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Posts posted by Farsight

  1. But if your conclusions don't match up with observations, then you conclusions must be wrong.


    Are you being mendacious here? Take a look at the opening paragraphs of Time Explained again. We observe colour, heat, sound, et cetera, but we do not conclude that these things are fundamental properties of the world. Look again at the transverse metre rule, where you measure time t for light to travel distance s and I measure the same value. You calculate c=s/t=300,000km/s as do I, but we know that your t is different to my t by a time dilation factor. We see a similar time dilation factor on the surface of the planet. So what do we conclude Edtharan? Here's a clue: c=s/t and t ain't what it was.


    This means that if we have a variable value of C the equation E=MC2 must therefore be wrong.
    No it doesn't. Where did that come from? Did you even read my previous post? I always measure c to be 300,000km/s, and so do you. Come on now, pay attention, when an object falls to earth where does the kinetic energy come from? Here's a clue: E=MC2 and M didn't change.


    The Moon observer then uses the value of C to conclude that the energy release in the experiment would be:

    2 * 1.672 × 10^27 kg * 299,999km/s^2


    Where as the Earth Observer would conclude:

    2 * 1.672 × 10^27 kg * 300,000km/s^2


    This therefore give different and mutually exclusive results. Both the Moon Observer and The Earth Observer can not both be correct in their conclusions.


    Groan. This is utterly facile. You've taken a conclusion and turned it into a measurement, then back into a conveniently "mutually exclusive" conclusion. Why do you do this? Why duck and dive with all this dishonesty that can't explain anything and doesn't want to see anything explained? Come on, show me I'm wrong. Explain to me what gives if c=s/t and the t is dilated. Does s contract even though it was a transverse metre rule? Come on Ed, think. Yours is the circular argument, and as ever it's going nowhere.

  2. The most important difference is that if you don't have dimension that is not directly perceivable, then you don't have a perception of motion (just displacement), Time as we know it would not exist for you. Importantly, this means that the Dimension we call Time is not special (other than we can not directly perceive it), it could have been any dimension at all. All dimensions are treated equally.
    So we can't see it, we can't perceive it, it offers no freedom of motion, but we should treat all dimensions equally. And:


    Under QM, time is change...


    Thud, thud, thud. That's my head banging on my desk at how people talk themselves into misunderstanding and mystery.

  3. You are mixing frames of reference here. You are saying that the frame of reference where the explosion is on Earth is the same as it is in Space.
    No I'm not. That why I stressed the difference between conclude and observe.


    This is not so. It is experiencing a different gravitational field and is therefore in a different frame of reference. Because of this, you forgot that the fact that the Earth observer, now is a different frame of reference to the explosion will disagree and give mutually exclusive results to the observer in space. You haven't answered the question at all. all you have done is shift the frame of reference and ignored/not factored in the observer on Earth.
    I've forgotten nothing. You might be on earth and I might be in orbit, and because I need to keep adjusting my watch I conclude that your c is slower than mine. But if I try to measure your c I actually measure my c.


    Actually, from experiments done, the annihilation energy is the same, not different, so this rules out that C is variable. In different gravitational potentials, the annihilation energy is the same. In annihilations involving moving frames of reference, the excess energy is accounted for due to the energy put into the system (by particle accelerators). There is no unaccounted for energy that indicates a variable value of C.
    This is more of the same. You will not measure a different energy.


    Current particle accelerators exist at different altitudes, and therefore at different gravitational potentials. They do not need to adjust the values of their results to account for the different gravitational frame of reference. If C was variable, they would have to.
    You always measure c to be the same old 300,000km/s regardless of your reference frame. You never measure 290,000km/s. But you can conclude that there is a difference in absolute terms. Time dilation and the transverse ruler tells you this.


    But if I change my frame of reference to the event, then according to your proposition, I should see a different energy value than one who doesn't change their frame of reference...
    No, you won't see a difference. You conclude there's a difference.


