hi all, there is a large contingent in here who are aggressively pursuing higher edu & advanced readings in physics, has anyone looked into coursera courses for physics? surprised it hasnt been mentioned. its very good for cs stuff, an initial focus, but lately expanded into general areas eg physics. was looking for GR related stuff which is a big topic in here, found some. (may be better for SR.) some have complained about book prices in here & heres some alternative. coursera.org
I'm not a physicist and I only have a basic understanding of science but I was wondering if gravity can move between universe in string theroey wouldn't it be possible to prove multiple dementions buy waiting untell gravitational waves to be found that has no point of origin and therefor must be ...
While giving comments many times I wanted to add links to my comment. But instead of using some short hand like Read "this" (this is a link in this case) I had to write Read "http:\\www.somelink.com". I tried ctrl+L to insert a link to my comment but it was of no use.
I have seen people inserti...
The terms exotic matter and negative matter/energy mean the same thing. And yes, any form of FTL drive requires exotic matter. Hawking proved that in his paper on chronology protection.
Well, he proved that closed timelike curves require exotic matter, but since any FTL travel implies the existance of CTCs his theorem covers FTL travel as well.
It isn't clear whether the Alcubierre drive would work even if exotic matter were available. There are arguments based on some ideas in quantum gravity that it could not work.
@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ again, when the Six Nations is on I'll check the results. But I only rarely watch a match, and even then only on the TV never at the ground.
I think Dark Side of the Moon is genuinely a seminal album that really changed the music world. But Wish You Were Here and Animals mainly consisted of tracks written at the same time as DSOTM. So basically they did nothing new until The Wall.
And by the time they did The Wall the band was disfunctional.
I don't think they're that special live. When I see a live band I like to really get into the spirit i.e. dance about like a loon and scream and shout. The Floyd are more a band you sit down and listen to.
The very, very best live show I've been to was Ozzy Osbourne's first solo tour in (I think) 1979 or 80.
If you write $G = T$, where $T$ is the stress-energy tensor for a point object then you can get a moving object just by applying the same Lorentz boost to both sides. Yes?
I was trying to develop an argument that starts with a moving particle and shows that the metric is the same as for a stationary particle by Lorentz boosting into the rest frame of the particle.
https://www.youtube.com/watch?v=GT8iVYfg9eM
"Skeptical scientists keep dismissing the concept of reincarnation based on the fact that there is no need for a soul. However, one of the most skeptical and smartest scientists ever, Carl Sagan, the author of the best book on skepticism 'The Demon Hau...
My two cents on that question: There are many issues postulating that entanglement as the essence of a soul 1. Entanglement, as far we knew, is way too fragile to exist for more than brief periods of time 2. Saying that some atoms in one neuron is entangled with another in explaining a soulmate seemed to be a gross simplification of the complexity of the human brain. How can you be sure that basically such a massive dynamical system will not drown out any possible quantum effects that can manifest. Also with so many entangled pairs (in the billions) we expect them to last less than an alto…
@Secret I can't of course prevent you from taking the question seriously, but it does make me wonder if you really haven't got anything better to do :-)
My way of handling cases when scientific concepts is being mystified to explain esoteric concepts is not just to treat it a nonsense (and as it turns out, they often are), but to explain why it is nonsense and how their proposals will fail by highlighting the key areas that will fail. (well most of the time, such analysis mostly benefits myself in convincing myself why they are nonsense/not testable anyway, as ever since I become an agnostic (although still mostly in science), I are aware myself becoming quite suceptible to esoteric concepts thus I have to stay alert)
Especially if you actually look closely at art, science, esoterics, religion, business, culture, the way they use their jargon shares a very simialr structural pattern.
It is as if they are using completely different terms to describe the same thing, or that this similarity is basically highlighting how we humans use language
> but it does make me wonder if you really haven't got anything better to do :-)
The good thing is that typing the above wall of text does not take me much time, thus it is a reasonable investment
@0celo7 Actually, screw what I and Acuriousmind talked about yesterday, $b$ is ultimately found to be $0$ because $$b=b+0=0^2+0=0(0+1)=0*1=0$$
Now I am suspecting that as long both multiplicative and additive identities are present along with distributive law, it will guarentee there exists a unique multiplicative absorbing element 0. Wikipedia on absorbing element also said that if 0 ever absorb itself, then that algebraic structure can only have 0 as th absorbing element
Purpose
The user base of Physics Stack Exchange is rather diverse in terms of cultural background, career progress, field of expertise, and personality.
While the discussion in hbar is lively, it is mostly focused on physics, with some side conversations on a variety of topics from cooking to fa...
