Editing Flux (section)

==== Properties ====

These direct definitions, especially the last, are rather unwieldy {{Citation needed|date=May 2025}}. For example, the arg{{nnbsp}}max construction is artificial from the perspective of empirical measurements, when with a [[weathervane]] or similar one can easily deduce the direction of flux at a point. Rather than defining the vector flux directly, it is often more intuitive to state some properties about it. Furthermore, from these properties the flux can uniquely be determined anyway.

If the flux '''j''' passes through the area at an angle θ to the area normal <math>\mathbf{\hat{n}}</math>, then the [[dot product]]
<math display="block">\mathbf{j} \cdot \mathbf{\hat{n}} = j\cos\theta.</math>
That is, the component of flux passing through the surface (i.e. normal to it) is ''j''{{nnbsp}}cos{{nnbsp}}''θ'', while the component of flux passing tangential to the area is ''j''{{nnbsp}}sin{{nnbsp}}''θ'', but there is ''no'' flux actually passing ''through'' the area in the tangential direction. The ''only'' component of flux passing normal to the area is the cosine component.

For vector flux, the [[surface integral]] of '''j''' over a [[Surface (mathematics)|surface]] ''S'', gives the proper flowing per unit of time through the surface:
<math display="block">\frac{\mathrm{d}q}{\mathrm{d}t} = \iint_S \mathbf{j} \cdot \mathbf{\hat{n}}\, dA = \iint_S \mathbf{j} \cdot d\mathbf{A},</math>
where '''A''' (and its infinitesimal) is the [[vector area]]{{snd}} combination <math>\mathbf{A} = A \mathbf{\hat{n}}</math> of the magnitude of the area ''A'' through which the property passes and a [[unit vector]] <math>\mathbf{\hat{n}}</math> normal to the area.
Unlike in the second set of equations, the surface here need not be flat.

Finally, we can integrate again over the time duration ''t''<sub>1</sub> to ''t''<sub>2</sub>, getting the total amount of the property flowing through the surface in that time (''t''<sub>2</sub>&nbsp;−&nbsp;''t''<sub>1</sub>):
<math display="block">q = \int_{t_1}^{t_2}\iint_S \mathbf{j}\cdot d\mathbf A\, dt.</math>