source: http://en.wikipedia.org/wiki/Vector_field
I just learned about vector fields and I want to shortly, hopefully correctly, describe them. They are a great technique to model phenomenas in physics, fields where on each point a certain force is being applied (force is a vector quantity). The space and system of all the points with the forces then is being called a field. It is either electromagnetic field, gravitational, electric etc. Vector field is a space where to each point (x0, y0) a vector (with certain magnitude and direction) is being assigned. A two dimensional, two variable vector function would look like this:
The i and j unit vectors set the x and y direction of the vector. So we have a function F of two variables x and y. For every point in the xy plane, there is a vector with its scalar and directional component, both dependent on the M and N function. An example of a vector field can look like this.
This is another example of vector field, where the vector function = F (x, y). A techniques and methods such as line integrals are useful in these kinds of model, calculating e.g. the work done when moving on a curve in a certain force field. The line integral is also useful for the proof of conservation of energy via the fundamental theorem of calculus.
So if we want to calculate the total work done, when moving in a field (if we think of the problem as in physics), then we need to calculate the line integral. It looks like this:
source: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/3.-double-integrals-and-line-integrals-in-the-plane/part-b-vector-fields-and-line-integrals/session-58-geometric-approach/MIT18_02SC_notes_28.pdf
We can think of the formula this way. It we take a small piece of the curve c (dr = (dx, dy), small displacement) and make a dot product with the vector function describing the field. Why a dot product?
A work is multiplication of force (in the direction of the movement of object) with the distance. We can draw this kind of picture:
But we need to find force in the direction of r. That's the component of F in the direction of r, which is magnitude(F) * cos(theta). There we write:
So this is why the line integral is a dot product between the force field and the infinitely small displacement. The result of the line integral is a scalar, a sum on the interval of the curve c makes is the result to this integral. The other formula is an alternation where it is derived from dr/dt = T ds/dt, where ds is arclenght and T is tangent unit vector.
Now I will try to describe the fundamental theorem for line integrals which describes the calculation of line integrals for the gradient vector functions. This applies only for the gradient fields.
source: http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/3.-double-integrals-and-line-integrals-in-the-plane/part-b-vector-fields-and-line-integrals/session-60-fundamental-theorem-for-line-integrals/MIT18_02SC_notes_29.pdf
We can rewrite the equation into a differential, which implies the logical sense of this theorem and proofs it.
Because the result of the integral is a subtraction of two points we can deduce two absolutely crucial things. One of them is the the path-independence of the curve in the field. If the end points of the curve are the same, then the integral (work done in the certain field) will be the same for any kind of curve with the same end/start point. The other one is the conservative force of the field. If the end/start points are the same, then the over all work done is equal to zero.
But we didn't look at the most crucial question of how can we determine wether the field is or isn't a gradient one? We can conclude it from these calculations.
We have a vector function F = (M, N) where M and N are partial derivatives of a function f(x, y). We know that the equation (on the right) with second derivatives is right, do if we differentiate M to the other variable than it is already partial derivative of, we get the equality. The final equation equal only if the vector field is gradient field. Another criterium though is that the vector function has to be continuous at every point of the field.
It's hard to imagine, but if let's have a gradient field. Then the absolutely crucial question is, what are each of the vectors gradients of? We call these functions potentials. They can be calculated through computing two line integrals (because the gradient functions are path independent in two points) or through antiderivatives and correct interpretation of fx and fy and their integrations and partial differentiations.
Importantly (I literally don't know nothing about it for now) is the Green's Theorem, putting into an equation line integrals on a closed curve and double integral of function's curl.
Object curl is very hard to describe (for me it's really hard to imagine it). If we have a vector function F = (M, N), then we can calculate it this way.
So if the curl of a function F is 0, then we can say that it is a gradient field. If it's not zero, then it describes how "curly" it is, the amount of rotation in the field (vague explanation). This is better explanation of the curl.
"In vector calculus, the curl is a vector operator that describes the infinitesimal rotation of a 3-dimensional vector field. At every point in the field, the curl of that field is represented by a vector. The attributes of this vector (length and direction) characterize the rotation at that point."
source: http://en.wikipedia.org/wiki/Curl_(mathematics)
And this is the equation of the Green's Theorem expressed in different way (and more detailed) than the one above.
source: http://en.wikipedia.org/wiki/Green's_theorem
This is shortly some interesting concept I came across when reading about vector fields. The more I studied the more complex it got and when I am trying to visualize curl or gradient field and it's proofs, applications I have really hard time. It's definitely challenging, because it has great applications and it's fun studying something so complex (at least for me).
Lukas Cerny, 11. 7. 2014
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