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;-)

Many thanks! Trying to be a science translator. ;-)

Much appreciated! I'm glad it helped. Hmmm.....I'll give some thought to how I can link the material to applications. Thanks for the input.

Hi! Their masses are different. Let's call those masses m1, m2, and m3. Here's the tricky part: the neutrino with a mass of m1 is not, say, the electron neutrino. The neutrino that has a mass of m1 is a combination of the 3 flavors of neutrinos. Similarly, the neutrino with mass m2 is also a combination of the 3 flavors, as is the neutrino with mass m3. A flavor of neutrino is determined by how it interacts with a charged lepton and a W boson. So, for example, if a W- decays to an electron and an anti-neutrino, that anti-neutrino is an electron anti-neutrino. The analogous is true for W- decays containing muons and taus. But, as we said above, this doesn't correspond to a specific mass of the anti-neutrino. If you have a bunch of W- bosons decay to an electron and an anti-neutrino, and you measured the anti-neutrino's mass, sometimes you'd get m1, sometimes m2, and sometimes m3! If that sounds weird, it's because it is. The fact that the neutrinos with definite masses do not have definite flavor (and vice-versa) is at the heart of neutrino oscillations. Understanding it pretty much requires knowledge of quantum mechanics. If you've got a bit of background on that, let me know, and I'll try to explain. Thanks!

Hi! If the errors are gaussian (and, for practical situations, that is actually a big if!) then the 8.33 and sigma are one and the same. In that case, the scientists doing the measurement are taking the sigma from the gaussian error distribution and quoting that as their error bar. One can ask, "How do they determine the value of sigma?" In particle physics, the answer to that is usually a lot of very hard work. Often, determining the error bar is more difficult than measuring the central value (here, the 649.2). Basically, physicists will try to figure out every important source of error in their measurement that they can think of, and try to get a good estimate of how large each of those sources of error will typically be. This means that they have to study their own scientific equipment, to see how precise it is. They have to do extremely complex simulations of both the physics processes they're trying to study and the apparatus they are using. They need to take into account statistical errors caused by the fact that their dataset is not infinitely large. Etc. Then they take all of these sources of error, and combine their effects to get the overall error bar. Is that error bar infinitely precise? Nope. It is always possible that some of the estimates that go into the uncertainty calculation turn out to not be good enough. Sometimes an important source of error is missed entirely. I can only think of one example I have here which demonstrates how good or bad these uncertainty estimates turned out to be. You might want to check out the video How the Measured Lifetime of the Tau Lepton Changed Over 30 Years: ruclips.net/video/xrLllaBiSWM/видео.html Here, we can look back on many measurements of the tau lifetime, with their quoted error bars, and see if those error bars were actually reasonable. Those results were produced by many different groups over many years, and they did a pretty darn good job of estimating their uncertainties. Let me know if that answers (or starts to answer) your question. Thanks!

At no point does she make any reference to a specific particle - this is a lecture on statistics in physics, not physics itself, and she is speaking in general terms to illustrate important features about testing these models (and does a good job of it!). There is no illusion here that a model tells you everything about the world, and what she is teaching is exactly how we go through the process of challenging a model and its predictions experimentally. I add that she is a professional physicist, speaks knowledgeably on the subject, and you only discredit yourself by arguing in such a condescending and infantile way.

@@aidenpeleg2789 You do realize your argument is a heap of errors in logic, don't you? The last one being Argumentum Ad Hominem. Not even once did you touch the actual objection.

@@Kraflyn I did not “touch the actual objection” because my only point was your objection did not apply to this video and is thus not in question. The “ad hominem” would only be a fallacy if I were using it as a part of my “argument”, but it is just an addendum here. Nonetheless, I don’t want to drag out a comments war on such a nice video.

@@aidenpeleg2789 There is no excuse for errors in logic... Finding an excuse is another error in logic! :D

Much appreciated!

Thank you!

Oh, wow! Sorry, I just noticed your comment and donation (and also on the other video)! Apologies, it's been a hectic few weeks! Much appreciated.

Hi! Are you just asking about the numerical result? It's approximately 1/(3.5 x 10^6), which comes out to about 3 x 10^-7. I'm not sure I've answered your question, though.

Hi! I don't think I'm quite understanding your question, but let's see how far I get..... You're setting alpha = 0.4. Then, under the null hypothesis, in 10% of trials, you'll get a result that is above the expected value of the null hypothesis, but close enough that you do not rule the null hypothesis out. Am I reading that correctly? If so, what is the question you're asking toward the end? Can you explain this a little more?

Many thanks!

