The h Bar

The area of the paper that is illuminated is then $A/cos\theta$, where $\theta$ is the angle measured between the beam and the normal to the paper surface.

user228700

@JohnRennie I don't get what this means, I'm so sorry :-(

user116211

Are you people talking about Iluminati? Hmm.

@MAFIA36790, will you help me think of a new domain name for the problemsolvingnetwork?

user116211

11:05 AM

Didn't you already name it?

yeah i want to change it because it is too similar to stack exchange

@Mew: problemscafe.com and problems-cafe.com are available. Also .net .org .us etc.

not bad

any other ideas?

@Kaumudi if you take a single light ray you can define a direction perpendicular to that light ray. So far so good?

Balarka Sen

What's new

11:08 AM

@BalarkaSen we are tring to come up with a name for the new community

user116211

@BalarkaSen 0celo7 is banished in Phantom Zone.

user228700

@JohnRennie Yes.

it is a community for problems olving

and question and answers

or problems and answers (panda)

@MAFIA36790 he got himself the ban he asked for?

Balarka Sen

@MAFIA oh well

11:09 AM

@Kaumudi so you can draw some small vector of length $dr$ normal to the light ray. Then you can rotate this vector keeping it normal to the light ray and it sweeps out a disk of area $dA = \pi dr^2$.

user116211

@Sanya Well, other than mods, no one can know about that.

@JohnRennie, how come physicsoverflow is allowed to be so close to stackoverflow?

@MAFIA36790 :< ... too bad

user116211

He asked to ban himself multiple times in the past.

user116211

Was he serious every time?

user116211

11:10 AM

Don't think so.

user228700

@JohnRennie Yes...

@Kaumudi Good. So that's what we mean by an area $dA$ normal to the light ray.

user228700

@JohnRennie Right.

I mentioned vector area a little while ago, and what I've just described is a vector area $d\hat{A}$ that is parallel to the light ray. If talking about vector area is going to get confusing shout at me and I'll stop :-)

user228700

@JohnRennie :-) No, I get what u're saying, do go on...

user228700

11:14 AM

(Vector area is not out of my syllabus or anything so it's about time I learnt it properly)

Anyhow, if the light rays are all parallel, e.g. like a laser, then our area elements $dA$ are all in the same direction so we can just add up all their areas to get a total area that the beam of light passes through.

user228700

Yeah, OK...

That's the area I've indicated by the vertical dashed line in ...

47 mins ago, by John Rennie

user228700

Uh huh...

@EmilioPisanty a potential-free gauge only works in the time-dependent theory, does it?

11:17 AM

And that's the area we need to use in the equation $P = IA$ when calculating the power flow through some area $A$.

user228700

Ah, OK...

Which is what you called the effective area

what about domyhomework.com?

oh not available

user116211

@Mew No.

user228700

It was difficult for me to visualize why this effective area was different from the original one until u put it this way.

11:18 AM

just joks

@Mew that sends entirely the wrong message

user116211

Like commanding someone.

just joking lol

user116211

Hey Hey @0celo7!

@Kaumudi does it make more sense now?

user228700

11:19 AM

@JohnRennie It does...

kool kaumudi

We can easily extend the idea to measure the power through curved surfaces, if you're interested ...

user228700

Yes, yes...and BTW, @Mew asked about if the surface was moving. What then?

@Kaumudi suppose I'm throwing one tennis ball per second at you

As long as you're standing still you catch one tennis ball per second.

user228700

OK before u proceed, can u PLS tell me how u do the thing where u post a message and THEN edit it to link it to another message?!?!

user228700

11:22 AM

@JohnRennie Boy, we've been doing a lot of stuff :-P At one point, we event went to space!

@Kaumudi Just click the edit link for the message, and while you have the editor open click the reply link on the previous message.

user116211

@Kaumudi yes; get ready for a change in the rate of tennis ball; protect yourself.

The editor will fill in the link and you just save the changes.

user116211

Hello master @loong.

user228700

@JohnRennie Ah, guess it doesn't work on mobile? Dammit Idk if 0celo7 got it to work on his mobile or not and now I can't know :/ Anyway, so OK, tennis balls.

