Halc

The velocity (both speed and direction) of an object like an asteroid is very frame dependent. So you seem to envision the frame of the sun say, where the asteroid is on a path not towards Earth, but one that will cross Earth's orbit exactly when Earth gets there. You hit it straight on in a direction opposite its motion in that frame. It slows, and the mostly unaltered paths still cross, but the objects are at that point in a different times. Now do that from Earth frame (the frame from which the rocket was launched). In that frame, Earth is stationary and the the asteroid is heading straight for us. If we hit it straight on in that frame, it will slow, but still get here a little bit later. Point is, a straight bullet shot fired from Earth isn't going to divert it. In that frame, to get it to miss, you need to apply lateral momentum to it. This involves sending the rocket on a curved path, wasting fuel compared to the straight path. And only the fuel expended for the lateral motion will effect the asteroid in a deflecting way. So you need a lot of fuel. One batch to get there (all unusable for deflection) and a whole separate batch to apply laterally to the thing. All this kind of presumes flat inertial motion of both Earth and asteroid, which is accurate if the thing is pretty close, but the idea is to get it when it isn't so close since it takes less effort to divert something far away. A smaller deflection is needed to effect a miss. So maybe our computers predict this collision on some prior orbit and we can manage to hit the thing slightly on some prior pass. Idea is to not hit it when it crosses Earth orbit since no defection there will prevent it from returning to Earth orbit repeatedly.

Speed is not a vector. You seem to be saying that moving 2.8 meters in 2 seconds is 1 m/sec. That's simple arithmetic that you're getting wrong. You also are still labeling a speed in meters where you say v3h=0.7m Do I have to point this out every post? The vertical component of each velocity vector is 1 and -1 per second respectively, not 0.7. Ditto with the h component, which is the speed of the road which you said is 1, regardless of whether ped2 knows about it or not. I tire or repeating this. You seem bent on just asserting the same wrong numbers over and over no matter how many times the mistakes are identified. You seem to be doing this deliberately, perhaps in order to argue some mathematically inconsistent point that you think has some unfathomable bearing on relativity. It doesn't. Relativity doesn't support your numbers at all, nor does it give different numbers for this scenario than does Newtonian physics.

I understand what you seem to be trying to say. I am trying to help you use consistent terminology to do so. You've been saying ped1 is stationary and immediately saying he has a nonzero speed. That's a contradiction, and it has to be fixed. Motion is relative, and ped2 is your designated reference. If motion is specified relative to anything other than ped2, you need to explicitly say this, else the statement is not even wrong. It renders further discussion impossible. Your scenario seems to take place at an airport with moving sidewalks. Ped2 is standing outside of one of them, and ped1 is standing on the sidewalk. Sometimes the sidewalk is on (cases 2 & 3) and sometimes off (case 1). The points in space (A-F) are all locations relative to Ped2, not relative to anything else. Oh, that's much better. You added 'relative to the road', which is the alternate frame reference that was missing before. Now it's not a contradiction. This part is still contradictory. You said it takes 2 seconds to do the two diagonals, one second apiece. Now you assert a different time of 2.8 s. That's a contradiction. It takes 2 seconds to go 2.8 meters at 1.4 m/sec which is the speed ped1 is moving relative to ped2. You also get units wrong. You multiply 1.4 seconds by two, which should get 2.8 seconds, but you call it 2.8 meters. I suspect the unit thing is just a typo. This is still wrong. Ped1 moved from A to D in one second, and that distance is 1.4 m, so his speed relative to Ped2 is 1.4 m./sec, not 1. You did not show your calculation of from where this 1 m/s comes. It really doesn't matter if he's aware of how ped1 is accomplishing his motion. Point is, he sees him 1.4m away after 1 second, and his speed is thus 1.4 m/sec, regardless of what the road is doing. You seem to think it does matter, and that a ped3 doing the same thing with a jetpack, but always being at the same location as ped1, is somehow moving at a different speed relative to Ped2. This is wrong. If you insist otherwise, well then perhaps this topic needs to be moved to strange claims section. This is really hard to parse, but apparently you've imposed a speed limit of 1, and the road moves incredibly slow. I think this is an attempt to an analogy of using natural units for speed rather than m/sec, but given your inability to describe simple Newtonian motion above, I think I won't go there and take your description as worded. "The pedestrian one is 0.7m wide" : What does that mean? He is now on a somewhat narrower super-slow moving sidewalk? Legal, but why? The road speed cannot be 0.1m since meters is not a unit of speed. You perhaps mean 0.1m/sec, so it moves perceptibly fast now. I think a typical airport sidewalk goes about 2 m/sec, but we're in the slow and safe airport today. Why the speed limit? Relative to what? Ped2? What if ped2 decides to walk left? Can nobody put a box on the sidewalk now? What prevents it? How very non Newtonian. The abrupt change of velocity at point D is straight downward (south), so the jet pack should be pointing south, not to the southeast as drawn. Better to just put a wall at D and have him bounce off it. No jet pack needed at all then. He can just be an inertial object in freefall the whole time.

