Particles passing through a surface

Today is the day I ask silly questions :

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...

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

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.

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.

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 $\vec{u}$

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 $dS$

• 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 $dS$ . Oct 9, 2014 at 14:30

I don't understand... Those two cylinders have the same volume $S.L$:

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...