- #1

JRUniverse

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Hi everyone, I'm new to PhysicsForums. I don't really know if what I have to offer is anything new to anyone or if it is simply meaningless numerological garbage, but it just looked too beautiful for me to ignore, and if this is common knowledge, then I humbly apologize. I did a search on this site and on google to see if this had already been noticed and didn't find anything. Hopefully I'm not mistaken. If I am, please forgive me and point me in the right direction.

I don't really have the physics background everyone that I've seen posting seems to have, so bear with me. I went to RIT as an undergraduate in electrical engineering a few years ago, but I dropped out halfway for various personal reasons. I know basic things about relativity and quantum theory from some courses I took, as well as books I have read like The Elegant Universe and The Fabric of the Cosmos by Brian Greene. So, if you respond to this post with something overly technical, I may or may not understand it. However, I will most definitely try my best.

I recently became very interested in Koide's Formula. It is fascinating to me. From what I've read so far on the forums, it seems to be pretty well known, but just in case you don't know what it is, it can be found on wikipedia here (http://en.wikipedia.org/wiki/Koide_formula). To reiterate what the website says,

[tex] \frac{m_e + m_\mu + m_\tau}{(\sqrt{m_e}+\sqrt{m_\mu}+\sqrt{m_\tau})^2}\approx\frac{2}{3} [/tex]

The article suggested a geometric interpretation to this equation. It has to do with the cosine of the angle between the [itex] <1,1,1> [/itex] vector and [itex] <\sqrt m_e, \sqrt m_\mu, \sqrt m_\tau> [/itex]. So, I figured that quarks may have something very similar. I grouped the quarks' masses up in terms of charge and made vectors out of them similar to the vector previously stated. Then, just for kicks, I took their magnitudes and compared the down-quark vector's magnitude to the up-quark vector and got a ratio of about 4.99825, which is very close to 5! Now, I rechecked my work and realized I had mistakenly used GeV units in the up-quark vector versus MeV units in the down-quark, but had I not, I would never have seen such a simple ratio. After correcting the mistake, the ratio was determined as [itex] 1/(2\sqrt 10) [/itex], which I would have never seen if I didn't make that initial mistake. This theoretical value is about 99.9651 percent accurate to the calculated real value. I'm terrible with sig figs, so anyone please feel free to correct me on any and all percentages I give.

To make more clear what I did, let,

[tex] \vec a_1 = <\sqrt m_e, \sqrt m_\mu, \sqrt m_\tau>\approx<\sqrt{0.511}, \sqrt{105.7}, \sqrt{1777}> [/tex]

[tex] \vec a_2 = <\sqrt m_u, \sqrt m_c, \sqrt m_t>\approx<\sqrt{2.4}, \sqrt{1270}, \sqrt{171200}> [/tex]

[tex] \vec a_3 = <\sqrt m_d, \sqrt m_s, \sqrt m_b>\approx<\sqrt{4.8}, \sqrt{104}, \sqrt{4200}> [/tex]

All vectors have units of MeV.

[tex] \frac{|\vec a_3|}{|\vec a_2|}\approx\frac{1}{2\sqrt{10}} [/tex]

The values I got for the masses were from the wikipedia article on elementary particles (http://en.wikipedia.org/wiki/Elementary_particle). I verified that wikipedia's numbers were accurate using this other site just to make sure (http://pdg.web.cern.ch/pdg/2010/tables/rpp2010-sum-quarks.pdf0 [Broken].

I did the same thing for the third and first vectors and found a nice little ratio involving four numbers. If you want it, PM me. I don't want to post it because it might be taken as overly numerological, even though it is 99.9166 percent accurate.

Getting back to Koide's Formula, I plugged the wikipedia numbers in and found that this well known ratio was about 0.6666323, which is about 99.9949 percent of 2/3. I used this as a baseline to determine whether what I found next was at all valid, and this is what blew me off my feet. I believe I must have checked and rechecked the numbers I'm about to give more than thirty times. Can someone please verify this to make sure that I'm not making a simple mistake over and over? The first equation's theoretical value is 99.9859 percent accurate and the second equation 99.9218. So, here it goes...

[tex] \varphi=\frac{\sqrt 5 + 1}{2} [/tex]

This is the golden ratio. And here are the two equations I discovered.

[tex] \frac{m_u + m_c + m_t}{(\sqrt{m_u}+\sqrt{m_c}+\sqrt{m_t})^2}\approx\frac{2}{3}\ast\sqrt{\varphi} [/tex]

[tex] \frac{m_d + m_s + m_b}{(\sqrt{m_d}+\sqrt{m_s}+\sqrt{m_b})^2}\approx\frac{1}{\sqrt{5}}\ast\varphi [/tex]

I have a gut feeling this is important, but I simply don't have the means to verify it scientifically. If I algebraically manipulate Koide's Formula using the [itex] <1,1,1> [/itex] vector, the cosine of the angle of Koide's Formula becomes [itex] \sqrt{2}/2 [/itex], which has an angle of [itex] \pi/4 [/itex], which is known. Doing the same with the above two equations is a bit more difficult to obtain a simple angle. Instead, all I got was a ratio of cosines for the second equation and nothing for the first. Anyone who has access to Matlab would probably have better luck than me for finding simple angles. The ratio I got could be seen as either [itex] \cos((3\pi)/10)/\cos(\pi/6) [/itex] or [itex] \sin(\pi/5)/\sin(\pi/3) [/itex]. They are the same thing. I'm going to refrain from hypothesizing what this means. Instead, I'd like to hear what all of you have to say. I don't know if I'm seeing something that's not there or if this significant. Can someone please help me?

