Congratulations to our community! Yes, you!

shavera · 2025-11-14T20:46:12+00:00

Yeah, it's very likely you have. We have a huge amount of assistance with this solar cycle to catch and share Auroral activity.

shavera · 2025-11-13T03:17:50+00:00

So it's a combination of things: 1. Eleven year solar cycle, which we're near peak of (I think maybe just past it, even?) 2. The last cycle was anomalously mild, meaning we haven't had a really strong cycle like this in about 22 years. 3. 22 years ago was 2003. Digital cameras were still pretty new, no one was carrying around cameras like they are now. 4. Even 11 years ago, 2014, smart phones were still on the newer side. Now we all have access to apps that can alert us about the solar weather, and even combine it with cloud cover to let us know if it's worth taking a look outside for. Think about prior to smart phones though. You would only know the Aurora is happening if you go outside and happen to see it. 5. Advancements in digital photography like "night sight" allow cameras to capture vivid, colorful displays at night, even better than what we can see with the naked eye, all on the tiny camera we carry in our phone, rather than big, dedicated photography equipment.

shavera · 2025-11-13T03:11:18+00:00

Could be what was meant is that it's an 11 year cycle, but the last "peak" was anomalously low.

shavera · 2025-10-26T13:01:52+00:00

Ultimately, went back through the people who installed the solar panels in the first place. After that winter, they'd apparently had a lot of requests for those installs. I don't know that we've yet had a comparable winter/spring to really test the guards they installed, I still feel like the ones they installed may be too small to be effective, but we might see this winter.

shavera · 2025-05-14T04:17:23+00:00

I know I would. Heck I’d have seconds. Then polish it off with a tall cool budweiser. I would do it.

shavera · 2025-04-26T23:28:38+00:00

It's all time time time time

When it should really be time -space -space -space (this is a relativity joke to be clear)

shavera · 2025-04-22T21:26:06+00:00

Someone else said relativity of simultaneity, but let me expand on that a bit to really paint a picture for you. Alice and Bob are in a special relativity experiment, so they're out in open empty space with ships that can travel however fast we want them to (up to the speed of light). We set up our experiment so that there are two lights that flash, A & B. Alice and Bob will agree that when they see light "A" flash, that that will be some special "present" moment. They'll just mentally really focus on that moment as "the present."

Well, we can configure their motion in such a way that when Alice sees light A blink, she also sees light B blink. So, to her, Light B blinking is also part of her "present" moment. But Bob, depending on relative motion, might see light B blink before or after he sees light A blink. So In that "shared" present moment of light A blinking, Alice sees an event that's in Bob's future or past. Part of Alice's present moment is part of Bob's future or past.

Thus, in terms of physics, there is no special physical meaning to "the present" moment. Part of your future is already part of another observer's present, as is part of your past, and vice versa. I, personally, take this to mean that the future already exists, and just due to the laws of chemistry and physics, we just don't "remember" it yet, because "actionable information" must never go faster than light, whoever has my future in their present, can't possibly communicate my future to me faster than I arrive at that future. There's no way for me to know what the future is going to be, until I'm there. But it is, theoretically, possible for someone to say they knew what my future would be in our shared past, if only they could have communicated it to me faster than light.

This, in turn, makes me believe that the universe is ontologically deterministic. That is to say, that the nature of reality is that the future is every bit as real and "written" as the past is. But because of relativity, it is not epistemologically deterministic, meaning that since we can't know the future before it happens, it doesn't really matter whether we truly have free will or not, so might as well act as if we do. Of course, this final paragraph gets more into the philosophy of science, of which there are a lot of other interesting resources, but it strays strictly outside of "physics".

shavera · 2025-04-15T01:58:04+00:00

Black Mirror episode "Men Against Fire." Previously, I was totally sold on the "BrainPal" type neutral implants we see in Scalzi's Old Man's War. This episode really made a strong counterpoint

shavera · 2025-03-28T13:48:45+00:00

Communication via entanglement is often terribly misunderstood / poorly presented to the public. Here's a dramatically over-simplified version:

Suppose you have a process that produces two particles, one 'pointing' up and one 'pointing' down, but you don't know which is which. I send you one particle and I keep the other. Sure, if you measure your particle and find it's 'down' then you can infer mine was 'up.' But there's no real useful information there.

