Recent content by JanEnClaesen

In other words, when we take for potential function instead of F the square root of (6F/6x)²+(6F/6y)² (in the particular case of two-dimensions). Does this lead to anything interesting?

That's all right, it's because the book doesn't seem to be well known that I posted about it, it's also out of stock on Amazon, so I haven't been able to order myself.

One of the advantages of not all speaking the same language is that a great, dead but legendary mathematician might still author a book. It's only recently translated, apparently: https://www.amazon.com/dp/1470417014/?tag=pfamazon01-20

Two spaceships approach an observer from an equal distance and from an opposite direction with an equal speed in the observer's intertial reference frame O. The speed of a spaceship in the intertial reference frame of the other spaceship S is 0.8c , what is the speed of one of the spaceships in...

Essentially, I was wondering whether a vector field could be considered as a surface or something more unitary in general. For example (x,y,(x²+y²)^(0.5)) is a cone and a (partial) vector field in space.

For example the surface (x,y,x²+y²), can for example surfaces be considered as one abstract 'vector' in some abstract 'vector'-space? The ' ' because surfaces might not be a vector space. For surfaces we can exceptionally define normal vectors at every point.

So it doesn't make sense to think of a scalar field as being 'conservative', in that the line integral between two points is path-independent? EDIT: a scalar field is probably almost never path-independent, unless it zero everywhere.

It's interesting that you call divergence the only scalar-valued first-order differential operator. This means that divergence is a necessary consequence, as opposed to an arbitary construct. Is there a way to see why divergence is the only possible first-order differential operator? Vector...

In more concrete terms: does it make sense to speak of the potential of a scalar field?

To every scalar field s(x,y) there corresponds a 'constant' vector field x = A s(x,y) and y = B s(x,y), where A,B are direction cosines. The vector field is only partially constant since only the directions, and not the magnitudes, which are equal to |f(x,y)|, of the field vectors are constant...

Are there smooth manifolds (excepting mirroring)? Basically you cut a shape in two parts and glue theme on another one. Generalising your construction: construct a shape with complementary protrusions (sort of a hermaphroditic shape), cut another shape along the protrusion plane and fit the two...

Is this so? I cannot think of a counter-example and it is too general a statement to prove.

It's a valid point, but if the first of a linear series of balls all traveling at the same velocity colllides with something immovable, it bounces back at the same velocity, thus bouncing with the second ball, however, when two equally heavy balls collide it's as if they go through one another...

Perhaps because y and y' are the values of y and y' at the boundaries of a path, in which case they do seem independent. Hence a natural follow-up question, instead of binding the positions, why don't we bind the velocities?

Why are y and y' treated as independent variables, while they are not? Another slightly related question: if ' = d/dt then df'/dg' = df/dg because f' = df/dg g', but if we differentiate f' to g' we implicitly assume that df/dg is independent of g', is it?