Problem Solving Strategies

John Rennie

5:49 AM

Deleted

@YuvrajSingh... moved! :-)

Yuvraj Singh...

@JohnRennie move this messages to chit chat.

John Rennie

Deleted

M. Guru Vishnu

6:21 AM

@JohnRennie, Good morning sir. Are you free now?

John Rennie

@M.GuruVishnu hi, yes I'm free.

M. Guru Vishnu

Sir, in the above diagram, what differences will occur if I merge the two batteries into one?

John Rennie

Merging the batteries would mean adding this extra wire:

i.e. you are connecting the two +ve terminals of the batteries together.

The question is whether any current flows through the red wire, because if no current flows through the wire you can add or remove the wire without affecting the circuit.

Yuvraj Singh...

It will short.

No current through red wire.

M. Guru Vishnu

@JohnRennie Yes sir. I understand no current will flow because ends are at different potential. I was referring to the following case:

I think it'll not make any difference, but I'm not sure about all of the consequences

John Rennie

6:34 AM

What you've drawn is the same as my diagram with the red wire

The red wire joins the two batteries in parallel.

M. Guru Vishnu

@JohnRennie, Thank you sir. I thought converting the two battery system to one might affect some parameters. But now I realised the battery are in imaginary parallel configuration due to the red wire.

@JohnRennie, Sir, may I ask one more doubt?

John Rennie

If no current flows through a wire ina circuit then you can add or remove that wire without affecting the circuit. This can be a useful trick to know.

@M.GuruVishnu yes, go ahead.

M. Guru Vishnu

@JohnRennie Thank you sir.

Could you please reply to the following message which I asked yesterday when you were not here?

17 hours ago, by M. Guru Vishnu

Is there any physical meaning of "Electric displacement vector" - $D=\epsilon_0\vec E+\vec P$?

John Rennie

Isn't it the total field inside a dielectric including the field due to the polarisationof the dielectric, or something like that?

M. Guru Vishnu

@JohnRennie Yes sir. $\vec E$ is the vector sum of original field and field due to polarization

John Rennie

6:42 AM

I don't remember to be honest ...

M. Guru Vishnu

@JohnRennie No problem sir. Did the following gave any hint?

16 hours ago, by M. Guru Vishnu

My thoughts on this: It has the dimensions of length^(-2) X charge i.e., similar to surface charge density. So, I think $\vec D$ must be something related to surface charge density, but which one is it referring to?

This concept seems very abstract. And I think it will make no difference if I skip this as it's easy for me to use to usual form of gauss law instead provoking the displacement vector

John Rennie

I don't think I've ever used the displacement vector, so I'm not sure when or how it could be useful.

M. Guru Vishnu

@JohnRennie Ok sir. Thank you :-)

M. Guru Vishnu

8:34 AM

@JohnRennie, Hi sir. Are you free?

John Rennie

@M.GuruVishnu hi, yes I'm free

M. Guru Vishnu

*Question:*A thin metallic plate is inserted between the plates of a parallel-plate capacitor of capacitance C as shown in the figure below. What is the capacitance of the setup? *My reasoning:*The given setup is no longer a capacitor and hence it's capacitance is zero. *Book's answer:*Indeterminate (with no explanation) *Final result:*I'm confused why is it not zero and is indeterminate. Could you please explain this sir?

:52956810 Hi. Yes. What about you?

John Rennie

I'd agree with you. Assuming the middle plate makes electrical contact with the top and bottom plates it's no longer a capacitor.

Ajay Mishra

@M.GuruVishnu I just wanted to ask a question , I thought John Rennie is not free.

M. Guru Vishnu

@JohnRennie But why should it be indeterminate sir? Is there any reasoning behind this?

John Rennie

8:37 AM

Maybe your book is saying it is indeterminate because it is not a capacitor and therefore it makes no sense to ask what the capacitance is.

Ajay Mishra

Because metals have infinite dielectric constant @M.GuruVishnu

Hey @JohnRennie I am just struggling with the expression of kinetic energy of the ball.

M. Guru Vishnu

@JohnRennie So a device which is not a capacitor doesn't have capacitance=0 but it's indeterminate. Am I correct sir? Is this how we define some quantities for something which do not exist?

Ajay Mishra

I am aware of the fact that the easiest way to solve the question is by considering the conservation of energy.

