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Due process does it the other way round.The action taken (stripping her citizenship) is manifestly illegal under international law. Well, strictly speaking... While surrounded by a bunch of people who may well be sympathetic to ISIS, she still says that she supports them. However, she is asking for passage back home to the UK. Who can tell what she will say when she's no longer in among that bunch? What Process? The one which the Home Secretary has ignored? Except she's not. She's British. No Odd as it may seem, you need a trial before you have a conviction. Unless, of course, a court remembers that she was a minor when she made that decision and has not subsequently been in a position to change her mind. She's now trying to do so. But, for some reason, you seem to have decided to forbid that option. The government of Bangladesh had never heard of her. She is not, nor has she ever been a bangladeshi citizen. She has never visited the place. Imagine it was the other way round- she was born to british parents in Bangladesh and now wanted to go back there. Imagine the Bangladeshi government was trying to claim she was British. How would that play out in the UK? Because, if we decide to ignore the rule of law, we might as well invite ISIS to run the place for us. Don't you see that we have to show that we are better than them? Handing him over to the Americans, knowing what would happen, makes Canada responsible for what happened to him at their hands. They shouldn't have done it; they didn't have to. The difference is that what the Americans did was legal under US law, but what Canada did was illegal under Canadian law. Fundamentally it looks like she's a criminal. We should treat her the same as any other criminal who went abroad and is now wishing to return. Bring them back, and take them to court.

Sorry, if I sounded too harsh in my previous reply.. you know.. there is Saturday night.. (8th beer etc. etc.)

Answering asked questions is never annoying, dealing with false assertions is. edit: lets qualify that slightly it can get annoying to repeat/repeat/repeat.... the same answers to the same person who makes no effort to learn from those answers. lol not to indicate yourself but in general.

Thank you for your explanations! I know it must be somewhat of an annoyance for you to read proposals from fools like me, and then have to explain the science. I imagine it happens quite frequently. I never meant to be annoying. I just thought I had a unique idea. An that must also be annoying for you. That we think, you didn't already consider that! Once again, thank you for taking the time to enlighten me!

Well your right in so far as photons do contribute to expansion, it is part of the current radiation density term. It simply isn't a major contributor in our current universe time slice. Every particle species contributes to a certain degree, albeit some species less than others. The dominant contributor however in our current time slice is [latex] \Lambda [/latex]. However as explained in previous threads the standard model of particles contribution will continuously reduce as the volume increases, while [latex] \Lambda[/latex] stays constant to the point where any deviational evidence isn't sufficient to show a variation due to volume change. Hence as the universe expands [latex] \Lambda [/latex] becomes more and more the driving contributor to expansion. Eventually reaching the point where it will be the only contributor where it can be determined.

I appreciate your fascination..

! Moderator Note When you have a model, then you will have fulfilled the requirement of the speculations forum. You can contact a mod to re-open this thread. But until you have some substance to discuss, and not just hand-waving conjecture, this is closed.

Your terse posts are really starting to annoy, Dimreepr. Care to explain what exactly, is prejudicial about the sentence in my post which you highlighted ?

Dangerous Political Climate for Democrat Leaders is a much better title if you wanted to limit it to that. I was on topic otherwise. Not a direct one. Definitely not. There may be extreme media that I am not aware of. Legally no. Morally yes. But you're getting what was advertised.

It was little more than a poorly veiled attempt to snipe at / about swansont, and clearly not posted for the purposes of genuine discussion. Such threads do not belong on this forum.

Generally you could use integral notation, and in the continuous case, like for the normal distribution, you should do so. Integral notation covers both cases. I don't remember how to work out the expected value of the median estimator for \(\theta.\) For the average estimator it should be fairly straightforward. With \(T=\bar{x}=\frac{1}{n}\sum_{i=1}^n x_i\) we get \[ E(T) = \int_{x_1,\ldots,x_n} \frac{1}{n}(\sum_{i=1}^n x_i) \prod_{i=1}^n \frac{1}{\sqrt{2\pi}\sigma} e^{\frac{(x_i-\mu)^2}{2\sigma^2}}dx_1\cdots dx_n.\] Actually after simplification it is just the sum of the expected values of each \(x_i\) divided by \(n\). Since \(E(x_i)=\mu,\) we get \(E(T)=\mu\). I hesitate to work out the details now, because I am not at home and only have my little notebook available.

