beejewel 6 Posted October 31, 2014 Share Posted October 31, 2014 "Humpty Dumpty sat on a wall Humpty Dumpty had a great fall, All the kings horses and all the kins men, couldn't put Humpty together again." The well known rhyme mentions nothing about how tall the wall was, or even if Humpty ever hit the ground, what if there was no ground, and Humpty fell right through into the abyss? Let us imagine the wall was a doughnut shaped structure around a black hole with the mass of the earth, with a radius equivalent to the earth. What was Humpty's potential before the fall and what happened to him after the fall, making it so difficult to reassemble him? Steven Link to post Share on other sites

elfmotat 324 Posted October 31, 2014 Share Posted October 31, 2014 What does his initial potential have to do with anything? It's difficult to reassemble him because he fell into a black hole, and he can't come back out. Link to post Share on other sites

beejewel 6 Posted October 31, 2014 Author Share Posted October 31, 2014 What does his initial potential have to do with anything? His ultimate velocity is a function of his initial potential, and what happens to him as he reaches the horison of the black hole is not very well defined, does he fall through the horison and increase the mass of the black hole, or is something else happening at that point? Link to post Share on other sites

elfmotat 324 Posted October 31, 2014 Share Posted October 31, 2014 His ultimate velocity is a function of his initial potential, and what happens to him as he reaches the horison of the black hole is not very well defined, does he fall through the horison and increase the mass of the black hole, or is something else happening at that point? His velocity is irrelevant though. The point is that he falls in, isn't it? A BH as small as the one you're considering (with the mass of Earth) would only be about the size of a small coin, so Humpty will be torn apart by large tidal forces before he even reaches the event horizon. His remnants will fall through and add to its mass. Link to post Share on other sites

J.C.MacSwell 453 Posted October 31, 2014 Share Posted October 31, 2014 (edited) "Humpty Dumpty sat on a wall Humpty Dumpty had a great fall, All the kings horses and all the kins men, couldn't put Humpty together again." The well known rhyme mentions nothing about how tall the wall was, or even if Humpty ever hit the ground, what if there was no ground, and Humpty fell right through into the abyss? Let us imagine the wall was a doughnut shaped structure around a black hole with the mass of the earth, with a radius equivalent to the earth. What was Humpty's potential before the fall and what happened to him after the fall, making it so difficult to reassemble him? Steven "Humpty Dumpty sat on a wall Humpty Dumpty had a great fall, All the kings horses and all the kins men, couldn't put Spaghettified Humpty together again." But the BH did... Edited October 31, 2014 by J.C.MacSwell Link to post Share on other sites

beejewel 6 Posted November 1, 2014 Author Share Posted November 1, 2014 (edited) "Spaghettified" This term comes from pop science and teaches little about what's going on. Before we worry about Humpty's reassembly, let us first look at what is happening before he reaches the SR radius. Humpty is falling from ground potential through the ultimate potential to the SR radius, now how can we use the known laws of physics to work out his starting potential? [latex]mgh=\frac{mv^2}{2}[/latex] (BTW has anyone pondered on why the number 2 pops up in most of these equations?) Another way to find [latex]U_p[/latex] is to integrate from r=SR to R=Earth radius, but in the dynamic case, involving Humpty's elevation from the SR radius, the standard equation, [latex]U_p=\frac{GMm}{r}[/latex] does not hold, because such extreme change in potential affects Humpty's own mass. Fortunately a correction can be made, so the equation becomes relativistic. [latex]U_p=\frac{GMm}{r\sqrt{1-\frac{2GM}{rc^2}}}[/latex] The corrective term in the denominator is a dimensionless factor numerically equivalent to [latex]\gamma[/latex] in SR, and using this equation we find out that the integral from the SR radius to infinity is equal to [latex]2mc^2[/latex]. (there is the number two again). Yes, the number two pops up all over the place, because matter is always created alongside antimatter, and because our classical understanding of physics only deals with matter we find ourselves dividing everything by two. Anyway, it turns out that we can find Humpty's starting potential by an even simpler method. We can simply convert Humpty's mass to energy in eV (electron volts) and divide by the number of nucleons ( ie. drop the c's and drop the e's), this will give us the potential directly in terms of Volts, which is a unit we are all familiar with. And you may want to check this for yourself, that Humpty's potential when sitting on the wall is a whopping 930,000,000 Volts. The key to unlocking the secrets of the Universe is in the following equation which I suggest should be called the "Rosetta Equation" for obvious reasons. [latex]\gamma=\frac{1}{\sqrt{1-\frac{v_2}{c_2}}}=\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}=\frac{1}{\sqrt{1-\frac{\phi^2}{\Phi^2}}}[/latex] Where [latex]\phi[/latex] is ground potential (observers potential) and [latex]\Phi[/latex] is the potential of a single free proton. This equation can be solved for any of the terms, opening a new window into the world we live in. So far as I know, no new physics has been introduced above, nor have I made any speculations, I believe each point I have made can be defended with a rational argument. Steven Sesselmann -edited thanks to Endy0816 who pointed out the typo Edited November 1, 2014 by hypervalent_iodine Advertising link removed Link to post Share on other sites

