I'm reading Feynman's lectures and I have a question regarding volume I chapter 11.

In this chapter, Feynman says that not every group of numbers is a vector. He says that if we apply an operation to some vectors the result is a vector if it's invariant towards a rotation. So if I define an operation over, say, vectors $\vec a$ and $\vec b$, like $\vec a + \vec b$, and get $\vec c$, I can check if $\vec c$ is a vector by rotating it to get $\vec {c'}$ and then rotating $\vec a$ and $\vec b$, to get $\vec {a'}$ and $\vec {b'}$, and do the operation to get $\vec {c'}$ again. If both $\vec {c'}$ are the same, we're good to go.

The thing that's bothering me is that considering just a rotation of the coordinate system the usual vector algebra is indeed invariant. But it's not if I translate it. What I find is that vector addition is not invariant to equations 11.2 in Feynman's text:

$x' = x - A$

$y' = y$

$z' = z$

If I add, for example $\vec a = (1,2)$ and $\vec b = (2,1)$ I get $ \vec a + \vec b = \vec c = (3,3)$. If I apply the above transformation to $\vec c$ I get:

$\vec c' = (0,3)$

On the other hand, if I apply the transformation to $\vec a$ and $\vec b$ before adding, I get: $\vec a' = (-2,2)$ and $\vec b' = (-1,1)$. If I add them, my result is:

$\vec c' = (-3,3)$

As you can see, the transformation is not invariant towards vector addition.

Is there a reason why translation is not important Feynman's argument? In the first part of the lecture he seems to imply that both spatial and rotational symmetry are important.


  • $\begingroup$ What are you doing when you say "I translate it"? Show us why that isn't invariant. $\endgroup$ – Bill N Mar 15 '17 at 18:06
  • $\begingroup$ Generally, it's a good idea to include an extract (within reasonable limits) of what you are basing your question on. It makes it easier to people to see the equations, and answer your question. A lot of people don't like clicking off site. meta.math.stackexchange.com/questions/5020/… $\endgroup$ – user146020 Mar 15 '17 at 18:16
  • $\begingroup$ Thanks both for the tips. I have edited the question to make explicit what I mean by "translate it", and to include the equations from Feynman's $\endgroup$ – Lgallego Mar 15 '17 at 21:39

Strictly speaking, vectors can't be translated. Translation is not defined in vector spaces. All vectors have their tails at the origin. This is clear from the way we write vectors: $$\vec{A} = A_x\hat{x} + A_y\hat{y}+ A_z\hat{z}$$

How do I translate that? I can multiply by a scalar. I can form dot and cross products. I can calculate a magnitude. I can rotate it. But I can't translate it. It's tail is implicitly fixed at the origin.

The fact that physicists can usefully translate vectors is a peculiarity of Euclidean space that is outside of the mathematical nature of vectors. What we are doing without knowing it is defining a vector space at every point in space so that we can define vectors anywhere. But then we need a rule that tells how to move a vector from one vector space to another. The rule for Euclidean space is so simple that we usually don't mention it: the components at the new location are the same as the components of the old location. But all this is outside of the mathematics of vector spaces.


Ah, I think You must misunderstand Mr.Feynman.

He says "Do these form a vector? “Well,” we might say, “they are three numbers, and every three numbers form a vector.” No, not every three numbers form a vector! In order for it to be a vector, not only must there be three numbers, but these must be associated with a coordinate system in such a way that if we turn the coordinate system, the three numbers “revolve” on each other, get “mixed up” in each other"

For any vectors, there exists a (maybe standard) Orthonormal basis, and the vectors can be expressed using it, e.g. in Cartesian coordinate system, a vector $\vec{a}$ can be expressed as $\vec{a}=a_x\hat{x}+a_y\hat{y}+a_z\hat{z}$. Assume $\vec{b}=b_x\hat{x}+b_y\hat{y}+b_z\hat{z}$ is another vector, so, $\vec{a}+\vec{b}$ must be a vector since it is a linear transformation. We define a new vector $\vec{c}=\vec{a}+\vec{b}=(a_x+b_x,a_y+b_y,a_z+b_z)$. Clearly, it is exactly a vector and satisfy any rules a vector must be satisfied, e.g. rotation. So, we say the groups of numbers$(a_x+b_x,a_y+b_y,a_z+b_z)$ constitutes a vector as listed in the order(the order of the three numbers is important). But, not every three numbers form a vector, such as $(a_y+b_y,a_x+b_x,a_z+b_z)$. So What does Feynman mean is the order of the three numbers in the parenthesis is non-communicative, and nothing more. Generally, any group of numbers can form a vector.

  • $\begingroup$ I don't think I understand you. Are you saying that $(a_y+b_y,a_x+b_x,a_z+b_z)$ is not a vector? I think it's a vector, just not $\vec a + \vec b$ $\endgroup$ – Lgallego Mar 16 '17 at 10:23
  • $\begingroup$ Yes, it is a vector. So, what does Feynman mean for "not every three numbers form a vector"? He just want to emphasize the three numbers is non-communicative for a given vector. $\endgroup$ – Feynman Mar 16 '17 at 11:12

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