OK. And all this is 'as observed by' ped2, which means ped2 is by definition stationary, and is with ped1 the whole time. The road moves to the right (as the arrow in the picture shows). Both pedestrians are at location A the whole time, since you said they were stationary. Locations A-F are coordinate locations relative to ped2, our definer of the frame. The locations are not painted on the road as I had earlier guessed. No it isn't. You said he was stationary. If the road is transporting him anywhere, he isn't stationary. You're being inconsistent. If you want him to ride with the moving road, then don't say he's stationary. As I said, ped2 seems to be there to define the frame. He cannot remain at (0,0,0,0) since after 1 second he will be at (1,0,0,0). But he starts at that event in all systems. I am presuming location A is at x=y=0. Apparently you've assigned the x to vertical (across the road, to B) and y horizontal (along with it), towards C,E and such. This is a contradiction. Ped1 is stationary relative to ped2 or he's moving. You have to pick one or the other. Scenario 3: OK. He's moving faster than 1 m/sec then because he's getting to D (1.41 away) in only one second. Your OP description said he was moving at 1/sec, not 1.4 That last line is wrong. You show time 2 above in the E coordinate, not 2.84. 2.84 is not computed from the above values as you assert. Maybe you should fix all these problems before moving on to the silly parts. You can't. You said the max speed was 1 (harsh cops, I tell ya), and then immediately suggest going faster than that.

This whole topic seems to be related to Galilean relativity and Newtonian physics. There is no relativity theory going on in any of this. I am assuming that points A through F are painted points marked on the sides of the path/road. If you mean otherwise, it isn't indicated. How does ped1 move from A to E if he's not moving? This is a contradiction. Given the pic, the road is moving to the left, and spot E painted on the road moves left to the pedestrian over 2 seconds. Anything other than that and the ped is not stationary. I take it ped2 is always with ped1 in this scenario? If not, then ped1 would be moving, and you say he isn't, as observed by ped2. So ped2 stays at point A, the road is stationary, and ped1 takes this diagonal path to D, then E. The distance is under 2.83, but close enough. Speed is as you say. If ped2 is not staying at A, then none of the above works. Ped2 is your reference frame. You have an orange line going from A to E but that's apparently not ped2, it's just a line. No it isn't. You have ped1 stationary in system 2, but you also contradict that. I am presuming that we're using the frame of the road for system 3. The speed values make no sense otherwise. Thus, since ped2 defines the frame, ped2 stays at location A of the road. Yes, it takes 2.83 seconds to go 2.83 meters at 1 m/sec. This is news? It isn't a difference in observed time, it is a difference in path taken. ped1 goes straight across the meter path, or he takes a diagonal, a longer path. All observers will measure 2.83 seconds in this 3rd scenario. Changing observers (frames) will change the measured speed and distance, but it won't change the 2.83 seconds. Not at these speeds anyway. A car might go by and measure ped1 moving at 100 m/sec relative to the car. That guy will still see it take him 2.83 seconds to go from A to E.

