Say that there was a single atom alone in a vacuum without any particles for it to bounce off of; would it have a temperature?  As we all know motion is relative, however with nothing to bounce or collide with, is the atom moving?  Or if it's not moving, then is absolute zero achieved?  Or if it is moving, then will it "slow down" as it gives off thermal energy, producing thermal energy?  note that i am not talking about : particle speeds in idealized gases where the particles move freely inside a stationary container without interacting with one another, except for very brief collisions   i am talking about no collisions with anything therefore the Maxwell–Boltzmann distribution is a different thing than what im talking about therfore Can a single molecule have a temperature is a totally different question
28356
A show on  the weather channel said that as a water molecule ascends in the atmosphere it cools. Does it make sense to talk about the temperature of a single molecule?
90933
So I was taught that:  Kinetic Energy (of electron) = Energy (of photon) - Ionization Energy  If E(photon) = IE, then KE=0 of the electron.   What does this physically/theoretically mean?   My current thoughts/interpretation is that enough energy/force is applied to ionize the electron so it is 'sufficiently far' from the atom, and then I guess it just moves with whatever speed it is moving at with natural kinetic laws, since no more energy/force is being applied....?  Any clarifications would be greatly appreciated, thanks in advance.
552
I don't have a lot of knowledge of special relativity and associated topics; some of the few things I know are that "all motion is relative" (that is, there is no 'stationary reference frame'), and the speed of light in vacuum ($c \simeq 3 \cdot 10^8~\mathrm{m~s^{-1}}$) is the absolute asymptotic speed limit (asymptotic meaning that you can never equal it, only get arbitrarily close). What escapes me is how those concepts work together - to my naive understanding, an object would never move in its own reference frame (and so would never reach $c$). What reference frame is $c$ measured against? (Is it measured against a reference frame?) Or am I looking at this the wrong way?
55172
I was studying a book about thermodynamics of nanosystems and I got stuck with this question in my mind which I couldn't find an answer for.  For instance, does a single water molecule have a state, like being liquid or something? Or is one molecule of TiO2 solid or gas? I would appreciate if anyone could clarify this.  I should state that I am not considering a macromolecule in my case.
13753
Are there experiments which show that single photons (not classical em waves) travel exactly at $c$ in vacuum?  What is the error bar in that case? The question is posed due to the fact that loop quantum gravity predicts variations in the speed of light.
72052
I understand that the average energy of each degree of freedom in a thermodynamic system is $\frac12kT$. And so, for an ideal monatomic gas, there are three degrees of freedom associated with the translational components of each atom, which gives $E$ = $\frac32kT$ for each atom in this system.  For an ideal diatomic gas, there are the three degrees of freedom from the translational components plus two more degrees of freedom associated with the rotational components of each molecule. So we end up with $E$ = $\frac52kT$ for each molecule.  My question here is why did we ignore the rotational degree(s) of freedom in case of a monatomic gas ? Like why isn't there another degree of freedom for the rotational component of a single atom ? Also, if that is neglected for some reason, then when does it become significant and can no longer be ignored ? For example, shouldn't it be included in calculations under extreme conditions like temperature of millions of degrees ?
53679
It's an old-new question (I found only one similar question with unsatisfactory (for me) answer: )  It's strange for me how all books simply pass by such an important question, and mentioning strange and mathematically unsupported reasons such as:   orbits are stationary (while as I know this is just idealization, there is no stationary orbits in reality even for Hydrogen)  electrons are actually not localized due to uncertainty principle, thus they have no acceleration (while obviously in a non-spherically symmetric orbits a kind of "charge acceleration distribution" always exist) vacuum fluctuations play a major role (according to QED).   I'm not interested in how Bohr or Schroedinger explained it, I want to see a rigorous proof with QM, QED or maybe even the standard model as whole. I would like to see how this question was closed.
37335
In many presentations of statistical mechanics where we have a system of particles having mass, such as the molecules of an ideal gas, the temperature is often equated to the average relative velocity of the particles. However no such velocity can exceed that of light. Is there therefore an absolute upper bound to temperature as well as an absolute zero?
47714
Just to get the thoughts rolling...  Consider a two state system with discrete energy levels $E_1$ and $E_2$ where $E_2 &gt; E_1$ which contains $N$ particles.  We can easily deduce that the state of maximum entropy is when energy level $E_1$ and $E_2$ each contain $\frac{N}2$ particles.  Based on the (flipped-upside-down) thermodynamic definition of temperature, $$(\frac{\partial E}{\partial S})=T$$ we see that when energy as a function of entropy increases, temperature also increases.  This means, for our two state system, every time a particle populates level $E_2$, there is a corresponding increase in temperature because that population increases entropy.  What about when we go beyond the point of maximum entropy where there are $\frac N2$ particles in $E_1$ and $E_2$?  Well, then we are adding energy to the system, but the entropy is decreasing.  Based on the Boltzmann factor, this can't be possible unless the temperature of the system is negative. That is, $$\frac{P_1}{P_2}=e^{-\frac{(E_1-E_2)}{kT}}$$ which approaches infinity unless one says the system has a negative absolute temperature.  Now we're getting to the point: this increase in energy which is connected to a decrease in entropy is what has been called a . This is most often connected to .  Forgive me for saying things you already know.  All of this raises a lot of questions though, the most striking to me is that there does not seem to me to be a good physical explanation for how this can be possible. For instance, when people say that the negative temperature system is "hotter" I get what that means as far as saying that heat will flow from a negative temp. system to a pos. temp system.  I'm fine with that except that it invokes the idea of a fairly macroscopic quantity--hotness--for a purely quantum phenomenon. So, is there any macroscopic physical meaning to this statement that the negative temperature is "hotter"?  Concerning population inversion in lasers: doesn't this mean that the negative temperature required for that laser to exist ought to be so "hot" no mere mortal can handle such a device.  After all, negative absolute temperatures are often explained as being "hotter than infinite temperature." (I've heard this other places, but also see the neg. temp. Wikipedia article for quote.) I understand that again from a mathematical perspective, but what is the physical connection? A statement like that must surely manifest itself in some way that is noticeable beyond the quantum level.  I guess I really want to understand how it could be this "hotter than infinite temperature" system isn't perceptible in the macroscopic realm? I mean, lasers are cool, but they aren't just melting through everything as if they have infinite temperature.    My bad for the obvious misuse of the word "physical" when I probably meant something more like "macroscopic". I hope the point is clear anyways.  Thoughts?
