elfmotat

But it's clear from the context of his post that that's not what he meant.

I want justice, not exaggerated charges. She should be in prison for not more than 11 years, or fined, or both, as per the law.

18 USC §793. Malintent isn't required for prosecution, only negligence. See for example the case of General Petraeus, who was prosecuted in 2015 for mishandling classified materials. 18 USC §1924. She stored classified materials on an unauthorized computer. It seems like you're reaching to find technicalities that excuse what she did. Of course there are conspiracies to usurp law and justice. There are many confirmed cases of just such a thing, as a quick google search will demonstrate for you. It's almost as if powerful people do shady things to stay in power . Who? Clinton is the establishment personified. Name a single person who you think has considerably more power and influence. Also note that I never claimed the Clintons don't have powerful enemies. It doesn't say that, nor did I claim it did. I provided that quote to establish that the decisions regarding her punishment were clearly politically motivated.

Oh, and as for this: -18 USC §793 Gross negligence in the handling of classified materials. I'm not sure how this can be argued. -18 USC §1924 Unauthorized retention of classified materials. She stored classified information on an unauthorized private server. As Comey said himself: "To be clear, this is not to suggest that in similar circumstances, a person who engaged in this activity would face no consequences. To the contrary, those individuals are often subject to security or administrative sanctions. But that is not what we are deciding now."

"In various interviews, Clinton has said that "I did not send classified material, and I did not receive any material that was marked or designated classified."[87] However, in June and July 2016, a number of news outlets reported that Clinton's emails did include messages with classification "portion markings".[88][89] The FBI investigation found that 110 messages contained information that was classified at the time it was sent. Sixty-five of those emails were found to contain information classified as "Secret"; more than 20 contained "Top-Secret" information.[90][91] Three emails, out of 30,000, were found to be marked as classified, although they lacked classified headers and were only marked with a small "c" in parentheses, described as "portion markings" by Comey. He also said it was possible Clinton was not “technically sophisticated” enough to understand what the three classified markings meant.[92][93][94] According to the State Department, there were 2,093 email chains on the server that were retroactively marked as classified by the State Department at the "Confidential" confidential level.[95][96] Of the 2,100 emails that contained classified information, Clinton personally wrote 104 and her aides wrote hundreds more.[44][97]" https://en.wikipedia.org/wiki/Hillary_Clinton_email_controversy#Classified_information_in_emails "Bill Clinton’s CIA Director Was Pardoned During Plea Negotiations for Storing Classified Data on Home Computer: John Deutch, CIA director under President Clinton, was found to have classified information on a government-owned computer in his home several days after he left the CIA. He had to be pardoned in the middle of plea negotiations by Hillary’s husband." Yes.

Yes. Have you read any of her leaked emails? She's like Machiavelli's wet dream. https://therationalists.org/2016/03/27/hillary-clinton-versus-the-world/ The GOP is a shell of its former self. It was easily co-opted by the shallow buffoon Trump. I don't think anyone has ever made the claim that Clinton is dumb. She's undoubtedly very smart. It's her sociopathic behavior that should make people afraid. The emails are criminal. What she did was criminal. If she wasn't Hillary Clinton she would be in prison, and rightfully so. http://www.thepoliticalinsider.com/hillary-clinton-email-10-punished-less/

Because (A/B)*(B/A) = (A/A)*(B/B) = 1*1 = 1. Try it out with numbers. If you have [(x+y)/(x-y)]*[(x-y)/(x+y)], pick random numbers for x and y. For example, x=6 and y=2: [(6+2)/(6-2)]*[(6-2)/(6+2)] = [8/4]*[4/8] = [2]*[1/2] = 1. https://www.google.com/search?q=z%3Dx-y+plot&rlz=1CASMAE_enUS631US631&oq=z%3Dx-y+plot

Well of course, doesn't everyone always get participation trophies for everything they participate in?

It's pretty hard to argue at this point that consciousness is anything other than some emergent property of information processing in the brain. And I agree, quantum phenomena is only relevant at scales much smaller than that of neurons.

For example? Usually this comes down to misunderstanding what is meant by "observer." I don't think consciousness has ever seriously been considered as relevant to QM.

No.

1992.

James (The Amazing) Randi. Not so much for the magic itself, but for everything he's done for skepticism and rational thought.

You could try being honest with her. Write her a letter so that you can organize your thoughts, and tell her basically what you told us: that her divorce from reality is causing a rift in your relationship, you find it unhealthy, and you've considered cutting her out of your life because of it. Maybe she'll do some self-reflection, maybe not. Worth a try though.

It depends on the type of thermistor you're considering. See the wiki page for more information: https://en.wikipedia.org/wiki/Steinhart%E2%80%93Hart_equation

Why? Is your mind immune to evidence? I realize you're joking, but we have a word for this and it's called being 'closed-minded.' You're setting an unreasonable standard of proof to avoid having to change your mind. Successful case of what? Time dilation? You've been given examples of experiments that successfully detect time dilation. Beliefs should be based on evidence. If they aren't, they're delusions. You've been presented with evidence. Science should not be considerate of feelings; yours, mine, or anyone else's. To what rabbit hole are you referring? The hole of following the evidence and trying to understand nature? I'm not sure why that's a bad hole to go down.

