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 Discussion on question by telion: Acc…

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benrg
Jun 7, 2024 15:59
I'm voting to close this as needing more focus, because it amounts to asking for a list of every possible justification of moral behavior other than "gods said to do it".
 

 Discussion on answer by Jon Purdy: Ho…

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benrg
Nov 25, 2023 20:08
@Steve There is a binary operator in lambda calculus, namely function application. You could argue that it's a two-argument function. With equal justification you could argue that $f(x)$ in standard mathematics is an application of a two-argument function. You could also argue that every mathematical function takes only one argument: an element of its domain. I think this philosophical issue is unrelated to the lambda calculus, and is largely just a matter of definition.
 

 Discussion on answer by KGM: What par…

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benrg
Jun 24, 2023 03:16
Aside from the absurd fuel requirements (even with a relativistically-optimal rocket engine), the ship is going to collide with just as many micrometeoroids on its trip as it would if it had been moving slowly, but compressed into an orders-of-magnitude shorter time frame, and each one will hit with enormously higher energy.
 

 Discussion on answer by Mark: Does In…

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benrg
Jun 6, 2023 09:15
One thing I don't understand about this answer is the repeated mentions of "materialism" as though it's the antithesis to belief in intelligent design. While that's consistent with the "wedge document" mentioned in another answer, it doesn't seem to actually make sense. The answer asks us to consider a bicycle. Is the designer of the bicycle not made of matter? What difference does it make? "Naturalistic" has the same problem. However it's defined, I don't see how "humans evolved and then made the bicycle" could fail to qualify as a naturalistic explanation. Is that not intelligent design?
 

 Discussion on question by Matthew Chr…

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benrg
Apr 21, 2023 18:42
@FlatterMann Note there are some major problems with that game, though not really relevant to this question. I don't see how one could get a better understanding of the twin paradox from the game. In principle sure, but in practice there's not enough going on in the game world. There's no way to set up a twin-paradox-like experiment.
benrg
Apr 21, 2023 18:42
It's unclear to me what you want to illustrate. Your other question says, e.g., 'the clocks attached to the destination star ten light years away "jump" from displaying "zero" to "ten years"', but I think you know enough about reference frames to know that's wrong. The clocks notionally attached to the star, representing the inertial rest frame of the star, always show that inertial frame's coordinate time. The notional clocks of the post-acceleration rest frame of the ship are flying past the star at almost the speed of light (and always have been). Is that what you want to illustrate?
 

 Discussion on answer by MaskedMagicia…

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benrg
Sep 29, 2022 21:06
If you don't put @benrg in your comments, I'm not notified of them. I told you that $\dot a^2$ isn't a second derivative. Then I asked if you understand the difference between $\dot a^2$ and $\ddot a$ because you seemed not to. You were offended and said you do. Now you claim I told you that $\ddot a$ isn't a second derivative, which I didn't. Are you sure you understand the difference? I don't think I can help you any more.
benrg
Sep 29, 2022 21:06
If the metric is $ds^2=c^2dt^2-dx^2-dy^2-dz^2$, then $g=\text{diag}(c^2,-1,-1,-1)$, and $T_{00} = T^{00} g_{00} g_{00} = T^{00} c^4$. You can give either $T_{00}$ or $T^{00}$ your preferred units, but the other one can't have the same units.
benrg
Sep 29, 2022 21:06
Do you understand the difference between $T^{00}$ and $T_{00}$?
benrg
Sep 29, 2022 21:06
If $T_{00}=ρc^2$, then $T^{00}=ρc^{-2}$. Generally I agree re units, but in GR there is nowhere that factors of $c$ "should" be, since space and time are unified (as another answer to this question said).
benrg
Sep 29, 2022 21:06
If $g_{00}=\pm c^2$, and $g$ is diagonal, and $T^{00}=\rho$ (a mass density, not an energy density), then $T_{00}=ρc^4=\rho_E c^2$. This is a good example of why trying to track factors of $c$ through calculations in GR is not a very good idea....
 

 Discussion on answer by JEB: What hap…

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benrg
Sep 18, 2022 18:59
@JEB That diagram isn't ideal because it doesn't show the beginning of the horizon as a point at $r=0$. Maybe this diagram or this one would be better.
 

