# Continuous vector field $F$ with $F(\vec x) - F(\vec y)$ parallel to $\vec x - \vec y$

In Goldstein's classical mechanics, he makes an interesting claim, that if there is a continuous vector field $$F$$ where $$F(\vec x) - F(\vec y)$$ is parallel to $$\vec x - \vec y$$, then $$F$$ must be a constant field.

We can attempt a proof by contradiction. If such a non-constant field exists, it's clear that we can first choose some point $$\vec x$$ and decompose our vector field, into a component $$F_{\vec x}^1$$ that always points towards (or away from) $$\vec x$$ and another vector field $$F^2_{\vec x}$$ that is constant and equal to $$F(\vec x)$$. We can then repeat the construction with some other point $$\vec y$$.

I've used most of the information from the problem hypothesis, except continuity, and I'm not so sure how continuity and the above paragraph will yield a contradiction.

Goldstein's claim is from his chapter on rigid body motion, in a discussion on angular velocity. The claim appears just before equation $$5.1$$ of the third edition.

• Let $F$ be a vector field with a basis ${e_1,...e_n}$ and let $F=f_1e_1+....+f_ne_n$, suppose that $F(X)-F(Y)$ is parallel to $X-Y$, so that there exists a real number $q$ such that $F(X)-F(Y)=q(X-Y)$, you can equate the components and each function difference is equal to the difference of their arguments times a constant, I'm not sure how to follow from here though. Jul 16 '20 at 17:18
• Also the continuity of $F$ implies the continuity the each function ${f_k}$ for all $k$ less than or equal to n. Jul 16 '20 at 17:22
• @Arsene1412 I think that it would have to be $q(\vec X, \vec Y)$ as the amount of scaling could vary between different pairs of vectors. The difference $F(x) - F(y)$ needs to be parallel to $x-y$, not a fixed multiple of it.
– user802859
Jul 16 '20 at 17:26
• I meant that at those specific points $X$ and $Y$, there is a constant such that the difference is a multiple of the difference of their arguments, I didn't imply that it was true for all possible pairs $X,Y$, seems like a good starting point. Jul 16 '20 at 17:29
• Also at those points $X,Y$ ones gets $f_k(x_k)-f_k(y_k)=q(x_k-y_k)$ Jul 16 '20 at 17:32

The claim is false, e.g. $$F(\vec{x}) = \vec{x}$$ satisfies the condition but is clearly not constant.
• Hmm... this is interesting. As it turns out, Goldstein made a different error in the second edition, which was corrected in the N. A. Lemos article. Goldstein proceeds to copy a line out of the Lemos article ($(\omega_1 - \omega_2) \times \mathbf{R} = 0$) and "correct" the error in the new, "revised," third edition, by making the mistake above! In fact, he even cites the article by N. A. Lemos at the bottom of the page.