11:24 AM

whoever thinks of a good name for physics.qandaexchange.com will be an instant moderator

I want physics.<something>.com

user116211

@Mew ;))

instead of qandexchange

user228700

@Mew xD You sound desperate af.

yeah bro

I can't think of anything

except the ones i've already thought of

@Kaumudi OK, suppose you're 3m away and I'm throwing the balls once a second at a speed of 1m/s, then at any time there will be 2 or 3 balls in the air on their way towards you.

user228700

11:25 AM

@Mew I will try to come up with something after areas, I promise!

ok ty :)

user228700

@JohnRennie Yeah, OK...

Loong

@MAFIA36790 hi

Now you sudden run towards me at a speed of 3m/s. It takes you one second to reach me, and in that one second you are hit by all the balls in the air. At least 2 and possibly 3 or them.

So the number of balls per second that hit you has increased from 1 to 2 or 3.

Now swap the balls for photons in a light ray, and the number of photons per second hitting you has increased, which means the light intensity has increased.

And you know all this of course, because it's just the Doppler effect.

user228700

@JohnRennie Possibly 3? How?

user228700

11:29 AM

@JohnRennie Ah, OK...

@Kaumudi it depends on the timing, whether you start running just before or just after I throw a ball. Actually, now I think about it I probably meant 3 or four balls. Anyway it doesn't matter - the point is that more balls hit you per second when you're running towards me.

user228700

@JohnRennie I start running after the 3rd second, yeah?

Oh no, now I've distracted you with the precise details of my rather dubious analogy :-)

Shall I draw one of my legendary diagrams?

user228700

Yes, I'm afraid u have :-P Not to worry tho, I get ur point.

Still interested in flux through curved surfaces?

Or is that something for another day?

user228700

11:33 AM

One second, I'm drawing one of my legendary diagrams :-P

user228700

@JohnRennie How much time dyou have, sir..?

There is nothing in my diary for today except drinking coffee

Mondays start with a panic while I sort out servers that have died over the weekend, then once that's done I chill for the rest of the day :-)

user228700

:-) Sounds good, then. It's just that I need to go for like, 15 mins in...26 minutes and then I'll be back-is that OK?

Fine.

Fine then.

user228700

11:35 AM

:-P @JohnRennie: Gimme like 2 mins while I figure out that tennis balls thing.

@JohnRennie, what is your job?

user228700

@Mew: What is your job? :-P

actuary

@Mew I'm basically retired. I work a couple of hours a day for an IT company managing their server monitoring software. I was an industrial colloid scientist.

cool

11:39 AM

@Sanya depends what you mean by 'time-dependent theory'.

user228700

@Mew Oh, wow, how old are u..?

why?

You can have time-independent force fields but time-dependent potentials.

user228700

@JohnRennie: OK! Curbed surfaces.

ok i'm going to delete my personal info now

user228700

11:40 AM

@Mew I figured that most everyone here is a student, that's why :-P

oh i can't delete my job

@Kaumudi, cos ur a student?

user228700

How does it matter?

how self centered

it doesn't amtter

i just like to reamin anonymous

and there's not that many in our profession u know

user228700

@Mew :-P No, there really are a lot of students here.

Thus any static electric field $\mathbf E = -\nabla \phi$ can be rewritten as $\mathbf E = -\frac{\partial \mathbf A}{\partial t}$ where $\mathbf A = (t-t_0) \nabla \phi$.

11:41 AM

@Kaumudi go back to where I was talking about the area element we get by considering a small vector normal to the light ray.

lol

Is that "cheating"? You tell me.

user228700

@JohnRennie OK.

i guess it is physics, nerly everyone is a student or a teacher

user228700

@Mew :-P

11:42 AM

That area element is really a vector area parallel to the light ray.

user228700

Yes...

user228700

Pointing outward, yeah?

how old are you girl?

And the power passing through that area is $P = I \cdot A$ where $I$ is now the energy flux as a vector.

user228700

@Mew U can find this information on my profile. Idc so much for anonymity :-P

11:44 AM

For light $I$ would be the Poynting vector, but that's probably getting too involved. It's just a vector that represents the energy flux.

user228700

@JohnRennie OK, one second...

