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June 9[edit]

Help me![edit]

On the page covering the topic on Covariance and contravariance of vectors, https://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors, it has been said contravariance vectors have components that "transform as the coordinates do" and inversely to the transformation of the reference axes. What are the difference between (or the definition of) components of the vectors, coordinates of the vectors and reference axes of the vectors? Please help me with...

granzer92 — Preceding unsigned comment added by Granzer92 (talk • contribs) 12:38, 9 June 2018 (UTC)[reply]

First, I'm not really an expert on this but no one else seems to be answering. Second, what I do know comes from a math viewpoint and physicists seem to have a somewhat different take on this. Covariance and contravariance is not a property of the vectors themselves but of they way are represented as a row or column via a basis of the vector space. You should also understand something about duality of vector spaces, so I'll start there. If V is a vector space then its dual, V*, is the vector space of linear mappings from V to R. You can think of V being column vectors and V* being row vectors and the mapping associated with a row vector being matrix multiplication of the row and a column to get a number. Although V and V* have the same dimension, there isn't a natural isomorphism between the two, though there is a natural isomorphism between V and V** so generally you can think of V and V** as being the same thing. (I'm assuming everything is finite dimensional here since the infinite dimensional case adds another layer which is best avoided on the first pass.) An example of a dual vector is the gradient of a function of several variables, since given a direction vector the gradient turns a direction vector into a directional derivative, in other words the gradient maps a vector to a number so it's in the dual space. Now when you learned multivariable calculus they taught you that the gradient is a vector, but this is only true because of the dot product. Basically, with a dot product, you can turn any vector into a dual vector; for a vector w, define a dual vector w* by the mapping w*(v)=w⋅v. This is fine as long as you're in Euclidean space where you have a natural dot product, but in a more general context the definition of dot product is somewhat arbitrary and it's more natural to think of a gradient as belonging to the dual space. If you choose a basis of V, then you can represent any other vector v in V as a column vector of coefficients. If you multiply every element of the basis by some constant, the result on the coefficients will be that they are divided by the same number. In other words, the coefficient vector is transformed in the opposite way as the bases. So the column vector of coefficients is a contravariant representation since it varies the opposite way when the basis is varied. On the other hand, you can take the coordinates of a dual vector to be the result of applying it to each of the basis vectors. You would normally represent this as a row vector and get the nice property that you get the value of the map w*(v) by multiplying the representations of w* and v. In this case, if you multiply every element of the basis by a constant then the coordinates of a dual vector will also be multiplied by the constant. So the coefficient vector is transformed the same way as the basis and you say the it's a covariant representation. Generally you get vectors ↔ columns ↔ contravarient and dual vectors ↔ rows ↔ covarient, so there are multiple related ideas where that are hard to pick apart; consult a textbook if you want the formal definitions. I hope this explanation isn't too long but to me you can't really do just one third of the picture without going into the other two thirds. --RDBury (talk) 12:47, 10 June 2018 (UTC)[reply]

Hello, I have prepared a bunch of tutorials on GR and address the very question that you ask, from a somewhat different perspective from RDBury above. You can see it at https://www.youtube.com/watch?v=3N_s7NY4Trg&index=12&list=PL9_n3Tqzq9iWtgD8POJFdnVUCZ_zw6OiB

Best wishes, Robinh (talk) 20:39, 10 June 2018 (UTC)[reply]

All paths of fixed length between two points[edit]

Given two points $p_{1}=(0,0),p_{2}=(1,0)$ on the Cartesian (2D) plane and $d\geq 1$ , what is the shape formed by the envelope of all possible paths from $p_{1}$ to $p_{2}$ of length $d$ ? 68.0.147.114 (talk) 21:07, 9 June 2018 (UTC)[reply]

I think it would be simply the ellipse (and its interior) with foci at p_1 and p_2, with the fixed sum of distances to the foci being d. Any straight path from one focus to any point on the ellipse to the other focus is a valid path, while any point outside the ellipse is too far away. And all points within the ellipse are on valid paths, since a path through any such point can be squiggled enough to make it long enough. (If d=1, the ellipse collapses to the line segment between the two given points.) Loraof (talk) 23:26, 9 June 2018 (UTC)[reply]

To add to this, the ellipse is centred at (1/2,0) with semi major axis d/2 and semi minor axis sqrt(d^2 - 1)→86.176.219.234 (talk) 19:01, 11 June 2018 (UTC)[reply]

I believe the semi-minor axis b is actually half that—namely (1/2)sqrt(d^2–1). Consider the triangle with vertices F(0, 0), center C(1/2, 0), and highest point B(1/2, b). FB = d/2 by the defining property of being on the route of length d; and FC = 1/2. Then the Pythagorean theorem gives my result for b. Loraof (talk) 20:55, 11 June 2018 (UTC)[reply]

Agree, I forgot to show the half→86.176.219.234 (talk) 10:46, 13 June 2018 (UTC)[reply]