The paper calls them 'black stars' and seems to posit a complete lack of event horizon, and that the matter all piles up just outside where the EH would be. This is very similar to Schmelzer's alternative generalization of LET (relativity, but denying both premises of SR). https://arxiv.org/pdf/gr-qc/0001095 He calls the 'frozen stars' since all the matter piles up outside the nonexistent EH. It's a presentist interpretation of relativity, and it equate absolute time with coordinate time. Just my thoughts. The alternative is that the matter gets 'inside' and is somehow prevented from going 'forward' which is the same as positing that if you put enough force on matter, you can push it into the past where there's more room.

For instance, yellow-orange laser light has a frequency of about 5x10**14 Hz. The binary light frequency is equivalently 1x10**110001 Hz. How this makes a difference is of course a mystery, but adding mirrors to the situation is just going to add to the delays since the path is less straight. Yea, about 6 to 45 minutes depending on how far away Mars is at the time. 6-45 minutes later is not 'before it was sent'. Zero justification for the suggestion otherwise has been given.

The collapsing star is a dynamic state, and quite different from a free-falling object. There's dense stuff, and then not even Pauli-exclusion can support it. OK, so there's even more squish, at least at first. A spatial dimension rotates and is replaced by time, and that time dimension is bounded. What was the time dimension rotates out to a spatial one, one with nearly infinite extension at that. Lots of new room to spread out, but the causal light cones don't allow arbitrary travel down this big space, so I cannot say the compression ends. An example of the space available inside a black hole, Sgr-A and the black hole at the center of Andromeda share a common singularity. They're the same black hole, a region of 4D spacetime bounded by a 3D event horizon hypersurface, and in that case, the same (connected) hypersurface. It's only in a slice of coordinate space at a given time that the one object has multiple cross sections, manifesting as a pair of black holes to us, for now. None of the above is particularly an answer to the question of if there is compressed matter in an established black hole. In coordinate time, yes, it's very dense, but that's more like length contraction than pressure. None of the matter actually reaches the event horizon in coordinate time, and yes, in that state, it (the original collapsing star matter) is very much under compression. The singularity condition exists, but isn't described, precisely because the physics there is singular, sort of like asking what the perspective of a photon is like. Got a link? That sounds like pop nonsense. Is it peer reviewed? That makes more sense. Still, to be matter, it has to persist, no? Agree to all. There's also a naked singularity. You can for instance just keep dropping electrons into a black hole until the charge is more than the gravity and no more (isolated) electrons can be added by any means. Similar issues if the infalling matter adds too much angular momentum. A given mass can only have so much of that. These are examples of frame independent singularities not obscured by the coordinate singularity of the EH. Ditto with the LLM answers, which is just massaged google results. Anyway, thanks for the post. Good informative stuff in there.

