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6:20 AM
did something happen to the [lorentz-symmetry] tag?
oh nvm it was made a tag synonym of [special-relativity]
 
 
2 hours later…
8:48 AM
user image
4
 
"It makes a ton of sense for time to be affected by gravity" - and that's why the dinos didn't survive :P
just had no common sense
 
and yet here we are
also with that nonsense
 
 
1 hour later…
9:59 AM
0
Q: Feature suggest writing elements on physics.se

ButtonwoodOn occasion, questions / answers / comments on physics.se deal with chemical elements. An example for this may be to discern between the isotopes of carbon, C-12, C-13, and C-14. I would like to suggest an expansion of the acceptable syntax to write, e.g., $^{12}\mathrm{C}$. At present, I know...

 
 
1 hour later…
11:16 AM
0
Q: Getting a better understanding of how and why off-topic homework-like questions get answered before they can be closed

Emilio PisantyBackground This is a follow-up to What happened with the experiment to lower the vote-to-close threshold?. (See that thread, and the initial proposal here, for background.) The experiment is currently running on a set of (other) Stack Exchange sites that need it more, and need it for other reason...

 
 
3 hours later…
2:01 PM
Can someone explain why the relabel of integration variable doesn't introduce a minus sign?
 
123
Hi All....
 
"EEG and MEG are direct and noninvasive measures of brain function. While EEG measures the small electrical currents resulting from postsynaptic potentials, MEG measures the magnetic fields induced by these currents."
Doesn't EEG measure the electrical current by measuring alternating magnetic fields?
 
2:15 PM
"During the procedure, electrodes consisting of small metal discs with thin wires are pasted onto your scalp. The electrodes detect tiny electrical charges that result from the activity of your brain cells."
how does it detect electrical change non-invasively without measuring change in neuron magnetic fields
 
2:45 PM
@JingleBells by measuring voltage differences
if there's more charge in one part of the skull than another, then two electrodes at different points of the skull will have a differential voltage between them
 
123
Hello @ACuriousMind
Does rectangular cartesian components are dependent in sense of lagrangian mechanics?
 
@ACuriousMind so just like a voltmeter, it measures the electrical potential difference? Between which two points tho? there are many electrodes
 
0
Q: Should the moon eventually collide with the Earth if no force besides the earth's gravitation enacted upon it?

RandomUserIn due time, would both of them collide if they were affected solely by the gravitational force of each other?

has to be a duplicate?
 
@JingleBells between all of them
you get one voltage for each pair of electrodes, and then they do some analysis magic to get a (very low resolution) picture of the charge distribution in the brain over time
@EmilioPisanty yes
@123 I don't know what that means. Lagrangian mechanics is based on generalized coordinates, which may or may not be ordinary Cartesian coordinates for position
 
@ACuriousMind Got it thx, does your picture reflect your mood?
 
123
2:58 PM
@ACuriousMind I have listen one lecture in which he says cartesian is dependent on each other. But i have learned x-y-z axes are orthogonal they are independent. That's why i asked you.
Where he is going to prove lagrangian formulism.
 
and doesn't the EEG measure voltage of scalp skin instead of neurons?
 
@123 I don't quite know what "dependent" in this context is supposed to mean. The generalized coordinates are generally independent of one another. If you can't choose coordinates that are independent for your system, then you have to do annoying stuff with constraints
but the (in)dependence depends on the physical system you're considering
the (Cartesian or other) components of position of a free particle are independent of one another as generalized coordinates
 
123
@ACuriousMind What is the benefit of removing contraints from generalized coordinates.
 
the components of position of a particle constrained to move on a shell with $\lvert r\rvert = r_0$ are not independent of one another - knowing all but one determines the last component
@JingleBells I'd like to think I'm not quite as weary of the world nor quite as old (nor quite as, uh, broken) as the person it shows :P
@JingleBells voltage is a property of a pair of positions in space - it's the difference of the electric potentials at these points - not a property of a substance
 
123
@ACuriousMind I have seen lectures where they proved lagrangian formulism from D'Almbert principle. but in all they are always many steps which is not understandable. Pls share any book name or youtube lecture where systematically proved lagrangian.
 
3:02 PM
@ACuriousMind excellent find =)
 
the hope is, of course, that changes in the voltage are due to charge moving in the neurons of the brain and not to something weird going on in the skin
 
@ACuriousMind So if I was in empty space (vacuum) and I had a phone floating, and I put a voltmeter to measure the voltage without touching the phone, it would still somehow capture the electrical potential difference?
 
