joigus 201 Posted July 25 (edited) 6 minutes ago, studiot said: This is not what Martoonsky says. Sorry, we almost x-posted. You didn't see my edit. The only thing that makes sense to me for the red line is the g(r) field strength. Edit: The only thing that doesn't check with me is the sign. It should be negative (inward pointing). Edited July 25 by joigus Addition 0 Share this post Link to post Share on other sites

studiot 2025 Posted July 25 (edited) 7 minutes ago, joigus said: Sorry, we almost x-posted. You didn't see my edit. The only thing that makes sense to me for the red line is the g(r) field strength. The fact remains he has messed up his primary equation and it is not up to us to fix it. What makes more sense is the fact that the air pressure (whatever its function of r may be) must be a maximum at the centre and decrease towards the surface. Thus if the equation were to be written P_{r} = P_{0} - f(r) which says in words that "the pressure at radius r is the pressure at the centre minus some function of the radius (which could be a suitable integral)" As Janus says Pr must equal the surface pressure of the atmosphere at the surface, and thereafter follow perhaps a different function in the atmosphere outside the Earth. Edited July 25 by studiot 0 Share this post Link to post Share on other sites

joigus 201 Posted July 25 1 minute ago, studiot said: What makes more sense is the fact that the air pressure (whatever its function of r may be) must be a maximum at the centre and decrease towards the surface. Thus if the equation ere to be written P_{r} = P_{0} - f(r) which says in words that "the pressure at radius r is the pressure at the centre minus some function of the radius (which could be a suitable integral)" True. There's also the problem of T(r), because that's not a constant with r. I suggest to let @Martoonsky speak. 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted July 25 (edited) @studiot, if we used the constants [math]\rho\left(R\right)[/math] and [math] g\left(R\right) [/math] then it would be true that [math] \int_{0}^{R} \rho\left(R\right)g\left(R\right)dr = \rho\left(R\right)g\left(R\right)R = P_{R} [/math]. However, this is not the case. When [math] \rho = \rho\left(r\right) [/math] and [math] g = g\left(r\right) [/math] then in general, [math] \int_{0}^{R} \rho\left(r\right)g\left(r\right)dr \neq \rho\left(R\right)g\left(R\right)R [/math]. Note also that [math] \int \rho\left(r\right)g\left(r\right)dr \neq \rho\left(r\right)g\left(r\right)r [/math]. If you still disagree with the math, then please show your own working out. @joigus, I was not trying to represent any particular formulas in my graph. I was just trying to point out graphically that two different functions will not have the same value of definite integral over the interval [math]r = 0[/math] to [math]r = R[/math] just because their endpoints have the same value. Edited July 25 by Martoonsky 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 9 Well, if anyone is following this, I've been working on it. I think I have a method worked out, but I'm still working on the details. More later. 0 Share this post Link to post Share on other sites

joigus 201 Posted August 9 Was thinking about this some 24 hours ago and wondering what had become of you and your problem. I'll keep an eye on it. It's interesting. 0 Share this post Link to post Share on other sites

studiot 2025 Posted August 9 On 7/25/2020 at 5:44 PM, Martoonsky said: @studiot, if we used the constants ρ(R) and g(R) then it would be true that ∫R0ρ(R)g(R)dr=ρ(R)g(R)R=PR . However, this is not the case. When ρ=ρ(r) and g=g(r) then in general, ∫R0ρ(r)g(r)dr≠ρ(R)g(R)R . Note also that ∫ρ(r)g(r)dr≠ρ(r)g(r)r . If you still disagree with the math, then please show your own working out. I neither agree nor disagree with you maths I just can't follow it. What do you mean by the definite integral of the product of two constants, from the limits zero to a constant R, with respect to an variable r ? 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 10 (edited) 23 hours ago, studiot said: I neither agree nor disagree with you maths I just can't follow it. What do you mean by the definite integral of the product of two constants, from the limits zero to a constant R, with respect to an variable r ? For constant [math]a [/math] and variable [math] r [/math], [math]\int a dr = a \int dr = ar + c \\[/math] However, for a function [math]f\left(r\right), \int f\left(r\right) dr \neq f\left(r\right)\int dr [/math] Thus [math]\int \rho\left(R\right)g\left(R\right)dr = \rho\left(R\right)g\left(R\right)\int dr [/math] and [math]\int_{0}^{R} \rho\left(R\right)g\left(R\right)dr = \rho\left(R\right)g\left(R\right)\int_{0}^{R}dr = \rho\left(R\right)g\left(R\right)[r \vert_{0}^{R}] = \rho\left(R\right)g\left(R\right)R.[/math] However, when [math] \rho [/math] and [math] g [/math] are the functions [math] \rho\left(r\right) [/math] and [math] g\left(r\right)[/math] then [math]\rho[/math] and [math]g[/math] may not be removed from the integral and the functions will affect the result. For instance, if [math]\rho = 6 r^3[/math] and [math] g = \frac{1}{r^2}[/math] then [math]\rho g = 6r[/math] and [math]\int_{0}^{R} \rho g dr = \int_{0}^{R} 6r dr = 6\int_{0}^{R} r dr = 6[\frac{r^2}{2} \vert_{0}^{R}] = 3R^2.[/math] But this is not equal to [math]\rho\left(R\right)g\left(R\right)R = 6R^3 \left(\frac{1}{R^2}\right)R = 6R^2.[/math] Thus when [math]\rho[/math] and [math]g[/math] are functions of [math]r[/math] they make a difference in the value of the integral. Edited August 10 by Martoonsky 0 Share this post Link to post Share on other sites