The equations are how everybody knows. No links were provided, so I googled the question and the first 8 hits (NASA, Smithsonian, various you-tubes, reddit) all suggest matter is compressed without bound. Much of this list of bad hits is due to my search terms of "black hole infinite density". First correct answer came from of all places Quora, a site known for severe wrongness of replies. Question was: Do black holes have infinite density? Answer by Toth: "The equations that describe some of the simplest black hole solutions, including the Schwarzschild black hole are (drum roll, please)… equations of general relativity in the vacuum. Yes, that’s right. The vacuum. There is no matter. The density is zero everywhere. The Schwarzschild solution is the simplest, spherically symmetric, static vacuum solution of Einstein’s field equations." Next hit was probably the most respectable forum I can name. https://physics.stackexchange.com/questions/246061/are-black-holes-very-dense-matter-or-empty Rennie (I think) replies specifically about the Schwarzschild metric, which wasn't technically the question: "The archetypal black hole is a mathematical object discovered by Karl Schwarzschild in 1915 - the Schwarzschild metric. The curious thing about this object is that it contains no matter. Technically it is a vacuum solution to Einstein's equations. There is a parameter in the Schwarzschild metric that looks like a mass, but this is actually the ADM mass i.e. it is a mass associated with the overall geometry." The Kerr metric is also a vacuum solution, which differs only by a nonzero angular momentum. There is an Oppenheimer Snyder metric that is an 'unrealistically simplified' solution to the formation of a black hole, but it fails to describe conditions at the singularity. I was hoping at least for some indication of the whole compression vs. tension distinction. None of these metrics seem to include Hawking radiation, so they describe black holes that exist for infinite coordinate time. Rennie continues: "[Observers falling with the star collapse] see the singularity form in a finite (short!) time, but ... the Oppenheimer-Snyder metric becomes singular at the singularity, and that means it cannot describe what happens there. So we cannot tell what happens to the matter at the centre of the black hole. This isn't just because the OS metric is a simplified model, we expect that even the most sophisticated description of a collapse will have the same problem. The whole point of a singularity is that our equations become singular there and cannot describe what happens. All this means that there is no answer to your question, but hopefully I've given you a better idea of the physics involved. In particular matter doesn't mysteriously cease to exist in some magical way as a black hole forms." So my post seems to be based on information about static metrics (Schwarzschild, Kerr, others), the geometry of which shows an end to time and no matter at all, but neither do those metrics show the end to the matter that made them since these kinds of black holes are not 'made'. They exist for eternity. So Op-Sny is probably a better metric despite being 'unrealistically simplified'. A coordinate system that isn't singular at the event horizon (like Kruskal–Szekeres coordinates) shows worldlines of infalling particles just ending in time at the singularity, not persisting with the other matter persisting there. The worldline of compressed matter would not end, but only join all together with the worldlines of other particles. As you (as an observer) fall into one, tidal forces pull you apart, not compress you. This doesn't stop at the EH. So compression ever happens, then the naive description would be when you smack into that physical singularity there where everything else has gathered. None of the metrics describe that. At best they just don't answer the question at all, and on those grounds, I am reneging on the authoritarian tone of my prior replies without suggesting that the 'high density matter' description is a better description. Learned stuff today, which makes this a win topic. I hope we all have. Yes, I've seen places that compute that mass. A moon mass is still going to take an awfully long time to radiate away at CMB temperatures. Infinite time actually, at least until the CMB radiation stops adding mass as fast as HR bleeds it off.

Yes. A black hole near end of life has almost no mass remaining. Gravitons as in gravitational waves, not any sort of force carrier. Gravitational waves carry information about changes to spacetime geometry, and an evaporating black hole is such a thing, so it has to generate such waves, whether or not those waves can be broken down into quanta. Light is energy. Any radiation reduces the mass of the thing radiating it. Light also has momentum. For small black holes, sure, but for larger ones, the odds of something like a positron escaping is incredibly low. Most would fall right back in due to gravity. Gravity can't pull back light if it's going in the correct direction (straight up). There's no matter in a black hole. A Schwarzchild black hole is a vacuum solution. Nothing gets squished in there. Things falling in actually get pulled apart. The singularity is not a location in space where there is matter squished together unreasonably. It's a line/plane/fuzzy region where time just ends. This comment suggests dense material in there somewhere. This is a misconception. Yes, better. Energy & mass are equivalent. The mass doesn't exit the black hole, but is created outside by separation of virtual particles, with the one with negative energy falling in and adding that much negative energy to the BH. The vast majority of the time, both virtual particles are thus pulled in, netting zero energy to the BH. The odds of one escaping becomes larger with the small holes. I don't know where the limit is, and what it means for mass to not be able to support an EH. I think a unified theory would really help give real answers to this. My statement of 'a few grams' might be way off, but classically there is no minimum mass, and at sufficiently low mass, the radiation becomes significant enough to qualify as an explosion. There is still nothing actually from inside the black hole escaping. There is no matter in there.

