Imagine a fundamental particle at rest with respect to ground frame and has temperature $x$ degrees celsius and imagine a 'similar' particle moving at a very high speed (take any value which would affect temperature significantly). So the particle moving has temperature greater than $x$ degrees celsius or exactly equal to $x$ degrees celsius?
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, $K_u =  {1 \over 2} m \overline {(v+u)^{2}}$).
96641
Because an object's temperature is inversely proportional to the wavelength of blackbody radiation which it emits, physicists have theorized the existence of  at around $1.4×10^{32}$ K.   Does this imply that temperature must be discrete? Meaning, the temperature could never be $1.0×10^{32}$ K, because that would imply a wavelength which is a non-integral multiple of the Planck length.  If it does imply this, does that mean that a system, if approaching the Planck temperature, and given more energy, would suddenly and radically change states from $0.7×10^{32}$ K to $1.4×10^{32}$ K? (perhaps much like a )  Pardon me if this question is bizarre, as I am not a physicist by profession (except maybe of the armchair variety).
93499
I am near the end of the book  by Steven Weinberg. I am reading it in order to get a better picture of the early universe in the Big Bang model. But one thing I am having trouble with and wanted to get some clarity. Namely: Question What is meant by &quot;temperature&quot; if the  in the Big Bang model was mostly radiation, or pre-inflation, something more basic than radiation? Explanation Below I will try to explain why I am getting confused. Please let me know if it needs clarification (but that's also part of the question, I haven't been able to clearly understand temperature in this context yet). Since the  is about the movement of massive particles: $$T   =   {m\overline{v^2}\over 3 k_B}$$ How do you think of temperature when there is no mass, or when there is mostly radiation in the very early universe? Here is where I get confused.. Radiation is moving at the speed of light $c$, and since the speed of light is constant, and since radiation doesn't have mass, it seems like the universe would be &quot;maximally&quot; hot (particles moving at the speed of light). But intuitively that's not true because the temperature formula requires mass. But then if the early universe was mostly radiation, how does it have a temperature? If mass slowly increased to be the majority ~400,000 years after the Big Bang, it seems the temperature would increase rather than decrease. But then taking into account the expansion, it makes sense that even if more matter was forming, the temperature would decrease. But how could it decrease if there was less matter before? To summarize, in trying to pinpoint temperature, I seem to be stuck in cyclic reasoning and am looking for a better way of understanding it. I am working toward learning the purely mathematical form of these theories, but have a ways to go. For now, looking for a description to broadly make sense of &quot;temperature&quot; during the evolution of the few moments after the Big Bang. Examples of how &quot;temperature&quot; is used In the book above, Steven Weinberg talks a lot about the temperature and radiation of the early universe, and how the early universe was dominated by photons/radiation, and ~400,000 years later became dominated by mass-particles when the temperature cooled. Some other quotes from the web include:  In the first moments after the big bang, the universe was extremely hot and dense. As the universe cooled, conditions became just right to give rise to the building blocks of matter – the quarks and electrons of which we are all made. () ...the temperature was so high that the four fundamental forces ... were one fundamental force. () ...the temperature of the universe was still too high to allow quarks to bind together to form hadrons. ()
30490
(One of examples that potential energy contributes to mass.)  Does hot object weigh more than cold one?  (One of examples that kinetic energy contributes to mass.)  If these are true and justified by mass-energy equivalence principle, could the underlying principle be justified?  I've learned and derived $E=mc^2$ but only for the case that $m$ is interpreted as rest one. Could I generalize this argument to include kinetic and potential energy? :  $$mc^2=E =∑m_0c^2+∑K+V=∑\gamma m_0c^2+V$$
99658
I remember the first time I saw this formula, $F = ma$, and it seemed perfectly correct and natural to me (in the sense that it is very straight-forward and logical, not that it is easy by any means). But then, I wondered, what if an object is moving with a constant speed? Maybe a car. If it travels with constant speed and hits me, it still hurts. So it's force must not be $0$.  One more thing related to this problem. If the same car is pulling another car behind it, if the car goes with constant speed, it implies that the first car doesn't exert any force to the second car at that moment, since it is not technically pulling it anymore (espacially if we don't take into consideration the friction).  So my question would be the following: How do these two phenomenons describe the same law of motion?   PS: If gravity wasn't behaving as some kind of acceleration pulling me constantly to the Earth's core, then if I jumped upwards, would I start slowly getting away from Earth?
