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I have studied that the net force on a charged object moving with velocity $v$ under both electric and magnetic fields in given as $\vec{F}=q(\vec{E}+\vec{v}\times \vec{B})$. I have also been told by my teachers that the net force given by this equation is frame independent for inertial frames.

Regarding this I have two doubts:

  1. Is the force frame independent in only magnitude or both magnitude and direction?
  2. Why is either magnitude or direction frame independent at all, since the force is dependent on velocity which is not frame independent. For example, if we assume a constant electric field $E$ and a constant magnetic field $B$ and two observers $O_1$ and $O_2$ where $O_1$ is at rest and $O_2$ is moving with constant velocity $v$. Now if we project a charged particle also with velocity $v$, $O_1$ will observe the particle's velocity as $v$ and hence the force on the particle observed by $O_1$ is $q(E+v\times B)$. $O_2$, however, observes the particle to be at rest as there is no relative velocity between O2 and particle. So shouldn't $O_2$ observe the force as only $qE$? The net direction of the forces observed by both observers should also be different.

How do we resolve this?

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  • $\begingroup$ Hello! I have edited your question using MathJax (LaTeX) math typesetting. For future questions, you can refer to MathJax basic tutorial and quick reference. Thanks! $\endgroup$
    – jng224
    Commented Aug 28, 2021 at 12:02
  • $\begingroup$ Thank you very much $\endgroup$
    – utkarsh
    Commented Aug 28, 2021 at 12:03
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    $\begingroup$ Have you actually tried to compute this force in two different frames? Don't forget that $E$ and $B$ are not frame-independent, either, so velocity is not the only thing that changes between frames! $\endgroup$
    – ACuriousMind
    Commented Aug 28, 2021 at 12:05
  • $\begingroup$ That's interesting, I hadn't thought about E and B changing with frame. If we have assumed constant fields, can we say that these will remain independent of frame? How do you think they will change? $\endgroup$
    – utkarsh
    Commented Aug 28, 2021 at 12:18
  • $\begingroup$ @Jonas : Congratulations for your successful efforts to keep the questions well-formatted (especially those post by new contributors). $\endgroup$
    – Frobenius
    Commented Aug 28, 2021 at 12:39

4 Answers 4

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Let $\bf p$ be the 3-momentum of a particle of charge $q$ situation in an electromagnetic field ${\bf E}$, ${\bf B}$. Then the following formula is fully correct in special relativity; it holds at all speeds: $$ \frac{d {\bf p}}{d t} = q ({\bf E} + {\bf v} \times {\bf B}) $$ In this formula, $t$ is the time as registered in some inertial frame in which all the quantities are being observed.

Now the question is whether, when we work out $d {\bf p}/d t$, we will get an answer that is frame-independent, i.e. the same no matter which inertial frame has been adopted for the purpose of saying what we mean by $\bf p$ and $t$. The answer is no when high speeds are involved, but approximately yes when the speeds are small compared to the speed of light. That is, in Newtonian physics the force and acceleration do not change from one inertial frame to another, but in special relativity they do.

What I expect the teacher wanted to make clear is that you can have cases where $\bf E$ is non-zero in one frame, and zero in another, but the magnetic field $\bf B$ will change in such a way that the force is almost unchanged (with corrections only at the level of $(v/c)^2$.) And similarly you can have one frame with $\bf B$ and another with no $\bf B$, but $\bf E$ will change so as to leave the force almost the same. (There are also cases where both fields are non-zero no matter what frame you pick). For speeds of order one metre per second, the difference between the force in one frame and another is at the level of a part in $10^{16}$, i.e. tiny. Since almost all of science is less precise than this, it is fair for a teacher to assert that the force is "unchanged".

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    $\begingroup$ Just for completeness if permitted: Following § 3.1 from the book Gravitation of Misner, Thorne and Wheeler the truly (at all speeds) frame independent force is $\frac{d\boldsymbol{P}}{d\tau}=\gamma\,q(\boldsymbol E+\boldsymbol v\times\boldsymbol B)$ (in fact this is only the spacial component of the four force). $\tau$ is proper time and $\gamma$ the well-known Lorentz factor. $\endgroup$
    – Kurt G.
    Commented Aug 28, 2021 at 15:55
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    $\begingroup$ @Kurt yes I see what you mean in that this is the spatial part of the four-force. The four-force is a four-vector and one may say that a four-vector is what it is and, as a four-vector, does not depend on inertial frame. But its components do depend on inertial frame: they transform as given by Lorentz transformation. $\endgroup$ Commented Aug 28, 2021 at 16:00
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Good question. You have stumbled upon the fact that Maxwell's equations are not invariant under a Galilean Transformation. Galilean Relativity says that two objects separating with a relative velocity $\vec v$ will calculate the same force on a particle: $$x'=x-vt$$ therefore $\ddot x'=\ddot x$

So first of all let us make this clear: The force is in fact unequal since Galilean Relativity is wrong and Maxwell's equations are correct.

To prove this: Start with the conjecture that the force is equal in the two frames and you will see that this condition does not reproduce Maxwell's equations.

This was in fact one of the problems that got Einstein interested in Special Relativity back in 1905.

