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Question on wording for paper

  1. Jun 14, 2010 #1
    Question on wording for paper (re-explained in a second post)

    In a paper (undergrad thesis type paper) on GR, I have the statement:

    A vanishing Ricci Tensor, i.e., [tex] R_{\mu \nu}=0 [/tex]

    is not enough to explicitly define a flat space-time. In the case where the manifold being investigated has only three-dimensions, proving that the Ricci tensor vanishes is enough to assume a flat space. However, in a four dimensional space-time, the case of [tex] R_{\mu \nu}=0 [/tex]

    is not sufficient proof of flatness. In a four-dimensional space-time, there are in fact geometries containing curvature that contain vanishing Ricci tensors. These correspond physically to gravitational waves.


    QUESTION:
    This isn't a direct quote (I don't have the file on this computer), but my question is if I can generalize the statement about four dimensions to any dimension higher than three?
    e.g., can I say "The use of a vanishing Ricci tensor as proof of flat space is valid only in three dimensions, and cannot be generalized to dimensions of four or higher."??

    I feel really uncomfortable any time I attempt to explain mathematics with "words." I'm always afraid I'm going to choose the wrong words and make a totally erroneous statement.


    If it makes a difference in the response, section immediately prior to this statement talks about the vanishing of the Christoffel symbols (or Affine connection....again, semantics has me uncomfortable) in a flat space causing the covariant derivative to equal the normal derivative. It also includes the expression


    [tex]
    R_{\mu \nu}=\Gamma ^{\alpha}_{\ \mu \alpha ,\nu}-\Gamma ^{\alpha}_{\mu \nu ,\alpha}=0 [/tex]

    I'm trying to say that just showing that [tex]
    R_{\mu \nu}=0
    [/tex] isn't enough to "prove" a non-curved space-time in four dimensions......and want to know if I can generalize that to any dimension higher than 3. (Perhaps I'm wrong on three dimensions too....I'm stating that [tex] R_{rs}=0 [/tex] IS enough to assume flat space in three dimensions (and again....I don't know if this is generalizable to dimensions less than 3 as well)


    I am just an undergraduate working on a thesis type paper and the professor I'm working with is gone for the week. I don't want to bother him with any more emails than necessary. lol

    *Okay....I give up.....I can never get the "R" to show up in LaTeX form. It happens every time I try to post math. I don't know what I'm doing wrong, but the third equation is supposed to start with (R_uv), not (uv). My apologies. If I'm doing something stupid, can a mod (maybe the one that changed my post to include the R last week) tell me what it is?
    Maybe I should keep it as my calling mark?
    I'll be known as that "stupid 'no-R' guy."
     
    Last edited: Jun 14, 2010
  2. jcsd
  3. Jun 14, 2010 #2
    I do not think that is correct. Consider for example the Schwarzschild spacetime, it has a vanisching Ricci tensor but the spacetime is stationary or, even more restricted, static. That implies, no gravitational waves.

    That depends. In the realm of general relativity it is more or less arbitrary how to define a slice of spacetime as space. However in the general case, even one with more than 4 dimensions, we are clearly outside of the realm of GR and thus what would you define as a spatial slice there?
     
  4. Jun 14, 2010 #3

    nicksauce

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    As mentioned above this is false. The Schwarzchild spacetime is a static spacetime with vanishing Ricci tensor.


    This should be clear if you consider the number of independent component of the Riemann tensor, whose vanishing does imply flatness. In 3 dimensions, both the Riemann tensor and the Ricci tensor both have 6 independent components, while in higher dimensions the Ricci tensor always has fewer independent components than the Riemann tensor.
     
  5. Jun 14, 2010 #4

    George Jones

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    Right. Troponin, can you use Einstein's equation to give a general statement on what the the vanishing of the Ricci tensor implies physically?
    I am not sure, but I think Troponin used the term "space" in the mathematical sense, so that a spacetime is an example of a space. This is how incksauce has interpreted it. Tropnin, this is an example of where mathematics and physics terminologies are not consistent; "space" in general relativity and cosmology often is used in sense that Passionflower has used it.
     
  6. Jun 14, 2010 #5
    For the record, I am NOT saying that a flat space-time, which is known beforehand to have no disturbances of any kind, will have any value for the Ricci tensor other than [tex] R^{\mu \nu}=0 [/tex]. In a Euclidean geometry, or a geometry that we know BEFOREHAND is flat, we can say that definitely, YES, [tex] R_{\mu \nu}=0 [/tex].

    But, simply getting a solution of [tex] R_{\mu \nu}=o [/tex] is NOT enough to say that ABSOLUTELY, in ALL cases, you can state (by definition) that your space must be flat. The definition is not correct in reverse.
    I'm saying that there ARE geometries of four dimensions of space and time in which [tex] R_{\mu \nu}=0 [/tex], and the geometry is NOT flat. That is, a curved four dimensional space time can have a vanishing Ricci tensor, in specific cases.....which correspond physically to gravitational disturbances in an otherwise empty space.

