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Everything posted by ralfcis

  1. Light is held by gravity? I thought the speed of light was un-affected in that it took a near infinite time to traverse a near infinitely contracted space that still works out to c.
  2. " (V2/C2) thus ultimately it is taking the velocity relative to C." Ok that's highly confusing. A velocity relative to c is not the same thing as expressing a velocity as a fraction of c. You can approach a star at .5c but your relative velocity to the light of that star is always c and your relative velocity to the vacuum along the way is 0c. For tony's point. If you were in a car approaching a siren (sound relative velocity to still air 750 mph) at 50 mph, your relative velocity to the road is 50 mph, to the air is 50 mph, to the sound is 800 mph but the sound's relative velocity to the r
  3. "the speed of light is constant as the measurement of the speed of sound in an airplane" I made the same mistake when I first started out. The difference between sound's fixed relative velocity to its medium and light's fixed relative velocity to its medium is that you can have a relative velocity to sound's medium but you can't have one to light's medium (which is the electromagnetic properties of a vacuum). You can move air (wind). You can move relative to air. So while sound's relative velocity is fixed to air, your relative velocity to sound, through its medium, is not fixed. The MMX pro
  4. Tony I'm going to try to understand your math by re-writing it into a format I understand. γ (V) = sqrt(1-V^2/C^2) translates to Y= c/sqrt(c2-v2) (Try using the blue bar above to write your equations) The rest I can't translate. What is K? The only relative velocity to c is c unlike what I think Victor is saying. The last equation is γ (v) = γ (- v) = sqrt (1-v^2 / C^2) does not conform? Why not? -v2 = v2.. Actually never mind I'm not interested.
  5. Yikes, you're right Popeye. I have done this same mistake many times before screwing up who's the observer but this time was really bad. I mean that's relativity 101.
  6. Umm according to relativity you can't have relative velocity to the vacuum. Light has a relative velocity to the vacuum so by the transitive property, you can't have relative velocity to light which is backed up by the relativistic velocity combo law.
  7. Yes but that's not my question. The relative motion of the ships whether they're parked on earth or both moving parallel to each other moving at .9c away from earth should be irrelevant because the ships are in zero relative velocity frames. Yet, the fact that in the parked scenario the gun is pointed at 90 degrees and in the "moving" scenario it's angled to compensate for the aberration of light means those two zero velocity frames are not equivalent. This seems to indicate earth has a low relative velocity to background space which indicates a preferred frame which is verboten in relativity.
  8. This post has faulty reasoning which was later corrected by Popeye. The aberration of light is not seen from the ships' perspectives, only from the earth's perspective of the ships going away from it at .9c relative velocity. Sorry I tuned out of this conversation so I don't know what you guys have been writing about but I woke up this morning with a related question of my own. Suppose two WWI biplanes were flying parallel to each other at some distance apart so their relative velocity is zero. One tries to shoot the other one down so he sets the horizontal angle of his gun at 90o to his p
  9. Ok like everyone who has come before you, you are not reading what I'm writing. I've already explained your story. You are stuck on your ideas which are based on almost total ignorance and you're neither willing to admit this or do anything about it. P.S. Maybe you should stick to just one thread for your stuff.
  10. I'm totally committed to the physical phenomena of relativity but I don't accept the theory that tries to explain those facts. Time does not relatively slow, nor does length relatively contract. If the earth explodes, the ships still have their atomic clocks which have universal accuracy and their clocks were all set to earth time when they started.They can calculate their relative velocity from the doppler shift ratios. From this they can calculate why their clocks are different. There is no absolute center of space or time but the ships have agreed to a common relative center which was the e
  11. Ok let's take your simplest diagram first. B1 reaches planet x in 4.35s of its time and 10s earth time where 5.65s of earth's time is the alignment of B1's perspective of the end of the race and the earth's perspective of the end of the race due to relativity of simultaneity. Oh I see what you're saying in the 2nd diagram. The earth's velocity gives B1 a push of .9c in the vertical direction while the thrusters give B1 a horizontal direction of .9c. This will seemingly make B1 follow a longer diagonal path to planet x. The problem with this reasoning is the thrusters give B1 a relative vel
  12. Maybe if you give me your answers I could figure out what you're trying to say. Relativity says there's no absolute still, only relative still where the doppler shift ratio =1 meaning rate of time looks at the normal rate for both. Now, distant stars look like they're not moving because they're so distant but the earth can measure its relative velocity to them by the red or blue doppler shift but that's not important to your question. If earth and x are relatively stationary then B1's relative velocity = .9c to earth or A and -.9c relative to x since it's approaching x. To figure out the rel
  13. 1. You're showing planet x has moved away from where it was and B1 is moving to where it was so it's not reaching where planet x is now. 2. A and B1 have the same velocity relative to each other. As I explained, time does not fly faster or slower in constant relative velocity. However, in relativity, each one "sees" the other's time as moving equally slower due to reciprocal time dilation. This cannot be seen in real time, it can only be calculated and it's an illusion of perspective. It's the same illusion of perspective as length contraction and, as I've explained in my Minkowski diagram, i
  14. I'd like to help but don't understand most of your sentences. a. "and send spacecraft to different directions of the Earth" ? Are you saying 3 ships fly from earth in different directions at .9c relative to earth? b. L=9ls (light seconds) not 9c c. "Let's make the Earth and Planet X have speed V" ? What does this have to do with 3 ships from earth? 1. Each ship covers 9ls in 4.35s of its time or 10s earth time. Relativity says these seconds are not the same but I say they are. The rate of time does not change for constant relative velocity; it ticks the same for all frames. So why are they d
