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2

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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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 ...

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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: $$V=\int^{t_{begin}}_{t_{end}} \: S \: \vec{(v ... 0 The way I see it, with larger \theta (as long as it's not \pi/2) it is still cylinder of volume dS.v.dt that will pass through the surface dS 2 Look at large \theta : fewer particles per unit time will reach the surface because their perpendicular velocity is much less. When \theta = \pi/2  zero particles cross the surface. 0 When a cuboid is immersed in water, the molecules at each face of the cube repels the water molecules making the cuboid having a definite boundary and that causes it to have a definite volume. But, how do this analogy works with an electron? Hence, the volume at quantum level is indefinable. 2 The "volume in which it is contained" is an ill-defined notion for quantum objects. Consider the simplest atom, the hydrogen atom, and look at the wavefunctions for the electron states. You find that, roughly,$$ \psi(r) \propto \mathrm{e}^{-r} so the probability to find an electron at a certain distance $r$ from the nucleus decreases with increasing ...

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