@Slereah Gell-Mann was obviously an accomplished scientist, but I didn't really like his attitude towards other physicists. I went through snippets of his interview on youtube where he talks about other physicists, and he comes off as envious and vain.
Here's a quote about Gell-Mann: "I had barely sat down when he began to tell me... that science writers were "ignoramuses" and a "terrible breed" who invariably got things wrong: only scientists were really qualified to present their work to the masses. As time went on, I felt less offended, since it became clear that Gell-Mann held most of his scientific colleagues in contempt as well." - John Horgan, The End of Science (1996) p. 212
Interface and colloid science is an interdisciplinary intersection of branches of chemistry, physics, nanoscience and other fields dealing with colloids, heterogeneous systems consisting of a mechanical mixture of particles between 1 nm and 1000 nm dispersed in a continuous medium. A colloidal solution is a heterogeneous mixture in which the particle size of the substance is intermediate between a true solution and a suspension, i.e. between 1–1000 nm. Smoke from a fire is an example of a colloidal system in which tiny particles of solid float in air. Just like true solutions, colloidal particles...
I worked in all sorts of areas. It was a fascinating job, though I did get bored with it eventually and switch to working in IT.
You'd be surprised how intellectually demanding IT jobs can be. ACuriousMind used to work in string theory, which is about as complicated as physics can get and now he works as an analyst/programmer.
Looking back, I think I got as much satisfaction from the IT work as I did from working as a colloid scientist.
Possibly more, since very few scientists see any great change as a result of their work while some of the IT stuff I did made a huge difference to the company I was working for.
My main job was writing software to monitor servers. By the time I retired I had to monitor about 800 servers and there is no way you can check that many servers manually.
So I wrote a big suite of software that monitored all the servers and alerted me if there were any problems.
It was fascinating because it require a detailed knowledge of how the operating system worked at quite a low level.
@cOnnectOrTR12 The unique advantage of software is that once it works, it works. If you train a dog to do something, then you can't just make 1000 other dogs do the same with no effort. If you write a program that runs on one computer, it will run on others like it with no change at all.
All the skill and knowledge of the programmer goes into the program once, and then it is available to everyone else, for eternity. (yes, it's not quite as simple, but it is a massive improvement over transmitting the ability to do certain things only from human to human or from human to book to human)
"Just calculation" misses the point - while deep down everything is just numbers and bytes and bits, computers handle all sorts of structured data for us. Images, formulae, personal records, etc. The power of symbolic computation is that if something is data, you can make a computer do work on it
@cOnnectOrTR12 but updates don't just "come"! Someone programmed that update and decided it is acceptable that it might break existing programs and then you decided to install it
don't blame the computer for poor human decision-making
@cOnnectOrTR12 my point is that its reliability is something the humans programming it determine. It is not an inherent fact about software that updates have to break things.
@cOnnectOrTR12 and that's the magic - it does exactly what the programmers tell it!
it doesn't need to "feel" to crunch the terabytes of data the LHC produces, or to send the words you're typing into this chat around the world to make them appear on my screen
It’s like you are the thinker and you employ a human calculator to calculate and memorise for you simply because you can’t handle such big calculations.
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This is live now on MSE for testing - please report any issues as answers on this questio...
To edit units, some editors use \mathrm.
eg.\mathrm{m/s}$\to\mathrm{m/s}$
Also some use \text.
eg.\text{m/s}$\to\text{m/s}$
Recently I edited a post using the first method, and I found that someone had suggested an edit with the second method. Which one should we actually use?
I'm not using the computer to do a "computation too big for my brain" when I use the internet. I'm using it to do something fundamentally impossible in a world where all work on data had to be done by humans.
Well it’s all technology so instead of celebrating computers we should be putting our mind on ai. That can well one day become smart as humans and probably more generous. To help us in our quest of finding big questions.
