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A vector field is a function which associates a vector to every point in $\mathbb{R}^3$. In other words, a vector field is a function which takes every vector in $\mathbb{R}^3$ as input and outputs another vector in $\mathbb{R}^3$.

In physics, however, vector fields can have different "densities". Let's consider for example the Lagrangian description of fluid flow. If we take a snapshot of the flow at a given moment in time, we can associate a velocity vector to every fluid particle.

These vectors, however, wouldn't be associated to every point in space, but would instead be more spaced out in regions of lower density and closer together in regions of higher fluid density.

Is there a mathematical, rigorous way of representing a discontinuous vector field like this and of giving a value to this property which I colloquially called "density"?

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    $\begingroup$ This question doesn't make sense. A proof can be rigorous or not but a definition is never "rigorous", and you're asking for some kind of definition. $\endgroup$
    – Suzu Hirose
    Jul 19, 2022 at 20:34
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    $\begingroup$ I would interpret asking for a “rigorous” definition as asking for a precise definition. I think there’s nothing wrong with asking for a precise definition — it’s a genuine question and a genuine attempt to understand math. $\endgroup$
    – littleO
    Jul 19, 2022 at 20:49
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    $\begingroup$ Probably what you want is simply a vector field $V$ paired with a locally integrable density function $\rho : \mathbb{R}^3 \to [0,\infty)$. $\endgroup$
    – diracdeltafunk
    Jul 19, 2022 at 20:51
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    $\begingroup$ When modeling fluid flow in physics, I thought that typically we do in fact associate a velocity vector with each point in space. $\endgroup$
    – littleO
    Jul 19, 2022 at 20:51
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    $\begingroup$ Seems like you want the distribution function, which gives a density of both particle positions and velocities. $\endgroup$
    – eyeballfrog
    Jul 19, 2022 at 20:52

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After reading the comments I suppose that instead of considering the continuous distribution of mass (density function) and assigning to any point in the area the velocity, you prefer to consider each particle and associate them it's mass and velocity.

If so, then maybe you just need to consider the discrete subset $A\subset \Bbb R^3$ of particles, a function of velocities $\vec v\colon A\to \Bbb R^3$ together with a function $m\colon A\to \Bbb R$ of masses of particles. Then any particle $a\in A$ weighs $m(a)$ and has velocity $\vec v(a)$.

If you'd like to consider the dependence of time there may be better to enumerate all the points by some index set $I$ (for example $I=\{1,2,\ldots,N\}$ where $N$ is a number of particles) and to consider the mass function $m\colon I\to \Bbb R$ and a vector function $\vec r\colon [0,\infty)\times I\to \Bbb R^3$ of positions. Therefore the $i$-th particle has mass $m(i)$ (I assume the mass doesn't change, doesn't depend on time), the position in time $t$ equals $\vec r(t,i)$ and the velocity $\frac{\partial}{\partial t}\vec r(t,i)$.

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