John Rennie

@M.GuruVishnu I suspect that's what your book means, though it's a silly question and I hope a question that bad would not appear in the JEE.

M. Guru Vishnu

@JohnRennie Fine sir. Thank you :-)

John Rennie

8:41 AM

@AjayMishra to answer the question you just need to show that $\ddot\theta = -k\theta$ for some constant $k$.

Ajay Mishra

Ya, I am aware of that, my problem lies in finding the expression of the kinetic energy.

John Rennie

The force is proportional to $\theta$ (in the small angle approximation) and the linear acceleration $a$ is related to the force in the usual way for a rolling object. Then the angular acceleration is just $\ddot\theta = ra$.

Ajay Mishra

But there would be frictional component which won't be constant, as the ball is rolling.

John Rennie

@AjayMishra there is no friction in rolling.

Well, not in ideal rolling.

Do you mean you just want the linear KE, instead of the sum of linear and angular KEs?

Ajay Mishra

I want the whole Energy expression, then differentiate it with respect to time, as the energy is constant, then perhaps, I can find a way through it.

user8718165

8:48 AM

Hello :-) @JohnRennie

John Rennie

I'm not sure I see where the problem is. You could either use conservation of energy to get the KE, or use your solution for $\theta(t)$ to calculate $v = r\dot\theta$.

@user8718165 hi :-) Classes over for the day?

Ajay Mishra

Well, the probelm is I got $PE = mg(R-r) \cos \theta$ and $KE = \dfrac{1}{2} m v^2 + 1/2 I \omega^2 , \text{Where I = \frac{2}{5} mr^2} $ , when I differentiate the expression and simplifiy I am getting $ \omega = \sqrt{\dfrac{5 gr }{7(R-r)^2} }$

But the given answer is $ \omega = \sqrt{\dfrac{5g}{7(R-r)}}$

John Rennie

Isn't the angular velocity going to depend on the initial displacement?

Ajay Mishra

9:04 AM

Oh, sorry! As usual physics is out of symbol. $\omega$ in the KE expression is angular velocity but the $\omega$ which is not in the KE expression is angular frequency.

I took angular velocity to be arbitrary.

John Rennie

> but the ω which is not in the KE expression is angular frequency.

Do you mean the frequency given by $\omega = 2\pi f$

I'm afraid you've managed to thoroughly confuse me now ...

Ajay Mishra

@JohnRennie Yes and this is used in the expression which does not involve KE.

Like in $\omega = \sqrt{\dfrac{5 gr }{7(R-r)^2} }$

John Rennie

How do you get the period by differentiating the KE?

Ajay Mishra

Let $U = KE + PE$ , for SHM $ \dfrac{\partial U}{\partial t} = 0$ , On using this in the equation I would get, $ mg (R-r) (-\sin(\theta)) \omega + \dfrac{7}{10} m r^2 (2 \omega) \dfrac{d \omega}{d t} = 0$

If I divide both sides by omega and use the approxiamation $ \sin \theta \approx \theta$

I would get something of this form

30 mins ago, by John Rennie

@AjayMishra to answer the question you just need to show that $\ddot\theta = -k\theta$ for some constant $k$.

John Rennie

Is your PE term correct?

Ajay Mishra

9:18 AM

I think something must be wrong, but I took the lowest point of the spherical bowl as my reference.

I just couldn't figure out the wrong step.

John Rennie

If we measure distance from the pivot point then at the bottom the distance is $R-r$ and at an angle $\theta$ it is $(R-r)\cos\theta$

So the change in height is $(R-r)(1-\cos\theta)$

And you'd use the approximation $\cos\theta \approx 1 - \theta^2/2$

Ajay Mishra

Oh, thanks. But it still it won't matter much. My answer would still be wrong.

John Rennie

Oh wait, sorry, you're calculating $d/dt(-\cos\theta) = \sin(\theta) \dot\theta$

Ajay Mishra

Yes.

$mg (R-r) (\theta) + \dfrac{7}{10} m r^2 (2) \dfrac{d \omega}{d t} = 0$ I will get <- this expression, which would yield the same answer.

Yuvraj Singh...

Typical jee question. But yes interesting.

Ajay Mishra

9:30 AM

I ain't a JEE aspirant. :P That's the main problem, I guess.

Yuvraj Singh...

@AjayMishra, yes.