    (CAPS are not meant to indicate YELLING, only emphasis for the particular word. I have no passionate stance on the matter)


    Just to clear up where I got the idea from, it's from this link:
    http://www.youtube.com/user/StanfordUniversity#p/c/6C8BDEEBA6BDC78D/10/XG02Yw7u2mc

    I got bored with going hours with no sounds other than my fingers hitting the computer keys, so I clicked a random video from Stanford's youtube account. It was a lecture on general relativity by Leonard Susskind. The specific digression is at about 1:20:15 or so.

    The specific quote(s) given by Susskind is: (bold or CAPS are where he placed emphasis when speaking)

    "The right statement is that, in regions of space and time in which there is no energy and momentum, the Ricci tensor is equal to zero.
    Now, what does that say?
    In three dimensional space, or two dimensional space, the Ricci tensor being zero is enough to say the space is flat. In four dimensional space time, the Ricci tensor being zero is not enough to tell you space time is flat. There are too many components. Too many independent components.
    What kind of things can you imagine that is going on....let's imagine a space-time that is completely empty of matter. Meaning mass......a space-time that is everywheres [sic] completely empty. Does that mean that it is flat?
    No....it means that the Ricci tensor is everywheres [sic] zero.

    Are there geometries which have Ricci tensors equal to zero everywheres [sic] but are not flat?
    Yes indeed there are.
    ***The bold in this case is my emphasis, not Susskind's***

    Anyone have an idea which those geometries correspond to?"

    {student suggests Euclidean geometry}.......

    "....oh oh..no..no..yes, of course, yes....no..no....geometries that are not FLAT. Euclidean geometries are flat. Are there geometries....are....ahhh....What might they correspond to physically?

    {receives no response from the audience}

    (it corresponds to) Gravitational waves.
    Just gravitational waves moving by. Waves...waves...wave like disturbances in space time which are not trivial.
    For example, there are plane wave gravitational excitations in which the various components of the metric wobble and wiggle, but where the Ricci tensor is zero but the Riemann tensor is not.
    So, there are solutions.....they are usually called vacuum solutions. They are called vacuum solutions because they don't contain any ordinary energy and momentum, but they do contain gravitational waves.....disturbances. Real disturbances."

    *NOTE* He re-affirms the statement, and re-explains it at about 1:35:00 in the video.

    I believe he is talking about Ricci-flat manifolds?





    I've already taken the digression out of the paper. I really dislike using 'words' as a way to define anything, which it seems is being done here. I also don't think the statement is one I could effectively argue (with a full understanding of it) in defense of......


    So, hopefully that clears up what I was saying. I may have worded it poorly in the first post (which I admitted was something that I hate having to deal with....and can easily/often do), but I wasn't speaking from the point that having a flat space tells you that the Ricci tensor vanishes.
    I was presenting the reverse....that a vanishing Ricci tensor in four dimensions did not, in EVERY case, indicate a perfectly flat space time. Gravitational waves could be present, causing the space to no longer be flat.....but where [tex] R_{\mu \nu}=0 [/tex]
     
    Last edited: Jun 14, 2010
  7. Jun 14, 2010 #6
    I am attempting to speak with "physics" terminology (which, unfortunately, may not be good for my case. lol).
    My understanding of GR and Differential Geometry is primarily from physics texts on GR. I have studied mathematical texts on the subjects, but the VAST majority of my understanding has come from physics texts.....so that is where I would be coming from with "words." (which I hate using....because I can never seem to give the same importance to realizing which words are "ambiguous" and which are "proper" as I would with the mathematics......I'm not much of a talker)
     
  8. Jun 14, 2010 #7

    nicksauce

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    You seem to have the right ideas, but just to be perfectly clear, here are the facts:

    -Minkowski spacetime, in any dimension, has vanishing Ricci tensor, and more importantly a vanishing Riemann tensor, and is called 'flat'.

    -A vanishing Ricci tensor in 2 or 3 spacetime dimensions (i.e., 1+1 or 2+1 dimensions) implies a flat spacetime

    -A vanishing Ricci tensor in 4 or higher spacetime dimensions (i.e, 3+1 dimensions) does not necessarily imply a flat spacetime

    -A spacetime with gravitational waves is an example of a non-flat spacetime with vanishing Ricci tensor in 4 spacetime dimensions

    -A non-flat spacetime with vanishing Ricci tensor does not imply gravitational waves, the Schwarzchild spacetime being an obvious counter-example
     
  9. Jun 14, 2010 #8
    Yes, exactly. Thank you for organizing it so concisely.

    My original concern (assuming everything you listed is true), is can I make the generalizations that




    "A vanishing Ricci tensor in 3 OR LESS spacetime dimensions implies flat spacetime."

    ***The emphasis is on the generalization to one dimension, I guess. Susskind listed 2 and 3 dimensions, so assuming his statement is correct, I can already include those***




    And also the generalization that



    "A vanishing Ricci tensor in 4 OR GREATER dimensions of spacetime does not necessarily imply a flat spacetime."

    ***Susskind only gives the example of 4 dimensions. I would assume that it could be generalized to all spacetimes with dimensions greater than 4 (I have no desire to discuss those "spacetimes"), I'm just curious about whether or not the generalization holds for them***
     
    Last edited: Jun 14, 2010
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