12
$\begingroup$

Most of the descriptions I have come across related to dot products between two vectors start off by stating/showing the vectors have a common point* to begin with - makes the notion of an angle between the vectors very easy to deal with in the final formula for dot products.

I am showing below two vectors A and B, with no common point.

enter image description here

Is it sensible to talk about obtaining a dot product between them (preferrably without translation)?

a) If yes, how do I go about finding the dot product?

b) If not, is there a mathematical expression that shows a contradiction and hence proving that such a context's mathematical possibility fails?

*By saying common point, means the tails of both vectors are fixed to a common point

Similar question: https://math.stackexchange.com/questions/136157/dot-product-of-two-vectors-without-a-common-origin (very difficult for me to understand)

| cite | improve this question | | | | |
$\endgroup$
  • 20
    $\begingroup$ This question is ill-defined. A (finite-dimensional) vector, by definition, is just an element of $\mathbb{R}^n$ for some $n$. We tend to draw these as arrows with a common origin, but that does not mean that just drawing two arrows with different origins somehow defines two vectors. It is not clear what mathematical situation (e.g. two elements of affine space, or two elements of a tangent vector space at different points) your drawing is supposed to represent, but the answer highly depends on it. $\endgroup$ – ACuriousMind Apr 3 at 17:04
  • 1
    $\begingroup$ @ACuriousMind I can edit my question based on the deficiences it has - but doing so may rob me of an opportunity for gaining knowledge on the two aspects you suggest. Apologies if I am asking for too much, can you please add answers to each of the aspects (two elements of affine space or two elements of a tangent vector space) addressed separately? $\endgroup$ – far_see Apr 6 at 8:58
16
$\begingroup$

Yes, a dot product is possible between vectors based at different points.

As an example, the Lorentz-invariant mass $m$ of a system of two particles with energies $E_1$ and $E_2$ and momenta $\mathbf{p}_1$ and $\mathbf{p}_2$ is given (in units where $c=1$) by

$$m^2=(E_1+E_2)^2-(\mathbf{p}_1+\mathbf{p}_2)^2$$

which involves the dot product $\mathbf{p}_1\cdot\mathbf{p}_2$ even though the two particles are not at the same point.

You can calculate a dot product using

$$\mathbf{a}\cdot\mathbf{b}=a_xb_x+a_yb_y+a_zb_z$$

regardless of where each vector is based.

A cross product, such as $\mathbf{r}\times\mathbf{p}$ for angular momentum, can similarly involve vectors based at different points.

When the vectors live in a curved space rather than a flat one, things get trickier. In general, you have to then “parallel transport” one vector to the other to have a sensible product between them. But in Euclidean space or Minkowski spacetime, there is no problem because this parallel transport doesn’t change the components.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Refer to the URL imgur.com/28SQ4uo I tried applying what you suggest and got the dot product 2. Just to double check the answer I get is right, I apply the idea that projections are just "shadows" by shining a torch below vector B. The distance 'b' (marked in light blue) is what I should have had from the dot product as you suggested. Where did I go wrong? $\endgroup$ – far_see Apr 3 at 8:33
  • 1
    $\begingroup$ You can interpret dot products in terms of projections when the vector you are projecting onto is a unit vector. In your case $\vec A$ is not a unit vector. You have to multiply your projected length (1) by the length of $\vec A$ (2). $\endgroup$ – G. Smith Apr 3 at 16:31
  • 1
    $\begingroup$ In effect your answer is that you can inner product two vectors at different places if you can map (the parallel transport, the connection) one vector to the location of the other, or both to a common location. This makes me fee that your answer is that you cannot inner product two vectors at different places. $\endgroup$ – Ponder Stibbons Apr 4 at 0:29
  • $\begingroup$ @PonderStibbons I don’t agree. We do it all the time in Newtonian physics, in Special Relativity, in particle physics, etc. We don’t do it in General Relativity, but I don’t think the OP had curved manifolds in mind. $\endgroup$ – G. Smith Apr 4 at 0:50
  • 1
    $\begingroup$ I agree in the sense that the transport is trivial in Newtonian mechanics in a Cartesian coordinate system. And if that is the only context, then it is best not to introduce complications. But the OP said " Is it sensible to talk about obtaining a dot product between them (preferrably without translation)? ", which makes me feel that instinctively they are noticing that there is something important happening with the translation issue. And translation is still important in Newtonian Mechanics with gauges. $\endgroup$ – Ponder Stibbons Apr 4 at 5:10
38
$\begingroup$

Vector by definition only has magnitude and direction. Origin is not part of the vector. In particular, you can unambiguously define a vector between two points $A$ and $B$ by finding the distance between these points and direction from one to another.

