- #1

SebastianRM

- 39

- 4

My question is, where does this equation comes from?

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- Thread starter SebastianRM
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- #1

SebastianRM

- 39

- 4

My question is, where does this equation comes from?

- #2

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A little more detail: A parametrized path through space is a function that given a value of the path parameter ##s## returns a point in space. With coordinates ##x, y, z##, this path can be described by three functions, ##x(s), y(s), z(s)##. The "tangent vector" to the path is what would be called "velocity" if the parameter ##s## were time: ##\frac{d\mathcal{P}}{ds} = V## where ##V## has components ##V^x = \frac{dx}{ds}##, ##V^y = \frac{dy}{ds}##, ##V^z = \frac{dz}{ds}##. The rate of change of ##U## as you follow path is given by:

##\frac{dU}{ds} = \frac{\partial U}{\partial x} V^x + \frac{\partial U}{\partial y} V^y + \frac{\partial U}{\partial z} V^z##

which is also called the "directional derivative" along ##V##.

The total derivative ##dU## is an operator which given a vector ##V## returns the directional derivative of ##U## along ##V##:

##dU(V) = \frac{\partial U}{\partial x} V^x + \frac{\partial U}{\partial y} V^y + \frac{\partial U}{\partial z} V^z##

The basis one-forms ##dx, dy, dz## in differential geometry don't mean what they mean in calculus, which would be an infinitesimal change in ##x, y, z##/. Instead, they are viewed as operators themselves, with the sort of trivial definition:

##dx(V) = V^x##

##dy(V) = V^y##

##dz(V) = V^z##

So ##dx## for example, is the special case of ##dU## where ##U(x,y,z) = x##

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