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I am a science laymen. I just saw the Veritasium video about entanglement and Bell (https://www.youtube.com/watch?v=ZuvK-od647c) and have a few questions ... which hopefully will not also be entangled. He says no matter how far apart the entangled electrons they flip "at the same time". How do they know? ... how does one measure both at the "same time"? Can multiple events actually occur at the same time?...or must there be some minimum time difference between any two events (measured events? .. macro? events?) (Is there a 'quantum time' that is different than a 'relative/macro? time'?) Dr. Don Lincoln says (https://www.youtube.com/watch?v=JFozGfxmi8A) "we measure the second one in less time, in "quick succession", so that the 'communication time' must be less than c would need. So isn't THAT communication faster than light? ... that you can read the transfer of information faster than if it were transported over c? (6:15sec). In other words, if c has a limit why isn't time (events) held to that same limit? There's no such thing as time without distance, correct? ... such is space-time? Isn't the nature of distance is that it's two sets of coordinate with a different time stamp?... and vise versa? Furthermore, why is it called 'travel at a distance' ... if "travel" means 'bounded by light' then isn't it something else than 'travel' or 'transfer'? If you had a rod that was incompressible wouldn't the far end literally move at the same time as the near end is moved? (assuming there's no relativistic compression by the observation of transferring energy, right?) Is space-time like that? ... waves get compressed with motion, correct? What is the ƒ of the wave is infinite? ... yes, this is just a crazy thought experiment ... then is there no compression and you in-effect get faster than light communication like the thought experiment rod? (I'm trying to concoct some way for instantaneous action at a distance.) My main questions are bolded but I included these other thoughts as I have the fantasy they help explain what I'm thinking. Thank you, Randy And if they are truly

Hello, Imagine having exited an electron and it went on a higher energy atom orbit. Is it possible to predict when the electron will emits a photon to lower his energy? If it's hazardous , does it follow a probabilistic model like a gaussian curve or smth like that? ty for your answers!

Explaining why there is no communication involved is tricky. But, basically, all we can say is that the measurements made by two people, A and B, are correlated in some way. So, if Alice measures spin up (for example) then Bob will measure spin down. And vice versa. So let's imagine that Alice and Bob each have one of a pair of entangled particles. If Alice looks at her particle she has a 50% chance of seeing spin up or spin down. And Bob has a 50% of seeing spin up or spin down. Imagine Bob looks at his particle. It could be spin up or spin down. Let's say he sees spin up. What does that tell him? Well, it tells him that if Alice has already looked at her particle, then she saw spin down. Or, if she hasn't yet, then it will be spin down when she does. Now imagine Alice looks at his particle. It could be spin up or spin down. Let's say she sees spin up. What does that tell her? Well, it tells her that if Bob has already looked at his particle, then he saw spin down. Or, if he hasn't yet, then it will be spin down when he does. So they both know what the other has seen / will see, but they can't use this to send any information. They can exchange information "classically" (at no more than the speed of light) and end out that their results were always opposite. But they can't transfer any information knowing that. Indeed. There is no such thing as absolute simultaneity. That means that one observer might see Alice look at her particle before Bob, while another observer might see Bob look at his first. (And another might seem them look at the same time.) But you can take that into account if you want to calculate how much time elapsed between the two events. So it isn't directly related to entanglement. (Note that Einstein never really accepted that part of quantum theory.)

No

Multi-party not multipart. They're referring to more than one person being involved. https://en.m.wikipedia.org/wiki/Secure_multi-party_computation

Sounds interesting. Einstein had obviously realised what he needed to capture mathematically before he understood that tensors were the best tool for the job. I have probably mentioned this before, but John Baez has written a great introduction to the concepts behind the Einstein equations using relatively simple math: http://math.ucr.edu/home/baez/einstein/

Does this help: https://en.wikipedia.org/wiki/Priority_encoder Actually, it might not because the first paragraph is incomprehensible. But the second paragraph is clearer. Basically, it has a number of input bits which are in priority order, and it tells you the highest priority bit that is set. I'm afraid I am not going to try and work out what it does in that system. Presumably it chooses the next person to be processed.

'Copying' is more complicated process than simple reactions and is usually effected by genetic processes. Your wish for alchemical transmutations is just not possible and was abandoned hundreds of years ago. Why don't you learn some real chemistry, based on the real world experience of the last few hundred years? It is really fascinating.

