Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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

  • This question does not appear to be about physics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

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.

share|cite|improve this answer
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.