10 ms·

Question: is the analogy of thermal energy as particles flying around and bouncing into each other just analogy? At what temperature would the particles fly at

• sp332 8y ago
  You can't accelerate anything to the speed of light without an infinite amount of energy. It just takes more and more energy to get closer to that speed. https://en.wikipedia.org/wiki/Speed_of_light#/media/File:Lorentz_factor.svg https://en.wikipedia.org/wiki/Speed_of_light#/media/File:Lor...
  • louisswiss 8y ago
    Does that mean it doesn't take more energy to accelerate an electron to the speed of light than a tennis ball?
    • Symmetry 8y ago
      In the sense that if you multiple infinity by an arbitrarily large finite number you still have infinity, yes.
    • badosu 8y ago
      No hadron can achieve the speed of light. See: Ek=mc^2/√(1−(v/c)^2)−mc^2 For v<<c you have significant discrepancies on the energy required to accelerate to a specific speed with regard to mass. I.E. Newton: Ek=(mv^2)/2 As v->c it does not matter as much, the lorentz factor is much more significant, the mass operates just as a base multiplier and sum factor. As v->c, x->0 where Ek~1/x, i.e. tending to infinity with a division by zero when v=c. In conclusion, the speed is the relevant factor instead of mass when near speed of light, regardless of the object being an electron or the mount Everest. Of course, assuming the equation holds ;-)
      • louisswiss 8y ago
        Thanks!
      • mirimir 8y ago
        Another perspective is that the object's effective mass is going exponential as v->c. I mean, that's why we say "rest mass", right?
        • badosu 8y ago
          Since E=mc^2 -> m=E/c^2. For a moving object you could then m=(Er+Ek)/c^2, which creates the impression that the mass is variable (as the term Ek is zero when at rest and increasing with velocity), giving rise to the terms 'rest mass' and 'relativistic mass' respectively for the rest energy and total energy equations. This interpretation is somewhat outdated but the terminology rest mass maintains its legacy. One could refer to it as the `(proper |invariant |intrinsic )?mass` instead. The variable mass issue is then 'solved' by 'refactoring' the equation to use momentum where mass is coupled with velocity, over which the complexity of the lorentz factor is engulfed.
          • trentlott 8y ago
            So how does 'invisible' kinetic energy (say, that of a ball traveling on the planet) change its mass? If we cancelled out that movement (relative to the Milky Way), does the mass change? Is how much do the various relative movements affect our mass, and would it be possible to pull tem apart? (Solar System, Relative to galactic center, other galaxies, etc)
            • badosu 8y ago
              I am not sure, I am not a physicist. What I understand is that the contemporary conception is that the mass does not change. I just presented the argument for a notion of rest mass and relative mass.
    • ajuc 8y ago
      Neither of these things is possible, so comparison is meaningless.
    • asdfman123 8y ago
      It takes more energy to accelerate a tennis ball to 80% of the speed of light (0.8c) than it takes to accelerate an electron to 80% the speed of light. It takes more energy to accelerate a tennis ball to 0.999c than an electron to 0.999c. But the faster and faster you go, the more energy is required. As the number approaches 100%, the energy required for each tiny fractional step goes up exponentially, so much so that it is impossible to accelerate either object to 100% of the speed of light. It just requires more, and more, and more energy.
  • Phrodo_00 8y ago
    > You can't accelerate anything to the speed of light You can't accelerate anything _with mass_ to the speed of light. Although I guess that stuff with no mass already travels at the speed of light, so you wouldn't need to accelerate it.
    • rmwaite 8y ago
      Precisely. Photons always move at c.
• caf 8y ago
  There's no finite level of kinetic energy at which the speed of an object exceeds c.
  • jdmichal 8y ago
    I believe you mean, "There's no finite level of kinetic energy at which the speed of an object equals c." Exceeding c is, of course, not known to be possible at all, even with infinite energy.
    • deleted 8y ago
      [deleted]
    • adrianratnapala 8y ago
      Except for massless particles, which do go at the speed of light. And that's important: photons etc. are part of the thermal soup just like every thing else.
• komali2 8y ago
  Right, this was how I was taught about temperature, and I'm just realizing now it probably isn't the best analogy... Like, do particles shot through a particle accelerator have a super high temperature? They're moving awfully fast! Or does it have to be "vibration," in which case, "vibrating" relative to what?
  • seiferteric 8y ago
    no it can be just relative to the average ke of the particles. Like in ideal gas model, its just single particles bouncing around in a box and temp is relative to the average ke of the particles. When you start introducing diatomic and above particles, then you have to worry about vibration (of the bonded particles) as well as rotations as other degrees of freedom for it to store internal energy.
