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The Lorentz Equation

$\mathbf{F}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B})$

involves a velocity vector $\mathbf{v}$. What is this velocity relative to? Based on e.g. this resource, it seems that when the electron is moving through a conductor, the velocity $\mathbf{v}$ is the velocity of the electron w.r.t. this conductor.

What if the electron is moving through a vacuum? What is the velocity relative to?


After getting helpful answers, I need some clarification. For this I made two drawings of a conductor, a voltage source applying $U$ volts to the conductor, a magnetic field $B$, and an electron $e$ in the conductor.

a) In drawing a, we use the reference frame $C$ of the conductor itself:

conductor with a electron e moving through it

The relevant vectors are now:

$ E = \begin{bmatrix} 0\\ \frac{U}{l}\\ 0 \end{bmatrix}, v = \begin{bmatrix} 0\\ v_{i}\\ 0 \end{bmatrix} B = \begin{bmatrix} -B_x\\ 0\\ 0 \end{bmatrix} $ Where $v_i$ is the velocity of the electron inside the conductor.

Using the Lorentz equation divided into a $F_E$ and $F_B$, we get:

$\mathbf{F} = \mathbf{F_E}+\mathbf{F_B} = q\mathbf{E} + q(\mathbf{v} \times \mathbf{B})$

$ F = q\begin{bmatrix} 0\\ \frac{U}{l}\\ 0 \end{bmatrix} + q \left(\begin{bmatrix} 0\\ v_{i}\\ 0 \end{bmatrix}\times\begin{bmatrix} -B_x\\ 0\\ 0 \end{bmatrix}\right) = q\begin{bmatrix} 0 \\ \frac{U}{l}\\ -v_{i}B_x \end{bmatrix} $

b) In drawing b, we have the same conductor but now from a "world" reference frame $W$, through which the conductor is moving with a velocity $v_c$: conductor moving through a world reference frame

The relevant vectors are now:

$ E = \begin{bmatrix} 0\\ \frac{U}{l}\\ 0 \end{bmatrix}, v = \begin{bmatrix} 0\\ v_c+v_{i}\\ 0 \end{bmatrix} B = \begin{bmatrix} -B_x\\ 0\\ 0 \end{bmatrix} $ And the Lorentz force is now: $ F = q\begin{bmatrix} 0\\ \frac{U}{l}\\ 0 \end{bmatrix} + q \left(\begin{bmatrix} 0\\ v_{i}+v_{c}\\ 0 \end{bmatrix}\times\begin{bmatrix} -B_x\\ 0\\ 0 \end{bmatrix}\right) = q\begin{bmatrix} 0 \\ \frac{U}{l}\\ -(v_{i}+v_c)B_x \end{bmatrix} $

Seemingly, the force becomes larger when looking at it from the world perspective, how is this possible?

(It's entirely possible that I made some mistakes here, this was drawn up according to my best understanding. Kindly point me to any mistakes I made.)

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    $\begingroup$ With respect to the frame you are in $\endgroup$ Mar 12, 2023 at 12:40
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    $\begingroup$ Regarding your edit --- you are assuming E and B remain the same. But they won't. For example, what is creating the B? If it's a current, electrons in that wire have different speed in the new frame of reference so the B they create is different. $\endgroup$
    – Džuris
    Mar 12, 2023 at 23:26
  • $\begingroup$ There are some sign errors in E, v and B field calculations and in cross products. $\endgroup$ Jul 5, 2023 at 7:57

2 Answers 2

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If you measure $\vec{E}$, $\vec{v}$, and $\vec{B}$ in a particular reference frame, then the Lorentz force equation will give you the force on a charge in that reference frame.

In a different reference frame, of course, $\vec{E}$, $\vec{v}$, and $\vec{B}$ will all be different, and so the resulting force will not necessarily be the same as in the original frame. In particular, the electric and magnetic fields transform amongst each other (for a "primed" frame moving in the $x$-direction) as \begin{align*} E'_x &= E_x & E'_y &= \gamma(E_y - v B_z) & E'_z &= \gamma(E_z + v B_y) \\ B'_x &= B_x & B'_y &= \gamma(B_y + v E_z/c^2) & B'_z &= \gamma(B_z - v E_y/c^2). \end{align*} This means that in the example you give above, there will be a component of the electric field in the $z$-direction, which will cancel out the additional component of the force arising from the magnetic field.

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    $\begingroup$ Thanks for the elaboration! It's most helpful. I do not fully understand it yet, but will look at this thoroughly again next weekend. I'm sure I'll get there :) $\endgroup$
    – Chris_abc
    Mar 14, 2023 at 17:55
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The velocity is relative to whatever frame of reference you choose to use. In fact, since the electron is massive, you can even go to its frame of reference, in which the velocity is zero!

Of course, this does not mean that you can just set $\mathbf{v}$ equal to any value in the force equation. What happens is that both the electric and magnetic fields also depend on your frame of reference, just like the velocity.

Usually when a problem asks you to calculate the Lorentz force a charge feels in the presence of some given fields, the frame of reference is already implicitly determined by these fields. For instance, if your electron (of charge $q_e$) is moving through an electric field like \begin{equation} \mathbf{E}(\mathbf{x}) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{(\mathbf{x}_Q - \mathbf{x})^2} \frac{\mathbf{x}_Q - \mathbf{x}}{|\mathbf{x}_Q - \mathbf{x}|} , \end{equation} you know that there is another point particle of charge $Q$ at rest in position $\mathbf{x}_Q$. Being "at rest" is also a statement that depends on the frame of reference; if you were to be moving in relation to this charge (and therefore perceive it to me moving relative to you), the field would not look like this. So if you are given this field, you know that you are working in the frame of reference in which this other charge $Q$ is at rest, so the velocity $\mathbf{v}$ of the electron should be the one relative to it.

Of course, nothing forces you to work in a certain reference frame if you don't want to. There are formulas which allow you to find the form of the fields in any other frame of reference which is rotated and has a constant velocity in relation to the one you started with. These are called Lorentz transformations, and this topic is part of special relavity.

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