tolman law beginner

Tolman Law in a Nutshell

The Tolman law describes how the temperature in a fixed gravitational field depends on the position (see https://arxiv.org/abs/1803.04106 for a pedagogic review). Here I present a simple general derivation of the Tolman law. It’s generally in the sense that it is applicable to any (thermalized) physical degrees of freedom for which one can define Lagrangian and Hamiltonian. It’s simple in the sense that, in the derivation, one does not need to care about any details of those physical degrees of freedom, while general relativity is applied at a rather elementary level. To achieve simplicity, at some points I use heuristic arguments based on physical intuition, thus avoiding mathematical rigor. After the derivation, I briefly discuss a few elementary examples.

1. Derivation of the Tolman law

In a static spacetime, the metric ##g_{\mu\nu}## satisfies ##g_{0i}=0##, while ##g_{00}(x)=g_{00}({\bf x})## and ##g_{ij}(x)=g_{ij}({\bf x})## do not depend on the global coordinate time ##t=x^0##. Consider a local static observer at the position ##{\bf x}##. For any local physical quantity ##A##, I introduce the notation ##A_{\rm obs}## denoting the value of ##A## as measured by such a local observer. In particular
$$dt_{\rm obs}=\sqrt{g_{00}({\bf x})} dt .$$
For any physical system (e.g. a system of particles or matter fields) the action can be written in a schematic form
$$S=\int dt_{\rm obs}\, L_{\rm obs}$$
where ##L_{\rm obs}## is the observable Lagrangian. (For instance, if the Lagrangian is equal to kinetic energy, then observable Lagrangian is the observable kinetic energy as measured by the local static observer.) Combining the two equations above, we have
$$S=\int dt\, L$$
where
$$L=\sqrt{g_{00}({\bf x})} L_{\rm obs} .$$
Hence the canonical Hamiltonian ##H## derived from ##L## has an analogous form
$$H(q,p)=\sqrt{g_{00}({\bf x})} H_{\rm obs}(q,p) $$
where ##q,p## are the variables of the physical system in the phase space. For an arbitrary Hamiltonian system in thermal equilibrium, Boltzmann has shown that probability distribution in the phase space is proportional to
$${\rm exp}\left(-\frac{H(q,p)}{kT}\right)$$
where ##k## is the Boltzmann constant and ##T## is the temperature. The temperature ##T## is a constant in the thermal equilibrium. By writing it in terms of the observable Hamiltonian ##H_{\rm obs}## as
$${\rm exp}\left(-\frac{H_{\rm obs}(q,p)}{kT/\sqrt{g_{00}({\bf x})}}\right)$$
we see that ##T/\sqrt{g_{00}({\bf x})}## is naturally interpreted as the observable temperature
$$T_{\rm obs}({\bf x})=\frac{T}{\sqrt{g_{00}({\bf x})}} .$$
This last equation is known as the Tolman law.

2. Tolman law expressed in terms of Newton potential

It is useful to rewrite the Tolman law in terms of the Newton potential ##V({\bf x})##. In general, we have
$$g_{00}({\bf x})=1+\frac{2V({\bf x})}{c^2}$$
where ##c## is the speed of light. In the Newtonian limit we have ##2V({\bf x})\ll c^2##, so the Taylor expansion gives
$$T_{\rm obs}({\bf x}) \approx T \left( 1-\frac{V({\bf x})}{c^2} \right) .$$
In particular, near the surface of the Earth we have ##V({\bf x}) \approx V_0 +gz##, where ##z## is the vertical coordinate and ##g## is the acceleration of the free fall, so
$$T_{\rm obs}({\bf x}) \approx T \left( 1-\frac{V_0 +gz}{c^2} \right) .$$

