what is causing gravitational pull in between massive bodies? what is causing gravitational pull in between massive bodies?     according to books, what i learnt is graviton particle &amp; some books are saying quantum field is the cause, so which is causing gravitational pull?
What is the cause of gravitional force of attraction? Ok,this is a silly question. But for quite long I have been thinking about this. What is gravitation all about?? My book directly writes since force of gravitation is directly proportional to the masses and inversly proportional to the square of the distance ie. $$F = G\frac{M_1 . M_2}{r^2}$$ . So is this gravitation all about? What is the cause of it? The book then says     We are not in the scope in classical approach so as to discuss the cause of this. However,as a matter of fact that if there are more than one mass,then there is gravitation as it is their fundamental property.( Cause of gravitation can be attributed to exchange of non-classical particles between the masses) .    Then it ended the talkings. What are those non-classical particles? Why and how are they exchanged? So far I have studied physics , I never found such weird phenomenon like this. I want to know what is the cause of gravitation and what is the cause of exchange of those so-called non-classical particles? And how do they do it?
Make gedit embedded terminal colours the same as the default terminal I've just updated to 11.10 and now my embedded terminal in gedit does not match my default terminal. I hate looking at a white and black terminal. Does anyone know how to make the embedded terminal match the default terminal?  This is a screenshot:    I want my embedded terminal in gedit to match my default terminal. series of young adult science fiction books from early 1980s about last of a race of martial experts that sees the destruction of his home planet User suggested great corrective edit to my answer, but the edit was rejected. What can I do to approve it? How do we pass the field api name in lwc for a lightning data table? Describing groups with given presentation? $\langle x,y\ |\ xy=yx,x^5=y^3\rangle$ and $\langle x,y\ |\ xy=yx,x^4=y^2\rangle$. I'm trying to describe the groups with presentations $\langle x,y\ |\ xy=yx,x^5=y^3\rangle$ and $\langle x,y\ |\ xy=yx,x^4=y^2\rangle$. I have some problems getting a good picture of what they look like...   For the first one, $xy=yx$ allows us to say that any element in the group can be written as $x^iy^j$ for some $i,j$. More precisely, using $x^5=y^3$, we can always reduce $j$ to $0,1$ or $2$, whereas $i$ could have any value. So I was thinking about $\mathbb{Z} \times C_3$. But the other way around (reducing $i$ this time) it's also $C_5 \times \mathbb{Z}$. But then I should be able to prove that both descriptions are equivalent, which I can't do...  For the second, I proceed similarly and get to the same kind of problem...  Could you tell me where I'm going wrong?  Thank you very much in advance! What is the correct way to show tiered possession? When there are multiple places where possession is shown, how are the words structured?  A few examples:   Greg's son's laptop is on fire. Mary's friends' boat is sinking. Thomas's neighbor's cat ate a mouse. Apple's Stock App's stats are useful. Microsoft's Xbox 360's games are fun.   Are these the correct ways to express possession? I used to find color codes which I want to use for texts or background colors on htmls with color edit tool of MS paint which pre-installed on Windows. But now I use Mac instead of Windows. How can I find color codes with Mac? Is an software like MS paint pre-installed on Mac? What is an idiom/slang for "someone who pretends to be good when they're not"? This is not from real life, but from a movie on my local TV. A character in the movie is really bad, but when she talks with others, she pretends to be an innocent/ good woman. I want to know how to describe this kind of person. Area of intersection of two Annulus Given two separate annulus with centers $[C_1,C_2]$ and their corresponding radii being $[R_1,r_1]$ and $[R_2,r_2]$ respectively, larger radius being $R$. There are methods to look at whether they are overlapping or not. If they overlap how to find the overlapping area. Is there any close form expression for it?  equation of first annulus   $r_1^2&lt;(x-x_{c1})^2 + (y-y_{c1})^2 &lt;R_1^2$   equation of second annulus  $r_2^2&lt;(x-x_{c2})^2 + (y-y_{c2})^2 &lt;R_2^2$  If there are three circles overlapping, we get overlapping area as solution to set of 4 in-equations by least squares approach(Though we don't end up getting total area but, area within tangents of circles ). What happened to /etc/apt/apt.conf? I was browsing through the apt folder to try my proxy settings and I was unable to find the apt.conf file in Ubuntu 11.04. Is there any alternative file for it?  Adding proxy settings to 99-synaptic in apt.conf.d does not help as it gets overwritten everytime I restart the system. How many subgroups of order $n$ does $D_n$ have? How many subgroups of order $n$ does $D_n$ have?  