The h Bar

trilolil

16:14

Hello everyone

short confirmation needed here

If the curl of the electric vector field E = 0

that means that the voltage accross the whole field is constant, right?

Anonymous

@JohnRennie Did you have a look at the video and the doubt I asked you about one of your answers today morning ?

Anonymous

EMF Induced in a Moving Conductor in Uniform Magnetic Field-Motional EMF (GA_EMI06)

►

Anonymous

9 hours ago, by S007

Anonymous

9 hours ago, by S007

@JohnRennie I have a regarding about one of your answers http://physics.stackexchange.com/a/152695/135977. I recently read that increasing or decreasing magnetic fields through a certain area produce clockwise or anticlockwise electric fields in circular pattern by Faraday's law. But how is "charge" involved here in the generation of the electric fields (having a circular pattern) ?

ACuriousMind

@trilolil No, it just means the magnetic field is constant in time.

@DanielSank Another title for the hall of fame: Double delta potential makes me cry

heather

16:22

@ACuriousMind, I edited the title

trilolil

@ACuriousMind is it me or are the mathematical symbols for curl and rotor exactly the same?

ACuriousMind

@trilolil I don't know what the difference is supposed to be. I've not heard "rotor" as a technical term, but I would assume it's the curl if I heard it used.

trilolil

@ACuriousMind sorry I meant gradient in stead of rotor.

symbol curl = symbol gradient

ACuriousMind

The symbol is called "nabla". If you write $\nabla \phi$ for a scalar function $\phi$ it denotes the gradient, if you write $\nabla \times E$ for a vector field $E$ it denotes the curl, and if you write $\nabla \cdot E$ it denotes the divergence.

Slereah

NAMBLA you say

ACuriousMind

16:26

Heuristically, you may think of $\nabla$ being a vector that contains the partial derivative w.r.t. the coordinate in each coordinate.

John Rennie

@S007 I did look at the video, and I agree with their calculation, but I don't think that invalidates my answer.

Anonymous

@JohnRennie Could you please explain me how charges are involved when only the magnetic field is changing ?

Anonymous

I am still a bit confused

Secret

[Division by zero] Theorem 3: For any structure that include an idempotent element $a$ such that it behaves like a one sided identity to all other elements in the structure. If the * cayley table can be generated by acting $S_n$ (where n is the number of elements in the structure) onto a row or column of it, then its + cayley table is isomorphic to the null semigroups

Proof: To be written, but basically it has to do with the dominance property of the zero terms

Now that's why most of the division by zero algebra I have found so far always have that null semigroup in it. It seems the order…

John Rennie

@S007 Electric and magnetic fields are different aspects of the same thing. If you look at an electric field in a boosted frame it looks like a combination of electric and magnetic fields. Likewise, if you look at a magnetic field in a boosted frame then it looks like a combination of magnetic and electric fields.

The point of my answer is that the divergence of an electric field is always zero unless some charge is present.

DanielSank

16:37

@ACuriousMind yeah, saw that

Anonymous

I am a novice to this topic. Could you please explain what "divergence of an electric field" means ?

Anonymous

@JohnRennie

John Rennie

Suppose you have a bath full of water. You can swirl it around and make all sorts of interesting currents, however the stream lines always form loops. A stream line cannot begin or end.

Now turn the tap on. This is a source of water and streamlines can begin at the tap and flow on into the bath. Now pull the plug out. Now streamlines can end on the plug as the water flows out.

In this case the divergence of the water flow is zero unless either water is flowing in or water is flowing out. the tap acts as a source and the plughole as a sink.

An electromagnetic field behaves in a surpisingly similar way. If there are no charges then the field can swirl around (like the electric field loops in the video) but fields lines cannot begin or end anywhere.

In this case electric charges act as the sources and sinks for the field lines. We generally treat positive charges as the source i.e. field lines start there, and negative charges as the sink i.e. field lines end there. But note that this is just convention. Unlike my analogy of water field lines don't actually flow.