3. Tolman law around the black hole

For the Schwarzschild black hole we have

$$g_{00}(r)=1-\frac{2GM}{c^2r}=1-\frac{r_S}{r}$$

where ##r_s## is the Schwarzschild radius. Hence the Tolman law gives

$$T_{\rm obs}(r)=\frac{T}{\sqrt{1-\frac{r_S}{r}}} .$$

In particular

$$T_{\rm obs}(\infty)=T, \;\;\; T_{\rm obs}(r_s)=\infty .$$

If the temperature arises from Hawking radiation by the black hole, then ##T## is the Hawking temperature

$$T=\frac{\hbar c^3}{8\pi Gk M}.$$

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  1. DrDu says:
    No, it means the local quantity, temperature, is ##T(x)##, not ##T_0##. I don’t understand why you think this is a problem. Why wouldn’t the quantity that a local thermometer measures be "temperature"?

    I think the Unruh example shows that the reading of a thermometer depends on gravitational potential.

    What you are calling ##g##, the thing that appears in the formula for ##T(x)##, is actually gravitational potential, not gravitational acceleration. (If you look at the Insights article, you will note that the ##g## there has indexes–it is ##g_{00} (\mathbf{x})##, which is the metric coefficient that, in suitably chosen coordinates, corresponds to the gravitational potential.)

    No, I am refering to the equivalence of acceleration and gravitation locally in GR. Hence a constantly accelerated thermometer will show the same temperature as a thermometer in a gravitational potential.

  2. PeterDonis says:
    So basically, it boils down to temperature not being a local quantity any more

    No, it means the local quantity, temperature, is ##T(x)##, not ##T_0##. I don’t understand why you think this is a problem. Why wouldn’t the quantity that a local thermometer measures be "temperature"?

    As any ##g## can be viewed as resulting from an acceleration

    What you are calling ##g##, the thing that appears in the formula for ##T(x)##, is actually gravitational potential, not gravitational acceleration. (If you look at the Insights article, you will note that the ##g## there has indexes–it is ##g_{00} (\mathbf{x})##, which is the metric coefficient that, in suitably chosen coordinates, corresponds to the gravitational potential.)

  3. DrDu says:
    Yes, I think this is now more a semantical problem than a physical one. To me this sounds a bit like insisting to work only with fields and not potentials, as only the former can be measured locally.
  4. vanhees71 says:
    Sigh. Temperature does not need to be constant to make sense but as a local quantity. In relativistic physics by definition it’s a scalar field, to be measured with a thermometer comoving with the medium sticking at a specific material volume element within this medium. So it’s a proper local scalar (scalar field) quantity by definition (this holds within general as in special relativity).

    Another question is, how to measure the temperature of some object at a distance like a star. The only quantities we can measure within GR are local quantities. So to measure the temperature of a star (as our Sun) we use the spectrum of its emitted em. radiation and fit a Planck distribution to it. Then we get an "effective temperature", but this temperature is not the temperature as defined locally but the radiation is shifted due to the Doppler effect (if the star is moving relative to our local inertial frame) and due to gravity (indeed the gravitational redshift of the spectral lines of the Sun was one of the first predictions about general-relativistic effects by Einstein). This you have to take into account to get the "true" or "proper" temperature of the distant object. That’s also nicely discussed in the above quoted Eur. J. Phys. paper.

  5. DrDu says:
    So basically, it boils down to temperature not being a local quantity any more and that you have to multiply the locally measured T(x) ( I still hesitate to this "temperature") with ##\sqrt{g(x)}## to obtain the global temperature. How does the Unruh effect fit into this picture? If ##T(x)=0## for one x, it must be 0 for any other x. As any ##g## can be viewed as resulting from an acceleration, there should be no Unruh effect.
  6. PeterDonis says:
    But then, ##T(r) = T_0/\sqrt(g)##, so they both have the same temperature ##T_0##, don’t they?

    The temperature that local observers at each height measure is ##T(r)##. ##T_0##, as the arxiv paper says, is just an integration constant, not an observed temperature (except for the observer at infinity, who, if he is in thermal equilibrium with observers at other heights–no net heat flow–will locally measure temperature ##T_0##).

