Question on Spacetime dilation.
in [Relativity]
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From: Ken S. Tucker on 21 Dec 2008 10:38 Hi Alen. On Dec 20, 6:14 pm, Alen <al...(a)westserv.net.au> wrote: > On Dec 20, 12:21 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > [...] > > Maybe Alen is a math genious, and knows more > > than Tom does, but that's easy to do anyway. > > Regards > > Ken S. Tucker > > I don't think I have to be a genius, and don't see > what the fuss is about, although I know that Tom > always likes to insist on 'absolute' precision in > notation. Even so, suppose we represent a vector > x + iy as A^i and define > > A^iA_i = 1, > then we will have > A_i = (1/|A^i|^2)(x - iy) > > If you multiply them you will get 1. Therefore we > should be able to meaningfully say > A_i = 1/A^i > > since we can say > A_iX = X/A^i and, if X = A^i, > A_iA^i = A^i/A^i = 1 Careful, even vanilla Minkowski spacetime has sqrt(-1) all over the damn place. Set A_i = i * |A_i| and A^i = i * |A^i|, (Does the Einstein summation convention apply?). > There is therefore some kind of meaningful > 'inverse' relationship between A^i and A_i, > in this case, and to realise this is helpful to > complete an understanding of the nature of > their relationship > > Similarly, if we have > x^ix_j = d^i_j, with d for delta, > > we will have > x^ix_i = 1 Is the summation convention suspeneded there? > and this therefore 'must' always confer 'some' > kind of valid meaning on > x_i = 1/x^i > > 'whatever' kind of an object x might refer to. 'Some' > kind of inverse is always a conceptual counterpart > of the simple fact that x^ix_i = 1. Is that not so?? The notation is a bit vague, I'd be happy to see time + 1D such as 0,1 in spacetime. > Alen Regards Ken S. Tucker
From: John Polasek on 21 Dec 2008 23:00 On Tue, 16 Dec 2008 11:57:32 -0800 (PST), "Ken S. Tucker" <dynamics(a)vianet.on.ca> wrote: >A clock (in K) moving at 0.8c (relative to K') is >dilated 0.6 by t' = t*sqrt(1 - v^2/c^2), so that >t'=(0.6)*t. > >In GR that is generalized to be, > >ds^2 =g_uv dx^u dx^v , {u,v=0,1,2,3}, > >and then by association equatable to > >= dx_u dx^u , > >= dx_0 dx^0 + dx_i dx^i , {i=1,2,3} , Eq.(1). Firstly, you are incorporating .8c which is related to SR, and admixing it into GR. Secondly, if you retain GR's guv matrix, even with zero gravity, then your first term should be negative, that is, -dx0dx0. I think I'm right on this. Remember the signature -1 1 1 1. >I expect I should then obtain, > >dt' = ds = (0.6) dt, Eq.(2). > >What differential coefficients should be subbed >into Eq.(1) to yield Eq.(2)? > >TIA >Regards >Ken S. Tucker John Polasek
From: John Polasek on 21 Dec 2008 23:09 On Thu, 18 Dec 2008 14:12:06 -0800 (PST), "Ken S. Tucker" <dynamics(a)vianet.on.ca> wrote: >Hi XXein and guys.... >... >> xxein: Welcome aboard to both you and Allen. I hope you enjoy the >> ride. >I posted this to SPF... > >To Theo and all theoreticians who embrace the signature (+---), >I'm unable to understand it so far, but I'm trying! > >On Dec 18, 12:16 pm, Theo Wollenleben <alpha0...(a)yahoo.de> wrote: >> Ken S. Tucker schrieb: >> > On Dec 16, 2:50 pm, Theo Wollenleben <alpha0...(a)yahoo.de> wrote: >> >> Ken S. Tucker schrieb: >> >>> ds^2 =g_uv dx^u dx^v , {u,v=0,1,2,3}, >> >>> = dx_u dx^u , >> >>> = dx_0 dx^0 + dx_i dx^i , {i=1,2,3} , Eq.(1). >> >>> dt' = ds = (0.6) dt, Eq.(2). >> >>> What differential coefficients should be subbed >> >>> into Eq.(1) to yield Eq.(2)? > >> >> g_uv = diag(1,-1,-1,-1) >> >> dx_0 = dx^0, dx_i = -dx^i >> >> ds^2 = (dx^0)^2 - (dx^i)^2 >> >> ds = sqrt(1 - (dx^i/dx^0)^2)*dx^0 = sqrt(1-(v/c)^2)*cdt > >> > If "dr" is a spatial displacement *vector*, and >> > e^i and e_i are the 3D spatial basis vectors, the >> > text book suggests that, >> > dr = e_i dx^i = e^i dx_i . >> > and also >> > dr^2 = dr.dr , (that's a dot/scalar product) >> > = e_i.e_j dx^i dx^j = g_ij dx^i dx^j >> > = e^i.e^j dx_i dx_j = g^ij dx_i dx_j >> > does that seem reasonable to you Theo and all? > >> Where the scalar product is given by x.y=g(x,y). That is correct, though >> "being spatial" in Minkowski space is not an invariant concept, of >> course. If we choose another basis, the same vector could have a nonzero >> time component. > >Ok, I've studied metrics such as g_i0, however they are not popular, >and I have been advised to understand the signature (1,-1,-1,-1) >you (Theo) introduced above, that are conventionally used by most >relativists and what they mean by that. > >A minor problem I'm stuck on is this, if > >dr = e_i dx^i = e^i dx_i , and dx^i = - dx_i then e_i = -e^i . > >The tensor textbook requires that the Kronecker Delta " delta^u_v " >has values of 1,0 such as > >delta^u_v = {1,0}when {u=v , u=/=v}, defined by > >delta^u_v = e^u.e_v , but I doubt it. The LHS is second rank, and the RHS 0 rank. The equation has no value. >e^1.e_1 = -1 , e^0.e_0 = +1 , e^1.e_0 = 0 , etc. Only the pseudo-tensor gij can produce the above nonsense. The base vectors alone can't pull it off. >gives 3 values for the Kronecker Delta. >That's what I need to correct and better understand. >Thanks, comments appreciated. >Ken S. Tucker > John Polasek
From: Ken S. Tucker on 22 Dec 2008 01:05 On Dec 21, 8:00 pm, John Polasek <jpola...(a)cfl.rr.com> wrote: > On Tue, 16 Dec 2008 11:57:32 -0800 (PST), "Ken S. Tucker" > > <dynam...(a)vianet.on.ca> wrote: > >A clock (in K) moving at 0.8c (relative to K') is > >dilated 0.6 by t' = t*sqrt(1 - v^2/c^2), so that > >t'=(0.6)*t. > > >In GR that is generalized to be, > > >ds^2 =g_uv dx^u dx^v , {u,v=0,1,2,3}, > > >and then by association equatable to > > >= dx_u dx^u , > > >= dx_0 dx^0 + dx_i dx^i , {i=1,2,3} , Eq.(1). > > Firstly, you are incorporating .8c which is related to SR, and > admixing it into GR. "admixing", SR is a GR limit. > Secondly, if you retain GR's guv matrix, even with zero gravity, then > your first term should be negative, that is, -dx0dx0. WHY? We want logic, NOT idle pronouncements. > I think I'm right on this. Prove it. >Remember the signature -1 1 1 1. That's a conjecture I want to see proven. Regards Ken S. Tucker
From: Eric Gisse on 22 Dec 2008 02:00
On Dec 21, 9:05 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: [snip] > >Remember the signature -1 1 1 1. > > That's a conjecture I want to see proven. > Regards > Ken S. Tucker Open an introductory textbook. |