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3:11 AM
maybe it's just me, but I think Becker, Becker, Schwarz is kinda rushed, especially at the beginning with the bosonic string
I feel like they're skimming over a lot of subtleties in the quantisation sections
4 hours later…
7:08 AM
Storm Christoph has arrived in Chester. The weather forecast is for 48 hours of continuous rain. Lovely.
Perhaps it's a good opportunity to do experiments on buoyancy.
2 hours later…
8:51 AM
Hey all ... Just checking. Does the equivalence principle imply the affine connection ?
Hello guys.
Hi @JohnRennie sir
What is the electric field and magnetic field of beam of electrons? As in cathode ray tube.
As wire with current does not have electric field. What about beam of electrons.
@123 the reason a wire doesn't have an electric field is because is electrically neutral. Each conduction electron is associated with a nucleus and the negative and positive charges cancel.
An electron beam is not electrically neutral because it consists only of electrons so there are no positively charged nuclei to balance the negative charge of the electrons. So the electron beam behaves approximately like a line charge. It's electric field will be approximately the same as the electric field of a line charge.
Ooo I see.. Good explanation for wire.
What is the magnetic field due to beam of electron?
Does it have same magnetic field as in wire?
It's just the magnetic field due to a line current i.e. the same as the equivalent current flowing in a wire.
Oookay.. Thanks. @JohnRennie sir.
One more question how we create uniform magnetic field. Because there is no monopole of magnetic field.
If magnetic field does not have monopole it always bent to the other pole.
9:04 AM
Are you asking how we create a magnetic field when no magnetic monopoles exist, or how we create a uniform magnetic field? Those are two very different questions.
I know magnetic field can be created by moving electrons in straight line so magnetic field is circular loop. If we move electrons in coils it create straight magnetic field.
Magnetic field is also created by spin property which is considered as rotation of electrons by its own axis.
I know magnetic field can be created by moving electrons in straight line so magnetic field is circular loop. If we move electrons in coils it create straight magnetic field.
Magnetic field is also created by spin property which is considered as rotation of electrons by its own axis.
I know magnetic field can be created by moving electrons in straight line so magnetic field is circular loop. If we move electrons in coils it create straight magnetic field.
Am I correct @JohnRennie
Am I correct @JohnRennie
My question is about uniform magnetic field which has straight line and magnitude of field does not change with distance. How it is possible to create such magnetic field if magnetic monopole does not exist.
Should edits like these really be approved?
It is impossible to create a perfectly uniform magnetic field, but it's possible to create an approximately uniform field by summing the fields from many dipoled.
For example inside a solenoid each coil creates a dipolar field, and the field inside the solenoid is the sum of all these individual dipole fields. If the solenoid is infinitely long then the infinite sum of all the dipoles converges to a constant value inside the solenoid i.e. the field is uniform everywhere inside the solenoid.
Obviously a real solenoid can't be infinitely long and the field inside a real solenoid is only approximately uniform, though it can be an exceedingly good approximation.
@NiharKarve I wouldn't have approved that as I'd be annoyed if someone edited my answer in that way. However I don't see it as a big deal. It hasn't done any harm and Cort Ammon is free to roll back the edit if they want.
For the most part, the user makes decent edit suggestions anyway
9:20 AM
Thank you so much @JohnRennie give me your brain.
Does magnetic field has same effect of other magnetic field as in coulombs force.
Because two magnet attract / repel each other in straight line.
No. You need to calculate the force on a magnetic dipole in a magnetic field. The force varies as $1/r^3$ and it depends on the relative orientation of the dipoles.
9:35 AM
Yes I remember any dipole varies 1/r^3
But two magnets attract/repel each other in straight line. As the same behavior in electric field.
9:51 AM
Why does two magnets attract/repel in straight line. Why it does not follow Lorentz force?
Pls explain @JohnRennie sir.
Because I think when two wires has current flow both has magnetic field and those magnetic fields apply force on wire perpendicular as in Lorentz force.
When two magnets are placed both has magnetic fields also in this it should repel /attract perpendicular. But it is not the case
10:41 AM
Any ideas on solving the equation r^2(d/dr* (r^2 dv/dr)))=cV, for V=V(r), where c is a constant?