    This is the major flaw of your argument. It requires that two mutually exclusive results must both occur.
    No Ed. You've gone astray from the off here. Read what I said again.


    You seem to be equating potential energy as to being equivalent to a change in C. But you can have other sorts of potential energy. What about the potential energy in a chemical reaction, is this due to a change in C, or the potential energy in a spring. Potential energy is a kind of stored energy, a tension if you will. How does a change in C deliver a store of energy?..
    Sure thing. I thought you'd assume the gravitational. The change in c is caused by a local tension gradient orthogonal to mass/energy stress. See ENERGY EXPLAINED and MASS EXPLAINED.


    The other thing is that the difference in energy of a Matter/Antimatter annihilation will be different to the change in energy of a cannonball launched into orbit. This means that for the same frame of reference you can get 2 (or more) different values for the change in C. So even if your proposition was correct, the same observer in the same frame of reference will get different results of their calculated local value of C, depending on what they are actually observing.
    Everybody measures c to be the same. If they end up with some mismatch in their calculations they've done something wrong.


    So it even fails for self consistency. So not only does it give results that are internally inconsistent, it does not give results that match to reality. On these ground your proposition, as it stands and you presented, can not possibly be true in this reality.
    Sorry Ed, you've reached that conclusion because you've gone off at a total tangent. Try it again. Try to ask smaller questions instead of blatting out a whole proof that falls down because of a mistaken initial premise.
  4. Farsight, you assume that I did not read your thread. False. furthermore I read the previous thread. In the second one you start off calling the two that said they had read your post liars.


    Huh? Tycho didn't read it because I don't say time does not exist. Look:




    And Edtharan obviously didn't read it because his reply to a post was apposite to the essay.




    But whatever. OK you don't want any help, fine.

  5. I thought the infinite time dilation happened when you approached the event horizon, Albers. Which takes us back to the old "frozen stars". No collapsing star has finished collapsing yet, and never will. Result = no singularities. Black holes have no hair, and no heart. I don't know much about the dimensional collapse of the quark field or transactional interpretation. Or truth be known, black holes either. But I offer what I can.

  6. If C is variable, then two observed in different frames of reference will disagree over the energy (due to E=MC^2) contained in the matter of an object (the object does not have to be at relativistic speeds or anything).


    Not if they're smart they won't. Think about that cannonball travelling at 1000m/s. It has considerable kinetic energy. When you change frames to travel alongside it, the kinetic energy is now zero. But you know the cannonball didn't change at all, so you know that the energy content didn't really change.


    Specifically in this scenario: An observer (A) in orbit around Earth is in a lower gravity gradient than one on Earth (B). According to your claims, Observer A will have a higher value for C than Observer B. Because Observer B is in a higher gravitational field than Observer A, Observer B will have a lower value for C.


    Yep, no problem.


    This means that a matter/antimatter explosion that occurs near Observer B on Earth will produce different energy values.


    No. The amount of energy contained within the matter and antimatter masses at that location is what it is. It's like the cannonball.


    As observer A sees "Time" moving slower for Observer B and concludes that c is lower relative to himself, in the equation E=MC^2, C will have a different observed value to what Observer B would see (as you said Observer B will see C at the same 300,000km/s locally as Observer A would measure locally. However Observer A is not measuring the value of c locally, but as it is at Observer B relative to himself and so gets a different value of c than he would if he measured it locally.


    OK, he concludes that c is lower on the surface.


    This means that the energy that is observed should be different for what Observer A sees and what Observer B sees.


    You're getting confused between observed and concluded here. If observer A is smart enough to calculate a lower value of c, he will also be smart enough to calculate that the annihilation energy released on the surface is less than the annihilation energy that would have been released in orbit. He will calculate that the difference equates to the kinetic energy of the masses falling from position A to position B. He will realise that the kinetic energy doesn't come from nowhere. It doesn't come from the "gravitational field" or from the earth. It comes, via the reducing c, out of the falling masses themselves.


