Pete Posted January 21, 2009 Share Posted January 21, 2009 More fundamentally, I think you can to an extent think of things "bending in time" viz [math]R_{tt}[/math]= "amount of bending in time" say, i.e. the time-time component of the Ricci tensor. Why not? The term bending is not one that is well defined in the context of general relativity. It cannot be said that matter curves time because no meaning can be applied to the curvature of 1-dimensional manifold. However I've seen relativists use the term "bend" to refer to gravitational time dilation. That usage also applies in the absense of spatial curvature. Link to comment Share on other sites More sharing options...

ajb Posted January 21, 2009 Share Posted January 21, 2009 The term bending is not one that is well defined in the context of general relativity. It cannot be said that matter curves time because no meaning can be applied to the curvature of 1-dimensional manifold. If you pick local coordinates [math]\{t, x,y,z\}[/math] then why not say something about curvature in the [math]t[/math] direction or something like that? My only objection is of course this would not be an invariant notion. And of course that casts doubt on to any usefulness. Link to comment Share on other sites More sharing options...

moth Posted January 21, 2009 Author Share Posted January 21, 2009 thank you all for your replies. i'm starting to see why bend in time is not the best terminology. interesting universe we live in. i can't believe i cut and pasted that formula( 2*pi*r*(1-v^2)^-1/2)to make sure i got it right but it's still miswritten i gotta be more careful. Link to comment Share on other sites More sharing options...

Pete Posted January 21, 2009 Share Posted January 21, 2009 If you pick local coordinates [math]\{t, x,y,z\}[/math] then why not say something about curvature in the [math]t[/math] direction or something like that? Can you please provide an illustrative example? Link to comment Share on other sites More sharing options...

ajb Posted January 21, 2009 Share Posted January 21, 2009 (edited) Can you please provide an illustrative example? Do I need to? You suggest some familiarity with GR. Have a look at the PlanetMath areticle on the Ricci tensor. Edited January 21, 2009 by ajb Link to comment Share on other sites More sharing options...

moth Posted January 22, 2009 Author Share Posted January 22, 2009 thanks for the planetmath link ajb, i've been looking for something like that. i recently read about Hinton (inventor of the tesseract) who came up with words to describe motion in 4 dimensions ana and kata (up and down or left and right)maybe i can invent a word that means the curve that matter gets from momentum(other than bend) and take it on the road. Link to comment Share on other sites More sharing options...

Pete Posted January 22, 2009 Share Posted January 22, 2009 (edited) Do I need to? Yes. You suggest some familiarity with GR. Have a look at the PlanetMath areticle on the Ricci tensor. I'll read more about that in the future but for now lets address your definition [math]R_{tt}[/math]= "amount of bending in time" Since the [math]R_{tt}[/math] = 0 in a vacuum then "amount of bending in time" = 0 in a vacuum. What is that suppposed to be telling us? [math]R_{tt}[/math] = 0 comes from Einstein's field equation written in the form (let [math]k = 8\piG/c^4[/math] [math]R_{\mu\nu} = -k(T_{\mu\nu} - (1/2)g_{\mu\nu}T)[/math] In vacuum [math]T_{\mu\nu} = 0[/math] and therefore T = 0 giving [math]R_{\mu\nu}[/math] = 0 and therefore [math]R_{tt}[/math] = 0. Edited January 22, 2009 by Pmb Link to comment Share on other sites More sharing options...

moth Posted January 22, 2009 Author Share Posted January 22, 2009 pmb "Since the R_tt= 0 in a vacuum then "amount of bending in time" = 0 in a vacuum." that seems okay if a vacuum means no energy.without ether what could bend in a vacuum? Link to comment Share on other sites More sharing options...

Pete Posted January 22, 2009 Share Posted January 22, 2009 pmb "Since the R_tt= 0 in a vacuum then "amount of bending in time" = 0 in a vacuum."that seems okay if a vacuum means no energy.without ether what could bend in a vacuum? By vacuum I mean [math]T_{\mu\nu}[/math] = 0 for all [math]\mu,\nu[/math]. Link to comment Share on other sites More sharing options...

ajb Posted January 23, 2009 Share Posted January 23, 2009 Yes. I'll read more about that in the future but for now lets address your definition [math]R_{tt}[/math]= "amount of bending in time" Since the [math]R_{tt}[/math] = 0 in a vacuum then "amount of bending in time" = 0 in a vacuum. What is that suppposed to be telling us? [math]R_{tt}[/math] = 0 comes from Einstein's field equation written in the form (let [math]k = 8\piG/c^4[/math] [math]R_{\mu\nu} = -k(T_{\mu\nu} - (1/2)g_{\mu\nu}T)[/math] In vacuum [math]T_{\mu\nu} = 0[/math] and therefore T = 0 giving [math]R_{\mu\nu}[/math] = 0 and therefore [math]R_{tt}[/math] = 0. I thought we were talking about non-vacuum solutions, i.e. the bending of matter! Link to comment Share on other sites More sharing options...

Pete Posted January 23, 2009 Share Posted January 23, 2009 I thought we were talking about non-vacuum solutions, i.e. the bending of matter!Who said that? They were talking about gravitational time dilation, right? For the sake of arguement consider a black hole. Everywhere outside the event horizon the spacetime is curved and there is no matter present. The Ricci tensor vanishes. I believe that the question about bending time applies to outside the event horizon. Link to comment Share on other sites More sharing options...

ajb Posted January 23, 2009 Share Posted January 23, 2009 I guess it is not completely clear if we are talking about a "test object" in the gravitational field of some source, i.e. in vacua or if we are talking about how the energy-momentum of an object "bends itself", principally in "time". If we are talking about the first the yes, the Ricci tensor vanishes. However, if we are talking about the latter then this is not the case. Try a little exercise. Convince yourself that on an Einstein manifold, (pick some coordinates) that the field equations give you that [math]R_{\mu \nu} = \lambda T_{\mu \nu}[/math] where [math]\lambda[/math] is a constant that you can calculate explicitly for a given Einstein manifold. Then, if we use my "definition" of time bending we have "Bending in time" = "Energy density". Which I thought was an ok "result". But I still would be very careful about how to really interpret what I have said. But the point being I can and have calculated something! Link to comment Share on other sites More sharing options...

Pete Posted January 23, 2009 Share Posted January 23, 2009 (edited) Actually I'm not even clear on what this about anymore since I can't imagine what "matter bent in time" could possibly mean. When defining new terms one starts with an idea and then forms a term around it. This is different. We are trying to start with a term and define an idea around it. Does this sound accurate? Edited January 23, 2009 by Pmb Link to comment Share on other sites More sharing options...

moth Posted January 23, 2009 Author Share Posted January 23, 2009 (edited) ajb "I guess it is not completely clear if we are talking about a "test object" in the gravitational field of some source, i.e. in vacua or if we are talking about how the energy-momentum of an object "bends itself", principally in "time"." i was thinking about the latter, but i can see that it's not so simple to swap time and space coordinates as though they are the same. Pmb bent in time would be like saying a vertical arch like over a door in a wall was bent in height,or a horizontal arch like a dam was bent in depth.it's clumsy, but it seems similar to what is going on in Lorentz contraction. Edited January 24, 2009 by moth Link to comment Share on other sites More sharing options...

Pete Posted January 24, 2009 Share Posted January 24, 2009 ... but i can see that it's not so simple to swap time and space coordinates as though they are the same.Absolutely not. Only in a mathematical sense are they treated on the same footing, not in a physical sense. Even mathematically there is a bit of a difference due to a negative sign in front of only one diagonal term in the Minkowski metric. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now