KJW

What are you talking about? Changing your IP address that external world sees, or changing your IP address within your local network? What you've written above seems to be about your local network. Not long ago, I had a problem in this area so I do have some experience, although it did take a change of modem to completely fix the problem. Your IP address that the external world sees is allocated to you by your ISP, and if you want to present a different IP address to the websites you visit, you could use a VPN.

Why do you say that freely falling charges do not radiate? A charged object sitting on a table doesn't radiate even though it is accelerating. I think the circumstances under which a charge radiates is not straightforward. This is a topic that interests me a lot and I've even been considering this issue quite recently. Part of the difficulty as I see it is that the mathematics that says an accelerated charge radiates is limited to the Minkowskian metric and needs to be generalised to arbitrary metrics. Another difficulty as I see it is how the Lorentz force relates to Maxwell's equations. I'll discuss this in more detail in the hopefully-not-too-distant future.

I think many websites are adopting this as their business model for ads:

What is a water wave without the water? In the case of an electromagnetic wave, without the medium, one still has the electromagnetic wave.

I knew this (that the distinction between x-rays and gamma radiation is how they are produced and not their wavelength, frequency, or energy), but had forgotten this (that the kinetic energy alone of the electron striking the nucleus could provide gamma radiation).

Hmmm. I'm actually somewhat surprised that in the fusion from hydrogen to iron, the first step to helium provides about 80% of the energy (or am I interpreting the diagram incorrectly?).

Do beta processes (any type) emit gamma radiation?

Thank you for your effort, but I was hoping for something I didn't already know (although a clarification of the distinction between global and local symmetries pertaining to Noether's theorem would be helpful). My particular difficulty is about the connection between a wavefunction, which describes probability of general objects, and electromagnetism, which is a more specific notion than the objects to which wavefunctions apply. And of course, there are the weak and strong forces, with their own symmetries. The symmetries of the electromagnetic, weak, and strong forces are often described as "internal symmetries", but I find this term unsatisfying. Another thing: How does [math]A'_{\mu} = A_{\mu} + \partial_{\mu} \phi[/math] from classical electrodynamics relate to [math]\psi\, ' = e^{i \phi} \psi[/math] from quantum mechanics?

What I meant was: to what symmetry does charge conservation correspond?

[My bold] To be fair, relatively few people are aware that charge is the result of a symmetry. Perhaps you could explain what that symmetry is.

You could use the endothermic reaction as the cold sink of a heat engine, with room temperature being the hot source.

I have to admit that the inclusion of the "exergonic" choice seems odd among the other four choices. But I'm not convinced that any of the other four choices are correct anyway. You say that tert-butoxide is a sterically hindered base, but I don't think it qualifies as a sterically hindered base. A sterically hindered base is one where the lone pair(s) are sterically crowded to the extent that the base is totally non-nucleophilic and is stronger due to the steric relief provided by protonation. Tert-butoxide is definitely nucleophilic. One other thing: In the question, the reaction is conducted in ethanol, which is a somewhat stronger acid than tert-butyl alcohol. Therefore, the base is predominantly ethoxide rather than tert-butoxide. This opens up the possibility of an SN2 reaction forming the ethyl ether product instead of an elimination reaction. This is stereospecific although no mention was made in the question that the starting material is anything but racemic.

Actually, it does because if the reaction was stereoselective then there would be two correct answers in contradiction of the given statement that there is only one (the reaction is exergonic). Ok, I was probably mistaken in my view that stereospecific excludes stereoselective, though I am perfectly aware that these two notions are conceptually distinct (as indicated by the textbook figure you presented).

No. There is only one answer allowed, and the reaction is most definitely exergonic, thus eliminating the other four choices as the answer. (BTW, E2 reactions are stereospecific.)

It seems to me that the new definition of AI is better than the old definition because it basically says that an AI system is intelligent, whereas the old definition can be satisfied by any old computer, intelligent or not.

An ideal clock always ticks at the same intrinsic rate regardless of its location and motion. Time dilation is always about the comparison of the length of a trajectory in four-dimensional spacetime of one ideal clock to the length of a trajectory in four-dimensional spacetime of another ideal clock. An important part of this is how the two ends of each of the two trajectories relate to each other.

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E. Exergonic Let's just say that this reaction will proceed irreversibly to the products.

