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Quote from: old guy on 05/10/2012 18:36:11There is no way a 20 meter probe will fit in a 10 meter bay. What more proof could you ask for?Yes it will... and no it won't!(Surprisingly, both are true!):http://en.wikipedia.org/wiki/Ladder_paradox
There is no way a 20 meter probe will fit in a 10 meter bay. What more proof could you ask for?
If the probe fits in the bay when they are both at relative rest, the probe will always fit in the bay.
The reason is that there better not be a relativistic difference in velocity – which would cause appreciable length contraction - as the probe enters the bay.
So with the relative velocities of the bay and probe near zero, there is no length contraction to consider.
So are we then left with an impossible choice: To redefine 'c' as a variable or to redefine Earth's diameter as a variable? How could the latter vary, given that Earth is in fact a solid, immutable (but for trivially) object?This challenge has consistently been ignored. "For this frame" vs "for that frame" does not answer it. It only addresses how it might me observed and measured variously... which is not the question.Science must not cop out and claim that what we know about Earth (its precise shape) varies with how we look at it. Physics tells us that a change in shape would require application of force, and length contraction theory does not claim to apply any forces.Also:If you insist that it would BE contracted if measured from a very high speed frame, how can physics account for such a shrinkage in a solid, rigid planet? This remains a very sincere and reasonable question requiring an explanation from physics besides “It would appear flattened from the extreme frame.” (No doubt it would.)
QuoteSo with the relative velocities of the bay and probe near zero, there is no length contraction to consider.Right, but the question was, will it fit in our 10 meter bay? We better know the answer before we send out our shuttle to capture it. (No.)
QuoteQuoteSo with the relative velocities of the bay and probe near zero, there is no length contraction to consider.Right, but the question was, will it fit in our 10 meter bay? We better know the answer before we send out our shuttle to capture it. (No.)Absolutely!Exercise due diligence. You know its relative velocity. You know its measurements judged from your frame. Crunch the numbers. You will know if it will fit or not. NO PROBLEM.
Does invariant 'c' require a variable Earth diameter?(Edit: The claim is that " because 'c' is invariant, length is not invariant.")If so, (ed: if Earth's diameter changes) how does physics explain the latter... or would it simply appear to vary from an extreme frame?
You can not have it both ways. That is why I set it up as I did.The probe *appeared to be* 10 meters long from Earth. Since it was approaching Earth at .866c, it was *measured to be* 10 meters, "length contracted" by 1/2 at that velocity. Of course, to retrieve any object in space, the shuttle must match velocities with the object... come alongside. Then the *actual length* of the probe was found to be 20 meters. So it simply will not fit. I have repeated this at least a dozen times. Do you get it yet?
You also know the probe will not fit in the 10m cargo bay of the shuttle because when the shuttle’s relative velocity is also .866c the cargo bay will only accommodate a 5m probe judged from relative rest.
I truly understand your misgivings about “is” and “appears to be”.
The way I reconciled it was, “It ‘is’ when judged from relative rest”.
Here is something else to ponder:Suppose inside of the probe there is a constant light projected from one end to the other end of the probe in the direction of motion. What is the difference in that light when judged from the shuttle, which is moving at the same relative velocity as the probe (.866c) and when judged from relative rest?
I can not get the "SR experts" here to address this challenge.
QuoteI can not get the "SR experts" here to address this challenge.Please restate exactly what the “challenge” is.Thanks,Butch
Regarding this from yesterday: "I just know that the image is not the object"...
I also know that a possible image of a flattened Earth (say with a 1000 mile diameter in the direction of an observer's very high speed travel) would/could only be an appearence, an image of a flattened Earth. We all know that Earth's diameter is not, has never been and will never be 1000 miles.