It's not so much that the universe isn't logical, it's that people are often too ignorant and/or unintelligent to understand it.
Nothing against anyone in particular, of course. None of us know everything.
Question for the science guys.
- Thread starter Threeofseven
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It's not so much that the universe isn't logical, it's that people are often too ignorant and/or unintelligent to understand it.
Nothing against anyone in particular, of course. None of us know everything.
That's sort of the point with the quote, at least how I read it.
We strive to make sense of it all, but for every answer and every piece of logic we uncover it always leads into yet another mystery... the mystery being the universe being one step beyond.
Can we go faster than light?
What would doing so look like?
What happens inside a black hole?
We want to make sense of it all and understand it so we can explain it, logically, but there is always another question that requires answers.
Really, so why have you not just floated off the earth into space.
Because the centrifugal forces of the Earths rotation on my very sexy body does not overcome the centripetal forces of gravity.
I'm simply more attracted to Earth than to space, even though my dreams are all out there in the cosmos
Many people don't understand the significance of what the speed of light means and represents. They just think that the speed of light (in a vacuum) is the fastest something can travel. While that's correct in certain terms, it's missing the larger picture regarding space-time dilation and why that actually is. And there are others who are just too stubborn to seem to even want to know those sort of things.
We call it speed of light, but it's really speed of causality. It's a small distinction, but an important one. Many things travel at speed of "light", e.g. radiation and radio waves.
That is an interesting way of conceptualizing it. Another thing people could ask is how old a photon is when it reaches the detectors inside a telescope from a galaxy halfway across the observable universe. Of course there's no direct way to measure a photon's age, at least none that I know of, but I think it's still an interesting and relevant question to ask.
And yes, gravitational waves also travel at the speed of light. I occasionally have my computer crunching through data to help detect them using Einstein@Home and BOINC.
And yes, gravitational waves also travel at the speed of light. I occasionally have my computer crunching through data to help detect them using Einstein@Home and BOINC.
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Well... Apparently photons experience neither time nor distance (proof: I read it on the Internet) so the question about age becomes philosophical.
The closer to speed of light you get the slower time moves. My understanding is that since the photon travels at the "speed of light" then time effectively stands still. For us it takes 8 minutes to observe the photon getting from the sun to earth, but for the photon this is instant. As soon as it leaves the sun it is already on earth.
How old something is, especially when taking time dilation into account, becomes a matter of who is observing.
If you travel at near speed of light then time moves slower for you. An hour for you might be a year for me.
So from my perspective you are +1 year old. From your perspective you are +1 hour old.
Who is right?
“Nothing travels faster than the speed of light with the possible exception of bad news, which obeys its own special laws. The Hingefreel people of Arkintoofle Minor did try to build spaceships that were powered by bad news but they didn't work particularly well and were so extremely unwelcome whenever they arrived anywhere that there wasn't really any point in being there.”
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This is a great question, and one that trips up a lot of physics students studying galilean relativity.
In your example, you compare the KE in 4 different reference frames, but since you studied physics, you know you cannot do this because ΔKE ~ Work = F*d which depends on Δd, which is defined by the reference frame. In your post, you effectively split the work done over 3 reference frames, but in doing so you recenter the starting point with each new frame, hence artificially reducing d each time. When you add up the distances required over each 1m/s acceleration then the final distance travelled, and hence the Work done is the same in the initial reference frame.
It's true as you say that the KE of a 1kg mass at 3m/s is KE=0.5mv^2 = 4.5J. But let's look at this in terms of Work done relative to the initial reference frame. Say you apply 1N force to a 1kg mass until it reaches 3 m/s, how far will travel before it reaches 1 m/s, 2 m/s, and finally 3 m/s?? Whip out your calc, and you'll see d traveled is 0.5m, 2m, and 4.5 m respectively. Hence the total Work to reach new frame from the initial reference is 1N x 0.5m = 0.5j, 1N x 2m = 2J, and finally 1N x 4.5m = 4.5J. Notice that final is the same as the KE of a 3 m/s object in the initial reference frame! You can increase the force applied and hence reduce the distance required to go from zero to 3 m/s, but the product of the W*d will always be the same: 4.5J.
