Why does div(v)=0 for a fluid means that the fluid is incompressible?

[asdf1]

why does div(v)=0 for a fluid means that the fluid is incompressible?

 

[Galileo]

The fluid doesn't diverge or compress, so that the density in a certain fluid element always stays the same.

Use the divergence theorem. Take an arbitrary volume region V of the fluid with surface S, then according to the divergence theorem:

$$\int_{S}\vec v\cdot d\vec a=\int_{V}\vec \nabla \cdot \vec v dV$$

Do you know how to interpret this for your fluid?

 

[asdf1]

$$\int_{S}\vec v\cdot d\vec a=\int_{V}\vec \nabla \cdot \vec v dV$$

Do you know how to interpret this for your fluid?

do you mean that the surface integral is the same as the volume integral?
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[Galileo]

No, the left integral is the fluid FLUX through the surface. The right one is a volume integral of the DIVERGENCE of the fluid.

Have you heard of the divergence theorem before?

 

[asdf1]

i looked the theroem up!
thank you!
so in other words, (the amount of fluid leaving)- (the amount of fluid entering)=0=incompressible...

 

[Galileo]

Right, that's basically it. Since div(v)=0 the flux through any closed surface is zero. The flux is a measure of how much fluid flows through the surface. Since the surface is closed there is no net accumulation of fluid in the volume.

 

[asdf1]

^_^
thank you very much!