 Discussion on answer by benrg: Cosmic…

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benrg
Aug 30, 2022 17:06
@vengaq Re #1, I think the resolution is that the energy you get is lower than the Newtonian prediction due to gravitational redshift, i.e. the same way this paradox is resolved. Re #2, there isn't any drag that would restart the expansion of a string. See the previous comment about a "frictional/Aristotelian effect". Anyway, a collapsing string would decay into ordinary matter.
benrg
Aug 25, 2022 15:20
@vengaq There is a perpetual tension, but you can't run a perpetual motion machine on perpetual tension (think of e.g. an electrically charged insulator). Even if $Λ=0$, there's a perpetual negative tension (pressure) in the tether due to gravitational attraction. You can extract work from it only until the tether's length reaches zero. If $Λ>0$, you can extract work until the tether's length reaches the Hubble length, then you hit the event horizon and the tether breaks. Reeling the mass back in before you hit the horizon costs as much energy as you extracted from letting it spool out.
benrg
Aug 25, 2022 15:20
@vengaq I'd recommend Christoph's answer to that question over pela's. Maybe I'll write one and try to dump some of what I've been saying here into it. In short, objects at relative rest may later gravitate toward each other, or (if $Λ>0$) away, but either way it's a force (= mass × acceleration) and is conservative. What doesn't exist is a frictional/Aristotelian effect that makes things want to go at a certain velocity. If you tether a galaxy, it doesn't "remember" its old speed and return to it, nor is it dragged back to it by galaxies you didn't tether, or by expanding-space aether.
benrg
Aug 25, 2022 15:20
@vengaq Page 64 is (according to the section name) a nonrelativistic argument. The nonrelativistic case ends up being very different because there's no event horizon. There are a number of old questions on this SE about why you can't lower things below an event horizon. Most are about black hole horizons, but it extends to other event horizon types. I don't mind writing more about this paper, but it seems way out of scope for this question. You could ask another. I think "Harrison says such and such, but benrg said blah blah in comments, who's right?" is a fine question.
benrg
Aug 25, 2022 15:20
@vengaq He says "The energy mined inside the Hubble sphere ... is finite" at the top of page 65... although he then goes on to say that you might get more by tethering a body past the de Sitter horizon, which is wrong (you can't fish things out from beyond event horizons). I didn't want to say too much about Harrison's paper in the answer since it seemed peripheral, but if you want me to pick it apart I can. I just realized that you may have mistaken Harrison's tethering strings (which are just rope) for cosmic strings (which he only mentioned briefly), and asked the wrong question.
benrg
Aug 25, 2022 15:20
@vengaq It works, but you can only extract a bounded amount of energy that way (as Harrison correctly says). That would normally suggest a conservation principle of some sort at work, but Harrison nonetheless argues that there isn't one, and I think his arguments don't hold water.
benrg
Aug 25, 2022 15:20
@vengaq 1: The energy is already there in the initial conditions of ΛCDM cosmology. Inflation can (allegedly) produce those initial conditions but I don't have a good story for where it ultimately comes from. 2: The strings don't gain more and more energy. There is no totally well defined conserved energy, but there are good enough approximate conservation laws to prevent that, AFAIK. 3: The energy is the mass of the cosmic strings; it's similar to QCD flux tube energy.
 

 Discussion on answer by benrg: What i…

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benrg
Jul 4, 2022 22:14
@Yukterez (3.7), (3.8) and (3.9) are all the same manifold in different coordinates. You can't change the shape of a manifold by putting a different coordinate chart on it. (3.9) is the metric of a sphere, and if you substitute in other coordinates—theirs or any others—it's still a sphere. You're right that even in chapter 3 they don't seem to understand that: they seem to think that this manifold is an oblate spheroid that happens to have the same metric as a sphere. That is just not a thing in GR. Until you understand the difference between geometry and coordinates, you don't grok GR.
benrg
Jul 4, 2022 22:14
@Yukterez In chapter 3 they derive the metric of a slice of the Schwarzschild horizon at $t=0$ where $t$ is a boosted coordinate, and the result is $ds^2=(2M)^2(d\bar θ^2+\sin^2 \bar θ d\bar\phi^2)$, i.e., a sphere (equation (3.9)). At the beginning of chapter 4 they do seem to make claims in English that contradict that formal derivation. I don't know what to say except that the claims in chapter 4 are wrong.
benrg
Jul 4, 2022 22:14
@Yukterez What I think you'll find if you work it out is that the contracted size of a sphere of reduced-circumference radius $r$ is $r\sqrt{1-(1-r_s/r)β^2}$. That's equal to $r_s$ at $r=r_s$ (i.e. uncontracted), and it's a monotonically increasing function for $r\ge r_s$, so there's no contracted sphere with a smaller axis than a nested one. If you substitute $r_s=0$ then the spacetime becomes Minkowski and the formula becomes the special-relativistic one.
benrg
Jul 4, 2022 22:14
@Yukterez There's no such thing as the geometry in the rest frame. Geometry doesn't depend on coordinates. For my argument, I chose coordinates in which the non-contraction of the horizon is easy to see. I'm free to do that because all coordinate systems are equivalent. You're trying to rederive the effect in a much more complicated way using formally Lorentz boosted Kerr-Schild coordinates. If you do it correctly you'll get the same answer, but it will be difficult not to make a mistake, especially if you don't understand the difference between coordinates and geometry.
benrg
Jul 4, 2022 22:14
@Yukterez The coordinates are scaled by $γ$. You need to look at the actual geometry.
benrg
Jul 4, 2022 22:14
@Yukterez The Aichelburg-Sexl ultraboost describes a light-speed particle, not a moving black hole. I showed that the event horizon doesn't contract; you can't show my argument is wrong by making an unrelated argument, unless GR is inconsistent. In your argument you implicitly used the Lorentz contraction formula, which is only valid in Minkowski spacetime. If you look at the geometry of concentric spheres in Schwarzschild spacetime when sliced by "diagonal" spacelike hyperplanes, you'll find it isn't as simple. Note you must look at the actual geometry, not the coordinates.
benrg
Jul 4, 2022 22:14
@Yukterez A sphere would have the metric $ds^2=-dt^2+r^2 dΩ^2$, in which the $t$ coordinate does appear as $dt$. In the metric for the event horizon, it really doesn't appear at all.
 

