1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Scalar Product

  1. Aug 3, 2016 #1
    Scalar Product is defined as ##\mathbf A \cdot \mathbf B = | \vec A | | \vec B | \cos \theta##.

    With the construct of a triangle, the Law of Cosines is proved.
    ##\mathbf A## points to the tail of ##\mathbf B##.
    Well, ##\mathbf C## starts from the tail of ##\mathbf A## and points to somewhere.
    Finally, ##\mathbf B## points to the head of ##\mathbf C##.

    Now someone defines a new operation such as ##\mathbf A \circ \mathbf B = | \vec A | | \vec B | \sin \theta##. Following the same steps of proving the Law of Cosines.

    ##\begin{align}
    \mathbf C & = \mathbf A + \mathbf B \nonumber \\
    \mathbf C \circ \mathbf C & = (\mathbf A + \mathbf B) \circ (\mathbf A + \mathbf B) \nonumber \\
    | \vec C | | \vec C | \sin 0 & = | \vec A | | \vec A | \sin 0 + | \vec B | | \vec B | \sin 0 + 2 | \vec A | | \vec B | \sin \theta \nonumber \\
    | \vec A | | \vec B | \sin \theta & = 0 \nonumber
    \end{align}##

    ##\theta = 0##, ##\mathbf A = 0## or ##\mathbf B = 0##?

    I am not quite understand what the result means. It seems the result is not consistent with the constructed triangle.
     
    Last edited: Aug 3, 2016
  2. jcsd
  3. Aug 3, 2016 #2

    fresh_42

    Staff: Mentor

    How is the angle defined? In the usual scalar product, this isn't important, because ##\cos (\theta)=\cos (-\theta).##
    This changes with the sine. I have the feeling that this could be a problem when adding ##|A||B|\sin(\theta)## and ##|B||A|\sin(\theta)## and pretend it would be commutative. Also the distributive property has to be proven.
     
    Last edited: Aug 3, 2016
  4. Aug 3, 2016 #3

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Which is impossible, because the operation is not distributive.

    The assumption of commutativity is the other mistake in post #1.
     
  5. Aug 3, 2016 #4
    This derivation is incorrect, because you assumed the operation you have defined ##\circ##, is linear, in order to go from the second step to the third. Instead we have the identity.
    $$
    \mathbf A \times \mathbf B=(\mathbf A \circ \mathbf B) \mathbf n
    $$
    where ##\mathbf n## is a unit vector perpendicular to both ##\mathbf A## and ##\mathbf B##, in the direction obtained from the right-hand rule (in the case ##\mathbf A=\mathbf B##, you may pick any ##\mathbf n##). Then, if ##\mathbf C = \mathbf A + \mathbf B##,
    $$
    \mathbf C \times \mathbf C = (\mathbf A + \mathbf B) \times (\mathbf A + \mathbf B)
    $$
    Now the cross product ##\times## is linear, but you also get terms ##\mathbf A \times \mathbf B## and ##\mathbf B \times \mathbf A##, which cancel by a property of the cross product. So in the end you get
    $$
    \mathbf C \times \mathbf C = \mathbf A \times \mathbf A + \mathbf B \times \mathbf B
    $$
    Which just says that ##0=0+0##.

    Another way to see that your conclusion does not hold, is simply to see that ##\mathbf A \times \mathbf B## is non-zero in general, so by taking the magnitude, we find that ##|\mathbf A \circ \mathbf B|## is non-zero in general, and hence ##\mathbf A \circ \mathbf B## is non-zero in general.
     
  6. Aug 4, 2016 #5
    Now it is defined as the angle between ##\mathbf A## and ##\mathbf B## where ##\theta## is smaller than or equal to ##\pi## and it is commutative.

    How do you find out that it is not distributive?

    Um... How are these two examples related to the derivation?
     
  7. Aug 4, 2016 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    By looking at an example.
    $$A = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, B = \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix}, C = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$$
    $$(A+B) \circ C = 0 \neq A \circ C + B \circ C$$
     
  8. Aug 4, 2016 #7

    fresh_42

    Staff: Mentor

    If you object, that you only need ##A \circ (A+B) = A \circ B## then compute it with ##A=(1,2)\, , \, B=(1,3)##.
     
  9. Aug 4, 2016 #8

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Huh?
    That works, but this is not the only operation that was used in post 1.
     
  10. Aug 4, 2016 #9
    So if we have to define something, we have to calculate these laws before applying using the basic arithmetic rules.
     
  11. Aug 5, 2016 #10

    fresh_42

    Staff: Mentor

    You may define whatever you want.

    1.) As long as it is well-defined.
    This wasn't the case in your first attempt, because you assigned two possible values ##|A||B| \sin(\theta)## and ##|A||B| \sin(-\theta)## to ##A \circ B##. A definition is not allowed to be ambiguous.

    If this is given, it is absolutely legitimate to define, e.g. an operation ##\circ##. But the next steps should then be to try and find out, which properties such an operation has. This includes algebraic rules it obeys. There might be new formulas, known ones that are partially true or even totally true, ones which would be nice to have but don't hold and so on.

    2.) Any "basic rule" has to be proven before it is allowed to be applied.
    This means that rules you may be used to by arithmetic from real numbers are not automatically true in the context of your new operation. Therefore they have to be proven first. E.g. matrix multiplication isn't commutative, squares of complex numbers aren't positive and so on.
    You gave an example yourself: ##A \circ A = 0## although ##A \neq 0##.

    3) You might not need the entire basic rule.
    E.g. you didn't need distribution in all cases for your argument. Only in some special cases. Therefore you didn't need to prove distribution, merely the ones you actually use. However, it doesn't hold even in these cases.

    4.) You may summarize the above as follows: well-definition - toolbox - theorems.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Scalar Product
  1. Triple scalar product (Replies: 2)

  2. Scalar product (Replies: 13)

Loading...