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Is the magnetic force a vector or a vector field?

The magnetic force is written without arguments: $$ \mathbf F=q \mathbf u \times \mathbf B \tag{1} $$ Does it mean that $\mathbf r=(x,y,z)\in\mathbb R^3$: $$ \mathbf F(\mathbf r)=q\mathbf u(\mathbf r)\times \mathbf B(\mathbf r). \tag{2} $$

Or is $\mathbf r$ a function of $t$, $\mathbf r(t)=(x(t),y(t),z(y))$: $$ \mathbf F(t)=q\mathbf u(\mathbf r(t))\times \mathbf B(\mathbf r(t)), \tag{3} $$

Or is $\mathbf u$ the derivative of $\mathbf r(t)$, $\mathbf u(t)=\dot{\mathbf r}(t)$: $$ \mathbf F(t)=q\mathbf u(t)\times \mathbf B(\mathbf r(t))=q\dot{\mathbf r}(t)\times \mathbf B(\mathbf r(t)) \tag{4} $$

Which one of the last three equations is mathematically correct?

Mathematical detalis:

In equation $(2)$: Here I consider two vector fields, $\mathbf B:\mathbb R^3\rightarrow \mathbb R^3$ and $\mathbf u:\mathbb R^3\rightarrow \mathbb R^3$. $\mathbf r$ is a constant vector, $\mathbf r\in\mathbb R^3$, i.e. $\mathbf r=(x,y,z)$. If so $\mathbf F$ must also be a vector field, $\mathbf F:\mathbb R^3\rightarrow \mathbb R^3$.

In equation $(3)$: I thought of two vector fields, $\mathbf B:\mathbb R^3\rightarrow \mathbb R^3$ and $\mathbf u:\mathbb R^3\rightarrow \mathbb R^3$. But $\mathbf r$ is a vector function, $\mathbf r:\mathbb R \rightarrow \mathbb R^3$, i.e. $\mathbf r(t)=(x(t),y(t),z(t))$. And we parametrize the vector fields with $\mathbf r(t)$, so $\mathbf B(\mathbf r(t))$ and $\mathbf u(\mathbf r(t))$. If so $\mathbf F$ must be a function of one variable, $\mathbf F:\mathbb R\rightarrow \mathbb R^3$.

In equation $(4)$: I thought of the vector field $\mathbf B:\mathbb R^3 \rightarrow \mathbb R^3$, but $\mathbf u$ is not a vector field, just a vector function of one variable, $\mathbf u:\mathbb R \rightarrow \mathbb R^3$, i.e. $\mathbf u(t)=(x(t),y(t),z(t))$. Also here we parametrize the vector field, $\mathbf B(\mathbf r(t))$. And from classical mechanics we know that the derivative of the position vector is the velocity vector, so $\dot{\mathbf r}(t)=\mathbf u(t)$. If so $\mathbf F$ must be a function of one variable, $\mathbf F:\mathbb R\rightarrow \mathbb R^3$.

Emilio Pisanty
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1 Answers1

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The standard understanding of the Lorentz force is as your equation $(4)$: the force is only specified once you know the position $\mathbf r(t)$ of the particle and its velocity $\mathbf u(t)=\dot{\mathbf r}(t)$ at that time (though note a probable typo in your notation - $\dot{\mathbf u}$ is the acceleration, not the velocity), as $$ \mathbf F(t) = q \dot{\mathbf r}(t)\times \mathbf B(\mathbf r(t)). $$

However, it is also possible to consider situations like the force field felt by some extended charge distribution $\rho(\mathbf r,t)$ moving with some velocity field $\mathbf u(\mathbf r,t)$, in which case the total magnetic force on the distribution at time $t$ is given by $$ \mathbf F(t) = \int \rho(\mathbf r,t) \mathbf u(\mathbf r,t)\times \mathbf B(\mathbf r,t) \, \mathrm d\mathbf r, $$ which is a close analogue to your equation $(2)$, particularly if you consider the force density acting on each individual chunk $\rho(\mathbf r,t)\,\mathrm d\mathbf r$ of charge.

Your equation $(3)$, on the other hand, makes very little sense - there isn't really a situation where $\mathbf u$ is a function of the trajectory but it isn't simply its derivative.

Emilio Pisanty
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