Why volume is conserved but not the surface area?

[Adesh]

TL;DR

Why mass and volume is conserved but not the surface area?

A water drop of radius $10^{-2}$ m is broken into 1000 equal droplets. Calculate the gain in surface energy. Surface Tension of water is $0.075 ~N/m$.

So, for the solution of the above problem we need to know how much surface area (combining all 1000 droplets) have increased from the parent droplet and then use the relation between change in potential energy and change in area, $\Delta U = S\Delta A$.

For the surface area of each daughter (I don't know why "son" will not be appropriate) droplet we need to know the radius of each daughter droplet and we can find it using this relation
$$
1000 \times \frac{4}{3} \times \pi x^3 = \frac{4}{3} \pi (10^{-2})^3
$$
Where $x$ is the assumed radius of each droplets.

But the above equation is fundamentally an equation of conservation, total initial volume = total final volume. I want to know why we couldn't use the equation of surface area for finding the radius of daughter droplets? For example writing
$$
1000\times 4 \pi x^2= 4 \pi (10^{-2})^2
$$
Why we just assumed that volume will be conserved but total surface area will increase? What actually happened when the parent droplet got divided into equal 1000 droplets? I know mass cannot be lost/gained in this process (no evaporation) and hence mass got evenly distributed with conservation, but why the volume? If I remember correctly from my 9th Grade then "volume is the amount of space an object occupies" but why it needs to be conserved? If it is conserved then why not the surface area?

I need your explanation, please provide it.

 

[Dale]

If mass is conserved and density is constant then volume is also constant. Do you expect the density of the water to change when you split it into smaller drops?

 

[etotheipi]

If the density of the water is assumed constant (seems reasonable here), then conservation of volume follows from conservation of mass (caveat, of course mass is not fundamentally a conserved quantity, energy is, but that's sort of irrelevant here ).

The surface area for a given volume depends on the shape of the 3D object (there was a thread about a similar thing very recently in the Math forum). For instance, a sphere minimises the surface area for a given volume. Surface area is then not a conserved quantity. Which is intuitive, since you expect smaller things to have larger surface area to volume ratios.

Edit: Beaten by @Dale

 

[Adesh]

If mass is conserved and density is constant then volume is also constant. Do you expect the density of the water to change when you split it into smaller drops?

No, I think density is one of the fundamental properties of any substance. What if our parent droplet were to be a hollow object? Imagine a hollow sphere of some given radius, then if it is divided into 1000 equal parts then how can we reason that the volume will be conserved?

 

[etotheipi]

No, I think density is one of the fundamental properties of any substance. What if our parent droplet were to be a hollow object? Imagine a hollow sphere of some given radius, then if it is divided into 1000 equal parts then how can we reason that the volume will be conserved?

Take your hollowed out "sphere of water" and pour all of the water into a measuring cylinder. The volume of water is still going to be the same. But hopefully you can tell that the volume bounded by whatever sphere you had before isn't a conserved quantity unless you also include the air that was inside and assume that there were no changes of state.

 

[russ_watters]

No, I think density is one of the fundamental properties of any substance. What if our parent droplet were to be a hollow object? Imagine a hollow sphere of some given radius, then if it is divided into 1000 equal parts then how can we reason that the volume will be conserved?

It doesn't matter how complex the shape is unless it is too complex for you to correctly analyze. The situation is exactly the same.

...however, the surface area relationship will change if the parent ant child objects are different shapes.

 

[kuruman]

The original question was "Why mass and volume is conserved but not the surface area?". A simple experiment that one can safely do at home is to cut a loaf of bread in half. The mass and volume will be the same but the surface area is increased by the two "inner" areas generated by producing the two halves. The more slices you cut the more surface area you generate for the same mass and volume. It's the same idea for droplets except that they are spherical instead of rectangular.