@NiharKarve personally I think sci journalism is in trouble. They only talk about hype and never the null result. The sci books which try to explain things to common man are some kinda Zen bullshit riddle. The person doesn't know anything about physics but he does learn alot of useless analogies.
I ve seen those books do a lot of damage to students where they think physics is about constructing rubbish analogies and learn that's not the case in their degree
I hate this letter $\xi$. I dont understand why we have to use this symbol. I mean there are many greek letters easier to draw. Do we use this letter in order to imply the level of difficulty in the topic it appeared? Anyway, I always wanted to ask how to draw this letter. Should I first draw an ...
I haven't written many answers, but after I have identified what the user is asking, I usually go with a conversational tone, going over whatever has a question mark over it and any context supporting it.
Optionally(if you have the time), recently my answer https://physics.stackexchange.com/a/666...
how do we calculate the highest limit of force exerted by a charge on the another (i.e: k q1 q2/r^2), this shows that the force exerted is inversely proportional to the distance squared . suppose when r approaches 0, the force exerted will also approach infinity, but this is not practical and not possible. so what is the highest magnitude of force, a charge can actually exert ? how we can calculate it ?(of course I believe this is not the exact formula for calculating the force by a charge)
I don't have any idea about the practical case of highest limit achieved in lab, but on theory, infinite magnitude force is possible whenever you bring two electrons closer enough, but yes, how much closer enough remains a question and planks limit and quantum effects come into play
force very close to infinity is quite possible if you bring in two electrons sufficiently near even in practical case.
I don't have much idea regarding that but the people here might help you with that.
Again I must warn, bringing two electrons close enough for almost infinite repulsion force is not feasible as it'd require a very large amount of energy to do so and measurement there after,,
@Slereah Yes, that's what I meant, quantum effects at planks scale, not limit actually
two macroscopic charged objects just have a size, and you can't get to $r=0$ with them - there's just some finite distance where you can't push them into each other further
for two particles or whatever, quantum mechanics becomes relevant - they aren't actually little balls, and $r=0$ can't "happen" in the way you probably imagine it
@Slereah sure, if you're unlucky you get black holes
At what age you discovered the nonsense physicist call Lagrangian mechanics is just the nonsense engineers call D'Alembert principle which is just a different same nonsense in different face
so the moment you try to "defeat" Coulomb's law by saying "the force exerted will also approach infinity, but this is not practical and not possible", you really also need to stop thinking about point charges
if you want to be practical, there indeed are no infinite forces, but also no point charges, so there is no problem
just removing one part of the idealization but not the other is inconsistent
here's something I know how to establish by brute force but not in a nice way
suppose I have the usual singlet state $|\psi\rangle$. Since this has total spin 0, we have $S^x_{12} |\psi\rangle = 0\implies S_1^x |\psi \rangle= -S_2^x|\psi\rangle$, and simialrly for $y,z$.
as such, the outcomes for $S_1^x$ and $S_2^x$ are anticorrelated and the same holds for $y,z$
i think it still won't work, because you'd have $\psi$ being annihilated by three different operators
but that's not quite precise enough, because the singlet state is annihilated by infinitely many such combinations
@ACuriousMind the context underlying this, btw. in quantum computing one usually sees the Bell state given by $|00\rangle+|11\rangle$, which differs from the singlet state $|01\rangle-|10\rangle$
the prof whose i'm grading for noted in lecture that the bell state displays perfect correlations in X and Z, and added in passing that it's also perfectly correlated for any observable
buuuuuut
that didn't seem right
easiest way to check that it's not: $|00\rangle+|11\rangle = (1\otimes -i\sigma^y)[|01\rangle-|10\rangle]$
the action of $-i\sigma^y$ on the second qubit will exchange the X2- and Z2-basis states, but leave the Y2-basis states unchanged
in terms of qubits, i'd say it means that a spin observable S displays perfect correlation on a two-qubit state if it's of the form $|00\rangle + |11\rangle$ in the S-basis, up to phases on the coefficients
and perfect anticorrelation if it's $|01\rangle +|10\rangle$ in that basis
so that's where your annihilation condition comes from - "both give the same result" means that $S^i_1 - S^i_2$ is measured to be 0 always, i.e. this is an eigenstate of $S^i_1 - S^i_2$ with eigenvalue 0 for all $i = x,y,z$
I assume "brute force" means writing the state out in a basis and explicitly deriving equations for the coefficients from each of the three annihilation conditions?