Hi! So, the reason that the Bayesian chose 1% is that they have been through the exercise hundreds of times, and they've never actually seen one of the novelty coins perform as advertised (ie, always coming up heads); it has always turned out to be a fair coin. So, basically, the Bayesian goes in with the belief that the probability that the coin always comes up heads is small. So small, in fact, that the probability that it's a fair coin that just happens to come up heads 5 times in a row is larger. (Note that we didn't consider other scenarios, like that the coin comes up heads 60% of the time, which does admittedly make our example rather artificial.) The Bayesian prior always has a subjective element to it, based on the beliefs one has before the experiment is performed. As an alternative example, the Bayesian might look at the coin and say, "Oh, wait. This novelty coin is made by a different company than all the ones that I've seen before. In that case, I will not assign a high prior probability to it being a fair coin. OK, let's call the prior probabilities 50-50." If they do this, the probability that the coin is fair after the 5 flips (all coming up heads) is (50%*1/32)/(50%*1/32+50%) = 1/33. While the Bayesian prior always has an element of subjectivity, treating all hypotheses as equally likely before the experiment is done also has its pitfalls. The xkcd comic referenced in the video description illustrates this very well. ;-)

@@ThinkLikeaPhysicist OK.. But if we don't know anything about the coin, there is no subjectivity in choosing the prior. It will have to be 50%. I think the Bayesian analysis adds the ability to introduce a prior knowledge on the model (for example probability the coin is fair or biased). I don't know if there's a way to do that in the frequentist approach. Is there? But even if we forget about the prior, and we take the likelihood function, normalize it and interpret it as a probability of a truth given a measurement (like the probability that the coin is biased given 5 heads,) that to me is a much simpler and transparent way to state the result than the frequentist way. So if you tell me that the probabilty of that coin being fair is 3% given five flips that are all head, I will believe it and ask for you for more flips :) In the case of the sun going supernova in the xkcd comic, I am perfectly fine with the interpretation that it's true at 97% IF we know nothing about how often the sun goes supernova. But that's not the case :)

@@phyzwiz Hi! As for adding prior knowledge into a model in a frequentist approach.....well, let's say your prior knowledge is another experiment you did. In the context here, it would be a set of coin flips you've done before. So, you've got the coin flips you did before, and the coin flips you're about to do. You can take both of those sets of information and combine them into one big set of information. You can then ask, under a certain hypothesis, how strange the combined result is. In that sense, you can add prior information in in the frequentist approach--you take all experimental data, and ask how likely or strange the whole set of data is under a specific hypothesis. But, if you're asking for the probability, given some result, that a hypothesis is true, you are, by definition, asking a Bayesian question. The only way I can tell you that the probability that the coin is fair is 3% after coming up heads 5 times is if I am going the Bayesian route. And that requires choosing a prior, which is up to the person making the choice. It simply is not the case that P(coming up heads 5 times | coin is fair) is the same thing as P (coin is fair | came up heads 5 times). We can't do that.

I just learned what this meant. ;-)

Much appreciated! My older videos are a bit less polished than my newer ones!

Yup, and if floods have a tendency to cluster, the headline is even less meaningful! Good observation, and thanks for the donations! Much appreciated!

Hi again!

@@ThinkLikeaPhysicist <3

Hi! I'm afraid that my plots are usually made using obscure software that almost nobody uses, so I'm afraid it won't be useful to you. But, if you have any questions about what formulas or parameters were used for certain plots, let me know, and I can probably round that up for you.

You are most welcome!

You're welcome!

Hi! Yes and no. It depends on specifically what characteristics of the photon you give to this heavy photon. The Z, like the photon, is a spin-1 particle. In that way, they are similar. And, we could imagine other (hypothetical) spin-1 particles which we might call "heavy photons". The Z gets its mass from its interactions with the Higgs field. (I'm guessing, from your question, that you already know this.) The photon does not interact directly with the Higgs field, and does not acquire a mass. So, what's the difference between the Z and the photon such that the Z interacts with the Higgs field and gets a mass, and the photon doesn't? So, these spin-1 particles ("gauge bosons" as we call these particles like the photon and Z) interact with other particles. How they interact are determined by the quantum numbers of the particles. In the case of the photon, the relevant quantum number is electric charge; the photon interacts with particles that are electrically charged (like the electron), but does not directly interact with particles that are not electrically charged (like the neutrino). As the Higgs boson is not charged, it does not interact with the photon. We can relate this to why the photon does not get a mass. The Z's interactions are determined by a different combination of quantum numbers, which the Higgs boson does have. And thus the Z interacts with the Higgs; this is related to why the Z does get a mass from the Higgs field. We could imagine another heavy spin-1 particle whose interactions are such that it does interact with the Higgs boson, and it gets a mass in a way similar to the way the Z does. So, if by "heavy photon", you mean a heavy spin-1 particle, then, yup, you can look at the Z or this other hypothetical particle as a heavy photon. On the other hand, if you mean a spin-1 particle that interacts with particles through their electric charge (and only their electric charge), then no. Does that help?

