Good Morning sir 😊😊 @JohnRennie

@user8718165 morning :-)

@JohnRennie what are you doing sir?

@user8718165 discussing thermodynamics with Yuvraj

@JohnRennie Sir do you know 1 thing...

Yes?

@JohnRennie yeah sir...I saw...everyone is talking about thermodynamics these few days. Yesterday I saw tpg2114 talk about heat death :)

@JohnRennie sir, yesterday night I fully understood 2nd law at 1 AM sir...It took me 2 full days to get it

Good :-)

@JohnRennie hello sir...can I ask you a question?

You can ask though I'm still chatting with Yuvraj so I may be slow to answer.

@JohnRennie okay sir...

@JohnRennie sir, can we write 3rd law like $$\lim_{T\to 0}S=0$$

No, the entropy can be non-zero even at absolute zero.

Your equation is true for an ideal gas though.

@JohnRennie hello sir

@JohnRennie why sir...Its written that in a perfect crystalline solid its 0...I have a feeling that it can't be but why is it written sir?

Suppose you have a crystal like brass that is a mixture of two elements. The atoms of the two elements can be arranged in a disordered way even at absolute zero and there is entropy associated with this disrdoer.

@JohnRennie okay sir

@user8718165 You can do magic in such time. Been there, done that.

@PolarBear Hopefully, thanks! Well, how is it going ?

@JohnRennie hello sir :)

@JohnRennie Hello!

I'm working at the moment. Back in half an hour or so.

okay.

@JohnRennie, Hi sir :)

If you are free, kindly clarify this doubt - Is work-energy theorem valid in non-inertial frames? :: In non-inertial frame, I know I need to include the work done by pseudo forces. But is it really necessary for the work energy theorem to hold good in such frames?

@JohnRennie, Are you busy now sir?

@Intellex hi. I'm free for the next hour or so.

@JohnRennie Thank you sir. Could you please reply to my above message which is now below?

I've never thought about it, but yes I think the work energy theorem will apply in a non-inertial frame provided you include the work done by the fictitious forces.

@JohnRennie Thank you sir. May I ask one more doubt?

Yes

Recently, I read about the difference between centre of mass and centre of gravity. In one of the articles on the internet, it is said when the gravitational field is non-uniform, the centre of mass and centre of gravity of an object are different. According to my understanding, the COG moves towards the side having higher field. So for an infinite gradient of the force (eg. black holes), will the COG come out of the body or will it stay close (very close) to the bottom?

For reference: This is the article I read - wordpress.mrreid.org/2014/09/12/…

@JohnRennie, Simply, can the COG be outside the body unlike COM which is destined to be inside the body or its periphery (like a ring)?

The COG won't be outside the body. As you increase the gravitational gradient it will approach the bottom edge of the body.

@JohnRennie Ok sir. Even when the gradient is infinity?

The COG is basically just like the centre of mass except that when we sum $mgx$ the value of $g$ is a function of $x$.

@JohnRennie Fine sir. As it's a weighted mean it will stay always inside the body. Now understood sir. Thank you.

Another doubt sir - When COG moves down from the COM, toppling will become more easy or maintaining stability will be a great challenge right sir?

The nearer the COG is to the base the harder it is to topple the object. That's because you have to increase the angle of lean to move the COG outside the base.

@JohnRennie Ok sir. Thank you very much :)

Made a mistake while thinking

@JohnRennie, Sir, icebergs are like giant physical pendulums right?

Do you mean if you tilt an iceberg its rocking motion will be simple harmonic?

@JohnRennie Not necessarily SHM but periodic.

as they are irregular in shape

I think the motion is a good approximation to SHM for small angles, but just like a pendulum it becomes anharmonic at larger angles.

@JohnRennie Fine sir. The net buoyant force can be considered to act at the COM right sir?

And yes the irregular shape will cause deviations from SHM.

@Intellex Or does it act at the centre of buoyancy?

@JohnRennie Hi :-)

@JohnRennie Aren't they same?

@Intellex No, I don't think so. To be honest I can't remember how the centre of buoyancy is defined.

@AdvilSell hi :-)

@JohnRennie Ok sir. I'm in rotational mechanics. When I move on to buoyancy, I will clarify this and further doubts :)

OK :-)

@JohnRennie I have a doubt in QM again

Thank you sir. Bye :)

@AdvilSell yes?

the problem referred : ocw.mit.edu/courses/chemistry/5-61-physical-chemistry-fall-2007/…

here^

OK ... ?

the expectation value of x is coming out to be a/2

which is mathematically correct

but for the 2nd energy state we have a node at a/2

and thus the probability must be zero ?

by 2nd energy state I mean for $\psi_2$

The expectation value of x is the value you get if you do multiple measurements of x and then average the measurements. So it's basically the average position of the particle.

@JohnRennie Do these measurement span through different energy states also ?

It's certainly true that for the first excited state the probability density goes to zero at the middle of the box, but the particle is still equally likely to be in the left and right halves of the box, so the average position is going to be in the middle.

@JohnRennie Ah ! okay

@AdvilSell if the particle wavefunction is $\psi$ then the expectation value of x is $\langle x \rangle = \langle\psi|x|\psi\rangle = \int\psi^* x \psi dx$

@JohnRennie Hi there, let me know when you are free.

So if the particle is ina different energy state $\psi$ will be different.

@JohnRennie yes

The integral is over the wavefunction the particle has, not over all possible wavefunctions.

In fact all the wavefunctions will give an average position in the middle of the box because that's enforced by the symmetry.

@JohnRennie Can you explain enforced by symmetry ?

The potential is symmetric about a line through the centre of the box, and the wavefunction has to respect that symmetry. So the probability of finding the particle in the left and right halves of the box is the same. So when you calculate the average position it'll always be in the middle of the box.

@JohnRennie Ah ! okay got you , Thanks !!

:D

@Dante hi, I'm just answering a question in another room. Do you want to ask your question now and I'll reply as soon as I've finished in the other room.

Yes, works.

works?

You meant that you'll answer my question after you're done answering another question in other room right?

Yes

Yes, I'm fine with that.

completing the sentence, it's "another approach, and in particular you must not assume H is zero just because there is no free current in sight."

I'm having some idea about what the author is trying to say

What I didn't understand is, if $M$ exists only in the body of the bar magnet, how can it's divergence be non zero outside the magnet

@Dante I don't think it's saying $\nabla\cdot M$ is non-zero outside the magnet. It's just saying the divergence is not zero everywhere. And clearly it isn't since it's non-zero at the two ends of the magnet.

I need to go now. I'll be back tomorrow as usual.

okay.

@JohnRennie Oh, I think I get you.