John Rennie

@AjayMishra so you'll get:

$$ \ddot\theta = -\frac{g(R-r)}{\tfrac75 r^2} \theta $$

Ajay Mishra

Yes.

John Rennie

$$ \omega = \sqrt{\frac{5g(R-r)}{7r^2}} $$

Ajay Mishra

Oh, sorry, I got the same answer just exponentiated the wrong thing, but still this is not correct. :(

I just've no idea how those guy got that answer.

I mean this answer

46 mins ago, by Ajay Mishra

But the given answer is $ \omega = \sqrt{\dfrac{5g}{7(R-r)}}$

John Rennie

9:41 AM

That answer makes sense because in the limit to $R\to\infty$ you'd expect $\omega\to 0$, and the expression we got doesn't give that behaviour.

@AjayMishra AHA!

The PE term is $mg(R-r)(1-\cos\theta)$ and $d/dt$ gives $mg(R-r)\sin\theta \dot\theta$

But $\dot\theta \ne \omega$

Because $\omega$ is the angular velocity of the ball, not of the overall motion.

user434058

@JohnRennie Hi, I have a question! Can I ask right niw?

user434058

Now*

John Rennie

@FakeMod yes, go ahead

user434058

Wait a sec I will upload a pic

user434058

9:53 AM

@JohnRennie I was on a mobile device so it took a bit long Sorry for that!

John Rennie

I have done problems like that, but it was a long time ago and I can't now remember how it's done.

The rigorous way is to minimise the propagation time i.e. use the calculus of variations.

user434058

Now my specific doubt here is that:- when the light grazes the surface, at that moment the $\sin\theta$ will be $1$ and the refractive index will be the highest it possibly could be! So $n \sin\theta$ will have the maximum possible value.

user434058

@JohnRennie But can't we straight away use Snell's law

user434058

Continuing my doubt... Then how can the light have the same value of $n\sin\theta$ at any other point? Because already that value is possible for only one point and that value is equal to its maxima

user434058

@JohnRennie ??

John Rennie

10:00 AM

To be honest I'm not sure what you are asking ...

Suppose the radius of the object is $R$, then the question is saying that the distance of closest approch is $R$.

user434058

Yup!

user434058

But is this closest approach even possible

Ajay Mishra

@JohnRennie Oh! 1 symbol 3 meaning! thanks! I got the answer now.

John Rennie

@FakeMod so your trajectory looks like this.

@AjayMishra cool :-)

So $n$ varies continuously from $1$ to $1+r_s/R$, where $r_s = 2GM/c^2$

For very good physical reasons $r_s < R$ so the refractive index varies from 1 to something less than 2 as the light ray approaches.

user434058

@JohnRennie yes I am following you. Please continue

John Rennie

10:07 AM

So you're calculating the path of a light ray in a medium with a smoothly varying refractive index.

Which is a fairly standard problem but I can't remember how it's done.

user434058

Yes and here I suppose we should work out this question using polar coordinates

user434058

@JohnRennie I don't think I need a solution, I just need to know that how can this even happen? Let me explain my doubt!

John Rennie

@FakeMod OK ...

user434058

In the image you created(btw, I always wonder how you do dat), at the point of grazing, the angle that the light ray makes with the normal is $90^\circ$.

John Rennie

You mean to the normal at the surface of the spherical object? That's true, but I don't see the relevance.

user434058

10:12 AM

Also the refractive index at that point is the highest that the light ever passes throughout its journey.

user434058

@JohnRennie I mean the normal between the changing refractive indices

John Rennie

@FakeMod OK ...

user434058

I am trying to find that because I intend to use Snell's law

John Rennie

Snell's law isn't any use here

You need the equation for the deflection by a refractive index gradient.

user434058

So now at the grazing point, we have $n\sin\theta=1+\frac{r_s}{R}$. Right?

John Rennie

10:15 AM

OK, yes ...

user434058

Where $\theta$ is the incident angle at that point

John Rennie

OK ... but if you're thinking of using Snell's law that isn't applicable here

user434058

Now if we find any other point on the light's path, then at that point the value of $n\sin\theta < 1+\frac{r_s}{R}$.

user434058

@JohnRennie Wait! i didn't know that! But wuy?

user434058

Why*

John Rennie

10:19 AM

Snell's law applies when the light ray hits a discontinuity in refractive index. Here there are no discontinuities in the RI. The RI varies smoothly.

user434058

But can't we approximate this scenario as infinitesimally thick discontinuities

John Rennie

I don't think that's a useful approach

user434058

Do you have any suggestions on how to approch this? If necessary then pleas provide links to external sources where I can get to know how to solve these kind of questions.

