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Heat is the macroscopic phenomenon related to the kinetic energy of the particles. So, I wonder if there is a way to express the absolute temperature as the volume integral of the kinetic energy of the particles over a control volume.

I remember having found something like in Wikipedia, but without a proper reference, and I can't find it again anyway.

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  • $\begingroup$ In theory, you could just define temperature as $T=\oint_V v^2(\mathbf{x})\ dm$ $\endgroup$
    – user195162
    May 17, 2018 at 21:41
  • $\begingroup$ That simple ? How do you manage the consistency of units between K and m²/s².kg ? $\endgroup$
    – Aurélien Pierre
    May 19, 2018 at 4:07

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Given some particle distribution $f(t,\,\mathbf x,\,\mathbf v)$, one can define the pressure as, $$ p=\frac{1}{3}\operatorname{Tr}\left[\int m\mathbf w\mathbf wf\left(t,\,\mathbf x,\,\mathbf v\right)\,\mathrm d^3x\right] $$ where $\mathbf w=\mathbf v-\mathbf V$ is the relative velocity where $\mathbf v$ is the particle velocity and $\mathbf V$ the bulk flow. We also define the particle density, $$ n=\int f\left(t,\,\mathbf x,\,\mathbf v\right)\,\mathrm d^3x $$ Then, in thermodynamic equilibrium, the distribution function is essentially Maxwellian with some $T$, thus we can write $p=nT$ or, $$ T=p/n=\frac{1}{3n}\operatorname{Tr}\left[\int m\mathbf w\mathbf wf\left(t,\,\mathbf x,\,\mathbf v\right)\,\mathrm d^3x\right] $$ which may or may not be convenient to use.

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  • $\begingroup$ I assume you write vectors in bold, but what are Tr and t here ? $\endgroup$
    – Aurélien Pierre
    May 19, 2018 at 4:04
  • $\begingroup$ Tr is trace, t is time. $\endgroup$
    – J.G.
    May 19, 2018 at 6:10
  • $\begingroup$ @AurélienPierre yes, vectors are bold, as is standard convention, $t$ is time and Tr is the trace. Probably could just do a scalar product of $\lVert\mathbf w\rVert$, but it's sometimes nice to be complete. $\endgroup$
    – Kyle Kanos
    May 19, 2018 at 10:08
  • $\begingroup$ Standard just in America ;-) $\endgroup$
    – Aurélien Pierre
    May 21, 2018 at 23:24
  • $\begingroup$ @AurélienPierre last I checked ISO 80000-2 standards are global, not American. $\endgroup$
    – Kyle Kanos
    May 21, 2018 at 23:26
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Yes there is. Temperature is related to the average energy per degree of freedom in a system. For a volume of ideal gas containing N molecules, each having three degrees of freedom, the energy integrated over the volume is $E = \frac{3}{2}Nk_BT$. $k_B$ is Boltmann's constant.

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  • $\begingroup$ By any chance, do you have the analytic form of the original integral that leads to this result ? $\endgroup$
    – Aurélien Pierre
    May 19, 2018 at 4:14
  • $\begingroup$ All you need to know is the total number of degrees of freedom in the volume. In the ideal gas case this is 3 per molecule, so the integral is simply $\frac{3}{2}kT\int{dV\rho}$ where $\rho$ is the molecular number density. $\endgroup$
    – my2cts
    May 19, 2018 at 7:47
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I think what you're misremembering is an integral over temperature, such as $U=\int C_V dT$ expressing total thermal energy in terms of heat capacity.

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    $\begingroup$ No, I remember this one. I'm really looking for an equivalence motion-temperature at molecular level over a control volume. Which seems reasonable since it's 2 different kinds of energy. $\endgroup$
    – Aurélien Pierre
    May 19, 2018 at 4:06

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