 

[Adesh]

The original question was "Why mass and volume is conserved but not the surface area?". A simple experiment that one can safely do at home is to cut a loaf of bread in half. The mass and volume will be the same but the surface area is increased by the two "inner" areas generated by producing the two halves. The more slices you cut the more surface area you generate for the same mass and volume. It's the same idea for droplets except that they are spherical instead of rectangular.

Wow! That’s really a very nice demonstration (I did it in my mind). Can you please give me some intuition of how the surface area got increased in case of spherical droplets? (For bread you explained it too well)

 

[Dale]

No, I think density is one of the fundamental properties of any substance. What if our parent droplet were to be a hollow object? Imagine a hollow sphere of some given radius, then if it is divided into 1000 equal parts then how can we reason that the volume will be conserved?

Density is related to mass and volume by $\rho=m/V$ so $V=m/\rho$. Since $m$ and $\rho$ are constant $m/\rho=V$ is also constant.

 

[jbriggs444]

Wow! That’s really a very nice demonstration (I did it in my mind). Can you please give me some intuition of how the surface area got increased in case of spherical droplets? (For bread you explained it too well)

Start with a spherical water blob. Slice it three ways -- north-south, east-west, top-bottom so that you have eight smaller and misshapen blobs. Allow them to each relax into smaller spheres.

By inspection, you now have eight smaller spheres, each 1/2 the radius of the original so that their mass and volume are each 1/8 of the original.

Surface area of the original: $A_{\text{orig}}=4 \pi r^2$.
Total surface area of the eight: $A_{\text{new}}=32 \pi \frac{r}{2}^2= 8 \pi r^2$

 

[kuruman]

Wow! That’s really a very nice demonstration (I did it in my mind). Can you please give me some intuition of how the surface area got increased in case of spherical droplets? (For bread you explained it too well)

More generally, let the parent drop have radius $R$. Its volume and surface area will be $$V_0=\frac{4}{3}\pi R_0^3;~~~~S_0=4\pi R_0^2.$$ Now let's say we break that up into $N$ identical daughters of radius $R_N$. Then since the mass, density and volume don't change, $$V_0=\frac{4}{3}\pi R_0^3=N\frac{4}{3}\pi R_N^3~\rightarrow~R_N=\frac{R_0}{N^{1/3}}.$$The total surface area of the daughters is $$S_d=N\times 4 \pi R_N^2=4 \pi R_0^2 N^{1/3}$$ and the ratio of the daughter surface area to the parent surface area is $$\frac{S_d}{S_0}=\frac{4 \pi R_0^2 N^{1/3}}{4 \pi R_0^2}=N^{1/3}.$$Check: When $N=8$, the ratio is $2$ which agrees with @jbriggs444's example.

 

[etotheipi]

Conceptually, you could also imagine taking your giant droplet of water and spreading it across a table to create a very fine film. You could then (ignoring details about how many molecules thick you could actually make the film) let the thickness approach zero and see that the surface area can become arbitrarily large. Then you can imagine chopping that film into lots of tiny pieces to get your little droplets.

 

[kuruman]

Conceptually, you could also imagine taking your giant droplet of water and spreading it across a table to create a very fine film. You could then (ignoring details about how many molecules thick you could actually make the film) let the thickness approach zero and see that the surface area can become arbitrarily large. Then you can imagine chopping that film into lots of tiny pieces to get your little droplets.

A method used to estimate the volume of an oil spill on water is to measure the area of the spill and use thin film interference to figure out the average thickness.

 

[etotheipi]

A method used to estimate the volume of an oil spill on water is to measure the area of the spill and use thin film interference to figure out the average thickness.

That's a nice application, it's good to know theory is being put to good use! I wonder how accurately they can judge the colours. It looks like the pretty rainbow thin film interference is limited to very thin oil slicks...

 

[Adesh]

Thank you everyone for your nice explanations.

