Expressing Feynman Green's function as a 4-momentum integral

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The discussion revolves around the confusion regarding the classification of (z', p) as a 4-vector, particularly in the context of Lorentz covariance. Participants highlight that while traditional 4-vectors are well-defined, the introduction of new variables raises questions about their legitimacy. The conversation touches on the role of the restricted Lorentz group, which allows the use of the step function ##\theta(x^0 - y^0)## to maintain time orientation. Ultimately, the integration variables are acknowledged, but their interpretation as 4-vectors remains contentious. The discussion emphasizes the need for clarity in how these variables are treated within the framework of momentum space.
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I am a bit confused on how we can just say that (z',p) form a 4-vector. In my head, four vectors are sacred objects that are Lorentz covariant, but now we introduced some new variable and say it forms a 4-vector with momentum. I understand that these are just integration variables but I still do not see how this is okay. The interpretation of z' now is different.
 
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realanswers said:
I am a bit confused on how we can just say that (z',p) form a 4-vector. In my head, four vectors are sacred objects that are Lorentz covariant, but now we introduced some new variable and say it forms a 4-vector with momentum. I understand that these are just integration variables but I still do not see how this is okay. The interpretation of z' now is different.
But where did it come from? ##\theta(x^0 - y^0)##, right? The restricted Lorentz group (identity-connected part) preserves time orientation, so it's ok to use ##\theta(x^0 - y^0)## in that circumstance. The rest just involves expressing it in momentum space.
 
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...