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No. Electromagnetic interactions already have infinite range because photons are already massless.

This is definitely wrong. Argument of periapsis (location of lowest point) can drift for a number of reasons. Most likely, you have some numerical errors. If you can post the portion of the code that you use to evaluate orbit, we can look for an actual error.

Minkowski isn't Eucledian. Their metrics have different signatures. Both are flat, yes, but you shouldn't confuse the term "Eucledian" to simply mean "flat". The space-time of General Relativity is a Pseudo-Riemannian Manifold. That means that while it isn't necessarily flat, you can always take a small enough region of space that locally it behaves just like Minkowski space. This is why any principle that applies in Special Relativity globally must apply to General Relativity locally. Speed of light is a global limit in SR, but a local one in GR. Causality is globally enforced in SR, but only locally in GR. And so on.

That's a somewhat antiquated look on things. Cosmological constant was introduced rather artificially, and it only makes sense in geometrical interpretation of Gravity, which is outdated by half a century. Modern GR is derived as a gauge theory on Poincare symmetry. That's, essentially, where all the talk about Quantum Gravity comes from. If you look at Gravity from perspective of gauge theory, all curvature must be consequence of stress energy density, because that is the conserved charge of the theory. As such, you cannot account for accelerated expansion by simply plugging in an extra term. There must be physical energy and pressure associated with it. Hence, the hunt for dark energy. Of course, once we go back to the geometry of things, if we assume that dark energy is uniformly distributed across the universe, we could fold it back into cosmological constant. But that's just the way math works out. Oh, and energy and pressure terms are completely different things. In a particular coordinate system, you can picture a rank 2 GR tensor as a 4x4 matrix. Energy is in the top left corner. Pressure are the 3 diagonal terms following it. As such, pressure isn't even a scalar quantity in GR.

That's more of an illustration than explanation of what really happens. All known wormhole configurations end up having an event horizon, and are therefore not traversable, whenever only positive energy is used. All traversable wormholes will have some negative energy density. This is purely a consequence of geometry being traversable. It's not really a matter of "pushing" anything.

The only reason we need polar solvent are the hydrogen bonds in our proteins and DNA. The only reason we need these is because we live at 300K, and weaker bonds fall apart. At 90K, you need about 30 times less energy in the bond. Hydrogen bonds are too strong at these temperatures even with a polar solvent. At these temperatures you actually want non-polar bonds and a non-polar solvent to go along with them. This is Kinetics 101, seriously. Methane is an ideal organic solvent for Titan's 90K environment.

No. Both pressure and energy are diagonal elements of stress energy tensor, but they are distinct. Would not help with the warp drive.

The larger the ship, the less heat shielding, proportionally, you'll need for aerocapture. I'm not going to promise anything, since first few manned missions to Mars are likely to be extra-cautious, and who knows what sort of developments we'll have by the time we turn flights to Mars into a routine. But if we wanted to establish regular traffic between Earth and Mars right now, with current tech, I would absolutely expect aerobraking to be a big part of capture. (Also, a cycler. But that's a separate topic.)

There are plenty of non-polar options. One of the reasons we need polar bonds for our life is "high" temperatures which we live in. You need strong, polar bonds to have molecules stay together at typical Earth temperatures, and that means strong, polar solvent to break them apart and rearrange them. At Titan's temperatures, methane is actually the perfect solvent. Granted, the metabolism rate at these temperatures won't be great, but that's why we don't exepct anything but the simplest bacteria there, if anything at all. So far, everything I've seen on Titan indicates that the ethylene levels in Titan's lakes are much lower than we'd expect at equilibrium. Either something's metabolizing it, or there are some geological processes going on that we have no idea about. Either way, studying Titan should be very high on our list of priorities. I'd also label it as most likely place we know of to have life that has evolved separately from our own. While Mars is a more likely place to (have) harbor(ed) life, we'd be hard pressed to exclude cross-contamination with Earth, since life here and there would have to be pretty similar. With Titan, if we find life there, we'd know that life evolved on two separate planets in one Solar system. Having life on Earth tells us nothing. Finding life on Mars would be awesome, and we'd learn a lot from it. Finding life on Titan, though? That would be the holly grail of cosmology. We'd instantly learn not just that there is life elsewhere in the Universe. We'd learn that universe is absolutely teeming with it. I'm not saying odds of finding life on Titan are great, but they are definitely not terrible. And given that finding life there would be by far the single greatest discovery we have ever made, I think it's worth checking. We can afford to send a few serious probes there, similar to what we've sent to Mars. And we'd be fools not to.

