The proof of mass vector.
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From: Phineas T Puddleduck on 4 Jan 2007 20:48 In article <1167961542.832028.325870(a)v33g2000cwv.googlegroups.com>, "Barry" <Sirdry(a)hotmail.com> wrote: > > While you're here, can I ask if you've made any progress with your > ideas that a sphere contracts to a point when it isn't moving and that > all the points on a sphere's surface can't exist at the same time? What? -- This space reserved for Jeff Relf's 5-dimensional metric. -- Posted via a free Usenet account from http://www.teranews.com
From: ItsMe on 5 Jan 2007 14:15 Phineas T Puddleduck wrote: > "Barry" wrote: > > > > > While you're here, can I ask if you've made any progress with your > > ideas that a sphere contracts to a point when it isn't moving and that > > all the points on a sphere's surface can't exist at the same time? > > What? > Forgotten already? It was just 2 days ago. I was referring to the following exchange in the "Frequency and invariance, some trivialities" thread.: _________________________ I had been obliged to explain carefully to some pedantic poster that: In the rest frame of the sphere, all points on the sphere are at rest wrt each other. Hence, according to SR, the coordinate temporal distance between all those points, in that frame, is zero. It follows directly from the L-F contraction. For v = 0, ds^2 = dl^2 hence dt^2 = 0 And you had disagreed, writing: If you mean v =0 then dx=dy=dz =0 ds^2 = - dt^2 ________________________ Which response was not correct. To say that in the rest frame of the sphere, all points on the sphere are at rest wrt each other doesn't mean that all points on its surface are located at the same spatial point. It means that they all share the same coordinate time in the rest frame. Do you really think that a sphere contracts to a point just because it isn't moving? In measuring the length (l) of an object that is stationary in an inertial frame, we measure the position of both ends of the object simultaneously, so dt^2 = 0 ds^2 = - dt^2 +dx^2 + dy^2 + dz^2 so ds^2 = 0 + dx^2 + dy^2 + dz^2 (because of the simultaneity) so ds^2 = dl^2 (because of pythagoras) Which is what I wrote. You might find that kooky, but it's pretty elementary Barry
From: Phineas T Puddleduck on 5 Jan 2007 15:46 In article <1168024531.777403.78880(a)i15g2000cwa.googlegroups.com>, "ItsMe" <Sirdry(a)hotmail.com> wrote: > _________________________ > I had been obliged to explain carefully to some pedantic poster that: > > In the rest frame of the sphere, all points on the sphere are at rest > wrt each other. Hence, according to SR, the coordinate temporal > distance between all those points, in that frame, is zero. > > It follows directly from the L-F contraction. > > For v = 0, > > ds^2 = dl^2 > > hence dt^2 = 0 > > And you had disagreed, writing: > > If you mean v =0 then dx=dy=dz =0 > > ds^2 = - dt^2 I think the confusion here is whether dx is the rate of movement of a position 'x' on the sphere or a position a small distance from x. If its a rate of movement then you should talk of dx/dt or dx/dtau. dx on its own is a small offset position. -- This space reserved for Jeff Relf's 5-dimensional metric. -- Posted via a free Usenet account from http://www.teranews.com
From: Phineas T Puddleduck on 5 Jan 2007 15:49 In article <phineaspuddleduck-BB1493.20461705012007(a)free.teranews.com>, Phineas T Puddleduck <phineaspuddleduck(a)googlemail.com> wrote: > In article <1168024531.777403.78880(a)i15g2000cwa.googlegroups.com>, > "ItsMe" <Sirdry(a)hotmail.com> wrote: > > > _________________________ > > I had been obliged to explain carefully to some pedantic poster that: > > > > In the rest frame of the sphere, all points on the sphere are at rest > > wrt each other. Hence, according to SR, the coordinate temporal > > distance between all those points, in that frame, is zero. > > > > It follows directly from the L-F contraction. > > > > For v = 0, > > > > ds^2 = dl^2 > > > > hence dt^2 = 0 > > > > And you had disagreed, writing: > > > > If you mean v =0 then dx=dy=dz =0 > > > > ds^2 = - dt^2 > > > I think the confusion here is whether dx is the rate of movement of a > position 'x' on the sphere or a position a small distance from x. > > If its a rate of movement then you should talk of dx/dt or dx/dtau. dx > on its own is a small offset position. Following on - what does this mean for an area vector? At the end of the day, it doesn't help you define an area vector in a general case? -- This space reserved for Jeff Relf's 5-dimensional metric. -- Posted via a free Usenet account from http://www.teranews.com
From: ItsMe on 6 Jan 2007 08:50
Phineas T Puddleduck wrote: > In article <1168024531.777403.78880(a)i15g2000cwa.googlegroups.com>, > "ItsMe" <Sirdry(a)hotmail.com> wrote: > > > _________________________ > > I had been obliged to explain carefully to some pedantic poster that: > > > > In the rest frame of the sphere, all points on the sphere are at rest > > wrt each other. Hence, according to SR, the coordinate temporal > > distance between all those points, in that frame, is zero. > > > > It follows directly from the L-F contraction. > > > > For v = 0, > > > > ds^2 = dl^2 > > > > hence dt^2 = 0 > > > > And you had disagreed, writing: > > > > If you mean v =0 then dx=dy=dz =0 > > > > ds^2 = - dt^2 > > > I think the confusion here is whether dx is the rate of movement of a > position 'x' on the sphere or a position a small distance from x. > I wrote it in English first: ________________ In the rest frame of the sphere, all points on the sphere are at rest wrt each other. Hence, according to SR, the coordinate temporal distance between all those points, in that frame, is zero. ________________ There's no room for confusion, since no "movement" is involved. Barry |