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11:00 PM
but not to B with respect to A
 
Is "B with respect to A" an inertial frame...?
 
@Slereah Octonion GR?
 
(answer: No.)
 
@punkerplunk Frames are not objects. Frames are choices of coordinates. "The earth is the inertial frame" is a non-sensical statement.
 
ACuri I agree
 
11:02 PM
His name is ACM or Bajoran
ACuri slanders the name of Madame Curie
She was no mathematician
 
I just don't see why B is allowed but nobody allows X
 
She respected the Evidence
 
@0celo7 I don't think you can have octonion GR
Octonions are linked to like...
Hm
What Clifford algebra has 8 even components
 
@punkerplunk So, I think I get what you mean. If S1 is an inertial spaceship, S2 is one triplet doing the accelerating weirdness, and S3 is another one. Your question is, why is "S2 in S1's frame" valid to look at the twin paradox, but not "S3 in S2's frame"?
There really is nothing at all less philosophically pleasing about special relativity than there is about Newtonian mechanics/galilean relativity.
 
the great thing about Clifford algebras is
 
11:08 PM
One is invariant under the galilean group, one is invariant under the poincare group.
 
It is basically what your math teacher told you to never do
 
@NeuroFuzzy sort of yes. More specifically, though, I'm wondering if we can just consider the earth as another spaceship, and say it exists in another 'inertial' frame.
 
@Slereah Huh?
 
You can ADD A SCALAR TO A VECTOR
 
Yeah, how does that work anyway
 
11:10 PM
@punkerplunk well but it's not inertial in quotes. It's a real physical assumption for the frame to be inertial. So if you consider the earth moving slowly (much less than the speed of light) then yeah, you can just consider the earth as another spaceship, and say that its reference frame is inertial (ie, you neglect the Foucault pendulum and tides and blah blah noninertial frame stuff) .
 
Well it's not really "adding"
It's more that you are defining a space with basis vectors that are both
 
Huh
 
But what if the earth is moving at 0.99 the speed of light, and nobody notices because they're sharing that frame of reference
 
@punkerplunk Well, it is, with respect to some frame of reference. So what?
 
So, than we allow X
 
11:13 PM
?
 
Xandra
 
what do you mean "allow"?
 
I mean "fail to disprove"
 
"Xandra" isn't a logical statement within special relativity. I have no idea what you mean.
 
I mean, allow a journey such that someone can leave at some arbitrarily fast velocity, and return older
the opposite of the twin paradox
 
11:15 PM
Clifford algebra is like
You take all the n-forms of your space
And you make a vector space out of them
 
So long as they enter that 'exotic' frame inwhich the earth is in motion
 
@punkerplunk We do not allow that, no, that is not satisfied by any timelike path, as proven by the triangle inequality that I linked you.
 
@NeuroFuzzy shows you haven't read HE
 
@punkerplunk can you phrase things in terms of a Minkowski diagram and post it here?
 
Hm wait
 
11:17 PM
@0celo7 IN SPECIAL RELATIVITY DOOD
 
Clifford algebras have some business with the metric tensor
What was it again
 
@NeuroFuzzy I WAS JOKING DOOD
you need to calm down tonight
 
@0celo7 OH
 
I'm the sick one, you're killing my blood pressure
::eats another cough drop::
 
Oh right
 
11:18 PM
wonder if you can OD on these things
 
Clifford algebra uses the weird ass product
 
@everyone sorry if I'm filing the chat with newbie cancer
 
are there any non weird ass-products?
 
$uv = g(u,v) + u \wedge v$
 
wtf
@ACuriousMind explain pls
 
11:19 PM
So a scalar and a 2-form
 
I think @Slereah is mixing up "geometric algebra" and "Clifford algebra" here.
 
@punkerplunk It's fine! By "phrase it in terms of a Minkowski diagram" I mean, define a bunch of $(ct,x)$ points, and lines between the points.
 
Yeah I'm a bit fuzzy on Clifford algebras
 
@ACuriousMind IIRC cliford algebra has some crazy shit in it
 
While "geometric algebra" works with the Clifford algebra, it defines a bunch of other, indeed weird-ass, operations on it
 
11:20 PM
@punkerplunk with a line for the Earth and lines for the spaceships/observers, etc.
 
yeah
actually someone did it for me:
 
The Clifford algebra itself just has the product $v^2 = g(v,v)$.
 
@punkerplunk you'll find that there is no bunch of $(c\Delta t,\Delta x)$ that violate that triangle inequality. You just can't do it.
 
My only book on Clifford algebras is a weird ass book
 
@ACuriousMind please explain
 
11:21 PM
 
Okay, and the triangle inequality I linked is satisfied in that picture.
 
That picture, only this time replace blue line with "earth" and replace "earth" with X
 
Is that picture in an inertial frame?
Then some cataclysm must have happened to accelerate Earth to relativistic speeds
 
@0celo7 It means you take the free algebra on the space (i.e. just take the basis vectors $e_i$ and define additional objects $e_i\cdot e_j$,$e_i\cdot e_j\cdot e_k$ and so on, and then you impose the constraint $v^2 - g(v,v) = 0$.
 
No, you;re just bias because your'e on the earth
 
11:23 PM
Note that this is very similar to how one obtains the exterior algebra by imposing $v^2 = 0$ on the free (tensor) algebra.
 
o.o Bajoran you've lost me again
@ACuriousMind what
::cries::
what is a two-sided ideal
 
Halp
Too many algebras
 
@0celo7 Okay, lets start the other way around. The exterior algebra comes from imposing $v\otimes v = 0$ on the tensor algebra
 
@punkerplunk okay, I'm out, we're going in circles. The bottom line is inertial frames are special. They're special in special relativity, they're special in newtonian mechanics, they're absolute. There's a big difference between inertial and noninertial frames, and the laws of physics in inertial frames are different than in noninertial ones. You cannot apply the same laws in noninertial frames as you can in inertial ones.
 
