4 ms·
It's really really tiny and rather incredibly small. https://arxiv.org/abs/2102.00110 https://arxiv.org/abs/2102.00110 " Collaboration data to extract the r.m
by iamerroragent 4y ago
It's really really tiny and rather incredibly small.
https://arxiv.org/abs/2102.00110 https://arxiv.org/abs/2102.00110
" Collaboration data to extract the r.m.s. mass radius of the proton Rm=0.55±0.03 fm. The extracted mass radius is significantly smaller than the charge radius of the proton RC=0.8409±0.0004 fm. "
- gus_massa 4y agoSo 0.55/0.84 = 0.65, i.e. 35% smaller.
- dr_dshiv 4y agoCharge radius is 53% bigger, got it.
- pacaro 4y agoIt's almost as if the charge radius is in miles and the mass radius is in kilometers. It's imperial vs metric all the way down
- plank 4y agoIf charge is surface* effect, and mass a 'volume' effect, you might expect a ratio of 1.0/(0.5^[1/3]). This is 1.26 (or 0.79). Does not seem to fit experiment, even when fiddling with error bars. OK, so no volume vs surface effect then. *Suppose that both gluon and quarks are really in the exact same region, but that the 'effective' behaviour is "on the surface" for one of them, while "in the whole volume" for the other. In three dimensions, the "effective" radio would differ, in one it would be a factor (0.5)^(1/3) smaller.
- hibbelig 4y agoSo the charge radius is about 150% of the mass radius. Thank you.
- jcims 4y agoFor scale, the mass radius of the proton is roughly to one millimeter as one millimeter is to the diameter of the sun.
- mywittyname 4y agoThis is very helpful to understanding the sheer magnitude of the scale. I previously had no concept of the size of a ten thousandth of a femtometer.
- jcims 4y ago(Not a big deal but the .03 is the tolerance not the power.)