The h Bar

enumaris

00:01

is it rigorous or more of a pop sci book?

Kaku writes a lot of pop sci...

GR > QFT

no need to learn QFT

danielunderwood

It's a full textbook. I think a person would be quite frightened if they picked it up as a pop sci book

enumaris

I see

Maybe his textbook is good

he seems good at explaining stuff

but I've never read a textbook by him

maybe I'll check it out at some point cus I really was hoping to find a good QFT book lol

welp, time to head home

laters

danielunderwood

It's a bit old, but it's this one smile.amazon.com/Quantum-Field-Theory-Modern-Introduction/dp/…

I'll see how it goes

later dude

vzn

02:43

@bolbteppa nice find, on further look that recent intervierw seems to be the most written about Atiyahs physics almost anywhere incl him, think maybe its worth delving into deeper to some degree + deconstructing (if not verboten) uv.es/~azcarrag/pdf/…

user7777777

03:35

-5

Q: Iit is about network theorems

Please help me with this questions. I need the answers urgently ,anyone ready to help me out???? It was an assignment giving to us to submit tomorrow. Please help me out .I am on my knees pleading.

network

I honestly find this question hilarious (no offense)

user351417

04:16

Any 10k users about? I get the feeling that physics.stackexchange.com/questions/431888/… is identical to physics.stackexchange.com/questions/431896/…

user351417

It's the same user name, and I'm sure that I made some mathjaxifying edits on that yesterday.

Akash. B

hi guys

user351417

o/

John Rennie

@Chair yes, it's the same question. The OP has deleted then reposted it. I don't think that's a problem though. It's not a awful question.

Akash. B

@Chair 0/?

what do you meant by that?

user351417

04:20

@Akash.B That's hi...

user351417

Like a person waving

Akash. B

oh

user351417

the 'o' is a head, the / is an arm :P

Akash. B

@Chair O/

user351417

@JohnRennie Oops I hope that my comment on the deleted one wasn't too misleading. Did the OP respond to that?

user351417

04:21

I think the OP left an answer saying that the formatting on their question was terrible, and I told them to delete the answer. Maybe they went and deleted the question instead. Oh well

user351417

If the question wasn't negatively scored/closevoted, it's OK-ish for a confused new user to repost it.

user351417

@Akash.B Me no like. The head is too big :P

user351417

Lower-case 'o's are better.

Akash. B

Wireless power transmission using Tesla coils

►

guys check it out

user351417

-1

A: How is Hue, Saturation and Brightness of colours explained via EM and QED?

42 is the answer to everything. Just ask a famous book writer.

user351417

04:27

hehehe hehe he

John Rennie

@Chair no, your comment is fine. There was no response from the OP.

user351417

@JohnRennie I think I did try to emphasize that the delete button is below the text "My post looks terrible"... well, now the new version is in the close review queue (for migration to math SE). Can you see close votes on deleted posts?

John Rennie

@Chair I can't see any close votes on the deleted question, though I suppose it's possible close votes get annulled when the question is deleted.

user351417

Coolio. But I guess that with 1 rep, the OP wouldn't be seeing the close votes either way. Anyways, thanks! If the old version wasn't downvoted/closed, I guess there's no point investigating too much.

Sir Cumference

user7777777

05:06

Lmao

That's hilarious as well

John Rennie

@SirCumference has that been deleted? I can't find it.

jeea

Please someone help with a simple electromagnetic question regarding H,B,M

The pink colored cylinder carries current I. Yellow colored is coaxial cover with different permeability, same with blue covering. Will H at black point outside still be I/2(pi)(r) ?

pink part and outer part have permeability same as space

will B then at black point be (mu_0)(I)/(2(pi)(r)) ?

user7777777

Current is in which direction?

jeea

What about M

@user7777777 In upper direction or z.

user7777777

Oh. Is this supposed to be a cross section of the cylinder?

jeea

05:20

Yes

all are cylinders

user7777777

Does it have infinite length or finite length?

jeea

infinite

user7777777

alright, let me take a look.

jeea

only then we can apply ampere law i guess

user7777777

yes, correct.

it should still be the same

the enclosed current stays the same, doesn't it?

jeea

05:23

Yes it is constant at 12 A

user7777777

so I think it should be the same as that for a simple infinite wire.

jeea

How will we calculate M

user7777777

Can we use $\mathbf{B} = \mu \mathbf{H}$?

jeea

M = (x)H so is M zero

@user7777777 Yes thats the question

user7777777

I mean, is the medium linear, isotropic and homogeneous?

