# Derivation of Bernoulli's equation via Newton's second law

1. Dec 27, 2014

### Shinaolord

In the derivation on Wikipedia, it says the following
$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v = \frac{d}{dx}[\frac{v^2}{2}]$

How do they go from the second to last to last equation? I've been trying to understand, but I think I'm just looking over something incredibly simple.

Last edited: Dec 27, 2014
2. Dec 27, 2014

### DarthMatter

Hi,

first they replace $\frac{dx}{dt}$ with $v$, by definition. Then they use the chain rule of differentiation 'backwards'.

3. Dec 27, 2014

### Shinaolord

Yes I know where the v comes from, I just don't see how they get the 1/2

Could you perhaps show what you mean mathematically? The chain rule backwards, that reminds me of integration by parts, but I don't really seem to understand what you mean.

Specifically I don't understand how $\dfrac{dv}{dx}v =\frac{d}{dx}\frac{v^2}{2}$

4. Dec 27, 2014

### DarthMatter

Its because $\frac{d}{dx} v^2 = 2 v\cdot \frac{dv}{dx}$. So you have to take care of the 2.

5. Dec 27, 2014

### Shinaolord

Oh wow that is easy. I knew it was something simple, thank you!

6. Dec 27, 2014

### DarthMatter

You're welcome.

7. Dec 29, 2014

### vanhees71

Well, it's a bit inconsistent, because on the one hand they use $\vec{x}$ in the sense of Euler coordinates of fluid dynamics and on the other they use it as the trajectory $\vec{x}(t)$ of the material fluid element. The correct definition of the "material time derivative" of the velocity (and of any other quantity in the Euler notation)
$$\mathrm{D}_t \vec{v}=\partial_t \vec{v} + (\vec{v} \cdot \vec{\nabla}) \vec{v}.$$
So the correct derivation is as follows
$$\mathrm{D}_t (\vec{v}^2)=2 \vec{v} \cdot \mathrm{D}_t \vec{v}.$$
For an incompressible fluid you have
$$\rho \mathrm{D}_t \vec{v}=-\vec{\nabla} p.$$
Multiplying this equation by $\vec{v}$ gives
$$\frac{\rho}{2} \mathrm{D}_t (\vec{v}^2)=-\vec{v} \cdot \vec{\nabla} p.$$
If now you are in a static situation, where the pressure is time-independent you get
$$\frac{\rho}{2} \mathrm{D}_t (\vec{v}^2)=-\mathrm{D}_t p.$$
Since $\rho=\text{const}$ for an incompressible fluid, you finally get by integration over time
$$\frac{\rho}{2} \vec{v}^2 + p=\text{const},$$
which is Bernoulli's Law in the most simple case.

The most general extension of this law is the energy balance of a fluid, which expresses the energy-conservation law.