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  • $\begingroup$ Thank you for the answer. However, I'm still in high school and am not familiar with Galilean transformation or relativity. Is there any way to resolve the contradiction intuitively (ambitious, i know) or should i just wait a few years? $\endgroup$
    – utkarsh
    Commented Aug 28, 2021 at 12:20
  • $\begingroup$ For now it should be enough to know that the forces are not equal. $\endgroup$
    – Physiker
    Commented Aug 28, 2021 at 12:23
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    $\begingroup$ @utkarsh hmm how about a simple case? suppose particle moving in a frame where it has speed $v$ the field on it is $E + v \times B$ , in it's own frame, it has 0 velocity, let us call the electric field in this frame $E'$. Since the laws of physics are invariant in frames: $$ E' = E+vB$$ $\endgroup$ Commented Aug 29, 2021 at 8:11
  • $\begingroup$ Using this you can calculate the force on particle and you will see it agrees with old frame by definition $\endgroup$ Commented Aug 29, 2021 at 8:12
  • $\begingroup$ Thank you very much, it slipped my mind to consider the fields as frame variant also $\endgroup$
    – utkarsh
    Commented Aug 29, 2021 at 11:51
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(Three) Vectors cannot be invariants. Thus the Lorentz force is not an invariant and neither are the vectors (the electric field, velocity and magnetic field) that define it.

All that is unchanged is the equation $$\vec{F} = q( \vec{E} + \vec{v}\times \vec{B}), $$ which is true in any inertial frame, as long as all the contributing vectors are transformed correctly. i.e. For a different inertial frame

$$\vec{F}' = q( \vec{E}' + \vec{v}'\times \vec{B}'), $$ but, in general, $\vec{F}' \neq \vec{F}$ in either magnitude or direction.

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  • $\begingroup$ Thank you for the answer, I had originally failed to consider the variance of the fields themselves. :) $\endgroup$
    – utkarsh
    Commented Aug 29, 2021 at 16:46
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Since you're in high school, the best thing to do is look at a simple case. Suppose you have an infinite line of charge (linear density $\lambda$ along the $z$-axis) and a charged particle $q$. The electric field around the linear charge goes as:

$$ \vec E(r, \phi, z) = \frac{\lambda}{2\pi\epsilon_0 r} \hat r$$

The magnetic field is $\vec B = 0$.

So for a charge sitting at $r$, the Lorentz force law says:

$$ \vec F = q(\vec E + \vec v \times \vec B) =q\vec E = \frac{q\lambda}{2\pi\epsilon_0 r} \hat r$$

Now in a frame moving along at $\vec v = -\beta c \hat z$, things look different. The charge is moving at $\beta c$ along $+\hat z$ at $r'=r$.

The linear charge density is Lorentz contracted:

$$ \lambda' = \gamma\lambda$$

with $\gamma=\frac 1 {\sqrt{1-\beta^2}}$,so the electric field is:

$$ \vec E'(r', \phi', z') = \frac{\lambda'}{2\pi\epsilon_0 r'} \hat r'$$ $$ \vec E'(r', \phi', z') = \frac{\gamma\lambda'}{2\pi\epsilon_0 r'} \hat r'$$ $$ \vec E'(r', \phi', z') =\gamma\vec E $$

so it is stronger in any moving frame. The charge is also moving, with a current :

$$ \vec I = \gamma\beta\lambda c\hat z'$$

which creates a magnetic field circulating around the wire:

$$ \vec B'(r',\theta',z')= \frac{I}{2\pi\mu_0 r'}\hat{\theta}'=\frac{ \gamma\beta\lambda c}{2\pi\mu_0 r'}\hat{\theta}'$$

which can be written as:

$$ \vec B'(r',\theta',z')=\frac{-\gamma}{c^2}\big(\vec v \times \vec E\big) $$ $$ \vec B' =\frac{-\gamma\beta}{c}E\hat{\theta}'$$

The force on the charge is now:

$$ \vec F' =q(\vec E'+\vec v\times \vec B') $$

$$ \vec F' =q\gamma E(\hat r'-{\beta}^2(\hat z'\times\hat{\theta}') $$ $$ \vec F' =q\gamma E(\hat r'-{\beta}^2\hat r' $$ $$ \vec F' =q\gamma E(1-{\beta}^2)\hat r' $$ $$ \vec F' =q\gamma E\frac 1 {\gamma^2}\hat r' $$ $$ \vec F' =qE\hat r'/\gamma =\vec F/\gamma $$

or:

$$ \frac{d\vec p'}{dt'} = \frac{d\vec p}{dt}\frac 1 {\gamma} $$

where now the Lorentz factor that was from Lorentz contraction and charge becoming current now accounts for time dilation so that:

$$\frac{d\vec p'}{d\tau'} = \frac{d\vec p}{d\tau}$$

tl;dr: space and time mix, electric and magnetic fields mix, and charges and current mix, all in a manner such that the Lorentz for law is true in any inertial frame.

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  • $\begingroup$ Thanks a lot, this helped very much. I have accepted your answer $\endgroup$
    – utkarsh
    Commented Aug 30, 2021 at 9:26
  • $\begingroup$ I'm glad it helped. Sometimes "concrete" needs to precede "abstract". $\endgroup$
    – JEB
    Commented Aug 30, 2021 at 17:01

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