See this: en.wikipedia.org/wiki/Magnetic_susceptibility#Definition

jeea

05:29

not stated, only given that mu is uniform everywhere, and different in pink yellow and blue regions

pink and white regions have mu same as that of space

user7777777

Alright, I think that we can use $\mathbf{B} = \mu \mathbf{H}$

The susceptibility is just $\mu - 1$, and then it can be solved.

jeea

@user7777777 Thanks a lot for help ^.^

user7777777

no problem.

Avantgarde

0

Q: Why are subway stations so windy?

Why are subway stations so windy? What causes that powerful movement of air at the subway stations?

pressure

^Fluid mechanics

user7777777

05:58

Removed

Captain Bohemian

07:54

last afternoon I waked from witnessing a fierce storm.

Anonymous

08:36

@CaptainBohemian I enjoy storms :P (as long as I'm not getting drenched i.e.)

Lozansky

Need help interpreting a result

Say I have a current $I$ going through a loop described by $-a\hat{y}+t\hat{x} \cup a\cos \phi \hat{x}+a\sin \phi \cup a \hat {y} - u \hat{x}$ where $a\in \mathbb{R}$, $t\in (-\infty, 0)$, $\phi \in [-\pi/2, \pi/2]$, $u \in [0, \infty)$

Hm maybe that doesn't help

Like this

And it's supposed to be a half circle, I just can't draw

So I calculated the magnetic field and then integrated $\int_{\mathbb{R}} \mathbf{B}(z \hat{z}) \cdot \hat{z} dz$ and got $I\mu_0$

Which is Ampères law

For a closed path $\mathcal{C}$ which "encloses" the current

Does this answer make sense?

Anonymous

@Lozansky Is that the magnetic field at the origin?

Anonymous

There sure needs to be a $a$ (radius) term

Anonymous

Or else your integration was erroneous

Lozansky

@Blue It's the magnetic field integrated over $z$-axis

Oh and I realize my loop isn't closed

So let the lines extend very very far

And then close

Anonymous

08:53

@Lozansky Which is your closed path $\mathcal{C}$ in the diagram?

Lozansky

@Blue I'm not integrating over a closed path, I'm integrating over $\mathbb{R}$

Anonymous

@Lozansky Sure, but I thought you intended to compare the result with Ampere's law ?

Anonymous

I don't think there's a valid comparison to be made here

Anonymous

15 mins ago, by Lozansky

Which is Ampères law

Lozansky

@Blue Right, so I guess any $\mathcal{C}$ which encloses the path of the current does it

Anonymous

09:11

@Lozansky For $\mathcal{C}$ which encloses the path of the current, you can apply Ampere's law to get $$\oint \vec{B}.d\ell = \mu I$$ Okay. But then, how do you intend to compare it with the result you got by integrating the magnetic field over $z$-axis? The former is not equivalent nor derivable from the latter and vice versa (as far as I understand)

Lozansky

@Blue Yeah I agree. But it's the result I got and I'm not sure I am intepreting it right

Anonymous

How are you interpreting it?

Lozansky

Like the integration path is going in an infinite semi-circle

So it encloses the current $I$

Going from $z=-\infty$ and then somehow connects to $z=\infty$ in a closed loop

Anonymous

I don't think that makes much sense

Lozansky

:(

Anonymous

09:16

The path of current $I$ lies on the plane of your integration path

Anonymous

So Ampere's law doesn't apply

Lozansky

It does if my path encloses part of the circuit?

By Stoke's I guess?

Anonymous

@Lozansky Even then, the current $I$ wouldn't pass "through" the $\mathcal{C}$, isn't it?

Lozansky

No, it's enclosed by $\mathcal{C}$

Anonymous

It's sorta difficult to know whether the path(s) you're visualizing is same as the path(s) I'm visualizing

Anonymous

09:19

So better to draw a diagram to explain your point

Lozansky

Beware of my bad drawing skills

Anonymous

@Lozansky Ah, okay, I see what you're thinking. Took some re-reading

Lozansky

@Blue Something like this

Anonymous

You mean whether that infinite semi-circle is comparable to a closed path

Anonymous

Interesting

Lozansky

09:25

More like

Can the integral $\int_{\mathbb{R}}$ transform into a closed loop as per the picture (without affecting the result)?

And thus validating the interpretation

Anonymous

Right, gotcha

Lozansky

It's really the only way to make sense of my result, which I'm pretty sure is correct

Anonymous

13 mins ago, by Lozansky

Going from $z=-\infty$ and then somehow connects to $z=\infty$ in a closed loop

Anonymous

I think we can connect that path ^, by connecting the end points, to form a closed loop

Anonymous

Think of it like going from $z=-r$ to $z=+r$ and returning back to close upon itself

Anonymous

09:30

Ampere's law holds for that

Anonymous

Now there's no reason why it should not hold in the $r\to \infty$ limit

Lozansky

@Blue Alright cool, thanks!