Endy0816 457 Posted November 1, 2014 Share Posted November 1, 2014 (edited) Well the initial statement is incorrect. What you are defining either needs to have a smaller event horizon or a greater mass to be a black hole. That 'Rosetta equation' uses an incorrect definition of the Lorentz factor. ie. I think you are mixing and matching equations too, but it is late Halloween night here so I could well be mistaken on that. Edited November 1, 2014 by Endy0816 Link to post Share on other sites

elfmotat 324 Posted November 1, 2014 Share Posted November 1, 2014 Humpty is falling from ground potential through the ultimate potential to the SR radius, now how can we use the known laws of physics to work out his starting potential? [latex]mgh=\frac{mv^2}{2}[/latex] The left hand side of that equation only applies when the gravitational field is approximately constant. The right hand side only applies to nonrelativistic bodies. Since we're discussing black holes we need to take into account relativity. (BTW has anyone pondered on why the number 2 pops up in most of these equations?) The factor 1/2 comes from the series expansion of the relativistic energy: [math]E= \frac{mc^2}{\sqrt{1-v^2/c^2}}= mc^2+ \frac{1}{2}mv^2+ \frac{3}{8} \frac{mv^4}{c^2}+ \frac{5}{16} \frac{mv^6}{c^4}+...[/math] when v is much smaller than c (i.e. when things are nonrelativistic) all of the terms after mv^{2}/2 effectively go to zero. Another way to find [latex]U_p[/latex] is to integrate from r=SR to R=Earth radius, but in the dynamic case, involving Humpty's elevation from the SR radius, the standard equation, [latex]U_p=\frac{GMm}{r}[/latex] does not hold Okay, looks good so far. because such extreme change in potential affects Humpty's own mass. This is just plain wrong. The new factor comes from the metric and it has nothing at all to do with Humpty's mass. Fortunately a correction can be made, so the equation becomes relativistic. [latex]U_p=\frac{GMm}{r\sqrt{1-\frac{2GM}{rc^2}}}[/latex] Okay, looks good. The corrective term in the denominator is a dimensionless factor numerically equivalent to [latex]\gamma[/latex] in SR, and using this equation we find out that the integral from the SR radius to infinity is equal to [latex]2mc^2[/latex]. (there is the number two again). Yes, the number two pops up all over the place, because matter is always created alongside antimatter, and because our classical understanding of physics only deals with matter we find ourselves dividing everything by two. You shouldn't say it is equivalent to gamma, because it's not really except very superficially. Just because they both have square roots in the denominator doesn't mean they are equivalent. The factor of 2 here comes out of the derivation of the Schwarzschild metric itself. It has absolutely nothing whatsoever to do with anti-matter. I have no idea where you got that idea from. Anyway, it turns out that we can find Humpty's starting potential by an even simpler method. We can simply convert Humpty's mass to energy in eV (electron volts) and divide by the number of nucleons ( ie. drop the c's and drop the e's), this will give us the potential directly in terms of Volts, which is a unit we are all familiar with. And you may want to check this for yourself, that Humpty's potential when sitting on the wall is a whopping 930,000,000 Volts. This makes absolutely no sense. Dividing by "# of nucleons" simply gives you "energy per nucleon." You can't just divide the units by e - this isn't an electrostatics problem! Talking about this in terms of volts makes literally zero sense. The key to unlocking the secrets of the Universe is in the following equation which I suggest should be called the "Rosetta Equation" for obvious reasons. [latex]\gamma=\frac{1}{\sqrt{1-\frac{v_2}{c_2}}}=\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}=\frac{1}{\sqrt{1-\frac{\phi^2}{\Phi^2}}}[/latex] Where [latex]\phi[/latex] is ground potential (observers potential) and [latex]\Phi[/latex] is the potential of a single free proton. Like Endy says, you're just mixing and matching equations together without any real understanding of what they mean. The first one is the Lorentz factor. The second one only applies to non-moving things in Schwarzschild spacetime, and the third one is made up nonsense. This equation can be solved for any of the terms, opening a new window into the world we live in. So far as I know, no new physics has been introduced above, nor have I made any speculations, I believe each point I have made can be defended with a rational argument. Steven Sesselmann I count four things you made up: "This equation can be solved for any of the terms, opening a new window into the world we live in." The "Rosetta Equation." The voltage bunk. The antimatter bunk. There are members here who are very knowledgeable on the subject of GR. It's not so easy to fool us with nonsense equations. The latter half of your post is quite obviously speculative hokum. Link to post Share on other sites