Anonymous

@JohnRennie Oh! Thanks a ton. That was a great analogy. I have another doubt. What actually causes the electric field to coexist with magnetic field ?

DanielSank

divergence: amount a vector field points away from a point.

John Rennie

16:48

Maxwell's equations describe the way that electric and magnetic fields are linked. It all gets a bit complicated, but look at it this way ...

A static charge generates a static electric field - all very straightforward.

But if I start moving relative to that static charge then in my rest frame the charge is moving, and a moving charge is a current.

And a current creates a magnetic field. So what looks like a static electric field in one frame looks like a mixture of electric and magnetic fields in a moving frame.

Anonymous

@JohnRennie Ok I get that moving charge is a current and that in turn creates a magnetic field. But is there any reason why it creates a magnetic field ?

John Rennie

The question why is always an awkward one in physics. Experiment tells us that moving charges create a magnetic field, and Maxwell's equations predict the magnitude of the field.

But if you're asking why does a moving charge create a magnetic field I have to say I don't know. You'd have to ask whoever created the universe.

Slereah

There are many explanations one could give, but in the end there is no reason

Because you could always ask "But why is that?"

It is not a productive line of inquiry

John

Is F=BIL the force on a wire or on a single charge in a wire?

Anonymous

@JohnRennie Okay I get it :-P It seems to be a experimentally verified phenomenon. Anyway thanks a lot for all the explanation :)

Anonymous

16:56

@John Whole wire

Slereah

In the end the real reason is that it's a model we produced because it fits the experimental data

John

So why can we use F=BQv as the force on a single charge

Anonymous

@John you can use

John

Surely it should be BQv*no of charges

Anonymous

@John BIl= BQv * no of charges....see the derivation in your textbook or web

Anonymous

16:58

@Slereah I get it. Thanks :)

John

My textbook says BIL is the force on a single charge or on the whole wire

Which is from where my confusion stems

Anonymous

@John It is on the whole wire

John Rennie

Physicists Develop Highly Unstable Quadriscuits

Anonymous

Basically evB * nlS = BIl where n is the electrons per unit volume @John

John

BIL = B*(dQ/dt)*vdt where dt is the time taken for a charge to move the length of the wire at velocity v. So BIL=B(dQ)v not BQv. dQ is the amount of charge flowing passed as fixed point in time dt, not just a single charge. So BIL = BQv*no of charges flowing past a point in time t

Would I be correct in thinking that

Flowing past*

Anonymous

17:06

@John No. dt cannot be taken as time to travel length of wire

Anonymous

dt is elementary time in calculus

Anonymous

2 mins ago, by S007

Basically evB * nlS = BIl where n is the electrons per unit volume @John

Anonymous

Do you know neSv=I (currrent ?

Anonymous

v is drift velocity

John

I'm not aware of that formula

Anonymous

17:08

try to derive it

John

What is S?

Displacement?

Anonymous

@John S is area of wire cross section

John

Oh ok

n is number of electrons?

Anonymous

per unit volume

Anonymous

yes

John

17:10

Oh you already said, sorry

I think I've got it

Anonymous

@John Could you explain how you got it ?

John

Sorry I had to go, I'll write it out now

heather

what's a good starter book on QED?

Bernard Meurer

@heather Poke me when you're around, I have a task for you

Aha

heather

@BernardMeurer, (::poke::)

Bernard Meurer

17:21

That was a sync :P

I need help organising my books folder :D

heather

=D

and you want me to organize it or give suggestions for how to do so?

John

I=Q/t, Q= eno of charges in volume = enumber per unit volume * volume = en*Sx = ensvt therefore I=ensv

Anonymous

@John yes you got it :)

John

@S007 ah cool

heather

@BernardMeurer, would you be interested in the book "How to Solve Physics Problems" by Oman?

Bernard Meurer

17:27

ĥe

Bloody keyboard layouts

heather

@BernardMeurer ?

