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A Simple 1D kinematic exercises with metric tensor

  1. May 24, 2017 #1
    Hi All

    I would like to know if there is a way to produce simple one dimensional kinematic exercises with space-time metric tensor different from the Euclidean metric. Examples, if possible, are welcome.

    Best wishes,

    DaTario
     
  2. jcsd
  3. May 25, 2017 #2

    pervect

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    I suppose you could compute the relativistic rocket equation, the equations of motion of a rocket with constant proper acceleration. It's a kinematic problem, and you can use space-time tensor methods - see for instance the treanment in "Gravitation" (Misner, Thorne, Wheeler).
     
  4. May 25, 2017 #3
    It would be an exercise with the relativistic metric, isn't it? What about exercises with a more general 2x2 metric in 1+1 dimensions?
    Is there any chance of this being an easy problem?
     
  5. May 25, 2017 #4

    pervect

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    It depends on your definition of "easy". At the moment, I have no idea what your background is, so I'll try to describe things that I think are needed to do what I suggested. I'm not sure I will make a complete list, but I'll try. If you have all the necessary background, the actual derivation isn't that hard, getting the necessary background may be hard (or at least a lot of work and rather abstract) if you don't have it. I'm assuming that learning on one's own topics that are generally treated in college level undergraduate and graduate courses seems "hard".

    Let's work backwards, and attempt to explain what's needed. If you want to use the space-time metric, which is a tensor, you need some general familiarity with tensors. This is usually a graduate level subject.

    The introductory mathematical topic you'd need to consider learning about tensors would be linear algebra - and an abstract treatment in terms of vector spaces would be highly recommended. This is typically a college level undergraduate course.

    To use tensor methods, you'll need to express not only the space-time metric as a tensor, but the velocity and acceleration also as tensors. The 4-vector form of velocity and acceleration have the necessary tensor properties to work with the metric tensor. The specifics of 4-velocity and 4-accleration are probably undergraduate level, and part of special relativity. They're not particularly "hard", but they seem unfamiliar to a lot of readers on Physics Forums.

    For a textbook treatment of 4-vectors, I'd recommends "Space-time physics" by Taylor and Wheeler. This treats them without much of the formalism of tensors, but you won't necessarily know why you are learning about them unless you have some understanding of tensors. The way I see it, you'd either need to "just go along with it" and learn them because someone says you'll need to, or else you'd need to gain a certain familiarity with tensors to know why you're learning this stuff. It's hard to tell what people know and don't know unless they're willing to talk (and few people want to admit they don't know everything already), but my impression is a lot of people get stuck here, they don't see the need to learn about 4-vectors, because they don't have the even more advanced background to understand why they need to learn about them yet, so they don't learn them, which keeps them from learning why they need them.

    Once you have the background, the equations that describe a uniformly accelerating observer aren't hard. You basically need to know that the magnitude of a 4-velocity is always -1, and that that by assumption, the magnitude of the acceleration 4-acceleration is constant - because that's assumed in the problem defintion. Knowing these two equations is all you need, and the ability to solve them of course. Getting them requires you to know how to compute the magnitude of a 4-vector, which would be written in tensor notation as ##g_{ab} u^a u^b##, where g is the metric tensor, and ##u^a## is a vector. Making sense of this notation requires understanding tensors, again.

    There are some ways to make the solution easier, as I recall MTW does a short proof that the 4-acceleration is always perpendicular to the 4-velocity as part of their derivation of the 4-velocity and 4-acceleration of a uniformly accelerating observer. In tensor notation this is ##g_{ab} u^a a^b = 0##. Again we see the need for understanding some general things about tensors.

    There are a few other minor things along the way. When I say the magnitude of the 4-velocity is always minus one, I am assuming a certain "sign convention". Concisely, I can say that I'm using a "-+++ sign convention", doing a 4 dimensional treatment.