    Imagine being on earth and kicking the cannonball into orbit. You gave it kinetic energy, and this disappears into "potential energy" within the cannonball, commensurate with the higher value of c up there. The same sort of thing happens in special relativity when there's no gravity: the canonball's relativistic mass increases when you kick it to some higher velocity with respect to yourself.

  7. Don't take too much of Farsight's discussions on the topic as being generally accepted by the scientific community, nor confuse this issue with science. "Is time just a concept" is a philosphical/metaphysical discussion. It's not germane to relativity. If you want to discuss it, try the speculations/metaphysics section.


    It was however germane to Albert Einstein during his Princeton years:



  8. Edtharan: I don't feel I've got your full attention regarding the material I've posted or linked to. And yours is just about the only attention my essays have received here. There have been other comments, but many indicate no attention to the subject or my responses. So I've decided to spend my time more productively elsewhere. Apologies.

  9. I do so agree with Special Relativity. But I also see what it's telling us. I see why the postulates hold, what they mean, and how it evolves. To the best of my knowledge I think the same as Einstein did. Not in 1905, but in 1949.


    In his response to Godel's paper in the Schilpp volume, Einstein acknowledged that "the problem here disturbed me at the time of the building up of the general theory of relativity." This problem he described as follows: "Is what remains of temporal connection between world-points in the theory of relativity an asymmetrical relation (like time, intuitively understood, and unlike space), or would one be just as much justified to assert A is before B as to assert that A is after B? The issue could also be put this way: is relativistic space-time in essence a space or a time."

  10. But the external observer would measure a different value of C for you.


    Yep. Let's set Twin's Paradox symmetry to one side for a minute. If we had some kind of magic instant viewscreen that showed us the inside of Swanson's black box, we'd see his light going seven times slower than yours.


    And as C is used in E=MC^2, then if a person in a gravity well (or travelling fast) annihilated some matter and anti matter, then the person doing the annihilating will measure one value for the energy released and the distant observer will measure a different value.


    Yep. Look up electron volts. An electron at rest will annihilate with a positron releasing gamma rays of 511keV apiece. A relativistic electron annihiliating with a relativistic positron will release gamma rays of more than 511keV as far as the stationary observer is concerned. Don't get stuck on rest mass here - I take the view that something that has energy has mass. For example, a photon has mass of hf/c2.


    If this was a bomb, then the observers should see a different sized crater (not to mention the different trajectories of all the ejecta - so one observer might see a rock crush their house where as the other observer would not see it crush the house).


    No they wouldn't. You're going off on one here. If a rock hit a stationery observer it would cut his head a bit, but if it hit a relativistic observer it would take his head clean off.


    So, here again, we have two different, and mutually exclusive results occurring if we accept a variable value for C. No we don't. If we accept everything else to be relative, then there is no problem. If we accept Time to be a physical dimension as is Space and that, using geometry in the 4 Dimensions, then we do not need to have a variable value of C. In fact to get any sensible results, we need to have a constant value for C.


    I can only try to explain to the best of my ability. Look, here's the situation. The image below presents an illusion. The illusion is that square A is a different colour to square B. The reality is that square A is the same clour as square B. You're stuck thinking that the illusion is the reality. You can learn how to see that the squares are the same colour. And you can learn to see the simple truth of what I'm saying.




    See http://www.echalk.co.uk/amusements/OpticalIllusions/illusions.htm and use the swatch to prove to yourself that A and B are the same colour.


    Can you mathematically prove that? The maths and experiments in relativity indicate that this is not the case. The mathematics that underpin relativity do not come to this conclusion, so could you show us how it could?
    Maybe. But it's a lot of hard latex, and it would be easier for me to find something somebody else has already done. But you haven't read any of the papers I've linked to, so maybe not.
  11. Let's try it another way:


    Swanson, I give you a metre-wide light-measuring device that you hold transversely while a black box takes you for a trip. Edtharan, you stay in the lab with a similar device.