 

[vanhees71]

Of course all these arguments hold in the approximation that you consider water as being incompressible, which is the case with very good accuracy for any every-day situations of not too high pressure gradients. One has to keep in mind that the hydrodynamical equations consist of

(a) momentum conservation balance ->Euler (ideal flow) or Navier-Stokes (non-relativistic viscous flow)
(b) mass conservation -> mass conservation (continuity equation)
(c) an equation of state relating pressure and density

For an imcompressible fluid you assume the utmost simple equation of state, assuming
$$\rho=\text{const}.$$
The consequence of (b) then is
$$\partial_t \rho + \vec{\nabla}\cdot (\rho \vec{v}) = \rho \vec{\nabla} \cdot \vec{v}=0\; \Rightarrow\; \vec{\nabla} \cdot \vec{v}=0.$$
That the latter equation means that the volume of any material fluid element is conserved, can be seen as follows:

Define a fluid element $\mathrm{d}^3 x(t)$ at time $t$ and consider the same particles at an infinitesimal time later, $t+\mathrm{d} t$. The volume then is given by
$$\mathrm{d}^3 x(t+\mathrm{d} t) = \mathrm{d}^3 x \mathrm{det} [\partial_i (x_j+v_j \mathrm{d} t)] = \mathrm{d}^3 x \mathrm{det}(\underbrace{\delta_{ij}+ \partial_i v_j \mathrm{d} t}_{J_{ij}}).$$
Now expand the determinant up to order $\mathrm{d} t$. To get this think about the determinand being defined by successively expanding in terms of rows (or columns). Now our matrix has diagonal elements $J_{ii}=1+\partial_i v_i \mathrm{d} t$ (NO EINSTEIN SUMMATION) and off-diagonal elements $J_{ij}=\partial_i v_j \mathrm{d} t$. Thus the only contribution to order $\mathrm{d}t$ to the determinant is in the product of the diagonal elements, i.e.,
$$\mathrm{det} J_{ij} = \prod_{i=1}^{3} J_{ii} + \mathcal{O}(\mathrm{d} t^2) = \prod_{i=1}^3 (1+\mathrm{d} t \partial_i v_i)=1+\mathrm{d} t \sum_{i=1}^3 \partial_i v_i + \mathcal{O}(\mathrm{d} t^2).$$
From that we get
$$\mathrm{d}^3 x(t+\mathrm{d} t)=\mathrm{d}^3 x(t) (1+\vec{\nabla} \cdot \vec{v} \mathrm{d} t) + \mathcal{O}(\mathrm{d} t^2).$$
So finally
$$\mathrm{d}_t \mathrm{d}^3 x(t)=\mathrm{d}^3 x(t) \vec{\nabla} \cdot \vec{v}.$$
Thus $\vec{\nabla} \cdot \vec{v}=0$ implies that the volume of a given set of water molecules doesn't change with time, i.e., that the fluid is considered incompressible.

 

[A.T.]

Can you please give me some intuition of how the surface area got increased in case of spherical droplets?

For any 3D shape that is scaled uniformly, volume is proportional to length cubed, while surface area is proportional to length squared. So halving the droplet volume will not halve their surface areas, but the number of droplets will still double.

 

[vanhees71]

Take a simple example: Compare a sphere's surface to a cube's surface with the same volume. For the cube you have $V=L^3$ and the surface area $S_{\text{C}}=6 L^2$. A sphere of radius $a$ has $V=4 \pi a^3/3$ to make it equal in volume to the cube you thus have $4 \pi a^3/3=L^3$ or $a=[3/(4 \pi)]^{1/3} L$. The surface is $S_{\text{S}}=4 \pi a^2=(36 \pi)^{1/3}L^3 \simeq 0.8 S_{\text{C}}$.

BTW: a sphere has the smallest surface of all bodies with a given volume.

 

[Adesh]

Take a simple example: Compare a sphere's surface to a cube's surface with the same volume. For the cube you have $V=L^3$ and the surface area $S_{\text{C}}=6 L^3$. A sphere of radius $a$ has $V=4 \pi a^3/3$ to make it equal in volume to the cube you thus have $4 \pi a^3/3=L^3$ or $a=[3/(4 \pi)]^{1/3} L$. The surface is $S_{\text{S}}=4 \pi a^2=(36 \pi)^{1/3}L^3 \simeq 0.8 S_{\text{C}}$.

