Yes, the wavelength of photons will be preserved if they travel through non-expanding space. If the photon is emitted by a source that’s in motion with respect to a detector, there could still be redshift or blueshift from the relativistic Doppler effect. This would only depend on the relative velocity between the emitter and observer, and not on the distance the photon traveled between them.
Someone's already given an answer for a non-illuminated structure, but the necessary brightness of a light to be visible is also an interesting question.
We'll assume the light is located on the dark portion of the Moon. From experience, the dimmest stars clearly visible with the naked eye when right next to the Moon are around magnitude 1, which is about 3.6x10^9 photons/sec/m^2.
If we focus the light on the near hemisphere of the Earth (which has an area of 2.5x10^14 m^2) we need to produce 9x10^23 photons/sec. A green photon has an energy of around 3.7x10^-19 joules, so the total power output is 9x10^23 x 3.7x10^-19 = 333 kW.
For reference, this is roughly comparable to the wattage of the fastest electric car chargers. It's a lot of power, but well within the capability of a small lunar solar farm.
Geographically big, if you need to see it as more than a point. 20/20 human visual acuity is around an arcminute, or 1/60 of a degree. The whole moon is about a 1/2 a degree across as seen from Earth. Below that, it’s a matter at how good you are at picking reddish grey out from normal grey. Red is also scattered and absorbed pretty strongly by the atmosphere IIRC.
If it’s allowed to emit light, you can go a lot smaller. I’d guess an LED array with an emitting area of 10 m^2^ would be visible from Earth as a sort of star when in shadow, without doing any actual math on it. When competing with sunlight it becomes a visual processing problem again.
If it’s allowed to focus at you, too, a 1 cm aperture should be able to resolve a single kilometer on Earth, and at a kilometer of distance you can see a candle flame if it’s dark enough. It’s just a beefy laser pointer at that rate.
I'd say slice it first and freeze it while it's fresh. Put it in a plastic bag and it won't dry out noticably. However, it might get a bit limp(?) after defrosting. You can toast it slightly to make the crust crispy again. If your bread has a crispy crust in the first place. For sandwich, it's definitely fine.
I suppose we're talking about widely varying things. Here in Europe I think they have both pre-packaged bread and sandwich, which I suppose comes on the same truck as the other stuff. And sth like a Deli section with bread and small pizzas and stuff. That probably comes as frozen/refrigerated dough and is baked there. I'm not really sure if they do it like that at Aldi, but they do that in other stores. And I mean Aldi is just one supermarket chain.
As far as I know Aldi and Lidl try to use their established business procedures everywhere, because it's cheaper for them. I've never been to the US. But everywhere I've been, the stores all look very similar. I don't know if that applies to bread, but the store layout and product assortment is similar and I bet they also use the same IT infrastructure. Including maybe ePaper price tags that arent that common in some other countries. There are some differences. Some products vary and they don't ship them across the ocean but replace them with local products. And there are cultural differences. For example you can't rip open the plastic wrapper and pull out one bottle of water in some countries. But I bet a lot of it is the same across the world. Especially in Europe.
Product assortment for Aldi is significantly different in North America vs Europe/Germany. But again Aldi isn’t alone in doing this, it was just an example of shipping things frozen.
Bread baker here. I always freeze loaves that I bake but will not be able to consume before they’d go bad sitting on the counter. After removal from the freezer and defrosting just by sitting out on the counter, I find no difference between the frozen loaves and not frozen. It’s a great way to preserve bread for future consumption if you’re making multiple loaves at once.
hello, bread baker, your profession is one of the best in the world <3 please give me some advice on how to defrost bread faster than just letting it sit on the counter. my wife takes the bread out of the freezer, waits a bit, separates a piece from the frosted pre-sliced loaf, puts a piece on a pancake pan, warms, and sliiightly chars the sides, but randomly, or if the piece is eaten too slowly, it gets mushy. thank you.
I’d say this is probably thermal cycling. Some polymers can degrade with repeated thermal cycling (ie tires ‘cycling out’). Google scholar has a few papers on it.
Ok, stretchy isn’t a great word for that because basically all bread bags match what you did mean, though mine don’t tend to get crinkly in the freezer.
I am not familiar with the bags you are talking about (our bread bags aren’t stretchy), but this has a fairly straightforward explanation. Things that are elastic usually get stiffer when cold. This is part of why winter tires exist. There’s literally less molecular movement.
You did say they stay crinkly/brittle even after warming up, though. This is likely due to another mechanism. When a solid is created from a liquid, there is typically some type of crystal structure (with notable exceptions like glass). A material can have multiple crystal structures due to how the molecules line up. Often, the crystals are tiny, so you don’t see them, but you can have large crystals if something is cooled slowly. That’s how you get gems.
When crystals start to form, they start to incorporate as much of the surrounding material as possible. When they run into a neighboring crystal, they run out of material. Unless they just so happen to line up perfectly, they will remain separate. The space between them, called a grain boundary, can be a weak spot in something like a diamond. In metals, more grain boundaries actually make things stronger, usually. This is because metal crystals can slide along the plane of the crystal. This is why blacksmiths will quench stuff; the rapid cooling leads to smaller crystals, which leads to more grain boundaries.