Dot product is defined in terms of the two properties of vectors — magnitude and direction. In particular, it's equal to product of magnitudes and cosine of the angle between the directions of the vectors. Notice how there's no reference to any origin in this definition.

Now, of course there are situations where we assign vectors to some points in space. And of course, in some cases it may be meaningless to calculate dot products between the vectors assigned to different points. But this all depends on the exact problem context, there's no universal rule for this.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Apr 4 at 14:57
  • $\begingroup$ @user76284 Copying the message from there: This room has been automatically deleted for inactivity $\endgroup$ – Ruslan Apr 26 at 4:35
21
$\begingroup$

There are different ways to answer this question, but a natural one for the purposes of physics is with differential geometry. Some background:

(1) Every vector lives in some vector space. A vector space is a collection of vectors which obey particular axioms. Adding two vectors produces another vector, you can multiply the vectors by scalars (numbers), etc. These should be relatively familiar to any undergrad in physics even if you haven't had any formal training in linear algebra. In particular, every vector space contains a zero vector, which defines a unique "origin".

(2) The dot product of two vectors is an example of an inner product. An inner product is any map which assigns to every pair of vectors in a vector space a scalar, $\left<\mathbf{a},\mathbf{b}\right> = c$.

(3) A manifold, at its most basic, is a set of points and some set of coordinate functions which assign to every point a coordinate. In your example picture, you have drawn a Cartesian coordinate system on the manifold $\mathbb{R}^2$, 2-dimensional Euclidean space, which assigns to each point in the manifold the coordinate $(x,y)$. Note that a manifold is not necessarily a vector space, and in general does not have an origin. In that sense, a manifold is somewhat more akin to an affine space. The points in a manifold are not vectors! You cannot add them or multiply them by scalars.

(4) A (real) function on a manifold is a map that assigns a real number to every point in some region (subset) of the manifold. As an example, consider the one dimensional manifold $\mathbb{R}^1$, the real line, with natural coordinate $x$ assigning to each point the corresponding real number. A function $f(x)$ such as $f(x) = \sin(x)$ or $f(x) = \tan(x)$ assigns to each point in some subset of the manifold a real number. Note that in the second case, the function is not defined everywhere. Note these functions are differentiable everywhere they are defined.

(5) Consider a point $p$ in a manifold. Consider some sequence of points $\gamma(s)$ (a curve) passing through $p$, parameterized by some real number $s$. That is, for each $s$, say $s\in(-1,1)$, $\gamma(s)$ is a point on the curve. Let's say that $\gamma(0) = p$. One can then define the tangent vector to $\gamma$ at $p$, $\mathbf{v}_p$, as the directional derivative of functions along the curve $\gamma$ at $p$, that is $$\mathbf{v}_p(f) \equiv \left.\frac{\partial f(\gamma(s))}{\partial s}\right|_{s=0}$$ The important lesson of this definition is that vectors do not live on the manifold, instead, they live in tangent vector spaces which are defined at every point $p$ in the manifold. As an example, consider the surface of a sphere, $S^2$, which is a non-trivial two-dimensional manifold. If you imagine the sphere embedded in $\mathbb{R}^3$, you see that the tangent vectors do not lie on the sphere at all! At each point on the sphere, there is a tangent vector space which is a plane tangent to the sphere.

enter image description here

Thus vectors naturally live in the tangent spaces at each point. In the image you drew, the two vectors live in different tangent spaces. In general, there is no apriori way at this simple level of defining how one should go about comparing vectors in different tangent spaces. One could define one in terms of coordinates, but it will depend on the coordinates we use.

Okay, so all of that is the background. I won't go into any more detail, but the lesson is that in general, tangent vectors in different tangent spaces cannot be compared. In order to compare them, one must define a connection, a mathematical object which tells us how to connect different tangent spaces in a way that is independent of coordinates. However, as long as we restrict ourselves to flat space, such as the manifold $\mathbb{R}^2$ which you are considering, we actually secretly already have a natural connection, the flat connection. Effectively, this connection tells us we can just take any two vectors we like (expressed in the same cartesian coordinate system) and freely take their inner product (the dot product).