  15. Like in that Star Trek episode where a nascent universe started inside our own.
  16. I was ready to just give up on my quest to overturn relativity as I concluded no one understands relativity enough or understands anything I'm saying. But I reviewed my initial questions on the PSX (actually I've only been asking the same question under different guises) and found Dale not only knows his stuff but has been understanding and answering my questions all along. I just didn't know enough until today to recognise this. I just have to go to another forum he's on because the PSX is just not the place to discuss things with all the trolls and moderators on it. I tend to drive people cr
  17. Non one reads this thread anyway so I dump my conversations from the PSX here before they delete them there. I should have included that the invariant length dsds is the length ds2=dx2+dy2+dz2−dt2ds2=dx2+dy2+dz2−dt2 of the 4-vector (dx,dy,dz,dt)(dx,dy,dz,dt) between two spacetime points. In the rod's rest frame take these two spacetime points to be at the ends of the rod at the same time (so dt=0dt=0). Then ds=Lds=L. In a boosted frame the 4-vector between these same two spacetime points will be (dx′,dy′,dz′,dt′)(dx′,dy′,dz′,dt′) and ds′=ds=Lds′=ds=L still. But dt′≠0dt′≠0. So the observer
  18. Finally I got a nibble from someone with physics credentials after all these years: Your word "illusionary" seems synonymous with any measurement of the rod's length not done in the rod's rest frame. The idea that the rest frame length of the rod is invariant under boosts is well known. You are correct, the invariant length of the rod (or invariant distance to the star) does not change just because we look at it from a boosted frame. In this sense, measurements of the rod's length in other reference frames is illusionary, and we must return to the rest frame of the rod to measure its "real"
  19. Since my comments that involved a lot of math have disappeared silently and anonymously on the PSX once in the past on my Loedel question before a reopen review, I'm going to address Gary Godfrey's answer here. There seems to be a confusion between Galileo's principle of relativity and the illusion of perspective relativity. When we get in the car and drive down the road, we can't say we're stationary and our wheels are turning the Earth beneath us like a giant treadmill. Similarly, we can't say the earth is revolving around a proton in the LHC. The sun does not orbit the Earth daily. It's orb
  20. Here's a nice little article that shows relativity doesn't need to be interpreted as Minkowski spacetime: http://www.timeone.ca/losing-time/#sthash.yEsvwfhl.gg3iXMb9.dpbs I've been arguing this on the PSX but their minds are absolutely closed shut ignoring other mathematical interpretations: There is an obsession on wikipedia of whether some physical ruler could be held up to something long whizzing past and being able to determine that the moving object has physically shrunk due to length contraction. There is all kinds of supposed evidence that the results of certain particle collision
  21. Wow, so they, on the PSX, hold a reopen vote on my Loedel age difference question but some unknown person deleted all my math supporting my argument just before the vote. No record of it ever having existed and all my avenues to repost the math have been closed. Do they really fear anyone was going to look into my math and actually be convinced? Only in movies are mathematicians able to read anothers math as if it were text or 006 hieroglyphics. No one knows algebra anymore.
  22. I've figured out the formula for the hysteresis of simultaneity around the Loedel perspective. For the example above, A=B=2 And you're trying to find the resultant intersections on the velocity lines b or a. d = Y of ( relative velocity - perspective velocity) / (Y of perspective velocity). So b=Ad and a=B/d. For example if B =2 and we want the value of a for a perspective velocity of +7/9c (Y = 1.599), Y of (.6c - 7/9c) = Y of (-1/3c) = 1.06066. So b= 2(1.06066/1.599) = 1.33. So the age difference between Alice and Bob from the +7/9c perspective of A-b = 2-1.33 = .67yrs.
  23. Here's an Md of the hysteresis of simultaneity at the top and bottom triangles of a return trip at .6c. photos.app.goo.gl/rFd42uvoNWJ9r1EP7 The age difference from sample perspective velocities from -c to +c is the same for both top and bottom if you you subtract the Loedel age difference which is constant and without hysteresis during constant relative velocity. This is a constant age difference from which all other perspectives of the age difference between Bob and Alice can be calculated using a formula for the hysteresis. Hence, the perspective age differences are unimportant. This is huge
  24. Here's a muon-like example using the Loedel perspective. photos.app.goo.gl/fhSGp1WVdvRiNoh17 A ship comes in at .6c and can make various velocity changes at 3 ly away from earth. The swing in Loedel simultaneity lines occurs between 4 and 6 Earth time. No change in velocity results in no Loedel age difference. A stop results in the ship ageing 1 yr more but a velocity change to .8824c results in the ship ageing .8 yr less than Earth for the final Loedel age difference.
  25. Here's a sample of the awesome power of the Loedel velocity perspective: photos.app.goo.gl/TXD5j6HzvhvUDEvt9. The 1/3c Loedel lines of simultaneity extend up from the original .6c velocity and then intersect the light line from the turnaround. Then new -1/3c Loedel neon green lines of simultaneity (due to the new -.6c velocity at turnaround) extend to the -.6c velocity line. A new Loedel line of simultaneity (thick green line) extends back from there to the stationary perspective's point where the 1/3c lines of simultaneity started. This method can be employed from any original velocity to
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