BTW: a sphere has the smallest surface of all bodies with a given volume.

I think there is a typo when you wrote $S_C = 6L^3$, it should be $L^2$

 

[vanhees71]

Of course, I corrected it.

 

[etotheipi]

$$\mathrm{d}^3 x(t+\mathrm{d} t) = \mathrm{d}^3 x \mathrm{det} [\partial_i (x_j+v_j \mathrm{d} t)]$$

I wondered if I understood this step correctly; is $J_{ij} = \partial_i (x_j+v_j \mathrm{d} t)$ a component of the transformation/enlargement matrix, in which case the determinant of $J$ is the enlargement scale factor for the volume element in the time $dt$? It is not obvious to me how such a matrix is derived from the velocity field at any given point in the fluid

 

[vanhees71]

You can make this more rigorous by introducing Lagrangian coordinates $\vec{x}_0$ for the fluid elements. Then
$$\vec{x}=\vec{x}(t,\vec{x}_0), \quad \vec{x}(0,\vec{x}_0)=\vec{x}_0$$
defines the trajectory of a fluid element consisting of a fixed piece of matter (a "material fluid element"). This is usually assumed to be a one-to-one differentiable mapping (a diffeomorphism) at any time, i.e., there is the inverse function
$$\vec{x}_0=\vec{x}_0(t,\vec{x}).$$
The velocity of each material fluid element is of course given by
$$\vec{v}_0(t,\vec{x}_0)=\partial_t \vec{x}(t,\vec{x}_0).$$
The relation to the usual field description is in terms of Eulerian coordinates, i.e., $\vec{v}(t,\vec{x})$ tells you the velocity of the fluid element which is at time, $t$ at position $\vec{x}$. The relation between these different descriptions of the flow (i.e., the mapping between Eulerian and Lagrangian coordinates) are
$$\vec{v}_0(t,\vec{x}_0)=\vec{v}[t,\vec{x}(\vec{x}_0,t)] \; \Leftrightarrow \; \vec{v}(t,\vec{x})=\vec{v}_0[t,\vec{x}_0(t,\vec{x})].$$

Now you can ask, what's the volume $\mathrm{d}^3 x$ a fluid element occupies which was at $t=0$ at $\vec{x}_0$ having occupied the volume $\mathrm{d}^3 x_0$. For this you need the Jacobian determinant, i.e.,
$$\mathrm{d}^3 x= \mathrm{d}^3 x_0 \det \left (\frac{\partial x_j}{\partial_{x0k}} \right)=\mathrm{d}^3 x_0 J.$$
You have (Einstein summation implies)
$$J=\epsilon_{abc} \frac{\partial x_a}{\partial x_{01}} \frac{\partial x_b}{\partial x_{02}} \frac{\partial x_c}{\partial x_{03}}=\epsilon_{abc} \partial_{01} x_a \partial_{02} x_b \partial_{03} x_c.$$
Then you want $\dot{J}$. This you get by using the product rule.

It's too much to write though so rather we express the result by introducing the adjoints of the matrix defined as $(-1)^{j+k} D_{jk}$ the subdeterminants $D_{jk}$ of the Jacobian matrix $J_{jk}$, i.e., the determinant of the $(2 \times 2)$-matrix you get from erasing the $j$-th row and the $k$-th column from $J_{jk}$. Then you have
$$J_{jk} D_{kl}=D_{jk} J_{kl} =J \delta_{jl},$$
because that just describes how to calculate the determinant by

(a) (first expression) expanding it in terms of the $j$-th row (for $j=l$) or, for $j \neq l$ by expanding the determinant of a matrix where the $l$-th column is substituted by the $j$-th column giving a matrix with two equal columns and thus zero determinant

or

(b) (2nd expression) expanding it in terms of the $l$-th column (for $j=l$) or, for $j \neq l$ by expanding the determinant of a matrix where the $l$-th row is substituted by the $j$-th row giving a matrix with two equal rows and thus zero determinant.