Another way of saying this is that the metric is flat. The metric, as its name implies, tells us how we are to "measure" vectors. More specifically, it defines a unique inner product in each tangent space in a way that smoothly varies as we move through the manifold. It also tells us how to parallel transport vectors between tangent spaces so that they can be compared. Parallel transport on a flat manifold does nothing to the components of the vectors, they simply remain the same throughout the transport process. This is why we can take any two vectors and take their dot product in $\mathbb{R}^n$. This is not true on curved manifolds such as the sphere $S^2$, as demonstrated in the image below. Parallel transporting a vector around a closed loop back to its original tangent space actually changes the vector, and this is how we measure curvature! Transporting a vector along any arbitrary curve between two different tangent spaces will in general yield different results. Parallel transport is transport along a geodesic of the manifold, a curve of shortest distance. Curves of shortest distance in Euclidean space are straight lines which are always parallel and never cross, and this is effectively why we can freely transport vectors in whatever direction we like without changing their components.

enter image description here

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Thanks alot @Kai. This definitely gave me a deeper insight into the inner workings. $\endgroup$ – far_see Apr 25 at 19:08
5
$\begingroup$

YES, you can do that. I think answer in math.stackexchange.com is pretty much straightforward. Nothing stops you using Affine space instead of Euclidean. Such space has no origin in it, so vectors can't be associated to some specific starting point and all vector operations are translation-invariant. Take a look at this affine space picture :

enter image description here

A and B situations are identical in affine space.

| cite | improve this answer | | | | |
$\endgroup$
3
$\begingroup$

The short answer would be "the origin doesn't matter for the dot product". As far as the dot product is concerned, all vectors can be regarded as having the same "origin" if you insist, so your two vectors can both be shifted to "start" at the origin of the two axes you drew.

The two dimensional dot product in your case is simply $\mathbf{A}\cdot\mathbf{B}=A_xB_x+A_yB_y$ where $A_x,A_y,B_x,B_y$ are the lengths of the projections of each vector on the $x$ and $y$ axes -- lengths (which are scalars) do not depend on "where" your vector "origin" is in the plane, so the dot product doesn't either. Don't be confused by textbooks showing you that $A_x,A_y,B_x,B_y$ come from scaled basis vectors. They are scalars.

You can look up Affine Spaces after you go through a bit more linear algebra.

| cite | improve this answer | | | | |
$\endgroup$
1
$\begingroup$

Yes it is possible. Basically the vectors in coordinate system that we deal with are free vectors, which as the name itself suggests are free to move as long as the direction and magnitude aren't disturbed. So if you want, you can place the vectors in such way to coincide their head and tail or their tails and then perform calculations.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ In physics don't we assume a specific location for a vector? As an example, to know the effect of a force on a rigid body, we need to know where it interacts with the body to know the torque with respect to the center of mass of the body. $\endgroup$ – Not_Einstein Apr 4 at 0:41
  • $\begingroup$ @Not_Einstein not necessarily. Notice how you had to explicitly specify "torque with respect to". This is because torque, being a vector, isn't by itself concerned with the point with respect to which it was calculated. $\endgroup$ – Ruslan Apr 4 at 20:30
  • $\begingroup$ What would it mean if the force vector is located such that the rigid body is not along the line that the vector is pointing? If the line of the force vector intersects the body's c.m., then the body will translate. If it intercepts the body at a point other than the c.m., the body will rotate. So doesn't that mean that the location of the force vector matters? $\endgroup$ – Not_Einstein Apr 4 at 22:30
  • $\begingroup$ @Not_Einstein Visualisations like those only help us to make sure our vectors are defined correctly. In your example, saying that the force acts at some point helps us to define the radial vector $\boldsymbol{r}$. Then we could calculate e.g. the torque for the system, $\boldsymbol{\tau}=\boldsymbol{r}\times\boldsymbol{F}$. But note that "changing" the force vector to act through the CM instead of some other point actually amounts to changing how $\boldsymbol{r}$ is defined, instead; $\boldsymbol{F}$ itself does not change unless you are describing a force of different magnitude / direction. $\endgroup$ – hiccups Apr 6 at 5:53
  • $\begingroup$ @Not_Einstein there are two types of vectors as far as I know considering frame of reference , free vectors and localized vectors. Free vectors are open to change in location without any change in direction and magnitude. But in case of force,a localized vector, it does matter where it acts as you stated because if you change the orientation of where it acts the impact on the body it's being acted upon too changes. For example in case of a rod fixed at a midpoint and free to rotate, if you apply force on the top,it rotates in clockwise direction. When you apply the same forcein same directi $\endgroup$ – Satya Dheeraj Apr 7 at 4:13
1
$\begingroup$

The properties of vectors (as vectors in vector space) do not admit them having different origin. An inner product (or dot product) can be defined between vectors in the same vector space.

When we show vectors as displacements at different points, then we are talking of affine space, not simply vector space. In affine space, the vectors are still vectors, and displacement vectors at different points are identified by translation. So yes, you can still use the dot product, and it is done by translation.

In a differential manifold with curvature, parallelism is lost and you can no longer use translation. In this case vectors are defined in a different vector space at each point of the manifold (tangent space), and you cannot form a dot product between them.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.