0
$\begingroup$

In this question, we won't give two hoots about forces or acceleration in a fluid.

We'll just focus on velocity.

We have a given continuous velocity vector field that changes with time. However, the divergence is zero at any point in time and space.

We have a sphere of fluid of a uniform density within the same region of space.

The fluid consists of infinitesimally tiny particles and is effectively continuous.

The velocity vector field represents the velocities of all those infinitesimal particles within the sphere of fluid.

At first, the shape of the fluid boundary is spherical, as mentioned earlier.

However, as the fluid is moving, the shape of the overall boundary will change over time.

Since the divergence is zero, the density of the continuous fluid will never change.

So, with any given velocity vector field of zero divergence, how can we derive the equation for the shape and position of the fluid?

For example, a simple scenario would be a uniform velocity vector field whose magnitude increases over time. There is zero divergence.

The sphere of fluid in such a field would accelerate in one direction but its shape would not change.

$\endgroup$
  • $\begingroup$ Hi and welcome to the Physics SE! What are your thoughts? Please note that you are expected to have thoroughly searched for an answer before asking your question. And it's important to detail where you're stuck and why, in order to attract good answers. You can consider checking the advice on writing good questions. $\endgroup$ – stafusa Dec 26 '17 at 12:21
0
$\begingroup$

A typical way of doing this would be to put a set of imaginary particles at the surface of the sphere, and then predict the motion of each particle as a function of time using $$\frac{d\mathbf{x}}{dt}=\mathbf{v}(t,\mathbf{x})$$ This would involve numerical integration for each particle.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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