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  1. Re the wardrobe the force is not zero. It is counteracted by an equal and opposite force. I like this: Here is the .pdf link to above paper https://www.omicsonline.org/open-access/back-to-galilean-transformation-and-newtonian-physics-refuting-thetheory-of-relativity-2090-0902-1000198.pdf
  2. The above comes from this paper in the Journal of Physical Mathematics. https://www.omicsonline.org/open-access/back-to-galilean-transformation-and-newtonian-physics-refuting-thetheory-of-relativity-2090-0902-1000198.php?aid=80761&view=mobile PS I realised after I posted my previous that F = ma is a vector sum, and that to every force there is an equal and opposing force. Cheers from Edinburgh
  3. re : I go to a writers' group, and I told them about the F = ma issue. One cussed critic pointed out that F = ma isn't an equation (hence should not have an equals sign in it) and doesn't work if you are pushing on a wardrobe that won't move. My point is that F can never be applied in a vacuum, as in a region of nothingness, as no such place actually exists in nature. Space must contain something that allows light to pass through it as a wave, causes Lorentz' Law and the right hand rule to apply to a moving charge etc. Better men than me coined the term for this 'something' - the Aether. A
  4. Of course a differential of a whole charge was ridiculous, but qualitatively speaking, even a charge differential of 1 in 10^^100 would produce the same effect, no? Or are you talking about a residual electrostatic force without a charge differential? Thank you for the Lorentz link. I will look into Lorentz and Maxwell asap. (It's been a long, long time since I did double differentiation) 2 points strike me immediately from it 1. F = ma will not work with an aether since the aether will progressively resist motion relative to it. This effect would likely only become apparent at 'rela
  5. @ Andrew 1) Where exactly do Maxwell's equations become 'really nasty' under universal time? Does this mean UT is wrong, or Maxwell? 2) re residual gravity due to a charge differential between p and e What if an electron was -1 and a proton +2 attraction = 2x1 + 2x1 repulsion = 2x2 + 1x1 This indicates to me that a quantitative charge differential would result in anti gravity... ?
  6. Great to see your repsonse. I caught it late, being on holiday, and have taken a little time to formulate a reply. No doubt I will have left in a howler or two! One of my songs (I'm a musician/songwriter) admits as much "I'm not saying to every question I have all the answers But you can see that i take risks, and I know you're a chancer" The tune(s) must've worked, for I've been dating the subject for the best part of three years now :o) re: oxidation state For gravity to be the result of residual electrostatic attraction net repulsion must be slightly less than net
  7. I wondered about gravity being the result of residual electrostatic attraction. I reasoned that were this so, density should depend on oxidation state. AFAIK it doesn't. Pushing gravity always held an attraction for me :o) Having worked with liquid crystals the notion came to me that this might be the 'state' of the aether. Here's a new paper on Le Sage I just found FWIW. https://www.researchgate.net/profile/Glenn_Borchardt/publication/325746845_The_Physical_Cause_of_Gravitation/links/5b219dc3458515270fc6deb3/The-Physical-Cause-of-Gravitation.pdf edit to add: This experiment would be eas
  8. I guess by tacking you can do this, but I was thinking of sailing downwind. We are accelerating charged particles using a field whose speed of propagation is that of light. Hence lightspeed would be the asymptotic limit for speed of the charged particle. Probably a better analogy is a balloon in wind.
  9. Well this is a slow-moving thread lol. I saw a video of your solar collector, nice work! How is the paper on GR coming along? How are you dealing with the equivalent frame of reference problem? A rocket makes a round trip, the twin left behind ages more, yet we can just as well say say the rocket remained stationary and the planet made the round trip... Then there is this business of intrinsic red shift... ...which pulls the rug out from underneath Big Bang.... Lightspeed seems to me a limit only to anything accelerated by electromagnetism, since a yacht cannot sail faster than
  10. I used the non relativistic equation ΔV = ve ln m0/m1 http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation ve = 1000 m/s m0 = 1000 Kg m1 = 999 Kg (after 1 burst) ln m0/m1 = 0.00100 m1 = 998 Kg (after 2 bursts) ln m0/m1 = 0.00200 Velocity of rocket after first burst is 1.00 m/s Velocity of exhaust after first burst is between -1000 and -999 m/s average -999.5 m/s Mass of rocket after first burst is 999 Kg Mass of Exhaust after first burst is 1 Kg Velocity of rocket after second burst is 2.00 m/s Velocity of exhaust after second burst is between -1000 and -998 Km/
  11. Yes of course you are correct, energy having the velocity squared term loses its sign. And momentum is conserved. But... it still seems to me that energy as defined by E = 1/2 mv^2 is not. I think we can totally ignore relativistic effects at 2 mph! I have described the rocket with 1:1000 velocity gearing between the rocket and exhaust, I could just as easily have used 1:1,000,000 or any ratio you like. Obviously the greater the ratio, the smaller the discrepancies between the burns, tending AT THE LIMIT, as we get smaller, for each rocket burn to become identical, and for the rocket mass
  12. Calculus is a mathematical way of describing things which are not constant. Now you might think a 100Kg moving rocket in deep space is not constant, but unless it is firing its engine, it moves with constant velocity. In fact, without measuring its velocity relative to something else, there is no way of telling if it is moving at all. Now let's say it fires its engine for a tiny fraction of a second. It's velocity changes from x (unknown) to x +dx where dx is 1 mph. Meanwhile the rocket exhaust shoots off in the opposite direction at - 1000mph relative to the rocket's "frame of reference". N
  13. I think Andrew Ancel Gray makes a good case that there is no such thing as a photon http://www.scienceforums.com/topic/11645-7-reasons-to-abandon-quantum-mechanics-and-embrace-this-new-theory/
  14. Au contraire mon ami it matters a great deal. There is no way a crystal saturation process can produce a linear accumulation. Only single fired entities can do that. So there are 3 possibilities. 1) Your sensitivity calculations are wrong (this I doubt) 2) What is being released is a packet of ----ons mutually interfering in some weird way 3) The single ----on experiments are faked Do you have the wherewithal ie access to equipment to record an independent 'spot' v time count? edit I see you favour 2) http://scienceforums.com/topic/11645-7-reasons-to-abandon-quantum-mechanics-and-e
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