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?

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

Compare your first term to their second term

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

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

The implicit differentiation?

Oh

I see

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

@bolbteppa thanks for the insight :)

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

Yes.

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.

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

Hello World..

@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.

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

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

@JohnRennie Yes but why?

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.

Hi @JohnRennie Sir.

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

@JohnRennie

@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 :-)

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

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

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

Closed as "opinion-based"??

Reopen as "interesting-based" please :-)

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.

@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

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

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

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

@123 BF = C₂F + C₂B. Yes?

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

@JohnRennie you are using vector representations hee

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.

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

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

C_2 means point E

?

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

No @JohnRennie my bad writing it is point G

That point

Sorry about that

Ah, OK :-)

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

@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

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.

@JohnRennie Aah yes they are similar triangles.

How to proceed for G is equally divides BF?

I tried but nothing happened.

BG = BE cosθ. Yes?

Yes

And BF = BC cosθ

and we are told BE = ½BC

Yes

Yes okay

Is it obvious now?

Yes E is the midpoint of BC

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

And BF = BC cosθ so BG = ½BF

Yes

Let me see

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

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

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

The right hand rule is already pretty short ...

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

Yes

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

yes

@JohnRennie Thanks....

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

@JohnRennie Thanks

@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.

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

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.

@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 $q$.

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.

@JohnRennie

You mean what is 'its' value for m?

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?

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