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Why does the Schwinger parameter correspond to proper length? 
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#1
Dec2113, 03:30 PM

P: 5

I have just learned from nice article
http://motls.blogspot.com/2013/12/ed...yquantum.html that the propagator of a massive particle can be rewritten as an integral over the socalled Schwinger parameter t as $$ \frac{1}{p^2 + m^2} = \int\limits_0^\infty dt \exp(t(p^2 + m^2)) $$ In addition, in the blog article it is said that this Schwinger parameter p can be interpreted as the proper length of the propagator. I dont see this, so can somebody give a derivation/further explanation? 


#2
Dec2113, 03:49 PM

P: 1,368

Using Wolfram Alpha,
http://www.wolframalpha.com/input/?i...%5Bat%5D%29dt input, integral from 0 to infinity of ( exp[at])dt output, 1/a 


#3
Dec2113, 04:17 PM

Sci Advisor
HW Helper
PF Gold
P: 2,603

There's a nice clear presentation of the argument given in the 1951 Schwinger paper at http://www.thetangentbundle.net/wiki...time_formalism. Post back if that doesn't help.



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