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Streesenergymomentum tensor 
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#19
Dec3112, 08:38 PM

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The second "locally the metric is the Minkowski flat metric up to a second order term" shows that the EP is true only at a point, and only up to first derivatives of the metric. It's the second derivatives that are crucial in distinguishing curved and flat spacetime. This is why in writing the matter action that Pervect and I talked about in posts 14 and 15, the matter fields must be coupled to the metric and not its derivatives. This prescription from the equivalence principle is also called "minimal coupling". When you differentiate the full GR action L_{EH}+L_{matter} wrt the various fields, you then get the Einstein field equation with the correct definition of the stress tensor, and the equations of motion for the various matter fields. Because the equations came from an action in which matter was minimally coupled to the metric, the stress tensor derived in this way obeys the EP in the sense that in coordinates in which the metric (but not its second derivatives) have the same form as a Lorentz inertial frame, the components of the stressenergy tensor then have their SR meaning. 


#20
Dec3112, 08:46 PM

P: 665

LL pseudopotential is related to curvature via Einstein field equations with some modifications. More technically, it is related with Ricci tensor density and scalar curvature density through an identity derived according to Noether's prescription for finding the conserved quantities of a Lagrangian which is a scalar density. I hope this is enough. If you need more details, we can discuss it. P 


#21
Dec3112, 09:12 PM

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#22
Dec3112, 09:13 PM

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#23
Dec3112, 09:29 PM

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#24
Dec3112, 09:34 PM

C. Spirit
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#25
Dec3112, 11:30 PM

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#26
Dec3112, 11:49 PM

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You can see it says it kinda helps to understand how the tangent space may be seen as a 'linearisation' of the manifold. This makes sense if you think about a sphere and the tangent plane at the north pole. You can imagine that even if the only point of contact between the sphere and the plane is the north pole, you may describe very well the sphere around the pole by considering points on the tangent plane. This is my way of thinking about it. 


#27
Jan113, 12:06 AM

C. Spirit
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#28
Jan113, 06:06 AM

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*My exposure to both GR and DiffGeo was minimal a while ago (and has not improved significantly in the mean time), the course was based on the 75 page brochure by Dirac! :D 


#29
Jan113, 06:12 AM

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#30
Jan113, 07:49 AM

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pervect: Your post #17 was helpful.
In general I depend on you experts for current usage of terminology as well as physical interpretations of detailed mathematics. [I've been out of school way too long to try and catch up!] from your post: That term comes from PeterDonis and several old threads where you and DocAl and probably others were kind enough to help me understand spacetime 'gravitational curvature'. I came away with the understanding ['real'] gravitational observations relating to a rapidly moving massive body can be answered as if the body is stationary so that relative velocities are NOT considered as part of gravitational curvature... One aspect of THAT logic is the one I posted already: fast moving particles don't become black holes. A related perspective would be that for a single electron, as an example, the rest energy density of the electron is the only thing that causes spacetime gravitational curvature. The kinetic energy is framedependent, just as the velocity is....and does not contribute. So 'velocity' does not yet enter into my understanding of 'the amount of gravity'. Here is how docAl explained it from an old thread [ edited by me for brevity]: If a flat sheet of graph paper represents two dimensional space without gravity, with the introduction of gravitation the paper itself becomes curved. [Curvature that cannot be "flattened" without distortion.] Gravitational "spacetime curvature" refers to this curvature of the graph paper, regardless of observer, whereas visible/perceived curvature in space is related to distorted, nonsquare grid lines drawn on the curved graph paper, and depends on the frame choice of the observer...." So additional relative velocity DOES cause physical effects,as your quote shows, but THAT curvature was not considered 'gravitational curvature'...That is, the 'amount of gravity' ..... So while I believe the above is self consistent, I really do not know if such terminology is generally understood that way in these forums. And when I first posted in this thread I was unsure if the OP, got the answer he thought he did. PS: best thread on gravity in a while!! 


#31
Jan113, 08:46 AM

P: 5,632

kevin: You might find this discussion of interest:
Spacetime Curvature Observer and/or Coordinate Dependent? http://www.physicsforums.com/showthread.php?t=596224 April, 2012 "In general relativity, knowing all about the sources (the stressenergy tensor T) isn't enough to tell you all about the curvature." which complements the MTW quote. 


#32
Jan113, 09:36 AM

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#33
Jan113, 09:56 AM

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#34
Jan113, 10:31 AM

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Your first sentence induces the set inclusion {vector spaces} [itex] \subset [/itex] {topological spaces} which is not not correct (the axioms of a vector space don't mention a norm, so there wouldn't be any norminduced topology ). The right set connection is: {topological vector spaces} = {vector spaces} [itex] \cap [/itex] {topological spaces} 


#35
Jan113, 10:34 AM

P: 5,632

Let's consider an object moving in free fall relative to an observer following a particular path. If an identical particle at a different velocity, or an identical particle with the same velocity but also angular momentum then passes the observer, the particles will in general follow three different paths. That's the physics ,I think we can agree. In the view I gave, the second particle with angular momentum changes gravitational spacetime curvature relative to the first; the third particle with only additional velocity has a different 'visible curvature'....but it is not part of 'the amount of gravity'. That is just a convention, but one that seems to make sense to me regarding fast moving particles not becoming black holes. 


#36
Jan113, 10:49 AM

P: 123

Anyway, thanks for clearing that up for me. But that's troubling, as in the wikipedia article of the exponential map it is said explicitly 


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