# Why does the Schwinger parameter correspond to proper length?

1. Dec 21, 2013

### Dilatino

I have just learned from nice article

http://motls.blogspot.com/2013/12/edward-witten-what-every-quantum.html

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?

2. Dec 21, 2013

### Spinnor

3. Dec 21, 2013