Now to answer your somewhat philosophical question, "why does it take more energy to accelerate something going faster?":
Remember that the distance that you apply a force is a function of the velocity of the object in that reference frame hence,
Work = Force * d(v)
Something traveling faster will cover a lot more distance in the same amount of time, so as v gets bigger, d(v) also gets bigger for any given force applied in that frame. Or another way to put it:
ΔKE = 1/2mv[SUB]i[/SUB]2 -1/2mv[SUB]f[/SUB]2
Hence the work required to change the velocity grows with (v[SUB]i[/SUB]^2 - v[SUB]f[/SUB]^2) = (v[SUB]i[/SUB] - v[SUB]f[/SUB])(v[SUB]i[/SUB] + v[SUB]f[/SUB]), so if you plug larger values of v[SUB]i [/SUB]then the ΔKE required to change the speed by 1m/s increases as well.
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It's like having a motorway speed limit set so high only a Bugatti Veyron reaches it, and so all Veyrons travel at that speed. The speed limit might earn the nickname "The Speed of Bugatti", but the limit itself doesn't have anything to do with those cars. Other extremely fast cars can also end up limited at that speed.
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Great question, and again, it all comes down to the practicalities of Newton's 3rd law, and the fact that the mass in question has to be accelerated against some other mass.
Rather then even thinking in terms of 'mass' here, it helps to just think of it as inertia, as that's the key field. An inertia can only be accelerated or decelerated against some other inertia. The only exception would be relativistic mass / radiation pressure - that is, the 'effective mass' of a photon's energy per e=mC^2.
Ziljan's answer above is excellent (addressing the same basic question), my own perspective, as mentioned, is simply that if we're accelerating a mass against some other 'fixed mass', then its increase in speed relative to that fixed mass subtracts from the 'acceleration' component of the applied force (F=mA), and so to add a second, equal serving of velocity to an already-moving mass, we either have to apply the same force over a greater time & distance, or else raise the magnitude of the force over the same period / displacement.
Again tho, an external observer in another frame would see your 'other mass' counter-accelerating the other way; even if it's the Earth itself, and that counter acceleration, infinitesimal. The change in momentum of each mass in each direction will be equal, and so the system's net momentum unchanged.
Being able to maintain 500 mJ / kg-m/s up to any speed would be the creme-de-la creme of effective N3 violations. But N3 seems inviolable, because mass constancy. And thus N1 also. However i think it might be possible with my current theory, notwithstanding that i also think it's complete nuts and can't believe a word of it. However regardless of what i think it'll either work or not, and initially i'll be aiming for a fixed-rate deal of 96.23 Joules per 9.81 kg-m/s.
Obviously, paying basically 100 J for 10 kg-m/s divided between two 1 kg inertias seems like a rubbish deal at first. You could raise many times that much momentum for the same energy, via any conventional form of acceleration. But whatever the rate, provided it's constant with respect to rising velocity, then beyond some threshold velocity the system is over-unity. Below that it is under-unity, and in-betwixt those extremes, there'll be some speed at which the system is at perfect unity. So even if we're paying megajoule's per kg-m/s, provided that rate is truly speed-invariant, then an input of 5 MJ could yield a gain of 250,000 kJ.
Basically, when you have an effective N3 violation, you're creating a 'divergent' inertial frame, and equivalence with all other reference frames is broken. The runaway momentum's only objective reference frame is that of the fixed stars (a bit like a gyroscope, except angular velocity is absolute (as demonstrated by centrifugal / centripetal force), whereas linear velocity is (usually) exclusively relative).
This issue is one Roger Shawyer always seems to have dodged (i raised the same point on the NASA forums re. their EM drive tests); in response to questions about the apparent implications for CoE, Shawyer always seems to disingenuously offer assurances that the system isn't over-unity in terms of the work done within its own rest-frame (well duh!) - he never addresses the implications of measuring the system KE from any outside reference frame... and which imply that his system would by definition be OU if it actually worked, and moreso the faster it went..
