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Reviewing the Riddle of Relativity

Reviewing the Riddle of Relativity. by Ian McCausland. Herbert Dingle 1890-1978. Relativity. Maurits Escher. Andrew Lipson. The Central Question. Is there, or is there not, an internal inconsistency in Einstein’s special theory as described in his original paper on the subject ?.

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Reviewing the Riddle of Relativity

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  1. Reviewing the Riddle of Relativity by Ian McCausland

  2. Herbert Dingle1890-1978

  3. Relativity Maurits Escher Andrew Lipson

  4. The Central Question Is there, or is there not, an internal inconsistency in Einstein’s special theory as described in his original paper on the subject?

  5. The Polar and Equatorial Clocks • E1: If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be tv2/2c2 second slow. • E2: Thence we conclude that a balance-clock* at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions. • *Not a pendulum-clock, which is physically a system to which the earth belongs. This case had to be excluded.

  6. Dingle’s Question Applied to this example, the question is: what entitled Einstein to conclude from his theory that the equatorial, and not the polar, clock worked more slowly?

  7. Stadlen’s answer But the relative motion involved in this case, being circular, is non-uniform. I submit, therefore, that Einstein was wrong in saying that his prediction followed from the special theory, which deals only with the effects of uniform motion. This is not to say that the prediction was invalid. For Einstein was, intuitively, anticipating his later general theory, according to which the equatorial clock runs slower because of the centripetal force exerted upon it.

  8. Whitrow’s answer For a supporter of relativity, the essential difference between the two clocks is that relative to the centre of the Earth (which for the purpose concerned can be regarded as the origin of an inertial frame) the clock at the equator describes a circle and so cannot be associated with an inertial frame, whereas the polar clock is at rest and can be associated with an inertial frame for a period of time during which the curvature of the Earth’s orbit can be neglected.

  9. Maddox’s answer It seems now to be accepted that Einstein’s original argument was uncharacteristically loose. The point of the illustration is that a clock at the pole of rotation may be taken to be in an inertial frame which is nearly (but not quite) properly defined by the direction of the Earth’s motion around the Sun. The clock at the equator is in another. Einstein’s lack of clarity concerns the inertial frame of the observer of the two clocks.

  10. An Inconsistency in Special Relativity (1990)

  11. Good’s answer (1998) “In any case, Einstein’s statement was at least a slip in exposition. I call it ‘Einstein’s slip.’” “… Einstein’s slip, when taken literally, immediately contradicts KSTR. All of the inconsistency, claimed by McCausland, resided in Einstein’s slip; none of it resided in KSTR.” (KSTR = the Kinematics of the Special Theory of Relativity.)

  12. From the Abstract of Good’s last paper in the debate (2006) “What I previously called ‘Einstein’s Slip’ wasn’t a slip after all: … This correction doesn’t imply an inconsistency in KSTR, although it would appear to do so to those who regard time travel as logically impossible.”

  13. Properties of an inconsistent theory T p · q means p and q; p v q means p or q; ~p meansnot-p Suppose the postulates p and q of theory T are inconsistent, as shown by ~(p · q) p  q  ~(p · q) From the postulates p and q it follows that p ∙ q. Using the disjunctive syllogism: (p ∙ q)v s where s is any proposition whatever. ~(p · q) Therefore s

  14. How Einstein showed that a moving clock runs slow

  15. Dingle’s Equation Lorentz transformation: t' = γ(t –vx/c2) whereγ = (1 – v2/c2)-1/2 Δt'/Δt = γ – (γv/c2)(Δx/Δt) where ΔxandΔtare the space and time intervals between two arbitrary events

  16. Good’s Rebuttal Δt'/Δt = γ – (γv/c2)(Δx/Δt) “But actually both sides of the equation are equal to (1 – v2/c2)1/2 whatever the value of v and whatever events we choose, assuming of course that we are referring to a pair of clocks for which the one has velocity v relative to the other.”

  17. Jack Good1916-2009

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