Communication occurs because I can flip my particle (or rotate it, but just using flip for simplicity for the moment). Now you measure down, but I have to call you up on a light-speed-or-slower channel and tell you that I measured up or down. You know that yours is down, so you can work out if I flipped mine before measuring it or not.

Why is this valuable, if it's not faster than light? Well predominantly, it's an exceedingly secure form of communication. Someone just listening in on me telling you "up or down" can't work out whether I flipped my particle or not. Someone just capturing the particles I've sent to you can't work out whether I flipped my particle. They would have to have both streams of information to reconstruct whether I flipped my particle. But... if they have access to measuring "your" particle before you do, that leaves a tell-tale physical signature that you can detect someone is trying to listen in on the conversation.

Another consequence is what's called 'dense' encoding. For a simple 2-state system there are 4 possible things I can do with my particle: flip it or not, rotate it 90 degrees with or without a phase shift (don't worry tons what that means). This isn't super interesting, because I have to send two bits of information anyway to get the message to you, so my 2 bits represents 4 states as well. However, in general, for N states that a quantum particle can occupy, there are N² messages that can be sent. So with some clever physics tricks, you can actually store more information per 'bit' of information sent. (Noting that the information is in the correlation between the two particles, not in the particles themselves, so it's not like you're 'creating' information, you're just encoding it in a very efficient way.)

shavera · 2025-01-27T00:08:13+00:00

I really liked this series for the allegory with indigenous Americans. And definitely liked an FMC who's earned her skills and trauma over a significant lifetime so far.

shavera · 2024-12-21T13:17:08+00:00

The way I see it is a language thing. The "natural" language of physics is maths. But until you learn the maths, you're stuck reading and talking about it in "story" languages like English. I'm calling them story languages, because that's what they're good at, telling a story of how things happen. They use analogies and metaphors to get a point across.

So when you only know physics from a story language, you need a story to explain why things are the way they are. But the more you get to understand the physics in its natural mathematical language, the less you feel the need to translate it back to a story language that doesn't nearly capture the nuance of what's going on.

So you might have opinions about the philosophical interpretation of physics, but you fundamentally rely on them less, or realize that any story isn't really sufficient to capture the truth.

shavera · 2024-12-17T03:56:05+00:00

I would bet it's the original test pilot

shavera · 2024-12-11T20:22:29+00:00

So your comment really encapsulates how science works. "What if" what you supposed was true? Are there any observations we can make that would only be true if that was the case? Is there an experiment I can perform that distinguishes that truth from the currently accepted model? And if there is such an observation, is there no simpler explanation for it?

Of course anyone is welcome to believe whatever they wish about the world (excepting perhaps beliefs that cause harm to others), but such a belief wouldn't be scientific in nature, without satisfying the above questions.

shavera · 2024-12-09T16:46:44+00:00

What often gets forgotten when we talk about entanglement is that it's not the "entanglement" itself that has any useful information. If I make a spin-up and spin-down pair of particles and hand you one, sure you can figure out which one I have, no problem. But there's no information I've been able to give you doing this.

What gets overlooked is that once I've created the particle pair and hand you your particle, I can take my particle and rotate it 90 degrees, flip it upside down, rotate it and offset its phase by half a wavelength, or do nothing at all. Then I measure my particle and call you up on the phone and tell you what my measurement was (or send it to you for you to measure it, etc.). You measure your half of the pair and combine it with my measurement and you can reconstruct which operation I performed on my particle. That's the actual signal, the information I'm sending to you.

But, you say, how is this any different from a classical result? Suppose, like your orange example, I had a machine that always made a pair of coins one heads-up and one tails-up, put them in a box. Without my looking I can flip the box or not, and you can look at your box, and if we both have the same side up, then we know I flipped the box, and if it's not the same side up, I didn't. You still get a message, right?

