KJW

Of course you are right to mention that my proposition does not work in any frame - but i did mention explicitly that this construction requires to start from the preferred frame i.e. where the medium is at rest. Now, if we have an equation in one frame and need it in another, we can do the corresponding transformation. For the sound equation we would normally do that by Galilean trafos and hence get additional terms for the medium, right? But starting from the base frame we can now apply also a Lorentz trafo and get an equation without a medium - but in different coordinates. Before my previous post, as well as since your last post, I had spent considerable time over this issue. I must say that what you are saying here is correct. I had been misled by the known physics into believing that one required the complete wave equation in order to apply your modified Lorentz transformation. But in fact, your modified Lorentz transformation, as a coordinate transformation, can be applied to the incomplete wave equation that is valid in the particular coordinate system to which the modified Lorentz transformation is being applied. And in the new coordinate system, the incomplete wave equation is valid due to the invariance of the incomplete wave equation to the modified Lorentz transformations. But we know that for observers in frames of reference in which the medium is not at rest, the incomplete wave equation is not valid and that the observed speed of sound will depend on the speed of the medium relative to the observer. Thus, we can conclude that the new coordinates produced by the modified Lorentz transformations do not represent the space and time of any observer. The space and time of observers is governed by the speed of light in a vacuum. Also, it may be concluded that the principle of special relativity only applies to space and time and not to the coordinates produced by your modified Lorentz transformations.

Perfluorooctanesulfonic acid?

[math]\ket{\Phi}\ ^{\underrightarrow{\ket{\psi_1}}} \ket{\phi_1}[/math]

Did you read the second paragraph? "Quantum nonlocality does not allow for faster-than-light communication, and hence is compatible with special relativity and its universal speed limit of objects. Thus, quantum theory is local in the strict sense defined by special relativity and, as such, the term "quantum nonlocality" is sometimes considered a misnomer. Still, it prompts many of the foundational discussions concerning quantum theory."

Quantum teleportation uses a classical communication channel as well as a quantum channel to transfer information about the quantum state to the other location.

The press release does not mention "local" (or any word containing "local")

But Bell's inequalities do not imply non-locality in the sense of an interaction between distant locations. The no-communication theorem of quantum mechanics forbids this type of non-local interaction. In the absence of non-locality, Bell's inequalities do seem to deny counterfactual definiteness. The notion of counterfactual definiteness suggests that quantum wavefunctions represent a lack of knowledge rather than the reality itself, and therefore quantum mechanics would seem to deny counterfactual definiteness rather than deny locality.

The total force is 0 N because force is a vector and the vector sum of all the forces on the object is 0 N. By the way, have you ever seen this problem?

Do you have a proof of this? Whenever someone mentions something like this, I am reminded of the Burnside problem in mathematics. Normally expressed in terms of group theory, it can be interpreted in terms of a string of characters. The problem itself is quite general, but there is a specific case which remains an open problem: Suppose one has a string of characters from an alphabet of two characters (a binary number perhaps). Suppose also that no substring of any length repeats five or more times in a row. The question is: Is the string necessarily finite? An answer of "no" to this open problem invalidates the reasoning you applied in this thread for it would indicate that an infinite string doesn't necessarily contain all possible substrings, such as a substring that repeats five or more times in a row. I don't know the answer to this problem, but it does provide food for thought regarding the idea that an infinite space necessarily contains all possibilities.

No. I was not applying Birkhoff's theorem to energy-radiating processes. I was applying it to the case of a spherically symmetric object that expands, contracts, or arbitrarily pulsates due to its own internal processes, without absorbing or emitting energy. For any spherically symmetric spacetime such that outside of some sphere, the Ricci tensor is everywhere zero, the metric outside of this sphere will be the Schwarzschild metric of constant mass irrespective of anything that happens inside the sphere (provided spherical symmetry is maintained). The condition that the Ricci tensor is zero everywhere outside the sphere ensures that there is no addition of energy to or subtraction of energy from the inside of the sphere. That is not necessarily true. It may be that the release of gravitational binding energy as the system contracts is absorbed by a chemical reaction. And even if radiation is emitted, this emission is not generally concomitant to the release of gravitational binding energy. Thus, the decrease in total mass does not necessarily correspond to the change in gravitational binding energy. One form of radiation that is never emitted from spherically symmetric distribution of energy-momentum is gravitational radiation. I didn't make that claim with regards to the emission of radiation. In my second post, I said the opposite.