John Rennie

@FakeMod where are you in the educational system? School? University?

I ask because I need to know what maths you know.

user434058

High school, but don't worry I am always learning new stuff

John Rennie

10:28 AM

I don't think you can solve this type of problem with high school maths. It's done using the calculus of variations and you don't learn this until you get to university.

user434058

I know vector calculus(learnt it from Griffiths), a bit of multivariable calculus and a bit of variational calculus

John Rennie

Have a read through these notes

They describe exactly this sort of problem.

But I fear you'll find it hard going.

user434058

@JohnRennie please go on. I know EL equations, Beltrami identity, variation under constraints and all the basics

John Rennie

Alternatively, the problem is basically the gravitational deflection of light in the linearised limit, so Google for articles on calculating the gravitational deflection of light.

user434058

@JohnRennie I think that is a nice idea!

John Rennie

10:32 AM

@FakeMod it's because the coordinate velocity of light due to a body off mass M is:

$$ v/c = 1 - \frac{2GM}{c^2 r} $$

user434058

And do I have to do that☝️ question it from the basics.... Like applying the principle of least time and then using E-L equation to find that path??

John Rennie

So $n = c/v = \left( 1 - \frac{2GM}{c^2 r} \right) ^{-1} \approx 1 + \frac{2GM}{c^2 r}$ in the limit where $1 \gg 2GM/c^2r$.

user434058

Because I don't think that it was meant to be that hard

John Rennie

@FakeMod that's the way it's normally done, though there may be a simpler approximation to it.

user434058

@JohnRennie Hmm.. That's where it came from.. i see.

John Rennie

10:35 AM

That $2GM/c^2$ is the radius of a black hole of mass $M$.

user434058

@JohnRennie It is meant for high school students in India. And moreover this book isn't considere that tough

John Rennie

@FakeMod then there must be a simple approximation that I don't know.

user434058

@JohnRennie Ohhhh! So does light stop in a black hole

John Rennie

@FakeMod it's ... complicated :-)

user434058

Haha!!

John Rennie

10:38 AM

The velocity we are calculating is called the coordinate velocity and coordinate velocities don't have a physical meaning except at the location of the observer.

user434058

@JohnRennie Pleas go on!

John Rennie

14

Q: GR. Einstein's 1911 Paper: On the Influence of Gravitation on the Propagation of Light

Regarding the paper, what does Einstein means when he says: "If we call the velocity of light at the origin of co-ordinates $c_0$, then the velocity of light $c$ at a location with the gravitation potential $\Phi$ will be given by the relation: $c = c_0\cdot\left(1+\frac{\Phi}{c^2}\right).$ T...

Read my answer to that question for a decent overview.

user434058

That deleted statement was meant for some hardcore physics,, but i think i can handle this

John Rennie

Also see this if you're interested in more background:

19

Q: Does light really travel more slowly near a massive body?

It is a routine problem for beginners in general relativity to calculate the coordinate velocity of light for the Schwarzschild metric. Starting from the metric: $$ ds^2 = -\left(1-\frac{r_s}{r}\right)c^2dt^2 + \frac{dr^2}{1-\frac{r_s}{r}} + d\Omega^2 $$ We use the fact that light travels on a ...

general-relativity black-holes speed-of-light coordinate-systems observers

user434058

This all seems fascinating, but I have my exams coming and I need to be really focused :) However once I get through, I will surely learn these concepts.

John Rennie

10:44 AM

Yes, for now I wouldn't start worrying about general relativity :-)

user434058

Good bye! See you later!

John Rennie

@FakeMod Bye

user434058

Omg!! Sorry I forgot to THANK YOU!!

user434058

@JohnRennie thanks a lot

John Rennie

@FakeMod you're welcome :-)

user434058

11:05 AM

@JohnRennie How do you make these?

John Rennie

@FakeMod docs.google.com/drawings/d/…

M. Guru Vishnu

@JohnRennie, May I ask a doubt regarding our previous discussion on capacitance of the system where two plates of a capacitor are joined diagonally by another metal plate?

John Rennie

@M.GuruVishnu yes

M. Guru Vishnu

@JohnRennie, Is it wrong to consider the capacitance of the system to be zero?

