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

A Field transformation under an active transformation

  1. Apr 22, 2016 #1
    Under the infinitesimal translation ##x^{\nu} \rightarrow x^{\nu}-\epsilon^{\nu}##,

    the field ##\phi(x)## transforms as ##\phi_{a}(x) \rightarrow \phi_{a}(x) + \epsilon^{\nu}\partial_{\nu}\phi_{a}(x)##.

    I don't understand why the field transforms as above. Let me try to do the math.

    The Taylor expansion of ##f(x+\delta x)##, where the argument ##x+\delta x## is a ##4##-vector and ##f## is a scalar, is ##f(x+\delta x)=f(x)+\frac{\partial f}{\partial x^{\nu}}(\delta x)^{\nu} + \dots##

    So, ##\phi_{a}(x) \rightarrow \phi_{a}(x-\epsilon) = \phi_{a}(x) + (-\epsilon^{\nu})\partial_{\nu}\phi_{a}(x)##.

    Now, where did I go wrong?
     
  2. jcsd
  3. Apr 22, 2016 #2
    The translation you applied is ##x^\nu\to x^\nu - \epsilon^\nu##.
    That's the only difference.

    The field transformation you are trying to prove is ##\phi_a(x) \to \phi^{\prime}_a(x) \equiv \phi_a(x^\prime)=\phi_a(x+\epsilon)=\phi_a(x) +\epsilon^\nu\partial_\nu\phi_a(x)##.

    I find the treatment in Chapter 1 of Freedman and Van Proeyen's book very clear.
     
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: Field transformation under an active transformation
Loading...