studiot 2025 Posted August 10 (edited) Let us go back to the conditions you specified in the beginning of this thread. On 7/23/2020 at 4:03 PM, Martoonsky said: If there were a tunnel from one side of the earth, through the center of the earth, and on out the other side, what would the air pressure be in the center? I am assuming a spherical earth to take advantage of symmetry. On 7/25/2020 at 3:50 AM, Martoonsky said: I don't see this as much of a problem theoretically. You could make the shaft very narrow, or add air. If the atmosphere were depleted, it would reduce the value of the pressure at the center. I haven't yet tried to do a detailed calculation but the above graphic provides the answer. P-sub-zero is the pressure at the center (at radius r = 0). R is the radius at the surface. Rho = rho(r) is the density and g = g(r) is the acceleration due to gravity. I thought I would divide the integral into sections for the inner and outer cores, mantle and crust. I think I could come up with an average g for each section. I'm not sure what to do with rho yet since it varies with temperature and pressure. So the tunnel has two ends, both open to the atmousphere at r = R. The pressure at both ends must be equal and equal to the pressure of the atmousphere at the surface of the Earth. If they were not equal then air would move along the tunnel to make them equal. If they were not equal to the atmouspheric pressure then air would move into or out of the tunnel until they were equal. This pressure is denoted P_{R} So you have one value for P_{R}, given by the integral from r = R to r = the edge of space given by the conventional formula for the atmouspheric column above any point of the surface. You can choose a constant gravity and density or assume some meteorological function to get this figure. You get a second value for P_{R} by developing a formula for gravity inside the Earth, which must equal the first value, as already noted. So you can form an equation between these two for your answer. Edited August 10 by studiot 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 11 @studiot Yes, that's right. I was also planning on having the temperature be variable as we go down into the earth. @joigus I've run into a snag. I was planning on using the ideal gas law but I now realize that deep inside the earth, the air would be heated to a plasma. 0 Share this post Link to post Share on other sites

studiot 2025 Posted August 12 On 8/11/2020 at 4:37 AM, Martoonsky said: @studiot Yes, that's right. I was also planning on having the temperature be variable as we go down into the earth. @joigus I've run into a snag. I was planning on using the ideal gas law but I now realize that deep inside the earth, the air would be heated to a plasma. Within the Earth, the gravitational acceleration increases linearly as you move from the centre to the surface. So this has the form g = jr, where j is a positive constant. An interesting side effect of this is that if you dropped a capsule into one end of the tunnel it would execute SHM about the centre. Of course, above the Earth g diminishes in accordance with the nomal inverse square law. But you don't need this. I am working on the density ATM, so these relationships can be combined 0 Share this post Link to post Share on other sites

joigus 201 Posted August 12 On 8/11/2020 at 5:37 AM, Martoonsky said: I was planning on using the ideal gas law but I now realize that deep inside the earth, the air would be heated to a plasma. Yeah, that doesn't work. 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 12 5 hours ago, studiot said: Within the Earth, the gravitational acceleration increases linearly as you move from the centre to the surface. So this has the form g = jr, where j is a positive constant. This would be true if the earth were of constant density. I was planning on modeling the earth as a number of layers of different densities, which would complicate the formula for g(r). I had a strategy all mapped out, but now I'm having a bit of trouble dealing with the air molecules dissociating and becoming a plasma near the center. 0 Share this post Link to post Share on other sites