Thanks!

Ha! I got quite a good laugh out of this. ;-)

@@ThinkLikeaPhysicist I used this video to introduce a young person to this contentious issue. Since then they found a paper on the differing neurology between hearing and seeing which just happened to contain several instances of the term "Bayesian Heuristic" and they won't stop talking about it. It is a very well done video. :)

@@leo5961 That's great! Glad it was of use! Thank you.

Hi! Let's take a very long time T. The expected (average) number of times the event will occur in this time T is r*T. Let's divide this interval T into N sub-intervals. We're going to take N to be really big, so that each sub-interval is really, really short, and the probability of the event occurring more than once in any sub-interval is so small that we can neglect it. Then, if there are r*T occurrences of the event, r*T sub-intervals contain an event, and the rest will not. So, we choose a sub-interval at random; what is the probability that it contains an event? Well, that would be the number of sub-intervals that contain an event, divided by the total number of sub-intervals. That would be r*T/N, but T/N=delta_t, so it's r*delta_t. Alternatively, instead of trying to get p from r, we could try to get r from p. If we take p to be the probability that a given sub-interval has an event in it, then, if we have N sub-intervals, on average, p*N of them will contain an event. If those N sub-intervals, put together, make a time T, then the average number of events in time T will be p*N. But N is T/delta_t. So, the average number of events in a time T is p*T/delta_t. If we call r*T the average number of events expected in time T, then r*T = p*T/delta_t, which means r=p/delta_t. We can then turn this into r*delta_t = p. Let me know if either of those explanations helps! Thanks!

@@ThinkLikeaPhysicist Wow... wow... wow. I am truly speechless (literally screenshotted your response)! I have been struggling to understand that concept for a while now (as many just stated it as a given), but your explanation is beyond wonderful, I really thank you so much and applaud you for this clear and vivid explanation. I really can't thank you enough! Subscribed for sure, and even though I am no Physicist (I adore Physics though), I love your explanation. Thanks again

Yup, afraid so. The result comes in line with the SM prediction after they treat their backgrounds a little more carefully.

Hi! Which forward-backward asymmetry do you have in mind? Are you interested in a specific experimental result?

@@ThinkLikeaPhysicist A classic example is in electron-positron collisions resulting in the production of quarks. If more quarks are emitted in the direction of the incoming electron than in the opposite direction, this indicates a forward-backward asymmetry.

@@55846 Hi! I'm afraid I don't think I'll get a chance to make a video on it soon, but I'll keep it in mind in case I see a way to work it in. Thanks!

Great question! It's not easy. Here, I'm talking about a case where systematic errors can be estimated from detailed experimental studies. (To contrast, in statistics texts, one will often see the case where something is measured many times, and one can look at the spread in results. That's not what I'm talking about here. Here, I'm talking about a case where something is measured once, and an error bar is quoted.) OK, so, then where does sigma come from? It's strongly dependent on the case at hand, and requires detailed knowledge of the experiment. An experimentalist will basically try to come up with every important source of error that they can think of, and estimate how large those errors are likely to affect the final result. They may simulate making the measurement many times. For example, let's say you want to measure the mass of a particle that decays in a particle detector. You have to measure the energies/momenta of all of its decay products and then use them to calculate the mass of the particle you're interested in. But, your detector's measurements of those energies and momenta are not infinitely precise; hopefully you can figure out just how good those energy and momentum measurements are. (One way you can do this is by examining other particle physics processes occurring in your detector that are already well-understood--you can compare what you see in your detector with what is already known from previous experiments.) Once you know how well your detector measures those energies and momenta, you have to calculate how the errors on those quantities can filter through to your mass measurement. But something like this is just 1 source of error. In practice, there are several or many important sources of error that have to be studied. And the effects of those sources of error have to all be included to estimate sigma. I fear what I've written above is a bit too complicated and specific, so I'll try to summarize: the people doing the experiment need to understand their equipment and methods really well, and then they need to estimate how wrong their measurements are likely to be. This is often a very complicated process; calculating an error bar may be one of the hardest parts of producing an experimental result.