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I Scalar field

  1. May 16, 2017 #1
    Hello! Can someone explain to me how does a scalar field changes under a Lorentz transformation? I found different notations in different places and I am a bit confused. Thank you!
     
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  3. May 16, 2017 #2

    vanhees71

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    If ##x'=\Lambda x##, where ##\Lambda## is a Lorentz-transformation matrix, then a scalar field obeys by definition the transformation law
    $$\phi'(x')=\phi(x)=\phi(\Lambda^{-1} x').$$
     
  4. May 16, 2017 #3
    Thank you for your reply. This makes sense. However I found in Peskin's book on QFT a definition that is different from yours by a prime ( ' ) - I attached a screenshot of it. That is what got me confused. Do you know what does he mean by his notation?
     

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  5. May 16, 2017 #4

    George Jones

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    The expression that vanhees71 wrote is equivalent to the expression in Peskin and Schroeder, and both expressions are equivalent to

    $$ \phi'(Fred)=\phi(\Lambda^{-1} Fred).$$

    Why?
     
    Last edited: May 16, 2017
  6. May 17, 2017 #5
    Sorry I am a bit confused. What is primed and what is unprimed?
     
  7. May 17, 2017 #6

    vanhees71

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    Just read the equation as a whole. My final equation was
    $$\phi'(x')=\phi(\Lambda^{-1} x').$$
    Now rename ##x'## back to ##x##, and you get Peskin Schroeder's formula. You can name it "Fred" as suggested in #4 (although that's a bit unusual ;-)).
     
  8. May 17, 2017 #7
    What do you mean by rename? If we have a frame S stationary with respect to the field and S' moving with respect to the field (so ##\phi'## and x' are measured in S') then we have by definition ##\phi'(x')=\phi(x)## and by the Lorentz transformation we also have ##\phi(x)=\phi(\Lambda^{-1} x')##. This make sense. But in Peskin he has a mix of both ##\phi'(x)## and this is what confuses me. How does he get to a mix of primed and unprimed indices without a ##\Lambda## factor somwhere? And I am not sure how can you rename x' to x, once you decided which moves and which is fixed. (so to be clear, I understand that choosing prime and unprime as moving or not moving is arbitrary, but Pesking seems to mix them, which confuses me). Thank you for help!
     
  9. May 17, 2017 #8

    vanhees71

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    The formula of itself is unique. You can name the argument as you like. You can as well write the law as
    $$\phi'(y)=\phi(\Lambda^{-1} y).$$
    Of course, the prime at the field symbol on the left-hand side is crucial!
     
  10. May 17, 2017 #9
    But the argument of ##\phi'## is always seen from S', right?
     
  11. May 17, 2017 #10

    vanhees71

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    Yes, sure.
     
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