Question on Spacetime dilation.
in [Relativity]
|
From: Eric Gisse on 17 Dec 2008 12:42 On Dec 17, 4:29 am, Alen <al...(a)westserv.net.au> wrote: > On Dec 17, 2:43 pm, Eric Gisse <jowr...(a)gmail.com> wrote: > > > > > On Dec 16, 6:23 pm, Alen <al...(a)westserv.net.au> wrote: > > > > On Dec 17, 6:57 am, "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). > > > > > 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 > > > > c^2dt'^2 = c^2dt^2 -dx^2 -dy^2 -dz^2 > > > = c^2dt^2 -ds^2 > > > = c^2dt^2(1 - v^2/c^2) > > > so > > > dt' = (1/g)dt, where g = 1/sqrt(1 - v^2/c^2) > > > > I say, of course, that these are time dilation > > > equations only, and that the first equation has > > > been misinterpreted as a metric equation for > > > 100 years now! > > > Except it is a metric equation, regardless of your ability to > > understand. > > Won't you lot be surprised when you all have to eventually > face the prospect that it might not be a metric equation > after all!! The metric is the only quantity being used. So there really is no room to argue. [snip]
From: Ken S. Tucker on 17 Dec 2008 12:58 On Dec 17, 9:13 am, "harry" <harald.NOTTHISvanlin...(a)epfl.ch> wrote: > Hi Ken, > > "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote in messagenews:a9fd67c3-1282-4478-aca0-48bfa1fef236(a)i18g2000prf.googlegroups.com...> Hi Harald > > [...] > > > > >> I'm still not too sure what you mean. Perhaps the presentation? There are > >> roughly two ways of drawing such diagrams; just keep in mind that it's > >> never > >> more than a graphic representation of the math related to measurements. > >> - One presentation, as in that article, is symmetrical at the price of > >> all > >> non-orthogonal axes > >> - One is asymmetrical but the reference axes are orthogonal. > > > Yes, one can convert imaginary to nonorthogonal. > > >> The (older) presentation and a discussion of both is given in the > >> paragraph > >> just above Time dilation (Minkowski diagram in special relativity): > >> "For the graphical translation it has been taken into account that the > >> scales on the inclined axes are different from the Newtonian case > >> described > >> above. To avoid this problem it is recommended that the whole diagram be > >> deformed in such a way that the scales become identical for all axes, > >> eliminating any need to stretch or compress either axis." > > > Understood. > > >> >> *Note: quite some misunderstanding occurs due to poor phrasings such > >> >> as > >> >> in the above article that says that "the space itself is contracted", > >> >> and > >> >> Bell corrected that with his spaceship example: > >> >>http://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox > >> >> Harald > > >> > Thanks for the ref's. > >> > Seasons Greetings > > >> Happy holidays, > >> Harald > > > Well, for now I'm leaning toward unmoderated > > forums. I see your posts in SPF are being > > rejected because you may have said something > > that CF finds threatening to his "teleconnection" > > theory, I see SPF going down the tubes, because > > CF dominates all the other moderators, and has > > a serious rake (up his ...) and agenda. > > > Oh well, SPF was experimental. > > Regards > > Ken S. Tucker > > I doubt that you would have been allowed to post that on SPF, in contrast to > what CF is allowed to post there. ;-) Yeah, check below#, I posted a rebuttal that was rejected because I disproved Chucky , yet provides Mercury's Perihelion Advance. SPF was set-up to air alternative means of calculating reality, in a polite forum, but has now degraded to one that must support Chuckles teleconnection, SPF is going DOA. > Regards, > Harald The good news is, what's rejected by SPF is more interesting than what's accepted! Seasons Greetings Ken S. Tucker Onward!! # On Dec 15, 12:49 pm, Oh No <N...(a)charlesfrancis.wanadoo.co.uk> wrote: .... > There is of course no dispute that you can > use imaginary basis vectors with (++++) signature, but I think your > claim is that you are not doing that. In which case your metric > obviously fails to give you the scalar invariant mass from p^2 = m^2. On the contrary, I'll provide a brief demo of the Advance of Mercury's Perihelion using Modern SpaceTime, for you and fella's pleasure beginning with, p^2 = g^uv p_u p_v , where p_u = p U_u , U_u = dx_u/ds. U_i=0, Eq.(6) herein, http://physics.trak4.com/modern-spacetime.pdf p^2 = g^00 p_0 p_0 , where p_0 is rest mass, see http://physics.trak4.com/MST_Mass-Definition.pdf p ~ p_0 / sqrt(g_00) , g_00 = 1 - 2GM/rc^2 . mc^2 = p_0 dp/dr ~ -(GMm/r^2) * (1 + 3GM/rc^2) The last R.H.S. term is known as the "relativistic force supplement" and rotates Mercury's semi-major axis, and _evidentially_ disproves Charles remark. Ken S. Tucker
From: xxein on 17 Dec 2008 18:31 On Dec 17, 7:07 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > Hi Alen, et al. > > On Dec 16, 7:23 pm, Alen <al...(a)westserv.net.au> wrote: > > > > > > > On Dec 17, 6:57 am, "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). > > > > 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 > > > c^2dt'^2 = c^2dt^2 -dx^2 -dy^2 -dz^2 > > = c^2dt^2 -ds^2 > > = c^2dt^2(1 - v^2/c^2) > > so > > dt' = (1/g)dt, where g = 1/sqrt(1 - v^2/c^2) > > Yes that looks right to me. > > > I say, of course, that these are time dilation > > equations only, and that the first equation has > > been misinterpreted as a metric equation for > > 100 years now! You can write it in the form > > using g_uv, and it works mathematically, with > > g_uvs being 1 and -1, but all this doesn't really > > mean anything of any significance physically. > > Alen > > "significance physically" > It seems 4D spacetime is becoming deeply > ingrained into our equations, possibly > underestimating there complexity. > Ken S. Tucker- Hide quoted text - > > - Show quoted text - xxein: Welcome aboard to both you and Allen. I hope you enjoy the ride.
From: Ken S. Tucker on 18 Dec 2008 17:12 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
From: Alen on 19 Dec 2008 08:02
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 so you will always get delta^u_v = +1 only, or 0 otherwise? Alen |