Mathematical Wormhole pt2

[IkesLc]

People you aren't understading this video. First of all, this kind of wormholes is one kind of many. I can do one with the same kind of math only using real numbers. But that's not important, important is that real theoretical wormholes are solutions of Einstein Field Equations, not this kind of math. Second, is not necessary advanced mathematics for this. This is calculus and complex numbers. I am simply an autodidact and I can understand it. And third, I don't think this makes much sense.

[alliedtoasters]

Because your wormhole requires the use of imaginary numbers, the wormhole itself is also imaginary. But this is still an interesting result.

Is it possible to find a path in the imaginary plane that is even shorter than the specific path you chose, y=x^2+ix? Can we optimize to find the shortest possible path in the complex plane? Although we are defying the old adage, 6% is not t, at much, and I'm curious to see how useful this imaginary wormhole could be.

[alliedtoasters]

(hypothetically)

[clockw3rx]

the formula forgets multi-directional point to point .... meaning the % would fluctuate at their end points

[leetag]

Good explanation. Its a shame that 1st year university maths is required to understand most of what you said.

Just wondering, would this result be explained through the curvature of space. For example on the surface of a sphere or another curve the 'straight line' (as anyone standing on the curve would perceive it) that follows the curve is in fact not the shortest distance?

[gvsfgdf]

yo im in grade 12 and i understand it (but i took calc1&2 already) its soo cool my math teacher is fucked... im gonna trip him up with this tomorrow :D

[turtleman924]

yeah see. im only 12... so what da heck are u saying lol...

[DoubleDutchBust]

Mathematical Wormhole pt2

Microscopic Collaborations 3:32

Mathematical Wormhole pt2

Being Thousands of choices

within the cell.Understanding

each micro c and j wants to be selected.

It being intelligent to know not to release the inner most secret so easily.

Mathematical Wormhole

ThethinkingreY c j-j-c-j-j(q) finding-

[rimini000]

Now let y = ix.

The length of this curve is 0 between any two points. So the conclusion is that I can get to any point by travelling a distance zero along this curve....

I am simultaneously everywhere.

[b0bb0bs0n]

conclusion: time travel is possible; you just have to travel along a complex path.

????

[catherbi]

Not 100% convinced. Where does Y = x^2+ix come from? Are you assuming space is complex? There are lots of tricks in maths when you start using complex numbers!! QM uses complex numbers to get to a probability distribution, thus what does complex numbers in this QM physical mean?

[myusername20202020]

Why have you chosen y = x^2 + ix? Any physics associated with that particular choice?

[hazza59]

brain... cannot... compute

[breetak]

i agree with simplystimpy, its already given what c's length is by a2+b2=c2. I love parametrics.

[simplystimpy]

It might be clearer to mention not that x^2 goes to 0 faster than x (since as

bezvezeceda said, they both reach zero at the same time), but that if we make dx infinitely small, then dx^2 is infinitely small squared and is negligible compared to the other values in the expression. Similar to the Liebniz derivation of the derivative (y+dy = x+dx).

ps: kpate004, I don't get why you thought we needed to integrate c. That makes no sense at all.

[donylee]

Oh and yeah, just I WERE to use integration to write an expression for c, I will get the SAME expression. I won't bore you with the math, but its something like this.

dc² = dx² + dy²

dc = √(1 + (dy/dx)²)dx

Since straight line, dy/dx is always equal to constant m

Integrate w.r.t x from 0 to arbitrary x to get

c = √(1 + m²)x

Rearrange to get the same expression for c.

[kpate004]

your a dum ass... u did not integrate the c staright line... u integrated s curve but not the c curve... u just took the limit of c without integrating.... use the multiplecation integration rule... since u have x multiplied with another x term in the sq rt for c...

[donylee]

kpate004, here is the flaw in your argument.

Firstly, there is no multiplication integration rule. I believe you meant to say integration by parts or by substitution.

Second, I DO NOT need to integrate the c straight line because I am using simple geometry to write an expression for c. It is none other than a straight forward application of Pythagorus theorem. Totally valid since it is in the Cartesian plane.

[dragonlorder]

question:

1.when you put c= sqrt[x^4+2ix^2], if you hadnt taken the sqrt[x^2] out, the approximation would be different? and so is s=int[sqrt[4x^2+4ix]], if you put it s= 2/int[sqrt[x]sqrt[x+i]]....

Is there a standard to take approximation? Since it would be a different result, if you transform the expression in different way.

[LongShlong125]

Hey again, whats that dashed x means?

[AeroStew89]

You do mension that you are approximating though, i suppose that people might just hear 'taking the limit to zero' and put zero in every x :O... so you didnt make any mistake Dony as I realised what you were doing :)

[AeroStew89]

He is using is as an approximation, powers of x tend towards 0 faster than the others so you dont have to put 0 in every x expression ... i think :)

[donylee]

Hello Aero,

Yup, you are CORRECT. I made a small blunder where I said I was taking the limit. Instead, I should have said that I am using an approximation. Powers of X tends to 0 faster than X.

We then take the limit after.

[adb4]

that's all well and good in a two dimensional plane, but a lot of phenomena that work in one number of dimensions will not generalize to another (for example, there are a limited number of dimensions with a right handed cross product)

I would like to see a derivation of this idea using geodesics rather than just a straight line, though it would obviously be a little more involved

[donylee]

Hello adb4,

I approached this idea using fundamental calculus. I am sure that you are right in saying using Geodesics is might more involved and may give us more definite results. However, I have some problem comprehending the spacetime diagram that I can't even using Geodesics equations.

If you like to help me out, you can read my latest blog entry on my website to find out my troubles in relativity.

Still, thanks for watching.

[SamirShah25]

Wait I dont get so, so are you saying that the efficiency of Space and Time Travel depend on the dimension that the wormhole exists in?

[adb4]

no, I'm saying that a lot of mathematical "gospel" falls apart when things are not constrained to a single plane of motion in space-time

[enigmaparadigm]

ok just one concern i have with your math. that is at about 2:00 you evaluate the equation for "C" as "X dash" goes to zero, which is fine but you only evaluate one of the "X dashes" as it goes to zero... i dont know about you but i learned that when you evaluate the equation for a limit as a term goes to zero you have to do it for the entire equation not just part like you have done. it would be lovely if you could explain

[donylee]

Hello EnigmaParadigm,

Yup you are spot on in your reasoning. Watching my video again, I made a small mistake. The part where I said "X dash tends towards 0", we are NOT taking the limit yet. We are approximating c and s around the domain where X dash is close to 0. X^2 goes to zero closer than X so we can remove that term.

We will then take the limit of s/c at the end.

Good comment.

[bezvezeceda]

Sorry although i,m not close to my math, i do not understand why is it important that "X^2 goes to zero closer than X" if they arrive at zero at the same time? Sorry if it is a dumb question ... but then again they sat that the only stupid questions are unasked ones.

[myusername20202020]

I guess he simplified the calculus to make it easier to follow.

Without the approximation 'X^2 ->0 faster ...'

he would have had the following formula for s:

s = -1/8*I*Pi+(X*(X+I))^(1/2)*X+1/­2*I*(X*(X+I))^(1/2)+1/4*arcsin­h(2*X+I)

And then the s/c would behave near X=0 as:

series(s/c,X=0) = 2*sqrt(2)/3-1/30*I*2^(1/2)+O(X­^2) = 2*sqrt(2)/3.

Therefore his approximation provide the right answer and it is much easy to follow than if you don't do that approx.