Gravitational field and Minkowski spacetime

Discussion in 'Physics & Math' started by nimbus, Dec 13, 2013.

  1. nimbus Registered Senior Member

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    129
    I was wondering if someone could tell me if I’m understanding the following correctly?
    This is a quote from Einstein’s book ‘Relativity, the Special and the General Theory.’
    Page 155 under Appendix 5. Year 1952.

    To me, that seems to be saying… Minkowski-space (flat spacetime) can be considered a special case (type1) of a gravitational field.

    I understand, from elsewhere, the Schwarzschild metric goes smoothly into the metric of flat spacetime.
    So, is Einstein saying here, taking away the gravitational field (the Schwarzschild metric)would leave no space-time at all. Is my understanding right?

    The above Appendix 5 can be found here… Put in the www dot ( I’m still too new).
    relativitybook.com/resources/Einstein_space.html

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  3. paddoboy Valued Senior Member

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    27,534

    Remembering that I am some what of an amateur at this, I tend to agree with your interpretation.
    My mind also goes back to a question put to Sten Odenawald a while back that may help you......


    Can space exist by itself without matter or energy around?
    No. Experiments continue to show that there is no 'space' that stands apart from space-time itself...no arena in which matter, energy and gravity operate which is not affected by matter, energy and gravity. General relativity tells us that what we call space is just another feature of the gravitational field of the universe, so space and space-time can and do not exist apart from the matter and energy that creates the gravitational field. This is not speculation, but sound observation.


    http://einstein.stanford.edu/content/relativity/a11332.html
     
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  5. brucep Valued Senior Member

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    It's the special case where the physics is done in the tangent space associated with the Reimann manifold. I think you read what I said with respect to that. In EBH they tackle it this way. Using words in figure 2. Very good initial discussion. They derive it from the local geometry in the box on page 2-18. Now lets look at this in a practical way. Since there's matter in our universe there always exists local spacetime curvature. It's really small and over an area which we can call a segment of the tangent space, or express it using local Schwarzschild coordinates as in the box on page 2-18 we can approximate the spacetime as flat. Anywhere in the universe. There is always a segment of the path where the local effect of gravity can be, generally, ignored since it doesn't effect the results of experiments in a meaningful way. Two local experiments which can't ignore miniscule gravitational effects is the GPS and Gravity Probe B. For obvious reason. This turns out to be a big deal since the mathematics of SR is pretty easy to use.
    Chapter 2 Curving http://www.eftaylor.com/pub/chapter2.pdf

    The metric solutions to the EFE can include terms for curvature, angular momentum [rotation parameter], and charge. For instance the curvature component of the metric is 2M/r. When these components equal 0 the metric reduces to the metric of SR. The metric at 'boundary' and the one we use in the tangent space for the reason stated. IE the effect of gravity for the local physics is so small it's irrelevant to experimental results. The 'boundary' for EBH are the Schwarzschild remote bookkeeper coordinates 'far away' at rest in flat spacetime. All the metrics are for describing spacetime. The 'boundary' can be thought of as 'idealized coordinates' for doing the physics. IE a starting point on the manifold. The boundary is idealized as flat spacetime.

    I'll just add this for you. There's nothing in the relativistic theoretical model that defies common sense. It may seem like it but once you understand what it is it's pure common sense.

    A little more about the practical aspects of 'boundary'. Since local spacetime curvature is so small you can generalize 'boundary' to any local coordinates. For example the HST. EBH is great for understanding the great experimental tests of GR. It becomes the main focus of the book so the experimental models come into focus as you progress through the derivations from the metric.

    Another interesting thing going in is what is the total energy of a particle at 'boundary'. For Newton it's 0 so you need to find ways this is > 0 in the gravitational field of Newton. For GR, in geometric units, it's 1

    E=m

    so

    E/m=1

    Another big deal full of common sense. For Newton the energy per unit mass is 0? For GR it's 1. The way energy is conserved in GR is that initial energy is conserved over the entire natural path of an object to whatever r_shell arrived at. No energy exchange between the object and the gravitational field. This seems to be pretty confusing for folks who cut their baby teeth on Newton. It's a waste of time trying to find analogues between the theories where none exist.
     
