0
$\begingroup$

What does the {-y,x} in this equation mean?

.5 {-y,x}/(x^2+y^2)
Improve this question
$\endgroup$
4
  • $\begingroup$ Evaluate your expressions and see what you get. Maybe try {x,y}+5 to get a hint. Note that () and {} have utterly different meanings. $\endgroup$
    – John Doty
    Commented Sep 26, 2020 at 18:57
  • 3
    $\begingroup$ {-y, x} is a List. Many operations such as Times, Plus, Divide, etc. have the attribute Listable and exploiting this attribute can simplify code. Without this attribute, the expression would have to be written as 0.5 #/(x^2+y^2)& /@ {-y, x} or Table[0.5 t/(x^2+y^2), {t, {-y, x}}]or some other equivalent. $\endgroup$
    – Bob Hanlon
    Commented Sep 26, 2020 at 19:30
  • $\begingroup$ It probably represents a vector field in the plane. It is the velocity field of a uniformly rotating plate with an angular velocity of one radian per second. The whole expression is an irrotational (curl = 0) rotatory flow in which inner circles rotate faster than outer circles. $\endgroup$
    – Michael E2
    Commented Sep 26, 2020 at 21:47
  • $\begingroup$ Normalize[Cross[{x,y}]]/2 $\endgroup$
    – chyanog
    Commented Sep 27, 2020 at 7:38

1 Answer 1

Reset to default
1
$\begingroup$

You can use .5 {-y, x}/(x^2 + y^2) // TraditionalForm to see that .5 {-y, x}/(x^2 + y^2) can be use to express the vector value function.

$$\left(\frac{-0.5y}{x^2+y^2},\frac{0.5x}{x^2+y^2}\right)$$

and actually,{-y, x}/(x^2 + y^2) is the gradient of the function ArcTan[y/x] or ArcTan[x,y]

Grad[ArcTan[x, y], {x, y}]
Grad[ArcTan[y/x], {x, y}] // Simplify
(* {-(y/(x^2 + y^2)), x/(x^2 + y^2)} *)


GraphicsRow[{VectorPlot[{-(y/(x^2 + y^2)), x/(
    x^2 + y^2)}, {x, y} ∈ Disk[]], 
  VectorPlot[
   Grad[ArcTan[x, y], {x, y}] // Evaluate, {x, y} ∈ Disk[]]}]

enter image description here

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.