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 Tags Plancks Constants , schrodinger

 15th August 2009, 05:02 AM #1 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 Seen any equation similar to this? In the standard equation which was used to derive the many Planckian relationships given as $M^2=\hbar(c/G)$, i wondered if anyone had ever seen a similar equation to the following: $v^2 (\hbar c/G)= E^2/c^2 - p$ [1] Where the right hand side can be interpreted to be quite small, a quantization effort to gain the mass squared. In the appearance of velocity squared, we have the same dimensionality as what is found from derivations of the 4-momentum relativistically-invariant relationship describing the length of a vector on the left hand side. Normally, one replaces the energy and the momentum on the left hand side for $i \hbar(\partial / \partial t)$ and the momentum operator $-i \hbar \nabla$. But without inducing this, has this equation [1] in this form ever been used to anyone knowledge here?
 15th August 2009, 06:53 AM #2 sol invictus Philosopher     Join Date: Oct 2007 Location: Nova Roma Posts: 8,434 As written, it's not even dimensionally consistent. Perhaps you meant p^2. But even if so, the equation is wrong: the right side is a Lorentz invariant, and the left side is not.
 15th August 2009, 07:01 AM #3 I Ratant Penultimate Amazing     Join Date: Apr 2008 Posts: 15,305 Another note from the fringe wondering why 1+1=Rabbit.
 15th August 2009, 07:04 AM #4 Monketey Ghost Body of Work     Join Date: May 2003 Location: I'm on your screen! Posts: 14,814 Originally Posted by sol invictus As written, it's not even dimensionally consistent. Perhaps you meant p^2. But even if so, the equation is wrong: the right side is a Lorentz invariant, and the left side is not. So, no? __________________ The membership of this forum is henceforth to refer to me as potato-headed Bobby SSKCAS, member in long standing
 15th August 2009, 09:34 AM #5 Perpetual Student Illuminator     Join Date: Jul 2008 Location: USA Posts: 3,736 "...not dimensionally consistent." "...the right side is a Lorentz invariant, and the left side is not." __________________ It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman ξ
 15th August 2009, 09:34 AM #6 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 Originally Posted by sol invictus As written, it's not even dimensionally consistent. Perhaps you meant p^2. But even if so, the equation is wrong: the right side is a Lorentz invariant, and the left side is not. For M^2, we have \hbar c/G. v^2 on to this expression yields the value M^2v^2. According to a wiki source, (E/c)^2 - p^2 = M^2v^2. So i thought they where dimensionally consistent? Source: In the chapter ''Comparison with the Schrödinger equation'' http://en.wikipedia.org/wiki/Dirac_equation edit; yes it was supposed to be p^2 Last edited by Singularitarian; 15th August 2009 at 09:40 AM.
 15th August 2009, 09:39 AM #7 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 And to keep invariance true, its easy to apply gamma into the left hand side.
 15th August 2009, 09:41 AM #8 sol invictus Philosopher     Join Date: Oct 2007 Location: Nova Roma Posts: 8,434 P^2 Multiplying by gamma doesn't fix the problem.
 15th August 2009, 09:51 AM #9 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 Yes, i had already edited my post just above admitting to accidently leaving out the sqaured factor. Though i am curious, as to why the equation doesn't work? It about as abitrary as Einsteins equations for light which must be lorentz invariant $Mc - 2\Delta P= (M- E/c^2)v$?
 15th August 2009, 10:03 AM #10 ben m Illuminator   Join Date: Jul 2006 Posts: 4,712 Originally Posted by Singularitarian According to a wiki source, (E/c)^2 - p^2 = M^2v^2. That's nonsense; what Wiki did you get this from?
 15th August 2009, 10:05 AM #11 ben m Illuminator   Join Date: Jul 2006 Posts: 4,712 Originally Posted by Singularitarian Yes, i had already edited my post just above admitting to accidently leaving out the sqaured factor. You've "forgotten a squared factor" (or two) in basically every math-containing post you've made. Please proofread before you post. You want to know a good way to find these mistakes? Check your units.
