Particle Accelerator

Guru Vishnu

08:19

@JohnRennie: Hi sir :-)

John Rennie

@GuruVishnu hi :-)

Guru Vishnu

Are you free now?

John Rennie

Yes :-)

Guru Vishnu

Ok sir :-)

I'm having a doubt about fixing the internal pressure and radius of a bubble:

We know that pressure difference between the concave and the convex sides of a bubble is given by $2S/R$. The external pressure is fixed. So for a particular liquid, is the radius of the bubble well defined? I feel it does depend on the internal pressure. Is it not possible to fix the internal pressure based on the external pressure and the surface tension, sir?

John Rennie

The internal pressure is given by:

$$ P_{int} = P_{ext} + \frac{2\sigma}{R} $$

Yes?

Guru Vishnu

08:26

Yes sir.

John Rennie

And obviously $P_{ext}$ is always one atmosphere - well, unless you're doing the experiment in a pressure chamber or on a different planet :-)

Or at the top of a mountain I suppose.

And the surface tension is a constant for any given liquid.

Guru Vishnu

Ok sir. I thought of explaining the stability of the bubble by fixing the internal pressure under the given conditions - fixed external pressure and fluid (surface tension).

John Rennie

So the only thing that can vary on the right hand side is the radius $R$

Guru Vishnu

I thought a bubble bursts because of small distortions in the surface and the internal pressure tries to open the film and cause the bubble to burst. The two factors - size and pressure oppose each other, and hence I expected the bubble can't be too tiny and at the same time it can't be too large.

John Rennie

Bubble stability is a different matter.

Bubbles burst when something punctures the bubble, or when air currents or whatever stretch the bubble too much and it in effect fractures.

That isn't necessarily related to the internal pressure.

Guru Vishnu

08:32

Ok sir. Thank you for the clarification :-)

So, it seems we can't predict what radius a bubble would have unless it's formed.

John Rennie

No, because it depends on how much air it contains inside it.

Guru Vishnu

Ok sir. I understood that. Thanks :-)

Guru Vishnu

09:10

The following statement is from the section "contact angles":

> A liquid in equilibrium cannot sustain tangential stress. The resultant force on any small part of the surface layer must be perpendicular to the surface there.

Regarding the tangential stress part, I found the following Q&A which also says the same:

1

Q: Why can't liquids oppose tangential forces acting on them?

Also, when a glass of water is inclined at a certain angle, why does the surface of water continue to remain parallel to ground?

If a liquid can't sustain tangential stress, how can surface tension exist in the first place? Or what does "cannot sustain tangential stress" mean, sir?

John Rennie

The term tangential stress is a bit misleading. Liquids cannot sustain any shear stress because when you apply a shear stress they just start flowing and they continue flowing until the shear stress has become zero.

The only stress a liquid can maintain is a compressive stress.

Guru Vishnu

@JohnRennie Sir, doesn't both shear stress and tangential stress mean the same thing? This is how I learnt it for solid materials. Or is the definition different for fluids?

John Rennie

Yes, though as a colloid scientist I'm used to using the term shear stress.

Guru Vishnu

Ok sir :-)

But that still doesn't avoid the presence of surface tension, right sir?

John Rennie

If you consider a soap film, e.g. a point on a bubble then the surface tension acts tangentially to the surface so it is a tangential force. Yes?

Guru Vishnu

09:18

Yes sir.

John Rennie

But the force acts in all directions around any point in the surface, so when you vector sum the forces at the point they sum to zero.

So the net tangential force due to the surface tension is zero.

Guru Vishnu

Ok sir. So is this what meant by "can't sustain tangential/shear stress" sir?

Even if they exist, they must sum to zero...

John Rennie

Yes. If the net tangential force was not zero the bubble would change shape until the force became zero.

Guru Vishnu

Ok sir. This is my observation: while playing with bubbles, I remember seeing a lot of Brownian motion like movement of dust (?) particles. Earlier, I thought these were the effects of surface tension and the existence of some tangential force. Now it seems that's just the normal Brownian motion which takes place in a beaker.

Have you observed this sir?

John Rennie

Bubble surfaces swirl around, but that's the effect of evaporation and air currents causing local forces. The liquid in the bubble flows around to balance these out.

Guru Vishnu

09:27

Ok sir. Thank you. Let me go through this transcript and the topic completely and see if my doubt still remains. So far, I understood your point.

John Rennie

OK :-)

Guru Vishnu

If the liquid can't sustain a shearing stress, why can it sustain a compressive stress? I understand imbalanced normal forces are the reason for microscopic turbulence near the surface. It's said that the free surface of the liquid must be in such a way that the resultant force (adhesive+cohesive+gravity) must be normal to it.

John Rennie

If you put a liquid in a piston and apply a force to it then there is now way for the liquid to flow so it can remove the stress. Yes?

Guru Vishnu

Ok sir. But could you tell what does this mean "so it can remove the stress"?

The pressure and hence the stress increases throughout the liquid as per Pascal's law.

