Vector Fields, Divergence, and Curl
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- เผยแพร่เมื่อ 5 ก.ย. 2024
- We know about vectors, and we know about functions, so we are ready to learn about vector fields. These are like functions that take in coordinates and give back vectors, so there will be a vector associated with every single point in the field. There are two things we must be able to do with vector fields as well, which involve the del operator and either the dot product or the cross product with the vector field. These give us the divergence and the curl of the vector field, respectively. Let's learn all about these operations now.
Script by Howard Whittle
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This 15 minute video was more informative than the 3 hours of lecture videos my school posted for my online class. Thanks for making great content.
I feel like you’re one of my classmate
Can't be more true !
Sameee
So true. I have an exam on mathematical methods tomorrow morning and this helped me summarize three months of lecture in fifteen minutes 😭
you're lucky because our prof never taught this in school 😂
50 years ago I was wrestling with these functions, especially the visual concepts of divergence and curl. Never once was a concise summary like this was laid out. My career never called upon me to use these but I felt compelled to remaster them in my retirement. And next I will apply them. It's never too late to get the satisfaction of learning.
There is a minor mistake in 9:50 the x^2 is negative otherwise your tutorials are awesome ✨. Thanks for your great effort for making such amazing videos for students like us.
he really should make a correction or risk students getting lost & wasting time trying to verify their answer against the answer of an εxpεrt.
im glad i caught that and that you commented about it. small but crucial mistake
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This channel is a gold mine for MultiCalc
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in the same boat lol, gl with your 4th year
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there is a mistake in 9:50, the x^2
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9:40. I think for the vector k, its coefficient is (-1/y - x^2) rather than (-1/y +x^2)
yes i think so too
Yes I got that too
Hey, at 9:40 or so, why did the minuses in the square brackets become pluses when you did the partial differentiation?
bro, im a software engineer and i didnt study any maths past Calc 1 yet this explanation is really thorough that I could really follow along and understand the concepts. Your ability to teach is truly amazing
You understood this without learning linear algebra!? You must be a GENIUS!!!
@Prodigious147 You must have at least studied vector calculus, right? If not then I have every right to suppose that the age of geniuses is near.
@Prodigious147 Hmm... then all I can do is nothing but wish you best of luck for your mathematical journey.
PS: You can start by watching professor Leonard's lectures on calc. He has some of the best lectures around on the entire YT.
So clearly explained ! Thank you so much !
I got the comprehension right. del dot F = (d/dx of x + z^2)+(d/dy of y/xz)+(d/dz of zlny) = the scalar value and it is a measurement of divergence and a positive divergence means that more of it is leaving that converging, a negative divergence means that more of it is converging that leaving and a 0 divergence means none of it is diverging, all of it is converging. It is fluid or air.
del x F = taking the dimensional determinant involving i j k on the top row, d/dx d/dy and d/dz on the second row and x + z^2 y/xz and zlny on the third row and the answer will be the orthogonal curl vector. The magnitude length of the curl vector is the strength of the curl of the flow of it.
Thank you
Thank You, Sir. It is much more helpful then any 1 hour videos , I understand it finally after this video 👍😃
You’re a much better teacher than my college professor
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When you took the determinant, the K-hat component had +x squared, not -x squared. Was that intentional? If so, why?
@@larry23100 yes I got that too
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Well it’s just that my videos tend to accumulate views very slowly over several years, they don’t get much right away like normal channels. But yes more support is good, please tell your friends to watch and subscribe!
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Vector fields: defines a vector at each point in space. Made up of scalar fields.
Del: vector made up of differential operators.
The gradient of some function f is a vector field. -
If a vector field F can be written as a gradient of some function f, it is a conservative vector field and the function f is called as potential function for the vector field F.
Operations on vector field, F
Divergence: del dot F; results in a scalar field.
Curl: del cross F; only in 3 dimensions; results in another vector field; represents rotation of F - direction of curl = axis of rotation, mag of curl = mag of rotation.
Given that the second derivatives are continuous,
The curl of a conservative vector field is zero (zero vector).
The divergence of a curl is always zero.
The curl is fully able to exist outside of 3D (which should be obvious since reality is 4D). It just can't be represented as a vector field, but rather some other quantity. One way to generalize the curl to arbitrary dimensions is with the exterior or "wedge" product, which returns an oriented plane segment parallel to the two vector inputs rather than an oriented line segment orthogonal to them.
at timestamp 10:42, we assumed that "if f is a continuous partial derivative of second order" while at 11:00 we took "f" as first order partial derivative. Am I missing something in understanding it?
