3:20 AM
0

I went to the Physics Overflow website and it didn't give a tour. I'm afraid of being incompetent and causing trouble with the website and being handed over too much burden of responsibility to figure out all by myself not to cause too much trouble on it. I used to be a bad contributor to Stack E...

2 hours later…
5:07 AM
1

I know that there are several posts on the same idea, and I have read most of them, but still, my questions persist. I have listed the other posts on the topic at the end of this post. All the other posts say that the electric field inside an ideal wire must be zero because: As the potential dro...

@JohnRennie : Sir, Kindly explain to me in context of above question that how the electric field inside an ideal wire can become zero?

@DevanshMittal hi :-)

Hello Sir
🙏😊
If the electric field inside the wire is zero then how the current will flow?

The conduction electrons in a metal behave in a very similar way to a gas made up from electrons. This even has a name - it's called the free electron gas model.
So we can understand a lot about how electrons behave in circuits by making the analogy with a gas flowing in a circuit of pipes.

Yes Sir.

In real life there is drag when a gas flows through a pie due to the interactions between the gas molecules and the walls of the pipe. This is directly analogous to the resistance when electrons flow through a wire.

5:12 AM
Let's consider a simple circuit in which we have a battery and a resistance and they are connected by ideal wires, with the help of a switch. initially the switch is open. now at t=0, the switch is closed, now what motivates the free charged particles inside the ideal wires to move if the electric field inside the ideal wires has to be zero?

But let's assume we can make pipes that give zero resistance to flow. This would be the equivalent to having a circuit made up from zero resistance wires. OK so far?

Yes Sir.

OK :-) Suppose the gas is flowing at some speed v through the pipes. If there is no resistance then there is nothing to slow down the gas so it just keeps at constant speed i.e. we get a constant current.

My all the doubts originate from a single point that electric field inside an ideal wire is zero. I do not understand how then the current will be initiated in such a circuit.

And this happens even though there is nothing exerting force on the gas i.e. no battery.

5:15 AM
what gives the free charged particles inside the ideal wires the initial impulse for them to move with constant velocity?

In fact this really happens because we get exactly this in superconducting circuits where a current can flow in a loop when no battery or other source of EMF is present.

Yes Sir.

@DevanshMittal there are various ways the current could get started. The simplest way is to induce a current using an external magnetic field.
But I wouldn't worry too much about the details because this is just setting the scene i.e. explaining some basics.

Sir, I have to go to a meeting now so if you can respond to the questions I have asked in the thread then that would help me. thank you sir. I am sorry for this Sir.

OK, ping me when you're free again.

5:26 AM
Thanks Sir.

3 hours later…
8:08 AM
Who introduced the comma notation for partial derivatives used much in General Relativity??

@ManasDogra It was present at least as far back as Hawking-Ellis

@Slereah Was it present in Einstein's time?

I don't think so
It's not in Weyl's book, which is the oldest GR book
Hm, this 1963 book also uses it
Synge's 1960's general relativity book also uses it!

@Slereah This is the earliest I got.

I don't think I can go back much further with books, rly
There aren't a lot of big GR books between Weyl and Synge
After that it's mostly articles
Hm, what's a nice 50's paper

8:19 AM
@Slereah I haven't yet looked on Feynman lectures on gravitation or Dirac's small book yet.

those are both more recent than 1960 though

@Slereah Oh yes!

a Lanczos paper uses it in 1957
According to Lanczos that was already part of the Einstein notation
let's see the original paper of Einstein

2 hours later…
10:00 AM
How do passive RFID tags work? I mean, sure, they have no battery and current is induced by a parent device wirelessly but is the structure of the metal plates of the passive tag that provides it's unique detection?
I'm struggling to understand how inducing current into some passive metal plates with a weird shape and cuts can produce a unique response

10:41 AM
@JingleBells What do you mean by "some passive metal plates"? It's a microchip!
If you don't find it mysterious how the chips in your computer can encode unique logic behaviours, you shouldn't find this mysterious, either.

@ACuriousMind For example debit cards have this weird metal thingy:
I'm pretty sure that's not a microchip

11:09 AM
@JingleBells That's not an RFID card.
Those visible "metal plates with a weird shape and cuts" are the contacts of the EMV chip.

@JohnRennie : Hello Sir, Are you there? I am back now. I am free now for next 1.5 hours. May I have some of your time now?
https://physics.stackexchange.com/questions/563858/how-the-electric-field-inside-an-ideal-current-carrying-wire-can-be-zero

I wish to discuss on this question with you.

11:33 AM
@FadedGiant What those weird cuts mean? I see they are also on SIM Card and every company has it's unique own.
@DevanshMittal That's a good question, I remember asking this to my school teacher, didn't get any satisfactory response from him.

@abhas_RewCie: Thanks. What should we do now?
I think, nobody has a concrete answer.

@DevanshMittal I'm pretty sure @JohnRennie is a perfect man here and I doubt if there is hardly any physics he can't help us with. Usually, he's online after 9 AM (Morning) IST.

@abhas_RewCie: Yes, I believe the same. I wish to learn from him. Unfortunately morning times are not possible to me due to office.