    Last edited: Dec 13, 2013
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  7. nimbus Registered Senior Member

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    129
    brucep.
    From your description on the other thread, I see what you mean about tangent space representing a small local frame. IE, although the manifold maybe curved on the greater scale, the local frame at a point being to small to reflect this overall curvature.

    Would it be correct to think of the remote bookkeeper frame as a large tangent space, so large, it’s only way of ‘connecting’ with frames elsewhere on the manifold, is at oblique angles, that obliqueness representing curvature in the bookkeeper‘s records. IE, something like, tangents on the circumference of a circle overlap each other at angles?

    About the energy thing, I understanding that by looking for the path in the local spacetime metric, along which the ‘energy‘ of a rock ( what was Wheeler’s word, ’Momenegry’?) remains constant, you will be automatically tracing out a geodesic?
    Is that one way of defining a path of maximum ageing proper time for the rock or is it the other way round? Or something else?


    Paddoboy, It almost gets philosophical.
     
  8. rpenner Fully Wired Valued Senior Member

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    4,833
    For \(ds^2 = dx_1^2 + dx_2^2 + dx_3^2 - dx_4^2\) and \( ds^2 = g_{ik} \, d x_i \, d x_k \equiv \sum_{i=1}^4\sum_{k=1}^4 g_{ik} \, d x_i \, d x_k \) to both be true it follows that \(g_{ik} = \delta_{ik} - 2 \delta_{i4} \delta_{4k}\) where \(\delta_{\alpha \beta}\) is Kronecker's delta and is 1 only when the two subscripts are the same and 0 otherwise. So
    \(\sum_{i=1}^4\sum_{k=1}^4 g_{ik} \, d x_i \, d x_k \\ = \sum_{i=1}^4\sum_{k=1}^4 \left( \delta_{ik} - 2 \delta_{i4} \delta_{4k} \right) \, d x_i \, d x_k \\ = \sum_{i=1}^4\sum_{k=1}^4 \delta_{ik} \, d x_i \, d x_k - 2 \sum_{i=1}^4\sum_{k=1}^4 \delta_{i4} \delta_{4k} \, d x_i \, d x_k \\ = dx_1^2 + dx_2^2 + dx_3^2 + dx_4^2 - 2 dx_4^2 \\ = dx_1^2 + dx_2^2 + dx_3^2 - dx_4^2\)

    Or if you keep in mind that \(g_{ik}\) is always symmetric in General Relativity and thus only has 10 independent components we can write this as:
    \(dx_1^2 + dx_2^2 + dx_3^2 - dx_4^2 = ds^2 = \sum_{i=1}^4\sum_{k=1}^4 g_{ik} \, d x_i \, d x_k = \begin{pmatrix} dx_1 & dx_2 & dx_3 & dx_4 \end{pmatrix}\begin{pmatrix} g_{11} & g_{12} & g_{13} & g_{14} \\ g_{12} & g_{22} & g_{23} & g_{24} \\ g_{13} & g_{23} & g_{33} & g_{34} \\ g_{14} & g_{24} & g_{34} & g_{44} \end{pmatrix} \begin{pmatrix} dx_1 \\ dx_2 \\ dx_3 \\ dx_4 \end{pmatrix} = \begin{pmatrix} dx_1 & dx_2 & dx_3 & dx_4 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} dx_1 \\ dx_2 \\ dx_3 \\ dx_4 \end{pmatrix} \)

    As all the 10 components of \(g_{ik}\) are constants, and curvature is a fancy geometric function of the derivative of these components (it's complicated), it follows that Minkowski space (on the left) is a very particular form of space-time which is allowable in General Relativity, one for which there is no matter, no energy, no stress, no pressure and no cosmological constant. But the big take away from General Relativity is that if your experiment is small enough in spatial extent and fast enough (small in temporal extent) all space-time looks very much like flat Minkowski space-time. Curvature of a circle or space-time has a spatial (or spacial-temporal) scale and if you zoom in close enough to a curved line it looks flat. Thus learning about Minkowski space-time is step one to learning what about General Relativity says about space-time.
     