    When Swanson gets back, his device shows that light went back and forth across his transverse metre a trillion times. As far as Swanson is concerned it travelled a trillion metres. Edtharan's device says the light went back and forth seven trillion times. As far as he was concerned it travelled seven million metres.


    Swanson and Edtharan are both here now. But Swanson's light travelled a trillion metres instead of seven trillion metres. So Swanson's light travelled slower than Edtharan's. Swanson didn't notice it during the trip, because the devices are akin to light clocks. His whole body was similarly affected. He experienced less time because his light was running slower.


    Like I said, you can never measure your local c to be anything other than 300,000km/s. You only see the difference when you compare frames in experiments like this.


    Swanson's trip might have been a ride at .99c. Or it might have been to a high-gravity area. There's not much difference between the two, but in the latter situation there's less room for dispute about how far the light "really" travelled in some absolute reference frame.

  12. Wow' date=' this is just spectacularly wrong.


    You can't get away from length contraction issues in your measurement. Because of the motion, if you trace the path of the light that is transverse to the motion, it will have a component in the direction of motion, in the external viewer's frame. One object that would fit this description is the familiar light clock; the time is dilated, but the path is longer by the same factor.


    c is constant and the same in all inertial frames[/quote']


    If you're an external observer looking at a subject moving past you at .99c you see the lightclock like this /\/\/\/\. However if you're the subject you see it like this this |. You don't consider yourself to be moving. You measure c to be the same old 300,000,000 m/s. The transverse metre is unchanged, and the seconds are time-dilated sevenfold. Now if meters are the same but seconds are different, and c is metres per second, where does that leave c?


    If you consider time dilation to be real, you must consider the different c to be real too. Yes, you measure c to be the same in whatever inertial reference frame you're in. But the only way it's the same across these different reference frames is if you're measuring it in terms of motion through a preferred reference frame, an absolute space. Space instead of spacetime. That's the TIME EXPLAINED position.


    But let's park that for now, because we were talking about gravity as a tension gradient and a variable c. Edtharan, have you done any reading yet?

  13. So which one do you actually mean? Is C constant, or not?


    You always measure it to be the same whatever reference frame you're in, so in this respect it's constant. But it's actually different in different reference frames. It's just that you don't notice it. So it's not really constant.


    Imagine you blast off away from me and earth at .99c. I know from √(1-V2/C2) that your time experience is dilated by a factor of seven. I get you to measure the speed of light, in a tranverse direction to avoid any length-contraction complications. You find it's still the same old 300,000km/s. But your second lasts seven of my seconds. So your c is a seventh of mine.


    It's similar if you're in a high-gravity situation and I'm not. Your c is less than mine, but you don't notice it. And at all points between us c has intermediate values, hence the local gradient.

  14. I do declare I think I heard the word aether mentioned there, Martin. I share the sentiment expressed by fredrik. String Theory does not offer grasp. But I wouldn't single it out, the same might be said of Quantum theories, or even the Standard Model. More generally, physics arguably suffers from a surfeit of mathematical abstraction that struggles to provide that satisfying intuitive understanding that we all seek.

  15. What's the problem, h4tt3n? If you take this a little further, you can find something that tells you the energy of a photon is hf and its momentum is hf/c, so:


    energy E = hf


    momentum P = hf/c


    To get from mass m to momentum P=mv you have to multiply by velocity, which is c. And to get from momentum P to energy E you have to multiply by velocity c again. So to get from mass to energy you have to multiply by c squared. Which gives you:




    Great fun!

  16. Now, C and hence the speed of light, is less for me...
    No, the speed of light in one place is what it is. Your velocity means the light in your atoms and clocks is having to travel further, and your time experience is reduced. You measure c to be the same, but your seconds last much longer. Hang on, I've already said all this but you've paid no attention.


    You've written another long essay without checking up on anything I've said. What are you playing at? Are you even interested in physics, or are you just playing debating societies?










    I suggest you print out the above papers, sit down, and read them all. Also read MASS EXPLAINED. Also follow up the google link I posted earlier. Do some research then we can talk some more. All that's happening at the moment is that I'm doing the research, and you're sitting there in your blinkered ignorance saying no no no and telling me the earth is flat.