 15th August 2009, 10:27 AM #12 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 Originally Posted by ben m That's nonsense; what Wiki did you get this from? I thought that too, believe me or not. http://en.wikipedia.org/wiki/Dirac_equation
 15th August 2009, 10:53 AM #13 ben m Illuminator   Join Date: Jul 2006 Posts: 4,712 Originally Posted by Singularitarian I thought that too, believe me or not. http://en.wikipedia.org/wiki/Dirac_equation That says "m^2c^2", not "m^2v^2". If it ever said the latter, it was wrong. Please note that in this case "c" is not standing in for the velocity of some particle, it's standing for a universal constant whose value is 2.99e8 m/s.
 15th August 2009, 10:58 AM #14 ~enigma~ Banned   Join Date: Nov 2006 Location: Center of the universe Posts: 7,954 Originally Posted by ben m That says "m^2c^2", not "m^2v^2". If it ever said the latter, it was wrong. Please note that in this case "c" is not standing in for the velocity of some particle, it's standing for a universal constant whose value is 2.99e8 m/s. Your saying once again our resident troll is wrong?
 15th August 2009, 11:01 AM #15 Monketey Ghost Body of Work     Join Date: May 2003 Location: I'm on your screen! Posts: 14,814 Originally Posted by ~enigma~ Your saying once again our resident troll is wrong? He's never wrong, go read some books. I assure you, he knows what he's talking about. __________________ The membership of this forum is henceforth to refer to me as potato-headed Bobby SSKCAS, member in long standing
 15th August 2009, 11:04 AM #16 ~enigma~ Banned   Join Date: Nov 2006 Location: Center of the universe Posts: 7,954 Originally Posted by Magnifico2.0 He's never wrong, go read some books. I assure you, he knows what he's talking about. Read some books? You mean Singularitarian is correct although he once again omitted a relevant term? Maybe you need to read a few books...
 15th August 2009, 11:06 AM #17 Monketey Ghost Body of Work     Join Date: May 2003 Location: I'm on your screen! Posts: 14,814 Originally Posted by ~enigma~ Read some books? You mean Singularitarian is correct although he once again omitted a relevant term? Maybe you need to read a few books... __________________ The membership of this forum is henceforth to refer to me as potato-headed Bobby SSKCAS, member in long standing
 15th August 2009, 11:08 AM #18 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 Originally Posted by ben m That says "m^2c^2", not "m^2v^2". If it ever said the latter, it was wrong. Please note that in this case "c" is not standing in for the velocity of some particle, it's standing for a universal constant whose value is 2.99e8 m/s. I know. I was setting it for the special case of v=c, i thought that would have been obvious. For instance, as is found: pc^2=Ev reference: http://en.wikipedia.org/wiki/Special_relativity
 15th August 2009, 11:19 AM #19 ben m Illuminator   Join Date: Jul 2006 Posts: 4,712 Originally Posted by Singularitarian pc^2=Ev That equation is always true, not just in the special case v=c. Note that the true equation contains "v" on the right and a "c" on the left. They are different variables and they mean different things; "v" is the velocity of the particle whose energy is E and momentum is p. "c" is a constant. E^2 = m^2c^4 + p^2c^2 is also true. It does not contain v. "c" is a constant. You cannot substitute "v" in and get a true equation. If you have a particle with v=c ... well, picking this case and doing the substitution anyway is stupid; you didn't specify, and it was "not obvious"; your inclusion of M suggests that you didn't mean to specify that at all, in fact. Last edited by ben m; 15th August 2009 at 11:21 AM.
 15th August 2009, 11:36 AM #20 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 Originally Posted by ben m That equation is always true, not just in the special case v=c. Note that the true equation contains "v" on the right and a "c" on the left. They are different variables and they mean different things; "v" is the velocity of the particle whose energy is E and momentum is p. "c" is a constant. E^2 = m^2c^4 + p^2c^2 is also true. It does not contain v. "c" is a constant. You cannot substitute "v" in and get a true equation. If you have a particle with v=c ... well, picking this case and doing the substitution anyway is stupid; you didn't specify, and it was "not obvious"; your inclusion of M suggests that you didn't mean to specify that at all, in fact. Well, that is what is implied. Also, i usually use $M_0$ to represent rest mass.
