KJW

A long time ago on a different forum, I made the claim that only perfect black-body radiation, both in terms of its relative frequency distribution and its intensity, can have a single temperature attributed to it. Every other distribution of radiation has a pair of temperatures attributable to it. The lower temperature is defined as the temperature achieved by an absorption spectrum that minimises absorption and maximises emission, while the higher temperature is defined as the temperature achieved by an absorption spectrum that maximises absorption and minimises emission. Then one can construct an ideal heat engine based on these two temperatures to extract work from the radiation. One can't do that if the radiation is perfect black-body as this has only a single temperature.

I can define the temperature of an atom (as of that any material object) in thermal equilibrium with its surrounding environment as being the same as that of the environment which can be measured by a thermometer. It occurred to me how one might do that. One could define the temperature of a single particle in thermal equilibrium with its surrounding environment in terms of the kinetic energy distribution over time of the particle. Applying the ergodic principle transforms this to an ensemble distribution for which the notion of temperature naturally applies.

One can't dismiss a causative relationship between two quantities simply because the mathematical relation between them is an equality. One really needs to examine the logical aspect of the relationship to establish whether there is a causative relationship and in which direction it occurs. In the case of the Einstein equation, what one really has is a connection between the mathematical realm and the physical realm. Without the Einstein equation, all one has is Ricci Calculus. It is the Einstein equation that transforms this to General Relativity. The Einstein equation does two things: 1, it establishes that it is the mathematical Einstein tensor that corresponds to the physical energy-momentum tensor; and 2, it establishes a proportionality relationship between them, including the use of physical constants to ensure dimensional homogeneity.

Why does the 5 pots have the same temperature as the 3 pots? Why does the 7 pots have the same temperature as the 3 pots and not the 5 pots? One other thing: In the ideal gas law, pressure, which is proportional to temperature, depends only on the translational motion of the gas molecules. Neither rotational motion, nor vibrational motion affects the pressure and therefore the temperature of an ideal gas.

Suppose you add thermal energy to a substance with 5 degrees of freedom. The equipartition theorem says that equal energy will be added to each degree of freedom. This means that 3/5 of the total added energy will be added to the 3 degrees of freedom corresponding to translation, leading to an increase in the temperature. The remaining 2/5 of the total added energy do not contribute to the increase in temperature, but because the increase in temperature due to the energy added to the 3 translational degrees of freedom is concomitant with the energy added to the 2 non-translational degrees of freedom, the energy of the 2 non-translational degrees of freedom will still depend on the temperature. There will be 3 units of energy in the object with 3 degrees of freedom and 5 units of energy in the object with 5 degrees of freedom. But the two objects are at the same temperature. Therefore, only 3 units of energy in the object with 5 units of energy are producing temperature, the same temperature as produced by the 3 units of energy in the object with 3 units of energy.

But this conflicts with the equipartition theorem. Each degree of freedom contributes an energy of ½kT. The more degrees of freedom, the more energy. But this requires that only the ever-present three translational degrees of freedom contribute to the temperature. Otherwise, the energy would not be proportional to number of degrees of freedom. The increase in heat capacity as the available degrees of freedom increases indicates that the energy is increasing faster than the temperature. Thus, the 5/2 R for the inclusion of rotations indicates that the increase of R from 3/2 R to 5/2 R is not causing an increase in temperature beyond the 3/2 R arising from translations. If all degrees of freedom contributed to the temperature equally, the heat capacity of all substances at all temperatures would be the same regardless of the number of available degrees of freedom.

All degrees of freedom do contribute to the entropy. But not all degrees of freedom contribute to the temperature. If it did, then given the equipartition theorem, all substances would have the same heat capacity, and the above equation you posted would not hold.

I was actually thinking the same question. The answer I came up with was that vibrational motion within crystals is translational motion of individual molecules. The modes that do not correspond to temperature are rotations of individual molecules, vibrations within individual molecules, which are distinct from the vibrations between molecules within a crystal, and electronic states. The way I see it is if one were to place a quantity of helium, which has only translational modes, within a sample of a given substance that has an arbitrary number of modes, then how much energy is transferred from the sample to the helium or visa versa? At thermal equilibrium, the two temperatures must be equal, but what is the distribution of energy? I think the virial theorem is involved. I seem to recall that vibrational motion within crystals is not just translational motion but also potential energy. That is, two modes but only one corresponding to temperature.