Hi Dentre, Well, firstly Linear Programming refers to optimization while DP involves solving complex problems by breaking them down to simpler sub-problems using smart recursion. So to answer your query, to solve a LP problem using Dynamic Programming is by breaking a problem into sub-problems. Here are some great resources to help you understand DP problems. Dynamic Programming Dp Introduction - InterviewBit https://www.topcoder.com/community/competitive-programming/tutorials/dynamic-programming-from-novice-to-advanced/ Demystifying Dynamic Programming Hope this helps :)

I do not think they care. It is painfully obvious that xenophobia and nationalism was part of the agenda all along (in fact, aside from dismantling protective agencies, it is one of the few obviously consistent policies) to make America great white again.

If we take Bartlett first, then the purpose of the test is to figure out for several sets of data, and assuming that each set is normal distributed, whether they also have the same variance. If we think about the second table, it is possible that it represents four sets of data, one for each row, each set of data containing three values, for T1, T2 and T3. Or it is (more) possible that the table represents three sets of data, each containing four values. Let us say that it makes sense that the second table represents experiments in which for each of three temperatures T1,T2,T3 there were made four measurements. Then for T1 it means that values 2.42, 2.83, 2.25, 3.02 were measured, for T2 they were 3.05, 2.21, 2.18, 2.35, and for T3 it was 1.95, 2.23, 2.54, 2.56. We can calculate the estimate variances of each of these samples in the standard way, as 1/3 of the difference between the average of the squares minus the square of the averages. I trust that this is familiar to you? Then we have three estimated variances V1, V2, V3, one for each T. We also have to compute the estimate of the common variance V in case they were actually all equal. That will be \(V = 1/(12 - 3) \sum_{i=1,2,3} (4-1)Vi\), where the 3 means that we have three data sets, the 4 means that we have four data in each set, and 12 is the total number of data in the table. I have not made the computations, since I have no good calculator handy, and I would probably make confusing mistakes, sorry. Finally you have to compute the Bartlett testor itself. First we need the number \( D = (12-3)\log V - \sum_{i=1,2,3} (4-1)\log V_i.\) We can see from the formula above for V that it would be a pretty good match if all the V1,V2,V3 are the same, because then V would be equal to all of them, and this \(D\) would be zero. So \(D\) having a small value is good. To compute the final Bartlett testor we also need to have \( C=1 + (\sum_{i=1,2,3} \frac{1}{4-1} - \frac{1}{12-3})/(3(3-1)).\) The testor becomes \(B = D/C\). Now you have to check \(B\) against a \(\chi^2\) distribution with 3 - 1 = 2 degrees of freedom. Having typed all of that, maybe it is not as easy as I first thought. But try to compute as many of the numbers as you are able to.

! Moderator Note If you are hijacking your own thread with irrelevant nonsense, I think we are done here. Do not bring this subject up again.

The story was confirmed by the German government. Maybe you need to get your "BS meter" checked. The issues was not "what angers Germans". That is just an attempt to deflect the argument. It is almost as if you are biased.

But you think that about everything. I assume a conversation with you must be like: Me: Good morning! You: Oh no! Does that mean we are all going to die? Me: No. I was just saying hello. You: Oh but it sounded like you were saying the world was about to end. Me: ... They say that their measurements are consistent with previous measurements. (That is science-speak for "almost the same as") How on Earth can a small difference in the expansion rate in different directions (or even a large difference) be a bad outcome for us? It's not like anything has changed, apart from our understanding.

The thing to understand about telescopes is that they have effective ranges. Chandra doesn't have the range of Planck or Hubble. The limitation is inherent in the wavelengths it has been designed to collect. It's primary goal is to collect data on our local region of spacetime. Of which it does an incredible job. However one can have local anistropy without affecting the global distribution. Stars, galaxies etc are all examples. In cosmology homogeneity and isotropy is only affective at scales of 100 Mpc. The Milky way galaxy is only between 460 to 720 kpc. Exact value is difficult to determine as we cannot see directly it's diameter. Our local group has several nearby anistropies such as the great attractor. This causes other side effects on luminosity measurements etc. For example there is a study our local region may be underdense due to the great attractor. The first lesson a cosmologist learns is never trust a single dataset as fact. It is only after dozens of datasets come upon agreement do you develop a good confidence level by numerous independent studies and numerous independent equipment.

As with all the biased claims of Trump and therapy/vaccine. This too is BS. Who cares what allegedly angers Germans?

The us West Texas Intermediate (WTI) light crude futures contract for delivery in may ended the trading session on April 20 at minus $37.63 per barrel, according to data from us exchange operator CME Group. So the mass can also be negative.

Certainly, each particle of the Universe contains all the laws governing it as the image of the alive cell containing in its core all genetic information to be it living respective. This is what I seek to discover during my different works.