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Why does the Schwinger parameter correspond to proper length?

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Dec21-13, 03:30 PM
P: 5
I have just learned from nice article

that the propagator of a massive particle can be rewritten as an integral over the so-called 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?
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Dec21-13, 03:49 PM
P: 1,375
Using Wolfram Alpha,


integral from 0 to infinity of ( exp[-at])dt


Dec21-13, 04:17 PM
Sci Advisor
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PF Gold
P: 2,602
There's a nice clear presentation of the argument given in the 1951 Schwinger paper at Post back if that doesn't help.

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