Bernard Meurer

@heather I'd like you to:
1. Add that book to it, yeah
2. Look at the books we currently have and come up with possible categories
3. Think of a good naming scheme for the files ([AUTHOR] - [TITLE]? [AUTHOR] - [TITLE] - [ISBN]? Which one to use?
4. We discuss things and implement it making the book folder great again

heather

@BernardMeurer, I'll have to share the book with you and have you add it to the folder, as I'm unable to edit the book folder, just so you're expecting it

Bernard Meurer

@heather Nah, let me give you edit permissions

heather

@BernardMeurer, okay =)

Bernard Meurer

17:31

@heather Done

heather

how about this naming scheme: [AUTHOR] - [TITLE] - [EDITION] (last part if applicable)

I also added a copy of Calculus Made Easy that I had in the depths of my drive =)

it might also be worth it to have a pdf copy as well as the normal copy for each file (or not, I don't know)

@BernardMeurer, and then maybe we can have a Physics folder, a Math folder, a Coding folder, a general Computers folder, and then a Hacking/Security folder

Bernard Meurer

@heather Let's just make a "Computer Science" folder

and cram all computer crap there

I think [AUTHOR] - [TITLE] - [ISBN] is better, simply because then you can just search for the ISBN and always find your book

Right now it's easy but it will grow over the years and the template must continue to work well

user116211

17:52

@secret, heard of quasi-simple ring? It's a non-trivial ring containing no non-zero proper ideals. Now there is a theorem which goes like:

user116211

> If $A$ is a dense ring of linear operators of finite rank on a non-zero vector space $E,$ then $A$ is quasi-simple.

Secret

18:02

The term sounds unfamiliar.... So I am guessing when A is dense, there are no linear maps in A that have a nontrival kernel thus it can always map any basis in the vector space to any other basis, without being trapped in some subset of that vector space (as otherwise there will be a nonzero proper ideal.)

Have not read deep into rings yet (once again stuff after group theory) but my impression on ideals is it behaves like a kernel and when things get in there, they never get out again

Emilio Pisanty

18:14

@BernardMeurer Can I ask as a polite favour that you keep the m$x$ga phrase to a minimum on this chatroom?

Bernard Meurer

18:31

@EmilioPisanty I don't understand what your asking me

Emilio Pisanty

@BernardMeurer I am asking you to keep "make $x$ great again" to a minimum on this chatroom.

2

Bernard Meurer

@EmilioPisanty AH, oh sure, I don't have any problems with that :)

Emilio Pisanty

@BernardMeurer cool. appreciated.

heather

19:24

@BernardMeurer, okay, sure, that makes sense.

here, I'll make the folders and start sorting.

nevermind; you did that =)

Kyle Kanos

19:45

@EmilioPisanty What about making things grrrrreat!

bolbteppa

20:16

My god, what a journey, utter genius:

0

Q: Spinor Lorentz Transform via Vectors - Cross Product Issue

The Lorentz transformation operator acting on an undotted, i.e. right-handed, spinor can be expressed as $$e^{-\frac{1}{2} \sigma \cdot \mathbf{\phi} + i\frac{1}{2} \sigma \cdot \mathbf{\theta}}.$$ There is a very cool, almost childlike, derivation of this expression in Landau Vol. 4 S. 18, I've...

So simple

Every single other book makes it seem so complicated, ridiculous, Srednicki is probably the worst!

Or Peskin, or Tong, just telling you to memorize $S^{\mu \nu}$ and verifying that it works, should be mocked

DanielSank

I have two answers with >=90 score. When they break 100, where do I go to cash in for fame and/or glory?

Kyle Kanos

@DanielSank Your nearest Starbucks

bolbteppa

@0celo7 What do you think?

Can anybody tell me why one would even think to set $\mathbf{J} = \frac{\mathbf{\sigma}}{2}$ apart from being told the math works for some weird reason

Emilio Pisanty

@DanielSank you get a shiny gold badge and that's it

Kyle Kanos

@DanielSank Curious that one of those has 2 -3 answers

Emilio Pisanty

20:21

@DanielSank That's a nice version of the Feynman ratchet btw

That Messenger Lectures talk is at the very top of my best physics talks ever list

DanielSank

@KyleKanos haha

@KyleKanos eh?

