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Is there any particular reason on why I keep getting flat space?

  1. Sep 2, 2014 #1
    I have been deriving multiple Riemann tensors for the past few days (in 2D for simplicity sake) and for some reason, no matter what functions I put in the vector of transformation properties, I always get all 0's for my Riemann tensors.

    Note: By vector of transformation properties, I am referring to the vector R that you differentiate with respect to all the axes in order to derive the basis vectors that you dot product to derive the elements of your metric tensor. For example, the vector of transformation properties for 2D polar coordinates is R= <rcos(θ) , rsin(θ)>. This is what I mean.


    I have tried all sorts of functions within my vector. I have tried:

    R=<rθ , r/θ> where x1 = r and x2= θ

    I've tried:

    R=<sin(r), cos(θ)>

    I've tried:

    R= <er, eθ>

    I have tried:

    R= <sin(θ)cos(ø), sin(θ)sin(ø)>

    etc...

    No matter what, my Riemann tensor comes out all 0.

    Is there a reason for that? Is it because I am choosing relatively simple functions and need to forsake simplicity if I want to see some curvature? Is there some property of the Riemann tensor that guarantees that it will be all 0 under certain conditions that I don't know about?

    Just what kind of functions would it take for me to put inside of that vector for me to be able to derive a Riemann tensor with non zero elements?

    I heard about this object called a 2-sphere that apparently yields curvature of a 2 dimensional manifold. What are the transformation properties that I could use to derive a spatial geometry like that of a 2-sphere?

    It just seems very unlikely that it is a coincidence that I always seem to get 0 matrices for my Riemann tensors.
     
  2. jcsd
  3. Sep 2, 2014 #2

    PeterDonis

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    2016 Award

    Staff: Mentor

    I'm not sure why you would care about the "vector of transformation properties" when you are trying to derive the Riemann tensor. The Riemann tensor is derived from the metric tensor. Can you show some of the calculations you have done?
     
  4. Sep 2, 2014 #3

    pervect

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    Staff Emeritus
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    I don't follow what you mean here, sorry.

    As was previously mentioned, the 2d surface of a sphere isn't flat. So if you compute the Riemann tensor of the surface of a sphere, you should get nonzero values. I'm not sure what you are computing, you may be computing the Riemann tensor of a plane in various coordinate systems and/or basis. It won't matter want coordinates and/or basis vectors you use, a flat plane is - well - flat. Rather than go into what you are doing, I will try to motivate the metric of a surface of a spehre so you can calculate the Riemann tensor and (hopefully) get something that is nonzero.

    Suppose you use the lattitude (##\varphi##) and longitude coordinates (##\lambda##) as we define them on the Earth. (You can use other schemes if you like, I'm choosing these in the hope that they have physical meaning to you. I'm using the notation that Wiki uses, http://en.wikipedia.org/wiki/Longitude, http://en.wikipedia.org/wiki/Lattitude.

    Using radians, we can write the north-south length of a change in lattitude ##d\varphi## as ##R d \varphi##.
    The east-west length of a change in longitude is more complex, it depends on the lattitude. It's equal to ##R \cos \varphi d \lambda##

    Locally, north-south and east-west distances are orthogonal, so we can write the square of the distance between two nearby points (also known as the metric) as the sum of the squares of the length of the east-west distance and the north-south distance which we calculated above.

    ##ds^2 = R^2 d \varphi^2 + R^2 \cos^2 \varphi \, d \lambda^2##

    From this, you should be able to calculate the Riemann, I'm not quite sure what technique you are using so I won't go into those details - it is possible there is some problem in your technique, in that case I don't see any alternative to posting your work.

    For definitness I'd say calculate the Riemann in a coordinate basis, if not do tell us what basis you are using (even if you have to look it up to answer the question).
     
  5. Sep 3, 2014 #4

    pervect

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    Using some automated software, I can tell you what you should get for the Riemann tensor for the line element from my previous post (hopefully you can compare to your work to determine if you get the same results).