A metal won’t completely form crystals from every available molecule. Every process happens over time, and cooling a metal down extra cold causes it to shrink, which can cause any straggler molecules to join up with the crystals, which makes the metal stronger. That’s why some metal objects are “cryohardened”.
The last factor is that changing temperatures can change the most energetically favorable crystal structure. Tin pest is a famous example where in really cold weather, tin can change from its useful form to a brittle crumbly useless form, and it can only be fixed by remelting it.
It’s all a bit weirder with plastics cause they can be crystalline, non crystalline, or a mix, but my guess is you’ve changed the structure of it.
The follow up question would be the opposing force which keeps them in orbit(als)? This balance of force was called the planetary model which has this shortcoming that electrons might fall into the nucleus.
If electrons actually followed such a trajectory, all atoms would act is miniature broadcasting stations. Moreover, the radiated energy would come from the kinetic energy of the orbiting electron; as this energy gets radiated away, there is less centrifugal force to oppose the attractive force due to the nucleus. The electron would quickly fall into the nucleus, following a trajectory that became known as the “death spiral of the electron”. According to classical physics, no atom based on this model could exist for more than a brief fraction of a second.
What this means is that within the tiny confines of the atom, the electron cannot really be regarded as a “particle” having a definite energy and location, so it is somewhat misleading to talk about the electron “falling into” the nucleus.
the electron cannot really be regarded as a “particle” having a definite energy and location, so it is somewhat misleading to talk about the electron “falling into” the nucleus
Good way to put it. And if I recall correctly, electrons in “s” orbitals actually do spent a certain fraction of their time inside the nucleus.
There’s kind of alot going on, but the shortest answer is “the electrostatic force between the positive nucleus and negative electron creates orbits in the same way that gravity allows a moon to orbit a planet”. The electron is moving fast enough that it just “misses” the nucleus. At least, from a classical lens.
It gets more complicated when you introduce orbital angular momentum and start considering the magnetic effects of moving charges, and that’s what leads to the funky non-spherical orbital shapes.
As I understand it, it’s the quantum part of quantum mechanics.
Electrons can only have fixed energy states, they can only radiate or accept fixed sized packets of energy - a “quantum” of energy. So an electron that is hit with the correct sized quantum of energy can be excited up to the next orbital, and it will emit the same sized packet of energy when it returns to its ground state. So they can’t gradually emit radiation and fall into the nucleus.
Eventually electrons should spontaneously decay but that’s predicted to be in 10 to the power of 40 years or something like that.
Presumably there is a transformation of charge to energy which is then carried away by the photon, but all of this is beyond my understanding of the theories involved.
Charge conservation would indeed be violated, which is why this decay is not expected. Dave is mistaken: the half-life they’re referring to is an experimental lower-bound, not a actual expected value.
The electron orbits are quantized, with a lower limit. As found by Neils Bohr. Classical theory would have the electron attracted into the nucleus in a tiny fraction of a second. His theory was not liked at the time and he faced a lot of push back, but they eventually accepted it as it’s true.
The language of forces and balancing forces comes from classical mechanics.
The particles interact via electromagnetic waves. You simply have to apply quantum mechanics and solve the Schrodinger equation. It’s a different kind of thing.
There are a lot of good answers here already, but I’ll try to attack the question from a new angle.
Firstly, yes: they experience an attractive force from the nucleus, and would in principle have their lowest possible potential energy if they were located exactly in the nucleus. An equilibrium state is the state with lowest energy, so why aren’t they exactly in the nucleus?
Consider that an electrons position and speed cannot be exactly defined at the same time (uncertainty principle). So an electron with an exact position could have any speed. If you compute the expectation value of a particles kinetic energy, when the particle can have any speed, you’ll find that it’s divergent (goes to infinity).
So: Because an electron with an exactly defined position must have infinite kinetic energy, the equilibrium state cannot be an electron with an exactly defined position, and so cannot be an electron exactly in the nucleus. So what do we do?
We have to make the electrons position “diffuse”. Of course, that means it is no longer exactly inside the nucleus, so it gains some potential energy, but on the other hand it can move more slowly and has lower kinetic energy.
The equilibrium state is the state we find where the trade off between kinetic and potential energy gives us the lowest total energy, which is described as a 1s orbital. The electron is “diffuse” enough to have a relatively low kinetic energy, and “localised” enough to have a relatively low potential energy, giving as low total energy as possible.
Once you start adding more electrons you need to start taking Pauli exclusion into account, so I won’t go there, but the same manner of thinking still essentially holds up.
Yes, there are an infinite number of velocities you can use, but if you look at their distribution, you’ll find that it quickly goes to zero somewhere around 1-2 m/s, so the expectation value of the velocity is convergent.
If you have an object with a velocity taken from a distribution that doesn’t approach zero sufficiently fast as the velocity goes to infinity, the expectation value diverges. A simple example would be a person that would be half as likely to get up at a velocity of 2 m/s as 1 m/s, and half as likely to get up at 4 m/s as 2 m/s, etc.
The more mathematical version of the same argument is to compute the kinetic energy of a particle whose wavefunction is a delta pulse (i.e. a particle whose position is exactly defined), and you’ll find that the particle has infinite energy.
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