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Today is the day I ask silly questions :

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The book says the particles passing through the surface $dS$ are the ones contained in the cylinder of volume $dS.v.dt.cos(\theta)$ but I really don't see why.

For me, all the particles in the volume $dS.v.dt$ (which is bigger than $dS.v.dt.cos(\theta)$) will pass through the surface. I don't understand the need of putting $cos(\theta)$ in the equation...

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  • $\begingroup$ what does $\vec{u}$ mean? It seems like the velocity is changing? $\endgroup$
    – ROIMaison
    Oct 9, 2014 at 11:20
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    $\begingroup$ $\vec{u}$ is just the unit vector normal to $dS$ $\endgroup$
    – mwa1
    Oct 9, 2014 at 11:29

5 Answers 5

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Look at large $\theta$ : fewer particles per unit time will reach the surface because their perpendicular velocity is much less. When $\theta = \pi/2 $ zero particles cross the surface.

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The volume of the shape drawn there is $dS\,v\,dt\cos(\theta)$. $dS$ is not a cross section, it is at an angle to the axis $v$ is aligned with.

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You can't simply multiply $dS$ with $\vec{v}$ to obtain the volume. If we assume a coordinate system $x$ perpendicular to $dS$, and $y$ in the plane of $dS$.

The volume is defined as: $$V=\int^{s_{begin}}_{s_{end}} \: S \vec{dx}_{\perp S}$$ However, if we want to express it as a function of $v dt$ we get:

$$V=\int^{t_{begin}}_{t_{end}} \: S \: \vec{(v dt)}_{\perp S} \: =\int^{t_{begin}}_{t_{end}} \: S \: \vec{v}_{\perp S} \: dt $$

This shows $\vec{v}_{\perp S}$ which is in this case $\vec{v} \: cos(\theta)$ which is equal to $\vec{u}$

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The way I see it, with larger $\theta$ (as long as it's not $\pi/2$) it is still cylinder of volume $dS.v.dt$ that will pass through the surface $dS$

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  • $\begingroup$ But what is the size of $dS$ as $\theta$ gets large? I believe the original question is to find the flux per unit area, not per $dS$ . $\endgroup$ Oct 9, 2014 at 14:30
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I don't understand... Those two cylinders have the same volume $S.L$:

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So all particles exactly on (or beyond) surface $S$ at point $A$ and time $t$ will be exactly on (or beyond) surface $S$ shifted at point $B$ by time $(t+dt)$ assuming they all move at constant velocity $\vec{v}$. I don't see how the orientation of the surface changes anything.

I can't believe I'm stuck on something like this...

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