Ken Fabian 130 Posted August 12 (edited) I admit this one does have me scratching my head. I wouldn't expect heat to change the pressure once temperature equilibrium is reached but I'm not sure of it. But if you have an imaginary borehole you can imagine it being insulated and then consider the heat separately. On reflection it is likely the heat does matter - by changing the density of the air (or plasma), which changes the weight of the air column. My first impressions left me thinking the pressure would be the same as surface air pressure but on reflection it would be SAP plus the pressure from the weight of the air column within the Earth from core to surface - and that weight will vary along that column according to changed gravity gradient - and would be less than a gravity gradient based on a point mass at the core. Maths for that is beyond me. Edited August 12 by Ken Fabian 1 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 12 17 minutes ago, Ken Fabian said: But if you have an imaginary borehole you can imagine it being insulated and then consider the heat separately. Yeah, if I decided the borehole was lined with a perfect insulator, that would make life easier. I decided to take temperature into account. I'm pretty sure it would make a difference - it is part of the ideal gas law. 0 Share this post Link to post Share on other sites

studiot 2025 Posted August 13 (edited) 8 hours ago, Ken Fabian said: I admit this one does have me scratching my head. I wouldn't expect heat to change the pressure once temperature equilibrium is reached but I'm not sure of it. But if you have an imaginary borehole you can imagine it being insulated and then consider the heat separately. On reflection it is likely the heat does matter - by changing the density of the air (or plasma), which changes the weight of the air column. My first impressions left me thinking the pressure would be the same as surface air pressure but on reflection it would be SAP plus the pressure from the weight of the air column within the Earth from core to surface - and that weight will vary along that column according to changed gravity gradient - and would be less than a gravity gradient based on a point mass at the core. Maths for that is beyond me. Ken, +1 for the insulator idea. Conventional meteorological models convert the compression to temperature to perform the integration. But they also mostly consider gravity constant. Apparently this assumption makes less than 0.5% difference in the atmosphere (Yuan 1970) 8 hours ago, Martoonsky said: Yeah, if I decided the borehole was lined with a perfect insulator, that would make life easier. I decided to take temperature into account. I'm pretty sure it would make a difference - it is part of the ideal gas law. I am not convinced the air would reach plasma temperatures. Even without the Earth's rotation a plume of hot air would issue from the ends of the tunnel. The rotation makes it important to specify the direction/position of the tunnel. Edited August 13 by studiot 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 13 I think it's reasonable to expect some plasma development in the earth's core. The temperatures there are hotter than the photospheres of some stars. It's a matter of degree, though, and I'm hoping that the ideal gas law may still be accurate. I don't think there would be a jet of hot air from the borehole. This would evacuate the air from the interior, creating a vacuum and eliminating the pressure which would drive the flow of air. I believe the air would settle into hydrostatic equilibrium, as it does in stars and the earth's atmosphere. It would likely be very rarefied near the center. I don't plan on considering the earth's rotation for now, but that's a good point. The plasma issue is looking a bit thorny to figure out, so I think I'll go ahead for now just assuming that the ideal gas law applies throughout. 0 Share this post Link to post Share on other sites

studiot 2025 Posted August 13 37 minutes ago, Martoonsky said: I don't think there would be a jet of hot air from the borehole. This would evacuate the air from the interior, creating a vacuum and eliminating the pressure which would drive the flow of air. I believe the air would settle into hydrostatic equilibrium, as it does in stars and the earth's atmosphere. It would likely be very rarefied near the center. There must be. No vacuum either. You would get countercurent flow of colder air, just as you do in the atmosphere more generally. The system would most certainly not be in equilibrium, any more than the atmosphere or interior of the Earth is. Stars are constantly ejecting high temperature material. Yes I think it is a good idea to develop this in stages, refinign as a result of information gained at each stage. Because you tunnel goes right through the Earth, the function P(r) must be symmetrical about the centre ie an even function. So a power series expansion would contain only even powers. My model so far has a product of two functions of r so involves integration by parts. Are you up for that? I will be interested to see what temperatures you come up with with a static solution. 0 Share this post Link to post Share on other sites

Ken Fabian 130 Posted August 13 (edited) 3 hours ago, studiot said: Even without the Earth's rotation a plume of hot air would issue from the ends of the tunnel. I don't think that would be true; there is no source of air at the core to support a plume. Nor for flinging air out. I think it would reach pressure and temperature equilibrium and become effectively static. Air movement from convection can occur within the tunnel and changing weather based pressure differences at each end would generate air movements. Given the distances those air movements might be more like sloshing waves along the tunnel, generating their own small transient pressure variations. Edited August 13 by Ken Fabian 0 Share this post Link to post Share on other sites