    You asked about a 2-dimensional treamtent, one time and one space. The sign conventions here would be either "-+" or "+-" in the 2d case. It's just a matter of dropping the uneeded spatial dimensions. The sign convention isn't particularly "hard", but it can cause confusion as different texts use different conventions. Basically, we will always be using diagonal metric tensors at this simple stage, the spatial elements will always have an opposite sign from the time element. The sign convention is simply which one (time or space) is positive, and which one is negative.
     
  6. May 25, 2017 #5

    PeterDonis

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    What do you mean by "the relativistic metric"?

    Why would a "more general" metric not be "relativistic"?

    If you genuinely want a metric on a 2-dimensional spacetime instead of a 4-dimensional one, the possibilities are very limited, because the curvature tensor in 2 dimensions only has one independent component.

    If, however, you want to look at a 2-dimensional subspace of a 4-dimensional spacetime, there are more possibilities. For example, you could look at the restriction of the Schwarzschild metric in Kruskal coordinates to the t-r subspace (i.e., leaving out the angular coordinates). This is how the standard spacetime diagram in Kruskal coordinates is drawn (an example is on the Wikipedia page I linked to). Or, more generally, the t-r subspace of any spherically symmetric spacetime, since the spherical symmetry means the angular coordinates play no role in the geometry.
     
  7. May 28, 2017 #6
    PeterDonis, thank you for pointing out the weaknesses of my construction. In your first question I would say that I was trying to refer to the -1+1+1+1 diagonal matrix, which seems to me as the metric tensor representative of the empty space in relativity, when free from gravitational sources.

    In your second question, I believe you are completely right. I have commited a mistake. I was trying to refer, in this case, to metric matrices which are non diagonal, for instance.

    The main reason of my question is that my physics course was very poor in General Relativity, although some acquaintance with tensor was provided (mostly for use in mechanics), in such a way that I have concluded my physics course with the feeling that the failure in the system (educational) was due to the absence of easy exercises on this "moving on the geodesics" stuff. We often hear sentences like: the moon is doing approximately a uniform circular motion but it is free. It is just moving along the geodesic, curved by the presence of the earth, but which makes the moon with its motion an example of the first Newton´s law.

    Another conception I have (perhaps a misconception) is that the metric tensor, when used to just produce this kind of kinematics exercises, is useful only to generate quadratic forms like in ## ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2 ##.
     
  8. May 28, 2017 #7
    Thank you, pervect.
    Could one say that in the lowest level of complexity the metric tensor in GR is basically useful to produce quadratic forms as in ##ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2 ## ?
     
  9. May 28, 2017 #8

    PeterDonis

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    It would probably be helpful to you to work through an introductory GR textbook. Sean Carroll's online lecture notes are here:

    https://arxiv.org/abs/gr-qc/9712019

    They discuss tensors and geodesic motion.

    The metric tensor is a quadratic form. What you wrote down is called a "line element"--an expression for the length of an infinitesimal line segment in spacetime.
     
  10. May 28, 2017 #9
    But, PeterDonis, the quadratic form is the polynomial where as the metric is the tensor, the matrix. As in:
    ##x^T\, A\, x = \sum_{i=1}^n \sum_{j=1}^n a_{ij} \, x_i \, x_j ##.

    Is it correct?
     
  11. May 28, 2017 #10

    PeterDonis

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    This is a matter of terminology. The term "quadratic form" in this context could mean either the "line element" (the expression for ##ds^2##) or the "metric tensor" (the object ##g_{\mu \nu}##).

    The matrix is a representation of the tensor. The tensor itself is an abstract object in linear algebra. IIRC the Carroll lecture notes discuss this distinction.
     
  12. May 29, 2017 #11
    Thank you, PeterDonis.

    Anyway, if one can show me an easy kinematic exercise (with answer, please) on General Relativity (using basically a non trivial metric tensor) I would appreciate.

    Best Regards,
    DaTario
     
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