Well, it isn't. I'm not arguing that MWI must be false because it is unbelievable in this way. I am merely pointing out that most people do indeed find it unbelievable, and that is one of the main reasons that it has remained a fringe theory instead of commanding a consensus. That's an argument from incredulity. And your interpretation seems to rely on MWI being untenable, so your reliance on an argument from incredulity is especially problematic. Well, I believe in some form of MWI because it provides a genuine explanation for the intrinsic randomness of quantum mechanics. As I see it, a measurement of a quantum state, which can be regarded as a superposition of basis states corresponding to the possible values of the observable being measured, is an interaction between the quantum state and the macroscopic measuring device such that the measuring device responds differently to each of the different basis states, producing a superposition of macroscopic measuring device states that are in quantum entanglement with the superposition of basis states of the quantum state. When we observe the superposition of measuring device states, the resulting superposition of conscious states became quantum entangled with the measuring device states and hence also with the basis states of the quantum state. Each conscious state of the superposition subjectively experiences a single measuring device state of the superposition because the individual states of the macroscopic superposition are orthogonal, and orthogonal states do not exhibit interference. Note that macroscopic states are almost always orthogonal because arbitrarily chosen vectors in a high-dimensional Hilbert space are almost always orthogonal. If the problem is that there is no consensus, then that is a problem you will never solve.

Did you not see my post before the one you quoted? The post you quoted was connected to the one before it. In that post, I was saying that in general relativity, when a spherically symmetric distribution of matter expands or contracts or arbitrarily pulsates, the external gravitation does not change, implying that the total mass does not change. The second post was merely to clarify an aspect I overlooked in the first post.

I should remark that if the radius decreases, the change in gravitational binding energy would ultimately result in a corresponding change in thermal energy, which would eventually radiate away from the spherical object, resulting in a decrease in total mass corresponding to the change in gravitational binding energy. I suppose it is this change in total mass due to gravitational binding energy that the OP is referring to rather than energy-conserving transformation of gravitational binding energy to thermal or other energy during gravitational collapse.

According to general relativity, this is incorrect. Birkhoff's theorem states that any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat. This means that the spacetime outside of any spherically symmetric distribution of energy-momentum must be given by the Schwarzschild metric. It may be considered the general relativistic version of the shell theorem in Newtonian gravity. The significance of Birkhoff's theorem to what you wrote is that if any spherically symmetric object expands or contracts or arbitrarily pulsates, then the external gravitation does not change. In other words, the Newtonian gravitational binding energy, which depends on the radius of the spherical source of gravitation, does not produce a change in the total mass of the spherical source of gravitation as the radius changes.

