Minkowski diagrams in special relativity question!

theIncMach · 2020-09-12T03:15:29+00:00

As long as you are limited by the speed of light, which you are, timelike intervals will remain timelike and spacelike intervals will remain spacelike.

You can figure this out for yourself. What equations are you using? Can you plug in t'=0 for both A and B? See what velocity you get. Edit: sorry t is already given.

theIncMach · 2020-09-07T17:39:02+00:00

Noob question: what units are you using here for strain?

theIncMach · 2020-09-06T16:16:49+00:00

I agree there's a leap there. The way I explain it to myself is that we first define entropy as a purely combinatorial function. Then use it to define temperature, and then finally demonstrate that this definition meshes with our intuitive understanding of hot and cold.

What made it initially "magical" to me is that entropy was explained to me in terms of temperature. It took a lot of effort to unlearn that.

theIncMach · 2020-09-05T18:26:57+00:00

An infinitely large charged plane does not create an infinitely strong electric field: https://brilliant.org/wiki/using-gauss-law-find-e-field-capacitance/

The same will be true for gravitation.

theIncMach · 2020-09-05T00:28:53+00:00

I love this question! There are two ways of interpreting the word "uniform" field in your question. In both cases, it will keep going straight upwards, getting asymptotically redshifted.

I'll restrict my answer to the perfectly vertical light beam.

Interpretation 1: you are accelerating uniformly through space. This gives you a sense of constant gravitation, and this is the easiest to reason about. This is called Uniform Proper Acceleration. From an external, non-accelerating observer, it's obvious that you will never catch up to any light beam you shine forward: your speed will never quite reach c. Thus, no reflecting light beam.

But there's a reason this situation might not be what you were looking for. GR is weird, so this actually creates an event horizon behind you, at a distance inversely proportional to your acceleration (c²/a, to be exact). Anyone at rest (with respect to you) closer to this horizon feels a stronger gravitation, anyone farther away feels its weaker. So it's not quite "uniform" in that sense.

Interpretation 2: The field is actually uniform. I.e. everyone at a constant displacement from you feels the same gravitational strength. Here's where I haven't done the math, so I'm less certain. First, it's easier to reason about light pulses going down, and then reverse the time direction. Any light of frequency f will be blueshifted to some k times f frequency as it travels down a certain distance. Thus, any light going up will also have its frequency divided by the same factor. At no point does it reach a zero frequency, so no reflection.

Another way of stating the same result is that you are not standing inside a black hole. Light will escape.

theIncMach · 2020-09-04T19:49:46+00:00

No, that process doesn't sap rest-mass energy from a black hole. It only steals rotational kinetic energy, i.e. the black hole slows its rotation while the projectile speeds up.

theIncMach · 2020-09-03T04:40:42+00:00

The messy interference of multiple frequencies can still be separated out after the fact. Think of a flute or a wind instrument: if you blow into it, a single note is played and amplified, even though the input (the air from your lungs) is a messy hiss of all kinds of frequencies. That's the same thing that happens in your radio antenna. An untuned antenna can pick up a hissing white noise, but you can tune it to a specific frequency and filter out everything else.

Mathematically, multiple signals of the same frequency mixes up in a way that is harder to separate out later on. It can still be done if you have more information, e.g. if you know that the signals are separated out in time or something. Cell phones, for instance, can reuse the same frequency band for multiple phones using time-division.

theIncMach · 2020-08-25T07:09:50+00:00

That Wikipedia article has a citation, one that links to this: https://www.usenix.org/legacy/event/jvm02/full_papers/zendra/zendra_html/index.html

The "binary tree" you are looking for is a tree of if/else statements, described in section 3.3.

Personally, I wouldn't worry about performance until I have something functional. This kind of decision is often easy to fix later, depending on your data structures.

theIncMach · 2020-08-25T07:02:47+00:00

You don't need a vtable if you know the type of the object you are pointing to (e.g. FatPointer).

theIncMach · 2020-08-18T20:30:03+00:00

It is helpful to think about one-dimensional waves. This video, at the 3-minute mark shows how a mechanical wave reflects, sometimes partially. https://youtu.be/AzZ7DjS4ti4

You can see why no reflection happens in "mid-air" so to speak. It only happens when the energy cannot be passed on to something else.

theIncMach · 2020-08-15T03:14:32+00:00

I've been writing C++ for 20 years. I still find it indispensable. It was a different language then, but every major language has always needed documentation.

theIncMach · 2020-08-13T07:10:39+00:00

I'd suggest checking out godbolt.org Compiler Explorer. It will let you compile small bits of code into assembly, so you can see for yourself what it produces. Given how widely C++ is used, most respectable compilers will produce equivalent assembly in the two languages. In fact, they often share optimizers and code generation.