  • drostie 8y ago
    It's technically not 100% defined. See my comment below for more details about degrees of freedom and energy. Suppose you have a system with 100 degrees of freedom and 2 units of energy, spread out as (0.01, 0.01, ..., 0.01, 1.01). A bunch of its energy is in one of those hundred degrees of freedom. You can assign it two different temperatures: the temperature 0.01, which would describe how energy will right now flow into the system if you connect it to another system with a bunch of degrees of freedom with their own thermal energy (assuming that the 1.01 degree of freedom is "internal" and doesn't interact directly with the outside world), and the temperature 0.02, which would describe how energy will eventually be spread out and hence how it would eventually share freedom with the outside world. Temperature is ultimately defined in terms of how our uncertainty about the microscopic state a system is in changes as we add energy to that system. The higher this rate of change of uncertainty, the lower the temperature is -- this is why when you connect two systems of different temperatures, in the process of us becoming more uncertain about the fundamental state of the world, energy "spontaneously" flows from the higher temperature to the lower temperature: the certainty gained from stealing energy from the higher-T one is more than compensated by uncertainty created from pouring that same energy into the lower-T one. (In fact there is a family of systems of "negative temperature" which become less uncertain as you add more energy to them: they are "hotter than the hottest possible temperature" because they will gladly give their energy to any "normal" system in the process of us becoming more uncertain about the world.) The problem is that if we're certain that some degree of freedom has a given amount of energy that's "special", we have a bunch of different definitions of "temperature" depending on how "adding energy to the system" distributes between the "special" degree of freedom and the "thermal" degrees of freedom. So the usual process is to just totally separate those degrees of freedom as separate systems, the "thermal" ones have a temperature, the "special" ones do not.
• drostie 8y ago
  It's not just an analogy, and it contains the essence of the story, but it's also not the whole story. In physics we talk about the "degrees of freedom" of a system -- this is just the count of all of the independent ways that it can move. For each degree of freedom of a system you can calculate the average energy in that degree of freedom. By the equipartition theorem, at thermal equilibrium, all the degrees of freedom will have the same average energy, which will be T (if you measure temperature in units of energy). So if you think about dropping a bouncy ball in a tube and it bounces until it slowly comes to rest, it has these degrees of freedom -- the internal degrees of freedom of the atoms of the ball, the internal degrees of freedom of the atoms of the floor/tube -- and then two really obvious degrees of freedom, the center-of-mass position of the ball, which gains an energy scale due to the gravitational force, and the center-of-mass momentum of the ball, which trades energy with this position degree-of-freedom. Statistical mechanics says that as this system progresses, the location of the energy will slowly become more uncertain until it is on-average-evenly distributed across all of the degrees of freedom. That's why it bounces lower and lower: there is so much energy in the two "main" degrees of freedom -- maybe half a joule? -- whereas in the vibrations there is something closer to 10^-21 J of energy at room temperature. But the flip side of dissipation is always fluctuation -- this is in fact the subject of a major theorem! So the fact that this can randomly lose energy to these other degrees of freedom means that those degrees of freedom are also randomly kicking the ball. As you can imagine with ~20 orders of magnitude difference between the two, they don't kick this ball by all that much. But you have a lot of experience with a lot more tiny balls that are bouncing off the ground all the time. Take a deep breath. There they are. If everything were to come to its minimum energy configuration, why are these air molecules so stubbornly not falling to the floor? Well, they are trying to! But they are so light that they are being kicked back upwards by these random thermal kicks, so high that they can in principle go the many kilometers to the uppermost atmosphere. (Of course if they could go all that way in a single kick then air would have to be so non-interactive that we could not use it to talk to each other... the mean free path in air is actually about 68 nm, so in practice every air atom is getting its random thermal kicks from other nearby air atoms. But the ultimate origin of these random thermal kicks is the random kicks of the floor on the few hundred nanometers of air sitting above it, and that energy comes from the Sun and is mostly conserved as these atoms collide with each other -- but a tiny bit is often converted to little photons of infrared light that sometimes escape the atmosphere.) With that said as others have noticed, the free-particle energy relation in special relativity is E = γ m c². Famously, at rest, this factor γ = 1/√(1 − (v/c)²) is 1 and the energy of a particle at rest is E = m c². But as v gets closer and closer to c, v → c, this energy grows without boundary, E → ∞. So there is no finite temperature where a kinetic degree of freedom would exceed the speed of light. Indeed you can solve for v, as 1/γ² = 1 − (v/c)². So the velocity corresponding to any given total energy is v = c √(1 − (mc²/E)²). For a rest particle with E = mc² this is v = 0 as you would expect; or when the kinetic energy first gets to mc² we would have E = 2mc² and thus v = c √(3/4) = 0.866 c.
• pdonis 8y ago
  > At what temperature would the particles fly at the speed of light? Never. More precisely, never for particles of ordinary matter that have nonzero rest mass. Relativistic effects change the dependence of temperature on velocity (more precisely, the dependence of kinetic energy on velocity), so that kinetic energy/temperature increases without bound as the speed of light is approached. For a "gas" of photons, particles of light, the particles always move at the speed of light, because they have zero rest mass. But photons can have any kinetic energy, so a photon gas can have any finite temperature.