You are reading things into my post that aren't there! I'm not reading anything into your post other than that we are on different pages. You introduced the binary tree and I'm saying that I'm not going to accept any arguments from you involving the binary tree, and why. An ellipsis is convenient but is hardly a rigorous way to represent the infinite. And given that this topic is about hierarchies of the infinite, an ellipsis must not be a crucial part of a proof about the infinite. For every Cantor (v), there are 2v sequences in a finite list. Since v is an integer from the set N, there is no largest v. Thus the set of all finite sequences cannot be listed. The list has no end. It's simple. It's not simple. We wouldn't be on page 7 of this thread if it was simple. But I feel I need to clarify what I mean by an "infinite list". An infinite list is a one-to-one mapping between a given set and the set of natural numbers. That is, for every natural number there is an element of the given set to which it maps, and for every element of the given set, there is a natural number to which it maps. The set of sequences can't be listed because regardless of the mapping from the set of natural numbers to the set of sequences, there are sequences for which there is no corresponding natural number. But this is not the case for the set of finite sequences which can be mapped one-to-one with the set of natural numbers. I am NOT questioning Cantor's theorem. The subject is the 'diagonal argument'. You have rejected Cantor's theorem from the outset. Any discussion of the diagonal argument is ultimately about Cantor's theorem. Otherwise, what is the point of discussing the diagonal argument? That's actually not true. For example, relativity is much better understood by modern day physicists than it was by Einstein. That's because in the time since Einstein, physicists have had the opportunity to learn more about the theory than Einstein ever could. Sure, why not? I did say that everyone makes mistakes every now and then. By the way, why is Gauss in this list? Mathematics is not an idea/mental construct. "The set-theoretical axioms that sustain modern mathematics are self-evident in differing degrees. One of them – indeed, the most important of them, namely Cantor's axiom, the so-called axiom of infinity – has scarcely any claim to self-evidence at all" I said it was the validity of a mathematical theorem that is self-evident. I wasn't talking about axioms, which are often said to be self-evident truths. Thus, I am able to verify Cantor's proof to my own satisfaction and without reference to Cantor. Do you think repeating this multiple times will make it true? Do you think ignoring it multiple times will make it false? By assuming that the set of sequences is the same as the list of sequences, your argument against Cantor's diagonal argument is invalid as begging the question. Although Cantor proved that there are sequences in the set of sequences that are not in the list of sequences, even without such a proof, you need to recognise that this is a possibility that needs a proof one way or the other. So, by assuming that the set of sequences is the same as the list of sequences, you have made an error regardless of the validity of Cantor's proof. By contrast, I have not assumed that there are sequences in the set of sequences that are not in the list of sequences, that has been proven, and prior to a proof, it was something to be proven. Don't project your inability to comprehend infinity onto others. In particular, don't project your inability to comprehend infinity onto me.

But I'm suggesting that there was about 30 times more sodium than potassium in the earth's crust to begin with, and that over time this became depleted to the ratio we see today. In that case, why does sodium dissolve preferentially? Maybe sodium was in a more soluble form (eg chloride) than potassium (eg something else... I don't know what). While both sodium and potassium tend to form soluble salts, I think potassium is more likely to form a less soluble salt.

It could be that the earth's crust has already been depleted of its sodium by all the rainfall.

[math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{12 \phi^2 - 3 \phi^4}}{2 (1 - \phi^2)}[/math] From here, I took a shortcut and simply ignored the two [math]\phi^2[/math] outside the square root and the [math]3 \phi^4[/math] under the square root because I could see that these are not going to be a part of the final result. Thus: [math]1 + \alpha = \dfrac{4 - \sqrt{12 \phi^2}}{2}[/math] for [math]x \to 0[/math] (ignoring higher-order terms) [math]1 + \alpha = 2 - \sqrt{3} \phi[/math] However, one can verify that the shortcut leads to the same result as follows: [math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{12 \phi^2 - 3 \phi^4}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{12} \phi \sqrt{1 - \dfrac{1}{4} \phi^2}}{2 (1 - \phi^2)}[/math] The series expansion of [math]\dfrac{1}{1 + x}[/math] and [math]\sqrt{1 + x}[/math]: [math]\dfrac{1}{1 + x} = 1 - x + x^2 - x^3 + \cdots = 1 - x[/math] for [math]x \to 0[/math] (ignoring higher-order terms) [math]\sqrt{1 + x} = 1 + \dfrac{1}{2} x - \dfrac{1}{8} x^2 + \dfrac{1}{16} x^3 - \dfrac{5}{128} x^4 + \cdots = 1 + \dfrac{1}{2} x[/math] for [math]x \to 0[/math] (ignoring higher-order terms) Thus: [math]\dfrac{1}{1 - \phi^2} = 1 + \phi^2[/math] for [math]x \to 0[/math] (ignoring higher-order terms) [math]\sqrt{1 - \dfrac{1}{4} \phi^2} = 1 - \dfrac{1}{8} \phi^2[/math] for [math]x \to 0[/math] (ignoring higher-order terms) [math]1 + \alpha = \dfrac{1}{2} (4 - \phi^2) (1 + \phi^2) - \sqrt{3} \phi (1 - \dfrac{1}{8} \phi^2) (1 + \phi^2)[/math] [math]1 + \alpha = \dfrac{1}{2} (4 + 3 \phi^2 - \phi^4) - \sqrt{3} \phi (1 + \dfrac{7}{8} \phi^2 - \dfrac{1}{8} \phi^4)[/math] [math]1 + \alpha = 2 - \sqrt{3} \phi + \dfrac{3}{2} \phi^2 - \dfrac{7 \sqrt{3}}{8} \phi^3 - \dfrac{1}{2} \phi^4 + \dfrac{\sqrt{3}}{8} \phi^5[/math] [math]1 + \alpha = 2 - \sqrt{3} \phi[/math] for [math]x \to 0[/math] (ignoring higher-order terms)