All i know is this:
- i can buy 9.81 kg-m/s for 96.23 J, by applying a mutual acceleration between two equal inertias, one of which is prevented from accelerating by gravity. So all 9.81 kg-m/s is applied to one mass only.
- i can then mutually-decelerate them against one another - each mass thus ends up with 4.905 kg-m/s.
- per .5mV^2, the system now has 2 * 12.0295125 = 24.059025 J
- i spent 96.23 J, got 24.05 J back out, so 3/4 of my input energy has disappeared.
- all of my momentum has been conserved - none's been lost, as by dissipation - hence this KE loss is non-disspiative.
- a non-dissipative loss mechanism is precisely the same animal as a gain mechanism, only backwards. The tail-end of the same beast. It is 'under-unity' in precisely the same manner as its opposite twin would be over-unity.
So we now have two '1 kg' inertias (specifically we'd be using 0.125 kg-m^2 angular inertias, but same deal; i'm trying to keep things elementary-simple here). Basically, we have two equal inertias, both moving at a uniform 4.905 meters / sec.
- i can now invest in a second purchase of 9.81 kg-m/s, again only accelerating one mass, despite doing so by pushing / pulling against another one, yet whilst gravity prevents its acceleration..
- again, i can equalise the resulting speed difference with a mutual inelastic collision...
- each mass is now moving at 9.81 m/s, with 48.11805 J on each, so 96.2361 J total.
- i've now input 2 * 96.23 J, and i have 96.23 J, so now only 50% of my net input energy has been destroyed without trace! Get in!
- so again, i buy a third 9.81 kg-m/s reactionless angular acceleration, adding another 4.905 m/s to each mass. Net system speed is now 9.81 + 4.905 = 14.715 m/s.
- at 14.715 m/s, each 1 kg inertia now has 108.2656125 J of KE, so 216.531225 total.
- we've input 3 * 96.23 J = 288.69 J, we have 216.53 J, and dividing the latter by the former, 75% of our total input energy is present as system KE.
- finally, we'll add one more such acceleration, so paying another 96.23 J for another 9.81 kg-m/s, accelerating the net system by another 4.905 m/s:
- we were at 14.715 m/s, we've accelerated by another 4.905 m/s, so net system speed is now 19.62 m/s
- at 19.62 m/s, each 1 kg inertia has 192.4722 J, so total system KE is twice that; 384.9444 J
- we've spent 96.23 J, four times, so 4 * 96.23 = 384.92 J
Thus after precisely four cycles, we've hit unity.
We began 75% under-unity, then climbed to 50% UU, then only 25% UU, and then we hit unity. At this point, then, all three preceding cycles are redundant - we might just as well have accelerated the system to this state via any conventional means, it'd make absolutely no substantive difference whatsoever, aside from being a lot less hassle!
So let's simply take two '1 kg' inertias, and accelerate them up to a uniform speed of 19.62 m/s, using 384.9 J of PE (from wherever you like):
- now use our special trick to buy another 9.81 kg-m/s for another 96.23 J...
- each mass is now at 19.62 + 4.905 = 24.525 m/s.
- at 24.525 m/s, each 1 kg mass has 300.7378125 J of KE, so net system energy is 601.475625 J
- however, we've only spent 384.92 + 96.23 = 481.15 J
- !
- !!!!?
- 481.15 J in. Check it and double check it again. That's all we've spent. No more, no less.
- 601.47 J back out. Two 1 kg inertias, at 24.5 meters / sec. There's definitely no possible room for error. Both sets of values are equally valid, and mutually irreconcilable...
- 601.47 / 481.15 = 1.25 unity.
And each successive cycle thereafter raises the net efficiency another 25%, indefinitely...
So two types of build appear to be possible - i've come to thinking of them as "two-stroke" vs "four-stroke"; we could have a system which begins 75% under-unity, and so needs a good bump-start to get going, but as soon as it passes its unity threshold it'll take off, reaching 200% unity after 8 cycles, 300% after 12 and so forth... or else, we could use the "4-in-1" trick above to aim for a fixed 125% gain per cycle from the get-go, with potentially multiple cycles per revolution. This latter option is what i think Bessler was demonstrating..