@satan29 Isn't that just the radial part of the spherical coordinate version of the Laplace equation
alternatively, this is just $Del ^2 V =cV$
You can just use the Green's function of the lhs
using erm.. elementary methods?
i am an engineering undergrad and this was in our physics assignment
You can integrate $\frac{d}{dr} (r^2 \frac{dv}{dr}) = \frac{c V(r)}{r^2}$ then divide by $r^2$ then integrate again no
10:45 AM
that wont solve for v(r) ...
oh no i made a horrible mistake while posing the question
Any ideas on solving the equation r^2(d/dr* (r^2 dv/dr)))=cv, for v=v(r), where c is a constant?
another piece of information is that v goes to 0 at $\infty$
11:10 AM
@satan29 set $r = \frac{1}{t}$ and see what happens
$r^2 \frac{d}{dr} (r^2 \frac{dv(r)}{dr}) = c v $ is $r^4 v'' + 2 r^3 v' - cv = 0$ or $v'' + \frac{2}{r} v' - \frac{c}{r^4} v = 0$. This looks like a very complicated equation, first thing you do is look at it's singular points. $r = 0$ is clearly an irregular singular point, so already it's potentially scary, need to determine it's behavior about $r = \infty$ which is done by setting $r = \frac{1}{t}$. If I did it right (fast) it becomes very simple
In mathematics, in the theory of ordinary differential equations in the complex plane C {\displaystyle \mathbb {C} } , the points of C {\displaystyle \mathbb {C} } are classified into ordinary points, at which the equation's coefficients are analytic functions, and singular points, at which some coefficient has a singularity. Then amongst singular points, an important distinction is made between a regular singular point, where the growth of solutions is bounded (in any small...
11:29 AM
Small point of confusion, if $S[\Lambda]$ is the Dirac spinor rep, and I want to find the adjoint of this operator $(S[\Lambda]^\dagger)$, I take the adjoint of the exponential:


and use the identity $(S^{\mu\nu})^\dagger=\gamma^0 S^{\mu\nu}\gamma^0$, that's fine, but we then exponentiate the product $\gamma^0 S^{\mu\nu}\gamma^0$, which Tong says becomes $\gamma^0 S[\Lambda]\gamma^0$, and the only way I can imagine this working is if we Taylor expand the exp and pull out factors of $\gamma^0$ on either side (using $\gamma^0=\Bbb 1$
ah got it (the mathjax, not the question :p)
that should say $(\gamma^0)^2=1$
ah I've switched between $J^{\mu\nu}$ and $S^{\mu\nu}$ for the generators randomly, mb
@bolbteppa damn! $r^2 \frac{d}{dr} (r^2 \frac{dv(r)}{dr})= \frac{d}{dr/r^2} (\frac{dv(r)}{dr/r^2})$
ok I've screwed it up a lot but hopefully what I've written is somewhat decipherable
and then dr/r^2 is simply -dt !
which makes the equation d^2v/dt^2=cv
@Charlie what do you mean? Don't the neighbouring $\gamma^0$'s in the series just become $1$?
Only if $\gamma^0$ commutes with $S^{\mu\nu}$ which (unless I am completely mistaken) it doesn't
11:42 AM
I'm on a laptop atm, typing in mathjax is like pulling teeth sry I will make a tonne of typos
whoever decided to put the \ key where the alt key should be should be fired
$\exp \gamma^0 S^{\mu\nu} \gamma^0$ = $\gamma^0 S^{\mu\nu} \gamma^0 + \gamma^0 S^{\mu\nu} \gamma^0\gamma^0 S^{\mu\nu} \gamma^0 + \gamma^0 S^{\mu\nu} \gamma^0\gamma^0 S^{\mu\nu} \gamma^0\gamma^0 S^{\mu\nu} \gamma^0...$
you are a genius, I see
no probs
12:10 PM
@NiharKarve : Ha-ha :)
@Charlie That's nothing, on a German keyboard one has to press AltGr+ß to make a \ :P
12:32 PM
I am having a lot of confusion while making notes
Should I use a register
or making notes on separate pages
The disadvantage of register is that if I have to cut or ignore a page.I will have to cut it.Then , let us say if there are some important points that should be written after page 3 (on page 4 you had to write )but you cannot write it there because you have already written sth there now . Then writing that information on further pages makes it difficult to revise.
In pages , you can just attach that page which you needed exactly at the page number where you wanted to . But it is
Can anyone explain why we have isostrain condition for longitudinal loading?
And isostress condition for trnasverse loading?
