0 - 2020-04-13

Knight

02:42

@JohnRennie Sir, will you get free by 10:30 AM (UK time zone) :-) ?

In Sommerfeld’s Lectures on Theoretical Physics, Vol II, Chapter 2, Section 6, Page 43 we derive an expression for the equilibrium of liquids as $$ grad ~p = \mathbf F$$ Where $p$ is the pressure and $F$ is the exertnal force. Then he writes,
[ The equation above ]includes a very remarkable Theorem: equilibrium is only possible if the external force has a potential, that is, if $\mathbf F$!can be represented as the gradient of a scalar function: $$ \mathbf F = -grad ~U$$ Where the minus sign is prompted by the relation to the potential energy. The existence of the potential function $U$is n…

John Rennie

04:49

Consider the electric potential. This is a single values scalar potential and we get the electric field from it using $E = -dU/dx$.

Single valued means suppose we start at a point $x$ and transport a charge around a loop back to $x$ then the total work done on the charge has to be zero.

But now consider a magnetic field. Magnetic field lines form loops, and if we transported a magnetic charge round a loop and back to it's starting point the net work done would not be zero.

That means a magnetic potential is not single valued, so this would be an example of a scalar function that is not single valued.

Physically we take this to mean a magnetic potential cannot be defined and the magnetic field cannot be written as the gradient of some magnetic potential.

Knight

06:10

@JohnRennie Sir I’m so sorry but I couldn’t understand the explanation.

John Rennie

@Knight I'm working for a couple of hours - Monday mornings are always busy :-(

Knight

Yes sir. Hope you will get free before lunch :-)

John Rennie

08:37

@Knight hi, I'm free now

Math geek

09:31

-1

Q: What is the difference between $F_2$ and $W_A$?

Underlined statements are definitions of $F_1$ and $F_2.$ I am learning the basics of free body diagram. I am not able to judge the distinction between $F_2$ and $W_A$. $W_A$ is the weight of $A$ acting on $B$. $F_1$ is the reaction force of $B$ to $A$ as the consequence of Newton's th...

classical-mechanics

Math geek

10:39

sorry for the above link

in Mathematics, 2 mins ago, by Math geek

Knight

11:35

@JohnRennie Hello

John Rennie

@Knight hi :-)

Knight

How are you sir?

John Rennie

I'm good :-) How are you?

Knight

@JohnRennie I’m very fine. Can you explain me what that “single-valued” meant?

Please

John Rennie

Take the function $y = \arcsin(x)$

Knight

11:41

Okay

John Rennie

This isn't single valued because for every value of x there can by multiple values of y.

For example if $x=0$ then $y$ can be $0$, $\pi$, $2\pi$, $-\pi$ and so on. Yes?

Knight

Yes yes yes.

:)

John Rennie

In principle we could have a scalar field $U(\mathbf x)$ that was also multiple valued. Mathematically there's nothing wrong with this.

i.e. for every position $\mathbf x$ in space $U(\mathbf x)$ could have many different vaues.

The problem with this is that in physics it doesn't make sense to have a potential that can have multiple different values at the same point.

Knight

But sir in the above case we cannot call $arcsin ~x$ a function because a single $x$ results in multiple values of $y$, ha?

John Rennie

People have a tendency to use the term function in a vague sense to mean any form of mapping, single or multiple valued.

Knight

11:46

Yes, I understand.

1 min ago, by John Rennie

The problem with this is that in physics it doesn't make sense to have a potential that can have multiple different values at the same point.

^ is totally clear to me.

Math geek

I have confusion in taking free body diagram :(

May I get help

Knight

@Mathgeek Yes you will be helped but just a little later, sorry :-)

Math geek

okay sir

John Rennie

@Knight Sommerfeld is just saying that if you're going to write the force field as the derivative of a potential function $U$ then that function must be single valued.

Knight

@JohnRennie Doesn’t that always happen sir?

Means all he was saying is that $U$ cannot be any of the circular function and their inverses (to say a few)

?

John Rennie

11:50

The point I was making earlier is that if you attempt to define a potential for the magnetic field then you find the potential you end up with is multiple valued.

That is, you cannot write the magnetic field as the derivative of some scalar potential $U(\mathbf x)$.

Knight

If you’re not too busy (as I have understood the meaning of “single-valued”) can you please explain that magnetic field example of yours ?

John Rennie

This is going to get involved and I don't have time right now as I have to go soon.