Given these adjoints, you simply get
$$\dot{J}=\partial_{0k} v_{0j} A_{jk} = \partial_{0k} x_l \partial_l v_j A_{jk} =J_{lj} A_{jk} \partial_l v_j=J \delta_{jl} \partial_l v_j=J \vec{\nabla} \cdot \vec{v},$$
and thus
$$\mathrm{d}_t \mathrm{d}^3 x = \mathrm{d}^3 x_0 J \vec{\nabla} \cdot \vec{v} = \mathrm{d}^3 x \vec{\nabla} \cdot \vec{v},$$
which is the same result as gotten by the shortcut in my previous posting.

From both considerations it's also easy to derive the "material time derivative" of an arbitrary quantity. In the field-theoretical view with Euler coordinates if you have some quantity $f(t,\vec{x})$ and want a time derivative which describes how this quantity changes for a material element changes with time you have to take into account that within an infinitesimal time $\mathrm{d} t$ the fluid element which is at time $t$ at position $\vec{x}$ has moved to the position $\vec{x}+\vec{v}(t,\vec{x}) \mathrm{d} t$, and thus the material time derivative is
$$\mathrm{d} t \mathrm{D}_t f(t,\vec{x})=f(t+\mathrm{d} t,\vec{x}+\vec{v}(t,\vec{x}) \mathrm{d} t)-f(t,\vec{x}) = \mathrm{d} t [\partial_t f(t,\vec{x})+\vec{v}(t,\vec{x}) \cdot \vec{\nabla} f(t,\vec{x})],$$
i.e.,
$$\mathrm{D}_t f(t,\vec{x})=\partial_t f(t,\vec{x}) + (\vec{v}(t,\vec{x}) \cdot \vec{\nabla}) f(t,\vec{x}).$$
You get the same result with introducing Lagrange coordinates and define
$$f_0(t,\vec{x}_0)=f(t,\vec{x}(t,\vec{x}_0).$$
Then the material time derivative is simply
$$\mathrm{D}_t f(t,\vec{x}) = \partial_t f_0(t,\vec{x}_0)=\partial_t f(t,\vec{x}) + \partial_t \vec{x}(t,\vec{x}_0) \cdot \vec{\nabla} f(t,\vec{x}) = \partial_t f(t,\vec{x}) + \vec{v}(t,\vec{x}) \cdot \vec{\nabla} f(t,\vec{x}).$$
Both leads to the same result of course.

Now you can also express easily mass conservation of a material fluid element. You have
$$\mathrm{d} m = \mathrm{d}^3 x \rho$$
and thus
$$\mathrm{D}_t \mathrm{d} m = \mathrm{d}^3 x [\vec{\nabla} \cdot \vec{v} \rho + \partial_t \rho +(\vec{v} \cdot \vec{\nabla} \rho]=\mathrm{d}^3 x [\partial_t \rho + \vec{\nabla} \cdot (\rho \vec{v})]=0.$$
In Lagrangian coordinates
$$\mathrm{d} m = \mathrm{d}^3 x \rho_0 = \mathrm{d}^3 \vec{x}_0 J \rho_0$$
and thus
$$\mathrm{d} m = \mathrm{d}^3 x_0 (\dot{J} \rho_0 + J \partial_t \rho_0)=\mathrm{d}^3 x_0 J(\vec{\nabla} \cdot \vec{v}+\mathrm{D}_t \rho)=\mathrm{d}^3 x (\vec{\nabla} \cdot \vec{v}+\mathrm{D}_t \rho)= \mathrm{d}^3 x [\partial_t \rho + \vec{\nabla} \cdot (\rho \vec{v})]=0,$$
which is of course again the same result as the direct method with Euler coordinates only.