It's entirely possible that somebody either derived it or arrived at it experimentally before me, but I've never seen such use or derivation. Derivation does include some hand-waving as far as gravity turn goes, but I'll get to that in a moment. As far as I know, the only reason it's a poor approximation for real planets is because atmosphere isn't exactly exponential, rocket's drag/mass ratio changes as fuel as consumed, and drag isn't strictly quadratic at hypersonic speeds. Nonetheless, I'm not aware of any simple formula that does better, so it's a good starting point before you refine computations with a proper model or testing. The formula works for airless or nearly so bodies so long as you assume that TWR is not your limiting factor. With infinite TWR you escape from an empty world by accelerating to orbital velocity instantly. That way, there are no losses to gravity. Of course, that's not how real world works. So if you want to consider it a practical limitation, the approximation works so long as atmosphere is a bigger factor than TWR. Anyhow, onto derivation. Let us consider a purely vertical assent in which we attempt to escape the atmosphere. First approximation is that by the time gravity really starts to change, atmosphere is no longer a factor. So we will assume that g is constant. This is one of the very few approximations here that works even better in the real world than in KSP. This problem has an exact solution for quadratic drag. One must ascend at terminal velocity. Proof for constant density is trivial. The reason it's a bit more complicated with decreasing density is that you have to increase TWR past 2 to account for acceleration. Nonetheless, it's possible to prove that prescription of ascent at terminal velocity is still an optimal one. (Using Euler Lagrange Equation.) We consider special case of exponential atmosphere. This approximation is almost exact for KSP except for cutoff. It's considerably worse for real world. We can then write v(h) = vt exp(h/(2H)), where h is altitude. Clearly, we can look at it as a differential equation h'(t) = v(h(t)), and it has a relatively simple solution. h(t) = -2H ln((H - vt t / 2) / H) This is where things start getting exciting. You'll note that this equation diverges at t = 2H/vt. In other words, it takes that long to escape completely to infinity if you continue maintaining terminal velocity. Of course, you'll be moving infinitely fast, and that will require infinite acceleration and infinite fuel, but we aren't really trying to escape out to infinity. We just need to clear significant atmosphere. And naturally, acceleration diverges to infinity just as you clear atmosphere, meaning that nearly all of this t = 2H/vt is spent escaping the atmosphere. Final observation is that amount of throttle you need to apply to follow this trajectory is F(t) = 2mg + ma(t), where 'a' is acceleration and 2mg are your losses to drag and gravity. (Equally split at v = vt.) In vacuum and zero gravity, Integrating F(t)/m would give you your final velocity. So that's your real delta-V. However, the velocity you are actually getting is integral of a(t). That means integral of 2g over time it takes you to escape is your loss. And we know the amount of time it takes to escape. It's right above. So the total loss is 4gH/vt. There are several approximations up there, but they are all reasonable, and one can even do some estimates on the errors, and they do work out to be very, very small for KSP, and reasonable for real world. Hand-waving starts here. We don't have a full solution for gravity turn. There are numerical models, but no analytic solution we can take. At this point, I wouldn't call it an assumption. The conjecture is that gravity + drag losses during optimal ascent through gravity turn are approximately the same as during vertical escape. I do not have a satisfactory explanation for why it should be so. But it agrees with all of the numerical simulations on ascent, with all of the empirical numbers from KSP, including ascent from tallest peaks of Eve, and is even in the ballpark for real rockets.

That's not how entanglement works. At all. More importantly, ther is an actual theorem that states that entangled states cannot be used for communication. Event horizon is a type of coordinate singularity. What it means is that there are no time-like trajectories leading out of the black hole. There are space-like trajectories, which is the reason for Hawking Radiation. The discussion about information is really more about entropy. There are several arguments for why it's actually a bad thing. For details, see Cosmic Censor.

No atmosphere. Build a giant magrail on the ground. Use it for launches and landings. We should do this on the Moon, by the way. Soon-ish. Edit: But just for reference, a good estimate of atmospheric losses during launch is 4gH/vt, where g is surface gravity, H is scale height, and vt is terminal velocity at the surface. The derivation's a bit insane, and there are a lot of assumptions there, but it will get you within a ballpark for estimates.

Right. Atmo + gravity eats through about 1.5km/s. So it needs to be closer to 3.5km/s total. We can also do the math based on the stats. v1.1 is 505T off the pad. It has 5,885kN of thrust at 311s for 180s. That's 347T of fuel leaving 158T ascending at stage separation. Taking average ISP at 300s, I'm getting 3.4km/s of delta V from the first stage. So 2km/s at separation sounds reasonable to me, given aerodynamic losses. That leaves second stage with about 5.4km/s to orbit. (Earth's rotation FTW.) Similar math to above (801kN @ 342s for 375s) gives me 90T of fuel in second stage. With the 5.4km/s assumption above, this works out to a 22T rocket reaching orbit, of which 13T is the payload. That means empty, dry 2nd stage is 9T. Fueled with payload, 112T. And that leaves 46T for empty first stage. All entirely within reason. Disclaimer: I've ignored some finer aspects of the ascent, inclination of the orbit, and done quite a bit of rounding above. Still, it should be pretty close to actual values.

Not entirely true, even though it works out that way in practice. You can vary thrust of the SRB by varrying internal pressure. Unfortunately, it's complicated, less reliable than with liquid rockets, and comes at the cost of efficiency. At that point, the core advantage of SRB, which is mechanical simplicity, is completely gone. So it's much better to just use a liquid or hybrid engine. On the other hand, shutting down an SRB is relatively straight forward. You just have to relieve the pressure, which is sometimes as easy as opening up the top. I have no idea if Shuttle SRB used something like that, but I do know that some SRBs have emergency shutdown ability that works along these lines.