Anonymous

It's not fully over tho. I'm still thinking how to relate the $\int_{\mathbb{R}} \mathbf{B}(z \hat{z}) \cdot \hat{z} dz$ as a closed integral over a path like that :P

Anonymous

The point is that that path has to enclose the current carrying wire

Anonymous

So it can't be straight line path for sure

Anonymous

09:36

Maybe if we take the infinitesimal projections on the z-axis tho, its equivalent

GPhys

Announcement of the Nobel Prize in Physics 2018

►

Anonymous

09:52

@Lozansky In this case the most sensible closed contour would be the semicircle of radius $R$, closing back by the straight line $[-R,R]$

Anonymous

Then $R\to\infty$

Anonymous

Can you think of why $\int_{\mathbb{R}} \mathbf{B}(z \hat{z}) \cdot \hat{z} dz$ would be equal to the line integral of $\mathbf{B}$ over this contour?

Anonymous

This looks like a good math exercise, but unfortunately I don't have much time to look into this now

GPhys

Congratulations to Arthur Ashkin (1/2), Gérard Mourou (1/4), and Donna Strickland (1/4) for winning the 2018 Nobel Prize in Physics for their work on optical tweezers.

10

Ashkin for optical tweezers, Mourou and Strickland for chirped pulse amplification

Anonymous

@GPhys Nice. Optical tweezers are apprently quite useful in some quantum computing architectures.

Anonymous

10:05

I don't know what chirped pulse amplification is tho

Anonymous

Chirped pulse amplification

Chirped pulse amplification (CPA) is a technique for amplifying an ultrashort laser pulse up to the petawatt level with the laser pulse being stretched out temporally and spectrally prior to amplification. CPA is the current state-of-the-art technique which all of the highest power lasers (greater than about 100 terawatts, with the exception of the ~500 TW National Ignition Facility) in the world currently utilize. Some examples of these lasers are the Vulcan Petawatt Upgrade at the Rutherford Appleton Laboratory's central laser facility, the Diocles Laser at the University of Nebraska-Lincoln...

Lozansky

10:36

@Blue Hm, no not at the top off my head. Maybe it's possible to choose $\mathcal{C}$ such that it is orthogonal to $\mathbf{B}$ at every infinetesimal line segment $d\mathbf{l}$ along the "half circle"? I'm not too sure tho

Anonymous

@Lozansky It shouldn't matter which $\mathcal{C}$ you choose. It would be $\mu I$ for all of them. What I am concerned about is that if you take the loop I mentioned then only a part of the line integral by itself is $\int_{\mathbb{R}} \mathbf{B}(z \hat{z}) \cdot \hat{z} dz$ (which you claim is $\mu I$) i.e. the line integral over the $[-R,R]$. That implies that the line integral over the semicircle is $0$ (which doesn't seem quite right to me)

Anonymous

Are you sure about your calculation?

Anonymous

Maybe you're missing a $2$ somewhere

Lozansky

No, pretty sure. Remember that the semicircle is infinitely far way from the circuit , maybe this helps

Anonymous

Ah. The magnetic field due to the current carrying wire varies as $1/r$

Anonymous

10:44

Since the radius $R$ of the semicircle $\to \infty$

Anonymous

There's no effect on it

Lozansky

That's right

Think that does it?

Anonymous

So that does it :)

Anonymous

@Lozansky Yep

Emilio Pisanty

11:17

5

Q: How much light can pass through a point?

Analogy: an infinite number of lines can pass through a point. Is there a limit on the number of lasers that can pass through a point? Obviously, with lasers the “point” would be a sphere with the diameter of the beams.

optics

It is the tool to use if you want a truly intense laser beam

Laser physics was basically stuck in a rut for fifteen years until it came along

Avnish Kabaj

I was trying to figure out why does a ball when tossed away from me with a reverse spin will bounce back towards me

I came up with a shoddy formula using coefficient of restitution, frictional coeff

Basically friction will give a momentum towards me which should be equal to

$$\ (1 + e)\times(\text{velocity perpendicular to the ground}) \times \mu$$

If this quantity is more than the horizontal component of the ball then it should bounce back

The spin part isn't here and I don't think that spinning the ball will have a role as the angular momentum is conserved at the point of contact

And friction is the only force acting

But spin does play a role

From "experimentation"

What am I neglecting

Anonymous