@Physicsismylife At what stage of education are you? School, University, Research, etc.
@Qmechanic wow, qmechanic is a real person?
@satan29 No, s/he tries to convince us that s/he's one. Don't fall for that trap.
@FakeMod @Physicsismylife is preparing for JEE.
12:49 PM
Ah, in that case, @Physicsismylife just use a notebook a.k.a. a register. Stick/staple pages where necessary. Tear them off (or better, just strike out using a pen) when those aren't needed. Don't be very particular about notes of any subject other than chemistry. Also, consider creating a sort of handbook, which contains and summarizes all the major facts/formulae and stuff, which you could refer and complete reading within a day.
1:14 PM
When a Unidirectional Composite is loaded parallel the fiber direction, then the composite strain (εc) = matrix
strain (εm)= fiber strain (εf),
Why is this true someone please explain this to me..
1:45 PM
@FakeMod thanks
if thrust exerted by a body on a surface = weight of the body . Then ,
Here , F = 0 . Then we can’t weight of Body is 0 right
F = mg here. Then it won’t be 0.
Ohk thanks
2:05 PM
@Charlie this might help
@bolbteppa nice answer, but I think Charlie might be looking for something more representation-theoretic
I think this is the most representation-theoretic one will get, I just didn't use the word Schur to justify it :p
hi @kaylimekay, nice to see you here
Note the argument is independent of the dimension and signature whereas the argument from $\mathrm{SL}(2,\mathbb{C})$ isn't
@NiharKarve, hi, I lurk sometimes to see what other people are up to ;)
2:15 PM
Tried installing the Linux and the bloody thing bricked my hard drive >:c
2:26 PM
@bolbteppa Thanks for this, just reading through it, just one thing I think you have missed the gamma matrix on the end of this
I would type it out but it's painful writing mathjax on this cpu
writing anything directly on a CPU is painful :P
@Charlie good lord
what is that font
for some reason on laptop the font is weird for me
it actually doesn't look that bad
2:34 PM
everything is worse on laptop
what do you usually use?
pc :p
I use my old laptop to do work to separate my work/pc areas
I think trackpads alone are enough of a reason to hate laptops
trackpads suck, but the, uh, nipple thingy in the center of the keyboard is great
my work laptop has one and I almost prefer it to using a mouse
2:50 PM
the mouse of a Mac computer cannot copy and paste. It cannot roll, either.
My keyboard came with no nips installed unfortunately, just the lousy trackpad I refuse to use
Got it thanks
A pointing stick (nub, nipple, or clitoris) is a small joystick used as a pointing device typically mounted centrally in a computer keyboard. Like other pointing devices such as mice, touchpads or trackballs, operating system software translates manipulation of the device into movements of the pointer or cursor on the monitor. Unlike other pointing devices, it reacts to sustained force or strain rather than to gross movement, so it is called an "isometric" pointing device. IBM introduced it commercially in 1992 on its laptops under the name "TrackPoint", patented in 1997.The pointing stick senses...
there are some interesting synonyms there
3:07 PM
That's a bit below the belt.
3:23 PM
I am still a bit mad about losing that SSD
"I used to be a theoretician, then I took a vector to the knee.."
Hi folks! I have a question. The Reissner-Nordström metric of a charged black hole has a singularity at r=0, but otherwise, the metric is defined everywhere, even inside the event horizons. The electric field scales as Q/r^2 so that the gravitational action, integrated from r=0 to \infty, is actually divergent. I have seen however that people integrate the gravitational action from the horizon, and obviously, they don't have any divergence.
There is a particular prescription to make the gravitational action convergent?
3:54 PM
I don't know about this specific case, but as far as I know the standard trick to make the gravitational action convergent is the GBH non-dynamical term?
In general relativity, the Gibbons–Hawking–York boundary term is a term that needs to be added to the Einstein–Hilbert action when the underlying spacetime manifold has a boundary. The Einstein–Hilbert action is the basis for the most elementary variational principle from which the field equations of general relativity can be defined. However, the use of the Einstein–Hilbert action is appropriate only when the underlying spacetime manifold M {\displaystyle {\mathcal {M}}} is closed, i.e., a manifold which is...
1 hour later…
4:56 PM
@Slereah the GHY term is there to reproduce the einstein equations
Yes, but you may observe that there are two terms here
I don't see what it has to do with the convergence of the integral
One for the boundary, and an extra term for convergence
I think Straussmann gives it a specific name, let me check
yes, but there is no boundary at r=0, so... how can it cure the divergence?