@123 The systematic proof from d'Alembert's principle requires some steps. I'm not sure what you're asking for. Most derivations I've seen do a good job of it, but in practice most people look at these derivations once and then more or less use Lagrangian (or Hamiltonian) mechanics as a new axiomatic starting point for physics just like Newtonian mechanics.
@JingleBells what does "the voltage" mean there?
you can measure the voltage between two points in space
this voltage is in principle dependent on the electrical charge distribution in the rest of the universe
but most things are overall electrically neutral, so far from them there's usually no measurable electric field and hence they don't contribute to voltages far away from them
 
@ACuriousMind the electrical potential difference between those two points in vacuum where the phone (or brain) stays between those points and does not physically touch the voltmeter pins
 
@JingleBells I don't know what it means for an "electric potential difference" to "physically touch" anything
There is an electric field $E(x)$ at each point in space, caused by charge distributions via Gauß' law
In the electrostatic approximation, we have that $E(x) = - \nabla \phi(x)$ for some function $\phi(x)$. The voltage between two points $x_1$ and $x_2$ is $\phi(x_2) - \phi(x_1)$
at no point in this definition is any "touching" involved
 
3:16 PM
No need for the equations, I don't know what they mean (still thanks tho). I'm trying to understand how electric potential difference is measured. When I hear that term I imagine an electron and a proton some space apart, and they naturally want to attract each other, they will have some electrical potential energy. I'm not sure how a voltmeter works, maybe that's why I'm struggling to understand how an EEG works as well
 
@JingleBells "How" the voltmeter works in detail depends on its construction, there are various ways to measure potential differences
the only thing that matters is that it's a device that can measure the electric potential difference between two points in space (often by connecting these two points with a conductor of known resistance and measuring the current that flows through it)
when you have a wire with known resistance $R$ and you can measure current $I$ (think "counting the electrons that pass by you"), then by Ohm's law the potential difference between the endpoints of the wire is $RI$
in the case of small electric fields (and not large fields like in a typical circuit) like that of the human brain you need some good voltage/current amplifiers to get this to work, but it still works
 
Ye, I read that the voltmeter has high resistance not to take much of the load's current (the load whose voltage is measured). Isn't that a problem, since you have (and know) the high R, you measure some current, but that current doesn't reflect the current of the load, therefore leading to inaccurate calculation of V?
And if EEG is just an amplified voltmeter, doesn't that mean that in order to calculate the different electric potential differences between points in the brain, the neuron current needs to reach and go through the electrodes?
 
you're not calculating the potential difference between points in the brain
you're calculating voltages on the surface of the skull, and inferring from that the distribution of charges in the brain
 
3:32 PM
Understood, thank you!
So an EEG is just an amplified voltmeter that measures the voltage between a bunch of pair points on the surface of the skull and then that data is used to infer the voltages of the same points if they were a bit inward the skull, closer to the brain?
 
that's probably wrong in all relevant aspects if you wanted to actually build an EEG machine, but that's the idea, yes :P
 
gotchya, thx, what about the question a bit above
 
@JingleBells This one?
 
The voltmeter having an R that's not equal to the R of the load of measurement, and therefore leading to different I, and different V calculated
@ACuriousMind ye
 
I don't know what type of "voltmeter" they use in practice in EEGs, so I can't answer that
 
3:38 PM
I meant generally in voltmeters, but it's okay if you don't know
 
not all voltmeters work with the "known resistance, measure current" principle
so your question can't be answered in general, it depends on the specific construction of your voltmeter
 
understood, thanks!
 
3:51 PM
0
Q: Has the PSE philosophy about homework-style exercises and check-my-work problems changed since 2012?

Bill NThe Community recycling algorithm recently brought this homework-style question asked in 2012 to the Top Questions list. It's interesting to me that It's clearly an elementary homework-style problem asking PSE to check the answer. John Rennie answered the question by confirming the answer with 4...

 
123
4:17 PM
@ACuriousMind I don't undertand few steps of proof of lagrangian from D'Alembert. I will share you in link.
@ACuriousMind 1drv.ms/u/s!AozWlUoG8z4tngZ5U3bIbz0b32zU?e=KNTYDv Pls see link. How this step done. How minus come here.
 
@123 it's just an application of the Leibniz rule $(fg)' = f'g + fg'$ (where the prime is differentiation)
 
123
1drv.ms/u/s!AozWlUoG8z4tngdZZ-avPByuk1eG?e=QvR2Rk How generalized force attached with both terms of left side.
@ACuriousMind Yes but they put minus sign instead of plus.
 
@123 Well...not exactly
you should be able to figure out what's going on if you look closer - they didn't "put minus instead of plus"
 
123
@ACuriousMind may be they used this way $(fg)' - fg' = f'g$
 
yes, and you can check for yourself - if you apply the ordinary product rule to the first summand in your link, you get back the previous line
 
4:29 PM
 
123
@ACuriousMind It means they use the product rule the way i shared. Is this correct? Because in other other lecture they used it this way.
 