    ##ds^2 = r^2 d \varphi^2 + r^2 \cos^2 \varphi \, d \lambda^2##

    Note that I've changed the symbol that represents the radius of the sphere (a scalar constant) to a lower case r to avoid confusion with the symbol R for the Riemann.

    The Christoffel symbols for this line element are:

    ##\Gamma^{\lambda}{}_{{\lambda}{\varphi}} = \Gamma^{\lambda}{}_{{\varphi}{\lambda}} = -\tan \varphi \quad \quad \Gamma^{\varphi}{}_{{\lambda}{\lambda}} = \sin \varphi \cos \varphi##

    Now onto the Riemann. First I'll give the results in a coordinate basis. I'm not sure what notation your textbook might use for basis vectors there is some variance. Nonetheless, it is important to know what your basis vectors are. One common notation is to identify the basis vectors with the coordinate partial derivative operators , i.e. the basis vectors would be ##\frac{\partial}{\partial \varphi}## and ##\frac{\partial}{\partial \lambda}##. The dual of these basis vectors (the dual is a map from a vector to a scalar, so it has a scalar value) are commonly written as ##\boldsymbol{d}\varphi## and ##\boldsymbol{d}\lambda##. The boldface notation isn't universal, the boldface ##\boldsymbol{d}## specifies an operator that is applied to a vector yielding a scalar result, the non-boldface version still yields a scalar result, but it isn't an operator, it's just a scalar result.

    We can then write the four nonzero components of the Riemann in the coordinate basis as

    ##R^{\varphi}{}_{\lambda \varphi \lambda} = -R^{\varphi}{}_{\lambda \lambda \varphi } = \cos^2 \varphi \quad \quad R^{\lambda}{}_{\varphi \lambda \varphi} = -R^{\lambda}{}_{\varphi \varphi \lambda} = 1##

    We can also write it with all lower indices as:

    ##R_{\varphi \lambda \varphi \lambda} = R_{\lambda \varphi \lambda \varphi} = -R_{\varphi \lambda \lambda \varphi} = -R_{\lambda \varphi \varphi \lambda} = r^2 \cos^2 \varphi##

    The components in an orthonormal basis are also interesting.

    Feel free to skip this section if it confuses you, though if you do not yet know about non-coordinate bases such as orthonormal bases, you will eventually want to learn for the physical insight they can provide.

    I find that it's more intuitive to express the basis one-forms, the cobasis (most textbooks will do this as well). The cobasis oneforms have unit length and are equal to ##\hat{\varphi} = r \boldsymbol{d}\varphi## and ##\hat{\lambda} = r \cos \varphi \, \boldsymbol{d}\lambda##. The "hat" notation may not match your textbook, it matches older textbooks such as MTW, and the notation used in vector calculus for unit vectors. The "hatted" vectors always have unit length, but could either be covariant or contravarant depending on how they are used.

    Then we can write in an orthonormal basis

    ##R^{\hat{\varphi}}{}_{\hat{\lambda} \hat{\varphi} \hat{\lambda}} = R^{\hat{\lambda}}{}_{\hat{\varphi} \hat{\lambda} \hat{\varphi}} = -R^{\hat{\lambda}}{}_{\hat{\varphi} \hat{\varphi} \hat{\lambda}} = -R^{\hat{\varphi}}{}_{\hat{\lambda} \hat{\lambda} \hat{\varphi}} = \frac{1}{r^2}##

    With all lower indices in an orthonormal basis there is no difference in the values of the components (a consequence of the basis being orthonormal), i.e.

    ##R_{\hat{\varphi}\hat{\lambda} \hat{\varphi} \hat{\lambda}} = R_{\hat{\lambda}\hat{\varphi} \hat{\lambda} \hat{\varphi}} = -R_{\hat{\lambda} \hat{\varphi} \hat{\varphi} \hat{\lambda}} = -R_{\hat{\varphi} \hat{\lambda} \hat{\lambda} \hat{\varphi}} = \frac{1}{r^2}##

    Note that the expression for the Riemann in an orthonormal basis make it clear that the curvature has units of 1/radius^2, and that it is constant for a sphere and doesn't depend on ##\varphi## or ##\lambda##.
     
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