@Kaumudi, when you're as rich and famous as me privacy is important

btw i can't see ur age on your profile

i'm not very good at this stalking stuff

@JohnRennie, do you think physicsoverflow is a violation of stackoverflow?

user228700

@JohnRennie And it points in the direction of light, yeah?

@Kaumudi Yes.

user228700

@Mew :-P I see. Well, I'm 17-18.

user228700

11:46 AM

@JohnRennie OK...

oh

thought u were younger

user228700

@Mew 'Cause I'm so dumb? Yeah :-P

just jokin

@Kaumudi We have $P = I \cdot dA$ which is $P = I A \cos\theta$, where $\theta$ is the angle between the light ray and the vector area.

So that's where the $\cos\theta$ comes in that you mentioned right at the start of all this ...

@EmilioPisanty I was thinking of no time dependence in the fields, potentials or charge/current densities

user228700

11:48 AM

@JohnRennie Yeah, OK...

"Classical" electrostatics/magnetostatics

@Sanya Yeah, but ruling out time dependence on the potentials is too restrictive

it's the same physical situation as long as the force fields and charge and current densities are time independent

it is the same physical situation

@Kaumudi this expression $dP = I \cdot dA$ works for any area element. If that element is at some angle to the light ray it just gets an extra $\cos\theta$ term.

@Sanya Exactly. So all the differences are in your head.

user228700

11:50 AM

@JohnRennie OK...

It leads to plenty of awkward results, but that's EM for you.

but the physical interpretation of the scalar potential as potential difference only makes sense if we restrict it to non-time dependent potentials

user228700

And then I integrate this or something?

@Kaumudi Yes, to get the total power we just add up all the elements $dA$, and of course adding up the $dA$s means we do an integral: $$ P = \int I \cdot dA $$

otherwise we just need to say there is no physical meaning in $A_{\mu}$ in general

11:51 AM

@Sanya, I see you asked a question on physics.qandaexchange.com

I edited some of the latex to make it render better

@Sanya does it?

why?

@Mew thank you

for some reason with lmits of integration you need to put an additional \ before each _ character

i'm working on fixing this

user228700

@JohnRennie Hm, OK...

@EmilioPisanty because if $\Phi=0$ everywhere at any point, then interpreting the value of $\Phi$ at a point is useless

11:52 AM

Actually doing that integral can be tricky unless there's some simplifying symmetry you can exploit, but that's how the calculation is done.

but I see that I'd rather delete that post again

@Sanya That's a bit of a leap. You can have time-dependent potentials that don't obey $\Phi \equiv 0$.

user228700

I see. Well, I'm pretty sure I'll come across this integral soon enough so pls. expect me to annoy you :-P

@Kaumudi I doubt you'll need to do calculations like this until well into your degree.

user228700

(I'm supposed to find electric fields at points due to charge distributions using Gauss's theorem. Though they will be simple, I guess...)

user228700

11:55 AM

Same principle, right?

@EmilioPisanty ok - so what you're saying is "I can find situations where even though $\varrho$, $\vec{j}$, $\vec{E}$ and $\vec{B}$ are time-independent while $\Phi$ and $\vec{A}$ are time-dependent, I can still in a meaningful way interpret the difference between two values of $\Phi$ at two different points in space at the same time as their potential energy difference"?

@Sanya exactly

It might not be that useful

@Kaumudi Yes, the flux through some element of your Gaussian surface $dA$ is $E \cdot dA$.

user228700

Yeah, so that's why I asked u to pls. expect me to annoy you :-P

For example, you might have a time-dependent potential energy accompanied by a non-conserved total energy, because your system is not time-independent (as seen by the particle's equations of motion)

11:57 AM

But at your level you'll be dealing with highly symmetric systems where it's obvious how to calculate this.

but you can still build such an interpretation

user228700

Still. U know how dumb I am :-P

ur not dumb kaumudi

i'm sure u can come up with a good website name

user228700

@Mew Haha xD

@Kaumudi Physics can make all of us feel dumb. Quantum field theory is currently making me feel like a preschooler.

user228700

11:59 AM

@JohnRennie Oh! :-P You're doing qft now?

what about problemdump.com

user228700

@Mew: How dyou even trust that I'm creative? I mean, I'm kinda creative, I guess, but how did u trust it blindly? :-P

user228700

OK, be back in around 15 mins.