@EmilioPisanty yes, I like those questions because they require understanding noise.

Kyle Kanos

@DanielSank This one; there are two answers that have -3 scores alongside your >90 one

heather

@DanielSank, where would be a good place to start learning multivariable calculus?

dmckee

@DanielSank No one sent me any fame or glory when I got any of my over 100 votes posts on the network.

I even have one over 1000 votes on StackOverflow and I haven't even gotten a t-shirt for it.

But then it is one of those posts that feel like it should have the score it does—too little effort.

bolbteppa

20:38

@0celo7 if they are still being children go into the math chat

heather

@bolbteppa, he's banned from all chat rooms on the SE network

@dmckee, that Congratulations is for you =P

dmckee

BTW @Daniel--I'm impressed at your having so many highly voted answer with so few answers in total.

That's a very high rate of blockbusters.

DanielSank

@heather a book?

heather

@DanielSank, book, video, tutorial...anything's good

DanielSank

@dmckee why not?

heather

20:42

video would probably be best, now that I think about it, but again, anything's good

DanielSank

@dmckee thanks. My claim to fame on the site is the highest fraction of accepted answers.

dmckee

@JohnRennie Well, in this case we can appeal to the relativistic construction the theory and say that boosts mix the off-diagonal elements of the field tensor so that those components that represent static electric fields in the electrons rest frame represent magnetic fields in other frames. But, of course, that is still founded on observation when you get right down to it.

Kyle Kanos

If I have a claim to fame, it's due to my awesome reviewing

dmckee

@DanielSank I think that is a good question and we should take it up with the management.

Lapel pins at 100 votes, decoder rings at 1,000 and monocles at 10,000?

And where do top hats fit into this picture?

DanielSank

20:55

@dmckee What post was that?

@heather The only thing I can recommend is the usual books.

You're comfortable with single variable calculus?

Kyle Kanos

1117

A: How can I replace a newline (\n) using sed?

Use tr instead? tr '\n' ' ' < input_filename or remove the newline characters entirely: tr -d '\n' < input.txt > output.txt or if you have the GNU version (with it's long options) tr --delete '\n' < input.txt > output.txt

DanielSank

@dmckee zomg I want a decoder ring!

@dmckee Top hats are complimentary.

dmckee

Notice that I don't actually answer the question asked in that sed post. It's piggy-backing on the fully deserved popularity of the accepted answer (that's purest black magic mojo, that).

And it's good advice as far as it goes, but still...

heather

@DanielSank, I think so, yeah.

Kyle Kanos

Well order has been restored to the universe: the Steelers get back in the win column

There are three questions asked in this post, anyone more knowledgeable able to discern if they should be three separate posts, or are they close enough to be one?

Kyle Kanos

22:55

@EmilioPisanty: It seems that the EM-drive is popping back up because White's paper passed peer-review

Emilio Pisanty

23:16

@KyleKanos that's good to know

have you read the paper?

dmckee

Meh. Link to the paper: arc.aiaa.org/doi/10.2514/1.B36120

And no paywall either. At least for the moment. Get it while you can.

The abstract says 1.2 milliNewton per kilowatt, while the simple photon drive results should be 3.3 microNewton per kilowatt.

Physiks lover

@dmckee have you come across the zero-component lemma?

Emilio Pisanty

@dmckee emdrive?

no thanks

dmckee

@Physikslover Not by that name. What's it about?

@EmilioPisanty Well, that's my reaction too. But the point is that they have something that is not easily explained right now.

That doesn't mean a break-through, but it is good that someone is looking at it.

The paper claims the apparatus they're using for the test can measure to the single digit microNewton level, so this isn't as simple as treating noise as data unless they really don't understand their instrument.

Emilio Pisanty

@dmckee oh, probably

dmckee

23:29

The part I still want to check out is how they are getting all that power to the instrument.

