On Fri, Dec 6, 2024 at 7:45 AM John Clark <[email protected]> wrote:
> On Fri, Dec 6, 2024 at 12:10 AM Alan Grayson <[email protected]> > wrote: > > *>> from the garage man's POV the garage's length does not shrink but the >> car's length does. In Special Relativity time is diluted by the factor γ >> which is equal to 1 / √(1 - v²/c²) ; and an object's length will be >> reduced by a factor of the inverse of γ. So Length contraction reduces the >> length by 1/γ, and Time Dilation increases the time interval by γ. For >> example, at 87% the speed of light length contracts to half its original >> rest length, and time dilutes by a factor of two.* >> *The bottom line is that when two observers are in relative motion, like >> the garage man and the car driver are, they measure space and time >> differently. An event has a position and a time, and the closing of both >> garage doors is an event, so they will not agree if that event happened >> simultaneously when the entire car was in the garage or not.* >> >> >> > >> *> I don't think your proposed solution works. We're assuming the rest >> frame length of the car is larger than the rest frame length of the garage.* >> > > > *As Jesse Mazer points out, if the car fits in the car driver's frame of > reference then it always fits in the garage man's frame of reference. > However if it doesn't fit in the garage men's frame of reference then it > won't fit in the driver's frame of reference either; this can happen if the > car is not going fast enough, and the asymmetry between the two viewpoints > occurs because when the car driver and the garage man and the car and the > garage are all in the same frame of reference (a.k.a. they are not moving > with respect to each other) then they both agree that the car is longer > than the garage. So there is never a contradiction, there is never an > occasion where one of them predicts the car will fit in the garage and the > other predicts it will not. * > You didn't really answer my question before about whether you think there is any way to define the phrase "fits in the garage" in a way that doesn't involve questions of simultaneity. If we do use a definition involving simultaneity, the natural one is to look at the two localized events A="back end of the car passes by the front door of garage" and B="front end of the car crashes into back wall of garage" (assuming the car does not brake so that everything is inertial up to the moment of the crash). In a frame where the crash B happens *after* the back end of the car entering the garage A, there will be some interval of time where the car is fully inside the garage and it hasn't yet crashed. In a frame where B happens *before* A, the car never fit in the garage because the front end crashed into the back wall before the back end had entered the garage. When you say there is never a contradiction, do you deny we can pick values for the rest length of the car and garage and their relative velocity such if we use the Lorentz transformation, we find that B happens before A in the car rest frame (so the car doesn't fit in that frame), but A happens before B in the garage rest frame (so the car does fit in that frame)? Or do you accept that point, but think there is some other way to define the notion of "fits in the garage" that doesn't involve questions of simultaneity? If the latter you need to provide your own criterion. Jesse -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/CAPCWU3%2B0SQgBdHCmOF8zQm4CA72PgKWGvLCY4NG38Qt%3DzgV9Tw%40mail.gmail.com.