Thus mass conservation is expressed by the continuity equation
$$\partial_t \rho + (\vec{v} \cdot \vec{\nabla}) \rho=0.$$

 

[Herman Trivilino]

Why we just assumed that volume will be conserved but total surface area will increase?

It's not an assumption. It's a conclusion reached from observation and experiment.

 

[etotheipi]

@vanhees71 very cool, I've never come across a derivation like that before! I have learned a lot about the Jacobian today. Does Sommerfeld also cover this sort of thing in his fluid dynamics lectures?

 

[vanhees71]

Yes, he does. You need the Lagrangian coordinates for the formulation of (ideal-)fluid dynamics in terms of the action principle, and it also allows for the most elucidating treatment of relativistic fluid dynamics in terms of the action principle, I've ever seen:

D. Soper, Classical Field Theory, Dover (2008)

Otherwise the best book on non-relativistic (and also with a brief chapter on non-relativistic) fluid dynamics in my opinion is vol. 6 of Landau&Lifshitz. Sommerfeld vol. 2 is great too of course.

 

[Adesh]

It's not an assumption. It's a conclusion reached from observation and experiment.

Okay, I want to know more about it. Is it that density experiment (means density would not change) or something else?

 

[etotheipi]

Yes, he does. You need the Lagrangian coordinates for the formulation of (ideal-)fluid dynamics in terms of the action principle, and it also allows for the most elucidating treatment of relativistic fluid dynamics in terms of the action principle, I've ever seen:

D. Soper, Classical Field Theory, Dover (2008)

Otherwise the best book on non-relativistic (and also with a brief chapter on non-relativistic) fluid dynamics in my opinion is vol. 6 of Landau&Lifshitz. Sommerfeld vol. 2 is great too of course.

Thanks, I'll take a look! I hadn't come across the Lagrangian specification of a field before, but it seems fairly intuitive (quite similar conceptually to the $\vec{\xi} = \vec{x}(t_0)$, $\vec{x} = \chi(\vec{\xi}, t)$ of continuum mechanics).

 

[jbriggs444]

Okay, I want to know more about it. Is it that density experiment (means density would not change) or something else?

There are a variety of simple tests that can be done.

One such test is to take a cup of water and pour it into another. Volume is conserved no matter how you do the pouring. Stream, large droplets or small.

Or you can pour a pint of water into two cups. And back.

You can make cups with measured rectangular sides and compute a number: length * width * height and find that it matches the volume measure that you have observed to be conserved.

Another test would be to immerse a rock in a full cup of water and measure how much water overflows. Then shatter the rock and immerse the rubble in another full cup of water to see how much now overflows.

All of these tests and more have been done and have been done many times and to high precision. The results are reliable enough to be used in commerce and medicine. [The fellow who runs the gas station knows that how much fuel he purchases and puts in his tanks will pretty well match how much fuel his customers pump into their tanks. He trusts this relationship well enough to set prices based on conservation of volume]

Perhaps more interesting are the cases where volume is not conserved. Such as high pressure, wide temperature variations or compressable fluids. Or mixtures where, for instance, water plus sugar mix to form a solution whose volume is not the sum of the volumes of the solutes.

 

[Adesh]

Or you can pour a pint of water into two cups. And back.

Okay, I can take a tea cup full of water, then using this cup I filled two big coffee mugs with equal level. I can repour from coffee mugs to tea cup for ensuring that volume has not changed, but of course the surface of water in coffee mugs are wider. Is that correct ?

 

[jbriggs444]

Okay, I can take a tea cup full of water, then using this cup I filled two big coffee mugs with equal level. I can repour from coffee mugs to tea cup for ensuring that volume has not changed, but of course the surface of water in coffee mugs are wider. Is that correct ?

Yes. Almost certainly, the surface area of the two separate blobs of tea was greater than the surface area of the one combined blob.