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Lorentz scalars - transformation of a scalar field

  1. Aug 2, 2015 #1
    When one considers a Lorentz transformation between two frames ##S## and ##S'##, such that the coordinates in ##S## are given by ##x^{\mu}## and the coordinates in ##S'## are given by ##x'^{\mu}##, with the two related by [tex]x'^{\mu}=\Lambda^{\mu}_{\;\;\nu}x^{\nu}[/tex] then a scalar field ##\phi (x)## relative to ##S## transforms as [tex]\phi '(x')=\phi (x)\qquad\text{or}\qquad\phi '(x)=\phi (\Lambda^{-1}x)[/tex] where ##\phi '(x')## is the form of the scalar field relative to the frame ##S'##. Clearly, as it is a scalar it is a Lorentz invariant quantity.

    My question (albeit a simple one, so apologies for that) is, when we transform the coordinates from those defined in ##S## to those defined in ##S'## (i.e. ##x\rightarrow x'##), am I understanding it correctly that the field transforms as ##\phi\rightarrow\phi '##, such that it has a different functional form in the two different frames, ##S## and ##S'##, but in such a way that when evaluated at a given point ##p##, then it has the same value in both frames? (That is, if a point ##p## has coordinates ##x^{\mu}## in ##S## and coordinates ##x'^{\mu}## in ##S'## then ##\phi '(x')=\phi (x)##, however, the functional form of ##\phi## will be different to the functional form of ##\phi '##)
     
  2. jcsd
  3. Aug 2, 2015 #2

    PeterDonis

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    Yes, this is how it works. Another way to put it is that the point ##p##, physically, is identified by the value of physical quantities there, such as ##\phi##; its coordinates may change but the physical values do not.
     
  4. Aug 2, 2015 #3
    Ok great, thanks for confirming this. Is what I put about ##\phi## and ##\phi '## being different functional forms of the scalar field (relative to the two different reference frames ##S## and ##S'## respectively), but having the same numerical value at each point correct?
     
  5. Aug 2, 2015 #4

    PeterDonis

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    Yes.
     
  6. Aug 2, 2015 #5
    OK, great. Thanks for your help! ☺
     
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