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Everything posted by Genady
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I think so. Some kind of a "key".
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The pets should vote as well.
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Last week I've made four ATM withdrawals. This morning I've noticed that my account balance went up compared to a couple of days ago. Then I've noticed that all four ATM transactions disappeared from my activities. That is the total amount by which the balance grew. What should I do about it, if anything?
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Yes. Although they might become relevant at the Planck's scales.
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After considering the last two comments, by @Janus and @Markus Hanke, I've realized that my attempt has failed. The spatial distance in the rest frame between the two measurement events has to be accounted for as well. Thank you.
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Concerning Infinity (of course)
Genady replied to Boltzmannbrain's topic in Linear Algebra and Group Theory
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Concerning Infinity (of course)
Genady replied to Boltzmannbrain's topic in Linear Algebra and Group Theory
Yes, you did. That what "when n goes to infinity" means. It just another phrase for "limit". -
Concerning Infinity (of course)
Genady replied to Boltzmannbrain's topic in Linear Algebra and Group Theory
That exactly means that the limit of the partial sums is 1. It does not mean that a partial sum is 1. -
Concerning Infinity (of course)
Genady replied to Boltzmannbrain's topic in Linear Algebra and Group Theory
No, it does not mean that the partial sums total 1. It means that the limit of the partial sums is 1. I emphasize again: partial sums do not total 1; limit of partial sums is 1. -
Concerning Infinity (of course)
Genady replied to Boltzmannbrain's topic in Linear Algebra and Group Theory
The (2^n-1)/2^n does not reach 1 and the definition does not imply that it does. -
This exercise is from Gravitation by Misner, Thorne, Wheeler. I am concerned here with the second part of it, marked with green: My answer for the electric field is (ej, u). For the magnetic field, (ek, ei), where i-j-k are in the cyclical order of 1-2-3. Is it correct? For the reference, here is a components representation of the Faraday tensor in a Lorentz frame:
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I'll try. Take a surface of a cone and attach a line to the tip. This space is two-dimensional 'under' the tip and one-dimensional 'above' it. No coordinates can pass the tip.
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The probe would read higher energy, but that reading will not be a temperature. To get the temperature from that reading, the peculiar motion will need to be subtracted. Like we subtract the peculiar motion of Earth to get the isotropic reading of CMB.
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For this, I can think of a space with different dimensionality in different parts. It would be a topological space, but not a manifold.
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Large Language Models (LLM) and significant mathematical discoveries
Genady replied to Ghideon's topic in Mathematics
We cannot predict the result, but we know how to get to it step-by-step. Thus, I don't think it explains "why at a high level we don't know how deep learning architectures produce their outputs." -
I think, neither. We assume that (Topological space - Wikipedia).
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It's average kinetic energy of a random motion of molecules. We cannot boil water by running around a kettle, unfortunately. Temperature measurements themselves are not affected. I want to point to the origin of this OP. It started with this exercise from MTW: My explanation above is my partial answer to the question at the end. I've added the conditions of uniform temperature in the OP to make the second term on the right vanish. OTOH, I've added in my explanation the gravitational time dilation effect, which is not considered in the exercise. Thank you for the last remark.
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It does not.
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Concerning Infinity (of course)
Genady replied to Boltzmannbrain's topic in Linear Algebra and Group Theory
You mean rather (2n-1)/2n. But we know that it does not. -
The only way to avoid confusion of what slows relative to what is to look at things which are frame independent. The fundamental frame independent things are events. So, let's take two events. Event 1: the proton takes a temperature measure. Event 2: 1 ns later on its clock, the proton takes another temperature measure. The change in temperature between the two events is 10 degrees (on some scale). The rate of cooling observed by the proton is 10 degrees/ns. For the astronomer on Earth, the proton's clock moves slow. While it has advanced 1 ns, the astronomer's clock has advanced 2 ns. Thus, the astronomer observes the same cooling of 10 degrees in 2 ns, i.e., 5 degrees/ns. So, the proton observes faster cooling than the astronomer. A gravitational time dilation adds to the effect. It does not depend on speed, but on the strength of gravity at time and place of the event. Both events occur in the gravitational field of the star, which perhaps is stronger than the gravitational field experienced by the astronomer. This makes the proton's clock run even slower. Let's say, while the proton's clock has advanced 1 ns, the astronomer's one has advanced 2.5 ns. Thus, the astronomer observes the cooling rate of 4 degrees/ns compared to the proton's 10 degrees/ns.
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Let's check each effect in turn. Doppler. No effect since the CR can measure the temperature "by touch" or in the direction perpendicular to its velocity. Velocity reversal. I don't see any effect. It does not affect the temperature reading. The velocity does not reverse relative to the astronomer. Anyway, time dilation depends on speed and does not depend on velocity. If it does not, we compare the observations it does before its demise. Perhaps, but it will only weaken the effect.
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Hmm ... I don't say if the answer is right or wrong, but I think that comparing the ray's clock with the object's or the dwarf's clock is confusing. The question compares the ray's observation with the external observer's observation, and I think that only their clocks should be compared. For definiteness, less say that the external observer is an astronomer on Earth, i.e., very far from the dwarf.