/sci/ - Science & Math

[Anonymous No.6705082]

Hi, /sci/. I'm having a bit of confusion over relativistic kinematics.
I'm reading to try to figure out how to get the velocities for the decay particles in pion decay, and in the relevant section I came across this (see picture).
It's been a long time since I've studied relativity, so I'm assuming this doesn't make sense to me due to a gap in my relativity knowledge. It seems to me like they're saying that the momentum squared of a particle is equal to its mass squared in this case. I don't understand why this would be at all. If you understand this, please let me know, and I'm sorry if I've asked a stupid question.

[Anonymous No.6705793]

>>6705082
Still don't understand this.
I'm also wondering another thing, so I'll use this thread to ask that, as well.
I'm trying to get the velocity of the muon that comes out of pion decay. Here's what I'm doing:
Taking the relativistic momentum to be <span class="math"> \frac{29.79*10^6}{c} [/spoiler] eV (method pictured), I use:
<span class="math"> p = \frac{m(rest)*v}{\sqrt{1-\frac{v^2}{c^2}} [/spoiler] to
<span class="math"> \frac{m(rest)^2*v^2}{p^2} = 1 - \frac{v^2}{c^2} [/spoiler] to
<span class="math"> v^2(\frac{m(rest)^2}{p^2} + \frac{1}{c^2} = 1 [/spoiler] to
<span class="math"> v^2(\frac{1.1025 * 10^{16} eV}{8.874*10^{14}eV*c^2} + \frac{1}{c^2}) [/spoiler] to
<span class="math"> v^2(1.38*10^{-16} + 1.111*10^{-17}) = 1 [/spoiler]
<span class="math">v^2 = 1/(1.491*10^{-16}) [/spoiler] to
<span class="math">v = 8.188*10^7 = 0.3c[/spoiler]
God, I hope I didn't fuck that up. I haven't used Tex in a bit.
Anyway, I'm sorry if I'm doing something dumb. Please let me know if this is all right and I'm getting the correct velocity.

[Anonymous No.6705796]

>>6705082
Still don't understand this.
I'm also wondering another thing, so I'll use this thread to ask that, as well.
I'm trying to get the velocity of the muon that comes out of pion decay. Here's what I'm doing:
Taking the relativistic momentum to be <span class="math"> \frac{29.79*10^6}{c} [/spoiler] eV (method pictured), I use:
<span class="math"> p = \frac{m(rest)*v}{\sqrt{1-\frac{v^2}{c^2}}} [/spoiler] to
<span class="math"> \frac{m(rest)^2*v^2}{p^2} = 1 - \frac{v^2}{c^2} [/spoiler] to
<span class="math"> v^2(\frac{m(rest)^2}{p^2} + \frac{1}{c^2}) = 1 [/spoiler] to
<span class="math"> v^2(\frac{1.1025 * 10^{16} eV}{8.874*10^{14}eV*c^2} + \frac{1}{c^2}) [/spoiler] to
<span class="math"> v^2(1.38*10^{-16} + 1.111*10^{-17}) = 1 [/spoiler]
<span class="math">v^2 = 1/(1.491*10^{-16}) [/spoiler] to
<span class="math">v = 8.188*10^7 = 0.3c[/spoiler]
God, I hope I didn't fuck that up. I haven't used Tex in a bit.
Anyway, I'm sorry if I'm doing something dumb. Please let me know if this is all right and I'm getting the correct velocity.
Take two

[Anonymous No.6705798]

>>6705796
Fuck, forgot picture.

[Anonymous No.6705918]

bump

[Anonymous No.6705930]

First answer:
the p^2 is a covariant multiplication of the 4-vector, 4-momentum, whose components are (Energy,momentum vector)
when you do the covariant multiplication it gets you:
p^2=(E,P)*(E,-P)=E^2-P^2=m^2
wherre P is momentum vector and of course c=1

second answer: I think the calculation is correct but there is a simpler way
V/C=|P|/E so check if it's the same

[Anonymous No.6705976]

>>6705930
Oh, I had no idea that was what it was. Why is it the covariant instead of just a regular dot product?

Ah, geez, that is a lot simpler. Let's see:
<span class="math"> E = \sqrt{(p*c)^2 + (m(rest)*c^2)^2} [/spoiler] to
<span class="math"> E = \sqrt{898.2MeV^2 +11164.0MeV^2} [/spoiler] to
<span class="math">E = \sqrt{12062.2} [/spoiler] to
<span class="math">E =109.8MeV [/spoiler] so
<span class="math">\frac{v}{c} = \frac{|p|}{E} [/spoiler] to
<span class="math">v = \frac{|p|*c}{E} [/spoiler] to
<span class="math">v = \frac{29.79*10^6eV*c}{109*10^6eV} [/spoiler] to
<span class="math">v = 0.271*c[/spoiler]
Pretty much exactly what I above (<span class="math"> \frac{8.188*10^7*c}{c} = 0.273*c [/spoiler]) so I'll put down the small difference to me rounding differently and since I kind of carelessly used approximations for the rest energies, while then switching back to the more accurate numbers on a whim.
Anyway, thank you for your help. One thing that I will need to get used to is this c = 1 concept. I've heard it from teachers and stuff, but always it's just them accidentally not writing a c in an equation. We were never taught it or anything, so it's still kind of a weird concept.
I don't really get it still, since if c = 1, then why did you put the C under the V in your equation? And if it's not, then what's going on with the |P|? I kind of expected the equation to be <span class="math"> V/C = p*c/E [/spoiler], and so I kind of treated it like that, but since it worked out I'm assuming that it's right. Is |P| defined as the magnitude of p*c or something?
I'm having a little trouble with both the natural units thing and the difference between p and P, I think.