  7. DrDu says:
    Only when the locally measured temperatures at the two objects match the function ##T(r)## will there be no net heat flow.

    But then, ##T(r) = T_0/\sqrt(g)##, so they both have the same temperature ##T_0##, don’t they?

  8. vanhees71 says:
    Another aspect is also that the gravitational interaction is special from the point of view of kinetic theory, because it’s long-ranged and unscreened.

    Of course also the em. interaction is long-ranged (after all it’s described by a massless field). Already in this case you have to take care of the long-ranged nature by first shuffling a "mean-field piece" of the collision term to the left-hand side of the Boltzmann equation, which leads to the Vlasov equation, if you neglect all the rest of the "collision term". Then you have a self-consistent theory for the motion of the charged particles and the em. field (a "collisionless" Plasma). This neglects dissipation since the mean-field part cannot lead to entropy production. Taking then the remaining interactions into account you get a usual collision term with short-ranged interactions, because you have Debye screening in the plasma and then the usual Boltzmann (or Fermi-Dirac) distributions as equilibrium states.

    This is different for gravity, because there is no screening and it’s always attractive, and that’s why the universe as a whole doesn’t simply look like a boring equilibrated Boltzmann gas but shows all the structure like galaxies, galaxy clusters, the characteristic fluctuations of the CMBR, and all that.

  9. PeterDonis says:
    I don’t see that the zeroth law fails in GR.

    It doesn’t fail, it just has to be formulated properly to take gravitational effects into account.

    Of course you can introduce some local T(r) with different properties, but I don’t see why it should be called temperature.

    Because ##T(r)## is what predicts whether there will be net heat flow. Suppose we have two objects at different heights that can exchange heat–say by radiation. If they are both at the same locally measured temperature, then there will be net heat flow from the higher object to the lower object, because the radiation from the higher object will be blueshifted when it reaches the lower object, so its locally measured temperature at the lower object will be higher and it will heat the lower object up; and the radiation from the lower object will be redshifted when it reaches the higher object, so its locallly measured temperature at the higher object will be lower and it will cool the lower object down. Only when the locally measured temperatures at the two objects match the function ##T(r)## will there be no net heat flow.

  10. vanhees71 says:
    Your have to distinguish between global and local thermal equilibrium. Global thermal equilibrium means you have a medium with the same temperature everywhere. Local equilibrium means that the medium is in equilibrium within macroscopically small but microscopically large space-time volumes. In this case hydrodynamics is a valid approximation to the more general description by the Boltzmann(-Uehling-Uhlenbeck) transport theory or even more general the Kadanoff-Baym equations, which are a complete many-body quantum-field theoretical description of an off-equilibrium many-body system.

    Relativistic thermodynamics and kinetic theory has been a subject of quite a lot of debate for some decades since 1905. At least in my scientific community, relativistic heavy ion collisions, where hydrodynamics is kind of a "Standard Model" to describe the hot and dense fireball created in such collisions, the consensus is to use the definitions, where the thermodynamic potentials, temperature, etc. are all described as scalar quantities, i.e., being defined in the one frame of reference that is preferred by the physical situation, i.e., the local rest frames of the fluid cells. In kinetic theory also the phase-space distribution functions are scalars. AFAIK also the relativistic astrophysicists, who are quite closely related to our research in heavy-ion collisions (neutron star models need the nuclear/QCD equation of state, gravitational waves from neutron-star mergers/kilonovae are new probes to learn about this equation of state and thus complements and extends the tool box we have in heavy-ion collisions on Earth), follow this concept within GR.

    It’s an easy understandable fact that the non-relativistic macroscopic equations for transport (heat conduction, dissipative hydro), i.e., first-order-in-gradients approximations taking into account dissipation beyond the ideal-fluid description, cannot be simply extended to a relativistic theory. For that you need to take into account at least the 2nd order in gradients. Using the relaxation-time approximation for the Boltzmann equation this leads to Israel-Stewart hydro. In the past 20 years there has been a lot of progress in understanding dissipative relativistic fluid dynamics in terms of using higher moments in their derivation from relativistic kinetic theory and their validity range compared to kinetic theory.