Attempt to break the proof Let elements a,b,c,d,e,f,g,h,i,0 Then: $$0^2=b=c+d=0(e+f)=0g=h$$ $$bg=0^2g=0(0g)=0h=i$$ Now to determine what constraints do a to h need to obey to ensure $b\neq 0$
(Above showed my engineer attitude to maths problem (try to rig it until the desirable outcome is acheived, or die trying))
How would the laws of physics change if the speed of light was not constant and the Michaelson-Morley experiment was instead proven to be true?
Obviously relativity would not be true, but on a small scale, what would the differences be? Would Maxwell's equations hold up properly? Would electric...
Line 4 suggests multiplicative identity cannot exist $$0g\neq 0$$ As otherwise we have at least one case where 0 is absorbing, and we will be screwed because the proof will survive
So it does seems, all it takes is additive identity and distributive law (and that the binary operation is only defined as the juxaposition of the two elements that form its argument) to guarentee the existence of an absorbing element for the smallest size semiring
The existence of even just one neutral element (an element a such that a+b=b where b is some elements in the proper subset of the algebraic structure in question) is enough to lock up nearly half of the entries in the + Cayley table. The existence of an additive identity constrains all elements in the + cayley table.
Also $0+0$, because of distributive law, will determine the value of $b+b$
Most importantly, any such structure must not also have multiplicative identity elements (or at least, cannot exist multplicative neutral elements e for 0 such that $0*e=0$) else it will immediately collapse into the trivial ring
typo: all elements involving 0 in the + cayley table
Since a multiplicative identity cannot exist in a distributive structure without implying the existence of the unique absorber 0, division by zero algebric structures MUST be non distributive, or has binary operations defined nontrivially (not as an element which is the juxaposition of its two arguments)
Nontrivial semirings without absorber can exist with distributivity and identities only if the binary operation is nontrivial, such as this one:
What is an instructive example of a set $X$ equipped with two monoid structures $(X,+,0)$, $(X,\cdot,1)$, such that $+$ is commutative, the distributive laws hold, but $0 \cdot x = 0$ or $x \cdot 0 = 0$ do not hold?
Notice that in case these two absorbing laws hold, one calls $(X,+,0,\cdot,1)$ a...
Hi! I am studying graph theory applied to feynman diagrams and amplitues. There is a chapter in the book relating to find divergent subgraphs. I call $G$ the graph and $l(G)$ it's internal lines. At a certain point, the author says "we can always restrict ourselves to consider subgraphs which are regular (no isolated vertices) and which furthermore correspond to subsets of internal lines of G. Hence there are exactly $2^l(G)$ such subgraphs in G. "
My question is: why $2^l(G)$? I tried to compute all possible subgraphs of the graph with two vertices and three legs for each one contracted one-by-one . And, I see there are 7 subgraphs (including the graph itself)
@ACuriousMind I'm still confused. We have our unit vector field $v$ with its zero at the origin, and we take two spheres $S_\epsilon$ and $S_\delta$ that work in the definition of the index. Sure $S_\epsilon$ and $S_\delta$ are diffeomorphic, but how does that induce a homotopy on the maps $v:S_\epsilon\to S^n$ and $v:S_\delta\to S^n$?
In particular, these maps have a different domain!
Let $f:S_\epsilon\to S_\delta$ be an orientation preserving diffeomorphism.
Let the map $v:S_\epsilon \to S^n$ be denoted $v_\epsilon$ and let the map $v:S_\delta\to S^n$ be denoted $v_\delta$. Then I'm willing to believe that $v_\delta \circ f$ and $v_\epsilon$ are homotopic and have the same degree.
But I'm still not sure why $v_\delta$ and $v_\epsilon$ have the same degree.
@ACuriousMind Oh wait, $\mathrm{deg}(f)=+1$, so if $v_\delta\circ f\sim v_\epsilon$ then $\mathrm{deg}(v_\epsilon)=\mathrm{deg}(v_\delta \circ f)=\mathrm{deg}(v_\delta)\mathrm{deg}(f)=\mathrm{deg}(v_\delta)$?
@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Now don't ever tell me there isn't someone out to get us.
@ACuriousMind Ah! Assume WLOG $\epsilon<\delta$ and let $\eta:=\delta-\epsilon$. Consider the map $g:S_\epsilon\times[0,1]\to\Bbb R^{n+1}$ which for fixed $t$ is the diffeomorphism $g_t:S_\epsilon\to S_{\epsilon+t\eta}$.
Thus it it smooth in $t$ and $g_1=f$ from above.
Then we have a homotopy $F:S_\epsilon\times[0,1]\to S^n$ between $v_\delta\circ f$ and $v_\epsilon$ given by $F(x,t)=(v\circ g_t)(x)$.
@ACuriousMind I would appreciate you letting me know if this is alright, I came up with this myself (I'm impressed with myself if it's correct)
I guess I should say $g_t$ is the diffeomorphism obtained by scaling the vectors pointing from the origin and thus it's smooth in $t$.
user116211
1:58 PM
A $r$-neighbourhood is always a uniform neighbourhood.