Thanks,

J Rivera

I don't really have the physics background everyone that I've seen posting seems to have, so bear with me. I went to RIT as an undergraduate in electrical engineering a few years ago, but I dropped out halfway for various personal reasons. I know basic things about relativity and quantum theory from some courses I took, as well as books I have read like The Elegant Universe and The Fabric of the Cosmos by Brian Greene. So, if you respond to this post with something overly technical, I may or may not understand it. However, I will most definitely try my best.

I recently became very interested in Koide's Formula. It is fascinating to me. From what I've read so far on the forums, it seems to be pretty well known, but just in case you don't know what it is, it can be found on wikipedia here (http://en.wikipedia.org/wiki/Koide_formula). To reiterate what the website says,

[tex] \frac{m_e + m_\mu + m_\tau}{(\sqrt{m_e}+\sqrt{m_\mu}+\sqrt{m_\tau})^2}\approx\frac{2}{3} [/tex]

The article suggested a geometric interpretation to this equation. It has to do with the cosine of the angle between the [itex] <1,1,1> [/itex] vector and [itex] <\sqrt m_e, \sqrt m_\mu, \sqrt m_\tau> [/itex]. So, I figured that quarks may have something very similar. I grouped the quarks' masses up in terms of charge and made vectors out of them similar to the vector previously stated. Then, just for kicks, I took their magnitudes and compared the down-quark vector's magnitude to the up-quark vector and got a ratio of about 4.99825, which is very close to 5! Now, I rechecked my work and realized I had mistakenly used GeV units in the up-quark vector versus MeV units in the down-quark, but had I not, I would never have seen such a simple ratio. After correcting the mistake, the ratio was determined as [itex] 1/(2\sqrt 10) [/itex], which I would have never seen if I didn't make that initial mistake. This theoretical value is about 99.9651 percent accurate to the calculated real value. I'm terrible with sig figs, so anyone please feel free to correct me on any and all percentages I give.

To make more clear what I did, let,

[tex] \vec a_1 = <\sqrt m_e, \sqrt m_\mu, \sqrt m_\tau>\approx<\sqrt{0.511}, \sqrt{105.7}, \sqrt{1777}> [/tex]

[tex] \vec a_2 = <\sqrt m_u, \sqrt m_c, \sqrt m_t>\approx<\sqrt{2.4}, \sqrt{1270}, \sqrt{171200}> [/tex]

[tex] \vec a_3 = <\sqrt m_d, \sqrt m_s, \sqrt m_b>\approx<\sqrt{4.8}, \sqrt{104}, \sqrt{4200}> [/tex]

All vectors have units of MeV.

[tex] \frac{|\vec a_3|}{|\vec a_2|}\approx\frac{1}{2\sqrt{10}} [/tex]

The values I got for the masses were from the wikipedia article on elementary particles (http://en.wikipedia.org/wiki/Elementary_particle). I verified that wikipedia's numbers were accurate using this other site just to make sure (http://pdg.web.cern.ch/pdg/2010/tables/rpp2010-sum-quarks.pdf0 [Broken].

I did the same thing for the third and first vectors and found a nice little ratio involving four numbers. If you want it, PM me. I don't want to post it because it might be taken as overly numerological, even though it is 99.9166 percent accurate.

Getting back to Koide's Formula, I plugged the wikipedia numbers in and found that this well known ratio was about 0.6666323, which is about 99.9949 percent of 2/3. I used this as a baseline to determine whether what I found next was at all valid, and this is what blew me off my feet. I believe I must have checked and rechecked the numbers I'm about to give more than thirty times. Can someone please verify this to make sure that I'm not making a simple mistake over and over? The first equation's theoretical value is 99.9859 percent accurate and the second equation 99.9218. So, here it goes...

[tex] \varphi=\frac{\sqrt 5 + 1}{2} [/tex]

This is the golden ratio. And here are the two equations I discovered.

[tex] \frac{m_u + m_c + m_t}{(\sqrt{m_u}+\sqrt{m_c}+\sqrt{m_t})^2}\approx\frac{2}{3}\ast\sqrt{\varphi} [/tex]

[tex] \frac{m_d + m_s + m_b}{(\sqrt{m_d}+\sqrt{m_s}+\sqrt{m_b})^2}\approx\frac{1}{\sqrt{5}}\ast\varphi [/tex]

I have a gut feeling this is important, but I simply don't have the means to verify it scientifically. If I algebraically manipulate Koide's Formula using the [itex] <1,1,1> [/itex] vector, the cosine of the angle of Koide's Formula becomes [itex] \sqrt{2}/2 [/itex], which has an angle of [itex] \pi/4 [/itex], which is known. Doing the same with the above two equations is a bit more difficult to obtain a simple angle. Instead, all I got was a ratio of cosines for the second equation and nothing for the first. Anyone who has access to Matlab would probably have better luck than me for finding simple angles. The ratio I got could be seen as either [itex] \cos((3\pi)/10)/\cos(\pi/6) [/itex] or [itex] \sin(\pi/5)/\sin(\pi/3) [/itex]. They are the same thing. I'm going to refrain from hypothesizing what this means. Instead, I'd like to hear what all of you have to say. I don't know if I'm seeing something that's not there or if this significant. Can someone please help me?

Thanks,

J Rivera

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