Consider the Minkowski metric in two-dimensions, for instance
d_k = (|x'-x|^k+|y'-y|^k)^{1/k}
One may define velocity as
Dr/Dt = ((Dx/Dt)^k+(Dy/Dt)^k){1/k},
where D denotes the change in a given quantity. I find it misleading when it comes to k = 1 space (taxicab geometry). In taxicab geometry, one may first go along the x-axis, then the y-axis, or first go along the y-axis and then the x-axis, or any combination of the segments along x- or y-axis. One segment may be longer than the other one and may take longer time to cover a given block.
I propose the following definition for velocity in k =1 space.
Dr/Dt = Dx/Dt1 +Dx/Dt2, where Dt = Dt1 + Dt2.
Interestingly this definition does not obey the distributive property.
1. Do you mean Dx/Dt1+Dy/Dt2? What are t1 and t2?
2. I didn't understand the problem. Originally, we have the velocity
vector Dr/Dt=(Dx/Dt,Dy/Dt) where r=r(t)=(x(t),y(t)) is a (smooth) trajectory.
The scalar |Dr/Dt|_k depends on the metric d_k. What's incorrect?
Dear Semenov, Thank you adding an answer. t1 and t2 are taxicab takes as move x-axis and y-axis. Where t = t1 + t2 is the total time the cab take to go from initial to final point.
Nothing is incorrect. I was just thinking to define a metric for any vector.
@Shahid Nawaz Yes, I see your intention. However, do you insist on the second term Dx/Dt2 in your definition or it is a misprint: Dy/Dt2 was meant? Next, will we be able to recover the 1-distance between two points knowing Dr/Dt (by integration over the total time t) or we restrict ourselves moving only horizontally/vertically with constant scalar velocity?