joigus 201 Posted August 13 Even if air at centre is not a plasma, certainly the equation of state to be used would have to be a power series in the density, extending the way that the Van der Waals eq. deals with the first terms. \[P=a\left(T\right)+b\left(T\right)\rho+c\left(T\right)\rho^{2}+\cdots\] This is very general for real gases even in conditions of high pressure, very far from ideal gases. High temperatures makes it behave more gas-like. But estimating the T-dependent coefficients is another story... Ideal gases doesn't cut it because already for Van der Waals you have 3rd power of density essential to account for phenomenology. Using the tools of my trade what I would do if I were desperate to solve it is depart from a simplified molecular model and calculate the partition function. And from there to the equation of state. The thing that I see difficult from first principles is that the way I see it you must have an enormous reservoir of air to fill in the hole while at the same time have the air compensate for the enormous pressures of the solid/plasma, hot Earth material, that would tend to squeeze the borehole to a mathematical line. I'm reading also @Ken Fabian and @studiot's ideas, as the OP. See if I can relate them to my reasoning... The comment I want to make is that it is entirely possible that the parametrics of the problem becomes ridiculously impossible. After all, we're asking the atmosphere to hold the Earth in place, which wants to recover this hole by squeezing it out. Does that make any sense to any of you? ------- PD: The thing that makes me very suspicious is that assuming exponential atmosphere with T at centre given by known temperature at centre of the Earth returns numbers so out of whack that I'm confused. The whole thing could have an implicit assumption that makes it thermodynamically rotten at its core (pun intended). Totally agree with comments by @Martoonsky and @Ken Fabian that situation is static (local equilibrium). Everything must grind to a halt. 0 Share this post Link to post Share on other sites

Ken Fabian 130 Posted August 13 (edited) 15 minutes ago, joigus said: the way I see it you must have an enormous reservoir of air to fill in the hole while at the same time have the air compensate for the enormous pressures of the solid/plasma, hot Earth material, that would tend to squeeze the borehole to a mathematical line. I think the question is pressure for a lined borehole, a tunnel with walls that hold back the pressure. Or else the question reverts to what is the pressure at Earth's core, sans borehole. Edited August 13 by Ken Fabian 1 Share this post Link to post Share on other sites

joigus 201 Posted August 13 2 minutes ago, Ken Fabian said: I think the question is pressure for a lined borehole, a tunnel with walls that hold back the pressure. Or else the answer reverts to what is the pressure at Earth's core, sans borehole. Very good point! +1 Answer in jest: That's cheating! Answer in earnest: I think you're right that constraints are needed to make the problem well-defined. Are you assuming the tunnel also insulating, so we can guess only in (local) equilibrium with itself? 0 Share this post Link to post Share on other sites

joigus 201 Posted August 13 1 hour ago, joigus said: Are you assuming the tunnel also insulating, so we can guess only in (local) equilibrium with itself? Sorry, I didn't read carefully above. You're assuming that. 0 Share this post Link to post Share on other sites

Martoonsky 1 Posted August 13 (edited) It is essential that the walls of the borehole be able to withstand the pressures exerted on them by the earth. There's no way the air pressure in an open tube would be able to do it. Also, the volume of the troposphere is 1E10 cubic kilometers whereas the volume of a one meter borehole would be 0.01 cubic kilometers, so the volume of air should not be an issue. Edited August 13 by Martoonsky 0 Share this post Link to post Share on other sites

studiot 2025 Posted August 13 3 hours ago, Ken Fabian said: I think the question is pressure for a lined borehole, a tunnel with walls that hold back the pressure. Or else the question reverts to what is the pressure at Earth's core, sans borehole. Yes but that is not all. The whole question was originally too short on detail. It never was a practical proposition. So yes we can allow superstrong adiabatic walls. But don't forget that the real Earth has a molten nickel iron core of some sort. Drilling a hole into this will produce the all time great gusher of a well. So this thought experiment must allow some magic to be allowed to proceed. As a matter of interest we have produced pressure in excess of those at the centre of the Earth but only in a so called diamond anvil. 16 hours ago, Martoonsky said: This would be true if the earth were of constant density. I was planning on modeling the earth as a number of layers of different densities, which would complicate the formula for g(r). I had a strategy all mapped out, but now I'm having a bit of trouble dealing with the air molecules dissociating and becoming a plasma near the center. I am using constant Earth density (not gas density in the tunnel) as a first stab. That should produce useful ball park information. 3 hours ago, joigus said: Totally agree with comments by @Martoonsky and @Ken Fabian that situation is static (local equilibrium). Everything must grind to a halt. That cannot possibly be because heat must be transported down the temperature gradient. 3 hours ago, Ken Fabian said: I don't think that would be true; there is no source of air at the core to support a plume. Nor for flinging air out. I think it would reach pressure and temperature equilibrium and become effectively static. Air movement from convection can occur within the tunnel and changing weather based pressure differences at each end would generate air movements. Given the distances those air movements might be more like sloshing waves along the tunnel, generating their own small transient pressure variations. You don't need a source of air. I said circulation cell and countercurrent flow. They can occur in straw. Both these are already known to operate in terrestrial processes. The helical generator version is the closest model to the Earth's dynamo we have within the core. 0 Share this post Link to post Share on other sites