In the general theory of relativity, the speed of light  in vacuum is $c$. In the special theory of relativity, a postulate is made that the speed of light is the same (called $c$) in all inertial frames. Consider the following statement: &quot;The speed of light is $c$ in all inertial frames but it can vary in accelerating frames.&quot; My question is: Does this statement violate the principle of equivalence or the special theory of relativity or any fundamental law of physics?
I'm studying special relativity and saw the 2 postulates of Einstein. The most remarkable one for me is the universal speed of light. Einstein postulated that the speed of light in vacuum is the same for all inertial observers, regardless of the motion of the source. But what if we are accelerating and we are changing from one inertial frame to another, how will we observe the speed of light? Is it still the same? I suggest it's not? Because I alread saw a small introduction to general relativity where the light will bend in a gravitational field and therefore also when you accelerate. So does the speed of light change because it bends?
How can I straighten ALL the vertical UV lines at once? Is there a hot keyshort cut? Or if I have to do them one by one .. is there at shortcut for that? Thank you! How to revert uncommitted changes including files and folders? How to get GNOME Shell "Applications overview" to show the full name of applications Making an Emission shader emit a different colour of light than the colour assigned to the object? Prove that the unit circle, $S^1:=\{(x,y)\in \mathbb{R}^2: x^2+y^2=1\}$, Is a one-dimensional manifold Linux, find all files matching pattern and delete Looking to find all files (recursively) which have an underscore in their file name and then delete them via command line. Why do clocks measure arc-length? Apologies in advance for the long question.  My understanding is that in GR, massive observers move along timelike curves $x^\mu(\lambda)$, and if an observer moves from point $x^\mu(\lambda_a)$ to $x^\mu(\lambda_b)$, then his clock will measure that an amount of time $t_{ba}$ given by the curve's arc length; $$   t_{ba} = \int_{\lambda_a}^{\lambda_b}d\lambda \sqrt{-g_{\mu\nu}(x(\lambda))\dot x^\mu(\lambda)\dot x^\nu(\lambda)} $$ will have elapsed where $g_{\mu\nu}$ is a metric on spacetime with signature $(-,+,+,+)$.    Why is this so?     Here is how I would attempt to justify this fact in special relativity with $g_{\mu\nu} = \eta_{\mu\nu}$.  Consider an inertial observer $O$ in $\mathbb R^{3,1}$, and suppose that this observer sees a clock, which I'll call observer $O'$ moving around on a curve $x^\mu(\lambda)$.  If $O'$ were also an inertial observer, then given any event with coordinates $x^\mu$ as measured by $O$, observer $O'$ would measure the coordinates of the event to be $x'^\mu = \Lambda^\mu_{\phantom\mu\nu} x^\nu + x_0^\mu$ for some Lorentz transformation $\Lambda$.  If $O'$ is not inertial, then this is no longer true, and there is some more complicated family of transformations, say $T_\lambda$ between events as seen by both observers.  I would argue, however, that if we were to partition the interval $[\lambda_a, \lambda_b]$ into a large number $N$ of intervals $I_1=[\lambda_a, \lambda_i], \dots, I_N=[\lambda_{N-1}, \lambda_b]$ with $\lambda_n = \lambda_a+n\epsilon_N$ and $\epsilon_N=(\lambda_b-\lambda_a)/N$, then on each interval $I_n$, $O'$ is approximately an inertial observer in the sense that $$   T_{\lambda_n} = P_n + \mathcal O(\epsilon_N), \qquad (\star) $$ for some Poincare transformation $P_n$.  