@DevanshMittal My best guess would be that PD in Ideal wires is infinitesimal rather than absolute Zero, which helps to specify the direction of flow of electron....
Or we should rather define Ideal Wires by $\sigma \rightarrow +\infty$
@DevanshMittal You'll get a notification when he'll answer you. So, you can read his response later on :)

Ohk. Thanks a lot, for your kind help and inputs.

11:46 AM
Glad to help :)

1 hour later…
12:58 PM
I have a question
Should I do hard question in physics textbooks
I find it waste of time.

1:13 PM
I found cambridge book harder than one which MIT use

1:24 PM
@FadedGiant Hmm okay

2:02 PM
It is often said that for an electric wire the laid along the x axis the following relationship applies:

$V(x) = V_0 * \frac{x}{L}$ where as $V$ the voltage (potentail difference) and $x$ the lenght at point x and $l$ the full lenght of the wire.

could someone please link me to a page where this is derived or show me how it is derived correctly?

Thanks!

2:14 PM
i got it nvm thx

@JingleBells Just to be clear: There is an RFID chip in most modern cards, but it isn't visible (it doesn't need to be visible because it works via electric and magnetic fields, not contacts!)

well, there is one visible difference compared to the picture above:

2:49 PM
@ACuriousMind IDK if it's the RFID chip, but my Visa is partially see through and you get to see some copper wires running under the surface to and from the EMV chip (which you can also see behind kinda)

3:00 PM
@DevanshMittal hi Devanesh. I had gone out when you posted. What time will you be around on Wednesday?

4 hours later…
6:33 PM
@JohnRennie: Sir, May we talk at 5:00 PM, Indian Standard Time? I live in India.
On Wednesday.

6:58 PM
Am I right in saying that we can work in GR (and by extension SR) with coordinate systems that don't correspond to any particular frame. For instance we can't work simultaneously in an inertial frame and polar coordinates in SR. And so when we for instance look at the Schwartzschild metric in Schwartzschild coordinates we aren't talking about any particular "frame", we're just labelling points on the manifold.

Sure.
I mean, you can always define an observer by considering a coordinate line
But that observer may not be a particularly realistic one

what's a coordinate line?

A line of constant coordinate except one

ah

but yeah, overall there isn't an obligation for coordinates to be observer-related
there are procedures to construct coordinates for observers, if you so need

7:01 PM
Does there exist for instance a coordinate system that corresponds to someone stationary on the surface of the Earth?

Sure

Even though this observer is being "held" in place by non-gravity related forces

The Christoffel symbols will simply be non-zero, and those will correspond to coordinate forces
ie centrifugal, coriolis

ah I see

you can check that a rotating frame in SR will reduce to the usual coordinate force in the classical limit

7:02 PM
just one final question out of curiosity, is there a name given to the "set of all admissible frames", is analogy to the set of lorentz transformations?
I haven't considered (or even seen in textbooks/notes) much coverage of rotating frames

If you mean the set of all coordinate transforms, it's called the diffeomorphism group
If you mean the set of all (local) frames, it's the sections of the frame bundle

And that is the set of all conceivable coordinate transformations, not just those between observer related coordinate systems
ah
I have seen that phrase before
thank you ^ :)

if you need a simple example just pick $\mathbb{R}$
Then the diffeomorphism group is just the group of all (smooth) invertible functions

Specifically functions $f:\Bbb R\rightarrow \Bbb R$ right?

yes

7:09 PM
@Slereah ::ahem:: you mean smooth invertible functions with smooth inverses

@ACuriousMind boo

I'm a bit unsure how those functions relate to coordinate transformations though

$\mathbb{R}^0$ would be even better but then you lose some features :p
@Charlie Well for a simple one, consider for instance the rescaling of $\mathbb{R}$
Our basic coordinate system is, let's say, the canonical $\mathbb{R}$ coordinates
$x \mapsto x$
Our coordinate transform is $f(x) = \alpha x$, $\alpha > 0$
This is an invertible function, $f^{-1}(x) = \alpha^{-1} x$
this just corresponds to picking coordinates with a different "size" of measurement
then using that function and its inverse, you can switch from one to the other

hmm
So the elements of $\Bbb R$ don't have labels unless we use a coordinate system?
I've never given it much thought

I mean $\mathbb{R}$ does because it's $\mathbb{R}$, but manifolds do not in general, no
There's no canonical coordinates

7:15 PM
oh so $\Bbb R$ is equipped by default with the coordinate system you gave above

@Charlie It's more like one of your $\mathbb{R}$s here has labels and the other is just a line without markings. The $f$ is a choice of how to label the unmarked line.

If you consider $\mathbb{R}$ as just a geometric line, there is no preferred point you can pick as the origin
or preferred scale

yeah I guess so

It's perhaps better to use a circle as an example if the names of points in $\mathbb{R}$ throw you off

(If we consider this line to be say our spacetime, and the line being the time, the coordinates will correspond to the choice of a clock, physically speaking)
The origin is the starting time of the experiment, the scale is the scale defined by the unit of the clock

7:19 PM
So in that example changing the scale could correspond to changing frame?

The frame roughly corresponds to the direction defined by the coordinates
Here it's just a vector field, a constant one if we pick the canonical coordinates we defined earlier

I'm not sure what "direction defined by the coordinates" means

@Charlie The vector fields defined by the derivatives w.r.t. the coordinates

Oh I see this is that new way of thinking about vectors as derivatives

intuitively it's just "the direction the coord. axes point in"