  17. No. There is no energy loss. It is converted from kinetic to potential energy. No loss. If an object was accelerated from the surface of the Moon into space, then it would have been given a certain amount of kinetic energy...


    Now imagine the situation where you take an object from the surface of the moon, and place it a million miles from the moon. With no kinetic energy involved.


    So you now have a different definition of motion as well. Ok, so what is your definition of motion then?
    An ongoing displacement in space that can be measured against other ongoing displacements in space. I think it's futile to discuss this further until you examine the axioms that you interpret as explanation and proof.


    I can accept that that is what you are claiming, but if you actually look at the claim that C is variable, then there are so many other effects that will occur. Another effect is that you should see a reduction in the velocity of light. As particles with 0 rest mass will travel at C, then if C is changed then these particles will also have to change their velocity.


    They do change their velocity. But remember it's a vector quantity. They swerve.


    If, as you say, C is reduced as you near a gravitational object then, an object (say an electron accelerated to near light speed) with a velocity of 299,999km/s moving into a gravitational field (which has a value of C as 299,998m/s) will then exceed the local value of C. This particle is now superluminal. We can now have objects travelling faster than the speed of light in a vacuum. Again, this is a conclusion that has never been observed.
    As you approach the gravitational object it's like you're accelerating, moving gradually from one inertial reference frame to another one where c is different. However this different c dictates your local time experience so you don't measure any local difference. You have to mentally step out of the frames and look at what distinguishes one frame from the other. Remember your relativity, c has a local value of 300,000 km/s. It always has, and nothing can move faster than the local value of c. Your electron is basically a ring of light - it can never move faster than light.


    A variable value of C produces results that have never been observed, even in the situation that you say that C must change. A variable value of C (as you have presented it) does not agree with actual physical evidence.
    Your misunderstanding is what's causing you problems, see above. Come on, check out those links instead of spending all your time trying to find something to argue about.
  18. Motion is defined as: Distance/Time. Distance is the displacement and it is divided by time. You use "motion", but then you don't seem to be using the concept of Motion (Distance divided by Time). If you have redefined what motion is...


    In a nutshell I'm saying motion defines time, not the other way around. Let's park it there and agree to differ.


    So if C can be changed by a gravitational field, then this has other implications. For instance: E=MC^2. If C is lower near a gravitating body then the energy contained in a given mass will increase if you take it away from the gravitating body... As these effects have not been observed, it is safe to say that this can not be occurring.


    No, the change of c is the gravitational field. You need to check out those links and that google page before making a pronouncement. If you have a mass on the surface of a planet it takes energy to take it out of the gravity well. Think about it.

  19. Edtharan: I don't understand the first portion of your post above. You seem to be saying motion is displacement in time therefore time is not dependent on motion, QED. You seem to be using an axiom to give a proof. Please clarify.


    C is a constant in all frames of reference, this has been experimentally confirmed. How can you have a gradient in C if it is a constant?


    It's constant in any one frame, but it isn't constant when you look at the big picture across multiple frames. Imagine you're in a spaceship travelling at .99c. You measure the speed of light to be the same as ever, but your time is dilated by a factor of 7. Now step outside that frame and ask yourself what is c inside that frame? The situation is similar to one where you're standing on the surface of a neutron star. The local c is lower than it is a light year out in space, and there's a gradient in the value of c between the two places.


    A refractive index arises due to the fact that light interacts with the matter of the object which slows the light down...
    I said akin to a refractive index. It isn't quite the right term. I'm not sure what is.


    You are talking about having a variable value of C. A gradient. It might be possible for a region of space that slows light down, but this does not mean that C has been reduced...
    I am, and it does, truly. That's what you get when you take time out of the equation. It's the cause of the "curved spacetime" effect. Search google and check up on this, you should be able to find something to vindicate what I'm saying.