The defining feature of a lightlike trajectory in spacetime is that at all points along the trajectory, in all coordinate systems: guv dxu dxv = 0 That is, guv dxu dxv is a constant over the lightlike trajectory and invariant with respect to coordinate transformations. And because the chosen lightlike trajectory is arbitrary, guv dxu dxv is also constant over all possible lightlike trajectories. guv dxu dxv can't be non-zero because that would mean the trajectory is either timelike or spacelike, depending on whether guv dxu dxv is greater than or less than zero. Quite simply, spacetime does not admit the notion of a variable c, and this has nothing to do with how units of time and length are defined. Furthermore, for guv dxu dxv ≠ 0, the magnitude of an interval on the trajectory depends on the chosen endpoints of the interval. This is not the case for a lightlike trajectory, for which the magnitude of the interval is always zero regardless of the chosen endpoints. This makes the speed of light in a vacuum quite special compared to other speeds (even superluminal speeds). One thing you seem to be overlooking throughout this entire discussion is that one can't directly compare two arbitrarily chosen spacetime intervals. To perform an indirect comparison, one uses a physical object such as a clock or ruler to transfer the magnitude from the location of one interval to the location of the other interval. This relies on the magnitude not changing during the transfer from one location to the other. In the case of a clock or ruler being based on the laws of physics, it relies on the laws of physics not changing during the transfer from one location to the other. Thus, the first postulate of special relativity states that the laws of physics take the same form in all inertial frames of reference. (Note that in general relativity, local physics is special relativity). Now, the last statement depends on you choice of a metric. i know this is not an easy one to understand, especially given out intuition, but with math we can do a lot of trickery that defies intuition. Let's recall what Lorentz transformation originally are: coordinate transformations. so let's forget all interpretation and look at reality entirely from the perspective coordinates give us. Let's start with sound waves in a frame x at rest to the medium and pick some other frame x' which moves with a velocity v. What would happen if we apply a Lorentz coordinate trafo from x to x' but using c_s, the speed of sound, instead in the transformation? how does the sound wave equation look like in the new coordinates x'? Alternatively we can get the same answers when we start with the sound wave equation and ask ourselves under what kind of coordinate transformations this equation will remain invariant under? I think I understand what you are saying. You are saying that because the wave equation for sound is the same as the wave equation for light but with the speed of sound replacing the speed of light, the wave equation for sound will be invariant to Lorentz transformations with the speed of sound replacing the speed of light, and all of relativity will apply with the speed of sound replacing the speed of light. This would be true except that your initial premise is not true. The wave equation for sound: ∂2φ/∂x2 + ∂2φ/∂y2 + ∂2φ/∂z2 − 1/csound2 ∂2φ/∂t2 = 0 is incomplete. It is incomplete because there is no mention of the speed of the medium. Thus, the equation only applies to the frame of reference in which the medium is at rest. Therefore, the above wave equation is not invariant to Lorentz transformations with the speed of sound replacing the speed of light, and the rest of relativity doesn't apply either. Within the acoustic metric, the formulas become identical. But they are not physically identical. Consider the transverse Doppler effect. For sound, there is no transverse Doppler effect, whereas for light in a vacuum, there is a Doppler effect corresponding to time dilation. An important property of tensors is that a tensor that is zero (or non-zero) in one coordinate system is zero (or non-zero) in every coordinate system. A corollary of this is that a tensor equation that is true in one coordinate system is true in every coordinate system. Ideally, the laws of physics are tensor expressions. This makes sense because physical reality doesn't come with a coordinate system, and therefore the behaviour of physical reality is independent of any coordinate system. General relativity is explicitly about tensors. The idea of removing a background indicates the background is not a tensor and therefore goes against the spirit of relativity. As I said above, the above wave equation for sound is incomplete and valid only in the frame of reference in which the medium is at rest. This also means that the complete equation that reduced to the above equation is not a tensor equation (three-dimensional velocity is not a tensor).

This looks to me like an argument from incredulity.

From what I have read (I haven't watched any of the videos), he did identify himself, and that claims that he didn't identify himself or that he was threatening were blatant lies. It was also suggested that if they're willing to lie about events that took place in a room full of reporters, then what wouldn't they lie about?

It took me a while to figure out that all one needs to do is paste the URL of the image directly into the post. It defaults to imbedding the image but provides the option to display the link instead. One can then manipulate the size of the image. There is also a "Media Options" button at the top-right of the image.

It would appear that you believe reality has a non-trivial topology. I'm actually quite ambivalent about whether or not spacetime has a non-trivial topology. Anyway, I do consider the question of whether two different descriptions are describing the same reality to be a fundamental question. However, one shouldn't be blasé about what characterises equivalent descriptions. It is something that requires careful consideration. I do believe that reality does have more structure than what is provided by topology. Yeah, and it is the same two experiments also with the old standard, as the old one standard of length is still based on a derivative of the speed of light and therefore has no potential of deviation. This is my what bothers me. ... Your line of thought only works if the alternative standard of length used in an experiments can be considered sufficiently independent of c. I was actually addressing the concern that I was measuring the speed of light using a standard of length based on the speed of light. Anyway, there was a time when the standard of length was based on a platinum-iridium bar. How is the length of a platinum-iridium bar based on the speed of light? Actually, you did suggest that because atoms are based on electromagnetism that their size is based on the speed of light. But you didn't explain precisely how the electromagnetism of the atom leads to the size of the atom being based on the speed of light. On the other hand, given the fundamental connection between space and time that is manifested by the speed of light, it may be that a standard of length that is not based on the speed of light is impossible. That is, you may have a problem with a standard of length being based on the speed of light, but if it is impossible for a standard of length to be independent of the speed of light, then this becomes problematic to your idea that the speed of light can vary. and this idea works for any wave, not just light. the acoustic metric is a perfect example of that. it shows that we can treat sound waves identical to light in a vacuum with curvature and using that special definitions of time and space we get all the familiar framework. One can't replace the speed of light in a vacuum with the speed of sound. One can't even replace the speed of light in a vacuum with the speed of light in water. In the measurement of c based on the measurement of the speed of light in both still water and moving water, the choice of using light in water was merely to provide a speed that is fast enough for the relativistic effect to be significant. In principle, one could choose the speed of any object to apply the relativistic velocity-addition formula. Although like the speed of light in a vacuum, the speed of sound in a medium is constant with respect to the speed of the source, unlike the speed of light in a vacuum, the speed of sound in a medium is not constant with respect to the speed of the observer. The speed of sound in a medium is constant relative to the medium and therefore does not depend on the speed of the source relative to the medium. But the speed of sound relative to the observer does depend on the speed of the observer relative to the medium. The important role played by the medium with regards to sound in contrast to the absence of a medium with regards to light in a vacuum manifest in the difference in the Doppler effect formula for sound and for light in a vacuum.