Option 2 is saying that general relativity is not the most fundamental theory of physics. The dimension of mass is not only associated with gravitation, but is everywhere in physics. The current definition of a unit of mass is not even based on being a gravitational source, but rather specifies a numeric value of Planck's constant. However, if mass is quantum mechanically producing spacetime curvature, then that spacetime curvature must also be consistent with general relativity. There was a time that I believed in option 1 but in more recent times I have felt that spacetime curvature is too ethereal to provide us with anything other than gravitation. On the other hand, I find it difficult to accept that there may be some physics that isn't entirely based on the properties of spacetime.

Well, chromium(VI) compounds can cause organic material to burst into flames.

One thing should be mentioned: Only the translational modes of molecular motion contribute to the temperature. Different substances have different heat capacities because they absorb energy in all their modes, but only the translational modes increase the temperature, thus the more modes that are available to the molecule, the more energy that is absorbed for a given increase in temperature.

I think the image of Nikola Tesla on the first video should be considered a big red flag as far as pseudoscience is concerned.

Yes, the SMT-VSL and the expanding universe theory are equivalent. This means that there are no observational or experimental differences between these two points of view. If you are seeking to observe differences, then you are really saying that the SMT-VSL and the expanding universe theory are not equivalent because any observable difference is a non-equivalence. If you are abandoning equivalence, then why would matter shrink in preference to an expanding universe? The size of atoms is governed by laws of physics, whereas the size of the universe is not, so one would not expect there to be a constraint on the size of the universe similar to the constraint on the size of atoms. Also, if you are abandoning equivalence, then where specifically is the non-equivalence? That is, what specific observation or experiment distinguishes these two theories? This actually requires you to look beyond the apparent equivalences to something not deducible by a mere change in the point of view.

Let [math]M[/math] be the mass of the non-rotating spherical mass (neutron star but assumed to be non-rotating), [math]R[/math] be the radius of the spherical mass, and [math]h[/math] be the height above the ground at radius [math]R[/math] from which the object of mass [math]m[/math] (measured at height [math]h[/math]) is dropped. Assuming that the collision with the ground is completely non-elastic, the energy [math]E[/math] (also measured at height [math]h[/math]) released is: [math]E = \left(1 - \sqrt{\dfrac{g_{tt}(R)}{g_{tt}(R+h)}}\right) m c^2[/math] where [math]g_{tt}(R)[/math] and [math]g_{tt}(R+h)[/math] are the [math]tt[/math]-components of the Schwarzschild metric at [math]R[/math] and [math]R+h[/math] respectively. Thus: [math]E = \left(1 - \sqrt{\dfrac{1 - \dfrac{2 G M}{c^2 R}}{1 - \dfrac{2 G M}{c^2 (R+h)}}}\right) m c^2[/math] Note that [math]\sqrt{\dfrac{g_{tt}(R)}{g_{tt}(R+h)}} = \sqrt{\dfrac{1 - \dfrac{2 G M}{c^2 R}}{1 - \dfrac{2 G M}{c^2 (R+h)}}}[/math] is the ratio of the mass of an object at [math]R[/math] to the mass of the same object at [math]R+h[/math], the object being at rest at both heights.

You are talking about the theoretical/abstract model. I am asking about how the mass actually does that. There is nothing "abstract" about spacetime geometry. Gravitational time dilation is physically real as is the curvature it implies. It seems to me that many people believe that general relativity is just a theoretical model that describes physics in only an abstract manner typical of theoretical models. But no, geometry is very much a physically concrete notion and the formulae from Ricci calculus very much describes this physically concrete notion. The difficulty arises because the spacetime curvature that describes human-scale physics is quite miniscule, so it becomes very difficult to observe and measure the physically real curvature of spacetime. There are two possibilities of how mass causes spacetime curvature: 1, mass doesn't cause spacetime curvature, but that mass is how we physically interpret the spacetime curvature that exists fundamentally; 2, mass causes spacetime curvature through a mechanism associated with something like quantum physics, the fundamental forces of nature, or perhaps something entirely different.