    Could you explain your reasoning for this. You made this statement but then didn't elaborate on why this is so. As it is different to accepted theories and you seem to be basing your theory on it, this does need more explanation.
    It isn't really my theory. Have a read of Mass Explained and follow the link at the bottom for the Robert Close paper:




    Current theory accepts this "curvature' as an effect of mass. There is not "action at a distance" as in Newton's theory of gravity. You have not actually explained what is different about your theory, just that it is different.
    The curvature is the effect of a gradient which is the effect of mass/energy. Read Energy Explained before you read Mass Explained, because that's where I introduce stress and tension. Energy is a volume of stress, and Mass is this tied down into one place. Gravity is an orthogonal tension. Get a "rubber" (a pencil eraser) and squeeze it in the middle. You're putting it under stress. The ends bulge out a little, because of the orthogonal tension you've introduced via that stress. As I said I've got to write it up properly, so that's all I can really offer for now.
  20. Swansont, if you're going to dismiss questions like What is Gravity? as mere metaphysics please do it somewhere else.




    1) If the existence of time is dependant on motion, then the object in the centre of the experiment will not experience Time until the influence of the experimenter reaches it.


    Yes I agree with this.


    2) If Time is not dependant on motion then the object in the centre of the experiment will experience time normally regardless of whether the influence of the experimenter has reached it or not.


    I don't think I agree with this. If there is no motion I can't see how there can be any time experience. There's no incoming photons, no oscillations, no weak force decay, nothing. I guess it's rather like being knocked unconscious. You have no time experience, then you wake up and find you've skipped five minutes. If you repeated this ad infinitum there would be no time experience while you were unconscious, eg while nothing was moving, but you could gain a time experience by looking at the clock every time you woke up. You'd see that the clock had moved. Which brings me back round to what I was saying. Something's got to move somewhere, and time is the measure of it.

  21. Edtharan, I've got to go so I'll answer just one point:


    Could you explain in more detail what you mean by "a tangent to the curve". You started off saying that there is no curve, but then call your gradient a tangent to the curve. Also, what is the Gradient? Is it of gravity, or some "fabric" of space? And how is it graded?


    Let's say that a mass is somehow going to steer left because of gravity, and will move towards a planet instead of continuing in a straight line. There is no magical action-at-a-distance "force" zapping through space like cartoon magnetism. Instead the left side of the mass experiences some different local condition to the right side of the mass. If you were to move the mass a little to the right, you'd still see a different local condition on the left and right. There is a left-to-right gradient wherever there is a gravitational "field". If there were no gradient, both sides of the mass would experience the same condition, and the motion would be in a straight line. The effect of this gradient is curved motion over time, but this "curved spacetime" is an effect not a cause. You get a better idea of the cause if you remove time and take a snapshot as at some given instant - or in mathematical terms take the the time-differential of your curved spacetime. The differential of a curve is a gradient. It's a gradient in c, akin to a refractive index. It exists because mass/energy is stress, and gravity is an orthogonal tension that decreases with distance and area. You'll have to wait for GRAVITY EXPLAINED for the full works, and I might have to write SPACE EXPLAINED and/or CHARGE EXPLAINED to set the scene.


    Newtonian gravity is much more easily understandable than Einstein's gravity...


    IMHO Newton and Einstein explained what gravity does, not what it is.

  22. I used to go night fishing. You spend time on the beach looking up at the glowtube on your rodtip, so you see things in the sky.


    I've seen some odd things, which were definitely not conventional aircraft, or satellites, or meteors, or the planet Venus. So when I hear of reports being dismissed as such, I feel certain that there is some ongoing cover-up.


    I'm also certain that there's a whole pile of nutters out there faking abduction reports and photoshopping pictures, plus conspiracy theory nutters and crop-circle freaks sneaking around at night. These guys really muddy the waters.


    As to what the truth is, I don't know. I don't buy Nick Cook's The Hunt for Zero Point. I think "Black Ops" is too pat, too convenient an excuse. I think there's more, but I don't know what.

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