Now that we don't have permeating Ether anymore, why don't we consider the omnipresent Cosmic background radiation in place of it? and measure Speeds with respect to the CMBR? In that way an object moving through the space, can be said to have at least a 'kind-of' absolute velocity w.r.t Background radiation? I.O.W is that radiation stationary?
OK...before everyone blasts this with references to the relativistic invariance of the physical laws, time dilation, etc let me add some context. Also, I am an amateur with an interest in physics, so I don't know the details of the physics. However, I have read enough about the cosmological principle and cosmic time to develop this "confusion" .  Based on evidence gathered to-date, it appears that the universe is both , as evidenced by observations and the . This principle, when combined with the FLRW metric, allows space-time to be divided into non-intersecting slices and therefore establish a "cosmic time."   Now, the existance of such a time in no way suggests that it is the "right" time or a true "now", hence violating relativity -- although some presentist philosophers (e.g., W.L. Craig) have tried to make this argument, but it is not generally accepted in philosophical circles.  However, this leaves me confused about an apparent disconnect between relativity and the cosmological principle:  Isotropy only holds if we are at rest relative to the cosmic microwave background radiation (universal rest frame), o/w anisotropy is present - yet each reference frame is supposed to allow equally valid observations. How can these two be reconciled when most reference frames would lead us to conclude that the universe is NOT isotropic?  It seems that only by appealing to the idea of being at rest relative to "universal rest frame" can we explain away any discrepancies from isotropy as due to "peculiar motion". However, doesn't this give this "universal rest frame" and its associated time an empirically privileged status, even though physical laws work just fine in every reference frame?   For example (pardon any abuse of astronomy): if look off in some part of space and see only quasars, and then in another part of space and see only brown dwarfs, but I measure both as apparently the same distance from me, then can I conclude that we are in motion, since otherwise we would have an empirical contradiction (kind of like finding dinosaur and human remains in the same strata)?  Any help on where I am going wrong would be helpful. Intuitively, I don't think there should be a way to establish absolute ordering between space-like separated events, but the above reasoning suggests otherwise.
How do I restore the sources.list file? Add new alias to Gmail without SMTP (forwarding-only address) I used to be able to go to Accounts and Import → Add another email address you own and add an email account that's only a forwarding address. That is, it's not a mailbox and I don't have SMTP access to it.  Gmail would send me a confirmation email to that address, I'd confirm it's mine, and done. Now it only gives me the option to add SMTP credentials.  Did they stop supporting the previous method? Is there any other way I can get that option back? "More clear" vs "Clearer": when to use "more" instead of "-er"? What stops other benders from mastering the four elements? Integral around unit sphere of inner product For arbitrary $n\times n$ matrices M, I am trying to solve the integral  $$\int_{\|v\| = 1} v^T M v.$$  Solving this integral in a few low dimensions (by passing to spherical coordinates) suggests the answer in general to be  $$\frac{A\,\mathrm{tr}(M)}{n}$$ where $A$ is the surface area of the $(n-1)$-dimensional sphere. Is there a nice, coordinate-free approach to proving this formula? Does magic missile hit silent image's illusory creature? Do the Faceless Men take more than just a face? iOS: Compare two dates Working in a group blender project Do ground seeds contain oil? How do I copy to the clipboard in JavaScript? There are 11 letters in the word.   M - 1 I - 4 S - 4 P - 2  so the number of different permutations is $\dfrac{11!}{1! 4 !4!2!}$  Is this correct solution? How can I start the Blood on the Ice quest? I am currently part of the Stormcloaks and just finished Rescue from Fort Neugard. The Jarl wants to make me a thane and is telling me to buy a house, but his steward is telling me the house is unavailable due to some unpleasantness. I have already completed the 5 quests to help people of Eastmarch.  From what I can see online, I need to begin the Blood on the Ice quest. It is supposed to begin by walking in the windhelm graveyard at night, however every time I go there I don't see anyone.  Someone online said it's possible to miss the scene that triggers the quest if you go into Windhelm and leave without triggering it, but I am hoping this isn't the case. I am in Windhelm a lot because the Enchanting table, Blacksmith, and Fence are all in one location, so find it very convenient. The lady who is supposed to be murdered in the Graveyard (Susanna) is still alive in Candlehearth Hall, so I think I'm not bugged.  