Knight

:-) Thank you very much sir

I apologise for being too demanding

John Rennie

Do you know what a conservative field is?

Knight

Yes, work depends only on net displacement and curl of the field must vanish everywhere

John Rennie

11:55

Yes. So the integral between two points does not depend on the path taken. Yes?

Knight

Yes

John Rennie

But in magnetic fields the integral between two points can be different for different paths.

Knight

Yes sir got you.

John Rennie

That's the problem. It means the magnetic field is non-conservative.

Physically it happens because magnetic field lines form loops.

Knight

$$\int_{x}^{x} U(t) dt = U(x) - U(x) \neq 0$$

John Rennie

11:57

And if you get loops that always means the curl is non-zero.

Knight

Yes

John Rennie

@Knight yes, exactly, and that has to mean $U(x)$ is multiple valued.

Knight

Yes, if the integrals depends only on end values, and even if the end values are same there are no assurance that the integral will vanish, one example of such $U$ is $\frac{1}{ \sqrt {1-x^2}}$

$$\int_{0}^{0} \frac{1}{ \sqrt {1-x^2}} dx = \sin^{-1} (0) - \sin^{-1} (0) = 2 \pi $$

Thanks sir, thank you so much!

John Rennie

:-)

Knight

Have a good lunch sir !

And healthy too :)

John Rennie

12:03

I fast on Mondays

So no lunch for me today!

Knight

Oho!

Have a good fast sir! And healthy too!

LOL

@Mathgeek You there?

Math geek

yes

@Knight

Knight

$F_2$ is same in magnitude as $W_A$ but we didn’t write $F_2$ as $W_A$ because $F_2$ is a contact force, it is caused by $A$ on $B$ while $W_A$ is the force caused on $A$ by the earth. Does it help ? @Mathgeek

Math geek

My question is applying the freebody diagram to object $A$. we get $F_1-W_A=m_a a_A$

@Knight let me read sir

Knight

Ask away!

Math geek

12:11

For this diagram. shouldn't we apply one more force like $F_2$ here?

Knight

Why? And which?

Math geek

$W_A$ is the force caused by earth. Force apply on ground , let it be $F$ and its reaction force $N$.

Knight

$W_A$ is force on $A$ caused by the earth

Math geek

@Knight Yes

Knight

So, we got just two force on $A$, one caused by the earth and the other caused by the ground (normal force)

Math geek

12:17

@Knight why there is three forces above $F_1, W_A, F_2$?

Knight

Where?

Which diagram ?

Math geek

May I take 1 minute for thinking sir?

Knight

Yes

:-)

Math geek

This figure is clear to me

Knight

That’s nice

Math geek

12:22

$F_1$ is the reaction force from $B$ to $A$

Knight

Yes

I will be back in 3 minutes, don’t go anywhere

Math geek

okay sir

while looking at the figure 2

What are the forces acting on $B$ by $A$?

sorry

What are the forces acting on $B$ by $A$?

Suvitruf - Andrei Apanasik

You know, you can delete your own messages)

Knight

@Mathgeek $F_2$

Math geek

@Suvitruf-AndreiApanasik I was late to judge the message was irrelevent.

Knight

12:36

@Suvitruf-AndreiApanasik Hello!

Suvitruf - Andrei Apanasik

@Knight hi)

Knight

@Mathgeek The force on $B$ by $A$ is $F_2$ which is in same magnitude as the weight of A

@Suvitruf-AndreiApanasik It really feels good when some very nice people like you still have an avatar of Monica. I salute thee

Suvitruf - Andrei Apanasik

@Knight never forget (;

Knight

@Suvitruf-AndreiApanasik Yes because next could be us

@Suvitruf-AndreiApanasik Which Are you from?

Suvitruf - Andrei Apanasik

@Knight (｡•́︿•̀｡)

@Knight Russia.

Or, you mean, site? )

Knight

12:40

No, I really meant your country

@Suvitruf-AndreiApanasik Keep always coming here.

@Mathgeek Is your problem solved?

Math geek

@Knight Can we take $F_2$ as $W_A$ without ambiguity in the second diagram?

Knight

In calculation we can replace $F_2$ by $W_A$

Math geek

Knight

Yes

Math geek

here, we know by intuition, force exerted on $B$ by $A$ is only weight of $A$

Knight

12:46

Yes

Math geek

due to that reaction force $F_1$ acted by $B$ on $A$ is $F_1$

this solves my problem. Thank you very much.

Knight

You are welcome :)

Math geek

:)

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$Mathematics$ 0