an example of a "frozen shoulder" on a giant

smbc-comics.com/comic/giants

@MoreAnonymous OK so I couldn't resist attempting an answer. You'll have to judge for yourself how successful it is :-)

Hello World.

Hi @JohnRennie sir

@123 hi :-)

Is it possible to have torque on coil or magnetic field phenomenon by vector addition of magnetic field due to rectangular coil and external uniform magneyic field

@JohnRennie I think it's a good answer :). P.S: Im at work now so will be only able to chat intermittently

Because torque equation only consider external uniform magnetic field and current on coil.

@123 you could do the calculation that way, but it would be complicated. The energy density of a magnetic field is ½μB² so you could vector add the fields then integrate the total field to get the total energy. Then differentiating this energy wrt the coil angle would give you the torque on the coil.

It would be far simpler to use the Lorentz force on the coil, or approximate the coil by a dipole field and use the equation for the torque on a dipole in an external field.

@JohnRennie Thanks a lot for clear the confusion. These answers really helped me a lot to understand the phenomenon.

As coulombs force can be explained by two charges and distance between them. We create the idea of electric field from coulombs force.

But in lorectz force we already have magnetic field proportional to force. Why it is the case?? Why we won't define idea of magnetic force and magnetic field separately as in electrostatic??

One more question about magnetic field. How do we know the F = q(v x B), magnetic force is perpendicular to both magnetic field and direction of flow of charge.

Because by experiment we can observe only magnetic force not the magnetic field. How do we know field is perpendicular to force??? Pls pls explain.

We can measure the field direction in many ways. The simplest is just to see how it affects a magnet dipole e.g. just put a compass needle in it.

@JohnRennie you are right but in the case of magnetic field it is perpendicular to the force. How do we know where needle of compass moving is not the direction of magnetic filed?

Because needle is moving due to magnetic force not due to magnetic field. In my opinion

Let there exist a multivariable function $f(\mathbf x)$, where it's domain is $D\subseteq \mathbb R^n$. Let there be an open set $S \subset D$. It is given that $\partial f/\partial x_i$ is defined at every point $P \in S$, for all $i$ from $1$ to $n$, where $x_i$ are the $n$ orthogonal coordinates forming up the $\mathbb R^n$ space. From the above given information, can we conclude that $f$ is continuous at every point $P\in S$? I think yes, we can.

@FakeMod Wouldn't MSE be a better place to ask this?

@MoreAnonymous I also asked this in that chat room. But many times I have found more satisfying and understandable answers form this chatroom regarding such questions. In case this is inconvenient for anybody, I will consider not posting such questions here.

@FakeMod Oh its not inconvenient at all ... It was just a suggestion.

@MoreAnonymous Yeah, no problem. Thanks but I am consciously making the decision to also ask here :)

How to install mathjax to use latex on mobile. What is the best to add mathjax

@FakeMod This has been resolved

@JohnRennie I'm kind of confused about this but in SR can a point like object spin faster than the speed of light? I mean intuitively I suspect it would lead to some kind of contradiction. I mean if $ds^2 = -c^2 dt^2 + r^2 d \theta^2 $ but when $r=0$ then $\dot \theta $ can be anything

I'm not sure what it means for a point to spin ...

@FakeMod I have installed mathjax on mobile how to use it there is no installation file. Only .Js and json files

@JohnRennie From the reference frame of the point I see the whole world going round me

If you're worried about rotations, look up the Born coordinates

Should've linked to this directly, but okay.

It explains how to do rotation in GR

@Slereah Rotations don't work in SR?

It does

Comparing velocities is complicated in GR, though

@Slereah Won't an inertial frame agree with what ever the point sees?

What is an inertial frame in GR

@Slereah Im thinking of it in the context of SR

@FakeMod No, existence of the partial derivatives is not enough: $f(x,y) = \frac{2xy}{x^2 + y^2}$ with $f = 0$ at the origin has all partial derivatives, but is not continuous at the origin

you need continuity of the partial derivatives for them to say something about $f$, and if all partial derivatives exist and are continuous, then $f$ is not only continuous but differentiable

@ACuriousMind True, but if I impose the additional condition of bounded partial derivatives, then I can assert the continuity of the function.

The discussion in the MathSE chat starts from here ^

@MoreAnonymous en.wikipedia.org/wiki/Ehrenfest_paradox?

@ACuriousMind But the existence of bounded partial derivatives will guarantee the continuity of the function over an open region.

@NiharKarve I don't think Born rigidity would apply to a point particle

ohh you're talking about a single particle nvm

@MoreAnonymous A point cannot "spin" classically. There is no physical difference between a point to which you attach a rotating frame and one to which you attach a non-rotating frame. Only extended bodies can spin, and there are no ideal rigid bodies in GR.