@Kaumudi I can just tell

I've been trying to get a handle on QFT for years. The usual books are remorselessly mathematical, but I'm trying to understand QFT in a non-mathematical way. This may well be impossible, hence my struggles :-)

12:00 PM

if QFT is true, it will be understandable in a non-mathematical way

maybe when u understand it you can write a book for the rest

@Mew not necessarily the case. Tread carefully, the attitude if I can't understand it then it's wrong is the path to crackpottery.

lol

@EmilioPisanty ok ... but we can agree that even if it is overly restrictive, asking for time-independence for all quantities involved would allow us the classical "meaningful" interpretation as potential energy difference; the question now is - can we find a better categorisation of the kind of situations in which there is a meaningful way to physically interpret $\Phi(x_1, t)-\Phi(x_2,t)$, i.e., is there a less restrictive set of requirements that still allows us to build this interpretation

to be honest, I've never thought about it

I believe it is true

and thus I think it is understandable

@Sanya I disagree that a time-independent $\Phi$ is the only meaningful case for interpretation as a potential energy difference.

A non-conserved total energy is still an energy.

12:02 PM

@EmilioPisanty no, this is not what I have written

@Mew QFT works i.e. the calculations successfully describe experiment. Whether that makes it true depends on your definition of true.

I say that with these requirements, we can definitely interpret it always that way

@Sanya what do you mean by "meaningful", then?

Well just because the calculations work doesn't mean the mechanicsms behind what produces the output is correct

I don't see any problem physically with a time dependent potential difference. It basically means the system is dynamic and energy is only conserve when everything is taken account of thus becomign a closed system

12:04 PM

For example John, the "ether" could explain special relativity

but clearly the "ether" theory isn't as good as special relativity

@Sanya Yeah, something like that. But with the emphasis on overly restrictive.

@EmilioPisanty whenever we can attach a physical meaning to $\Phi(x_1, t)-\Phi(x_2,t)$, e.g. as the work divided by charge for transporting a charge from one point to the other [but here I would be free for other ideas too]

@EmilioPisanty my problem is - I need to build up from overly restrictive to less restrictive to understand

I suspect the work will become path dependent due to the time evolution of the potential

@Mew no it couldn't. The ether theory doesn't explain SR unless we mean different things by the ether theory.

@JohnRennie, I thought the equations of SR were known before Einstein

but the interpretation was wrong

the lorentz transformations

Lorentz ether theory

What is now often called Lorentz ether theory (LET) has its roots in Hendrik Lorentz's "theory of electrons", which was the final point in the development of the classical aether theories at the end of the 19th and at the beginning of the 20th century. Lorentz's initial theory was created between 1892 and 1895 and was based on a completely motionless aether. It explained the failure of the negative aether drift experiments to first order in v/c by introducing an auxiliary variable called "local time" for connecting systems at rest and in motion in the aether. In addition, the negative result of...

12:10 PM

The equation for the Lorentz contraction had been proposed (by Lorentz) in 1903 or 4 I think. But it was an ad hoc suggestion.

@JohnRennie, "Because the same mathematical formalism occurs in both, it is not possible to distinguish between LET and SR by experiment."

"However, in LET the existence of an undetectable aether is assumed and the validity of the relativity principle seems to be only coincidental, which is one reason why SR is commonly preferred over LET."

we know that in general (Maxwell EM with explicit time dependence in fields, densities) without a chosen gauge we cannot give any interpretation to $\Phi$ and $\vec{A}$ while with the over-restrictive requirements we have an interpretation to $\Phi$; that's a starting point for me at least - at that point, the in-between is unknown territory. That was my way to think about it @EmilioPisanty - but nvm, thanks for the input :) EM is confusing me again and again

@Sanya Yeah, it's presented as this clear-cut thing and it's actually much more subtle and confusing

That said, I'm slightly less sure that you can rewrite the EOM quite that cleanly

it actually comes down to $$\frac{\mathrm d}{\mathrm dt}(m\mathbf v +q\mathbf A) = -q\nabla \phi +q(\mathbf v\cdot \nabla) \mathbf A$$

@Mew problemsbazaar.com or variants thereof ...

because of the difference between $\frac{\mathrm d \mathbf A}{\mathrm dt}$ and $\frac{\partial \mathbf A}{\partial t}$.

