@ChrisWhite Ah, thanks! I think I might talk to some professors about indep study in abstract algebra. I just actually crunched the numbers and the grad algebra course conflicts with my stat mech class.
Hey! I have got a doubt. I have been reading about special relativity these days and they say that your speed basically indicates the direction of your motion through space-time. But what does that mean?
I mean is space-time considered as an absolute entity which is always there as it was considered about space before 1905? (I haven't been familiar with general relativity so I wanted to answer my question within the domain of SR only, even if it is not adaquate to describe actual physical reality)
Now, it's easy to get the perpendicular direction from the lines, and you just need to project the distance between any two points on the lines onto that direction to get the distance of the lines.
Think about the perpendicular line. There is a plane spanned by it (and some other vector) such that the projections of the two lines onto that plane are parallels
And that rather immediately means that the projection of any lines connecting two points on the two lines onto the perpendicular line is the distance between the lines.
Think about the perpendicular line. There is a plane spanned by it (and some other vector) such that the projections of the two lines onto that plane are parallels
@0celo7 Let's first fix some terminology, otherwise this gets weird to follow: I'll take $\vec d$ for the perpendicularly connecting line, $\vec a$ for $\overline{RS}$ and $\vec b$ for $\overline{PQ}$.
Okay. But if you can, say, rotate $\vec a$ about $\vec d$ such that it becomes coplanar and parallel with $\vec b$, that means that the projection of $\vec a$ into the plane spanned by $\vec d$ and $\vec b$ is also parallel to $\vec b$.
And if you project such a line now onto the perpendicular direction, you get a perpendicular line connecting two parallels. The length of such a line is the distance between them.
So, if you now use that to first calculate $\vec d$, and then project, say, $\overline{PR}$ onto $\vec d$ after you normalized it, you get the distance.
(The resulting formula is the same as in that solution)
@0celo7 The volume of a parallelepiped is $\text{base surface}\cdot\text{height}$.
You choose as base the plane spanned by $PS'Q$, then the height is the projection of $\overline{PR}$ onto the direction perpendicular to that plane.
Dividing out the surface area, what's left is the projection of $\overline{PR}$ onto the direction perpendicular to $\overline{PQ}$ and $\overline{RS}$.
Still tho. $A=P-Q$, $B=R-S$, $C=Q-S$. Then $0=\partial_{t_2}(At_1-B t_2+C)^2=-2 B\cdot (At_1-B t_2+C)$. So $B^2 t_2=B\cdot A t_1 +B\cdot C$, and likewise $A\cdot B t_2=A^2 t_1+A\cdot C$. Works in any no. of dimensions! I'm just sayin' ^^
That that is the correct way to calculate the distance we saw already another way...but I'm not sure how one is suppsed to see it from the parallelepiped itself
I think that "solution" would have cost me points back in school when we did this because it's not obvious why it works :P
@0celo7: I've read a bit further, but I get the impression that BBS is singularly useless for learning anything of substance from it. Everything I encounter is either stuff I already know, or incomprehensible at the level of detail given.
@0celo7 Basically the entire chapter 3, but especially the section on "Witten's open string theory" where they define the "string wedge product" $*$ by a non-sensical diagram that's just a bunch of parallel lines and never really bother to tell you what it actually is.
If I add a comment, then realise I've forgotten to add the @fred to ping a user I can either:
edit my comment and put in the @fred
delete the comment and retype it with the @fred
Does Fred get pinged in both cases? Specifically, in case (1) if I edit an existing comment and add the @fred does...
Yes. If I find a question or answer you've commented on, add a comment, then edit my comment to ping you we'll see if it works.
I've just added a comment to:
someone's coursework found on google: large.stanford.edu/courses/2012/ph241/tilghman1 on such proposal $$$$ QUOTE "And regarding nuclear waste, you don't want things to break. If the turbine of the bottoming steam cycle, or some other part of the heat exchanging process, were to fail, it would be difficult to repair without getting close to the radioactive waste." $$$$ Just us you won't found anyone near the core of a nuclear power station, you won't want spent fuel waste placed near your home — SecretAug 18 at 6:11
Your question is unclear. $\vec\nabla\cdot\vec A$ is flux per unit volume. $\int\int\vec A\mathrm{d}\vec S$ is flux. Integrating $\vec\nabla\cdot\vec A$ over a volume gives flux. By the divergence theorem, both expressions for flux are equal. I have no idea what other than that you are asking about. — ACuriousMind1 hour ago
Ok, that's pretty clear. It would be nice to know if I edited it and changed the ping whether the new pingee got a ping. But I need a thread where two people currently present have commented ...
Why three colors? Sure, it's the obvious choice given the need for three-way Pauli exclusion, but that's a heuristic guess, not a proof of need. Can you adequately provide exclusion using instead only two colors and their anti's?