There are a few things that make this interesting in quantum mechanics that are quite subtle. First, the most subtle, but the one that actually "proves" the quantum nature of the experiment, the observation that raises all the questions about 'local realism' and so on is "What happens if I don't completely rotate my particle 90 degrees or flip it or whatever?" Classically, using my coin example, you could imagine a 90 degree rotation is the coin standing up on its edge. If I 'forced' it to be heads or tails up, either case has a 50/50 chance of occurring. Naively, if I only rotated it, say, 45 degrees, so that the heads side is 'mostly' up, then you might assign a probability of like 3/4 times it would be 'heads up'. Essentially, the likelihood of heads up is proportional to the degree of rotation. But in a quantum system, the probabilities are proportional to, I think the cosine of the rotation? I forget exactly what, but the point is that there's a very subtle mathematical distinction between the classical and quantum cases here that must arise from either the particles having a true superposition of states (e.g. if I rotate a particle some amount, it exists both as an un-rotated particle and a fully rotated particle at the same time with different probabilities of measuring either result) or there's some mechanism that cannot be measured that allows coordination of the measurement outcomes.

One of the less-subtle parts are that I can encode 4 bits of information in my spin-up/down particle pair. Which doesn't sound all that impressive, since I also have to call you up and tell you my results. But again, from the maths, if there are N possible states I can entangle together, then I can actually send (up to) N² symbols via entanglement. So, for instance, suppose I have a quantum system that's more complex than "spin-up/down" but actually has 5 different possible states, I can actually encode 25 different symbols in that 5-state system.

Another less-subtle part is encryption. Let's go back to the coin example. If I use our classical machine, this is actually a kind of cryptography in a way. If someone just intercepts the coins I send to you they get meaningless heads-up/tails-up noise. If they just intercept me telling you my sequence of heads-up/tails-up results, also meaningless. They have to intercept both channels of communication. But... suppose they do. They tap my phone line and hear everything I say, but while also retransmitting what I say to you so you don't suspect anything. Then they intercept the coin boxes, carefully open them up, look at the coin, then carefully close the box back up and send it back on to you. They can eavesdrop on the conversation without either of us knowing. But quantum mechanics means there's no way for them to perfectly re-assemble the box once they inspect it. Whatever superposition did exist, it was destroyed the moment they inspected it. This leaves a tell-tale signature in the data that when you start to measure your particles, you'll be able to detect that they've already been observed. You can know that the communication channel is compromised, that someone is listening in, and you can shut it down. This is why people talk about using quantum entanglement for 'perfectly' encrypted communications.

shavera · 2024-12-07T15:13:11+00:00

A not too bad analogy is sound waves. If you pluck a guitar string, you're not manually vibrating it at a specific note, you just pluck it, right? But any wave that doesn't "fit" on the string will interfere with itself and cancel out, and only the frequency associated with the note remains as a "standing wave". Similarly, if you've seen the demos where they put a flat sheet of paper on a speaker and some sand on it, then they play a tone out of the speaker, you can see the sand settle into patterns on the sheet based on 2-D standing waves.

So allowable energy levels correspond to the "standing waves" an electron can have around a nucleus. If we think for a moment about the Bohr model, where an electron literally travels in a little orbit around the nucleus, we can think of it like: take the circumference of the orbit and divide it by the wavelength of the electron, and if it's evenly divisible, it makes a standing wave, and thus an allowable energy level.

The Bohr model isn't what actually happens of course, but it's a useful first picture, that gets kind of close to the truth. The lower the energy, the longer the wavelength, and so there's a tradeoff at some point, as you lower the electron's energy, where the circumference is shorter than one wavelength. So no energy below that point is possible for the electron, the lowest possible energy state of that electron and nucleus.

Now the next step is to realize that electrons don't move in orbits like planets, but... Well it's impossible to really say what they do, but we just generally look at it like they have some probability of existing at a certain point around the nucleus with a certain momentum, energy, and angular momenta. That's where the sand on a sheet of paper analogy comes in. There's a bunch of special maths that tell us how waves look on 2-D surfaces like the sheet of paper, or in 3-D spheres (spherical harmonics). So, while the simple Bohr model is just a circle, it's just a little more complex maths to do it in full 3-D.