Then we would note that since $O'$ is stationary in his own reference frame, he measures his worldline to have the property $\dot x'^\mu(\lambda) = (\dot t(\lambda), \mathbf 0)$ so that $$   I_{ba}=\int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x'^\mu\dot x'^\mu} = \int_{\lambda_a}^{\lambda_b} d\lambda \, \sqrt{\dot t^2} = t(\lambda_b) - t(\lambda_a) = t_{ba} $$ On the other hand the integral on the left can be written as a Riemann sum using the partition above, and we can invoke ($\star$) above to get \begin{align}   I_{ba}  &amp;= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x'^\mu(\lambda_n)\dot x'^\nu(\lambda_n)}\right] \notag\\ &amp;= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x^\mu(\lambda_n)\dot x^\nu(\lambda_n)} + \mathcal{O}(\epsilon_N^2)\right] \notag\\ &amp;= \int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\mu} \end{align} Combining these two computations gives the desired result.  How do others feel about this argument?  I'm not completely comfortable with it because of the assumption $(\star)$ I made on $T_\lambda$.  I imagine that in GR a similar argument could be made by invoking local flatness of the metric. Exterior power "commutes" with direct sum I know that for vector spaces $V, W$ over a field $K$, we have the following identity :  $$ \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \simeq \Lambda^n(V \oplus W) $$ which holds in a canonical way : for $0 \le k \le n$, the maps  $$ \Lambda^k(V) \times \Lambda^{n-k}(W) \to \Lambda^n(V \oplus W), \\ (v_1 \wedge\cdots \wedge v_k,w_{k+1} \wedge \cdots \wedge w_n) \mapsto (v_1 \wedge \cdots \wedge v_k) \wedge (w_{k+1} \wedge \cdots \wedge w_n) $$ factors through the tensor product's universal properties, which gives us a map $\varphi_k : \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \to \Lambda^n(V \oplus W)$, and since the category of $K$-modules has biproducts, we get a canonical map  $$ \varphi = \bigoplus_{k=0}^n \varphi_k : \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \longrightarrow \Lambda^n(V \oplus W) $$  What if I replace $K$ by a (commutative ring with $1$) ring $A$? Does this map have a kernel? I know it is surjective (it suffices to "look at it", i.e generators of the codomain are obviously all hit by $\varphi$). I don't know if it helps, but I am working in integral domains. In the field case, this is obvious if $V$ and $W$ are finitely generated, since one can use dimension arguments. I wouldn't mind if the above identity held only in the finitely generated case either. If it worked in general it would be cool though.   Note that when trying to compute the inverse map with the universal property, there seems to be a problem... my guess would have been $$ (v_1 + w_1,\cdots,v_n + w_n) \mapsto \bigoplus_{k=0}^n (v_1 \wedge \cdots \wedge v_k) \otimes (w_{k+1} \wedge \cdots \wedge w_n) $$ and this map is obviously $A$-multilinear, but there is a problem with the alternating property ; I get some terms left. I'm guessing this is the wrong map... if there is one. Find $P(n+1)$ for a polynomial P Fix a nonnegative integer $n$. Let $P(x)$ be a polynomial of degree $n$ (over the real numbers) such that for all $k\in\left\{0,1,\ldots,n\right\}$, we have $P(k)=\dfrac{k}{k+1}$. Find $P(n+1)$.   My attempt:  I consider a new polynomial $Q(x)=(x+1)P(x)-x$. This $Q$ is a polynomial of degree $\leq n+1$ satisfying: $Q(k)=0\quad\forall k\in\{0,1,2,\ldots,n\}$. Therefore $Q(x)=\lambda x(x-1)\cdots(x-n)$ for a certain real $\lambda$. Hence, $Q(-1)=\lambda \cdot (-1)^{n+1}(n+1)!$, but also $Q(-1)=(-1+1)P(-1)+1=1$. Comparing these, we get $\lambda \cdot (-1)^{n+1} (n+1)! = 1$, thus $\lambda=\dfrac{(-1)^{n+1}}{(n+1)!}$. Since $Q(n+1)=\lambda(n+1)!$ then $Q(n+1)=(-1)^{n+1} $, and from $P(x) = \dfrac{Q(x)+x}{x+1}$ we deduce: $$P(n+1)=\dfrac{(-1)^{n+1}+n+1}{n+2}.$$ I doubted that the answer were $P(n+1)=\frac{n+1}{n+2}$ but the term $(-1)^{n+1}$ wasn’t expected. Is there any mistake here? Does someone have an alternative proof? I finally got fed up when wanting to read about bash's read and it's -s option with man bash. I found the right spot eventually (around line 4500), but it was a frustrating as usual, since both /read and even /\s-s\s searches have way too many matches.  So, the question is: How can I read long man pages efficiently, or get same information in other ways, locally? As a specific example, how to reach the relevant documentation after seeing read -s pwd in a shell script? A good answer could be a shell script snippet, or hint about some tool and how it is used, or something else entirely, as long as it helps in finding the right spot to read.  