Thanks professor! questions: whenever we are calculating the curl of a vector field, is it always not continuous if the curl is not zero?
No. What will make either of these concepts non-continuous, is if there are non-differentiable points or paths in the original function. You will see this visually as a kink or a cusp if you make a graph of the original component of the vector field.
As an example, consider the vector field:
F =
Along the line x=0, the x-component of the vector field is a non-differentiable function. The divergence and curl along this line, is undefined. There will be jump-discontinuities, when you take derivatives of the x-component of the vector field to calculate divergence and curl.
Thank You So Much Sir.
very coherent... thank you for sharing your vision
Great explanation. Thanks.
Great refresher video
These short videos put lenthy university lectures to shame.
Thank you so much
Thank you. It was very beneficial.
hereafter about green's theorem, line integral, stokes' theorem
those are all coming!
@@ProfessorDaveExplains i forgot to mention that of surface integral
Awsome explanation!
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at 10:00, should be (-1/y - x^2) for the k component
yes
...and at 8:53, should be -j[d/dz(x^2y-d/dx(xyz)]
I didn't understand much, due to the fact thta it lacks graphing, but formthe rest is a spectacular work.
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10:51 "If f has continuous 2nd order partial derivatives then the curl of its gradient is zero"
How can we prove that a Conservative Vector field's gradient function f : [F(vector)=del f] has continuous 2nd order partial derivatives???
Edit : Apologies. Didn't watched further that the very next point was the proof ❤
Amazin explanation! Thank you Dave!
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Thank you.
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hello Prof,
can these directional vectors of the vector field intersect each other?
Good question, short answer is yes; th-cam.com/video/rB83DpBJQsE/w-d-xo.html for anyone else wondering.
Universities should use your teaching style to model how professors should teach in lectures. Students would be less frustrated when learning new concepts, and education would be a lot more fun.
Sir ,there is a mistake in one question (in curl example ) it should be -1/y -x^2
yeah you're correct
How did you get to the 9:43. I can't see how.
great
Thanks bro
Terrific!
why... why is there an operation that only works for 3D, it makes no sense... the dot product and cross product are 2 operations extremely dependent on the number of dimensions you have. but I mean in 2D you could have the cross of a single vector that would give you back a perpendicular vector, or if you're taking the cross products between 2 vectors in R^4 it'd return a whole plane perpendicular to the 2 vectors at the same time, which could be broken up further into 2 perpendicular vectors for the plane.
PS. I got no answer but I figured if anyone reads:
It's called wedge product. This is the real: vector = dot + wedge. (Aka. Parallel part plus orthogonal part)
Dot product and divergence work no matter how many dimensions you have. Dot product means multiply corresponding components, and add up the results. Divergence is the differential operator that is analogous ot a dot product.
Cross product and subsequently curl, are calculations that only work in 3 dimensions. Since we live in a 3-d universe, there are plenty of applications of these concepts to physical principles that govern our lives.
You can take a curl of a 2-dimensional vector field, and the result will be exclusively in the third direction, perpendicular to both of the original dimensions of the vector field.
The wedge product is indeed a viable alternative to the cross product. It returns an object usually called a "bivector," which acts as an oriented plane segment/area. In adding the dot product to the wedge product, I see you've discovered the geometric product, which between two vectors effectively gives an object that acts like a complex number in 2D and like a quaternion in 3D. (It does _not_ act like an octonion in 4D.) With this product, the divergence and curl of a vector field can be combined into a single complex-like object that I've seen called the "vector derivative." It also gives the shortest version of Maxwell's equation(s) that I've seen: ∇F = J
The change in the electromagnetic field is equal to the source density.
Why do you add k but minus j in the determinant?
When we expand a determinant we alternately use + and -
you are the best!
what's the difference between a scalar function... and an "ordinary scalar function?"
No difference. Just an adjective to emphasize that it isn't a vector field.
These are really good videos! Thank u a lot
Thank u! :*
What is the unit of curl and divergence?
It depends on what the vector field represents.
Let's assign an arbitrary unit of u, to the quantity represented by the vector field. Assume x, y, and z are all spatial dimensions measured in meters.