I mean I'm not 100% sure that there's much you can do in general to remove divergences if the space you're integrating on isn't compact
AdS has constant curvature so it will be divergent too
Or do you mean any radius around $r = 0$?
I think $r = 0$ is a boundary since it's singular there
although then again, it's a boundary of measure 0, so not important
5:01 PM
I mean that there should be a prescription to remove these divergences and still having a physical result
as we do in QFT with renormalization
the r=0 is indeed a UV divergence
Also wait
Does the action of the Reissner metric actually diverge?
Since its matter term is the EM field and the EM field has a zero trace stress energy tensor
Since $$R = - \kappa T$$
@Physicsismylife Please don't post your questions here directly after you asked them; interested people watch the main site anyway, and if everyone did it, the room would be flooded with new questions.
5:32 PM
@ACuriousMind ok.As you say
why is the direction of pressure for a box inside a liquid always perpendicular to the surface of box
The fluid is at rest here
5:54 PM
Q: 2020: a year in moderation

JNat As we say goodbye to the old year and welcome the new one, we have a tradition of sharing moderation stats for the past 12 months. As most of you here are aware, sites on the Stack Exchange network are moderated somewhat differently to other sites on the web: We designed the Stack Exchange netw...

@newUser in other words, the Lagrangian of electromagnetism diverges as $r \to 0$ when you plug in an electrostatic field
In this case it's equal to the Hamiltonian of the electrostatic field which diverges and is maybe the first occurrence of the concept of renormalization
Renormalization is a collection of techniques in quantum field theory, the statistical mechanics of fields, and the theory of self-similar geometric structures, that are used to treat infinities arising in calculated quantities by altering values of these quantities to compensate for effects of their self-interactions. But even if no infinities arose in loop diagrams in quantum field theory, it could be shown that it would be necessary to renormalize the mass and fields appearing in the original Lagrangian.For example, an electron theory may begin by postulating an electron with an initial mass...
6:19 PM
@bolbteppa yes I know the same problem arises in electromagnetism. Just wondering how you deal with it when computing gravitational actions
Yeah it's worth asking as a question, it seems weird that the action is defined everywhere while the solution minimizing it diverges at the origin...
7:23 PM
Isn't that one of those things that Schreiber talks about
With the gerb of Lagrangians or something
so that you only define Lagrangians locally with transition functions
This may not be a smart question, but the spin index that labels the spin of the two plane wave solutions of the Dirac equation, $v^s(\vec p)$ and $u^s(\vec p)$, the fact that there are only two of them, does this imply that a classical Dirac particle begins with spin aligned along, say the $z$-axis, and the spin remains aligned along this axis indefinitely? Assuming no frame change?
The classical equivalent of the spin is the polarization vector
You may remember that, given appropriate combinations of polarization vectors, it may spin
I'm basically unsure why we can't have spin in an arbitrary direction, why it has to be either $s=+1/2$ or $s=-1/2$, unlike in QM for instance where the the spin state changes in time
sry I was in the middle of typing that and god distracted
Well they're solutions
You can combine them
...why is god distracting you?
7:26 PM
Since it's a free field, the sum of two solutions is also a solution
he was trying to explain qft to me but he says it doesn't make sense :p
@Charlie the eigenvectors of spin in any direction form a basis of the space of all spin states
the u/v-things are just a specific choice of such a basis
oh wait it's almost exactly like the $\Bbb C^2$ used in qm
just in C4
You can express, say, a spin pointing to $z$ in the $x$ basis
They just won't be eigenvectors
7:28 PM
@Charlie yes
ie you need a combination of them
yeah, duh, what am I doing
Don't try to construe too much the two components of the spinor as "directions"
it doesn't quite work
if you want a direction you have to use the gamma matrices to cast them as vectors
7:44 PM
All the gerbs in the world wont address the singularity at the origin
I mean
For any point of a manifold, there's a neighbourhood where the action is finite
Which is good enough to define the EOM at all points
2 hours later…
10:05 PM
Q: Please Consider Re-opening This Question

MitchMy question at Evidence that Time is a Natural Phenomenon? was closed as a duplicate of What is time, does it flow, and if so what defines its direction? I read the related question and I don't see that my question is answered. Related discussion on that thread presumed knowledge of the answer t...


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