Question on Spacetime dilation.
in [Relativity]
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From: Ken S. Tucker on 19 Dec 2008 10:53 Hi Alen. On Dec 19, 5:02 am, Alen <al...(a)westserv.net.au> wrote: > On Dec 19, 9:12 am, "Ken S. Tucker" <dynam...(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 > > > e^1.e_1 = -1 , e^0.e_0 = +1 , e^1.e_0 = 0 , etc. > > > gives 3 values for the Kronecker Delta. > > That's what I need to correct and better understand. > > Thanks, comments appreciated. > > Ken S. Tucker > > I would think that, if > > delta^u_v = e^u.e_v > > then e_v = 1/e^v Wow Alen, that's an original way of writing and looks correct to me, haven't seen it done like that before! > so you will always get delta^u_v = +1 only, > or 0 otherwise? That's the way it should be, but I cannot find anyone who can produce a logically consistent solution - seamless to general tensor calculus. Fortunately superior methods exist. > Alen Seasons Greetings Ken S. Tucker
From: Tom Roberts on 19 Dec 2008 14:46 Ken S. Tucker wrote: > Alen wrote: >> I would think that, if >> delta^u_v = e^u.e_v >> then e_v = 1/e^v > > Wow Alen, that's an original way of writing > and looks correct to me, haven't seen it done > like that before! There is a very good reason for not having seen it before -- it is nonsense! This is what happens when you use notation like a black box, without understanding what the symbols actually represent. In particular: * you are confusing indexes that label components of a tensor with indexes that label basis vectors. * you are confusing the Kronecker delta (which is a matrix of real numbers) with the mixed metric components (which are the components of a rank (1,1) tensor). Both are sometimes notated as delta^u_v which probably contributed to your confusion (though the mixed metric components are more commonly notated g^u_v). * that division written above has no real meaning (e^v is a basis vector, which is not an element of any division algebra). Ken, you do this all the time, and it completely invalidates >90% of what you write around here. I rarely respond, as you clearly do not understand the issues, and prefer to rant and rave rather than learn something. Hint: write down what EVERY symbol means, including what type it is and the type(s) of its argument(s), if any. In particular, what is the "." in your "delta^u_v = e^u.e_v"??? -- there is no normal dot product defined between a vector (e_v) and a covector (e^v). And what are the "/" and the "=" in "e_v = 1/e^v"??? ["=" cannot be equality when applied to incommensurate types.] Tom Roberts
From: Ken S. Tucker on 19 Dec 2008 20:21 Hi Tom. You have limited reading comprehension skills, (sorry the truth hurts), about on my level when I was in Grade 7, try to improve. On Dec 19, 11:46 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > Ken S. Tucker wrote: > > Alen wrote: > >> I would think that, if > >> delta^u_v = e^u.e_v > >> then e_v = 1/e^v > > > Wow Alen, that's an original way of writing > > and looks correct to me, haven't seen it done > > like that before! > > There is a very good reason for not having seen it before -- it is > nonsense! Once again Tom, your limited exposure and under- standing of REAL math becomes apparent, as you failed to account for the "domain of applicability", i.e. unit orthogonal basis vectors. Myself I would NOT do it that way, but I've seen some genius mathematicians who know more than I do, and way more than you Tom, take that type of approach, it's called initiative. > This is what happens when you use notation like a black box, > without understanding what the symbols actually represent. In particular: > * you are confusing indexes that label components of a tensor > with indexes that label basis vectors. > * you are confusing the Kronecker delta (which is a matrix of > real numbers) with the mixed metric components (which are the > components of a rank (1,1) tensor). Both are sometimes notated > as delta^u_v which probably contributed to your confusion (though > the mixed metric components are more commonly notated g^u_v). > * that division written above has no real meaning (e^v is a basis > vector, which is not an element of any division algebra). > > Ken, you do this all the time, Tom, you FAILED to read what I wrote, read "haven't seen it done