If I am right then the E field vectors at points don’t tell about the strength of field. It’s the flux of field that tells us where the field is stronger and weaker.
In griffiths it is explained like this. You draw field vectors at points around the charge. The vector lengths decreases as we move away from the charge. Then it says join the vectors and we get field lines.
But doing so have we lost the information about the field strength. No , the strength of field is determined by how close the lines are . Near the charge it’s closer so more strength and away the field lines are apart so field is weaker. So the density is what tells us about the strength and that is defined as flux. This flux is directly proportional to charge enclosed by the surface and that is gauss law.
field lines are a convenient visual representation. but ultimately they're not the actual physical thing. the physical thing is $\vec{E}(x,y,z)$
@acm what i have: |00> + |11> is (perfectly) correlated in X and Z, anti-correlated in Y, |00> - |11> is correlated in X, anti-correlated in Y and Z, |01> + |10> is correlated in X and Y, anti-correlated in Z, |01> - |10> is anticorrelated in all 3.
for the four Bell states
the fact that some combos of correlated/anticorrelated are missing is probably due to the usual bell states avoiding $i$'s
@cOnnectOrTR12 we keep telling you to not focus on the "field lines" so much, and everytime you come back and it's a field lines argument again. At this point I have nothing to add.
i said my combinations wrong above. should have been:
|00> + |11> is (perfectly) correlated in X and Z, anti-correlated in Y, |00> - |11> is correlated in Y and Z, anti-correlated in X, |01> + |10> is correlated in X and Y, anti-correlated in Z,
so if we start with a perfectly anti-correlated state, every possible non-trivial operation on it is some combination of rotations on the 1st qubit and rotations on the 2nd qubit
@MoreAnonymous Bogolyubov, Sheldon Glashow and Einstein. Seems like he always explicitly stated something he didn't like about other physicists (at least these 4, which are the ones I saw in that video).
but it's not entirely trivial I think, since if you put yourself in their situation, it's not obvious how boiling water would be very different from burning wood
metal is an interesting case, b/c in metallurgy you work so much with alloys
and i could imagine that people thought that heating a metal up effectively "added" fire to it
(come to think of it, you could probably think of distillation like that. add fire to water, converting it into steam, and then remove the fire to get back water)
@Semiclassical can't tell if you then solved it, but perfectly correlated can be understood in this context e.g. as maximal mutual information when performing local projective measurements. |00>+|11> is perfectly correlated just as |01>+|10> is. A maximally entangled state is indeed maximally correlated wrt any local projective measurement, your example also shows max correlation as it gives |00>+|11>
a simple way to see that this is the case is the fact that any local unitary operation, applied to a bipartite state, doesn't change its Schmidt coefficients (i.e. the singular values if you think it as a matrix)
another way to state this more generally is that the eigenvalues of the partial trace of $(U\otimes V)\rho(U^\dagger\otimes V^\dagger)$ are the same as those of $\operatorname{Tr}_2\rho$, for any bipartite $\rho$ and unitaries $U,V$
@glS I suspect the terminology might be throwing me a little, then. The distinction I was using was perfect correlation vs perfect anti-correlation, ie, are Alice’s values always equal to Bob’s or always opposite to Bob’s
But you could also view those both as “perfect correlation, just with positive vs negative correlation
I definitely concur that all four Bell states are maximally correlated in the sense that knowing Alice’s value for an observable suffices to predict Bob’s value
So to combine the points: the Bell states are all maximally correlated, but none of them display positive correlation for all observables