3:00 AM
Hello World... Of PhysiKs..

3:51 AM
Guys , how can we draw the structure of light. Like there are photons inside of it for example. On a pen and paper , how can we draw it.
How should the labelling be

4:33 AM
@ZeroTheHero Feel free to ping me any time in here when you want to talk dispersion/dissipation! I'll try to make sure to pop in once in awhile so I can stay pingable

4 hours later…
8:29 AM
I know that mainly supply and demand cause a stock price to fluctuate, but how is the actual price determined? Like, is there a guy sitting on a desk averaging some numbers every millisecond to determine what price to set for the stock of that company?

8:59 AM
@SlebLagnej at any moment in time there will be people wanting to buy shares in company xxx, and those people place buy orders saying what price they are willing to pay. Other people holding the stock will look at those buy orders and decide whether to sell them or not.
If the sellers are not happy with the prices offered by the buyers they won't sell and no trades will take place. If they are happy then they'll sell, and the stock is then sold at that price.
The prices you see on the ticker boards are calculated by taking some time period and averaging the prices for the transactions that happened during the time period.
So the prices are not set by anyone. The prices that transactions take place at depend on how desperate people are to buy and how desperate people are to sell.

9:21 AM
Um, that's not how it is done at the stock exchanges that I know.

No ?

Yes, the potential buyers say what price they are willing to pay. But also the potential sellers say what price they are willing to accept. Then the stock exchange will fix a price so that the maximum number of shares will be traded.

There are dozens of different types of orders and explaining them all would turn my answer into a multipage essay.
The point is that every transaction has an associated price, and the published share price is an average (of some sort) of the actual transactions that took place in some time period.
I say of some sort because I don't know exactly what averaging method is used, but I willing to bet it's complicated :-)

Ah yes, that point is also different at the stock exchanges that I know. They don't show some kind of average, they show the exact transactions.
so on Friday at 17:35:40, 73 shares were traded at a price of 75,67 €. That's the exact price, not an average.

9:47 AM
I read that there's been a 27.5% shift in the market wealth, world wide, since the pandemic started; namely, towards the most wealthy.

But when I buy or sell stock, the price is already set. Is there some special people that get to say what price they want to buy and at what price they want to sell?

@SlebLagnej you and I are statistical noise when it comes to the stock market.
Your stockbroker will tell you the price they think you can sell at or you need to pay to buy at. You place the order with your stockbroker but you may or may not get that price.

@SlebLagnej When you place the buy or sell order, the actual price is not set yet. The price will be made by the stock exchange when they look at all buy and sell orders.

Although my experience is that you do get the quoted price or very close to it.

Gotchya, but where is the board of who offered to buy at what price and who offered to sell at some price?

9:55 AM
That's managed by the software in the stock exchange.
The software matches up buy and sell orders automatically.

So my broker (be it a website or an individual) uses software from NYSE or NASDAQ to put my price on the "board" and if there's a match/agreement on price with someone else, the transaction takes place? And at the end of some period, the software calculates and averages all transactions to put up a price on the ticker?

Yes

Gotchya, thx
index funds are awesome

I like to watch where the big money goes.

average annual return for S&P-500 is 10% and if you have 1 mil in it, u'll be making a good salary by doing nothing lol

10:02 AM
A couple of decades ago I used to buy and sell shares, but I discovered that I was doing about as well as an index fund. So eventually I gave it up and just bought index funds instead.

yeah, best to go with the flow

Hmm, but when we say that return is 10%, that includes the change in price, so for me to cash out, and get my 10% "profit" I'll have to sell my shares, which will prevent me from making money next year, is that correct?

@SlebLagnej You cannot have your cake and eat it.

:(
I can eat only about 1.5% of my cake without losing it (dividends)

If you start the year with $100 worth of shares then at the end of the year you have the same number of shares but they are worth$110. Then you can sell one eleventh of your shares so you are back to $100 of shares and$10 cash.

10:07 AM
yes, my bad, I actually calculated that but forgot lol

You need to be careful about making assumptions about the long term return from shares.
It is true that in the US the stock market has grown at about 8% averaged over the last century, but this is no guarantee that this growth will be the same over the next century.

Sorry, gotta go, will be back later

cya

btw if you read my name backward you can figure out who I am (hehe)

@SlebLagnej wow
if you read that backwards, youll figure out my reaction

7 hours later…
5:15 PM
@ACuriousMind how's life

5:36 PM
Are Dp-branes embedded into spacetime in the same way the fundamental strings are? Or are they already sitting in spacetime?