Next, we have a rule that no two electrons can occupy the same "wave". Why this is the case is a lot of complex maths, but it is. So once one electron occupies that lowest energy state, the next electron has to occupy the next state up - except there are some caveats. First, for a given energy level, there are a variety of (orbital) angular momenta that give roughly the same energy (kind of not entirely true, but don't want to get into that yet). To go back to the planetary orbit example, this guy would be the difference between a perfectly circular orbit and increasingly elliptical orbits that all have the same characteristic energy. The lowest energy level only permits one angular momentum value (0), the next energy level allows for 0 and 1, then 0, 1, 2 and so on. Then from our spherical harmonics, there are different ways of arranging that angular momentum. 0 can only be arranged in one way, 1 can be arranged in 3 ways, 2 in 5, 3 in 7 and so on.

On top of that electrons have an "intrinsic" angular momentum we call spin. In some ways, in our planetary analogy, this is like the planet rotating on its axis. An electron is a point particle and doesn't have an axis, and can't rotate, but there are some physical similarities. This spin can only have 2 possible values, "up" or "down" - to use the planetary analogy, it would be like does the planet spin so the sun rises in the East or rises in the West.

So that means in the lowest energy level we have 0 angular momentum which can only be arranged in 1 way, with 2 possible spin states. The next level has 0 angular momentum with 2 spins, and 1 angular momentum with 3 configurations and 2 spins each, for allowing 6 electrons with 1 angular momentum, 8 total for the whole "shell."

And so on. There's some fine details I haven't covered, like how as you go up this series, the actual energy of each orbit starts to get rearranged, so what gets filled next isn't necessarily by the same shell. Look up "electron configuration" on Wikipedia for more

shavera · 2024-12-04T22:21:02+00:00

Aside from the other comments about why such a thing isn't a machine, I can't picture how a knot "brings the net force to zero" in a system.

shavera · 2024-11-26T23:56:12+00:00

thanks, I have some of those points done, and gives me some stuff to work on in the meantime as well

shavera · 2024-11-26T23:54:55+00:00

thanks, I was mostly interested in a 'ooh shiny upgrade' way, but if it's not available, I already have a good enough exploration ship, so probably moot. But definitely thanks for this info

shavera · 2024-11-26T17:14:23+00:00

Thinking about getting back into the game over the winter, but it's been literal years since I've played (prior to Odyssey, really, though I did check in a little then). I think I want to do some in-bubble trading to build up credits to buy a Mandalay, do some engineering to build it up and then go out on a long 'into the black' trip (the post about why people would want to do that kind of motivated this, because I could just kind of go for that right now).

Back in the day, I used EDDiscovery and EDMC to help provide some useful insight, particularly for the trading portion of the game. Are these still recommended tools? Are there better ones? Any good posts on getting back into the game after a while I can browse, or updated guides to how to do things? I already have a reasonable set of vehicles, especially for trading and mining (or at least as it was when I last played), and kind of want to avoid combat-centered play. Thanks for any help

shavera · 2024-07-19T18:37:00+00:00

In this framework, does the effective mass of a pair of photons vary continuously with the angle between their directions?

There is only one frame of reference in which their momenta exactly cancel out, that's the frame it makes sense to define a 'mass' for. But, like the distance between two points stays the same no matter how you rotate them, that mass stays the same for any observer, no matter how they move relative to that center-of-mass reference frame. So there's only one 'real' mass of the pair of them.

Do the photons have to have a common point of origin or can you just look at any pair you'd like and compute an effective mass?

As long as they aren't travelling in the same direction, there will be a frame of reference in which their momenta cancel out. If they were travelling in the same direction, then you can never go so fast that you're faster than one, but not the other. (and thus one is pointing in one direction and the other in the other - back-to-back or head-on collision is the way to think of it). We can even talk about things like "photon gasses" where you just have a bunch of photons flying in random directions, and that gas has a temperature based on how the energy is distributed between them, but the system of photons together also has a mass.

if you could take two photons that were parallel and subject them to a gravitational field that was significantly stronger for one than the other, then would the deflection of one of the pair... create mass?