@ACuriousMind it does?
 
11:25 PM
The Clifford algebra is just a modification of that - we impose $v\otimes v = g(v,v)$ instead.
 
You might have to slow down for us dummies :p
 
@0celo7 Of course! That's what antisymmetrization is
 
Why $v \otimes v = 0 $ generate exterior algebra
Oh
 
@NeuroFuzzy the laws are not different! there are just terms that fall out for inertial frames
 
If I cannot assign the inertial, than neither can you.
 
11:26 PM
But wait
 
isn't it $v\wedge v=0$
 
While if it's antisymmetric, it will be 0, is the reverse true?
 
@0celo7 potato potato
 
@NeuroFuzzy more like correcto confusingo
 
@0celo7 Yes, one calls the product in the resulting thing $\wedge$ instead of $\otimes$, no big deal.
 
11:26 PM
@ACuriousMind I SEE
so $\otimes$ is not the standard tensor product
 
@Slereah Unless you are over a field with characteristic two, yes.
 
it's just some new product
 
that is a math problem
Not enough symbols to go around
 
oh god we've gone into math mode again. the mathematicians have invaded
 
@0celo7 Well...not after you have imposed $v\otimes v = 0$...
(or, formally, divided out the ideal generated by that)
 
11:28 PM
alarm: evidence will be erased if we don't get rid of them
@ACuriousMind what's an ideal
 
@0celo7 Informally, the "ideal generated by $X$" is the set of all multiples of $X$.
 
@0celo7 annoying/distracting from the point -.- anyways I'm off.
 
@NeuroFuzzy no, it's confusing to say that they are not the same laws of physics
the laws are exactly the same
> set
 
@ACuriousMind : Basically you are only considering elements v such that v x v = 0?
 
In this case, the ideal generated by the relation $v\otimes v = 0$ is the set of all tensors such that they can be written as $v\otimes v \otimes \dots$ for some $v$.
(We must take the ideal because the only things you can divide out of rings such that the quotient is still a ring are ideals)
You people need to take some sweet abstract algebra ;)
 
11:34 PM
@ACuriousMind next year
@ACuriousMind what
@ACuriousMind eh, I'll believe you on that one
 
@0celo7 for example, in introductory courses you say the laws are newton's laws. Well, the form of Newton's law F=ma changes. So the law changes from frame to frame. Of course that's not true in the lagrangian formulation though, because that transforms with the coordinate system. That's the sense in which I mean it, and it doesn't confuse things to say that.
It elucidates that in newtonian mech/spec rel, inertial frames are real/physical/tangible/important. Different definitions give different theorems.
 
obe
Can you do newtonian mechanics in generalized coordinates?
 
yes
it's called Lagrangian mechanics
 
obe
:(
 
I thought you read Arnold
 
obe
11:40 PM
Really...
lagrangian mechanics is not an extension of newtonian mechanics.
it's entirely different.
f = ma with generalized coordinates doesn't give you the euler-lagrange equations does it?
What does it do though?
 
did you do the problem set
you should do 6) right now
no QM needed
 
obe
I'll do it later.
This will end up being the last night before.
 
oh god it's like @Secret but worse
über massif sounds interesting
 
What is it with people posting incomprehensible drawings here lately? :P
Also, are these swirls eyes staring at us?
 
11:46 PM
dei mutti ist über massif
 
They somehow resemble hearthstone tokens
all by accedent
I'm trying to satisfy the constraints of 'inertial' frame
 
Random math question here. When solving Laplace's equation to find spherical harmonics, the expression$$\frac{1}{\Phi}\frac{\mathrm{d}^2\Phi}{\mathrm{d}\phi^2}=-m^2$$is always used. Why $\Phi$, and not $\Theta$?
 
obe
Inertia frames satisfy $\frac{d\vec{e_r}}{dt}=0$
 
(latex isn't encoding correctly)
on my browser :( I'm not getting the full glory of yoru equations!
 
@HDE226868 Because we only use other Greek letters when we have run out of all the phis ;)
Seriously, some people use $\phi,\varphi,\Phi$ before using another letter
(They should be burned at the stake)
 
11:51 PM
@ACuriousMind Haha. I meant why is that particular coordinate ($\phi$) used, and not $\theta$?
 
obe
It doesn't matter does it?
 
therefore light is not moving, and it is our apparatus which is moving 'into' the standing wave. Our entire discipline of science is backwards
 
@HDE226868 Uh...you did assign the meanings of the angle variables $\phi$ and $\theta$ somewhere prior, I hope, and the convention is to measure the latitude as $\phi$, no?
 
@ACuriousMind I did indeed.
Oh. . . Now I think I see. $\theta$ and $\phi$ are expressed differently in definitions relative to other coordinate systems. I should have seen that.
 
Well, the naming in the standard solution is kinda pre-cognitive because the author knows that $\Phi$ will be a function of $\phi$ and $\Theta$ a function of $\theta$.
I consider it good style to make a comment when one does such naming in anticipation of the result, because people are often left wondering how one could have known that at the step of naming it, when the simple fact is one couldn't
 
11:56 PM
There's a decent chance we're talking about completely different things, but you helped, so thank you.
 
@HDE226868 Aren't you talking about the naming of the variables the Wikipedia article also uses?
Regardless, good I helped ;)
 
@ACuriousMind No, I was talking about why $m$ is defined based on $\phi$, and not $\theta$.
 
@HDE226868 Uh, you could as well take the equation for $\theta$ as its definition, unless I'm misunderstanding something.
 

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