Simply, covariant and contravariant refers to tensor indices (upstairs index is contravariant, downstairs index is covariant). For a full understanding of what they mean, I recommend a textbook like Schutz's General Relativity textbook. He gives a great indroduction to tensors and the notation involved. Invariants are scalars which are the same in every reference frame. Variants (although I've never really seen this term in practice before) are simply scalars that change with reference frame. EDIT: I realized examples would probably help. Contravariant: An example of a contravariant tensor (rank-1, aka vector) would be velocity: [math]v^\mu = \frac{dx^\mu}{d \tau}[/math] An example of a covariant tensor would be the gradient of the electric potential [math]\phi[/math]: [math]\partial_\mu \phi = \frac{\partial \phi}{\partial x^\mu}[/math]. One of the useful aspects of this notation is that summing over upper and lower indices of vectors produces invariants, i.e. objects that don't depend on your choice of reference frame: [math]\sum_\mu q \, \partial_\mu \phi \, v^\mu = power[/math]. (Also, the electric potential was probably a bad example because it's actually part of the electromagnetic 4-potential. As such it will also transform under a coordinate transformation, so the above example isn't actually true. I chose it because it's the most familiar. Substitute it with a true scalar field for an accurate example.)

Presumably, but since it's unobservable we could never know for sure.

No, you were right the first time . In Newtonian gravity "r" is a distance, whereas when describing a Schwarzschild black hole "r" is merely a coordinate. Radial distances need to be found via an integral of ds. The r-coordinate only asymptotically approaches a measure of distance as it increases. So, even though the two equations look identical, they mean different things.

Yes, typically when one speaks of a "background" they are referring to a specific spacetime geometry. You could, for example, do field theory on top of a Schwarzschild background spacetime.

Generally a "source" is something whose presence implies some nonzero field. A field is defined on a manifold whose dimensionality is already determined. I.e. space and time already exist, they are not fields, and nothing acts as a "source" of them. Energy-momentum acts as a source of (Einsteinian) curvature. Zero energy-momentum does not imply no spacetime, it implies flat spacetime. Hopefully that clears up any confusion, but experience tells me it probably won't .

I'm not exactly sure what your question has to do with transverse and longitudinal mass. Proper acceleration is the magnitude of four-acceleration. So if you have a path (in this case a circle), you can parametrize it, find four-acceleration, then take its norm to find proper acceleration: [math]x^\mu (t)= (ct,Rcos \, \omega t,Rsin \, \omega t)[/math] [math]u^\mu (t)= \frac{dx^\mu}{d \tau} = \frac{dt}{d \tau} \frac{dx^\mu}{dt} = \frac{dt}{d \tau} (c,-R\omega sin \, \omega t,R\omega cos \, \omega t)[/math] we know that: [math]d\tau^2 = dt^2 -c^{-2}( dx^2 + dy^2)[/math] or: [math]\left (\frac{d\tau}{dt} \right )^2 = 1 -c^{-2} \left ( \left ( \frac{dx}{dt} \right )^2 + \left ( \frac{dy}{dt} \right )^2 \right ) = 1 -c^{-2} \left ( R^2 \omega^2 sin^2 \, \omega t + R^2 \omega^2 cos^2 \, \omega t \right ) = 1-\frac{R^2 \omega^2}{c^2}[/math] So: [math]u^\mu (t) = \frac{1}{\sqrt{1-R^2 \omega^2/c^2}} (c,-R\omega sin \, \omega t,R\omega cos \, \omega t)[/math] Then we can compute four-acceleration: [math]a^\mu (t) = \frac{dt}{d \tau} \frac{du^\mu}{dt} = \frac{1}{1-R^2 \omega^2/c^2} (0,-R\omega^2 cos \, \omega t,-R\omega^2 sin \, \omega t)[/math] And from there we can compute proper acceleration: [math]a_{prop}= \sqrt{a^\mu a_\mu} = \sqrt{\frac{R^2\omega^4 cos^2 \, \omega t+R^2\omega^4 sin^2 \, \omega t}{(1-R^2 \omega^2/c^2)^2}} = \frac{R\omega^2}{1-R^2 \omega^2/c^2}[/math] So, given the circular path's radius and the object's angular velocity, the proper acceleration felt by the object is [math]R\omega^2/ (1-R^2 \omega^2/c^2)[/math]. Note that the object's tangential speed [math]v=R \omega[/math] can also be plugged into this equation to get: [math]a_{prop} = \frac{v^2/R}{1-v^2/c^2} = \frac{\gamma^2 v^2}{R}[/math] EDIT: Whoops, forgot to square the demoninator .

All observables have corresponding operators, and the non-commutativity of two operators is what defines the uncertainty relation between these observables. Since there needs to be uncertainty for there to be entanglement, the answer to your question would be: yes.

Disregarding higher-order effects, if the gravitational potential energy between two masses was [math]+GMm/r[/math] instead of [math]-GMm/r[/math] then gravity would behave akin to like charges repelling each other (but much weaker). The result of which would be a universe with no large structures, and definitely no black holes. If you want to keep gravity the way it is, except have it be repulsive at very short distances, the formulation of such a theory would be much more complicated. (I suspect it's probably been done before, but I haven't seen it). However, the shorter the scale at which it becomes repulsive, the more irrelevant the effects would be due to how weak gravity is in the first place compared to the other forces.