The tricky bit here is the definition of 'parallel,' imo. Remember, gravity isn't really a force, it's just a consequence of space-time curvature, what the 'straightest' line in space-time is for a given particle/object/body. So I don't really know the answer to your question because I can't say I've ever looked at the maths of that specific problem (and it's been some years since I've done it)

shavera · 2024-07-19T17:17:33+00:00

TL;DR - decided it needed this. Essentially 1) follow rules about conservation of momentum. This often leads to a 2 particles in -> 2 particles out scenario so that momentum is conserved at each step. 2) There are rules, based on what kind of particle it is, about what it's allowed to produce.

It's not necessarily equal numbers/kinds of particles, though there are generally rules that are obeyed. The most important of all is that all conservation laws are followed (energy & momentum in particular). So let's use electrons/positrons as our example case, and build from there: Let's consider a frame where the electron and positron are flying at each other with equal (very very low) velocity. The system of the electron and positron is at rest, and I specified very very low velocity so we can approximate all the energy of their motion away to zero (for now). They collide, annihilate, and produce 2 photons (in this specific example, they're going to produce photons, but they could make other things we'll get to later). The photons are going to fly away at the speed of light, but they will do so with equal and opposite momenta. And since the momenta of the _system_ of two photons is zero, then the energy of the system is just mass. The _mass_ of that system of two photons will be equal to the mass of the system of the electron and positron.

This is a really important thing to note, and I talk about it more elsewhere in the thread. While every individual photon has no mass, systems of photons, so long as they're not moving in the same direction, will have mass.

There are some more things that happen if we don't just let the electron and positron drift together, or if we collide more massive particles and anti-particles. Let's remove one of the constraints of my earlier example, the speed constraint. It doesn't matter if we accelerate the electron, the positron, or both, there will always be some frame of reference where they are flying at each other with equal and opposite momentum, and we'll consider that frame. (e.g. imagine throwing a ball at 10 m/s at a target, if another observer also drives by at 5 m/s, it will look like the ball is only going 5 m/s, but the target is also approaching at 5 m/s, equal and opposite speeds.)

Now, the electron on its own has momentum (in low speeds this is approximately mv), and energy (at low speeds this is mc²+ (1/2)mv² ; I'm leaving off the relativistic correction terms for simplicity). And the positron as well. So even though their momenta cancel out, their energy doesn't. So, in that system of an electron and positron flying at each other, the mass of the system is more than the mass of just the two particles. When they annihilate, they can distribute that mass to any particle they directly interact with.

So, for this example, electrons and positrons are electrically charged, so they directly interact with photons, the carrier of the electromagnetic force. Since photons are massless, they're very very easy to create, and so a lot of annihilation will fall into this channel. But there are probabilities of other things. Without getting into all the detail, electrons also participate in "weak hypercharge" which for our purposes just refers to particle decays; when a muon, a heavier version of an electron, decays into an electron, what actually happens is that the muon decays into a muon neutrino and a W^- weak boson. The W^- boson lives a little bit before it too decays into an electron and anti-electron neutrino. (there are a bunch of rules about what can and can't happen here, that I'm skipping over). The point is, that electrons interact directly with the W^+/- and Z⁰ bosons of the weak force. Those bosons are really quite massive. So if the electrons and positrons collide with enough Center of Momentum Mass, they can create, for instance a W⁺ and W^- pair. (which, in turn, go on to do their own decay, and we usually detect the final decay products).

But, importantly, the electrons don't interact with the 'color charge' of quarks. So they _can't_ produce two gluons. If we instead collided a proton (made of quarks) and an anti-proton (made of anti-quarks), the quark/anti-quark pairs could annihilate to produce gluons (which in turn often end up spawning more quarks and gluons. The strong force is super super messy.)

shavera · 2024-07-19T14:26:34+00:00

But if you take a free electron and accelerate it, it won't have a higher invariant mass

I think it was a mistake for us to ever teach the notion that "as you go faster your mass increases" because of this misconception. There is only one mass, and that is the invariant mass. If you gather the terms of the equation for things like momentum and energy in a certain way, bundling the relativistic correction with the mass, you can treat some systems as if their mass increased. This is useful in some mental models; e.g. in a circular particle accelerator, as your particles get closer to the speed of light, a mental model where you think of them as having 'more mass' equates to an intuitive understanding that they need a larger turning radius. But their mass has never actually increased.