40878
The Maxwell-Boltzmann velocity distribution in 3D space is $$ f(v)dv = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 \exp\left(-\frac{m v^2}{2k_B T}\right)dv$$ It gives the probability for a single particle to have a speed in the intervall $[v,v+dv]$. But this probability is not zero for speeds $v &gt; c $ in conflict with special relativity.  Is there a generalization of the Maxwell-Boltzmann velocity distribution which is valid also in the relativistic regime so that $f(v) = 0$ for $v&gt;c$ ? And how can it be derived? Or can a single particle distribution simply not exist for relativistic speeds, because for high energies, we always have pair-production meaning the particle number is not conserved and a single particle distribution can not be defined in a consistent way?
18170
Could someone provide me with a mathematical proof of why, a system with an absolute  (such that of a spin system) is  than any system with a positive temperature (in the sense that if a negative-temperature system and a positive-temperature system come in contact, heat will flow from the negative- to the positive-temperature system).
103087
The thought randomly occurred to me that a circular particle accelerator would have to exert a lot of force in order to maintain the curvature of the trajectory.  Many accelerators move particles at fully relativistic speeds, and I want to ask how that affects things.  Why does this matter?  Well, if I understand correctly, a particle in the LHC moving at $0.999 c$ would be dramatically more difficult to keep moving in the circle than a particle moving at $0.99 c$.  Reading Wikipedia, I was delighted to find a fully specified problem within a single paragraph.       The LHC lies in a tunnel 27 kilometres (17 mi) in circumference, as deep as 175 metres (574 ft) beneath the Franco-Swiss border near Geneva, Switzerland. Its synchrotron is designed to collide opposing particle beams of either protons at up to 7 teraelectronvolts (7 TeV or 1.12 microjoules) per nucleon, or lead nuclei at an energy of 574 TeV (92.0 µJ) per nucleus (2.76 TeV per nucleon).   I can (or Google can)  the proton case to come to $0.999999991 c$.  In order to correctly calculate the force that the LHC must exert on it, do I need to start from $F=dp/dt$, or can I get by with $v^2/r$ times the relativistic mass?  I'm not quite sure how to do the former.  Question: Say I have 2 protons, both moving so fast they're going almost the speed of light but one has twice the energy of the other.  They move side-by-side and enter either an electric or magnetic field that causes them to accelerate perpendicular to the velocity vector.  Is the radius of curvature mostly the same for the two, or is it different by a major factor (like 0.5x, 1x, or 2x)?  I want to read some comments from people who understand these physics well.  The particles beyond a certain energy are all going almost exactly the same speed.  Is it still okay for me to say they have the same "acceleration" too?  They just happen to have absurdly huge inertia compared to the same particle at a more modest fraction of the speed of light.  Is this the correct perspective?
79992
In the book of The First Three Minutes by Weinberg, on pages 106-107, it is stated that  SECOND FRAME. The temperature of the universe is 30,000 million degrees Kelvin [...] The nuclear particle balance has conse- quently shifted to 38 per cent neutrons and 62 per cent protons. [...] THIRD FRAME. The temperature of the universe is 10,000 million degrees Kelvin. [...] The decreasing temperature has now allowed the proton-neutron balance to shift to 24 per cent neutrons and 76 per cent protons.  What is the reason for this balance shift between neutrons and protons? And what determines the rate of change of the neutron/proton ratio?