In any case, you can rewrite it through Hamilton's equations, and see what comes out

12:17 PM

@JohnRennie, that's a really good one

the EOM is always the main stop there

@Secret I think if we take a gauge transformation $\Lambda = a t^2$ where $a \in \mathbb{R}$, then $\nabla \Lambda = 0$ (thus $\vec{A}$ is unchanged as well as $\vec{E}$) but the value of the potential changes linearly in time even though everything is time-independent physically speaking - I think that should work

@JohnRennie, definitely going onto the short list

@Secret That can be the case, but it depends on the setting. If you just set $\mathbf A = (t-t_0)\nabla \chi$ then there's no problem.

If you do something crazier then of course you can have a problem.

@EmilioPisanty I'm not quite sure I'm following you :| you want to look at the EOM to regain something like a force law (force equals gradient of potential)?

12:23 PM

Ultimately, the work is a physical quantity (since it's defined in terms of the force fields) so it can't change with a gauge transformation. What can change is the relation of the potential(s) to the work preformed.

@Sanya What I'm saying is: if you want to interpret $(\phi,\mathbf A)$ as anything, then your first stop should be re-writing the EOM in terms of those potentials.

That's what you do in the standard electrostatic case: $m \dot{\mathbf v} = -q\nabla \phi$. That enables you to dot that equation with $\mathbf v$ and then integrate over time to get an expression linking changes in $\mathbf v^2$ to changes in $\phi$.

If you have a more complicated situation, and you want to try for an interpretation, repeat the same procedure and see what you can get.

I see the point

seems like it will get messy ... thanks for the explanation :)

I have just computed the $\Lambda$ suggested by Sanya, and indeed because the spatial derivative vanishes and the time derivative of $\Lambda$ is linear (making V to have an extra linear time dependence), what you end up for such gauge is that both $\vec{E}$ and $\vec{B}$ remains the same.

$$\vec{E}=-\nabla (V-2at) + \frac{\partial A}{\partial t}=-\nabla V + \frac{\partial A}{\partial t}$$
$$\vec{B}=\nabla \times \vec{A}$$

However it is not very clear how to interpret this gauge if $\vec{E}$ and $\vec{B}$ are from a time independent system. Might have to play with the symbols a bit to ch…

@Secret what I'm doing there is just instead of choosing the "energy origin" once, I rechoose it again differently any given instant of time - I don't think that has much of a meaning, let alone any physical implication to be honest

Setting a gauge is kinda like a more generalised version of setting where your zero reference point is, thus I suspect under this gauge the potential at infinity will have a linear time dependence. Thus all $\vec{A}$ and $V$ will gain a time dependence which is cancelled out in order to result in the physical system of time independence

@Mew Surprisingly physicsbazaar.com is also available. I assume you want a domain like physics.something.com in case you want to add other subjects later.

12:32 PM

so my rough guess is that the baseline becomes time dependent

@JohnRennie yeah i was hoping to leave the option open to add more

So it should work, but I am not sure what advantage will have by shifting the time dependence onto the baseline

12:43 PM

@Secret Yes. This is a special case of a more general case, where you transform by $\phi \mapsto \phi +\partial \chi/\partial t$ and $\mathbf A \mapsto \mathbf A + \nabla \chi$, as usual, and you give $\chi(\mathbf r,t)$ an arbitrary time dependence, but no space dependence.

(possibly with some muddled signs there btw)

@Secret it's just creating an artificial time dependence as a proof of concept

You question does cause me to wonder further about a possible problem solving method on finding a gauge so that the gauge absorb the time dependence of the system thus using this gauge effectively can convert a time dependent system to a time independent one

Currently playing with the symbols to see if anything sensible pop out

Q: Improve your chances of an answer by writing your question properly

1:04 PM

@obe I will not discuss moderation issues in off-site channels. He'll have to use meta/contact SE like everyone else.