The end might produce a few grams of actual matter in a brief burst of radiation. "Explode" makes it sound bigger than a wink. The vast majority of the original mass radiates away as massless particles, mostly photons and gravitons, neither of which is a building material for stars.

for the differential aging in the twins' paradox, not the acceleration, nor the frame change. Yes, when the orbiting twin turns his gaze around, he will appear to be approaching instead of receding. So the redshifted view of the tower clock will change to blue shift, the difference being purely Doppler effect in both directions. The dilation is due to speed, and speed isn't affected by where anybody is looking, so the dilation is unchanged at the far side of the planet. Yes, there is acceleration, but all of it orthogonal to motion, so since the 'twins' are at the same potential, the dilation is constant for the entire orbit. It is objective. The orbiting twin will be younger when the meet again, just like the one that goes out and back to the distant star. To do this in special relativity, the planet can have no mass, and the 'orbiting' twin would need to curve his path via say a string tied to the center of Earth to get him to curve is path like that. Rockets also works, but the engineer in me hates to waste fuel when there's a better way.

The station isn't necessary. All that is needed is a frame reference. One can say that each rocket is moving at .9999c relative to frame arbitrary abstract inertial frame S which happens to have nothing stationary in it. The lack of a stationary object in S would make it difficult for any rocket guy to directly measure that speed, but it can still be computed. A frame is, after all, an abstraction, not a physical thing. So if X is moving west relative to S at 0.9999c and Y is moving east relative to S at 0.9999c, then X is moving at about 0.999999995c relative to Y and V-V. (.9999 + .9999)/(1 + 0.9999*0.9999) = 0.999999995 In that case the velocity of the two respective departure points relative to each other needs to be specified. Without that, there is no way to compare the rocket speeds relative to each other since there can be no common frame. Usually, in the absence of an explicitly specified frame, an observer on any object (a rocket say judges his own speed to be zero.

So it seems. Imagine how many posts I've made that are then wrong. My definition of proper velocity seems right, but that's different than rapidity. I must admit I need to digest that page better to figure out what the rapidity actually is, how it is useful.

The way I've heard 'rapidity' used is as a proper velocity. It is proper acceleration integrated over proper time, and thus it adds the normal way. Two obvious examples: I accelerate my ship away at 1g in the same direction for 3 years ship time. That gets me up to a rapidity of a bit over 3c, meaning I travel 3 light years (Earth inertial frame) for every year I age. At that rapidity, the ship has a velocity relative to Earth's inertial frame of about 0.995c Second example is recession rates of galaxies. Gnz-11 (no longer the most distant object seen) is increasing its distance from us at a rate of about 2.3c. That's a rapidity. Rapidities add the normal way. If galaxy X recedes from us at 1.2c and galaxy Y (further away from us on the same line) recedes from X at 1.3c, then Y recedes from us at 2.5c. If an inertial frame was valid over such distances (such as it would be in say a zero energy Milne metric), then that rapidity of GNz-11 would correspond to a velocity of 0.98c. This underscores the fact that recession rates are not expressed relative to any inertial frame. Anyway, since the usage comes from Mordred, he may be using the term in a different way than I tend to see it.

Non-sequitur. All numbers have equal chance of winning, so no choice, popular or not, alters your odds, but as swansont points out, unpopular numbers yield better average payouts. You seem to presume N is less than 64, which is not always the case. Even if it is the case, your conclusion is again a non-sequitur. Picking five random numbers under 32 gives the exact same odds of winning as picking 5 random numbers in the range of say 33 to 40. This is the simplest of mathematics: Every possible choice has the same odds as any other if the draw is random.

