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I have a very little knowledge of cross product of vectors and don't really understand why there is a need to calculate it.

I know that the cross product of two vectors gives a resultant in the direction which is perpendicular to both the vectors.

In a right-handed Cartesian coordinate system there are three unit vectors, i.e. $\hat{i}$, $\hat{j}$, and $\hat{k}$.

Now since each is perpendicular to the others, the cross product of any two should give the third one.

My question is that if we calculate the cross product of $\hat{i}$ and $\hat{k}$, then it gives $-\hat{j}$ i.e. $$\hat{i}×\hat{k}=-\hat{j}\,.$$ But why can't we get positive $\hat{j}$ since it is also perpendicular to $\hat{i}$ and $\hat{k}$. I am totally confused here. Can anyone please tell me what am I thinking wrong here.

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    $\begingroup$ Since you are using right-handed coordinate system you must use the right hand rule. $\endgroup$ – Deep May 31 '18 at 15:49
  • $\begingroup$ Thats what am I asking why to use right hand rule because the positive and negative j cap are both perpendicular to i cap and k cap. $\endgroup$ – The Mathemagician May 31 '18 at 15:52
  • $\begingroup$ First we need a unique answer, i.e. we need a cross-product (which usually represents some physical quantity, such as torque) to be one unique vector. Among the two perpendicular vectors, which one you pick to represent the cross-product is an arbitrary choice on your part. Left-handed system is as legitimate as right-handed system; most people follow the latter. $\endgroup$ – Deep May 31 '18 at 15:58
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    $\begingroup$ If this whole left-handed, right-handed thing bothers you, there is something called geometric algebra in which this whole issue is neatly sidestepped. $\endgroup$ – Deep May 31 '18 at 16:06
  • $\begingroup$ I voted to migrate this to math.SE. $\endgroup$ – AccidentalFourierTransform Jun 1 '18 at 0:55
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You are correct that there are two different unit vectors that are perpendicular to a given plane. As such, in order to have a well-defined cross product, we must make a choice, by convention, of which of these two vectors we will use. The only requirement on this choice is that it must be consistent everywhere we use it; we aren't allowed to switch from one choice to the other arbitrarily. There are two possible choices for the above product; the conventional choice is $\hat{i}\times\hat{k}=-\hat{j}$, which we call the right-hand rule because the direction can be easily calculated using the right hand. This choice also fixes all other unit vector products in such a way that $\hat{i}\times\hat{j}=\hat{k}$, $\hat{j}\times\hat{k}=\hat{i}$, and $\hat{k}\times\hat{i}=\hat{j}$ (basically, any cyclic permutation of the unit vectors in "alphabetical order" gives a positive result). Since the cross product is antisymmetric (not commutative!), swapping the two unit vectors in any of the above examples gives you the negative of the original result.

The unconventional choice is $\hat{i}\times\hat{k}=\hat{j}$, which is sometimes called the left-hand rule, for the same reason. This choice gives you opposite signs on all of the unit vector products above (i.e. the "alphabetical order" choices are negative, and swapping the two inputs in any one of them gives you the positive result). We generally don't ever use this one, but physics would be no different if we did, as long as our choice for the definition of the cross product is consistent.

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  • $\begingroup$ Antisymmetric or Anticommutative ? $\endgroup$ – Elements in Space Jun 1 '18 at 7:32
  • $\begingroup$ @ElementsinSpace Those are cinnamons. $\endgroup$ – AccidentalFourierTransform Jun 1 '18 at 15:45
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To invoke George Mallory, we calculate it because it's there--or more accurately: because it exists, and it acts like a vector under rotations.

Given 2 vectors $\vec a$ and $\vec b$, there are only a limited number of things that we can create that transform nicely under rotations:

There are the scalars (which are the same in all reference frames):

$$ ||\vec a||,\ \ ||\vec b||,\ \ \vec a \cdot \vec b $$

The vectors (under rotations):

$$ \alpha\vec a \pm \beta\vec b, \ \ \vec a \times \vec b $$

And finally a pure rank-2 tensor:

$$ \frac 1 2 [\vec a \vec b + \vec b \vec a]-\frac 1 3 (\vec a \cdot \vec b){\bf I} $$

Note the complicate from of the rank-2 tensor arises because I excluded the part that has the same information as the cross product:

$$ \frac 1 2 [\vec a \vec b - \vec b \vec a] $$

and then subtracted the part that has the same information as the dot product:

$$ \frac 1 3 (\vec a \cdot \vec b){\bf I} $$

These are all the geometric objects that two vectors can make. One approach to physics is write all physical laws in a coordinate-free geometric manner--where everything is a relation between such objects.

