# Isn't this wikipedia equation of navier-stokes actually wrong? [closed]

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:

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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
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
Ehm, what do you think is wrong? You should be specific. – Bernhard Jun 27 '13 at 13:22
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
@tpg2114: It was a suggested edit from an anonymous unregistered user. – Qmechanic Jun 27 '13 at 15:06

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.
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