@ACuriousMind I suppose you can't make a point particle spin but you can start with an inital condition where you just happen to find a spinning point

@MoreAnonymous What does a "spinning point" mean? Angular velocity is proportional to $r\times v$, where $r$ is the distance to the center of rotation and $v$ the velocity of the rotating object, and $r=0$ always for a point w.r.t itself as the center of rotation.

@ACuriousMind I was just thinking kinematically $ds^2 = - c^ dt^2 + dr^2 + r^2 d \theta^2$ when $r= 0$ $\dot \theta $ can be anything

@MoreAnonymous What is $\dot{\theta}$?

$d \theta /dt$

You wrote down a metric, $\theta$ is one coordinate, $t$ another, and they don't depend on each other, so differentiating $\theta$ w.r.t. $t$ doesn't make any sense

it's as nonsensical as writing $\partial y / \partial x$ for the $(x,y)$-coordinates of 2d space.

What if I had $ds^2 = - c^ dt^2 + dr^2 + r(t)^2 d \theta^2$ and $r(0) = 0$ and $r(t \neq 0) \neq 0$ and it is continuous then can I take limits at $t=0$?

you can do with metrics whatever you want but I don't see what this has to do with a spinning point :P

but at $t=0$ it is a spinning point :P

@MoreAnonymous a metric can have an angular momentum. An obvious example is the Kerr metric.

Or a metric can be written in rotating coordinates.

@MoreAnonymous What does the metric you wrote down have to do with a point spinning? You just wrote down a generic FLRW metric in hyperspherical coordinates.

well, with $t$ dependence instead of $r$ dependence, but I don't understand why you think this makes anything spin

The Kerr metric would be the nearest thing I can think of to a spinning point, though the singularity is not a point.

So $ds^2$ is the geodesic in my eyes. So I have a particle spiraling

@JohnRennie Indeed, and the ring-shape of the singularity is precisely because otherwise there would be nothing distinguishing the axis of rotation!

Yes, so I don't think you can have a spinning point in GR.

You can obviously use rotating coordinates, but then the angular momentum is zero.

@MoreAnonymous It's the metric. You have to solve the geodesic equation to find the geodesics. Are you trying to run (=find edge cases to the formalism of GR) before you can walk (=properly apply the formalism of GR) again?

@ACuriousMind I guess your right I should revise my GR more

before i ask this

Also just for reference I was thinking of $ds^2 = 0$

@ACuriousMind Well, you can make an ideal rigid body in GR

It's just gonna violate some energy conditions :p

This is apparently a 'canonical' reference for the time problem of time stuff

126 pages

mama mia

"the diffeomorphism group is analogous to the gauge group in Yang-Mills theories" ::twitches::

@Slereah that's a big problem :-)

Another one of these inconclusive debates on here :p

@ACuriousMind Well, "analogous" is a pretty broad term

@Slereah sure, but the author seems to think this means only "Diff(M)-invariant quantities" are observable, although he also mentions the nature of an observable is contentious - but I think it is uncontroversial that the values of a scalar field would be observable, even though it is clearly not invariant under a diffeomorphism like translations

Well, the true observable is the value of a field as measured by a specific apparatus

And that truly is diffeomorphism invariant :p

@Slereah well, "observable" in GR here certainly cannot mean the same as "I can build an apparatus that measures this"

Doesn't it

People don't talk about "observables" a lot in GR, but when they do, it's usually centered on a specific observer

there's lots of apparati you can build that measure non-invariant quantities like energy, no?

I mean depends how you define everything, I suppose?

yes, everything depends on how you define everything!

I'm just saying I'm not convinced this paper actually presents a coherent picture of what it wants to talk about

But I would say that energy measured by an observer is an observable

my fave paper on the topic remains https//link.springer.com/article/10.1007/BF02923262

you're missing a : there :P

oops

did you type the 'https' by hand or something?

link.springer.com/article/10.1007/BF02923262

Me see star

@ACuriousMind we'll leave it to the reader to determine what "symmetry" means

Diffeomorphism group is certainly one in the sense that it's an automorphism of the bundle that leaves the action invariant

but it is not a principal gauge bundle

I still think what I wrote here is essentially correct, and the confusion arises because the actual gauge transformations and the action of a diffeomorphism just look really similar

just use this handy diagram

@JohnRennie I have provided a few updates on yhe problem we were discussing

@Slereah is that nLab

I understood the id

yes

Hey everybody

I have a quick question

There's an object with mass m. This object only moves along the vertical axis.

I call that axis y-axis.

This object is being thrown at y=0 with speed $V_0$

There is gravity and air resistance.

Air resistance is given by $\vec{F}=-m\alpha \vec{V}$ and $\alpha$ is reel number.

So the net force acting on this body is $\vec{F}_{net}=-m\alpha\vec{V}-mg\hat{y}$, right?

