# Transport Theorem

1. Dec 30, 2011

### SubZer0

Hi all,

Can anybody explain the Transport Theorem (http://en.wikipedia.org/wiki/Rotating_reference_frame#Relating_rotating_frames_to_stationary_frames) in more non-physicist terms? I simply can't wrap my head around the visual of this theorem, which has the gist of:

d/dt f = df/dt + (angular velocity) x f(t)

Where f(t) represents the position in time.

I simply cannot visualise the result. If I use an angular velocity vector of [0, 1, 0] (rotation around y axis), and a position of [0.5, 0, 0], the resulting derivative is [0, 0, -0.5]. I would at least expect a time derivative with changes of x, and z, instead of just z. Am I not understanding the actual *meaning* of the result? If df/dt = [0, 0, 0], I'm assuming that the rotation should result in a derivative vector of something like [x, 0, z], where x and z are some non-zero values?

Thanks!

2. Dec 30, 2011

### Andy Resnick

I'm more familiar with the Reynold Transport theorem (http://en.wikipedia.org/wiki/Reynolds_transport_theorem), but your post seems to be a special case.

The Reynolds transport theorem simply accounts for changes in the coordinate system (or changes to the shape of a volume) in addition to changes in a physical quantity. I suspect the same holds for you- the second term accounts for the rotating coordinate system.

3. Dec 30, 2011

### SubZer0

Thanks for the reply, Andy. I looked into some physics texts, and I think what was throwing me off was the fact that a rotation of a particle [1, 0, 0] around y [0, +/- 1, 0] in the equation (what you were referring to above as the derivative accounting for the rotation of the frame), ended up with a velocity which was [0, 0, +/- 1], which, when you think about it, a rotation with an angular velocity of [0, +/- 1, 0] *would* result in a velocity of the particle of [0, 0, +/- 1], ie, the particle is accelerating instantaneously directly forward/backward on the X/Z plane. I think what threw me was the fact that at that instant, I was assuming there would be a velocity which was *turning*, with both X and Z components. Realistically, this is not how the velocity would look at that instant, as the tangent of a particle [1, 0, 0] moving on a circular orbit is [0, 0, 1] at that instant.

So that seems to make sense to me now.