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

  1. Dec 27, 2014 #1
    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. jcsd
  3. Dec 27, 2014 #2
    Hi,

    first they replace ##\frac{dx}{dt}## with ##v##, by definition. Then they use the chain rule of differentiation 'backwards'.
     
  4. Dec 27, 2014 #3
    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.
    Thank you for replying though!

    Specifically I don't understand how ##\dfrac{dv}{dx}v =\frac{d}{dx}\frac{v^2}{2} ##
     
  5. Dec 27, 2014 #4
    Its because ##\frac{d}{dx} v^2 = 2 v\cdot \frac{dv}{dx} ##. So you have to take care of the 2.
     
  6. Dec 27, 2014 #5
    Oh wow that is easy. I knew it was something simple, thank you!
     
  7. Dec 27, 2014 #6
    You're welcome.
     
  8. Dec 29, 2014 #7

    vanhees71

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    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.
     
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