If $\vec{V}=V(t)\hat{y}$ then $\vec{F}_{net}=-m\alpha V(t)\hat{y}-mg\hat{y}$

Thus, $\vec{F}_{net}=-m(\alpha V(t)+g)\hat{y}$

So the acceleration of this mass is $\vec{a}=(\alpha V(t)+g)\hat{y}$

I tried to find the $\vec{V(t)}$ of this body but I couldn't.

here we go again :-)

any internet sleuths around?

en.wikipedia.org/w/…

the Wikipedia page on the new SI used to have a nice table listing the impact on all sorts of constants

but it's now gone from there

does anybody have any idea where it went?

ah hah

tagged on en.wikipedia.org/w/… as original research

@EmilioPisanty What does impact of a constant mean?

@RenatoRenatoRenato impact on a constant.

@EmilioPisanty Ah sorry, so you meant how the constants were affected by the new SI conventions? My misreading surely didn't help my understanding :)

removed on en.wikipedia.org/w/…

@RenatoRenatoRenato yes.

relevant talk section en.wikipedia.org/wiki/…

@rob do you know what's the latest published CODATA set?

i.e. is CODATA 2018 available yet?

I have question. Consider two wires at some distance.

Case-1 : First wire has current and 2nd wire no current what is the effect of coulombs force of wire to other.

Case 2: Both wires has current what is the effect of colomb force on each other.

In case-2 how do we know the force effect is due to magnetic force or coulombs force.

Pls pls explain.

@JohnRennie sir pls see my questions.

@123 Wires are electrically neutral when viewed from the outside, why would there be a Coulomb force at all?

@ACuriousMind thanks for reply. Pls consider one wire and another free electron shoot parallel to wire at some distance.

@NiharKarve I've resisted making that exact comment at all of these questions going "as we all know, <completely wrong explanation follows>" :P

Case-1 wire has current and another free static charge. What is the effect of colomb force on free charge.

@123 The wire is still electrically neutral, so...why would there be a Coulomb force at all?

you need at least two net charges attracting/repelling each other to have a Coulomb force

Aaahh I see. All the electric field cancel each other.

In case of electric current in wire.

@ACuriousMind what is the effect of magnetic field if wire has current and we shoot free electron parallel to the wire. What is the trajectory or motion of electron

@123 The magnetic force is $q v\times B$, so just plug in the velocity of the electron and the magnetic field from the wire and there's your answer

The trajectory is circular??

no, the magnetic field is circular

the force is perpendicular to both the field and the velocity

How the electron is moved under this.

if you don't know, look up the "right hand rule" for how to figure out the direction of cross products

Yes I know the right hand rule.

@123 how do you think it moves? you know the force, and you should know Newton's laws, so what are you uncertain about?

I am confused how do know the force feel by electron not due to Columb force which is some other force we said magnetic force

what does the origin of the force matter for how it moves under the influence of that force?

I am really confused with idea of magnetic field and magnetic force. How we observed experimentally magnetic field and magnetic force.

you already mentioned a classic experiment: take two wires and run currents through them, then observe that there's a force between them

What if both electric and magnetic force are there. What force feel by electron. And how do we know which is electric and which is magnetic.

in general, you have both electric and magnetic forces - the full electromagnetic force is $F = qE + qv\times B$

you can distinguish them by observing that the magnetic force depends on the velocity of the test charge, while the electric force does not

i.e. if you have a charge at rest in a magnetic field, there won't be any force on it, but once you start moving it, there will be one

Ookay.. Hm.. Is there any experiment which determined the magnetic field of wire due to current. Experiment should not determine magnetic force

I don't understand the question

Because we developed the idea of magnetic field and magnetic force separately.

And both are perpendicular to each other. How do we know where is magnetic field. Force can be easily determined by push pull effect.

How magnetic field determined by experiment.

by measuring the magnetic force on some current running through it - our best instruments for measuring magnetic field do that indirectly via the Hall effect

the fields are essentially defined by the forces they exert on test charges/current - you measure an electric field by observing the force it exerts on charges at rest inside it, and you measure a magnetic field by observing the force it exerts on current inside it

Tha

Many many thanks @ACuriousMind

Electric force and electric field is very easy and simple to understand. Because both act in same direction. But magnetic field and magnetic force is too much confusing.

@ACuriousMind your and Qmechanic's trademark phrases are rubbing off on me: I actively resist trying to say things like "What makes you think that...?" in a colloquial setting.

Plus I also ask myself, "Who are we?" at 2 in the morning

...good to know that that's one of my trademark phrases :P

> the latest available. Termed the "2018 CODATA recommended values," they ... became available on 20 May 2019 and replaced the 2014 CODATA set.

@Slereah you might find this interesting arxiv.org/abs/1209.3985

@RyanUnger Looks pretty neat