John Rennie

I think you might be reading too much into what was a rubbish question.

If something isn't a capacitor then asking what is its capacitance is arguably a meaningless question.

M. Guru Vishnu

11:11 AM

@JohnRennie I think you're right sir :-)

@JohnRennie Is there anything like "capacitance of a conductor"?

John Rennie

Yes, and that is called the self capacitance

Capacitance

Capacitance is the ratio of the change in an electric charge in a system to the corresponding change in its electric potential. There are two closely related notions of capacitance: self capacitance and mutual capacitance. Any object that can be electrically charged exhibits self capacitance. A material with a large self capacitance holds more electric charge at a given voltage than one with low capacitance. The notion of mutual capacitance is particularly important for understanding the operations of the capacitor, one of the three elementary linear electronic components (along with resistors...

M. Guru Vishnu

@JohnRennie Thank you sir. I didn't know that (or my book didn't tell me). I'm going to learn it now.

John Rennie

Self capacitance is a very simple concept. If you put some charge $Q$ on an isolated object and the voltage produced on the object (relative to infinity) is $V$ then the self capacitance s just $C = Q/V$.

The point is that it applies to single isolated objects not pairs of plates.

M. Guru Vishnu

@JohnRennie Thank you sir. I thought it something like electric displacement vector. But now it seems so simple (I think due to your explanation!).

John Rennie

Cool :-)

user8718165

11:23 AM

@JohnRennie hello sir :)

John Rennie

@user8718165 hi :-)

user8718165

@JohnRennie sorry! I didn't reply to you.

John Rennie

No problem :-)

user8718165

@JohnRennie then I was just in the middle of a class :) I had a chem class after that :)

John Rennie

Is that it for the day now, or are there more classes to come?

user8718165

11:25 AM

@JohnRennie nooooo

John Rennie

@user8718165 :-)

At least once the exams are done all this will be a thing of the past.

user8718165

@JohnRennie yeah sir...could we just move to the general room (these chats too) :-)

user8718165

4:34 PM

@JohnRennie hello sir :-)

@JohnRennie Is lunch finished sir :)

John Rennie

@user8718165 yes :-)

user8718165

@JohnRennie sir, can I ask you a little question?

John Rennie

@user8718165 yes, though I reserve the right to defer it until tomorrow if it's hard.

user8718165

@JohnRennie Sure sir :-) No worries :-)

@JohnRennie Sir, what is pressure energy? Is it simply the pressure?

John Rennie

There is no such thing as pressure energy.

user8718165

4:41 PM

@JohnRennie sir $P=F/A=\frac{Fx}{Ax}=$ energy/volume

@JohnRennie hello sir...

John Rennie

Because something has the same units as energy doesn't mean that something is energy.

e.g. both torque and work have units of Newton metres. It doesn't mean they are the same.

user8718165

@JohnRennie sir, I found this in Bernoulli's principle....

John Rennie

Though it's true that $dW = PdV$, and $dW$ is an energy.

user8718165

@JohnRennie yeah sir :-)

John Rennie

So $P = dW/dV$

user8718165

4:47 PM

@JohnRennie okay sir... yeah sir...it's written...

@JohnRennie so it's basically saying about this... Is it?

John Rennie

Actually that's right isn't it, because if we make $x$ in your equation small we get $P = Fdx/Adx$

And $Fdx = dW$ while $A dx = dV$

user8718165

@JohnRennie yeah sir... It's working out... But I'm not getting how is actually pressure related to energy... I mean it isn't energy by itself...right sir?

Mr. Xcoder

20

Q: What is Pressure Energy?

While deriving Bernoulli's Theorem, our teacher said that the sum of KE, PE and Pressure Energy per unit volume remains constant at any two points. $$P + \rho g h + \frac{\rho v^2}{2} = \text{Constant}$$ In this, he stated that the first term is Pressure Energy per unit Volume. What exactly is ...

fluid-dynamics pressure bernoulli-equation

user8718165

@Mr.Xcoder I saw this BTW....It helped.

@JohnRennie hello sir :) Did I make sense ?

John Rennie

@user8718165 pressure isn't energy. But it is the rate of change of energy with volume.

user8718165

4:54 PM

@JohnRennie thank you very much... Got it sir :-)

Transcript for

Problem Solving Strategies