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submitted 2 years ago by fossilesque@mander.xyz to c/science_memes@mander.xyz

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[-] nicetriangle@kbin.social 11 points 2 years ago

I'm pretty bullish on science investments, but I've heard multiple arguments that this thing is probably not worth the money. The most prevalent argument I've heard to the contrary is basically "we could discover something that might be interesting." But like very little in terms of concrete measurable returns on investment for it.

This article does a good job of arguing against it I think. https://www.scientificamerican.com/article/the-world-doesnt-need-a-new-gigantic-particle-collider/

My mind isn't made up on the topic, so like can anybody explain to me why this thing is actually worth 30+ billion dollars?

[-] jadero@mander.xyz 9 points 2 years ago

Off the top of my head, I can't think of any advance that didn't at some point depend on people just dicking around to see what they could see.

"What happens if we spin this stick really really fast against this other stick?"

"Cool! What happens if we put some dried moss around it?"

"That's nuts, man! Hey, I wonder what happens if we toss some of our leftovers in there?"

"C'mon over here, guys. You gotta taste this!"

At worst, a project like this keeps a lot of curious people in one place where we can make sure they don't cause harm with their explorations. At best, whole new industries are founded. Never forget that modern electronics would never have existed without Einstein and Bohr arguing over the behaviour of subatomic particles.

Say the actual construction cost is $100 billion over 10 years and operational costs are $1 billion a year. Compared to all the stupid and useless stuff we already spend money on, that's little more than pocket lint. We could extract that much from the spending of one military alliance and it would look like a rounding error. Hell, we could add one cent to the price of each litre of soft drinks, alcoholic beverages, and bottled water and have money left over.

[-] Sodis@feddit.de 9 points 2 years ago

Yeah, but you could also fund a lot of other research with this budget. The point is, physicists just don't know, if there are more particles existing. There is no theoretical theory there predicting particles at a certain mass with certain decay channels. They won't know what to look for. That's actually already a problem for the LHC. They have this huge amount of data, but when you don't know, what kind of exotic particles you are looking for and how they behave, you can't post-process the data accordingly. They are hidden under a massive amounts of particles, that are known already.

[-] Waraugh@lemmy.dbzer0.com 2 points 2 years ago

So why don’t they just use post processing to remove all the known particles and start looking at the particles that remain, discover a new one, remove it, continue until there’s none left?

[-] Sodis@feddit.de 1 points 2 years ago

There are multiple reasons for that. We don't know the decay channels of already discovered particles precisely. So there might be very rare processes, that contribute to already known particles. It is all a statistical process. While you can give statements on a large number of events, it is nearly impossible to do it for one event. Most of the particles are very short-lived and won't be visible themselves in a detector (especially neutral particles). Some will not interact with anything at all (neutrinos). Then your detectors are not 100% efficient, so you can't detect all the energy, that was released in the interaction or the decay of a particle. The calorimeters, that are designed to completely stop any hadrons (particles consisting of quarks) have a layer of a very dense material, to force interactions, followed by a detector material. All the energy lost in the dense material is lost for the analysis. In the end you still know, how much energy was not detected, because you know the initial energy, but everything else gets calculated by models, that are based on known physics. A neutral weakly interacting particle would just be attributed as a neutrino.