Physics Help Forum Vector differentiation

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 Jun 24th 2008, 09:38 AM #1 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 Vector differentiation I just read a book which talks about vector differentiation. There are words like binormal,principal normal,torsion, curvature actually I don't understand why vectors can be differentiated.Could anyone explain a bit,please? I only know that for curves, they can be differentiated to have their tangents plotted as a new graph, but for vectors in physics, they are in three dimensional, what are the physical meanings? Vectors can be differentiated so does it mean that vectors are like some curves?
Jun 25th 2008, 04:32 AM   #2

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 Originally Posted by werehk I just read a book which talks about vector differentiation. There are words like binormal,principal normal,torsion, curvature actually I don't understand why vectors can be differentiated.Could anyone explain a bit,please? I only know that for curves, they can be differentiated to have their tangents plotted as a new graph, but for vectors in physics, they are in three dimensional, what are the physical meanings? Vectors can be differentiated so does it mean that vectors are like some curves?
Vectors are quantities in a given direction. Typically in Physics we use them in association with a given coordinate system so we can decompose the vector into components in the directions of the coordinate axes.

In 3D space there are three coordinate directions so the vectors have three components. Typically these components depend on time, so we have for a vector v(t):
$\displaystyle \bold{v}(t) = \left ( \begin{matrix} x(t) \\ y(t) \\ z(t) \end{matrix} \right )$

Torsion and curvature are properties of curves in 3D space. I'm not sure what a "binomial principle norm" is. (But there are any number of ways you can define a measure of "distance," all of which have different properties. It is likely one of these. If you would like to know more about this and you have some knowledge of linear vector spaces, look up what a metric tensor is.)

-Dan
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 Jun 25th 2008, 08:43 PM #3 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 For vectors with x-,y-,z- coordinates depending on time, must the curves be continuous?
Jun 26th 2008, 04:49 AM   #4

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 Originally Posted by werehk For vectors with x-,y-,z- coordinates depending on time, must the curves be continuous?
For any spacial motion obeying the principles of (and behind) General Relativity at least both the position and its velocity functions must be continuous. In Quantum Mechanics it is the wave function and its first derivative that are usually required to be continuous. (There is the case of a "delta function" potential that can be solved using the Schrodinger equation that gives a discontinuous first derivative, but I have never heard of an instance where an actual delta function potential exists.)

In GR if the position and velocity curves are not continuous then we can displace an object by a measurable distance instantly. This is forbidden because the object would then be "traveling" across the space interval faster than the speed of light. The discontinuity in the wave function causes technical problems with the probability distribution, which I don't readily recall.

-Dan
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