@DanielSank Nice (wonder why that never gets questions onto the HNQ list)! Alas, no MathJax :(

Physics Meta

0

My background and full time job is based on sales and SEO related Internet work. Without false modesty, I am good at my job, and I want to use my experience to write a post that I can refer some users to, as regards why their questions are not being answered, or how they could be written to incr...

discussion

user228700

I'm sorry, that took waaay longer than 15 mins :|

user228700

I had to go buy some books :/

user228700

@JohnRennie Yes, omg, this is a good one.

user228700

U should've just asked JohnRennie @Mew.

1:14 PM

Hi Kauma

I've been waiting a long time

what is ur idea/

Caption: Well it seems for classical fields where the space and time components can be separated, it gives somewhat more symmetric expressions. However, it might not be a favorable choice of gauge given the integration involved.

physics.qandaexchange.com/?qa=233/domain-name-suggestions

(NB I was eating, thus I cannot access to my keyboard to type all of that out)
Nb2 Mistake: No that is not the D'Alembertian, cause it's missing a $c^2$ factor

^ please give any domain name suggestions here: physics.qandaexchange.com/?qa=233/domain-name-suggestions

user228700

@Mew "Kauma" is the weirdest short-name I've ever had lol :-P

1:17 PM

lol ty

user228700

Sigh, @Mew: Dude, I have like too many things happening in my brain at the same time. I mean, I'm not even that creative in the first place, if I'm not in the mood. (If I'm in the mood, I can come up with nice poems in 8 mins lol-true story :-P) I did think about it when I went out to buy books but yeah, I'm sorry? :-(

@Kaumudi no worries don't stress out

user228700

If I come up with something, I'll let you know. I'm sorry again :/

no problemo

:)

it's pretty hard

especially to find ones that aren't taken

user228700

Yeah :/ Good luck tho! It's amazing, what u're doing. I hope it goes well...

1:43 PM

Caption: I am going back to group theory, the fiddling stuff is not really making sense

Yes it's a proof of concept, but it seems to be more complicated than the common gauge fixing choices excep for special forms of the functions $\vec{E}$ and $\vec{B}$

I am going to stick with the common ones...

okay, time for a math question

I'm blanking on whether or not a general rank-1 tensor needs to be self-commutative. That is, I can't recall if $[k_\mu ,k_\nu]=0$ is necessary

@Jim Commutative? Are you talking about an operator-valued vector field in QFT or something?

@ACuriousMind More as a general rule, but yes that would be good enough

@Jim In general, you have that tensors have numbers as entries, so the components commute rather generally :P

sure, I guess for operators it's not necessary but there isn't a great way of showing it as a general statement

1:57 PM

@ACuriousMind if the underlying numbers are a field ... what about fermionic stuff?

I seem to remember difficulties

I thought it is an axomic requirement that all lie algebra has to be reflexive i.e. $[x,x]=0$ regardless of what object x is?

@Jim Also, no, the tensor doesn't have to fulfill that. The vector fields we have in QFT usually fulfill that because those are the canonical commutation relations we impose

@Secret it is, but do all tensors have to be part of a Lie algebra?

But as long as your r.h.s. is a proper rank-2 tensor, it doesn't need to be zero at all.

that I am not sure, those I have came across so far in quantum have commutator to induce the lie algebra

so QFT might be something completely different and I am not aware about

2:00 PM

@Secret We're not looking at $[x,x]$. We're looking at $[k_\mu,k_\nu]$. The bracket doesn't know that "$k_\mu$" and "$k_\nu$" belong to the "same object".

@Sanya Pfff, details, details ;)

hmm ok, I see

@ACuriousMind that's what I thought. Because I'm coming across a $k_\mu k_\nu$ in some math and I want to call that symmetric to reduce it to 10 equations, but I am doubting that this is a necessity

@Jim Yes, if $k$ is operator-valued that is not necessarily symmetric.

@ACuriousMind well, I guess this system is going to become overdetermined

barlop

this guy giving a ted talk, named Aaron O'Connell: , says he has done quantum effects on large objects, using very cold temperatures.. he doesn't give much info in the ted talk and doesn't sound scientific, he looks and sounds like a hairdresser, but he is a scientist apparently.. is it a fraud? youtube.com/watch?v=dvYYYlgVAao

it has 1100 upvotes and 105 downvotes so people are buying into it.. don't know if it's legit or not.