5:48 PM
@SlebLagnej eh.
@Charlie I'm not sure what the difference between those things is supposed to be
You can view every submanifold of a space(time) as being "embedded" in it
conversely the image of any embedding is a submanifold

Ah ok, that's fine

@ACuriousMind hm.
2 months and I finish high school, thank you god
I've been waiting for this moment for 12 years

6:26 PM
@DanielUnderwood May I ask a question to an answer of yours on the main site?
It regards:
3

The difference between propagating and evanescent is most obvious when looking at the intensity functions of the wave. You are probably familiar with a propagating wave, as its intensity is oscillating like a sine or cosine wave; this oscillation is what causes them to propagate or travel. An eva...

@SlebLagnej I hope it will be as great as you imagine it :)

6:43 PM
@ACuriousMind Thanks XD

@ACuriousMind Yeah, you're right

Go for it!
Though I answered that 8 years ago...now I feel old

No worries :)
Well, I'd like to know, if very simplistically, propagating waves are all non-evanescent waves? Or are there more waves beyond propagating and evanescent waves to consider when studying lenses?
Also...
Evanescent waves do kind of propagate too, right? If one has a planar interface between two media, then they oscillate along the boundary, and decay exponentially into the other medium.
The answer is very clear, by the way. Just wanted to air these additional questions.

what does "an operator has a nontrivial kernel" mean?

In the context of linear algebra, the kernal of an operator is the set of all vectors that are mapped by that operator to the zero vector. "trivial kernal" means that the only vector mapped to the zero vector by that operator is the zero vector itself.
In mathematics, the kernel of a linear map, also known as the null space or nullspace, is the linear subspace of the domain of the map which is mapped to the zero vector. That is, given a linear map L : V → W between two vector spaces V and W, the kernel of L is the vector space of all elements v of V such that L(v) = 0, where 0 denotes the zero vector in W, or more symbolically: ker ⁡ ( L ) = { v ∈ V ∣ L...

6:58 PM
@schn sure
reality can of course have fields that are neither a propagating nor an evanescent waves, but if you're looking at waves and lenses and using the words "propagating" and "evanescent" to begin with that's likely not relevant to the situations you're considering

Knowing the content of the kernel of an operator is useful in all sorts of contexts because often it can be a condition for a map to be a bijection or an isomorphism

Well I'm glad ACM had an answer. Unfortunately I don't remember enough for "evanescent wave" to have much meaning beyond having an exponential decay
I'm curious of where I even came up with the knowledge for that answer. I think we covered optics a bit in one of our intro courses, but didn't really touch it again in later years

No worries, and thanks @ACuriousMind. Propagating waves seems to be a term used in optics but haven't come across it in electromagnetism. I guess there is not much rigor to it, but are called that way since if the x-axis is the optical axis, then the evanescent waves would propagate along the y-axis and so would be "non-propagating".

@schn no, the point is what Daniel says in his answer - evanescent waves fall off exponentially in intensity with distance and hence cannot make it "far" in whatever direction they might technically be "travelling", while the propagating waves are your usual waves with no intensity decline (plane) or only quadratic decline (spherical). Far away from wherever these waves are being emitted, only the propagating ones will be relevant
another way to think about this is that there is no energy transport (=Poynting vector) associated with the evanescent "waves", they do not "propagate" because they do not transport energy away from the source

7:13 PM
@ACuriousMind Consider the following diagram:
From Griffith's Introduction to Electrodynamics.
Equation 9.201 describes the evanescent wave. Is it not propagating in the x-direction?

it is, but it is not a "propagating wave", you're getting hung up on terminology. Call the two types "far-travelling wave" and "near-field wave" or something like that if you're uncomfortable with the double meaning of "propagating" here

Yeah, makes sense
But the evanescent wave has the possibility to travel far into the x-direction right? :)
Its intensity will oscillate in that direction, but not go to zero as it would in the z-direction.

yes
(but your material will typically be finite and it'll die off where the material stops)

true

3 hours later…
9:53 PM
@Slereah If a magnetron is delivered higher and higher and higher voltages will it get to the point where it will emit father than just microwaves?

10:13 PM
At 14:57 (which the link leads to), is it a typo in the expression for $\epsilon_r''(\omega)$? In the numerator inside the integral, should it not be $(\epsilon_r'(\Omega)-1)$? youtu.be/W3ExJCJ019s?t=897