Another way to say this is that, in this kind of spacetime and under geodesic motion, shapes (ie angles) are preserved, whereas volumes and surfaces are not. You and I discussed this in the thread about the cosmological expansion of time as well as space. By coordinate-transforming to the manifestly conformally flat metric, it becomes easier to derive cosmological redshifts due to the simplified geodesic equation for light as a result of the preservation of angles (speeds in spacetime). This is true for homogeneously and isotropically distributed sources. What about other "smeared" sources? The Weyl tensor is not necessarily zero, but the curvature is not away from the source. A non-zero Weyl tensor can coexist at the same location as a non-zero Einstein tensor. But the energy-momentum tensor corresponds only to the Einstein tensor. I felt the need to distinguish between the curvature that is generated by an energy-momentum source and propagated to remote locations, and the curvature that corresponds to the energy-momentum itself. It's worth mentioning that inside a spherically symmetric shell of matter, the spacetime is flat. That is, while gravitational curvature propagates from the shell of matter to outside of the shell, it does not propagate from the shell of matter to inside of the shell.

You might be wondering where the formula [math]a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] comes from. I've already provided the clue that it comes from the Killing equation. Here's the derivation: Let [math]T u^\alpha = T \dfrac{dx^\alpha}{ds}[/math] be a timelike Killing vector field, where [math]u^\alpha = \dfrac{dx^\alpha}{ds}[/math] is a unit tangent vector field of timelike trajectories of stationary observers or other objects in stationary spacetime. Note that [math]g_{\alpha\beta}\ u^\alpha u^\beta = 1[/math] The Killing equation: [math]\nabla_\mu (Tu_\nu) + \nabla_\nu (Tu_\mu) = 0 \ \ \ \ \ \ \ \ \ \ \text{(}\nabla_\mu \text{ is the covariant differential operator)}[/math] [math]\nabla_\mu T\ u_\nu + T\ \nabla_\mu u_\nu + \nabla_\nu T\ u_\mu + T\ \nabla_\nu u_\mu = 0[/math] [math]u^\nu (\nabla_\mu T\ u_\nu + T\ \nabla_\mu u_\nu + \nabla_\nu T\ u_\mu + T\ \nabla_\nu u_\mu) = 0[/math] [math]\nabla_\mu T\ u^\nu u_\nu + T\ u^\nu \nabla_\mu u_\nu + u^\nu \nabla_\nu T\ u_\mu + T\ u^\nu \nabla_\nu u_\mu = 0[/math] Considering the individual terms: [math]\nabla_\mu T\ u^\nu u_\nu = \nabla_\mu T = \dfrac{\partial T}{\partial x^\mu}[/math] [math]T\ u^\nu \nabla_\mu u_\nu = T\ g_{\nu\sigma} u^\nu \nabla_\mu u^\sigma = T\ \nabla_\mu (g_{\nu\sigma}\ u^\nu u^\sigma) - T\ g_{\nu\sigma} u^\sigma \nabla_\mu u^\nu = -T\ g_{\nu\sigma} u^\nu \nabla_\mu u^\sigma = 0[/math] [math]u^\nu \nabla_\nu T\ u_\mu = \dfrac{\partial T}{\partial s} u_\mu = 0 \ \ \ \ \ \left(\dfrac{\partial T}{\partial s} = 0 \ \ \text{ by definition} \right)[/math] [math]T\ u^\nu \nabla_\nu u_\mu = T\ b_\mu[/math] Therefore: [math]\dfrac{\partial T}{\partial x^\mu} + T\ b_\mu = 0[/math] [math]b_\mu = -\dfrac{1}{T} \dfrac{\partial T}{\partial x^\mu}[/math] (Note that [math]a^\mu = c^2 g^{\mu\nu}\ b_\nu[/math]) [math]a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math]

In another thread, I showed that the flat-space Friedmann–Lemaître–Robertson–Walker (FLRW) metric can be coordinate-transformed to a manifestly conformally flat metric. This means that the Weyl conformal tensor is zero, and therefore according to my way of looking at this, there is no gravity (although there is still the curvature associated the Ricci tensor and scalar).

One thing I should mention: I distinguish between the curvature directly associated with the energy-momentum and the curvature that is away from its source energy-momentum, and tend not to use the term "gravity" to describe the curvature directly associated with energy-momentum. The curvature associated with gravity is called the Weyl conformal tensor. It has the same algebraic structure as the Riemann tensor, but its contraction is zero (—> zero Einstein tensor). In the [math]{C_{\alpha\beta\gamma}}^\delta[/math] form, it is invariant to conformal transformations: [math]g_{\mu\nu} \longrightarrow \varphi g_{\mu\nu}[/math] for arbitrary scalar function [math]\varphi[/math]