 15th August 2009, 11:41 AM #21 ben m Illuminator   Join Date: Jul 2006 Posts: 4,712 Originally Posted by Singularitarian Well, that is what is implied. That is what you claim you meant to imply. Fine, whatever. Let's rewrite your original post, then: Originally Posted by editing for correctness i wondered if anyone had ever seen a similar equation to the following: v^2 (\hbar v/G)= E^2/v^2 - p^2 = E^2/c^2 - p^2 = 0 You see, Sing, if v=c, then E^2/c^2 = p^2 and the difference is zero. The "m" in the famous E^2 = m^2 + p^2 equation is, in fact, the rest mass.
 15th August 2009, 04:20 PM #22 Raze Thinker     Join Date: Jul 2009 Posts: 237 Don't know whether to laugh or to sigh.
 15th August 2009, 04:31 PM #23 Wowbagger The Infinitely Prolonged     Join Date: Feb 2006 Location: Westchester County, NY (when not in space) Posts: 13,673 Has anyone seen any equations similar to this? a = b = 1 a2 = ab a2 - b2 = ab - b2 (a - b)(a + b) = b(a - b) a + b = b a = 0 1 = 0 __________________ WARNING: Phrases in this post may sound meaner than they were intended to be. SkeptiCamp NYC: http://www.skepticampnyc.org/ An open conference on science and skepticism, where you could be a presenter! By the way, my first name is NOT Bowerick!!!!
 15th August 2009, 04:55 PM #24 Raze Thinker     Join Date: Jul 2009 Posts: 237 Ah, that's a classic, lol. Originally Posted by Wowbagger Has anyone seen any equations similar to this? a = b = 1 a2 = ab a2 - b2 = ab - b2 At this step you have 0 = 0 So it's done. You can't go further. Quote: (a - b)(a + b) = b(a - b) a + b = b a = 0 1 = 0
 15th August 2009, 06:41 PM #25 ben m Illuminator   Join Date: Jul 2006 Posts: 4,712 Originally Posted by Singularitarian Though i am curious, as to why the equation doesn't work? It about as abitrary as Einsteins equations for light which must be lorentz invariant $Mc - 2\Delta P= (M- E/c^2)v$? Why doesn't the equation work? I have a guess. You are making specific type of mistake that is very, very common in intro physics classes; we call it "equation mining". You're looking up random equations on Web pages; you notice that you can find two equations, both containing (for example) an "m" term, and you figure you solve either equation for "m" and substitute it into the other. You do this with very little understanding of what the equation represents and what its terms are. For example, M=sqrt(hc/G) isn't some general property of masses. The terms on the right are constants, so this is simply calculating some particular mass; it's the definition of the Planck mass. What does, for example, E^2=m^2c^4+p^c^2 mean? It means that if you are considering a particle with rest mass m and momentum p, E is is its total energy. So when you "substitute" sqrt(hc/G) for m, you are just stating "if you have a particle whose rest mass is the Planck mass and whose momentum is p, its energy is E". Nothing universal, nothing generally interesting. Want to keep plugging? Go ahead, there are other true equations you can apply to "a particle whose rest mass is the Planck mass", if you want to ... but not all equations so apply. v=c, an interesting special case in other circumstances, does not apply here for example. E=mc^2 does not apply, since that's a different E than the one in the E^2 = m^2 + p^2 equation. And so on.
 16th August 2009, 04:16 PM #26 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 Well yeh, but i never surfed the net intentionally as you are making out. I was already aware of the relationship $GM^2=\hbar c$. It wasn't until investigating the length of a vector did i come across a dimesional relationship: The equation in the OP was just an accident of other equations i was writing, so it had nothing to do with me generally surfing the net. Also, you must remember ben, the form you have given above is indeed a planck mass value, however, i never intend to create this value when in the original equation in the post. Under the special condition of v=c, the equations dimensions are perfectly fine. As i said, when i define rest mass, i write $M_0$ usually. (And please, ben, don't talk to me about the energy-momentum equation as if i am some idiot. I know fine well what it is all about, probably more than you).