like that before!" what does that mean?? > and it completely invalidates >90% of > what you write around here. LOL, Since you're obviously illiterate that is likely to occur, and benefits me. >I rarely respond, as you clearly do not > understand the issues, How would you know Tom, you can't even read. > and prefer to rant and rave rather than learn > something. Where is this so-called "rant and rave"? > Hint: write down what EVERY symbol means, including what > type it is and the type(s) of its argument(s), if any. > In particular, what is the "." in your > "delta^u_v = e^u.e_v"??? -- there is no normal dot product Of course there is, you want to understand a simple definition of the metric, g_uv = e_u.e_v g^u_v = e^u.e_v , Kronecker delta. g^uv = e^u.e^v . > defined between a vector (e_v) and a covector (e^v). And > what are the "/" and the "=" in "e_v = 1/e^v"??? ["=" > cannot be equality when applied to incommensurate types.] Again I'll stress, within "the domain of applicabilty" Alen's notation looks ok to me. In GR it's fairly common to approximate g^00 = 1/g_00 , g^11 = 1/g_11 that works 99.9% of the time in weak fields, and is where Alen's suggested notation will lead to. Unlike you Tom, I'm NOT a goose-stepping idiot who automatically assumes an unfamiliar suggestion is verboten, on the contrary, I respected my fellow poster's (Alen) suggestion, and after consideration, and wide experience with variations of notation, find it reasonable, that's my expert opinion. A few century's ago the sqrt(-1) was considered near cranky, I can provide many examples like that in the evolution of mathematics. Using Tom's attitude and his ilk, only Roman Numerals would be allowed if I allowed him to stifle progress, but I won't. Maybe Alen is a math genious, and knows more than Tom does, but that's easy to do anyway. Regards Ken S. Tucker
From: Koobee Wublee on 19 Dec 2008 20:55 On Dec 16, 11:57 am, "Ken S. Tucker" 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}, OK. As a reference, this is equation [1]. > and then by association equatable to > > = dx_u dx^u , > > = dx_0 dx^0 + dx_i dx^i , {i=1,2,3} , Eq.(1). No. > 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)? Try to divide both sides of equation [1] by the time coordinate x^0. In which, you get (ds/dx^0)^2 = g_uv dx^u/dx^0 dx^v/dx^0 Voila! Thats it. You can also write it as (ds/dx^0)^2 = g_00 (1 + g_uv/g_00 dx^u/dx^0 dx^v/dx^0) For example, if discussing Schwarzschild spacetime where ds^2 = c^2 (1 2 U) dt^2 dr^2 / (1 2 U) r^2 dO^2 Where ** U = G M / c^2 / r ** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2 ** dx^0 = c dt ** dx^1 = dr ** dx^2 = dLongitude ** dx^3 = dLatitude ** g_00 = 1 - 2 U ** g_11 = - 1 / (1 2 U) ** g_22 = - r^2 cos^2(Latitude) ** g_33 = - r^2 Then, (ds/dt)^2 / c^2 = (1 2 U) (1 (dr/dt)^2 / (1 2 U)^2 / c^2 r^2 (dO/dt)^2 / (1 2 U) / c^2) In flat spacetime, U = 0 You end up with (ds/dt)^2 / c^2 = 1 (dr/dt)^2 / c^2 r^2 (dO/dt)^2 / c^2 Is this what you want?
From: Ken S. Tucker on 19 Dec 2008 21:08
On Dec 19, 5:55 pm, Koobee Wublee <koobee.wub...(a)gmail.com> wrote: > On Dec 16, 11:57 am, "Ken S. Tucker" 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}, > > OK. As a reference, this is equation [1]. > > > and then by association equatable to > > > = dx_u dx^u , > > > = dx_0 dx^0 + dx_i dx^i , {i=1,2,3} , Eq.(1). > > No. Yup:-), just vanilla summation. Ken > > 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)? > > Try to divide both sides of equation [1] by the time coordinate x^0. > In which, you get > > (ds/dx^0)^2 = g_uv dx^u/dx^0 dx^v/dx^0 > > Voila! Thats it. You can also write it as > > (ds/dx^0)^2 = g_00 (1 + g_uv/g_00 dx^u/dx^0 dx^v/dx^0) > > For example, if discussing Schwarzschild spacetime where > > ds^2 = c^2 (1 2 U) dt^2 dr^2 / (1 2 U) r^2 dO^2 > > Where > > ** U = G M / c^2 / r > ** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2 > ** dx^0 = c dt > ** dx^1 = dr > ** dx^2 = dLongitude > ** dx^3 = dLatitude > ** g_00 = 1 - 2 U > ** g_11 = - 1 / (1 2 U) > ** g_22 = - r^2 cos^2(Latitude) > ** g_33 = - r^2 > > Then, > > (ds/dt)^2 / c^2 = (1 2 U) (1 (dr/dt)^2 / (1 2 U)^2 / c^2 r^2 > (dO/dt)^2 / (1 2 U) / c^2) > > In flat spacetime, > > U = 0 > > You end up with > > (ds/dt)^2 / c^2 = 1 (dr/dt)^2 / c^2 r^2 (dO/dt)^2 / c^2 > > Is this what you want? |