Still, as i say, the maths of this are completely trivial... the bigger question, as to where this energy gain or loss is coming from or going to, is the really pressing issue. We have a non-dissipative loss from the very first cycle. Yet it is utterly water-tight and irrefutable. Conversely, we hit unity after four cycles - and obviously, there's nothing at all contentious about unity. Energy out = energy in. But only for precisely four cycles? If the solution for four cycles is right, then by the same logic, so is the solution for three cycles, or five. Equally, if the solution for just one cycle is correct, then so is the solution for four, or forty... Everything here is fully consistent with all known laws, at every step... Indeed, the gain is made possible precisely because of the conservation laws, and not in spite of them! It depends upon CoM and CoE holding precisely as they're supposed to, at every stage..
Each in their own right, gravitational and inertial interactions are fully conservative, and immutable. Yet implicit within the maxim that 'what goes up, comes back down' is the proposition that the thing that goes up and comes back down could be reaction matter; in fact, as demonstrated, provided the internal force is precisely equal to gravity, no upwards-acceleration is even required in the first place; gravity can passively cancel counter-momentum, preventing it from ever manifesting during an otherwise-mutual acceleration..
Replace gravity here with almost any arbitrary source of force, (force is force, F=mA) and we get the same net result.
So instead of 96.23 J per 9.81 kg-m/s, we note that 96.23 is simply 9.81 * 9.81... hence more fundamentally, we're paying 9.81^2 J per 9.81 kg-m/s of momentum.
Since that "9.81" value obviously applies to gravity's acceleration, we can derive the value for any force field by inverting F=mA to A=F/m and giving A^2 J / A P (where P=mV = kg-m/s). But since we're not talking about acceleration, but rather an energy / momentum constant, the fields are e^2 / P. However this ignores the numerical equivalence of A^2 / A, so i've defined a single variable 'N' (for 'number'), to get ((N^2 J / N P) / second), which i've come to regard as 'Bessler's constant'.
(1 * ((N^2 J / N P) / s)) = 1/8mV^2 (25% unity)
(2 * ((N^2 J / N P) / s)) = 1/4mV^2 (50% unity)
(3 * ((N^2 J / N P) / s)) = 3/8mV^2 (75% unity)
(4 * ((N^2 J / N P) / s)) = 1/2mV^2 (unity)
(5 * ((N^2 J / N P) / s)) = 5/8mV^2 (125% unity)
(6 * ((N^2 J / N P) / s)) = 3/4mV^2 (150% unity)
(7 * ((N^2 J / N P) / s)) = 7/8mV^2 (175% unity)
(8 * ((N^2 J / N P) / s)) = mV^2 (200% unity)
Etc.
Again tho, then net sum of (N P) has also been applied to Earth, in the opposite direction ('upwards', relative to our system), via gravity... so this is an incredibly dirty and dangerous system! The planet has simply never been subject to this form of pollution before, and so far from being our "Star Trek moment", we could be looking at a compelling variable in the Drake equation and Fermi paradox..
Think about that for a second - i've noticed this, simply by doing a thought-experiment (apply a mutual 1 G acceleration between two masses in free-fall). Bessler apparently noticed it 300 years prior. Long before the industrial revolution....
...maybe most civilisations discover it too... in their infancies.. and so that's why it's so quiet out there. We've dodged an almost statistically-inevitable bullet by missing this pseudo-N3 break. We discovered "positive displacement" engines (steam, ICE's), then electromagnetism.. We've gone on to tame nuclear energy, LENR's up and coming, we're into the information age and rapidly approaching Kurzweil's singularity... but maybe we only just made it out of the 18th century, by the skin of our teeth!?
Maybe most civilisations discover it much earlier, try and adopt it for agriculture, mining and eventually transport, and inadvertently destroy their planets before realising, too late, what they're doing..
Kerazy stuff eh...