 16th August 2009, 04:25 PM #27 Wangler Master Poster     Join Date: Feb 2008 Location: OH State Posts: 2,230 Originally Posted by ben m You want to know a good way to find these mistakes? Check your units. I had a great undergrad physics professor, who taught me that valuable problem solving skill. He would always have us work with the variables, never numbers ("they get in the way")...solve the problem, check units....units o.k.? answer seems reasonable? you are likely right, then. plug in numbers to your hearts content.
 16th August 2009, 04:31 PM #28 ben m Illuminator   Join Date: Jul 2006 Posts: 4,712 Originally Posted by Singularitarian Also, you must remember ben, the form you have given above is indeed a planck mass value, however, i never intend to create this value when in the original equation in the post. Under the special condition of v=c, the equations dimensions are perfectly fine. As i said, when i define rest mass, i write $M_0$ usually. Two questions, then: since v=c only for a massless particle, why did you write the energy-momentum equation for a particle of mass M_Planck? Second: why did you use the variable v at all? Quote: (And please, ben, don't talk to me about the energy-momentum equation as if i am some idiot. I know fine well what it is all about, probably more than you). I know! Why don't you post a relativity quiz? Oh, wait ...
 16th August 2009, 04:39 PM #29 ben m Illuminator   Join Date: Jul 2006 Posts: 4,712 Originally Posted by Singularitarian Under the special condition of v=c, the equations dimensions are perfectly fine. No, after you fixed the p^2 the dimensions were fine. v has the same units/dimensions as c. v != c doesn't break the unit consistency; as Sol pointed out, it breaks the Lorentz invariance of the left-hand side of your equation. Quote: As i said, when i define rest mass, i write $M_0$ usually. First: bad idea; no one else in physics does this. The rest mass is m. If you want to write the "relativistic mass" (which is rarely used anyway) just use E/c^2. Second: well, you took the thing that you called "M" and you plugged it into the energy-momentum equation in a place that called for a rest mass. If you did not intend M to be the rest mass, you should not have plugged it into the "M" of M^2 = E^2 - p^2.
 16th August 2009, 04:58 PM #30 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 Well, i missed the sqaured momentum by accident. But it was really what i meant originally. And i don't know about the last. I've had lecturers telling me not to use $\gamma M=m$.
 16th August 2009, 05:00 PM #31 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 Originally Posted by Wangler I had a great undergrad physics professor, who taught me that valuable problem solving skill. He would always have us work with the variables, never numbers ("they get in the way")...solve the problem, check units....units o.k.? answer seems reasonable? you are likely right, then. plug in numbers to your hearts content. I have to admit, yes, i never checked it over when i wrote it out for any errors, but to be honest, i know my units.
 16th August 2009, 05:42 PM #32 ben m Illuminator   Join Date: Jul 2006 Posts: 4,712 Originally Posted by Singularitarian And i don't know about the last. I've had lecturers telling me not to use $\gamma M=m$. Yep---nobody calls (gamma x rest mass) the "mass" anymore. You'd never guess from reading pop-science accounts of Special Relativity, which frequently report the long-dead early-20th-century usage. Anyway, I am not asking you to use "gamma M = m" (where M=rest mass and m="relativistic mass"). I'm asking: you plugged some sort of mass into the rest mass slot in M^2 = E^2 + p^2, then specified that v=c which implies zero rest mass. Why?