Emilio Pisanty

I just mean that I'm not gonna be the one to try and explain it

Physiks lover

@dmckee the zero-component lemma states that if a component of a four-vector is zero in every frame, the whole four-vector is zero

dmckee

It seems too obvious for them to not think about, but I'd like to see what they are thinking.

Emilio Pisanty

@dmckee how much power are they actually putting in?

looks like in the tens of watts from skimming it

dmckee

@EmilioPisanty They show data up to 80ish watts and 80ish microNewtons.

Kyle Kanos

23:32

@EmilioPisanty No, I've not.

dmckee

@Physikslover OK. Seems fair. If any component was non-zero there would be some boost that shifts part of it into the individual component you're looking at.

But I should like to remind people of the pioneer anomaly and the blow up at OPERA. There are a lot of way for something to be 'unexplained and weird' that don't lead to a science fiction future.

Kyle Kanos

Don't forget BICEP2...

dmckee

@KyleKanos Well, that was an 'explained' thing that wasn't so. But, yeah.

Kyle Kanos

It was still something published that was later retracted as incorrect

dmckee

Then there is the saga of the petnaquark. Published, retracted and then confirmed. Isn't science fun?

Physiks lover

23:37

@dmckee it seems obvious when pointed out to people. But it means that it can be used to prove that if a component of a four-vector is composed in all frames, then the whole four-vector is also conserved. So conservation of energy implies conservation of momentum also.

Emilio Pisanty

@KyleKanos this doesn't need to be retracted for the concept to be shown to be bollocks

dmckee

@Physikslover Yeah. That's cute.

Emilio Pisanty

> The test campaign included a null thrust test effort of three tests performed at vacuum at 80 W to try and identify any mundane sources of impulsive thrust; none were identified.

ACuriousMind

@KyleKanos I commented, and I already advised OP before to split up an even broader question.

Kyle Kanos

@ACuriousMind Yay! I did more good!

Emilio Pisanty

23:40

@Physikslover yeah, that's pretty cute. But keep in mind that it just follows from the symmetries. Time-translation invariance plus boost invariance also gives you spatial translation invariance.

You get this with some frequency.

e.g. in classical mechanics, take any two conserved quantities, then their Poisson bracket is also conserved

Sir Cumference

Are there anymore general relativists here?

Emilio Pisanty

say, $L_z$ and $L_y$ (which, if conserved, yield $L_z$ as a conserved quantity too)

Sir Cumference

Chris White was the go-to person back in the day :(

Emilio Pisanty

but then rotation invariance on $x$ and $y$ also requires rotation invariance on $z$

Physiks lover

@EmilioPisanty right, but it's astonishing that most relativity books don't mention the zero-component lemma. I first came across it in this book: amazon.co.uk/Relativity-Made-Relatively-Andrew-Steane/dp/…

and it seems the author got it from Rindler's Book from the 70s

ACuriousMind

23:44

@DanielSank Nowhere. I have three answers with >100 score and I only think one of them is genuinely good, and by far not as good as some others I've given.

Physiks lover

And Rindler has written an article on it here: scholarpedia.org/article/Special_relativity:_mechanics

ACuriousMind

You'll get a shiny meaningless gold badge for them when they pass 100, though

Kyle Kanos

Gold badges are far from meaningless

They represent how awesome you are

Physiks lover

to be exact, here: scholarpedia.org/article/…

ACuriousMind

@KyleKanos I don't feel awesome for having posted this :P

Sir Cumference

23:47

@ACuriousMind Woah

Kyle Kanos

I feel awesome for having done this

I got a shout-out

(10k only)

ACuriousMind

@KyleKanos How do you know that "Kyle" is supposed to be you? :P

We also have KyleOman, after all

Argon

As I understand, to find the matrix components of a QM operator, I use $[a_{ij} = \langle{i} |O| {j}\rangle $. What happens, however, if $i,j$ can take on negative values? Am I missing something?

ACuriousMind

@Argon How do you define $\lvert i \rangle$ in the first place?

And why would it be bad if they took negative values?

Sir Cumference

@KyleKanos I've also gotten a shout out before :P

Though it got removed :/

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