How would they differ in an inertial frame that moves in the $z$-direction with velocity $v_z$?

 

[etotheipi]

Can you be a little more detailed about your assertions of the expressions for the radius of curvature and the normal component of the acceleration? How did you get them? How would they differ in an inertial frame that moves in the $z$-direction with velocity $v_z$?

The first part is that a helix of equation $(x(t),y(t),z(t)) = (a\cos{t}, a\sin{t}, bt)$ has a radius of curvature $\frac{a^2 + b^2}{|a|}$; there's a derivation here.

The second is that in intrinsic coordinates, we have
$$\vec{v} = v\vec{e}_t$$ $$\vec{a} = \dot{v} \vec{e}_t + \frac{v^2}{\rho}\vec{e}_n$$ where $\vec{e}_n$ is the basis vector towards the instantaneous centre of curvature and $\vec{e}_t$ points in the tangential direction. Since the magnetic field does no work here, $\dot{v}=0$ and we must have $\vec{a} = \frac{v^2}{\rho}\vec{e}_n$.

This coordinate system is established in the same inertial frame as the Cartesian system, except its basis vectors vary with position in the base space. I don't know if it makes sense to boost these coordinates.

 

[A.T.]

This coordinate system is established in the same inertial frame as the Cartesian system, except its basis vectors vary with position in the base space.

But then they aren't inertial coordinates, so why do you expect the same accelerations as in the inertial frame?

 

[etotheipi]

But then they aren't inertial coordinates, so why do you expect the same accelerations as in the inertial frame?

I don't understand. The basis vectors of a polar coordinate system vary with position in the base space, but we can still establish a polar coordinate system in an inertial frame.

 

[etotheipi]

@A.T. I found this sound-byte from Wikipedia:

In a general coordinate system, the basis vectors for the coordinates may vary in time at fixed positions, or they may vary with position at fixed times, or both. It may be noted that coordinate systems attached to both inertial frames and non-inertial frames can have basis vectors that vary in time, space or both, for example the description of a trajectory in polar coordinates as seen from an inertial frame.[13] or as seen from a rotating frame.[14] A time-dependent description of observations does not change the frame of reference in which the observations are made and recorded.

I suppose intrinsic coordinates are weird in that the basis depends on the trajectory. I'm struggling to figure out how this coordinate system relates to a frame of reference.

For a polar coordinate system it is easy; if the origin is accelerating or the line $\theta=0$ is rotating w.r.t an inertial coordinate system, then those polar coordinates are not inertial. Otherwise, those polar coordinates are inertial. In both cases, the basis vectors at any given time vary with position.

It is not so easy to visualise for intrinsic coordinates. The origin appears to be fixed (the point $s = 0$), though, and it's not really obvious to me why there would be any fictitious forces. Indeed for circular motion, we could take $F_n = ma_n \implies F_n = \frac{mv^2}{\rho}$ and get the right answer, so there isn't a need for fictitious forces there.

I'm still pretty confused as to the OP

 

[etotheipi]

The tangent vector is $$\vec{e}_t = \frac{1}{\sqrt{(r_L \omega)^2 + v_z^2}} \begin{pmatrix}-r_L \omega \sin{\omega t}\\r_L \omega \cos{\omega t}\\v_z\end{pmatrix}$$ whilst the normal vector is $$\vec{e}_n = \frac{\dot{\vec{e}_t}}{|\dot{\vec{e}_t}|}= \begin{pmatrix}-\cos{\omega t}\\ -\sin{\omega t}\\0\end{pmatrix}$$ From the equation of the helix obtained in Cartesian coordinates, we have that $\vec{a} = r_L \omega^2 \vec{e}_n$. So the normal component of acceleration in intrinsic coordinates is $r_L \omega^2 = \frac{v_{\bot}^2}{r_L}$.

The question is they why $v_{\bot}$ feeds into that equation when we can otherwise derive $\vec{a} = \dot{v} \vec{e}_t + \frac{v^2}{\rho}\vec{e}_n$ with $v = |\vec{v}|$ as the magnitude of the total velocity?