Decelerating the rocket just wastes all the momentum it already had. If you have delta-V left over, accelerate more, not decelerate. This gives maximum momentum transfer to the thing, which is what is needed to deflect it. Of course, it's best to hit it more or less from the side, which is inefficent for something coming more or less straight at you. Indeed, but also the harder it is to tell if the effort is needed at all, or if the effort will actually make the trajectory worse, due to miscalculation. Look at all the news about some asteroid that's going to hit Earth, and then it misses it by a mere million km. You can't send a defection mission out to every big rock that might get that close, but by the time we know it will hit, it's too close that a small defection is enough. It's also harder to get something out to an incoming object quickly if its further away. Takes more delta-V to get out there, leaving less to actually impart momentum to the thing. I don't think nukes are very effective in a vacuum. It will leave a nasty stain and small crater and will defect almost not at all, unless you can get the thing to embed itself a ways in without destroying the mechanism in the process. There is armor-piercing technology that helps with that sort of thing. Look at the bunker-buster bombs they have, designed to penetrate a long way and still explode, sometimes even hours later. But those bombs are heavy and not too fast, hitting at far slower speeds than what would likely occur in a rocket/asteroid interception.

Yes, they'd be comparable. You're asking if a straight line (or a geodesic one) is a similar length to one with some minor curvature to it, and both are about 2000 years long, differing only in perhaps the 6th significant digit (a guess). The length of the actual worldline would be a tiny bit shorter than the interval between the two events. This is essentially the gist of the twins scenario where one twin has a shorter worldline than the one with the straighter worldline.

PBS space time is a pop science source. They can say what they want. Speed for one is relative, so for instance I am currently going very near light speed relative to a muon waiting for me to go by it. The universe is unaffected by my moving at this speed. The speed at which something goes relative to something else has zero effect on the geometry of the universe. It is possible, but a 3D torrid surface, sort of like the video game asteroids takes place in a 2D torrid surface. It means that there are preferred axis orientations, and that if you travel along any of these axes, you get back where you started. Any other direction and you don't. Your post seems to describe something like that. Such a universe would be unbouned, but finite in volume, very much like the surface of Earth, except retaining the flat curvature. The universe is much larger than the distance to the event horizon, and since one cannot reach the event horizon, one cannot traverse even one loop no matter the speed. Presuming that expansion wasn't accelerating (or happening at all, for simplicity), there would be no event horizon, and thus one could go all the way around. So in asteroids, the universe isn't expanding at all, and if one goes vertical, you get back to the staring point after 1 lap, in say 5 seconds. If you go diagonal at the same speed, it might take a minute to get back approximately to the center of the screen, the starting point. You seem to presume that one direction is longer than another, sort of like the asteroids screen not being square. That's fine, but it isn't necessarily the case. If two depart from the center at the same speed and time, the vertical guy gets back before the one going the long way. If the aspect ratio is 3:4, then it will take 3 or 4 laps for them to meet each other again. Length contraction plays no role in this. For one, it is a coordinate effect, not strictly a physical one. A length contracted ship going at 0.9c still gets 3.6 light years from here in 4 years (as measured by 'here', the frame in which he is contracted). It being a coordinate effect, one does not experience length contraction. One is by definition always stationary relative to ones self, and thus there is no contraction to experience. And a paradox isn't something experienced either. You've not identified any paradox. You just say that there is one. What each person would see, if moving fast relative to stars, is a bunch of stars moving fast relative to him. They both see that. There is no paradox identified.

