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#### Alexmahone

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- Jan 26, 2012

- 268

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- Jan 26, 2012

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- Jan 26, 2012

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How about

$$\int_{v_{0}}^{\infty}N(v)\,dv?$$

$$\int_{v_{0}}^{\infty}N(v)\,dv?$$

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- Jan 26, 2012

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But that has dimensions vxN.How about

$$\int_{v_{0}}^{\infty}N(v)\,dv?$$

As stated, the answer to the problem is infinity since we have to consider infinitely many velocities. (Note that I have a continuous distribution.) So I guess we cannot have such a graph.

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- Jan 26, 2012

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$$\int_{v_{0}}^{\infty}N(v)\,dv$$

still works. And if your $N(v)$ is to be at all realistic, it must die off fast enough for the integral to converge, since we believe there are a finite number of molecules in the universe!

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- Jan 26, 2012

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But I told you that N has units of [number of molecules]. With all due respect, N having units of [number of molecules]/[v] doesn't make any sense. Pretty sure you're mistaken.

$$\int_{v_{0}}^{\infty}N(v)\,dv$$

still works. And if your $N(v)$ is to be at all realistic, it must die off fast enough for the integral to converge, since we believe there are a finite number of molecules in the universe!

- Jan 26, 2012

- 890

On a continuous graph N cannot be a number it must be a number density.But I told you that N has units of [number of molecules]. With all due respect, N having units of [number of molecules]/[v] doesn't make any sense. Pretty sure you're mistaken.

CB

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- Jan 26, 2012

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What do you mean by number density?On a continuous graph N cannot be a number it must be a number density.

CB

Last edited:

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- Jan 26, 2012

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As I understand the situation you are trying to create, the x-axis is continuous and the y-axis is discrete so for any $N_i$ there should be a corresponding range of velocities \(\displaystyle v_a<v_i<v_b=N_i\). However, N can't realistically be non-integer though unless molecule means something different than I've ever seen.

- Jan 26, 2012

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Number of molecule in a unit speed interval (or rather \( \lim_{\Delta v\to 0} N(v-\Delta v/2, v+\Delta v/2)/ \Delta v \) where \( N(v-\Delta v/2, v+\Delta v/2) \) denotes the number in the speed range \((v-\Delta v/2, v+\Delta v/2) \) )What do you mean by number density?

CB

- Jan 26, 2012

- 890

This gives a plot that is independedent of assumptions about discrete or continuous (or for that matter mixed continuous and discrete) distributions (one has steps, the other does not).

CB

- Feb 13, 2012

- 1,704

The speed distribution of the molecules of a gas with absolute temperature T has been found by Maxwell and Boltzmann about hundred and fifty years ago...

$\displaystyle f(v)= \sqrt{\frac{2}{\pi}\ (\frac{m}{K\ T})^{3}}\ v^{2}\ e^{- \frac{m\ v^{2}}{2\ K\ T}}$ (1)

... where m is the mass of a single molecule and K is the 'Boltzmann's constant'. If N is the global number of molecules, the the number of molecules with $\displaystyle v> v_{0}$ is given by...

$\displaystyle N_{0}= N\ \int_{v_{0}}^{\infty} f(v)\ dv$ (2)

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

Of course in the formula (2) You consider $N_{0}$ as continuous variable and is $\displaystyle N_{0}= N\ P \{v>v_{0}\}$. In case You intend to consider $N_{0}$ as an integer, then the x-y diagram must be an histogram where on the X axis is reported an integer k and on the Y axis is reported the number of molecules $N_{k}$ having speed $\displaystyle k\ v_{u} \le v< (k+1)\ v_{u}$. In this case the number of molecules having speed $\displaystyle > v_{0}$ is...The speed distribution of the molecules of a gas with absolute temperature T has been found by Maxwell and Boltzmann about hundred and fifty years ago...

$\displaystyle f(v)= \sqrt{\frac{2}{\pi}\ (\frac{m}{K\ T})^{3}}\ v^{2}\ e^{- \frac{m\ v^{2}}{2\ K\ T}}$ (1)

... where m is the mass of a single molecule and K is the 'Boltzmann's constant'. If N is the global number of molecules, the the number of molecules with $\displaystyle v> v_{0}$ is given by...

$\displaystyle N_{0}= N\ \int_{v_{0}}^{\infty} f(v)\ dv$ (2)

$\displaystyle N_{0}= \sum_{k= \text{int} \frac{v_{0}}{v_{u}}} ^ {\infty} N_{k}$

Kind regards

$\chi$ $\sigma$