Split Complex Numbers and the Lorentz Boost

[@johnreid4830]

Excellent treatment linking split complex numbers, hyperbolic geometry and SR. Very enlightening, and many thanks for your efforts.

[@tricanico]

that was Beautiful! I've been subscribed to this channel for quite some years now and I hadn't seen this video. Just Amazing.

[@philipoakley5498]

Excellent. I like to think of "The angle of Time", especially for units of measure. Thus SI units have the simple Galilean angle of a paltry 3.3 nano radians (1 second, 1 metre). This helps indicate to those who haven't thought about the time cone that hard, just how easy it is to be fooled by everyday measures (SI ~= Imperial at this scale) . What's really missing is some natural real phenomena that we can see that follows the hyperbolic phenomena that we can name and explain.

[@ep9196]

Fantastic Series!

[@eduardkuegler3655]

Thank you very much for this video. I like to suggest to add the much more interesting case when you have comoving particles, where the relative velocity for small velocities is slightly higher, than the sum. See also „Julian Schwinger`s method of generalized velocities „Einsteins Erbe Die Einheit von Raum und Zeit“ p. 70“ in Spektrum Bd. 14. It is interesting, because in that case, two entangeled Fotons can be investigated.

[@fuseteam]

+mathoma a video on hyperbolic trig would be nice additional applications of dual and split numbers would be great too

[@thomasolson7447]

I noticed this a few weeks ago, so that's how I wound up here. I only finished grade 12, and I was high and drunk through most of that. I get my information from Khan Academy and Michel van Biezen. They haven't mentioned this, so far, in the stuff I came accross. γ=1/sqrt(1-(v/c)^2), I seen this, it looked familiar. cos(arctan(y/x))= 1/sqrt(1+(y/x)^2) I thought to myself, "all I need to do is throw an 'i' in there and that positive will become a negative." cos(arctan(i*v/c))=1/sqrt(1-(v/c)^2), I mean... If we're going to do this, there might be a better way. cos(arctan(i*v/c))= *c/sqrt(c^2-v^2)* , guess we should do the other side. sin(arctan(i*v/c))=(i*v/c)/sqrt(1-(v/c)^2)=(i*v/c)*γ=(i*v/c)* *c/sqrt(c^2-v^2)* = *i*v/sqrt(c^2-v^2)* I took a peak at the arctan power series. That was cool. Long story short: arctan(i*v/c)=ln(sqrt((c+v)/(c-v)))=i*ln((c+v)/(c-v))/2 Now back to cos and sin. cos(x)=(e^(i*x)+e^(-i*x))/2 sin(x)=(e^(i*x)-e^(-i*x))/(i*2) cos(arctan(i*v/c))=(e^(i*i*ln((c+v)/(c-v))/2)+e^(-i*i*ln((c+v)/(c-v))/2))/2=*(e^(ln((c+v)/(c-v))/2)+e^(-ln((c+v)/(c-v))/2))/2* sin(arctan(i*v/c))=(e^(i*i*ln((c+v)/(c-v))/2)-e^(-i*i*ln((c+v)/(c-v))/2))/(i*2)=*(i*e^(ln((c+v)/(c-v))/2)-i*e^(-ln((c+v)/(c-v))/2))/2* I didn't know anything about hyperbolic functions until a couple days ago. I noticed a a^2-b^2=1 in a thumbnail somewhere. I tried them out. cosh(x)=(e^(x)+e^(-x))/2 sinh(x)=(e^(x)-e^(-x))/2, it close, there is just an i in front I guess. i*sin(x) Oh, and look at this! (d/dv)((e^iiln((c+v)/(c-v))/2-e^(-i)iln((c+v)/(c-v))/2)/(i*2))=(ice^ln((c+v)/(c-v))/2+ice^-ln((c+v)/(c-v))/2)/(2(c^2-v^2)) ice is cool. So, what would the area of this thing be? (-i)/(8*((v+c)/(c-v)))+i*c/(8(c-v))+i*v/(8*(c-v))= *-i*c*v/(2(v^2-c^2))*

[@demr04]

Very nice. Thaks you

[@alejrandom6592]

This is beautiful!

[@byronwatkins2565]

Is there a way to do this using Clifford Algebra?

[@angeldude101]

Yes. The standard way seems to be with Spacetime Algebra, with the time dimension squaring to 1 and the space dimensions squaring to -1. The apparent velocities/distances/etc are then the bivector terms between a time-like and space-like axes. Naturally I can't possibly do it justice on my own, so I recommend looking it up. In short, the time-like dimension squaring to 1 makes it act very much like the split-complex numbers here causing the same kind of hyperbolic rotations. Those relative bivectors also behave just like the Pauli matrices if you're familiar with them.

[@mesutkaraman9677]

Is this playlist ordered with respect to learning progress?

[@Math_oma]

Yeah, you should check out the first two split complex number videos because I use some of that material in this video.

[@Math_oma]

Of course, feel free to ask questions, too.

[@mesutkaraman9677]

Mathoma thank you for definitely significant knowledges and please continue!