John Rennie

Suppose I have a cube of water and I put it down on a flat surface. Then there is a shear stress on it due to gravity. So the water starts flowing. The cube flows into a puddle.

Guru Vishnu

09:43

Ok sir.

John Rennie

The water stops flowing when the shear stress has fallen to zero. In the case of a puddle that happens when the surface tension at the edges of the puddle balance out the gravitational forces.

Guru Vishnu

I don't know whether it's right to call that force near the interface as "surface tension"; Isn't that a combination of adhesive and cohesive forces apart from surface tension, sir?

John Rennie

It's the force acting at the interface between the edge of the puddle and the surface the puddle is on.

Guru Vishnu

Ok sir.

Guru Vishnu

10:29

@JohnRennie: Hi sir. Are you free now?

John Rennie

@GuruVishnu yes

Guru Vishnu

Fine sir.

2

Q: Why don't we include the adhesive and cohesive force while calculating rise in a capillary tube?

The contact angle of a liquid solid interface is explained by saying that the liquid surface must be perpendicular to the resultant of adhesive cohesive and gravitational forces acting on it, since it cannot sustain shear stresses. However, once the contact angle is determined, the cohesive and...

newtonian-mechanics surface-tension capillary-action

Do you have an alternate answer for the above question sir?

Particularly, I feel, I'm having the same doubt as the following comment:

I understand your point, the adhesive and cohesive forces play a role in determining the contact angle but surely that is not the end of the story. While writing Newton's Second Law for the fluid column why are we justified in omitting them even though they are at work? In the energy approach you illustrated, there would be some potential energy associated with these forces as well which has seemingly been omitted. Why? — Gerard Oct 23 '15 at 5:33

John Rennie

I'm not sure I understand what Gerard is asking

Guru Vishnu

Did you read the question sir? I think the comment alone is not self understandable.

Or let me explain:

As per Jurin's law,

$$h=\frac{2S\cos\theta}{r\rho g}$$

Why aren't cohesive and adhesive forces involved in the derivation of the above formula?

How can it be self-contained in the contact angle term sir?

John Rennie

They are included. The angle $\theta$ depends on the cohesive and adhesive forces.

Guru Vishnu

10:40

Yes sir. I understand. But I don't see why is $S$ involved here. My book states the free surface attracts the tube and this is a consequence of Newton's third law.

John Rennie

Why wouldn't S be involved?

I just don't understand the problem.

Guru Vishnu

Because it's just the force within the fluid molecules. The force between the liquid surface and the tube is adhesive force and not surface tension as per my understanding.

John Rennie

Have you seen the derivation of the contact angle?

Guru Vishnu

Derivation? No sir. I thought it was experimentally determined.

John Rennie

Let me draw a diagram ...

Guru Vishnu

10:45

So is the $\cos\theta$ term inversely proportional to $S$ in some way?

John Rennie

Guru Vishnu

@JohnRennie Only thing I know about it is this - the contact angle must be in such a way that the net force (adhesive+cohesive+gravitational) must be perpendicular (normal) to the tangential plane of the free surface of the liquid.

John Rennie

All interfaces have a surface tension because the surface tension is just the interfacial energy in disguise.

So when we look at the point where the liquid, air and solid all meet there are three forces acting as I've drawn. OK so far?

Guru Vishnu

@JohnRennie Ok sir. I think I have a terminology doubt - Isn't the force between solid and liquid and liquid with non-surface liquid molecules called as adhesive and cohesive forces respectively rather than surface tension?

John Rennie

No. The terms adhesive and cohesive forces are rather vague while surface tension is precisely defined.

Or a colloid scientist might also use the term interfacial energy because the interfacial energy and the surface tension are the same thing.

Guru Vishnu

10:53

Fine. I didn't know that. This seems new to me.

John Rennie

I can prove it if you want ...

Prove that interfacial energy and the surface tension are the same thing I mean.

Guru Vishnu

@JohnRennie Another doubt before I say ok : Why are the two horizontal forces horizontal? In the explanation of contact angles, the adhesive/cohesive forces weren't normal or parallel to the interface.

John Rennie

Because surface tension always acts in the plane of the interface.

Guru Vishnu

Ok sir. Now I think cohesive/adhesive forces are much different than interfacial energy.

@JohnRennie But aren't they dimensionally inconsistent? One is force per unit length and the other is energy.

John Rennie

Dimensionally inconsistent eh? :-)

Shall we check?

Guru Vishnu

10:58

Ok sir. I think I pressed the wrong switch :-)

John Rennie

(interfacial energy is energy per unit area not just energy)

Guru Vishnu

I agree. Surface energy per area is surface tension. Yes it's consistent dimensionally. Misinterpreted energy as pure energy.

John Rennie

:-)

Guru Vishnu

Fine sir :-)
If you wish shall we proceed regarding the main question?

6 mins ago, by John Rennie

Prove that interfacial energy and the surface tension are the same thing I mean.