🙃🙃🙃
 
@123 I'm trying to say that you don't need me to confirm it - you can confirm it yourself
 
123
What about my second link. Where he applied generalized force with both terms. How it applied with both terms of left side to create L = T - V
 
when someone uses $f'g = (fg)' - fg'$, then using the "reverse" $(fg)' = f'g + fg'$ on the first term must give you back $f'g$
@123 I have no idea what I'm looking at in your second link
a bunch of out-of-context symbols is not something I can magically understand
@EmilioPisanty that's what happens when you procrastinate writing your paper by answering stuff on physics.SE :P
 
123
4:34 PM
@ACuriousMind In my second link. grad of generalize force is potential V at right side. When he take it to left side there are two terms. He put potential with both terms. How?
 
right side of what?
where is it being "taken to"? There are like 5 equations on that snippet, you have to be more precise
 
123
@ACuriousMind right side of generalized force.
 
can you just type out the specific step you're unsure about here in MathJax?
 
123
@ACuriousMind Yes my main question is that how in lagrangian potential applied with both term to make L = T - V. What is the rule.
 
I don't understand the question, sorry
 
123
4:36 PM
How last step derived?
 
$L = T-V$ is a definition, it's not a derivation.
you can't derive a definition, so I don't know what you're talking about
 
123
In second link first step is only T appear, how it come to L = T - V in last step. This is my question.
 
...that's exactly what it explains in the middle!
what do you think the steps in between are for? :P
 
123
@ACuriousMind What i understand in between he shows generalized force as $\nabla{V}$ . So generalized force become potential V. Then he take this V to left side.
am i correct or wrong. what is understand.
 
so what is missing?
there's really nothing hidden there, it's just straightforward algebra
 
123
4:44 PM
@ACuriousMind missing is that. There are two terms on left side. How this V appear with both terms. In my sense it only come with one term.
 
@123 What do you think they mention $\frac{\partial V}{\partial\dot{q}_i} = 0$ for?
 
123
@ACuriousMind This is what i think of. If this is zero we can write it wherever we want. Because it is zero. Am i correct?
 
123
@ACuriousMind Oooooo... I seeeeee... Thanks.
Thanks for the support @ACuriousMind
:-)
@ACuriousMind You and @JohnRennie are teachers by profession?
 
neither of us, no
 
4:54 PM
Is the determinant of a 2x2 matrix ad-cb or ad-bc?
 
ad - bc
also wait the two choices you have are the same
 
depends on whether you labeled it via rows or columns :P
 
Real numbers commute, remember
 
First row is (a, b) and second row is (c, d)
 
I just remember it as a shape rly
First one diagonal, then the other diagonal
 
4:56 PM
So in my case, the entries do not commute.
 
Are they submatrices?
 
more like quantum operators
 
ah yes
the point of the determinant is that it is invariant under basis change, so you can check which one is correct by using one
only one of them should be invariant if they are not equivalent
 
If this sounds very strange, it might be that I am way off the playing field. Okay, thanks, I will try that!
 
I'm not sure "determinant" for a matrix with entries in something that's not a commutative ring is well-defined
 
4:59 PM
so that $\det(U^\dagger A U) = \det(A)$
 
123
What are you doing @ACuriousMind . If you are not teachers. You helped us here.
 
I have a job as a software developer, mostly writing tools for other developers
 
123
@ACuriousMind you and John sir are physicist. But developers not match.
 
I don't know what that means
John has a PhD in chemistry, not physics, for starters :P
 
123
@ACuriousMind Ooo, why you have so interest in physics and math.
 
5:11 PM
I studied physics at uni, and I always found the rigor of math comforting so I took a bunch of math classes
it's not forbidden to have more than one interest/skill, you know? :P
 
123
I am post grad in chem but don't have in chem
Interest in chem
I found chem is not a rigor subject. Just remember the facts.
 
 
1 hour later…
6:36 PM
@ACuriousMind Do they allow you to take any course you want at university in your country?
and any number of them?
Btw can any one of you suggest me a book with solved examples on coordinate free approach to physics..specially differential geometry, topology tensors and all that...
 
7:25 PM
folks
omg
drop everything
Mathematica reference page for the Function function
there's a new shorthand for Function with a named variable
 
@ManasDogra it varies a lot between universities and faculties, but in math and natural sciences you often are supposed to get some credits from one of the other sciences or math. I think for us there were even some (one or two courses worth) of credits where they didn't care at all which field they came from (I did math for those, too :P)
 
 
2 hours later…
9:37 PM
@EmilioPisanty How does the Function function function?
 
10:27 PM
@SirCumference apart from the (commendable) alliteration, is that a real question?
 
11:25 PM
@123 Pauling didn't do chemistry in a rote-memorization fashion at least, I'd say there's a lot one could do that avoids the mind-numbing rote memorization aspect of it
 

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