 Discussion on question by don: Specif…

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benrg
Aug 27, 2021 09:34
I think that most moral strictures derive from empirical observations of problems and experimentation with behavior changes to fix them. The experiments are framed as rules of Correct Behavior to sell them to the masses, and over time they tend to ossify and the original motivations are forgotten, but it seems almost certain that rules against incest, eating pork, and such derive from observations of disease, birth defects, etc. I don't know if this works as an answer because I know nothing about Sam Harris's beliefs.
 

 Discussion on question by Leo B.: Wha…

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benrg
Apr 8, 2021 17:31
The Wikipedia article doesn't claim that SpaceMaker is the earliest; it's just the earliest one listed (at least among those that are dated at all). The blog post also doesn't claim it's the first executable compressor, just the first on IBM PC, and even that's only a guess, if a plausible one.
 

 Discussion on answer by Emilio Pisant…

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benrg
Mar 27, 2021 22:33
If this device is similar to a moving mirror whose time-averaged position doesn't change, then the time-averaged frequency-shift factor has to be 1 (unless you've invented time travel). What is the frequency shift as a function of time? When does it redshift the reflected light and when does it blueshift it?
 

 Discussion on answer by benrg: Warp d…

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benrg
Mar 4, 2021 17:12
@Joeseph123 Here's yet another way of looking at it: my construction has two tubes, one in the $y<\frac14$ half-space and one in $y>\frac14$. Each of these half spaces can be causally foliated: they're standard Alcubierre tubes in nonstandard coordinates, as you observed at the beginning. But there's no causal foliation of the whole space. There isn't enough room between the tubes to join up the hyperplanes from either side while keeping them spacelike everywhere. If there were, the mouths would be too far apart to reach the entrances from the exits, and there would be no CTCs.
benrg
Mar 4, 2021 17:12
@Joeseph123 Are you talking about my construction? There is no point on my CTC at which it reverses time direction. You seem to want the CTC to never go backwards relative to some globally defined time coordinate, but no spacetime could ever have a CTC by that criterion. Note that the entrance and exit of Alcubierre warp tubes are geometrically distinguishable. The boosted tubes can only be entered at $t=0$ and exited at $t=-1$.
benrg
Mar 4, 2021 17:12
@Joeseph123 See my edit to the answer.
benrg
Mar 4, 2021 17:12
@Joeseph123 I don't know if this is what you're asking, but the CTCs are $\bowtie$ shaped. The vertical edges are outside the warp tubes (traveling from one mouth to another) and the diagonal edges (which don't intersect each other) are inside. The mouths could be placed anywhere in the spacetime as long as they're reachable from each other by timelike worldlines.
benrg
Mar 4, 2021 17:12
@Joeseph123 Yes, there is no global Lorentz invariance, but there is "something resembling Lorentz invariance". You could think of it this way: if an Alcubierre warp tube is physically realizable, it must be possible to construct a machine that makes one. Can you construct two of those machines? Can you give them positions and velocities of your choice using conventional rocket engines before turning them on? What is the resulting spacetime geometry, supposing the machines are far enough apart that they don't interfere with each other's operation?
 

 Discussion on answer by Raffzahn: Wha…

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benrg
Feb 2, 2021 14:03
The answer is wrong as it stands. REP is a prefix, not an opcode. It doesn't repeat the following instruction, it just sets a bit that is tested by some instructions and ignored by others.
 

 Discussion on question by Gluon Soup:…

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benrg
Jan 23, 2021 12:04
There is absolutely no local effect whatsoever from the expansion of the universe. Local accelerations come only from locally present matter, moving however it's locally moving. See this answer.
 

