The h Bar

schn

00:25

In section 10.2.2 in Griffith’s when deriving the Jefimenko equations, the curl of the retarded $\mathbf A$ is taken. Why is there a minus sign?

$\nabla \times f\mathbf a =(\nabla f) \times \mathbf{a}+f(\nabla \times \mathbf{a})$.

bolbteppa

Compare your first term to their second term

schn

It is analogous, but there is a minus sign in front of it.

bolbteppa

What's the other difference between your first term and their second term besides the minus sign

schn

The implicit differentiation?

Oh

I see

It’s simply swapped, of course, and then the minus sign

@bolbteppa thanks for the insight :)

bolbteppa

$\mathbf{a} \times \mathbf{b} = - \mathbf{b} \times \mathbf{a}$

schn

00:32

Yes.

schn

02:17

In Griffith’s electrodynamics section 10.3.1, he makes the point that “In Maxwell’s electrodynamics, formulated as it is in terms of charge and current densities, a point charge must be regarded as the limit of an extended charge, when the size goes to zero.” Why? In electrostatics, there were also charge and current densities and point charges were accepted.

bolbteppa

I think it's just trying to say that a point charge has to be interpreted in terms of the $\rho$ expression so you can't bypass the issue of retarded time

123

04:05

Hello World..

user 85795

04:18

in Mathematics, 8 mins ago, by copper.hat

print("hello world!")

satan 29

05:08

@ACuriousMind i mean that seems to be true, but then that suggests that there is a remarkable reconciliation between Ohm's and Gauss's law

both of them together imply that total - charge = total + charge

which to me, then implies that coulumb's (on which Gauss law was based) and Ohm's law aren't just independent "axioms" or "assumptions" but rather something like "fundamental truths", since together they are implying something so obvious.

Jack Rod

@ACuriousMind I read tilting of saturn axis can u help why is it so?

John Rennie

05:26

@JackRod isn't that due to the interaction with its moons?

Jack Rod

@JohnRennie Yes but why?

John Rennie

To be honest I don't know. I doubt there is a simple answer.

Saturn has an equatorial bulge because of its rotation, so it isn't a perfect sphere.

That means other objects can exert a torque on it by pulling on the equatorial bulge. The Moon does this to the Earth and that causes the precession of the axis of rotation, but I don't think the Moon changes the angle of tilt by much - it does change it a little.

123

Hi @JohnRennie Sir.

Jack Rod

Some studies claim that its tiltation will be doubled as soon as time passes

@JohnRennie

John Rennie

05:45

@JackRod We probably read the same article on it.

There as an article posted about a week ago about this wasn't there?

Saturn has several large moons, so the orbital dynamics are really complicated. I suspect the calculation of the axis change was done by a numerical calculation.

@123 hi :-)

bolbteppa

06:26

People have to stop closing questions like (even if it is just another unfortunate mistake) this so quickly

0

Q: Is there a "square root" of Klein–Gordon equation which is different from Dirac equation?

Legend has it that Dirac arrives at the Dirac equation (all equations are in Planck units $c= \hbar = 1$ in this post): $$ i\gamma^\mu\partial_\mu\psi - m\psi =0 $$ by taking the square root of the operator $$ \partial^\mu\partial_\mu $$ of the Klein–Gordon equation which is the relativistic vers...

John Rennie

@bolbteppa I'm interested to see if there are any answers to it and I have voted to reopen.

bolbteppa

Took out the 'shocking' bit and added a point which makes the complex conjugate attempt even more daunting :p

Nihar Karve

Closed as "opinion-based"??

user 85795

06:46

Reopen as "interesting-based" please :-)

123

07:09

1drv.ms/u/s!AozWlUoG8z4tigjfSUb3eXV66nSl

Pls see the link how do I find in isoceles triangle midsegment of equal side and median of third side intersection is the midpoint of third side.

bolbteppa

@vzn I didn't watch much of the video about that experiment, it looks like they are hyping a single solitary experiment when one knows one needs repeated experiments before any conclusions can be drawn, not to say it looked like a very special model, or is there more to it

123

I am stucked in solving question pls help me to figure this out. Any theorem or hint can help???

John Rennie

@123 If C2F = ½BF then C2F = C2B, so you just need to show C2B = ½BF

123

@JohnRennie sir I don't understand pls explain more.

John Rennie

@123 BF = C₂F + C₂B. Yes?

123

07:21

I tried but I don't came to the conclusion you suggest.

@JohnRennie you are using vector representations hee

John Rennie

No. BF is a straight line, and the point C₂ divides it into two parts, C₂F and C₂B, so the length of BF is the sum of the lengths of the two parts i.e. BF = C₂F + C₂B.