89668
In  article it says:     The electrons can only orbit stably, without radiating, in certain orbits (called by Bohr the "stationary orbits") at a certain discrete set of distances from the nucleus. These orbits are associated with definite energies and are also called energy shells or energy levels. In these orbits, the electron's acceleration does not result in radiation and energy loss as required by classical electromagnetics. The Bohr model of an atom was based upon Planck's quantum theory of radiation.   My question is: how can this be if it is actually impossible? Any accelerating point charge should emit radiation, as explained , for example. It is not like I can posit the existence of an orbit where the electron is stable, just out of mid air. What am I missing here?
10231
Quite a few of the questions given on this site mention a photon in vacuum having a rest frame such as it having a zero mass in its rest frame. I find this contradictory since photons must travel at the speed of light in all frames according to special relativity.  Does a photon in vacuum have a rest frame?
8542
In (non-relativistic) classical physics, if the temperature of an object is proportional to the average kinetic energy ${1 \over 2} m\overline {v^{2}}$of its particles (or molecules), then shouldn't that temperature depend on the frame of reference - since $\overline {v^{2}}$ will be different in different frames?  (I.e. In the lab frame  $K_l = {1 \over 2} m\overline {v^{2}} $, but in a frame moving with velocity $u$ relative to the lab frame, $K_u =  {1 \over 2} m \overline {(v+u)^{2}}$).
58531
My knowledge of atomic dynamics is a little superficial. But to my understanding an electron travels an orbital path around a nucleus of an atom. "correct?" So let's say that if a hydrogen atom were to travel "extremely close to" the speed of light (relative to some laboratory system). With the electron circling the proton approximately already traveling "extremely close to" the speed of light, how is it that the electron can circle the proton without exceeding light speed? (or does it continue to orbit at all?)
71597
Does it just vanish into space leaving nothing behind or does it expel some material?    Also, talking more about black holes, as far as I understand the term "temperature", it is defined by the amount of kinetic energy of the particles of a body and from what I have read, there are no particles inside a black hole, I know that you can find the temperature of a black hole by analyzing its Hawking radiation but I cant really understand how the term "temperature" can be applied to a black hole. Does it mean that whatever is that it is inside a black hole has kinetic energy? Have physicist already experienced the death of any regular black hole (by regular I mean space-sized black holes not micro ones created in labs)? Also, where is the Hawking radiation form? Because if it is released by the black hole I bet it must have formed outside the event horizon, but how exactly?
2053
Time has many definitions per se, but the basic idea being it's "the measurement of change" so as we know, all matter looses it's ability of changing with the loss of kinetic energy. and the where it becomes zero is the absolute zero (−273.15°C). so my question would be, will time stop/freeze at Absolute zero? and does time get slow while reaching there?
31539
I have a volume with $N$ molecules; I need to assign to each particle a velocity vector: $$|\mathbf{v}_{i}|=[v_{x}, v_{y}, v_{z}]^{T}$$ for the $i$-th molecule; the velocities must follow the Maxwell-Boltzmann distribution; here is the question: I know the definition of the M.B. distribution, how do I extract the velocities that I have to assign to each molecule? Basically what I want to know is how to calculate $$v_{x}=?$$ $$v_{y}=?$$ $$v_{z}=?$$ for each molecule.
95455
"Now we turn our attention to what happens in $S'$, in which the particle is at rest and the wire is running past (toward the left in the figure) with the speed $v$. The positive charges moving with the wire will make some magnetic field $\ B'$ at the particle. But the particle is now at $rest$, so there is no magnetic force on it! If there is no magnetic force on the particle, it must come from an electric field. It must be that the moving wire has produced an electric field. But it can do that only if it appears $charged$-it must be that a neutral wire with a current appears to be charged when set in motion."  So i got this from Feynman's lectures in physics volume 2 pg 13-7 and 13-8. I find it hard to believe that the particle experiences no magnetic force due to the moving wire although it's stationary. Since this is classical and not quantum electrodynamics if we assume the charged particle to have inertia, then by Newton's second law, the particle will remain at rest until acted upon by an external force, which i assume is the $magnetic$ $wind$ created by the moving wire past the stationary charged particle. Could someone please clarify??