To complete the connection between Fourier transforms and differential operators with regards to conjugate variables that I started earlier in this topic: [math]\text{Let }F(\xi) = \displaystyle \int_{-\infty}^{\infty} f(x)\ \exp(-2\pi i\ \xi x)\ dx \ \ \ \ ;\ \ \ \ f(\pm \infty ) = 0[/math] [math]\text{Then }\displaystyle \int_{-\infty}^{\infty} \dfrac{d}{dx}f(x)\ \exp(-2\pi i\ \xi x)\ dx[/math] [math]= -\displaystyle \int_{-\infty}^{\infty} f(x)\ \dfrac{d}{dx}\exp(-2\pi i\ \xi x)\ dx[/math] [math]2\pi i\ \xi\ \displaystyle \int_{-\infty}^{\infty} f(x)\ \exp(-2\pi i\ \xi x)\ dx[/math] [math]= 2\pi i\ \xi\ F(\xi)[/math] [math]\text{Therefore }\dfrac{d}{dx} \equiv 2\pi i\ \xi[/math] [Please refresh browser window if the above LaTex doesn't render]

The Heisenberg uncertainty principle relates the standard deviations of conjugate pairs of variables. But a variable having finite standard deviation does not necessarily have compact support (eg Gaussian function). So the Heisenberg uncertainty principle doesn't really say anything about the support of conjugate pairs of variables. However, I have indicated that at least one of a conjugate pair of variables must be without compact support. How this impacts on the physics is unclear since it is reasonable to assume that physical variables are bounded in value.

"Compact support" means that the function is zero everywhere outside of some finite domain.

There is an inverse relationship between the standard deviation width of a function and the standard deviation width of its Fourier transform. The value of the product of the two standard deviations depends on the function (hence the inequality) with the minimum value (where the inequality becomes the equality) achieved by the Gaussian function whose Fourier transform is also the Gaussian function. It can be proven that the Fourier transform of a function with compact support does not have compact support. To prove this, note that any function with compact support is the product of some function with the rectangular function. Therefore, the Fourier transform of any function with compact support is the convolution of some function (with or without compact support) with the sinc function, and therefore does not have compact support. The converse is not necessarily true. For example, both the Gaussian function and its Fourier transform (also the Gaussian function) are functions without compact support.

This is a common misunderstanding of refraction. Refraction is easiest to understand in terms of classical electromagnetic waves. This can then be translated to the quantum picture provided the important aspects of the classical picture are maintained. When an electromagnetic wave passes through a medium, it exerts a force on the charges and charge dipoles of the medium. Depending on how easily the charges and charge dipoles of the medium can move in response to this force, the motion of the charges and charge dipoles of the medium creates an electromagnetic wave that combines with the original electromagnetic wave to produce a total electromagnetic wave that is delayed with respect to the original electromagnetic wave and therefore travels through the medium at a slower speed. Thus, the refractive index of the medium depends on how readily the charges and charge dipoles of the medium can respond to the passing electromagnetic wave. This depends on the frequency of the passing electromagnetic wave. Higher frequencies exert a greater force, but larger bulkier charges and charge dipoles respond more to lower frequencies. At visible frequencies, only electrons can significantly respond to the passing electromagnetic wave, and in this case, the refractive index depends on the polarisability of the electron orbitals of the medium and increases with frequency due to the increasing energy of the photons.

Thanks. +1 I have an interest in supramolecular chemistry, in particular, host-guest chemistry.