The wiki suggests that you have to "The quest will initiate after you entered Windhelm four times from another hold", however I have entered and left quite a few times during the course of my game and know I've been here way over 4 times.  I tried pickpocketing the key, however I don't automatically get the quest on entering the house, and there are a bunch of "Investigate InvestigationTarget" spots in the house. Also, I was reading the Butcher's journal and he was saying he almost caught Susanna when she left Candlehearth, however I have never seen her leave Candlehearth. Perhaps my glitch is she never leaves? I'll have to head to Windhelm on another character and check it out.  Also, I saw something online saying this quest won't trigger if you go past 245 in-game days. Would someone be able to confirm this for me? (Have the XBox version, so waiting 245 days is annoying)  Can someone tell me the exact steps to begin this quest? Antivirus software is breaking SE web page? (blocking CSS and JS)? What is the difference between lands and mana? ${\rm Aut}(G)$ is cyclic $\implies G$ is abelian I would appreciate if you could please express your opinion about my proof. I'm not yet very good with automorphisms, so I'm trying to make sure my proofs are OK. Proof: Since ${\rm Aut}(G)$ is cyclic, ${\rm Aut}(G)$ is abelian. Thus for any elements $\phi, \psi \in{\rm Aut}(G)$ and some elements $g_i \in G$, $\phi\psi(g_1g_2)=\phi\psi(g_1)\phi\psi(g_2)=g_3g_4=\phi\psi(g_2)\phi\psi(g_1)=\phi\psi(g_2g_1)=g_4g_3$. Hence, $g_3g_4=g_4g_3\implies G$ is abelian. The proof seems to be quite straightforward, but I'd rather ask for advice. Using conjugates to find a limit with a cubic root: $\lim_{h\to 0}\frac{\sqrt[3]{h+1}-1}{h}$ I currently have $$\lim_{h\to 0}\frac{\sqrt[3]{h+1}-1}{h}$$  Now, I know that when you have square root instead of a cubic root it's easy. You just multiply by the conjugate and simplify afterwards. If it were a sqrt I know that the limit is 1/2. I know that the result here is 1/3 but I can't seem to get there. I always end up with 1/4 due to getting something like: $$\frac{h+1-1}{h(\sqrt[3]{h+1}+1)(\sqrt[3]{h+1}+1)}$$  After simplifying and evaluating the limit you end up with $$\frac{1}{(\sqrt[3]{0+1}+1)(\sqrt[3]{0+1}+1)}$$  which turns out to be $1/4$.  What am I doing wrong? Can't log in using Firefox in Stack Overflow Here's what is happening; everytime I go to Stack Overflow (stackoverflow.com, NOT meta) with Firefox, it gives me the "Welcome back" tab that pops up from the top. Whenever I click in the "click here to refresh the page" link, all it does is refresh the page without logging me back in.  Even if I try to log in using the "log in" button at the top, all it does is pop up the tab that says "Welcome name here, you're being redirected." The problem is I'm not being logged in, not to mention that it keeps refreshing the page over and over again without giving me a choice to log in with Facebook.  For some reason, I was capable of logging in to Meta Stack Overflow using Firefox. I am still unable to log in to Stack Overflow. Any ideas? $A$ and $B$ are $n\times n$ matrices and $A+B=AB$. I have an interesting proof that this implies $A$ commutes with $B$, but the proof only works when $||B|| \lt 1$. I'm looking for a way to salvage the proof so that it works for all $B$.  $$A=AB-B$$ $$A=(A-I)B$$ by repeatedly substituting $A$ with $(A-I)B$ on the RHS, $$A= (A-I)B^N - \sum_{i=1} ^{N-1} B^i$$  Since the space of $n\times n$ matrices is Banach (hence complete) under the standard operator norm, the limit $\sum_{i=1}^{\infty}B^i$ is well defined given $||B|| \lt 1$. Hence, taking the limit as $N \rightarrow \infty$, $$A =  -\sum_{i=1}^{\infty}B^i$$  Now the LHS commutes with B, hence so does the RHS.    Is there some trick to extend the method to work with any $B$ where $||B|| \geq 1$? I know that the result is true due to another simple algebraic proof, but I'd like to see if this method can be salvaged. In a question on "along" vs "alongside", I have seen this question and the correct answer on the web site makes me confused.      The question is: I enjoyed my walk _________________ the river.  a) along b) alongside  I selected "...along the river" as the correct answer, but on the web site it is shown as wrong and the correct answer is "alongside".  How can that be? I still do not agree, because I have studied the difference between the two. "along" is just the right preposition for thin and long places like road, path, street, beach etc".   So, why would the correct answer be "alongside" which simply means "next to"? Do you not think "along" seems to be better fitting?  Regards,  Note: I would be more than gratefull, if somebody show it to me in a video, like "a walk along the river" and "a walk alongside the river". Maybe I can understand the how "a walk along the river" and "a walk alongside the river" would take place.