beejewel 6 Posted November 1, 2014 Author Share Posted November 1, 2014 "This equation can be solved for any of the terms, opening a new window into the world we live in." The "Rosetta Equation." The voltage bunk. The antimatter bunk. Yes, the equation solves new problems, yes I gave it a name (makes it easier to refer back to), yes the potential energy term is mine, and yes antimatter is produced whenever matter is produced, and I believe it gives rise to many of the terms that feature factor 1/2. I disagree that anything outside the textbook is "bunk" and therefore incorrect, your points 1 & 2 above are irrelevant as they refer to my writing style, so please let's focus on points 3 & 4, what's the issue here? Steven Link to post Share on other sites

elfmotat 324 Posted November 1, 2014 Share Posted November 1, 2014 Yes, the equation solves new problems, yes I gave it a name (makes it easier to refer back to), yes the potential energy term is mine, and yes antimatter is produced whenever matter is produced, and I believe it gives rise to many of the terms that feature factor 1/2. I disagree that anything outside the textbook is "bunk" and therefore incorrect, your points 1 & 2 above are irrelevant as they refer to my writing style, so please let's focus on points 3 & 4, what's the issue here? Steven The issue is that they are nonsensical. As in, they don't make any sense. You pulled them out of a foul-smelling orifice. Link to post Share on other sites

beejewel 6 Posted November 1, 2014 Author Share Posted November 1, 2014 The mass per nucleon, when expressed in eV is the potential energy of the matter, and when you drop the e's, becomes an expression of a bodies potential. We know of no stable particle heavier than a proton, the rational conclusion is that the potential of the proton is a constant and an upper limit to potential Steven Link to post Share on other sites

elfmotat 324 Posted November 1, 2014 Share Posted November 1, 2014 The mass per nucleon, when expressed in eV is the potential energy of the matter, and when you drop the e's, becomes an expression of a bodies potential. We know of no stable particle heavier than a proton, the rational conclusion is that the potential of the proton is a constant and an upper limit to potential Steven This isn't an electrostatics problem! You can't just divide by charge! Dividing by charge just gives you energy per charge! There's a gravitational potential, which you get by diving the potential energy by the mass m. There isn't any electrostatic potential in Schwarzschild spacetime. I don't know how to explain it any more simply than that. As far as there being no stable particles heavier than a proton, what about, for example, a helium atom? Link to post Share on other sites

beejewel 6 Posted November 1, 2014 Author Share Posted November 1, 2014 (edited) This isn't an electrostatics problem! You can't just divide by charge! Dividing by charge just gives you energy per charge! I never said to divide by charge, I said drop the e's because 1 eV is the energy gained by one electron when falling through a potential of 1 Volt. There isn't any electrostatic potential in Schwarzschild spacetime. Well there isn't any Schwartzschild in electrostatic potential either, so I guess that makes us even As far as there being no stable particles heavier than a proton, what about, for example, a helium atom? We were discussing mass per nucleon, and H measures in at 938 MeV all other elements have a lower, mass per nucleon, only the neutron exceeds the proton, but it isn't stable. 938 MeV is the upper limit to potential in our world, and [latex]\frac{938}{2}[/latex] is the lowest potential in our word and it is the potential at the SR radius. My argument for this is, all the kings men, and all this kings horses, or if you prefer neither the Americans, the Russians nor the Chinese, given all the resources of the Universe, will ever be able to create a potential higher than 938 million volts. The reason is straight forward and simple, take a cathode of pure electrons and an anode of pure protons, and the surface potential can never be higher than the potential of the individual particles it is made from. Edited November 1, 2014 by beejewel Link to post Share on other sites

elfmotat 324 Posted November 1, 2014 Share Posted November 1, 2014 (edited) I never said to divide by charge, I said drop the e's because 1 eV is the energy gained by one electron when falling through a potential of 1 Volt. "Voltage" or "potential difference" means the energy per unit charge difference between two points in an electric field. There is no electric field here. Dividing by charge doesn't give you a potential difference, it gives you energy per charge. I don't know how else to say it. Edited November 1, 2014 by elfmotat Link to post Share on other sites

beejewel 6 Posted November 1, 2014 Author Share Posted November 1, 2014 "Voltage" or "potential difference" means the energy per unit charge difference between two points in an electric field. There is no electric field here. Dividing by charge doesn't give you a potential difference, it gives you energy per charge. I don't know how else to say it.Sounds like you are contradicting yourself above. You make a statement that there is no electric charge here, but have you ever tried measuring it? Actually the potential gradient at sea level is as follows; [latex]v_{rel}=c(\frac{\Delta\phi}{\Phi})[/latex] [latex]v=\sqrt{2gh}[/latex] [latex]\sqrt{2gh}=c(\frac{\Delta\phi}{\Phi})[/latex] When you plug in the numbers and use 1 meter as the height and solve for [latex]\Delta\phi[/latex] it works out to 13.9 Volts per meter elevation. Steven Link to post Share on other sites