Yes, [math]T[/math] and [math]\dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] depend on [math]M[/math] and [math]r[/math]. I would prefer to say that the curvature associated with mass causes gravity, although the mass is remote from the gravity, unlike gravitational time dilation which is local to the gravity. Whether or not mass causes the curvature associated with it, or is the curvature associated with it is something I can't answer. @Genady, [math]g^{rr} = \dfrac{1}{g_{rr}}[/math]

[math]T[/math] is the relative magnitude of a timelike Killing vector at a given location in three-dimensional space. In terms of the metric, it is [math]\sqrt{g_{tt}}[/math], but Killing vectors are covariant, and the magnitude of a Killing vector is invariant once its magnitude at some location is defined (if [math]K^\mu[/math] is a Killing vector field, [math]\lambda K^\mu[/math] is also a Killing vector field for arbitrary constant scalar [math]\lambda[/math]).

Do you need to consider the stereochemistry of the hydrogenations?

[math]a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] For a stationary object in the Schwarzschild metric: [math]ds^2 = \left(1 - \dfrac{2GM}{c^2r}\right) c^2 (dt)^2 - \left(1 - \dfrac{2GM}{c^2r}\right)^{-1} (dr)^2 - r^2 \Bigl((d\theta)^2 + sin^2\theta (d\phi)^2\Bigr)[/math] [math]g_{tt} = \left(1 - \dfrac{2GM}{c^2r}\right) c^2[/math] [math]g_{rr} = -\left(1 - \dfrac{2GM}{c^2r}\right)^{-1}[/math] [math]a^r = -c^2 g^{rr} \dfrac{1}{\sqrt{g_{tt}}} \dfrac{d\sqrt{g_{tt}}}{dr}[/math] [math]a^r = -\dfrac{c^2}{2} \dfrac{1}{g_{rr}\ g_{tt}} \dfrac{dg_{tt}}{dr}[/math] [math]a^r = \dfrac{1}{2}\dfrac{dg_{tt}}{dr}[/math] [math]a^r = \dfrac{1}{2}\dfrac{d\left(c^2 - \dfrac{2GM}{r}\right)}{dr}[/math] [math]a^r = \dfrac{GM}{r^2}[/math] One might be surprised to see that the acceleration of a stationary object in the Schwarzschild metric is exactly Newtonian. But this is due to way the [math]r[/math]-coordinate is defined. The [math]r[/math]-coordinate is defined such that the surface area of a sphere of radius [math]r[/math] is [math]4 \pi r^2[/math]. Thus, for the acceleration field to be a conserved field, the surface integral of this field over the sphere needs to be constant with respect to the [math]r[/math]-coordinate, requiring that the acceleration field obey an [math]r^{-2}[/math] law exactly.

One thing I find intriguing is that by an oxidative process, ammonia can be converted to the strongest known reducing agent (azide ion, N3–). Consider boron trifluoride, BF3: A boron atom has three electrons in its outer shell. The three fluorine atoms each provide one electron, giving the boron atom of boron trifluoride six electrons in total in its outer shell. But boron atoms want to have eight electrons in their outer shell to complete the octet. Thus, boron trifluoride will react with a fluoride ion, F–, which supplies a lone-pair of electrons, to produce the tetrafluoroborate anion, BF4–, which has eight electrons in the outer shell of the boron atom, completing the octet.

I prefer to think of it this way: Oxidation is the loss of electrons. Reduction is the gain of electrons. Acids accept lone-pairs of electrons. Bases donate lone-pairs of electrons. It would seem that there is a correlation between being an oxidising agent and being an acid, and a correlation between being a reducing agent and being a base. However, they are distinct notions. The hydride anion, H–, is an example of both a base and a reducing agent. Indeed, the reaction: H– + H+ —> H2 is both an acid-base reaction and a redox reaction. By contrast, diborane, B2H6, is a Lewis acid that is a reducing agent. Whereas the hydroxide ion, OH–, is not really much different from any other base, the hydrogen ion H+, is singularly unique as a Lewis acid in having no electrons at all. As a proton, a subatomic particle, its charge density is so great that, within the context of chemistry, it will never be free. It will always be bound to some electron density, regardless of how reluctant that electron density is to bond to anything. Therefore, a Brønsted acid is conceptually distinct from a Lewis acid in that the hydrogen ion is always bound to a base, and the measure of the Brønsted acidity is really a measure of the basicity of that base (the weaker the base, the stronger the Brønsted acid).