Since heat is defined as the movement of molecules, and because of relativity time slows for faster moving objects, would a hot object be in a slower time frame then a cooler object, because the hot objects molecules are moving faster?
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asked May 27, 2016 at 13:44
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4$\begingroup$ As an aside, in contrast to the usage in chemistry, "heat" in physics is always an energy transfer (comparable to work) and is a property of interactions not of objects. The phrase you want is "thermal energy" or "internal energy" $\endgroup$– dmckee --- ex-moderator kittenCommented May 27, 2016 at 17:30
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3 Answers
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No, the hot object itself experiences no relativistic effects. The particles in the object however may well experience time dilation or length contraction. In fact in very hot gases and plasma, ideal gas laws may not always apply due to relativistic effects.
These effects are especially noticeable in gases once the kinetic energy or fermi energy (for fermionic gases) exceeds $mc^2$ by a significant amount.
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2$\begingroup$ Totally right re: special relativity. But note that the increased energy technically has GR effects. $\endgroup$ Commented May 27, 2016 at 15:38
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2$\begingroup$ If the object is composed of particles it does experience time dilation, but no length contraction since it evens out on the macro scale. $\endgroup$– 2501Commented May 27, 2016 at 16:49
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5$\begingroup$ How is it possible for every particle of an object to experience time dilation, but not the object itself? $\endgroup$ Commented May 27, 2016 at 18:15
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4$\begingroup$ @BlueRaja-DannyPflughoeft Imagine building a pendulum clock out of metal containing some trace of radioactive material. We use the usual mixed metals and central bridge procedure to insure that the pendulum doesn't length as the clock heats. In principle, the clock will keep accurate time over some range of temperatures, but the decay rate of the radioactive impurities as measured by a Geiger counter set nearby will depend on temperature. In practice, of course, you don't get enough dilation to care about in solids. $\endgroup$ Commented May 27, 2016 at 22:59
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1$\begingroup$ The hotter you make something, the more it tends to look like a photon gas. That is, quantum mechanics becomes irrelevant and you are left with a $\Gamma = 4/3$ ideal gas. $\endgroup$– user10851Commented May 28, 2016 at 1:42
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A hot gas will exhibit relativistic effects for its particles. A particular particle will have a time dilation effect between collisions with other particles. If its energy is $E = K + mc^2$, $K$ = kinetic energy, and mass is $m$ the Lorentz gamma factor for that particle is then $\gamma = E/mc^2$. This happens to all of the particles for their energy $E$ determined by a Boltzmann distribution.
What about the whole gas? Since the macroscopic object is stationary with respect to your frame there is no relativistic effect for the whole thing. In special relativity the Lorentz gamma factor is between the frame of a moving body and the frame of the observer. All of those gas particles are on different frames. They can't be said to define a relativistic effect as a statistical average.
answered May 27, 2016 at 19:33
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Just to add some influences of heat on a stationary clock on the surface of a star.
Heat is energy, and energy increases the energy-momentum tensor, so effectively increases the mass. With everything else remaining equal, this would increase time dilation.
On the other hand, heated bodies increase their volume. The same mass over a greater volume leads to less time dilation.
