When I was studying about vectors and their use in physics, I found something called zero vector. My physics textbook says it's

a vector whose initial and terminal points coincide is called zero vector, it has zero magnitude but an arbitrary direction, i.e. it cannot be assigned a direction

What is the significance of this zero vector? For example, if the force acting on a body has no magnitude then is there any meaning to say that the force has a direction? Thanks in advance

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    $\begingroup$ Be careful with wording here: "null vector" in Minkowski space can also mean one with a Minkowski norm. The zero vector is one of these, but all the nonzero vectors lying on the light cone are also "null" in this sense. $\endgroup$ – Selene Routley Oct 12 '16 at 13:25
  • $\begingroup$ @Ramanujan i think waiting for an answer here will be more appropriate,I appreciate you help but i can't get them. $\endgroup$ – Vidyanshu Mishra Oct 12 '16 at 13:27
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    $\begingroup$ Start with a special case. What would count as an answer to the question "What is the significance of the real number $0$?". $\endgroup$ – WillO Oct 12 '16 at 13:31
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    $\begingroup$ It is an additive identity( if i got it right). $\endgroup$ – Vidyanshu Mishra Oct 12 '16 at 13:36
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    $\begingroup$ Okay. So why isn't that also an answer to your question? $\endgroup$ – WillO Oct 12 '16 at 14:35

$\newcommand{\norm}[1]{\lVert #1 \rVert}\newcommand{\ip}[2]{\left<#1,#2\right>}$I want to be at first somewhat mathematically abstract and return to the physical interpretation later. In mathematics you define a vector space $V$ over a field (like real numbers etc.) $F$ to be as a triplet $(V,+, \cdot)$ where $+$ denotes the vector addition and $\cdot$ denotes the scalar multiplication, which has the following properties:

  • $u + v = v + u$
  • $u + (v + w) = (u + v) + w$
  • There exists an element $ 0 \in V$ such that $v + 0 = v$
  • For every $v ∈ V$, there exists an element $−v ∈ V$, called the additive inverse of $v$, such that $v + (−v) = 0$
  • $ \alpha(\beta v) = (\alpha\beta)v $
  • $ 1v = v$ where $1 \in F$
  • $\alpha(u + v) = \alpha u + \alpha v$
  • $ (\alpha + \beta)v = \alpha v + \beta v$

for all $u,v,w \in V$ and $\alpha, \beta \in F$. Note that the existence of the zero vector $0 \in V$ is an axiom, so without it you cannot have a vector space. Note however that the direction or the magnitude of a vector doesn't come up in the definition, so what is going on here? The thing is that we physicist like to have additional structures on vector spaces, such as a scalar product of two vectors and a magnitude of a vector. Furthermore we want to have that (in physicists notation) $ \vec v \cdot \vec v = \norm{v}^2$ where $\norm{\cdot}$ denotes the magnitude of a vector. So when we know what a scalar product is then we know what the magnitude of a vector is (in some sense). Then the question is this: What is the definition of a scalar product (I'll denote the scalar product as $\ip \cdot \cdot$ from now on.) Scalar product is a function $\ip \cdot \cdot : V\times V \to \mathbb R$, (I take a real vector space for simplicity) which obeys following axioms:

  • $\ip \cdot \cdot$ is linear in both of its arguments
  • $\ip vu = \ip uv$ for all $u,v \in V$
  • $\ip xx \geq 0$ and $\ip xx = 0 \iff x = 0$

So the question might be: Where is the $\cos \theta$ term, with which we defined the scalar product? The troubling part is that there is no $\cos \theta$ term. As it turns out a scalar product satisfies this wonderful inequality called Schwartz-Inequality:

$$ \norm x \cdot \norm y \geq \ip xy $$

note that for $x,y \neq 0 $ we can write this as:

$$ \frac{\ip xy} {\norm x \cdot \norm y} \leq 1 $$

now we can define $\cos \theta := \frac{\ip xy} {\norm x \cdot \norm y}$ since the term on the right is always less than 1. You now see that we define the angle to be $\frac{\ip xy} {\norm x \cdot \norm y}$ not the other way around. So what does this mean for the zero vector? Well nothing because you see that we have divided by $\norm x$ which is zero for the case of the zero vector. Thus it doesn't make any mathematical sense to talk about the direction of the zero vector.

So let's talk about the physical meaning of this whole mathematical abstract explanation and for the sake of the argument let's think about vectors as arrows. Take two vectors, one of which is pointing up and the other pointing left and let's make the length of these vectors to go to zero. Obviously in both cases we get the zero vector because remember $\norm x = 0 \iff x =0$. However there is an ambiguity about the direction of the zero vector. Since both of the vectors were showing in different directions and a vector (in this a the zero vector) should show in only one direction. At this point we say that the direction of the zero vector is ill-defined. The question is this: Does it matter in which direction it points to? Of course not! Think of it this way: If I push you from your back or from your side it makes a difference. You would certainly be able to tell the difference (Corresponding to a force with nonzero magnitude). However if I don't push you from your back or your side, I bet you cannot tell the difference. (This corresponds to applying a force with magnitude zero.) Another example is the velocity vector. It makes quite a difference when you are driving your car to the south or to the east but if you are not driving, well then you are both "not going" to east and south (which is why the direction is ill-defined) However note that in the absence of the zero vector you wouldn't be able to say that you are not moving. Another great analogy is just the number line. There are positive numbers and there are negativer numbers but what if I asked you what is the sign of zero. Well it is (vaguely speaking) both and neither positive and negative so we can say that the sign of zero is ill-defined or you can arbitrarily choose a sign however it makes you happy.

