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There is a wikipedia page about NS Existance and Smoothness

It seems to me that the Navier Stokes equations is wrong?
(because in one side of equal sign unit is $\frac {m}{s^2}$ but in other side it is $\frac {kg.m}{s^2}$) Navier Stokes Equation:

enter image description here

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closed as off-topic by Qmechanic Jun 27 '13 at 14:55

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Your second link doesn't work. What do you think is wrong with the NS equation on Wikipedia? –  Vibert Jun 27 '13 at 13:12
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I think this is off-topic, unless you can turn it into a much more specific question about the physical concepts involved. A better place to discuss it would be the talk page on Wikipedia. –  Nathaniel Jun 27 '13 at 13:14
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Ehm, what do you think is wrong? You should be specific. –  Bernhard Jun 27 '13 at 13:22
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This question appears to be off-topic because, as @Nathaniel said, the question belongs on the corresponding talk page of that Wikipedia entry. It is not the purpose of Phys.SE to correct all the typos of the Internet. –  Qmechanic Jun 27 '13 at 14:55
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@tpg2114: It was a suggested edit from an anonymous unregistered user. –  Qmechanic Jun 27 '13 at 15:06

1 Answer 1

Yes, it is incorrect. There should be a $\frac{1}{\rho}$ multiplying the $\nabla p$ term.

This form is the incompressible form where it is assumed $\rho$ is a constant. This allows it to be factored out of the derivatives on the left hand side, then both sides are divided by $\rho$. This is why there is kinematic viscosity, $\nu$ on the right and not molecular viscosity, $\mu$.

You must also define the force to be the body force (force*volume/mass), not the total force, otherwise there should be a $\frac{1}{\rho}$ factor there too.

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This is because the equations have been nondimensionalised, i.e. the density has been eliminated by scaling all the other units so that $\rho=1$. This is fairly common practice in fluid mechanics, particularly among mathematicians, who don't really care about the physical interpretation. –  Nathaniel Jun 27 '13 at 15:07
    
@Nathaniel I disagree, $\nu$ is defined on the linked page as the kinematic viscosity and nowhere does it say things were non-dimensionalized. I concede it is the non-conservative form which is why $\rho$ could be pulled out of the derivatives on the LHS, but there is nothing to indicate that it was non-dimensionalized. –  tpg2114 Jun 27 '13 at 15:10
    
Oh, ok. I did jump to that conclusion without really reading the page. –  Nathaniel Jun 27 '13 at 15:14
    
On the Wikipedia page for the Navier–Stokes existence and smoothness there is a note that addresses this discrepancy: "More precisely, p(x,t) is the pressure divided by the fluid density, and the density is constant for this incompressible and homogeneous fluid." –  OSE Jun 27 '13 at 15:56
    
Given the note, it is "correct" as written; however, I wouldn't present it that way as clearly it is confusing. But that's up to wikipedia to settle and outside of the scope we have here. –  tpg2114 Jun 27 '13 at 16:01

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