My work:  Since subgroups of $D_n$ are either cyclic or dihedral, the subgroups of order $n$ of $D_n$ are $\left&lt; r \right&gt;$ (cyclic) and $D_{n/2}$ (dihedral). However, this works only for $n$ is even and $n\geq 2$. I don't know how to do when $n$ is odd. Find when three numbers have the same remainder when divided by the same number If three numbers 112, 232, and 400 are each divided by the number D, each of their quotients will have the same remainder R. Find R where R>1  How should I approach this? Remove Gnome Notifications or set Permanent Do not Disturb How to view recently opened media files? In windows we have an option to view recently opened media files as my recent documents, I was  wondering does there exists something like that in Ubuntu?  via terminal or via GUI? Low current battery monitoring Subgroups of finitely generated groups are not necessarily finitely generated I was wondering this today, and my algebra professor didn't know the answer.      Are subgroups of finitely generated groups also finitely generated?    I suppose it is necessarily true for finitely generated abelian groups, but is it true in general?  And if not, is there a simple example of a finitely generated group with a non-finitely generated subgroup?  NOTE:  This question has been merged with another question, asked by an undergraduate.  For an example not involving free groups, please see Andreas Caranti's answer, which was the accepted answer on the merged question. Problem with installing Steam on Ubuntu 15.04+ A doubt regarding change of variables in Double Integrals. So I have recently learnt to calculate &quot;Double Integrals&quot; and I have a doubt on change of variables. Lets say we are evaluating a double integral of the form I = $\int\int _R F(x,y) dx dy $ , where $F(x,y)$ is an integrable smooth function and $R$ is the region of integration in the $xy$ plane. Now I want to convert to polar co-ordinates $x(r,\theta)=r \cos(\theta)$ and $y(r,\theta) = r \sin(\theta)$. Now the region $R$ in the $xy$ palne can be suitably converted to some region $R'$ in the $r \theta$ plane. The function $F(x,y)$ may take the form of some smooth and integrable $G(r,\theta)$. My main doubt is with the change of the area element. According to my textbook, $dx dy = \frac{\partial (x,y)}{\partial(r,\theta)} = rdrd\theta$. So the area element in the $r\theta$ plane is $rdr d\theta$. Now I want to experiment a little bit with my understanding : I write $dx = d(r\cos\theta) = -r\sin\theta d\theta + \cos \theta dr$  and $dy = d(r\sin\theta)=r\cos\theta d\theta + \sin\theta dr $. And now I multiply the two expression for $dx$ and $dy$ : $dxdy= -r^2\sin\theta\cos\theta d\theta d\theta - r \sin^2 \theta drd\theta + r \cos^2 \theta dr d\theta + \sin \theta \cos\theta dr dr$. Why isn't this expression equal to $rdrd\theta$ ? The $d\theta d\theta$ and $dr dr$ terms look very odd to me. Am I commiting some abuse of notation or something ? Also there are no higher order terms like $dr dr d\theta$ which I can ignore. I am totally confused with all this. Can someone explain this to me a bit clearly? Thanks in Advance :). EDIT 1 : I know differentials are not the &quot;usual&quot; kind of algebraic quantities and we can't just add and multiply them casually. I am just multiplying $dx$ with $dy$ with the spirit of multiplying some $\delta x$ and $\delta y$ and letting both $\delta x$ and $\delta y$ tend to zero. EDIT 2 : Is there a branch of Mathematics where addition and multiplication of differentials is very common? Is there something like differential of a differential ? How to find fixed points for DES weak keys $f$ is irreducible $\iff$ $G$ act transitively on the roots Let $F$ be a field and $f\in F[X]$ a separable polynomial. Let $K_f$ be the splitting field of $f$ and $G_f=\text{Gal}(K_f/F)$ its Galois group. Show that $f$ is irreducible $\iff$ $G_f$ acts transitively on the roots on $f$. My work $\implies :$ Let $f\in F[x]$ irreducible, $\alpha \in K_f$ a root and $\alpha _1,...,\alpha _n$ the other roots of $f$ (so $K_f=F(\alpha ,\alpha _1,...,\alpha _n)$). What I have to show (I think) is that for all $i$ there is a $\sigma \in G_f$ s.t. $\sigma \alpha =\alpha _i.$ Q1) Is it right ? Q2) Is $f$ the minimal polynomial of $\alpha $ ? (To me it is, but my teacher said that it can be not the case, but I don't understand why). Since $\alpha $ and $\alpha _1$ have the same minimal polynomial, there is an isomorphism $\sigma :\alpha \longmapsto \alpha _1$. We can observe that $\sigma : F(\alpha )\longrightarrow K_f$. Now $F(\alpha ,\alpha _1)/F(\alpha )$ is algebraic, then we can extend $\sigma $ to a morphism $F(\alpha ,\alpha _1)\longrightarrow K_f$. Now, we can continue like that and extend $\sigma $ as $\psi: K_f\longrightarrow K_f$ s.t. $\psi(\alpha )=\alpha _1$. Q3) The fact that $\psi(\alpha )=\alpha _1$ comes from the fact that $\psi|_{F(\alpha )}=\sigma $, right ? The bijectivity is obvious since it's a field homomorphism and that $\sigma (K_f)=K_f$. Q4) How to conclude that $G_f$ act transitively on the roots ? I don't have particular problem for the converse.