# Why is the kinetic energy of a fluid given as an integral?

The kinetic energy of a fluid occupying a region $$\Omega \subset \mathbb{R}^3$$ is given by

$$T = \frac{1}{2}\int_\Omega |v(x)|^2 dx.$$

I am looking for some physical intuition on where the above comes from given that the classical definition of kinetic energy is $$T = \frac{1}{2}mv^2$$. More specifically, I have two questions:

1. What happened to the mass $$m$$?
2. Why are we integrating?

My best guess is that the mass is taken to instead be a density, $$\rho$$, and we further assume that $$\rho = 1$$ (why is this justified?). In which case, for an infinitesimal amount of fluid, we find its average velocity $$v(x)$$ in that region and multiply the volume to get something like $$T = \frac{1}{2}mv^2$$. Repeating this over all infinitesimally small areas in $$\Omega$$ and summing them up amounts to taking the integral. If this is indeed correct, then my question reduces to why is taking $$\rho = 1$$ justified?

• Where did you find this equation? Setting $\rho = 1$ is unusual and suggests that the person who wrote this question was sloppy and simply forgot to write $\rho$. Commented Nov 11, 2022 at 22:07
• Why are we integrating? A fluid isn’t like one particle with one velocity. Fluids can have different velocities at different points. If you didn’t integrate, what point would you take the velocity at? And why that point instead of some other? Commented Nov 11, 2022 at 22:23
• The kinetic energy of everything is an integral. The simple equation is for point-like objects that don't exist in practice AND, in some cases, for other objects that can behave like point-like under certain circumstances. Commented Nov 12, 2022 at 16:44
• @knzhou Setting $\rho=1$ is extremely common. For incompressible flow the density is a completely uninteresting constant and we can get rid of it simply by declaring $P=p/\rho$ and we have just $d \boldsymbol{u}/dt= -\nabla P + \nu \nabla^2 \boldsymbol{u}$. When reasinoning about turbulent kinetic energy and similar, we usually just use the kinematic units end use $k = \boldsymbol{u}^\prime \cdot \boldsymbol{u}^\prime /2$ en.wikipedia.org/wiki/Turbulence_kinetic_energy It is just the tke per unit mass and not unit volume. The same thing in different units. Commented Nov 13, 2022 at 9:54
• @knzhou To be clearer, using the kinetic energy per unit mass is not the same as declaring $\rho=1$. I just said that setting it to one is often done, especially by mathematicians. Both approaches are very common and one uses the kinetic energy per unit mass even when density is not equal to one. Commented Nov 13, 2022 at 10:01

The kinetic energy of a fluid is the same as normal mechanics, $$T=mv^2/2$$. However, that's not generally useful as we don't usually have masses but densities, so we instead consider the kinetic energy density, $$\mathcal{T}\equiv\frac{1}{2}\rho v^2.$$ Then in order to get the total kinetic energy, you must integrate over all space, $$T=\int\mathcal{T}\,\mathrm{d}\mathcal{V}=\frac{1}{2}\int\rho v^2\,\mathrm{d}\mathcal{V}.$$ So to answer your enumerated questions, $$m$$ is absorbed by $$\rho$$ and you integrate over all space to get the total kinetic energy of the domain.

As to the subsequent question, I do not know why the source of your equation has neglected $$\rho$$. Presumably it was a typographical error, but that's not something someone can determine without seeing the source first hand. You may want to check other sources to confirm the equation.

1. The elementary kinetic energy, i.e. the kinetic energy of a small element of fluid is

$$dK = \dfrac{1}{2} dm |\mathbf{v}|^2 = \dfrac{1}{2} \rho |\mathbf{v}|^2 dV$$.

2. Kinetic energy is an additive physical quantity. To get the total kinetic energy of the fluid in the volume $$V$$, we need to add all the elementary contributions. The fluid in the volume $$V$$ is a continuous medium, and integration is the operation of summing over continuous functions.

Thus, the total kinetic energy of the fluid in volume $$V$$ is

$$K = \displaystyle \int_V \dfrac{1}{2} \rho |\mathbf{v}|^2 dV$$.

## Why are we integrating?

According to the no-slip boundary condition:

In fluid dynamics, the no-slip condition for viscous fluids assumes that at a solid boundary, the fluid will have zero velocity relative to the boundary.

Which means that if water flows in a long pipe, the outer droplets will have 0 velocity and 0 kinetic energy, while the center droplets will have the highest velocity and kinetic energy:

The only way to calculate the total kinetic energy of a given volume of water is to calculate the kinetic energy of every droplet. In other words: to integrate.