Skip to main content
9 events
when toggle format what by license comment
Dec 23, 2020 at 13:26 history edited Adomas Baliuka CC BY-SA 4.0
fixed error
Dec 17, 2020 at 20:30 comment added Philippe Malot Unless $a=b=0$, the set of $C^1$ functions $x$ on $[\alpha,\beta]$ such that $x(\alpha)=a$ and $x(\beta)=b$ is not a vector space.
Feb 22, 2017 at 9:22 comment added ACuriousMind Well, making the path integral rigorous is a wholly different cam of worms...the individual "parts" of the measure the physicist writes down don't actually exist in a straightforward manner, but at least in QM (not QFT), the Wiener measure provides a formalization.
Feb 22, 2017 at 9:18 comment added Adomas Baliuka The piecewise case certainly does come up. For example when one considers idealized singular potentials, such as for describing collisions or in applications to ray optics. Also there must be some reason why the function space is never specified in physics literature? I was hoping, someone would know a reason, other than perhaps there being no real reason to care about this in applications. Also the main motivation for the question seems to be path integrals and the above doesn't answer this.
Feb 22, 2017 at 9:00 comment added ACuriousMind How is this "way too restrictive" for most applications? Can you give an example in classical non-field mechanics where you'd need more?
Feb 22, 2017 at 0:38 vote accept Nate Stemen
Feb 22, 2017 at 0:38 vote accept Nate Stemen
Feb 22, 2017 at 0:38
Feb 21, 2017 at 22:40 history edited Adomas Baliuka CC BY-SA 3.0
added 47 characters in body
Feb 21, 2017 at 22:35 history answered Adomas Baliuka CC BY-SA 3.0