The units of divergence would therefore be u/m, and likewise for the unit of curl. The unit of second-order derivatives of the vector field, like the Laplacian, would be u/m^2
why did he have to consider P, Q, Q into those (x^2y, -x/y, xyz) when determining the curl?
They are placeholders so we don't need to write in the contents of the vector field's component functions. You could use any letters you want, but it is common for literature to use the P/Q/R trio in this context.
im learning this in multivariable calculus...before linear algebra :(
Finally!
How to find the scalar function If I know its gradient?
Integrate the x-component of the gradient. Call the arbitrary constant of integration C(y, z)
Integrate the y-component of the gradient. Cancel terms that are already common in the previous integral. Add terms that didn't exist in the previous integral, in place of C(y, z). Call the arbitrary constant of integration, D(x, z). Repeat for the z-component, and call the arbitrary constant of integration E(x, y). Add up the three results, cancelling terms in common as you do. Terms that are not in common, are terms that are part of the partial constant of integration functions, C(y,z), D(x,z), and E(x,y). When you get to the end of it, call the arbitrary constant of integration K, that is now no longer a function of x, y, or z. K can be any single number, that doesn't depend on any of our function inputs.
If the field is conservative, there will be plenty of terms that are common among each integral result. If the field is non-conservative, you will end up with contradictory terms.
As an example, suppose our scalar function is:
f(x, y, z) = x^2 + x*y*z + z*y^2 + z
Find its gradient, and call it F:
F = grad f(x, y, z)
F =
Integrate F's x-component:
int y*z = x^2 + x*y*z + C(z, y)
Integrate F's y-component
int x*z + 2*y*z = x*y*z + z*y^2 + D(x, z)
Notice that x*y*z appears in both of the above functions, which means we can cancel it in one of them, and add the two.
f = x^2 + x*y*z + z*y^2 + D(z)
Now integrate F's z-component
int x*y + y^2 + 1 = x*y*z + z*y^2 + z + E(x, y)
Combine the terms from all of the above, :
f = x^2 + x*y*z + z*y^2 + z + K
And you see we now have our original function, with the only difference being the arbitrary constant of integration K. There are an infinite number of potential functions for any given vector field, that all have an identical shape. This is why we have to define a datum of potential energy in physics, where potential energy is by definition zero, for it to be meaningful.
Pay close attention to the wording of the problem. If the problem simply says, "find *a* function f(x,y,z), such that grad f(x,y,z) = vector field", then it is OK to omit the arbitrary +K on the end. You can keep it there as a matter of principle, but you are technically correct if you omit it, or make up your own number to take its place. Because you found one function of the infinitely many possible answers. By contrast, if it says "find *the* potential function", then you need to include the +K on the end. The key difference be the article "a" vs "the", in the problem statement wording. Different books or classes may have different conventions for naming this constant. I learned to use K.
Most of the time when you use the potential function, you'll end up cancelling this K anyway. But there are some applications where it is of interest to keep it around, and solve for it via boundary conditions.
Can curl be taken in 7 dimensions?
The curl can be taken in any number of dimensions as long as you use an alternative to the cross product that generalizes nicely. Technically it's possible to take the curl in 1D, but it would always be 0.
Curl is ultimately a rotational measure, which looks like a scalar in 2D and like a vector in 3D, but behaves noticeably differently. One generalization of the curl gives its 4D version 6 components, which is notably different from the size of a vector in the same vector space.
Wow
لا فض فوك
cool
But i is a unit vector in x axis. How can it be with y? Accordingly how can you multiply x with j hat!?!?
A vector field in 2 dimensions in general, consists of two functions of both spatial coordinates, x and y.
So F = . Alternatively, F = P(x,y) * i-hat + Q(x,y) * j-hat
Both P and Q are functions of both spatial coordinates, and could contain either x, y, or a mixture of both in their definitions.
In his example, he is defining P(x, y) to equal y, and Q(x, y) to equal x. Thus, F = , or F = y * i-hat + x * j-hat. It is just a coincidence that P doesn't contain x, and that Q doesn't contain y.
Was that a flock of birds that flew over my head or...
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WIsh I'd found these 3 years ago when I was doing these modules in Uni. Now I've finished Uni and watching these for a recap 🥲
F**k my professor. And love you for breaking thse topics
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