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Vector Squared

  1. Jun 13, 2008 #1
    Hi guys, I understand that if A is a vector then A^2 = A.A = ||A||x||A||. Wot I don't understand is is this something that can be show or proved or is it more axiomatic. I can understand the step from A.A to ||A||x||A|| but why should A^2 = A.A?
     
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  3. Jun 13, 2008 #2
    Hi Galadirith

    [tex]\vec{A}^2=\vec{A}\sdot\vec{A}[/tex] is more or less what the ^2 means. Just write down the components and you'll see it's true
     
  4. Jun 13, 2008 #3

    D H

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    In a way, it is the other way around. The inner product norm is defined as

    [tex]||\mathbf A|| \equiv \sqrt{\mathbf A \cdot \mathbf A}[/tex]

    Squaring both sides of this definitional form,

    [tex]||\mathbf A||^2 =\mathbf A \cdot \mathbf A[/tex]

    One somewhat common convention is to write the magnitude of some vector as an unadorned symbol:

    [tex]A=||\mathbf A||[/tex] (using bold to indicate vectors)

    [tex]A=||\vec A||[/tex] (using arrows to indicate vectors)

    With this notation, [itex]A^2 = \mathbf A \cdot \mathbf A[/itex]. Note that this follows directly from the definitional form. Many people who do not use [itex]A[/itex] to represent magnitude still use [itex]A^2[/itex] to denote the square of the magnitude.

    One final note: This is widely considered to be bad form: [itex]\mathbf A^2[/itex] or [itex]\vec A^2[/itex]. Don't do that, please.
     
  5. Jun 13, 2008 #4
    uh oh... :uhh:

    thanks, I won't do that again. But this is correct [tex](\vec{A})^2[/tex] right?
     
  6. Jun 13, 2008 #5

    nicksauce

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    [tex](\vec{A})^2[/tex] is the same as [tex]\vec{A}^2[/tex]. Do you mean [tex]||\vec{A}||^2[/tex]?
     
  7. Jun 13, 2008 #6
    Thanks guys, ill remember not to use [itex] \mathbf A^2 [/itex] or [itex] \vec A^2 [/itex]. I should have put this more in to a context aswell. The reason for asking this is when the issue arrises in the expansion of [tex] ( \mathbf A + \mathbf B )^2[/tex] to result in [tex] \mathbf {AA} + 2\mathbf {AB} + \mathbf {BB}[/tex]. Why is it that the multiplication of the vectors results in their dot product and why not the vector product? its is simply by definition that [tex] \mathbf {AB} = \mathbf A \cdot \mathbf B [/tex]. I suppose also it nothing to do with them being equal, an it is literally an axiom, just as [tex] a \cdot b [/tex] is the same as [tex] a \times b [/tex] where [tex]a[/tex] and [tex] b [/tex] are scalars and the operation of multiplication is defined and not something that can be proved.
     
    Last edited: Jun 13, 2008
  8. Jun 13, 2008 #7
    No, I just wanted to find whether adding () wouldn't that bad of a form. It's rather inconvenient but still, is it correct?
     
  9. Jun 13, 2008 #8

    D H

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    It doesn't. That is why writing [itex]\vec A^2[/itex] is considered bad form. This could mean the inner product of A with itself, [itex]\mathbf A \cdot \mathbf A[/itex], or the outer product of A with itself, [itex]\mathbf A \otimes \mathbf A[/itex], or some other product of A with itself.

    Writing [itex]A^2[/itex] where [itex]\mathbf A[/itex] is a vector is OK; not great, but acceptable. Note that I used bold face in writing [itex]\mathbf A[/itex]. This is one of the standard ways of denoting that something is a composite object (e.g., a vector, or a tensor, or a matrix, ...) One can also use something like [itex]\vec A[/itex], which leaves no doubt that A is a vector. On the other hand, the 'A' in the expression [itex]A^2[/itex] has no adornments. This is short hand for denoting that the 'A' here is a not a composite object -- i.e., it is a scalar. What you are implicitly doing is writing the vector A as the product of a scalar and a unit vector:
    [tex]\mathbf A = A \hat A[/tex].
    With this implicit notation,
    [tex]\mathbf A\cdot \mathbf A = A^2 \hat A \cdot \hat A = A^2[/tex]
    since the inner product of any unit vector with itself is tautologically one.
     
  10. Jun 14, 2008 #9
    Oh k thanks DH, i deffinatly understand all that, I understand the many problems that arrise from use index notation with vectors, the problem I have then is how do you know what kind of vector multiplication to apply when two vectors are multiplied together. I know in most circumstances it wouldnt be hard to see, but for example the shortest distance between two lines, you could apply pythag and then you end up getting two vectors multiplied by one another. How would you know what type of multiplication to use, do you have to think about what type of result you want, weather you want a vector or scalar, and depending on that you use different types?
     
    Last edited: Jun 14, 2008
  11. Jun 14, 2008 #10

    HallsofIvy

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    There are only two types of vector multiplication, dot product and cross product, and they have different symbols: [itex]\vec{A}\cdot\vec{B}[/itex] and [itex]\vec{A}\times\vec{B}[/itex]. I would prefer never to use "[itex]\vec{A}\vec{B}[/itex]" but when you see it it means [itex]\vec{A}\cdot\vec{B}[/itex].
     
  12. Jun 14, 2008 #11

    D H

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    You forgot the outer product, [itex]\vec{A}\otimes\vec{B}[/itex], and some people do use [itex]\vec{A}\vec{B}[/itex] to mean the outer product. Note well: I disagree with this notation just as much as using [itex]\vec{A}\vec{B}[/itex] to denote the inner product. Both abuses of notation are equally bad IMO.
     
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