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RE: Dark blobs as stealthy dark matter
- meaning there is a lot more dark matter around than neutrinos ? (on the cosmic scale) so its more likely that it collides somewhere within a lighyear of lead , lol ? thats what it means?
and
- that's the opposite then, blobs with a mass of tons made of matter that's not matter would be nearly impossible to detect over large distances even if you go by the bending of light (it has a name , right? some kind of "something"-effect) because a few tons or thousand tons of dark matter wouldnt make a dent or a bend in the spacetime fabric (in that case) ?
how close am i to the theory ? :p
hm , something odd with the markdown numbering system 1) 2)
No, this means that dark matter may interact more, just because the interaction strength is stronger.
Yes, indeed. Except that with a few tons, you don't bend much so that this would not be noticeable.
okay, so i had question two right, does that mean i pass ? :) cos in most unis you dont pass on 50% , right ?
when i started OU for psychology i think the required score was about 70% or no bach lol
but part 1 leaves me hanging for a moment there
what exactly do you mean by interaction strenght ? i thought gravity pulled depending on mass, so a more massive object will be more subject to it ? thats why the moon revolves and a comet just slings by ?
or is there something else at work ? because right now variable interaction strength sounds to me a lot like variable gravity (ooh dont shoot me hhahah)
no really what i read is (please do correct me) the same amount of mass in a blob of dark matter or a zone of neutrinos will interact differently with the gravitational pull and therefor make a different "dent" in spacetime (thats regardless of volume ofcourse, that far i got already :)
When there is a dark sector, they may come with their own set of interactions that are different from the fundamental ones we know. Those interactions may have different strength. Not electromagnetic, not weak, not strong. Just a different one.
I hope this short answer clarifies.
yes, it actually does , thanks , its an unknown variable