# Is Velocity Really a Vector?

In non-relativistic physics, physical quantities $$Q$$ are characterized by how they transform under a Galilean transformation $$g \in \mathcal{G}$$. $$Q \rightarrow Q' = D[g]Q$$ where $$D[g]$$ is the linear representation of g.

Let $$r$$ be a rotation, $$a$$ be a spacial translation, $$t$$ be a time translation and $$b$$ be a boost.

Scalars take the trivial representation for everything: \begin{align*} D[r] = 1, D[a] = 1, D[t] = 1, D[b] = 1 \end{align*}

If we accept force $$\vec{F}$$ to be the model example of a vector, then it transforms under the following representations: $$D[a] = 1, D[t] = 1, D[b] = 1\\ D[r] = R, \text{for some R \in \mathcal{O}(3)}$$

But 3-velocity does not transform trivially under boosts; the boost velocity just adds up. Does that mean that velocity is not a vector?

Is there a representation where we can see that force and velocity are both the same kind of object? (vectors), or are they just different?

• Clearly the transformation is different; whether it's a vector depends on what you mean by "vector". The meaning of that word depends very much on the context. Commented Aug 7, 2019 at 12:40
• @Javier Since we're talking about 3d vectors, then in the context of general physics with Galilean relativity, where the laws should be Galilean invariant. So I would think it's natural to classify the type of the quantity based on how it transforms under Galilean transformations. In special relativity where the symmetry group concerned is the Poincaré group 4-force and 4-velocity transform exactly the same way, so they are the same type of objects (vectors). Commented Aug 7, 2019 at 15:29
• Would you elaborate on why you chose force as a model example of a vector? Why not displacement? For example, Griffiths writes "Formally, then, a vector is any set of three components that transforms in the same manner as a displacement when you change coordinates." Commented Aug 7, 2019 at 23:22
• @AlfredCentauri I just had to choose something. I believe Displacement and Force transform exactly the same way under boosts (trivially). But not Velocity. Commented Aug 7, 2019 at 23:48
• @AimanAl-Eryani - read this post and my answer, as well as the links provided in the answer. The treatment of forces and rotations as lines is part of "Screw Theory" => "The Geometry of Mechanics". Commented Aug 8, 2019 at 15:52

It boils down to this:

A Galilean transformation is only defined as a transformation of position vectors:

$$x'(t) = x(t) - vt$$

From this we can derive how time derivatives of position transform:

$$u'(t) = \frac{d}{dt} x'(t) = \frac{d}{dt}x(t) - v = u(t) - v$$

$$a'(t) = \frac{d}{dt} u'(t) = \frac{d}{dt}u(t) - 0 = a(t)$$

Then assume mass is Galilean invariant and define the force vector by Newton's second law. We get:

$$F'(t) = m'a'(t) = ma(t) = F(t)$$

• That's a way to show that force transforms with the trivial representation under boosts, but velocity transforms with a $-v$. Where you using the word "vector" in position vectors to just mean a 3-tuple? I was certain position was not a vector (because of how it transforms under translations). If not how do you see both position and force to be vectors? Commented Aug 8, 2019 at 9:22
• So I am using vector here to mean an element of the 3 dimensional vector space we call 'space'. Further postulate 3-space has a Euclidean metric to define distance then all 3-vectors transform the same under distance preserving maps of 3-space. i.e. rotations. Different vectors then transform differently under the larger Galilean group. Commented Aug 8, 2019 at 10:16
• That is distance preserving maps leaving the origin fixed. Should also include reflections in that but then get into discussion of pseudo vectors. So in summary vector in Newtonian physics refers to a specific representation of the rotation group. Commented Aug 8, 2019 at 10:27
• Positions would indeed be vectors if you consider only rotations. But why are rotations so special? They take you between inertial frames. But so do translations and boosts (the entire Galilean group). Commented Aug 8, 2019 at 10:45
• Well, pseudo-vectors/bivectors are not vectors. To exclude reflections would also be arbitrary, so I think they belong to this discussion. (Otherwise, maybe ignoring reflections can be justified by the fact that nature doesn't have parity inversion symmetry; e.g. there are left-handed neutrinos but no right-handed neturinos). Commented Aug 8, 2019 at 10:50

First of all, speaking of a vector in nonrelativistic physics, one usually refers to the properties of the given quantity under spatial rotations only. That is, a vector is defined as a certain representation of the rotation group.

That said, however, the use of the term vector clearly is just a matter of convention. Why vectors in nonrelativistic and relativistic physics are treated differently is to some extent a matter of sociology, but there can also be more concrete reasons for doing so.

First, the philosophy whereby physical laws are derived from their symmetries, and not vice versa, was historically only introduced along with relativity, hence the focus on manifest Lorentz invariance in relativity. Second, it is a mathematical fact that the classification of representations of the Galilei group is much more tricky than the same problem for the Lorentz group (the reason being that the Galilei group is not semisimple). See for instance this paper: the Galilei group has not only 3-vectors, but also two different types of 4-vectors, and 5-vectors, among others. The construction of invariants for a given representation is likewise a nontrivial task. I believe this is the main practical reason why representations of the Galilei group and their use are much less frequently discussed than those of the Lorentz group.

To answer the original question, velocity is a vector under rotations. But under Galilei transformations, velocity is just a part of a 5-vector.

• "vectors in nonrelativistic and relativistic physics are treated differently." can you elaborate a bit, please. Commented Aug 8, 2019 at 17:59
• @ja72 By this I just meant that in relativity, vectors are defined by their behavior under rotations and boosts (i.e. all Lorentz transformations), whereas in nonrelativistic physics only rotations are used, not boosts. This refers to the OPs original question. There was nothing deep in this statement, it's just a matter of terminology. Commented Aug 8, 2019 at 18:25
• Thanks! This addresses my question. the philosophy whereby physical laws are derived from their symmetries: Well, not necessarily derived from the symmetries, but the laws still have to respect them (or be invariant under them). The paper linked is walled, but I found it on arxiv: arxiv.org/abs/math-ph/0604002 . Just to allow some more time for others to answer if they wish - otherwise I'll settle with accepting this one. Commented Aug 8, 2019 at 18:39