123

@JohnRennie sir if possible pls use my points. It will be easier for me to calculate things.

John Rennie

I am using your points. The points F, B and C₂ are the point you have marked on your diagram.

123

C_2 means point E

?

John Rennie

C₂ is the point you've labelled as C₂ in the middle of the triangle

123

07:29

No @JohnRennie my bad writing it is point G

John Rennie

That point

123

Sorry about that

John Rennie

Ah, OK :-)

OK, then if GF = ½ BF that implies GF = GB = ½BF.

123

@JohnRennie I need to prove G is midpoint of BF by other information. First we don't know G is midpoint of BF

We just know AB = BC, DE Midsegment, BF Median. This is what we know

John Rennie

What I'm saying is that if you can show BG = ½BF then you have proved G is the midpoint of BF.

And that's easy because BGE and BFC are similar triangles.

123

07:34

@JohnRennie Aah yes they are similar triangles.

How to proceed for G is equally divides BF?

I tried but nothing happened.

John Rennie

BG = BE cosθ. Yes?

123

Yes

John Rennie

And BF = BC cosθ

and we are told BE = ½BC

123

Yes

Yes okay

John Rennie

Is it obvious now?

123

07:47

Yes E is the midpoint of BC

John Rennie

So substitute for BE in the equation BG = BE cosθ to get BG = ½ BC cosθ

And BF = BC cosθ so BG = ½BF

123

Yes

Let me see

@JohnRennie you rock..... :-) Thank you

John Rennie

I need to drop out for about an hour now to work.

123

08:47

@JohnRennie For moment direction do we always use right hand rule. Or is there any shortcut.

John Rennie

The right hand rule is already pretty short ...

123

In questions there are so many forces acting in different directions. We always use right hand rule?

John Rennie

Yes

123

@JohnRennie Thanks. Because by moment direction I decide for plus and minus sign.

John Rennie

yes

123

08:49

@JohnRennie Thanks....

John Rennie

Torque is a vector and adds like any other vector. So calculate all the torques from the different forces and vector add them.

123

09:11

@JohnRennie Thanks

123

09:42

@JohnRennie if three forces are in equilibrium. I seen lami's theorem for that.

But the book used forces proportional to the length of same side of triangle where forces.

John Rennie

Isn't that basically the same thing. It's because the three forces forma triangle that Lami's theorem applies.

123

Why book used forces proportional to the same length of same side for three forces in equilibrium.

1drv.ms/u/s!AozWlUoG8z4tiRGnqbh-1awLXWUJ

@JohnRennie pls see alternate solution at page-55

1drv.ms/u/s!AozWlUoG8z4tiRNqsb46LaSk10Fz

Here page-53 in last question started and diagram.

schn

11:24

@bolbteppa I see. In electrostatics, however, one started with $q$ and then went on to $\rho$. Now it seems to be the opposite. One starts with $\rho$ and can’t get $q$.

schn

12:20

In Griffith’s section 10.3.1, when proving why there is an extra factor in integrating over the charge density when it depends on the retarded time, he makes the argument that there can only ever be one point along the trajectory of the particle that “communicates” with the field point. Because if there were two such points, the component of the particles velocity towards $\mathbf r$ would be greater than $c$.

I don’t follow the geometric argument for this. How can one determine the velocity of the particle in a given direction?

The average speed from 1 to 2 would be the magnitude of the vector from 1 to 2 divided by the time it takes for the particle to go from 1 to 2.

user15072279

13:38

[![enter image description here][1]][1]

[1]: https://i.sstatic.net/2QejV.jpg

Note : S and S dash are just frames given where S is an inertial frame ( it is not accelerating ) .

From this proof of finding pseudo force , we can see that m$a_0$ is a pseudo force here. Now , I see pseudo force here is mass of S dash * acceleration of S dash with respect to S

. So , is it like Force on the S dash with respect to S. Then , what is value m for ?

user15072279

14:18

@JohnRennie

Prateek Mourya

16:52

You mean what is 'its' value for m?

M.N.Raia

17:59

Hello people! Hope you're all ok and healthy.

So, is it a problem when the Volume integral of Null energy condition is infinity when r-> \infty?

Physics Meta

18:11

0

Q: Is "mathpages.com" a trusted website?

It has occurred many times that, in my Qs/As, I refer the SE contributors to the contents published on www.mathpages.com. As far as I involved, the content is rational and plausible, however, I do not know to whom the website belongs or whether its content is unanimously trusted by academics. Is ...

discussion

user15072279

18:41

@PrateekMourya m has to be some body’s mass right ?

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