LOL definitely, that's the only possible explanation. Good ol' Wiki eh? Still, you can understand why this has been missed for so long. Even if you did it by accident, you'd simply lose 75% of your input energy. And most folks automatically assume that this KE loss is dissipative (apparently oblivious to the implied paradox that no momentum has likewise been dissipated). Everyone knows "perpetual motion" is impossible, a fool's errand, and so-called "gravity mills" are the bottom of the pile of unworkable machines. Gravity's constant, mass is constant, all forms of leverage or power conversion merely trade force for displacement, and the net integral of force and displacement for any closed loop trajectory through static fields can only be zero, by definition.
The only possible route to OU via classical mechanics is, as GraXXoR notes above, is via an 'effective' N3 violation (effective insofar as it can generate and accumulate momentum of one sign only, at least within an ostensibly-closed system); that is, a means of paying a fixed amount of input energy for momentum, irrespective of rising velocity. So Bessler was using this trick to generate reactionless angular momentum, re-investing the RKE gain as GPE and so giving an outward appearance apparently confirming common conceptions of what a working gravity mill would look like.
Although the maths are extremely basic, only people already-familiar with these terms will be able to follow their implications with any confidence - you can't teach people physics, and how to break physics, at the same time. So if you don't understand, the only sane conclusion is that i'm talking utter codswallop. You don't wanna be so open-minded your brain falls out..
But for anyone who already knows basic mechanics, you can understand the gain principle and how the cycle works, and do the maths yourselves.. if it is a viable solution for the Bessler case, then it's also a warp drive, and not-at-all 'free' energy, at least so long as it's bound to Earth...
If i get this thing working i may start a thread on it over in the general science forums..
Errr...... Not sure if you have said that quite right. When travelling at near light speed time will be or appear to be normal for the traveller. A stationary observer would note that you did not age as much as they did. So as I'm sure you are aware, they both experience time normally but they will observer each other as having time pass differently. So, your "who is right?" question is a good one and I guess the answers is ...... they are both right.
Mathematically that’s cool, like I mentioned. But what I don’t really feel in my gut is why applying 1N for 1 second imparts / requires more energy the faster the vehicle is going. I know it is so from a KE standpoint, I just don’t intrinsically feel it like I do pretty much every other concept that feels right to me once I go through all the maths.
Applying 1N for 1 second describes the Impulse which is related to but not the same as Kinetic Energy. KE is frame dependent because Work = Force x distance (kg*m2/s2), and the distance traveled depends on the frame of reference because the frame could be moving at different rates compared to the object. So Kinetic Energy required to change the speed is almost an arbitrary value based 100% on the relative speed of the object in that frame of reference.
The Impulse = force x time = Δ momentum (kg*m/s), which your gut is indeed correct is independent of the frame because forces and time are independent of any non-relativistic inertial reference frame. So you are right, applying 1N for 1s would result in the same impulse regardless of the frame.
Work and impulse are similar and related concepts, but they are very different.
For example:
An intuitive way to imagine this is have a 1kg toy car that is travelling 13m/s hit the brakes at a=1N/kg for 3 second until it is going at 10m/s. How far will travel before it is going 10m/s?
x = (13^2-10^2)/(2) = 34.5 meters => ΔKE = 1N x 34.5m = 34.5 J
Now imagine that the toy car is traveling at 3m/s on a moving walkway at an airport, and the walkway is moving forwards at 10 m/s. Once again we slow the car by 1m/s over 3 seconds. How far will the car travel before coming to a halt in the frame of the moving walkway? And what is its Kinetic Energy in the frame of the walkway?
x' = (3^2-0^2)/2 = 4.5 meters => ΔKE = 1N x 4.5m = 4.5 J
However if you take that same 2nd example and see how far the toy car traveled on the moving walkway relative to the stationary ground, you will get the same answer as before: x = 34.5m, ΔKE = 34.5 J
So you see, KE and distance traveled is completely dependent on the reference frame, even though the impulse applied in each frame is the same 1N for 3 seconds.
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You know those threads you wish you had never started reading?
Yes. The frame of reference is all important.
So I seem to be back at my original conundrum.
Let me reframe it.
If a stationary man is watching a car accelerating away from him at 10m/s^2 and assuming no losses due to friction, and calculates the KE rising quadratically, how does he explain where the extra energy is coming from. At some point the gain in KE gain will outstrip the power of the motor.
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