  11. DrDu says:
    The point I want to make is that equality of temperature of systems being in thermal equilibrium is not any property of temperature but defines temperature via the zeroth law (transitivity of thermal equilibrium). Hence temperature labels equivalence classes of systems being in thermal equilibrium. I don’t see that the zeroth law fails in GR. Of course you can introduce some local T(r) with different properties, but I don’t see why it should be called temperature.

    PS: This is more a critique of the arxiv article than the of the excellent Insights article.

  12. PeterDonis says:
    in the arxiv articlefrom the first post

    Ok, that clarifies what ##T_0## is (it’s basically the same as ##T## in the Insights article), but it doesn’t explain what you mean by ##T_0## being the "correct" temperature. The arxiv paper clearly states that ##T_0## is an "integration constant" and that the temperature that determines thermal equilibrium is what it calls ##T(r)## or ##T(x)## (and what the Insights article calls ##T_\text{obs}##), i.e., as the arxiv paper states, in a gravitational field "the equilibrium temperature of a spherically symmetric static distribution of a perfect fluid is not uniform". So it would seem to me that a single value, whether it’s ##T_0## or any other single value, can’t be "correct" as a description of this situation.

  13. DrDu says:
    In my understanding, T_0 is the correct temperature. Obiously, T_0 is also not unique, but empirical temperature isn’t either in classical thermodynamics. I think the main rational behind the local temperature concept is to have temperature still to be proportional to the mean energy of a photon gas. But I don’t think that one should give up a general definition to save some special material equations.
  14. PeterDonis says:
    the zeroth law, which demands that temperature has to be the same for persons being in thermodynamical equilibrium

    The problem with this is that it doesn’t work as it stands in a gravitational field. Two objects at different altitudes in a gravitational field must be at different temperatures if they are to be in thermodynamical equilibrium with each other. Or, if we have a continuous fluid in a gravitational field, the temperature of the fluid must vary with altitude if it is in thermodynamical equilibrium. That is the point of Tolman’s law.

    The zeroth law has to be modified to cover such cases, to make clear that it only holds locally, in a patch of spacetime small enough that the effects of spacetime curvature can be neglected. (Strictly speaking, the patch also has to be small enough that any path curvature, i.e., proper acceleration, of objects of interest can also be neglected.)

  15. DrDu says:
    I never understood the point of this argument. Rather the concept of a "local temperature" reminds me of early notions like "transversal" and "longitudinal" mass and the confusion they caused. I also don’t see what precisely is "naive" about defining temperature by the zeroth law, which demands that temperature has to be the same for persons being in thermodynamical equilibrium.
  16. vanhees71 says:
    The trouble with the first-order gradient expansion of transport equations around (local) thermal equilibrium (Navier-Stokes equation for viscous fluid, heat conduction, diffusion) is that it leads to parabolic PDEs, which lead to superluminal/instantaneous "signal propagation" and is thus unacceptable as a relativistic theory. The solution is, as shown by Israel and Steward for viscous fluid dynamics, is to go to higher orders in gradient expansion, including some memory effects. Coming from the transport equations this is intuitively understood in the relaxation-time approximation of the collision term, because the 2nd order gradient expansion includes the finite "relaxation time" and leads to hyperbolic equations, which can obey Einstein causality if the parameters fulfill some conditions such that no faster-than light propagation occurs.

    https://en.wikipedia.org/wiki/Relativistic_heat_conduction
    https://itp.uni-frankfurt.de/~greif/Bachelorarbeit.pdf

  17. martinbn says:
    What’s the heat equation in this setting? If it is not in equilibrium what is the law by which the temperature changes (of course, this is not phrased in relativistic language, but I hope it is clear what I am asking).
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