Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this tensor operation valid?

  1. Jun 20, 2008 #1
    Can anybody help me clear my following doubt?
    Suppose, I have a relation of the form
    [tex]
    f(p_{\mu},q_{\mu})=0
    [/tex]
    Then can I multiply the both sides by [tex]p^{\mu}[/tex] and then contract?
    [tex]
    p^{\mu}f(p_{\mu},q_{\mu})=0
    [/tex]
    After this I want to use the identity [tex]p^{\mu}p_{\mu}=m^2[/tex] as known in special relativity. So I am first multiplying both sides by a contravariant tensor and then using the summation convention. The question is am I doing it right? I feel this is a valid operation and I have tested one simple example where this is valid but I am not able to prove it in general. Can anybody help? You may refer to some books or website also where I can look up.
     
  2. jcsd
  3. Jun 20, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi arroy_0205! :smile:

    It would have to be fµ, not just f:

    [tex]f_{\mu}(p_{\nu},q_{\nu})=0[/tex]

    Then you can contract:

    [tex]p^{\mu}f_{\mu}(p_{\nu},q_{\nu})=0[/tex]

    Were you thinking of a particular function f? :smile:
     
  4. Jun 20, 2008 #3
    Hi tiny-tim,
    Thanks, and yes you are right it should be [tex]f_{\mu}[/tex]. In fact I was trying to consider a general form like say
    [tex]
    a_1p_{\mu}+a_2q_{\mu}=0
    [/tex]
    there is no constant term in the left hand side. But the proof is still not clear to me. This is sort of obvious but without a solid proof I hesitate to accept. May be I am overlooking something obvious and trivial regarding its proof.
     
  5. Jun 20, 2008 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi arroy_0205! :smile:

    Then, multiplying both sides by [itex]p^{\mu}[/itex] :

    [tex]p^{\mu}(a_1p_{\mu}+a_2q_{\mu})\,=\,0[/tex]

    so

    [tex]a_1p^{\mu}p_{\mu}\,+\,a_2p^{\mu}q_{\mu}\,=\,0[/tex]

    or

    [tex]a_1m^2\,+\,a_2p\cdot q\,=\,0[/tex] :smile:
     
  6. Jun 20, 2008 #5
    That is true, however by multiplying with [tex]p^{\mu}[/tex] I am also introducing summation, this is not like ordinary multiplication by a function but multiplication followed by summation. My doubt is there, whether this kind of operation is appropriate.
     
  7. Jun 20, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    No, it's fine! :smile:

    It is like ordinary multiplication by a function …

    it's four ordinary multiplications, added together …

    which is perfectly valid! :biggrin:
     
  8. Jun 20, 2008 #7
    Yes. Of course it is valid.
    Just think of this operation component-wise.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Is this tensor operation valid?
  1. Is this valid? (Replies: 4)

Loading...