The moral of the story is that you need the zero vector to describe (vaguely speaking) the absence of something. Concretely you need the zero vector in order to say that there is an inverse to a vector (see additive inverse in the way beginning). More like how you need the number zero.

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    $\begingroup$ i can only upvote it as my mind do not allow me to accept it.I got a massive amount of knowledge here but something is there which is beyond scope of my knowledge.I do not blame u as u tried to cover everything. $\endgroup$ – Vidyanshu Mishra Oct 12 '16 at 19:59
  • $\begingroup$ I haven't got to the physics part maybe that could be way :) $\endgroup$ – Gonenc Oct 12 '16 at 20:43
  • $\begingroup$ I hope the intuitive part makes more sense now. $\endgroup$ – Gonenc Oct 12 '16 at 21:09

Examples of null vectors are a bit funny like acceleration of a particle moving with uniform velocity, position vector of the origin w.r.t the same origin, displacement of a stationary particle etc. Note that in these examples, one cannot assign an unambiguous direction to the vector.

Probably you also know that null vectors are intrinsic to the defining structure of a vector space.

EDIT: In equilibrium, the net force on a particle is zero ( $\sum\limits_{i}\textbf{F}_i=\textbf{0}$) i.e., the resultant of all the vector forces acting on that particle has zero magnitude. And indeed, the particle is subjected to zero resultant force. But mathematically, when you add vectors, you cannot have a scalar or a tensor or a complex number. Physically, you might ask how can something which have no direction but still be a vector. This is why I said null vectors doesn't fit into the elementary definition of proper vectors.

All confusions arise because you are thinking in terms of the elementary definition of vectors (i.e., in terms of magnitude and geometric direction.) Scalars, vectors (and in general tensors) should be really defined according to their transformation properties under coordinate transformation.

  • $\begingroup$ I would say why a direction if there id no magnitude in any direction?? $\endgroup$ – Vidyanshu Mishra Oct 12 '16 at 13:16
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    $\begingroup$ Exactly. It doesn't have a direction. However, you must understand that when you subtract two equal vectors you don't get a scalar, you always get a vector. You cannot subtract 10 apples from 20 apples and get 10 tomatoes. Right? This doesn't make sense. $\endgroup$ – SRS Oct 12 '16 at 13:20
  • $\begingroup$ we are not subtracting 10 apples from 20,we are subtracting 20 from 20.And if there is nothing,how would anyone know that what was there (apple or tomato) $\endgroup$ – Vidyanshu Mishra Oct 12 '16 at 13:22
  • $\begingroup$ That is exactly what I'm saying. The resulting object doesn't have any sense of direction and very often, you would find a term called a proper vector which has both a magnitude and a direction to distinguish it from null vector. Well. You tell me what do you get when you subtract two equal vectors? $\endgroup$ – SRS Oct 12 '16 at 13:28
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    $\begingroup$ The only way in which the equality $\textbf{A}-\textbf{A}=\textbf{0}$ makes sense is that the RHS is also a vector. If you regard the RHS as a scalar number $0$ (instead a null vector), then one can write $\textbf{A}=0+\textbf{A}$. But what if the meaning of the RHS? Adding a scalar to a vector has no meaning. $\endgroup$ – SRS Oct 12 '16 at 13:30

If you know that it's the additive identity vector, then I don't see what your question is.

That's what it is. We generally have a bad habit of writing things like $\vec{F}_\mathrm{net} = 0$. Not such a good idea. In class I go out of my way to write $\vec{F}_\mathrm{net} = \vec{0}$.

$0$ is an integer or real number, and does not belong to the vector space. $\vec{0} = \left<0,0,0\right>$. It's a different thing. We can operate only on objects that exits in our space.

Physically it means no force (in the case of forces). If there is no force, there is no direction to be assigned.

Addition after comments

Consider how one finds the direction of a vector in two dimensions:$$\theta = \tan^{-1}\left(\frac{y}{x}\right)$$ In the case of the zero vector this would be $$\theta = \tan^{-1}\left(\frac{0}{0}\right)$$ Indeterminate/Undefined.

  • $\begingroup$ Since you said there is no direction to assign then why we use term arbitrary direction. $\endgroup$ – Vidyanshu Mishra Oct 12 '16 at 16:06
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    $\begingroup$ Aha. That's the issue. The statement you quoted is not carefully worded. The word "arbitrary" should be replaced with "undefined". Please tell us which textbook you found this in. $\endgroup$ – garyp Oct 12 '16 at 16:12
  • $\begingroup$ I just did a search on "zero vector" and "arbitrary direction". That phrase turns up a lot. In my opinion, it is a poor phrase. Misleading at best, incorrect at worst. $\endgroup$ – garyp Oct 12 '16 at 16:16
  • $\begingroup$ @VidyanshuMishra- Intuitively, if a "vector arrow" collapses to zero length and becomes a mathematical point, then the direction is indeterminate, and as garyp pointed out, the word "arbitrary" is misleading. $\endgroup$ – SRS Oct 12 '16 at 16:16
  • $\begingroup$ @garypI don''t think you know my book's name cuz its not so famous (like feynman lectures or resnich halliday) $\endgroup$ – Vidyanshu Mishra Oct 12 '16 at 16:18

If you apply 10N from left and 10N from right on an object, what is the resultant force?

You will say resultant is either null or 0.

So, 0 helps us tell mathematically that there is no net resultant vector on any object. Like this it is very helpful.

For direction, arbitrary means unknown. By arbitrary direction we mean there is no direction to that vector. So, zero vector maybe very well said as a scaler. You can hence say any scaler is a vector with arbitrary direction.


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