elfmotat 324 Posted November 1, 2014 Share Posted November 1, 2014 Sounds like you are contradicting yourself above. You make a statement that there is no electric charge here, but have you ever tried measuring it? Actually the potential gradient at sea level is as follows; [latex]v_{rel}=c(\frac{\Delta\phi}{\Phi})[/latex] [latex]v=\sqrt{2gh}[/latex] [latex]\sqrt{2gh}=c(\frac{\Delta\phi}{\Phi})[/latex] When you plug in the numbers and use 1 meter as the height and solve for [latex]\Delta\phi[/latex] it works out to 13.9 Volts per meter elevation. Steven There is no electromagnetic field in Schwarzschild spacetime, by definition. It looks like you're just stringing random equations together, and I have no idea where you're getting Volts from given there's no electric field. Link to post Share on other sites

ajb 1567 Posted November 1, 2014 Share Posted November 1, 2014 "Voltage" or "potential difference" means the energy per unit charge difference between two points in an electric field. There is no electric field here. Dividing by charge doesn't give you a potential difference, it gives you energy per charge. I don't know how else to say it. I have a feeling this has been discussed elsewhere already... Anyway, from what I can see beejewel has just derived some units of mass in terms of volts. Volts is energy per charge, you take an electron to be that charge, then you can say that an electron has mass 1 volt (c=1). Using the standard definition of electron volts as a mass unit would be better. Link to post Share on other sites

beejewel 6 Posted November 1, 2014 Author Share Posted November 1, 2014 (edited) Providing the following statement, [latex]\frac{1}{\sqrt{1-\frac{v_2}{c_2}}}=\frac{1}{\sqrt{1-\frac{\phi^2}{\Phi^2}}}[/latex], is true, we can solve for v [latex] v=c(\frac{\phi}{\Phi}) [/latex], Where [latex]\phi[/latex] is the observers potential and [latex]\Phi[/latex] is the proton potential. The relative velocity between two bodies of different potential then becomes, [latex] v_{rel}=c(\frac{\Delta\phi}{\Phi}) [/latex], It is known from classical mechanics, that a body in free fall at sea level reaches a velocity of, [latex] v=\sqrt{2gh} [/latex], therefore the difference in potential between radius r and r=r-1 must be, [latex] \sqrt{2gh}=c(\frac{\Delta\phi}{\Phi}) [/latex] [latex]\Delta\phi[/latex] which works out to 13.9 Volts. From my point of view, there is no longer a destinction between electrical potential and gravitational potential, it's all one and the same thing. This is a rational argument, and not a string of random equations as elfmotat suggests. Steven - I fail to see why this thread was moved to the speculations forum, so far it's just a discussion around the known laws of physics. Edited November 1, 2014 by beejewel -1 Link to post Share on other sites

ajb 1567 Posted November 1, 2014 Share Posted November 1, 2014 From my point of view, there is no longer a destinction between electrical potential and gravitational potential, it's all one and the same thing. However, it is not true that you can simply equate the electromagnetic field with the gravitational field, not even in the static limit. It is true that there are some formal similarities, again this is especially true in the (non-relativistic) static limit. Newtonian gravity does look like electrostatics. The problem is that this formal analogy does allow you to just equate the two. In fact a simple experiment using a charged test particle in the presence of a electromagnetic field and with just a gravitational field should convince you that gravity and electromagnetism cannot be the same thing. In fact this is a nice fact about the equivalence principal. Loosely it works because the inertial mass is the same as the gravitational mass which is the 'charge' for gravity. That is the property that allows the field to couple to our test particle. (More technically we need to include energy here also, but anyway...) However, for electromagnetic theory it is the electric charge that is the 'charge' for electromagnetism, and this is different to the inertial mass of the test particle. Link to post Share on other sites

beejewel 6 Posted November 1, 2014 Author Share Posted November 1, 2014 It may not be so straight forward, because the potential energy [latex]U_p[/latex] is stored in the atom itself, it is the separation between the peak and the trough of the electron/proton wave, which holds the stored energy, which in turn is a function of the observers potential. Outside the atom the stored electrical potential simply manifests itself as mass. Steven Link to post Share on other sites

swansont 7597 Posted November 1, 2014 Share Posted November 1, 2014 It may not be so straight forward, because the potential energy [latex]U_p[/latex] is stored in the atom itself, it is the separation between the peak and the trough of the electron/proton wave, which holds the stored energy, which in turn is a function of the observers potential. Outside the atom the stored electrical potential simply manifests itself as mass. Steven ! Moderator Note Let's try and keep this to one speculation at a time and not branch out. Any discussion of this should take place in another thread. Link to post Share on other sites

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