  9. brucep Valued Senior Member

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    4,098
    It's the local frames where the experiments are done in the tangent space. The most useful local frame for doing experiments is the laboratory frame of SR. Spacetime curvature is local phenomena and is very small when measured in the tangent space [pretty much what Gravity Probe B measured as the Geodetic effect]. So the local frames are where the invariant measurements are made including the speed of light. The remote bookkeeper frame [Schwarzschild coordinates] is a remote [global perspective] frame dependent observation/measurement. They discuss it in Chapter 2 of EBH [the chapter I linked earlier]. Since we're here Ill say something about that. If you look at figure 2 the 'long curved lines' represent the global view of the remote bookkeeper frame. The global view takes into account the spacetime curvature over the entire path. So you have a local view and a global view [this isn't very rigorous but generally true]. You connect these different views with components of the metric equations [derived transformation equations].

    In Schwarzschild coordinates and geometric units the remote coordinate speed of light is

    dr/dt_bkkpr = 1-2M/r [remote radial coordinate speed of light]. Notice the curvature component 2M/r. Think about what it means and think about where the tick rate is being measured.

    So for all local measurements of the [coordinate] speed of light

    dr/dt_shell = 1 [local radial speed of light]. This is the invariant one. c. Maxwell said so and when you solve the metric for 0 that's the answer you get. Theoretically if you could measure this at every point on a manifold, which models our universe, it would be c. Invariant. That's what relativistic physics predicts. So it becomes important to say that both invariant and frame dependent measurements are completely valid. For example I love to link Joe Dolan's first discovery of the 'dying pulse train' predicted by the remote bkkpr of GR. I think you'll find this interesting and 'dying pulse train' is covered in a chapter problem in EBH.
    'Death Spiral' Around a Black Hole Yields Tantalizing Evidence of an Event Horizon
    http://hubblesite.org/newscenter/archive/releases/exotic/black-hole/2001/03/text/

    A video re-enactment
    http://imgsrc.hubblesite.org/hu/db/videos/hs-2001-03-a-low_mpeg.mpg

    Including everything that rpenner wrote down.

    So look at this in the first chapter called 'speeding'. They derive the SR constants of motion for energy and momentum and assign M as an invariant. From there they build the relativistic energy equation.
    Chapter 1 EBH Speeding.
    http://www.eftaylor.com/pub/chapter1.pdf

    Should go over that in detail. The same method applies to finding constants of motion for natural paths through curved spacetime. Specifically energy and angular momentum.
     
  10. nimbus Registered Senior Member

    Messages:
    129
    brucep, It's making sense and I'm seemingly gone back to start the book again with abit more understanding. Thanks.
    Quicky.
    So,I think I can see why they say there's no Newtonian 'force', since one way was to define force as a change in momentum over time, but since energy and angular momentum remain constant on a natural path through a curved spacetime geometry model, there's no need in that model for 'force'. I think. Thanks

    rpenner, got some of that, but most...Ay caramba!

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  11. brucep Valued Senior Member

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    rpenner is amazing. Prof Taylor's genius was building a wonderful undergraduate model for teaching this amazing physics to folks like me and young folks who are still in undergraduate school. I read it when the first edition came out. I took my time and worked every derivation. I was about 48 when I decided I was going to learn the math so I could study GR. Two years later I'd just finished calculus when I discovered Prof Taylor's EBH. Amazing good fortune for me since it meant I could start right then instead of preparing to read Gravitation [MTW]. That's exactly right. Gravity is the result of spacetime geometry [spacetime curvature] not a force. Force in the gravitational field [g] is tidal acceleration which is the first derivative of the curvature component in the metric. Both are local frame invariant phenomena.
     

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