# Displacement after Time T in a Velocity Field

1. Mar 29, 2012

### Teg Veece

I have a velocity field that is static in time. At every location, x, there is a corresponding velocity vector.

I'm trying to work out the displacement of a particle after T seconds if I drop it into the velocity field at time t=0 and at location x_0.

I was thinking something along the lines of ∫v(x_0)dt between the limits of t=0 and t=T but it doesn't seem right because the integrand doesn't seem to depend on t at all (in fact v(x_0) is a constant so it can't be right). Maybe there's a double integral involved...

Anyone know what a suitable form for this equation would be?

2. Mar 29, 2012

### Hechima

For simplicity, you could consider a one dimensional case:

$\frac{dx}{dt}=v(x)$

This is a separable differential equation, so:

$\int_{x_0}^{x_f}\frac{dx}{v(x)}=\int_0^T dt$

But it gets you a solution for the final time, T.

$T = \int_{x_0}^{x_f}\frac{dx}{v(x)}$

If for a fixed x0, you can suppose that the total time is a function of the ending position.

$T = f(x_f)$, where this function is given by the right hand side of the equation above.

So, to get the ending position for a given total time, you would need to take the inverse of that function

$x_f = f^{-1}(T)$

In order to get a closed form for this, you need to plug in the velocity field. I don't think there's a way to get a general expression for an arbitrary velocity function.

Now, for a two dimensional problem you need to do this for each component of the position vector.