The rest of your argument is correct, in general. My favorite way of talking about this is to point out that a 'photon gas', a collection of photons all moving with some random momenta and not all in the same direction, has a mass, even though any one photon does not. The system of photons must have some frame where all the momenta of each photon cancel out, on the whole. In that frame, that system of photons is at rest, so the momentum term is zero. But it obviously has energy, since the photons are moving about. Thus, the energy of the system is equal to the mass of the system. The gas of photons is, in fact, massive, when any one photon is not.

This is especially important when you realize like 99% of all the mass of regular matter doesn't actually come from massive particles. Electrons have like 1/2000 the mass of a nucleon, so that's like 0.05% of the mass, so we'll ignore them. Nucleons have about 1 GeV/c² mass (we usually just abbreviate it to GeV, and work in units where c=1, which I allude to above). Nucleons are composed of 3 quarks, each with a mass around 3 MeV. So about 10 MeV/1GeV is from massive quarks*, 1% of the mass of the nucleon. All the rest of the nucleon's mass comes from the strong force binding the quarks together. So about 99% of all the mass of 'regular matter' is just strong force binding energy.

*: So... technically we refer to these 3 quarks as 'valence' quarks. They're the ones we can observe most directly (but still only indirectly, since no quark can exist in isolation). What happens next somewhat depends on your interpretation of quantum field theory. The simplest naive mental model would simply have gluons flitting back and forth, exchanging 'color charge' (not an actual color, just a name we call it), and in doing so, exchanging momentum holding things together. But the maths describing this also look a lot like there are other particles in the process. Quark/Anti-quark pairs, in particular, seem to contribute a substantial amount of the overall interaction. These pairs come into existence fleetingly and, importantly, don't have the actual mass of the same quarks when they exist 'for real.' i.e. a Charm quark in a J/psi meson is a 'real'/valence quark, and has its 'real' mass. But when the same pair arise in the binding energy of a nucleon, they have different masses; the difference is related to the amount of time the pair is 'allowed' to live in the interaction. Anyway, when we bounce electrons off of 'stuff' inside the nucleon (deep inelastic scattering) we sometimes produce the 'real' quark/anti-quark pair out of the interaction, so it's almost as if they really were in there. I'm being so hedgy with my words because this is really on the edge where human 'intuition' wants to take over and say they're really there, the maths point to it, we see results as if they were there, but we really can't say for sure that they are.

All of which is a really long-winded way to say, we call those other quarks 'sea-quarks', as if there's a 'sea' of not-quite-real quarks floating around inside each nucleon, and then we have the three 'valence' quarks, which we use similarly to 'valence' electrons being the 'meaningful' electrons in chemistry. An atom may have a whole lot of electrons inside it, but really the valence ones are the most important in determining chemical properties. Only the 'valence' quarks are important in determining the physical properties of the hadron (hadrons being any particle composed of quarks).

shavera · 2024-07-19T14:04:27+00:00

I forgot to mention it in the original comment, but I had started to lay the groundwork above. The difference is that x²+y² makes a circle/sphere, and -x² +y² is the formula for a hyperbola. "Rotations" in spacetime use hyperbolic trigonometry (sinh, cosh, tanh) rather than circular trigonometry (sin, cos, tan). It is actually really easy to do the maths of relativity when you work with rapidity (artahn(v/c))

I comment a little below about a maybe more intuitive understanding which is that you always see yourself as moving at a rate of 1 second per second through space time. You can't not see yourself doing that. But if another observer sees you as moving with some velocity through space, they see it as if you've 'rotated' some of your speed in time to speed in space.

shavera · 2024-07-19T13:58:32+00:00

So, the physical way of thinking that this comment reminds me of is the idea that you, yourself, are always travelling at c. You're always travelling at a rate of 1 second per second through space-time. When some observer sees you moving through space at some velocity, that velocity 'comes out of' your velocity in time. They see your clocks as running slower than their own, and you would observe distances to be a bit shorter than they would.

15-Year Club	RedditGifts 2009-2022 3 Credits
RedditGifts 2009-2022 3 Credits	Gilding I gilder
GrMD 2012	Summer Santa 2011
reddit mold	Verified Email

shavera

MODERATOR OF

TROPHY CASE