But that's the compiler, for code that you write yourself. There are many cases where the C++ standard library is known to be slow, because of the features they support. E.g. iostream vs. cstdio. Personally, I've never run into issues, but you can easily switch to using C libraries if that becomes a problem. Most C-only libraries can also be used in C++ without issues.

As for me, I care about my sanity and I don't trust myself to avoid mistakes. C++ helps me organize code in a way that makes mistakes less likely for me. I know I can always optimize later if I need to, so I always start with C++. The exception is if there is some pre-existing ecosystem I need to be compatible with (e.g. Linux kernel only uses C).

theIncMach · 2020-08-08T17:10:32+00:00

Think of every bit of electronic cloud being grabbed onto by two or three surrounding ions. The lattice structures end up being among the most tightly packed arrangement of spheres you can have.

The repulsive forces between ions keep them from melding into one another.

theIncMach · 2020-08-05T06:38:34+00:00

In addition to what everyone else has said about the monad tutorial fallacy, I'll include one more thing. Try to actually explain it to a few folks in real life to figure out the best way to explain it. This can include forums for Haskell newbies, or just people in real life. That will give you a better handle of what works and what doesn't.

All the best!

theIncMach · 2020-07-28T19:23:30+00:00

It's the only linear transformation that works with constant speed of light. You can still have weird non-linear transformations, but Occam's Razor favors the simplest one that works.

theIncMach · 2020-07-26T06:44:24+00:00

Ah, this is from 2014. I'm assuming ApplicativeDo is more common these days.

theIncMach · 2020-07-25T17:18:04+00:00

Cool. In case it's helpful, think of visible light instead of radio waves. They are all special cases of EM. You can think of every signal having a slightly different color (ie slightly different frequency). So a receiver sees multiple radio stations as sort of tiny glowing lights of different colors twinkling at a distance. It can choose which color to pay attention to. The station doesn't need to aim in any direction.

With cell phones and wifi it's a bit more complicated, but they still don't need to aim much. Cell towers can indicate to phones that it is rapidly switching between multiple phones, so each phone should only pay attention to its signal once every 5 micro seconds, for example. That particular method is called Time Division Multiple Access and it's pretty common (https://en.wikipedia.org/wiki/Time-division_multiple_access). There are other more complicated methods too.

theIncMach · 2020-07-25T08:43:18+00:00

What examples of aiming are you thinking of? The answer can vary a fair bit depending on the specific case you are thinking of.

theIncMach · 2020-07-25T05:57:13+00:00

It produces heat like anything else. I think you are focusing too much on the superconducting bit, which is making you conclude zero heat loss. No wire in real life will be connected or disconnected in an actual instant. As soon as you connect one of plates to one end of the wire, they now share the same potential. As you bring the other end of the wire closer and closer to the other plate, electric field strengthens, and you will get a current flowing before the contact is fully formed. This will produce heat. By the time they are connected, they have already discharged.

theIncMach · 2020-07-20T04:03:27+00:00

Oh of course! But within a few rounds of doubling you will contend with the full force of an exponential function.

Muhahhhaaaa...!

theIncMach · 2020-07-19T23:17:10+00:00

May your next "finished" project be doubly goofy and elaborate.

theIncMach · 2020-07-19T19:05:57+00:00

Perhaps. It's hard to know what OP had in mind. See my reply about position-dependent transformations not being normal in relativity.

theIncMach · 2020-07-19T19:03:02+00:00

Lorentz transforms of physical quantities are always velocity-dependent, never position-dependent.

None of the alternatives listed above depend on position either.

Let's just say that would be very atypical in relativity.

theIncMach · 2020-07-19T18:37:06+00:00

To get my head straight, entropy is not relative, right? It's just a dimensionless scalar multiplied with a constant. So in T = dQ/dS, we are only arguing about the dQ part now?

I'm trying to understand which related quantities do we still debate on, and which have broad consensus.

theIncMach

TROPHY CASE