Yeah, but they still have to become null-dust eventually to conserve momentum. And that's basically means a photon drive or equivalent. So while Q-thruster can and should work, its efficiency cannot exceed 300MW/N. There are experiments suggesting higher efficiency, but I find them very doubtful. We'll see, I guess, but I wouldn't expect anything out of it.

But... I mean... Negative volume? Honestly, never thought of it this way. Sounds insane, but given that we are dealing with a non-positive definite metric, maybe? My measure theory is rusty at best. I'm going to bug my friends from math dept. about this. Back to slightly less insane physics of General Relativity, in general (heh) negative energy does not mean negative mass. In Alcubierre metric specifically, however, you get both. It's negative energy density and negative mass density. But both are densities, so your volume comment still applies. Edit: Has anyone actually integrated the energy density in a warp bubble? I mean, has anyone actually verified that the total energy is negative? I mean, I know I didn't think of that, but somebody ought to have, right? Right? I'm going to need to do some serious math. After I talk to some serious people who know how to do serious math. Already looked through some analysis textbooks. They all assume positive definite metric. Gah. A little while later edit: Looks like I've managed to thoroughly confuse myself up there. The densities in GR are 3-volume densities, which, naturally, are frame dependent. That's why the 4-vector quantity containing energy, the 4-momentum, has a tensor density. It's sort of like the flow density, and we are interested in its intersection with a 3-volume in selected coordinate system. Since the 3-volume is always normal to proper time of the coordinate system, the 3-volume is always positive. Essentially, you have to be moving faster than light to begin with to get positive energy turn into negative energy. So for all sane purposes, negative energy density of the warp bubble translates to negative total energy, as well as negative total mass. Ergo, requirement for exotic matter stands.

It's the energy density that needs to be negative for the alcubierre drive, actually.

I am. Most of the mass is dynamically generated. We'd still have massive matter even without the Higgs Field. Now, weak interactions would be quite different, because with massless W and Z bosons they'd have infinite range.

Higgs field has almost nothing to do with mass, same as any other field. The only particles that gain mass due to Higgs are electroweak gauge bosons. For everything else, Higgs interraction is a tiny contribution to total mass, and it's always positive.

Yes, Forward Euler is notorious for causing energy buildup in harmonic problems. For small deflections, Implicit Euler or pretty much any 2nd order or higher RK method will conserve energy. For large deflections, where dynamics becomes non-linear, you will have energy fluctuations, but they should be random, so simulation should remain stable.

Orbits stop being elliptical as you get close to the event horizon. Once you cross event horizon, all trajectories lead inward. Newtonian physics simply does not apply here.

It's because US Engineers used pounds as a unit of both force and mass. When you do that, you end up with a factor of g in your equations. In USSR, they used metric system to define ISP from the start, so they had it in units of m/s and ISP is simply equal to effective exhaust velocity. No arbitrary factors!

Why? They don't use up any energy. Having something that exists forever is not the same thing as having a perpetual motion machine. The term's misleading, but if there is no net energy output and no net entropy decrease, then it is not a violation.

I wouldn't say it was proven. Hawking has shown that field theory suggests that black hole can emit a particle by interacting with particle-antiparticle pairs near the event horizon. That brought up a whole lot of questions, but it's a fairly accepted position now. Naturally, if black holes don't actually evaporate, then the question has no meaning, so we are having this discussion under assumption that Hawking is, indeed, right about this radiation. As far as experimental verification, Hawking Radiation remains unconfirmed. Though, there have been some experiments that have detected radiation which fits the pattern. So we might have confirmation of it soon as more data is gathered.

Hm. This is a very interesting question. Large black holes evaporate very slowly. In fact, they are growing faster than they can evaporate from background radiation alone. It takes a very tiny black hole to evaporate. But that's now. As the universe expands, both matter and radiation are going to become less available for black hole to grow, while evaporation rate remains constant. Eventually, there should be a point that even large black holes start to evaporate. The other point is that it takes a supermassive black hole for you to fall into intact. That spagettification starts to happen pretty early for smaller black holes. You'll be ripped into atoms before you cross the horizon. But supermassive black holes have quite reasonable tidal forces near the horizon. A ship can fall in intact. If universe expands without limits, any black hole will evaporate in finite time. Mind, it's going to be absurdly long time, one which makes current age of the universe seem like a fraction of an instance. But finite nonetheless, and there is no indication that there is some sort of a grand time limit on the universe. Even if protons start decaying, black holes should be fine. So as you fall into the black hole, you must pass that threshold. You must reach time of that black hole's eventual end. But therein lies the paradox. If this is supposed to happen in finite time, it will happen before you reach event horizon. That brings up a whole lot of problems. a) You can't actually fall into black hole, because it will evaporate before you get there, but you won't get out alive either because Tidal forces will increase as it shrinks, and if that's not enough c) It will blast you with an ever increasing dose of Hawking Radiation. None of this sounds right, so there must be a flaw in one of the assumptions. I will take another look at statement of Hawking Radiation, and see which coordinate system it's actually always finite in.