51531
(This is not homework; a friend shared with me this puzzler and neither of us can figure it out.) Suppose you are in a plane traveling at velocity $v_1$ relative to the ground. The flight attendent pushes a cart of mass, say, $m$, accelerating it from rest to $v_2$, relative to the plane (so relative to the ground, its velocity goes from $v_1$ to $v_1+v_2$). From your perspective, the work done is then $$W = \Delta E = \dfrac{1}{2}mv_2^2$$ but from the perspective of the ground, the work done is $$W = \Delta E = \dfrac{1}{2}m(v_1+v_2)^2 - \dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_2^2 + mv_1v_2.$$ Why are these quantities unequal?  I've seen some similar problems, like a body colliding with a wall in different reference frames, but I don't have the physics experience to tell if that question is analogous to this one. So, my apologies if this is a duplicate. :)
38664
I'm a newbie in physics. Sorry, if the following questions are dumb. I began reading &quot;Mechanics&quot; by Landau and Lifshitz recently and hit a few roadblocks right away.  Proving that a free particle moves with a constant velocity in an inertial frame of reference ($\S$3. Galileo's relativity principle). The proof begins with explaining that the Lagrangian must only depend on the speed of the particle ($v^2={\bf v}^2$): $$L=L(v^2).$$ Hence the Lagrance's equations will be $$\frac{d}{dt}\left(\frac{\partial L}{\partial {\bf v}}\right)=0,$$ so $$\frac{\partial L}{\partial {\bf v}}=\text{constant}.$$ And this is where the authors say   Since $\partial L/\partial \bf v$ is a function of the velocity only, it follows that $${\bf v}=\text{constant}.$$  Why so? I can put $L=\|{\bf v}\|=\sqrt{v^2_x+v^2_y+v^2_z}$. Then $$\frac{\partial L}{\partial {\bf v}}=\frac{2}{\sqrt{v^2_x+v^2_y+v^2_z}}\begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix},$$ which will remain a constant vector $\begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}$ as the particle moves with an arbitrary non-constant positive $v_x$ and $v_y=v_z=0$. Where am I wrong here? If I am, how does one prove the quoted statement?  Proving that $L=\frac{m v^2}2$ ($\S$4. The Lagrangian for a free particle). The authors consider an inertial frame of reference $K$ moving with a velocity ${\bf\epsilon}$ relative to another frame of reference $K'$, so ${\bf v'=v+\epsilon}$. Here is what troubles me:   Since the equations of motion must have same form in every frame, the Lagrangian $L(v^2)$ must be converted by this transformation into a function $L'$ which differs from $L(v^2)$, if at all, only by the total time derivative of a function of coordinates and time (see the end of $\S$2).  First of all, what does same form mean? I think the equations should be the same, but if I'm right, why wouldn't the authors write so? Second, it was shown in $\S$2 that adding a total derivative will not change the equations. There was nothing about total derivatives of time and coordinates being the only functions, adding which does not change the equations (or their form, whatever it means). Where am I wrong now? If I'm not, how does one prove the quoted statement and why haven't the authors done it?
82614
From      The temperature of an ideal monatomic gas is a measure of the average kinetic energy of its atoms.    Now if I remove all the particles from the box shown below will the temperature be zero?
95076
I was wondering if temperature is related to the average translational kinetic energy of the molecules then why does the average kinetic energy of say a moving object not affect the temperature? What is the mechanism regarding these vibrational motions that imparts temperature. Also do a gas and liquid at the same temperature possess the same kinetic energy? and a heavier molecule traveling at the same speed as a lighter molecule will have a greater temperature?
122275
Why is absolute zero considered to be ? Wouldn't regions such as massive gaps between galaxy clusters have temperatures of absolute zero?  I just do not see why our model must work the way that it does. I mean there have to be regions with no thermal energy out there, the universe is massive.
113510
Can a free electron with a constant velocity V то move eternally at this rate in the inertial frame? The question arises from the fact that a moving charge creates near its trajectory varying electric and magnetic fields. Whether kinetic energy of electron is spent for it?
35668
We know that in high temperature, resistivity in metals goes linearly with temperature. As temperature is lowered, resistivity goes first as $T^5$ due to "electron-phonon" interaction, and then goes as $T^2$ as temperature is further lowered due to "electron-electron" interactions.   My question is, is there an intuitive way of understanding the above power-law dependence at each temperature range physically?
118791
If object A is moving at velocity $v$ (normalized so that $c=1$) relative to a ground observer emits object B at velocity $w$ relative to A, the velocity of B relative to the ground observer is  $$ v \oplus w = \frac{v+w}{1+vw} $$  As expected, $v \oplus 1 = 1$,  as "nothing can go faster than light".  Similarly, $v \oplus -1 = -1$. (same thing in the other direction)  But what if object A is moving at the speed of light and emits object B at the speed of light in the exact opposite direction? In other words, what is the value of $$1 \oplus -1?$$ Putting the values in the formula yields the indeterminate form $\frac{0}{0}$. This invites making sense of things by taking a limit, but $$ \lim_{(v,w)\to (1,-1)} \frac{v+w}{1+vw}$$ is not well-defined, because the limit depends on the path taken.  So what would the ground observer see? Is this even a meaningful question?  Edit: I understand $1 \oplus -1$ doesn't make sense mathematically (thought I made it clear above!), I'm asking what would happen physically. I'm getting the sense that my fears were correct, it's physically a nonsensical situation.