# Tag Info

6

If you have heat, you have "random" movement of different molecules in your material. When two different molecules with different velocity vectors interact, they sometimes emit a photon carrying away some of their energy. This results in the relative kinetic energy of the two molecules being reduced (due to conservation of energy). This average relative ...

-2

This can be understood easily: If your temperature is higher than the surrounding temperature heat will flow out to the surrounding. It is analogous to electric current which moves from a higher potential to a lower potential. Similarly heat current flows from high heat potential(high temperature) to a lower heat potential(low temperature)

1

In the present epoch the universe is a long way from thermal equilibrium. It consists of a large number of isolated hot spots (a.k.a. stars) in a sea of background radiation which has an average temperature of just $2.7$ K. But stars have finite lifetimes (although this can be a very long time for white dwarf start) so eventually all the stars in the ...

9

Everything that is not 0 Kelvin radiates electromagnetic energy. In vacuum, this is the only relevant form of heat transfer. The hotter you are, the more energy you radiate (I believe the relevant equation is given here). The question whether you cool off or heat up in space depends on whether you absorbs more electromagnetic radiation than you give away. ...

42

You exchange heat with the objects around you. If the objects around you are hotter than you, you'll heat up. If the objects around you are cooler than you (neglecting the heat you're generating due to metabolic processes), you'll cool off. In space, the objects around you (mostly interstellar medium) is cooler than you so you radiate more heat away from ...

0

Because the potential falls off quickly with distance (like $r^{-6}$, characteristic of a van der Wasls interaction) the problem you encounter is essentially the same as the one you would encounter if you asked for the mean distance between two noninteracting particles. If any positive distance is allowed, then that average clearly diverges; the particles ...

3

let say you have this vector differential equations: $$\vec{\ddot{x}}=-\vec{f}(\vec{x},\vec{\dot{x})}\tag 1$$ and you want to check the stability at a stable point $\vec{x}_0$ according to Lyapunov theory, you have to linearized equation (1) and calculate the eigenvalues of the linearized system. we first transformed equation (1) to first order ...

2

Partial answer I know Lyapunov stability better from chaotic systems in one dimension like the logistic map and I can confirm that one can calculate the Lyapunov constant for non-stable points. The key thing we test with Lyapunov is whether two points close together move together or apart in a dynamical system. If they move together it is stable, if it ...

4

Here we will work in the Euclidean$^1$ formulation, where Euclidean time $\tau\in\mathbb{R}$ is real. The potential is assumed to satisfy $$V(q_0)~=~0~=~V(\sigma) ,\tag{A}$$ cf. OP's figure. The bounce solution satisfies the following boundary conditions (BCs) $$\dot{q}(\tau_i)~=~0\quad \wedge\quad q(\tau_i)~=~q_0\quad \wedge\quad q(0)~=~\sigma,\tag{B}$$...

1

I have tried to follow your sequence. It was not clear to me from the description if at the end you were bringing B together with C before or after they each were each in contact with A. But it turned out not to matter as either way heat will flow from C to B. If I followed you correctly, the sequences are as shown in the diagram below. The equilibrium ...

1

One kind of equilibrium is thermal equilibrium which generally means there are no temperature differences between a system and the surroundings driving heat transfer between the two. The system and the surroundings are in mutual thermal equilibrium.. But the system and surroundings may not be in mechanical equilibrium due to pressure differences between them....

1

The internal energy per unit mass (the specific internal energy) of a solid is proportional to temperature. The total internal energy equals the internal energy per unit mass times the total mass. Increasing the volume (and thus the mass) of the a solid, in the absence of any heat transfer from the surroundings to the solid, or a black body that generates ...

3

Let us consider a situation where we put one "blue" molecule in a gas of "red" molecules. We allow it to equilibriate with a heat bath of temperature T. In this context the "blue" molecule is certainly a single classical particle and it can be associated with a temperature. The important point here is what information do we know about the "blue" particle, ...

5

Thermal energy is the energy contained in the internal degrees of freedom of a system, so there are two problems with saying that the KE of a single particle gives its temperature. First, the system’s overall momentum, angular momentum, and so forth are by definition external degrees of freedom. Since they are external then by definition they are not ...

1

It is not really plausible, at least not to me and no more plausible than what it purports to replace namely Caratheodory's "axiom of adiabatic inaccessibility". The latter states that in an arbitrarily small neighborhood of equilibrium there are states that cannot be reached by an adiabatic process. To your question the answer is in Lieb's paper right after ...

0

Good question, I am trying to answer this question(partially) from symmetry considerations. If the system has the time reversal or inversion symmetry, then $\varepsilon_{-k}=\varepsilon_{k}$. From this $v(-k)=-v(k)$, and $f(\varepsilon_{-k})=f(\varepsilon_k)$. Since for any point $k$ in the BZ, we can always find its counterpart $-k$, therefore the ...

0

Building off of Rob Jeffries answer, I will try to write out a complete expression for the case of electrons with higher kinetic energy than the Blackbody thermal energy $\sim k_b T$. The starting point is the expression for the total power irradiated by an electron to a photon density $U_{\mathrm{photon}}$ through inverse Compton scattering (see the nice ...

2

If you are going to ignore electron interactions then the only processes going on are interactions between photons and electrons. If those interactions are limited to Thomson scattering (which is elastic), then the electrons cannot change their energies and so can never achieve a Maxwell-Boltzmann distribution. However, even for low electron energies there ...

-1

If you are ignoring the e-e interaction your system is effectively a free electron. This electron is moving in a background of thermal radiation. Inelastic Thomson scattering of photons will eventually decrease the initial electron momentum to a thermal value. I assume a nonrelativistic electron. The question is then at what timescale will its initial ...

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