I can think of at least three kinds of time, and the distinctions are important. What clocks measure is proper time, which, in GR terminology, is the spacetime interval along the time-like worldline followed by the clock, completely analogous to proper distance being measured by tape measure following a space-like worldline. Proper time is an invariant (it is not frame dependent). It is physical and very real. Getting beyond strictly physical, there is also coordinate time. Coordinate time is the time coordinate assigned to events by a particular abstract coordinate system. It is purely abstract (a mental thing, not a physical one). It is very frame dependent, but still physics, not metaphysics. If we get truly down to metaphysics/philosophy, we get to the third kind of time, which is the progression of the present moment. It is 'that which flows'. When somebody asks if time is real, they're probably talking about this kind. It probably has an official name but I don't know it. I call it teapot time since like the teapot purported to be in orbit around the sun somewhere beyond Jupiter, there is zero empirical evidence for it, but its existence also cannot be disproven. So to give an example of the three kinds, we decide to throw Bob (and his watch) into a black hole. In some coordinate systems (say that of a distant observer), Bob takes forever to reach the event horizon. In a different coordinate system such as Kruskal–Szekeres coordinates, he falls right in in some finite time X. That's all coordinate time. Proper time is easy: Bob's watch reads time P upon reaching the event horizon. That value is the same regardless of choice of frame. As for proponents of teapot time, the only theory I know that includes it is forced to deny the existence of black holes at all since events within cannot exist, so any talk about falling into one is unsupported. This of course provides a semi-empirical test for presentism. If you jump into a large (at least say 15000 solar masses) black hole and find yourself inside, you've disproved presentism, but of course you cannot publish your findings in a science journal any more than you can publish your discovery of an afterlife. One has to die to test it, and the guy inside the black hole is, while not yet dead, certainly doomed.

You're asking two different questions: What is the most likely value of s when all of the list has been selected at least once? At which value of s will it be more likely than not that all of the list has been selected? The answer to the first question is seems necessarily lower than the answer to the second. Do you see why? I don't see an obvious way to derive either value right away.

It's a top, simple as that. Not clockwork at all. We had these as kids. They went under the brand name 'whizzers' or some such. This one is decorated as a bug. The wheels underneath drive an internal vertical-axis flywheel. You'd get quicker responses to your questions if the title actually gave any clue as to what the topic is about. You seem to give the same title to all your topics, however unrelated.

You got it all wrong Bufofrog. What kind of fool is going to believe that? Everybody knows it was last Tuesday, not Wednesday.

People can produce enough thrust to fly here on Earth, but it takes an athlete. Don't know the current record, but somebody flew over the English channel with human-powered flight. So sure, if you can do it at 1g, you can do it at lower g with less effort. I think your mistake is presuming it is just a matter of strapping on winds and arm-flapping. But our arms are situated nowhere near our center of gravity, so instead of flying horizontally like the pictures depict, you'd hang vertically if all your weight was born by your arms. I suppose it could work, but the aerodynamics would be horrible. All gliding done by humans is a different setup designed to support you everywhere, hence the funny flying-squirrel setup that the base jumpers use. I suppose that one could generate thrust by a more 'flappable' version of one of those. That (gasoline, electric, etc) would be powered flight. We also have that here on Earth, so again, doing it at low g would be easier. You can also strap on a rocket, something else a human can do here on Earth. What if the pressure was much higher? How much easier/harder would it be to fly at say 1g or less with thicker air. It has more mass to support you, but more drag as well. If it's dense enough, the buoyancy alone would be enough.

The velocity of the universe would be the change in its position relative to something that isn't the universe. It really makes no sense for a universe to have a velocity.= Velocity per second per megaparsec would be something like v/t/d (velocity, time, distance) which is d/t / t / d which is 1/t², something with different units than velocity at least. As for special relativity, that only applies to flat Minkowskian spacetime with zero energy and mass anywhere. For this reason, it is simply inapplicable to our universe except locally. Your equation adds values of different units, which makes it meaningless. You can't add meters to Pascals, and (in your case), you can't add H (units 1/t) to V/γ (units d/t) Another criticism: γ is the Lorentz factor of what exactly? It should be the factor for some speed, but "the Lorentz factor which is used in incorporating special relativity into the equation and using the laws of special relativity" is just a word salad.