Danu

2:07 PM

@Shing cc @Obliv Nothing happened between me and 0celo7, and I'm not banned (I'm a (chat) moderator, in fact).

Electric motors (part 2) | Physics | Khan Academy

can some one help me?

plz watch this video from 0:00 to 3:40 if you are free

►

@Jim Jim, the $k_\mu$ in the question you just answered is just a momentum vector, not an operator.

@ACuriousMind Is it? I wasn't about to look up the reference

@Jim The propagator is a function of a standard-issue "classical" momentum vector, yes

also, it wouldn't be the first time I've seen a momentum vector defined in terms of $\partial_\mu$

and I needn't tell you that isn't necessarily symmetric unless the operand is continuous

point remains, however, that the OP had at least 10 equations when they were looking for 2. I'm not sure how they got as far as they did. Even plugging it in to the original statement still leaves two indicies to use to find A and B

2:29 PM

did anyone watch it?

...

2:39 PM

@MartianCactus where is his magnetic field going?

out of the drawing board?

no

from left of screen to right

now my question-

if the magnetic force is acting upwards and downwards(when the sheet is completely perpendicular) then won't it rip apart the board?

I think in the case of a very strong magnetic field and a very weak board I'd tend to agree - but most materials can stand a bit of tension, so I don't see a big problem there

let's however hope for someone else to come by and comment

it's always good to have more than one opinion on these things :)

also

due to the torque the sheet will spin to the other side too right

well

wont the magnetic force still be there acting in the same direction?

wont that push the board back where it came from

until its in a stable 90 degree turn zone?

no, because the direction of the currents changes

between 0° and 180°

so does the direction of the force

i dont see how it changes?

2:52 PM

well because the sheet is spinning and on one side of it, the current goes up, on the other down ... so once it has turned by 180°, it's not in the same situation as before, but the same situation with current directions reversed

sorry, I'll need to head out - maybe we can follow up the discussion later

oh ok

still not clear but maybe someone else may come

3:25 PM

anyone for help plz?

3:42 PM

what do you need help with @martian

watch that vid frm startin to 2:#0

3:30 min

do you have a hard time understanding what happens to the forces when the circuit rotates? @martian

no

did u watch it?

@MartianCactus Do you have a question about magnetic induction in motors?

yes, I don't see the question

3:47 PM

heres the thing-

when the board becoes 90 degree

becomes*

wont the force tear it apart?

what makes you think that

This is a rotating coil of wire in a uniform magnetic field?

The current loop feels a torque that makes it want to align its magnetic dipole moment with the external field.

If it's the wrong way, it tries to flip over.

but when it is 90 degress the forces act on the edges of the circuit

wont it tear apart the wires then?

There are forces on the wires the whole time. That's what makes the circuit rotate.

You can compute what those forces are. For the simple case, and order-of-magnitude for the complicated cases, it's $F=BIL$

... for field $B$, current $I$, straight segment length $L$.

so the forces are not strong enough to break the wire?

3:53 PM

Not usually.

You could design a device that way, but it wouldn't act like a motor.

oh

also when the coil has completely roated 90 degrees

he says that it will keep going the way it was

due to momentum

doesnt he?

@MartianCactus Yes, inertia is your friend.

oh

inertia yeah

I did an experiment once where we use a solenoid with a 10,000 amperes of current. Apparently when it was being initially tested, one of the coil supports failed and a coil moved a few inches.

The engineer involved called it a change-your-pants moment.

but wont the forces now affect the coil in the opposite way?

3:56 PM

But the supports were reinforced and we ran the thing for a year without any motion.

the direction of forces are still the same but wont they push the coil the other way after its turning more than 90 degrees?

@MartianCactus Yes, the magnetic force makes the coil want to align with the field.

Shin Kim

What's the thing

So, if you want a motor, you have to have some method for reversing either the field direction or the current direction.

Shin Kim

I mean what kind of experiment did you guys run with that 10,000 A coil

3:59 PM

Alternating-current motors are a little easier; DC motors use some sort of bushing so that the direction of the current in the coil depends on the orientation of the coil.