Note: I'm not tagging with  because I want the question to be about man page reading in general, even though that quite possibly is the most commonly encountered humongous man page. When no model comparison, should I use REML vs ML? I'm running LMM, and I will make no comparison of models. Could I ask which one should I use between REML and ML? Let $R$ be a commutative ring with identity, and $S$ a multiplicative subset of $R$ not containing $0$. Which step in my reasoning is wrong?   Every element of $S$ is a unit in $S^{-1}R$. Every element of $S$ that's a zero divisor in $R$ is a zero divisor in $S^{-1}R$. But zero divisors can't be units. Therefore it's not possible to localize $R$ at $S$. Undercooked beans in chili? I made a bit of a mistake, being a lazy bachelor and inexperienced at cooking.  I was making chili and when it came time to add beans I discovered I didn't have any canned beans.  I did however have some dried beans.  I decided to save time and add these directly to the pot of chili and add some water.  However, after much simmering the beans maintain a certain unpalatable crispness.  I have also read that undercooked beans contain a chemical lectin which can cause stomach discomfort.  I have also read that beans should be soaked, the water discarded to remove the lectin, and then boiled to cook them.  Is there any way to salvage this?  Should I just continue cooking the chili for hours?  Will I have to throw out my potentially delicious chili? The following happened playing Minesweeper:   How do I figure out which square to click? Is it possible to animate which characters of a text object are assigned to a material? What do we mean by hyperparameters? Can anyone give me full details about what we mean by hyperparameters, and what in the Dirichlet distribution are called hyperparameters? A practice example for the estimation of those parameters would also be useful. "State-of-the-art" and "technology" Ask question post button remains disabled I've noticed this twice in the last ~2 weeks or so. The &quot;post your question&quot; button can remain disabled for some reason even when your question is shown as &quot;ready to publish&quot;. I.e., you've gotten to the &quot;review&quot; page after the initial edit. If I remember correctly, one case was on Stack Overflow, and the other was on . I'm not sure what exact scenario leads up to this. I haven't paid attention to the button until the last step when I'm ready to click it. The workaround has been to reload the page. (I copied my content for safe-keeping first.) This works, and the button then shows up as active. Afterwards, however, on both occasions, it forced me through the &quot;are you a human&quot; verification.  Update - this question itself did not exhibit this, nor did it do the .  Update - I'm using Chrome on Windows and am not sure what version it was when this happened.  It has  installed, but there aren't any scripts for any Stack Exchange sites.  It has  installed.  The virus scanner on the PC I probably was using was How can I use $variable in a shell brace expansion of a sequence? I want to use $var in a shell brace expansion with a range, in bash. Simply putting {$var1..$var2} doesn't work, so I went "lateral"...   The following works, but it's a bit kludgey.    # remove the split files echo rm foo.{$ext0..$extN} rm-segments &gt; rm-segments source rm-segments   Is there a more "normal" way? has played the role of Wolverine as a major character in various X-Men films since 2000 and contributed highly to the brand of the X-Men movie series. But is he playing the Wolverine character for the last time in Logan?  I am getting to know from my friends that  is the last movie in which he is playing Wolverine and they all say that this is his last contracted film.  Are these rumors true? Is there any official information that confirms that this is Hugh Jackman's last movie as Wolverine?