Rather than challenging you again I will try to be helpful and provide more detail. This discussion is solidly founded in Mechanics. Mechanics can be reduced to relationships (Laws, formule etc) between three suitable quantities. The current international standards use Mass, Length and Time (although some disciplines also use Force, Length and Time, eg Fluid Mechanics) In the ISO these are realised as the kilogramme, metre and second. These are called units. For the purpose of dimensional analysis you simply need to work in consistent units. So we work in unitless dimensions M, L and T and expressed as a product of powers (including negative powers) of these three. The baic relationship connecting the last two is the definition of velocity Note I am using an equals sign for the defining relationship and an arrow to denote the derived dimensions. [math]velocity = \frac{{dis\tan ce}}{{time}} \to L{T^{ - 1}}[/math], This can be extended to acceleration [math]aceleration = \frac{{velocitychange}}{{time}} \to L{T^{ - 2}}[/math] These are both simple geomtrical relationships. Introducing a Physical law - Newton's second Law we define force [math]force = mass*acceleration \to ML{T^{ - 2}}[/math] and work (or energy since work is a particular type of energy) [math]work = force*dis\tan ce \to M{L^2}{T^{ - 2}}[/math] Now we are in a position to derive the dimensions of the Universal gravitational constant G, using Newton's Law of gravity. [math]Gforce = G\frac{{{m_1}{m_2}}}{{{d^2}}} \to G{M^2}{L^2}[/math] Newton's law is an equation and to be equal the dimensions must be identical on both sides of the equation so comapring [math]ML{T^{ - 2}} \leftrightarrow G{M^2}{L^2}[/math] [math]\frac{{ML{T^{ - 2}}}}{{{M^2}{L^2}}} \leftrightarrow G[/math] [math]{M^{ - 1}}{L^3}{T^{ - 2}} \leftrightarrow G[/math] Which yields mys tated dimensions for G. Now Avogadro's number is defined as [math]Avogadro'sNumber,{N_a} = \frac{{Molarmass}}{{molecularmass}} = \frac{M}{m} \to \frac{M}{M}[/math] remembering my earlier note that the molar mass and the molecular mass must be in the same units we find that the constant is mass/mass or Kg/Kg That is dimensionless. This is not different from say the dimensionless quantitly 'Reynold's number' which is the ratio of inertial forces to the viscous forces or The dimensionless quantity 'strain' which is the ratio of two lengths Or the dimensionless quantity 'duty factor' which is the ratio of two times. The first two of these other examples are variables and the third is a constant, demonstrating that dimensionless quantities can be constants or variables. There are many examples in Physics of such dimensionless ratios and I note you made no comment when I stated your 'reduced volume' was not a volume but also a dimensionless quantity since it is based on the ratio of two lengths. Be aware there is some disinformation about concerning the dimensions of Avogadro's Number, concerning moles. Just remember that the ratio must be in the same units.

There is no such thing as a charged photon. I don't know how likely it is for a photon to interact with a virtual particle. It is possible to "promote" a virtual particle pair into being real by adding energy, so there is a chance you could do this with a photon, provided the fluctuation had an energy deficit smaller than the energy of the photon.

Bollocks. She subscribes (present tense) to a belief that that is totally antithetical to what we hold as fair and just. What she has uttered is no way that of a person who sees the error of their ways. Her team lost, she wants to defect back.

To add: The glaciers accumulate mass at the highest level and basically slide down under their own weight to the warmer lower level and melt, supplying the rivers.

Yeah, I'm a sexy senior citizen... A new thread could probably be started on the morality of a woman having this woman's hair!

No shit, Sherlock...