[Dr. John D. Hollister Titor !SSDoCt0rTo No.6705980]

[Dr. John D. Hollister Titor !SSDoCt0rTo No.6705983]

>>6705976
look at a particle like a cone through a cylinder in a sphere

[Dr. John D. Hollister Titor !SSDoCt0rTo No.6705986]

>>6705976
you can also look at it this way:
make a finger gun with your index and thumb. the distance is E. point it and shoot an imaginary bullet.

you have two hands.

you can also look in the mirror; or does the mirror really look through you?

how can our eyes be real if mirrors aren't real?

[Anonymous No.6705987]

oh no

[Dr. John D. Hollister Titor !SSDoCt0rTo No.6705988]

in a gif, for intuition, OP

[Dr. John D. Hollister Titor !SSDoCt0rTo No.6705994]

>>6705986
when i meant distance, it was that between your finger-tips. but also between your two hands.
like a^2 + b^2 = c^2, pythagoras
http://en.wikipedia.org/wiki/Icosahedron
to be honest, base 10 sucks.
273 in base 2 is 100010001, in base 13 it is 180.

http://www.tauday.com/
My DNA has two sides to its ladder.
Ever wonder why your opposable thumbs are two-phalangeal and your eight fingers are three-phalangeal?
Ever wonder why NATURE LOVES COURAGE AND SYMMETRY?
(2^5)*(8^3) = 2^14 = (|..14.7.41..|)^2
Ever wonder why you have 12 strands of DNA, 10 fingers, yet only 2 active strands?

[Anonymous No.6705995]

>>6705082
momentum squared is by definition mass squared.

In relativity, mass is by definition the square root of the absolute value of four-momentum squared.

Burn this in your mind.

[OGWillikurrs No.6706007]

To the OG Anon who posted the mess of variables and exponents, I think your post is a bit over the heads of most of 4chan's lurkers.

[Dr. John D. Hollister Titor !SSDoCt0rTo No.6706014]

>momentum squared is by definition mass squared.

>In relativity, mass is by definition the square root of the absolute value of four-momentum squared.

>Burn this in your mind.

who pointed the finger gun of the universe? what started entropy? was it consciousness?

[Dr. John D. Hollister Titor !SSDoCt0rTo No.6706018]

>>6705986
>>6705983
>>6705986
i mean, ancient egyptians clearly had knowledge of advanced mathematics and pi/tau. their pyramids are very much proof.
http://www.bibliotecapleyades.net/thot/esp_thot_9.htm
http://www.hiddenmeanings.com/cosmos.html

[Dr. John D. Hollister Titor !SSDoCt0rTo No.6706020]

>>6706018
and if you look at DMT, serotonin, tryptamine, the indole ring, and how it relates to perception, the third eye, etc.

[Dr. John D. Hollister Titor !SSDoCt0rTo No.6706021]

>>6706020
even ancient china's seal of the dragon shows the structure of DNA.
it was only recently we learned about quaternions. so how do they relate to quantum mechanics, and decay of particles, OP? very, very closely.

[Anonymous No.6706025]

>>6705976
>Why is it the covariant instead of just a regular dot product?

For the same reason that we don't talk about things like E^2 - px^2 -py^2, leaving out pz. To use an ordinary dot product is to talk about something that is not a good relativistic quantity: you're arbitrarily choosing an axis (time) and ignoring it. The result is a quantity that behaves "badly": i.e., it will not be constant if you boost between reference frames. Mass is a constant, so it had better equal something that is covariant.

[Dr. John D. Hollister Titor !SSDoCt0rTo No.6706026]

>>6706021
i mean,
quaternions are ijk
3d cartesian grid is xyz
DNA is a double helix
/\/\/\/\/\/\
-A<=>T+
+G<=>C-
\/\/\/\/\/\/

[Dr. John D. Hollister Titor !SSDoCt0rTo No.6706040]

here is a number for you, OP
198343820949598453
198343820949598453/planck's constant) = 2.99338573e50 m^-2 kg^-1 s
198343820949598453 / 354895949028343891 = 0.558878797

198343820949598453
in base 13, 3B4B4050A21B94CA
in base 16, 2C0A8A77F183100

354895949028343891
in base 13, 6C19AB7514867326
in base 16, 4ECD7FC3E0F4040

[Anonymous No.6706041]

>>6705995
>>6706025
Shiiit.
Okay, thank you.
I probably should have worked harder to understand to the relativity portion of my modern physics class. Too busy bein' hyped for quantum. Definitely will have to hit my book again once I get back to my apartment.