 16th August 2009, 09:28 PM #33 ben m Illuminator   Join Date: Jul 2006 Posts: 4,712 Originally Posted by ben m M^2 = E^2 + p^2 (Typo alert: I meant M^2 = E^2 - p^2)
 20th August 2009, 07:48 PM #34 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 I refer you to the Springer Link journal paper, from the Rutherford Observatory, Columbia University, New York, N. Y by L. Motz, published under the Italian Physical Society, and the summery was thus described; ''Summary In a previous paper (referred to as I in the text) it was shown that the Weyl principle of gauge invariance leads to the relationship $Gm^2 =\hbar c$ for a particle of inertial mass $M$ obeying the Dirac equation, where G is the Newtonian gravitational constant. Instead of interpreting this equation to mean that $G$ takes on the extremely large value $\hbar c/m^2$ inside a particle like an electron (as we did in I), we now write it in the form $Gm^2/c = \hbar$ and treat it as a quantization condition on the square of the gravitational charge $\sqrt{Gm}$. We show that this same quantization condition can be obtained from an angular-momentum component in the general-relativistic two-body problem as well as from the Machian definition of inertial mass in a rotating universe by using the Dirac-Schwinger procedure for quantizing charge. From this quantization condition we now deduce that the fundamental particle in Nature (the uniton) has an inertial mass equal to about $10^{-5}g$. The possibility of using the uniton to shed light on the mystery of the « missing mass » in the Universe is discussed. Other cosmological implications of the uniton are also discussed and it is suggested that unitons can clear up the solar-neutrino discrepancy.'' Now, i know for a fact that Dirac's equation is relativistically invariant, so it seems now that the union of the two can be expressed relativistically, since the above is referring to the equation that obeys the Dirac Equation itself.
 20th August 2009, 08:52 PM #35 Reality Check Penultimate Amazing   Join Date: Mar 2008 Location: New Zealand Posts: 10,978 Originally Posted by Singularitarian I refer you to the Springer Link journal paper, from the Rutherford Observatory, Columbia University, New York, N. Y by L. Motz, published under the Italian Physical Society, and the summery was thus described; ''Summary In a previous paper (referred to as I in the text) it was shown that the Weyl principle of gauge invariance leads to the relationship  for a particle of inertial mass  obeying the Dirac equation, where G is the Newtonian gravitational constant. Instead of interpreting this equation to mean that  takes on the extremely large value  inside a particle like an electron (as we did in I), we now write it in the form  and treat it as a quantization condition on the square of the gravitational charge . We show that this same quantization condition can be obtained from an angular-momentum component in the general-relativistic two-body problem as well as from the Machian definition of inertial mass in a rotating universe by using the Dirac-Schwinger procedure for quantizing charge. From this quantization condition we now deduce that the fundamental particle in Nature (the uniton) has an inertial mass equal to about . The possibility of using the uniton to shed light on the mystery of the « missing mass » in the Universe is discussed. Other cosmological implications of the uniton are also discussed and it is suggested that unitons can clear up the solar-neutrino discrepancy.'' Now, i know for a fact that Dirac's equation is relativistically invariant, so it seems now that the union of the two can be expressed relativistically, since the above is referring to the equation that obeys the Dirac Equation itself. It is the particle that is obeying Dirac's equation (emphasis added). __________________ Real Science: NASA Finds Direct Proof of Dark Matter (another observation) (and Abell 520) "Our Undiscovered Universe" by Terence Witt: Review 1; Review 2
 20th August 2009, 09:17 PM #36 ben m Illuminator   Join Date: Jul 2006 Posts: 4,712 Yes, sqrt(hc/G) is an allowable rest mass for a particle. (So is any other value) Yes, a particle of any rest mass will obey the relativistic energy-momentum equation. (The relativistic energy-momentum equation, E^2 = m^2 + p^2, is not the same thing as the Dirac equation; do you have them mixed up? They do both occur in the same Wikipedia article. In any case, if your particle is a fermion it may also obey the Dirac equation.) No, such a particle cannot travel at v=c as you (eventually) specified. What in Motz's paper do you think contradicts that statement?
 21st August 2009, 09:43 AM #37 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 I had a few vodkas last night, and when i read it, it said something else. Sorry.
 21st August 2009, 09:44 AM #38 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 If it where a fermion though, i better start doing some math to try and prove that.
 21st August 2009, 09:46 AM #39 Singularitarian Banned   Join Date: Jul 2009 Posts: 1,008 For instance, you can derive not only the Dirac equation from above if it where a fermion, but you could also derive the schrodinger equation.
 21st August 2009, 07:52 PM #40 Olowkow Illuminator     Join Date: Oct 2007 Posts: 4,925 If we were where we were wired, where we're wary of wearing wiry war ware we wore at the whirring weir, why were warring wair wielding weary werewolves where we're weird? __________________ Our remedies oft in ourselves do lie, which we ascribe to heaven. --Shakespeare Last edited by Olowkow; 21st August 2009 at 08:02 PM.

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