John Rennie

OK, consider a soap film held inside a wire frame. I'll use this as an example because it's the sort of thing we've all played with so it should be intuitive. I'll draw a diagram:

@GuruVishnu there. The light blue area is the soap film.

Guru Vishnu

11:06

Ok sir. The blue rod is movable right?

John Rennie

The blue line is a sliding wire that we can pull outwards a distance $dx$ to stretch the film.

Guru Vishnu

Ok sir. I saw this setup to find the surface energy of a soap film.

John Rennie

In that case you've done the calculation!

Guru Vishnu

Yes sir. Now I see interfacial energy you meant is same as the surface energy I referred to. However:

13 mins ago, by John Rennie

Prove that interfacial energy and the surface tension are the same thing I mean.

John Rennie

That's what I'm about to do.

Guru Vishnu

11:08

Ok sir :-)

John Rennie

Let's call the length of the wire $\ell$. Calling it $d$ was a poor choice because that's going to get confused with the $dx$.

Guru Vishnu

Ok sir. Just to simplify things - Are we going to obtain $U/A=S$?

If yes, we can definitely skip this step.

John Rennie

So if we pull the wire a distance $dx$ we create a new area of soap film $dA = \ell dx$, and the increase in energy is $EdA = E\ell dx$ where $E$ is the energy per unit area. OK so far?

Guru Vishnu

I was just wondering how could $U$ and $S$ have the same dimensions if the interfacial energy and surface energy meant the same thing.

@JohnRennie Yes sir.

John Rennie

$U$ is the total energy?

i.e. in my notation $U = EA$ ?

Guru Vishnu

11:12

Increase in surface energy. I missed a $\Delta$ sir because HCV didn't mention it. I thought it was some kind of convention.

But he uses this interchangeably.

John Rennie

OK. Anyhow the working is trivial. The increase in energy has to come from the work done $E\ell dx = dW = Fdx$

Guru Vishnu

Finally he says: We see that the surface tension of a liquid is equal to the surface energy per unit surface area.

John Rennie

And $F = S\ell$

So we end up with $E=S$

Guru Vishnu

instead of increase in surface energy per unit surface area.

@JohnRennie Ok sir. So is $E=U/A$ or in English: Interfacial energy equal to the surface energy per unit area? I see it's just surface tension then. Just a new term to be stored in memory.

John Rennie

Yes

Guru Vishnu

11:16

Apart from the forces being along the surface, especially the horizontal ones, I find it's ok to proceed sir:

26 mins ago, by John Rennie

John Rennie

It's trivial from here. The horizontal component of the total force at the junction must be zero because if it wasn't zero the liquid would flow left or right depending on the total force.

So we get:

$$ \gamma_{sa} = \gamma_{ls} + \gamma_{la}\cos\theta $$

Rearrange to get $\cos\theta$

Guru Vishnu

Ok sir. Just for confirmation: Is $\gamma_{sa}$ adhesive, $\gamma_{ls}$ cohesive and $\gamma_{la}$ surface tension sir?

John Rennie

No

$\gamma_{sa}$ is the surface tension of the solid-air interface, which is probably easier to understand as the interfacial energy of the solid-air interface.

This is related to cohesion.

Guru Vishnu

Fine sir. It also solved the mystery why it's parallel to the surface :-)

John Rennie

Imagine taking a block of the solid and splitting it along a plane of area $A$. So where we had a solid-solid interface of area $A$ we now have two solid-air interfaces of area $A$.

Guru Vishnu

11:22

So all the $\gamma$ terms are just the surface tensions of appropriate phases on either sides like solid/gas or solid/liquid, and so on.

John Rennie

Yes.

Guru Vishnu

Ok sir. It seems we're just transforming the components of cohesive and adhesive forces to surface tensions. Similar to expressing 5 m/s of velocity in the first quadrant as 3i +4j. Is this a correct inference sir?

John Rennie

Basically yes

Guru Vishnu

Ok sir. Thank you :-)

But, the following doesn't seem to hold true:

42 mins ago, by Guru Vishnu

So is the $\cos\theta$ term inversely proportional to $S$ in some way?

It needs $\gamma_{ls}=S$ to be satisfied...

John Rennie

@GuruVishnu $S$ is what I've called $\gamma_{la}$. Yes?

So our equation for the contact angle is:

$$ \gamma_{sa} = \gamma_{ls} + S\cos\theta $$

$$ \cos\theta = \frac{\gamma_{sa} - \gamma_{ls}}{S} $$

$\cos\theta$ is indeed inversely proportional to $S$ if the other two surface tensions are constant.

Guru Vishnu

11:35

Ok sir. Fine. That was an error from my side when I switched from one set of terminologies to the other. I understood this completely. And the modified Jurin's law is as per my initial expectation, independent of $S$:
$$h=\frac{2(\gamma_{sa}-\gamma_{ls})}{r\rho g}$$
and has the cohesive and adhesive terms hidden inside the surface tension terms.

Thank you very much sir :-)

John Rennie

:-)

I have to go now.

Guru Vishnu

Ok. Bye sir.

Transcript for

♦ Particle Accelerator