 Discussion on answer by Andreas Blass…

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benrg
Dec 30, 2020 12:34
@freakish I looked at one paper, "A Modified AES Based Algorithm for Image Encryption", which caught my eye because it mentioned AES. It has >200 citations. It reads like the work of a high schooler who read some Wikipedia articles. In fact I'm now wondering if this whole field was spawned by people who thought the famous Tux ECB image on Wikipedia represented an open problem in cryptography. Surely you believe there exists such a thing as objectively poor work, and domain experts whose evaluations are more than opinion.
 

 Discussion on answer by doublefelix: …

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benrg
Oct 23, 2020 14:16
When I wrote the first comment I hadn't read your answer and I guessed you were an undergrad. So it wasn't meant to be condescending but I guess it was anyway. I hope you'll concede that your views about a Bohmian solution are at least non-mainstream and that most physicists probably believe this can be solved within standard QM, even if not in the way I say. I'll try to find a source that covers this well. Maybe I'll learn something.
benrg
Oct 23, 2020 14:16
If I understand correctly, your simulation modeled the detector as an "instant classicalizer" with no internal structure that applies Born+collapse at every moment of time. Of course that produces results inconsistent with experiment. The quantum nature of the detector can't be neglected in this problem. I assure you it isn't necessary to appeal to Bohmian mechanics to solve this. I'm aware of the problem of defining a time operator in canonical QM but you don't need a time operator for this problem either. You just need to model the detector as a quantum object like everything else.
 

 Discussion on answer by benrg: What t…

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benrg
Oct 20, 2020 15:48
@Dirac'stwin No – a one-to-one correspondence between eigenvalues and (normalized) eigenfunctions only exists if the eigenvalues are all distinct. I covered this in my last comment. The space of eigenfunctions of this operator with eigenvalue 0 is very large: it consists of every state that vanishes inside the detector.
benrg
Oct 20, 2020 15:48
@lucky-guess When you choose an eigenbasis, the eigenstates are linear combinations of basis states that have the same eigenvalue. When the eigenvalues are all distinct, the eigenstates are the same as the basis states (up to scalar multiplication). When the eigenvalues are not all distinct, there are far more eigenstates than basis states. The eigenvalues are never all distinct in real life. "Complete" measurements, where the post-collapse state is completely determined by the measured value, never happen in reality.
benrg
Oct 20, 2020 15:48
@lucky-guess All of these collapses are to eigenstates. What you may be missing is that there's a big difference between an eigenstate and a basis state when the eigenvalues are degenerate. They're extremely degenerate here. Every state that is zero inside the detector is an eigenstate with eigenvalue 0. / In the delayed choice quantum eraser, there is a collapse when the particle is detected at the screen before the delayed choice, and another later.
benrg
Oct 20, 2020 15:48
the sensitivity of my screen determines when the particle arrives? To an extent yes, see quantum Zeno effect. if you wait long enough the particle will be detected anywhere in space? Its momentum is uncertain so yes. localized the particle position without induced any wave function collapse. But null measurements do cause a collapse. or as many times as we like - no because you can't choose the measurement outcome. "Null" is just like any other outcome as far as QM is concerned, it gets no special treatment.
benrg
Oct 20, 2020 15:48
@lucky-guess It's nonzero inside before a measurement, zero after. The wave function spreads out and overlaps the detector again between measurements. The time between measurements is small but nonzero because of quantization of measurement time. Please don't ask me to construct a more realistic detector; this is the same level of abstraction as every other QM thought experiment.
benrg
Oct 20, 2020 15:48
@lucky-guess The operator is diag(1,1,...,0,0,...) in the position basis where the ones are the points inside the detector. It's more realistic than the usual position operator, but still not very realistic.
benrg
Oct 20, 2020 15:48
@lucky-guess The probability density is zero inside, nonzero outside, and at the boundary there's a discontinuity. As I said, it's unrealistic, but it's more realistic than the delta function you get after a standard position measurement, or the particle that's always detected at one slit or the other in the double slit experiment. It's just a thought experiment. All that really matters is that the measurement operator has two eigenvalues corresponding in some way to "inside" and "outside" (or click and no click).
 

 Discussion on answer by Richter65: Ph…

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benrg
Nov 7, 2019 15:12
This fails to answer the question. The basic units of physics are length, time, and mass. Real multiples of the base units have a physical interpretation. Complex multiples seemingly don't. Why do the complex numbers in the theoretical formalism disappear from the predictions? And if they disappear, why were they introduced in the first place? This question has a lot of specific answers in different contexts. "Complex numbers are used in math" is not an answer.
 

 Discussion on answer by CuriousOne: H…

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benrg
Aug 21, 2014 21:11
firtree is correct—this effect does not exist. That is, there is no experiment that could detect it, even in principle. The problem is not just that you've chosen the wrong coordinates but that you've chosen the wrong geometry. The FLRW geometry implies (by the GR field equation) the existence of dust expanding with the Hubble flow at arbitrarily small scales. This would have a detectable gravitational effect if it existed, but it doesn't.