More like, make up your mind, which is it. I actually said: "E0 must be a member of L". I didn't say that E0 is not a member of L. You incorrectly inferred that from: "When the sequences are placed into a list, the diagonal argument shows that although the negation of the diagonal belongs to the set of sequences, it does not belong to the list of sequences". E0 is a member of the set of sequences. E0 is not a member of the list of sequences. This is not a contradiction. The list of sequences is not the set of sequences. Cantor proved that the list of sequences is not the set of sequences. You are not going to convince me otherwise. The infinite binary tree is actually ambiguous. It can represent all possible sequences of 0 and 1, or it can represent all possible finite sequences of 0 and 1, which is still infinite. The latter can actually be made into a list. The former cannot be made into a list. Your argument becomes invalid if you assume the binary tree represents all possible sequences of 0 and 1 but treat it like it represents all possible finite sequences of 0 and 1. Due to the subtlety of the distinction between these two cases, I will not accept an argument from you based on a binary tree and will assert that a binary tree is not a list. That Cantor can define an E0 that is excluded from the list is what proves his theorem. I suspect that you think he has forced a result. It is true that regardless of how a list is constructed, even if it is formed from a set that can be listed, the E0 derived from this list will be excluded from the list. But E0 isn't just excluded from the list. E0 is also a member of the set of all possible sequences of 0 and 1. Therefore, the exclusion of E0 from the list proves that the set of all possible sequences of 0 and 1 cannot be listed. But earlier in this thread, I showed that the set of all finite sequences can be listed. I then applied the diagonal argument to this list to obtain an E0 that is excluded from the list. However, in this case the E0 obtained was also not a finite sequence, and therefore a proof using the diagonal argument that the set of all finite sequences cannot be listed fails. This is just plain silly and brings to mind the term "cargo cult mathematics". One point that you've overlooked is that one doesn't need to consider sequences to prove Cantor's theorem. One can prove Cantor's theorem using sets, and the mappings between their elements and their subsets. So, if you're focusing specifically on sequences, then that won't invalidate proofs based on alternative notions. But one can still make statements that are true for every element of an infinite set. I never suggested that Cantor was infallible. And everyone makes mistakes every now and then. So what? If there was an error in Cantor's proof, then the error isn't just Cantor's, it's all the people who have accepted Cantor's proof as valid, including myself. So, you have to do better than mention Cantor's fallibility. What about the fallibility of everyone else who accepted Cantor's proof? Of course it's irrelevant. We're talking about a mathematical theorem, not Mein Kampf. The validity of a mathematical theorem is self-evident, not based on the author.

[math]\times\!\!\!\!\phi^2[/math]

[Here you agree E0 is a member of L, which represents M.] [Here you say E0 is not a member of L, which represents M. ?] You think I contradicted myself, don't you? That's because you fail to distinguish between the set of sequences and the list of sequences. You fail to recognise that there are sequences in the set of sequences that are not in the list of sequences. By failing to recognise this, you are unable to understand Cantor's proof because this is precisely what Cantor is proving. By assuming that the list of sequences has all the sequences that are in the set of sequences and using this assumption in an attempt to disprove Cantor's theorem, you are begging the question. By "the list is incomplete", I mean that there are sequences in the set that are not in the list, and that is precisely what Cantor is proving. Having "no last member" doesn't come into it. But by assuming that there is only one infinite, you are assuming that Cantor's theorem is false in your attempt to disprove Cantor's theorem, again begging the question.