# Why is surface tension measured in units of milliNewtons per meter? Rather than square meter(s)?

Why is liquid surface tension written in units of mN/m, or milliNewtons per meter?

The related concept of surface energy for solids uses units of milliJoules per square meter.

## The units of surface tension

Well, just by inspecting your units closely:

$$1\; \mathrm{N} = 1\; \mathrm{J}/\mathrm{m}$$, so $$1 \;\mathrm{J}/\mathrm{m}^2 = 1\; \mathrm{N}/\mathrm{m}$$. They are the same units, and in this case they are measuring the same thing.

Measuring surface tension in force per unit length is the same as measuring it in terms of work per unit area. You can look at surface tension either as the work you need to do to create a surface of a given area, or the force that a film of liquid exerts on its free edges, like in this diagram from Wikipedia:

When we try to slide the right edge through an additional distance $$dx$$, we need to create a small extra area that is $$L dx$$ big. We need to perform a small amount of work $$dW$$ to do this: for a liquid, this small amount of work is proportional to the small area, with the coefficient of proportionality being the surface tension $$\gamma$$.

$$dW = \gamma L dx$$

What force would we need to apply through this distance? We should already know that work is force times distance:

$$dW = F dx$$

So the force we need to apply to stop the liquid film (like a burst bubble) from collapsing in on itself can be found from dividing out/cancelling the little extra distances $$dx$$:

$$F dx = \gamma L dx$$

$$F$$dx$$= \gamma L$$dx

$$F = \gamma L$$

so the force due to surface tension is proportional to the length of the edge $$L$$, and $$\gamma$$ must have units of force/length.

## A detour on bubbles

The thing that really confused me about this topic to begin with is we naturally think about surface tension as having something to do with pressure. A bubble is under some pressure, isn't it? We need to blow into it to blow it up. And if I carefully stick a straw into a bubble just right, the bubble will naturally deflate and blow air out of the straw.

If someone asked me why one needs to put effort (or work) into blowing up a soap bubble, and how they pressurise the air inside, I would probably casually say surface tension. But the real technical meaning of of "surface tension" is NOT "pressure created by a surface", it is "work needed to expand a surface".

So, if the real meaning "surface tension" is not pressure, what is the relationship between the two for my imaginary bubble?

Let's say we have a soap bubble floating in the air. I'm going to assume its spherical for simplicity, and write the radius $$R$$.

The actual soapy water in this bubble has two surfaces, an outside and an inside. The total area of soapy water in the bubble is

$$A = 2 \times 4 \pi R^2 = 8 \pi R^2$$

and the air inside has a volume

$$V = \frac{4\pi}{3} R^3$$

Let's slowly pump a small amount of air into the bubble, causing the radius $$R$$ to increase by a small amount $$dR$$. The total area of the bubble increases:

$$A \rightarrow A+dA, dA = 16\pi R dR$$

And a small amount of work $$dW$$ needs to be done on the surface to achieve this:

$$dW = \gamma dA = \gamma 16\pi R dR$$

Where does this work come from? The gas inside the bubble. Because the volume of the bubble is changing, the work that the gas inside does is $$p dV$$:

$$p dV = p 4\pi R^2 dR$$

(Really $$p$$ is the extra pressure above atmospheric, not the absolute pressure.) If the bubble is in equilibrium with the air inside (not wobbling -- no movement), all of the work done by the air inside goes into expanding the surface, so

$$dW = p dV$$

$$\gamma 16\pi R dR = p 4\pi R^2 dR$$

Cancel out common factors and you can solve for $$p$$, the excess pressure inside the bubble:

$$p = \frac{4 \gamma}{R}$$

Whenever we see surface tension resulting in a pressure, there is an associated length. Look: $$R$$ is on the bottom here, so the units of the right hand side are

(force/length) / length = force/(length x length) = force/area

Which is correct if we were expecting a pressure.

## Summary:

• Surface tension is the amount of work we need to do to create a surface.
• The units of surface tension are Energy/Area = Force/Length.
• Pressure is a force applied per unit area. Surface tension is not pressure.
• When surface tension causes a pressure to appear, the actual surface tension of the liquid will be divided by a length at some point.

Both are equivalent, since $$\text{mJ}=\text{mN}\cdot \text{m}$$ where $$\text{mJ}$$ is millijoules, $$\text{mN}$$ is millinewtons and $$\text{m}$$ is metres. Thus

$$\text{mJ}\cdot \text{m}^{-2}\equiv \text{mN}\cdot \text{m}^{-1}$$

You're correct that the two concepts are indeed representing the same quantity. If the surface tension of a surface is $$T$$ then it would need $$T l dx$$ amount of work to move its boundary of length $$l$$ by a distance $$dx$$. Since $$l dx$$ would be the increase in area $$dA$$, the amount of work needed to increase the area of the surface by a unit area would be simply $$T$$ (and this is what surface energy is, by definition, the amount of work needed to increase the surface area by a unit amount). So, the two quantities are the same.

However, there is nothing wrong in the units you mention. Since $$\text{Joule=Newton}\cdot\text{ meter}$$, the two units you mention are identical.

Notice, the surface tension of a liquid is the force acting per unit length of an imaginary line drawn on the free surface of the liquid (its unit is $$N/m$$). Furthermore, the surface tension force is small enough hence written in small unit $$mN/m$$. Example surface tension of water is $$72mN/m$$

The unit of surface energy is $$mJ/m^2$$ which is the work done to increase the free surface area by $$1$$ unit. It is measure in $$mJ/m^2$$ which is equivalent to $$mN/m$$