Excellent (and to be honest anticipated) answers, especially experiment. What bothers me is the possibility of subtle information-level equivalences that could be hidden by more elaborate structure (which is why experiment is the best answer).
Look at the tip-balanced fermion cube with two 2D layers of down, anti-up (or vice-versa) quarks. It's interesting just to ask "how much can I do with just that 3D space, and nothing more?" It may be a different physics... yet it's surprisingly rich.
If z is vertical, the two color/anticolor axes would of course just be x and y.
I maintain that your cube is just a visualization of the representation theory of SU(3) that for some reason reasonates with you very well. I'm not sure what "different physics" you're seeing there.
The resonance goes remarkably deep. By "different physics" I only mean "probably wrong or incomplete." But it does seem to capture more than anticipated.
There is another figure I just love for its symmetry: The imaginary-units quaternion octahedron, which when aligned within the cube has its own delightful mapping to alternating fermion layers. +1 products clockwise per face, -1 counterclockwise, each pointing to one of the cube/fermion vertices. Create a single consistent edges though all of the edges and the pro and anti layers pop out naturally based on those products.
I will show that he Dark energy is an artifact of the referential used.
We measure with atoms, ex: 1 kg is defined as an ensemble of atoms, lets say N, in Paris, such that one atom weights 1/N kg. Similar recursivity with length/time unit.
We live at Earth's surface and not knowing that it ro...
Editing seems to be off entirely for me today, so just to fix a word typo:
"There is another figure I just love for its symmetry: The imaginary-units quaternion octahedron, which when aligned within the cube has its own delightful mapping to alternating fermion layers. +1 products clockwise per face, -1 counterclockwise, each pointing to one of the cube/fermion vertices. Create a single self-consistent edge flow pattern for all of the octahedron and the pro and anti layers fermion layers pop out naturally based on those +1 or -1 products."
This question that was wrongly marked as duplicate by Carl Witthoft, ACuriousMind, Hritik Narayan, Kyle Kanos, CuriousOne 2 hours ago , to this question .
The marked one asks "How can the unstable particles of the standard model be considered particles in their own right if they immediately dec...
We got good news and bad news. Good news: antibiotics are working. Bad news: got a 96 on my first physics exam because I didn't draw enough coordinate systems.
@0celo7 And the start of each new academic year will bring more sickly freshmen carrying all sorts of plagues. Wise upperclassmen avoid freshmen for the first month of school.
I'm reading about mathematical foundations of quantum measurements: Did you know that for any observable (intended as a projection-valued measure, i.e. the spectral family, but the result is also valid for positive operator-valued measures) there exists at least one quantum measurement process? And that there is a repeatable measurement process only if the observable has discrete spectrum? I always thought there was something unclear about the foundations of quantum measurements...
but now I am rather convinced there is nothing fancy, just plain interaction between to quantum systems (with suitable properties)
@yuggib I did not know the thing about the repeatable measurement. I guess it is related to non-existence of the eigenstates for continuous "eigenvalues"?
@ACuriousMind In practice you measure the spectral projection on the interval of values $[x-x_0,x+x_0]$ for position; but that observable (the spectral family) is not discrete, so the measurement is not repeatable
@yuggib bells thm proves there is something "nonlocal" in QM. prior to that the view that the uncertainty principle is due to measurement apparatus was tenable, less tenable (some say untenable) after that.
@0celo7 hope you get well soon. are you getting enough sleep? lack of sleep inhibits immune system function & has other misc consequences. something not to be trifled with.
ok. stress can exacerbate vulnerability. didnt you just start university? are you eating well too? this is all personal ofc but there are a bunch of university students on here & hope you guys all take care of yourselves with all the intense edu rush going on.
it is not uncommon for new university students to have "sketchy" eating/ sleeping habits. new/ unfamiliar environment. eating on the run or less time to eat/ sleep. etc... just remember youll learn/ retain better with good food/ sleep.
If I add a comment, then realise I've forgotten to add the @fred to ping a user I can either:
edit my comment and put in the @fred
delete the comment and retype it with the @fred
Does Fred get pinged in both cases? Specifically, in case (1) if I edit an existing comment and add the @fred does...
I will show that he Dark energy is an artifact of the referential used.
We measure with atoms, ex: 1 kg is defined as an ensemble of atoms, lets say N, in Paris, such that one atom weights 1/N kg. Similar recursivity with length/time unit.
We live at Earth's surface and not knowing that it ro...
This question that was wrongly marked as duplicate by Carl Witthoft, ACuriousMind, Hritik Narayan, Kyle Kanos, CuriousOne 2 hours ago , to this question .
The marked one asks "How can the unstable particles of the standard model be considered particles in their own right if they immediately dec...