We need the cross product because it is one of these objects and appears in many physical laws, e.g.:

Angular momentum:

$$ \vec L = \vec r \times \vec p $$

torque:

$$ \vec {\tau} = \vec r \times \vec F $$

electromagnetism:

$$ \vec F = q(\vec E + \vec v \times \vec B) $$

and so on.

In your question you pointed out the sign ambiguity--why is it signed? This is very important. Note that the cross product has the same information as the antisymmetric tensor:

$$ {\bf A} = \frac 1 2 (\vec a \vec b - \vec b \vec a )$$

Well, the cross product basically is $\bf A$, and under a change of sign of all coordinates (a so-called Parity transformation), it does not change sign, unlike a regular vector $\vec a \rightarrow -\vec a$.

These type of vectors are called axial vectors or pseudo vectors--because the rotate like vectors, but reflect like tensors. It is a fundamentally different geometric object from a vector--in fact, in mechanics and electromagnetism, you will never see a vector depending on a axial vector, as that would violate parity symmetry: that the mirror image of the process differs from the process.

For instance:

$ \vec F = m\vec a $

is a relation between 2 vectors, while:

$ \tau = I\dot{\omega}$

is a relation between 2 axial vectors.

So in summary, the cross product introduces a new type of geometric object that is relevant all over physics.

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The anti-commutative nature of the cross product (that $\vec b \times \vec a = - \vec a \times \vec b$) is useful for representing some aspects of physical reality.

Not all operations in the physical world are commutative. Rotations in three dimensional space, for example, do not commute. Grab a book, give it a 90° rotation about one axis, then a 90° rotation about some other axis. The orientation of the book will be quite different if you apply the same rotations but reverse the order of the rotations (i.e., rotate about the second axis and then about the first).

The cross product is a useful tool with regard to some rotational operations precisely because of the anti-commutative nature of the cross product. The cross product is also useful in other arenas, and once again, it's useful because the underlying physical reality has a non-commutative flavor.

Note that I used "some" multiple times. We use the cross product where it makes sense. The cross product of any two arbitrary quantities that can be represented as three-vectors can readily be computed. That readily-computable result may or may not have any physical meaning. The cross product is useful if it does. If it doesn't, it's about as useful as is the concept of adding one kilogram of apples to one meter of wire.

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First off I think you have a typo (you meant -j cap). There are two perpendicular directions to the i-k plane. The cross product operation uses the right hand rule. crossing two consecutive elements of the orthonormal set {i, j, k} in order from left to right will give the third in the sequence. The sequence can be extended i, j, k, i, j, k etc. So i cross j gives k, j cross k gives i, and k cross i gives j. You need to respect the order of the vectors in this operation. You get +j from k cross i. Switching the order of the vectors being crossed introduces a minus sign.

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Suppose a parallelepiped's vertices are $k\vec{a}+l\vec{b}+m\vec{c}$ with $k,\,l,\,m\in\{0,\,1\}$. Its volume can be shown to be the determinant of the matrix whose columns are $\vec{a},\,\vec{b},\,\vec{c}$. We can write this as $\sum_{ijk}\epsilon_{ijk}a_i b_j c_k$, where $\epsilon_{ijk}$ is the fully antisymmetric symbol satisfying $\epsilon_{123}=1$. This implies $\epsilon_{ijk}=\epsilon_{jki}$, and the parallelepiped's volume is $\vec{a}\cdot\vec{b}\times\vec{c}=\vec{b}\cdot\vec{c}\times\vec{a}=\vec{c}\cdot\vec{a}\times\vec{b}$, with the usual cross product definition $(\vec{u}\times\vec{v})_i:=\sum_{jk}\epsilon_{ijk}u_j v_k$. Note the antisymmetry guarantees $\vec{u}\times\vec{u}=\vec{0}$, so $\vec{u}\cdot\vec{u}\times\vec{v}=\vec{u}\times\vec{u}\cdot\vec{v}=0$.

The physical cross product called angular momentum, $\vec{L}:=\vec{r}\times\vec{p}$, has already been discussed. This time, antisymmetry can be used to prove $\vec{L}$ is conserved for a radial force. We have $\dot{\vec{L}}=\dot{\vec{r}}\times m\dot{\vec{r}}+\vec{r}\times \vec{F}$. The first term vanishes because the arguments are parallel; for radial forces, the same logic also applies to the second term.

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