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Please help with Corrections

  1. Sep 13, 2005 #1
    Hello all. As most of you know my web site is located here


    I've got the time and the patience (due to good meds) and I'm in the mood to start collecting notes on errors that are in my web site. Please feel free to browse through them all and post whatever errors you find there. I will not correct them right away since I need to make a solid list first. That will take a week or so. Then I'll be gone for a while (They're yanking that disk out - Yay!) and will try to make the corrections when (1) my buddy re-installs MS Word on my machine an (2) my doctor says its okay to sit in this chair for extended periods.

    I will truly appreciate any and all corrections (not dislikes - that's a matter of discussion and requires opening a new thread for each dislike) of the site. Please add them to the list to be gathered here if anyone chooses to take a gander and find some errors. I'd also like to add to my web site more about the geometrical aspects of GR and the difference between tensors and Lorentz tensors from a geometric point of view. Thus suggestions for additions are also most welcome

    Thanks in advance all you great folks! :approve:

  2. jcsd
  3. Sep 13, 2005 #2
    Hi Pete,

    I took a quick look at your physics pages and have already noticed two typos. On page 12 of On the concept of mass in relativity, you have a paragraph that says, "In the 3+1 view one Weyl’s definition of mass is as follows: In all inertial frames of reference there are quantities, m_k, such that in a collision process, where there are n incoming particles with velocities u_1,…, u_n and p outgoing particles with velocities u_1+1,…, u_n+p respectively, …"

    Did you mean to say that "In the 3+1 view, one of Weyl’s definitions of mass is as follows:"?

    Also, I'm sure you meant to say "p outgoing particles with velocities u_n+1,…, u_ n+p". The first term u_1+1 is an obvious mistake.
  4. Sep 13, 2005 #3
    On the concept of mass in relativity

    Page 2

    > "There are also contradictory claims statements on certain points, which should be points of fact."

    You probably want to remove either claims or statements.

    > "For example; while some relativists hold the relativistic mass is the source of gravity others hold that it is not."

    Perhaps you want to delete the and insert that.

    > "The goal of this paper is to present a complete and thorough review of the concept of mass and to present arguments that it is relativistic mass and not proper mass that is the most reasonable concept of mass."

    The word the is missing.

    > tesors

    You mean tensors.

    > The former is referred to as the 3+1 view
  5. Sep 14, 2005 #4
    Page 3

    > Line 3: "article 4-vectors and 4tensors"

    A hyphen for 4-tensors would be nice.

    > Each of these are briefly described below

    Missing a period.

    > "Each such frame corresponds to a particular inertial observer where an observer is a collection of rods and clocks associated with the given frame of reference. Each such is referred to as a coordinate clock."

    I'm not saying this is wrong. I only wonder if it's a standard definition.

    > "The introduction of a Lorentz frame, and hence a Lorentz observer, is said to produce a “3+1” split which, for example, “splits” a 4-vector (tensor of rank one, to be defined below) into a 3-vector plus a scalar (tensor of rank zero)."

    This doesn't seem right to me. I have never heard of this “3+1” split and would like to see a precise mathematical proof.
  6. Sep 15, 2005 #5


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    The time coordinate of a 4-vector (relative to an observer frame) is a scalar (of course) but it's not a tensor of rank zero, i.e. it's not invariant. How can it be when it's different for each frame?
    Last edited: Sep 15, 2005
  7. Sep 15, 2005 #6
    First let us be clear on this "invariant" thingy. See


    Note in Eq. (13) and Eq. (14) they system being used to calculate the scalar is different in each case. Now notice that R*i is an invariant in the sense described. This can be very tricky stuff if not worded properly.

    Also - What results after the spilt into space and time are not 4-tensors, they are 3-tensors, i.e. Cartesian tensors.

    The 3-vector is a Cartesian vector. The time component of a 4-vector is Cartesian scalar.


    ps - Thanks for all the help guys. Great stuff.
    Last edited: Sep 15, 2005
  8. Sep 15, 2005 #7
    There is no proof to splitting a 4-vector into a Cartesian 3-scalar and a Cartesian 3-vector. Once an observer is defined so are these vectors. MTW speak about it but Thorne's new text speaks of it too.


    Start reading after Eq. (1.67) on page 25, or simply do a search on the term "split".

  9. Sep 15, 2005 #8


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    It may help to clarify this.

    An observer with unit timelike 4-velocity [tex]u^a[/tex] can decompose a 4-vector [tex]V^a[/tex] into temporal and spatial parts in his frame. That is to say, one can write [tex]V^a[/tex] as the sum of two 4-vectors, one parallel to [tex]u^a[/tex] and the remainder orthogonal to [tex]u^a[/tex].

    Using the metric (assuming the [tex]+---[/tex] signature so that [tex]u^ag_{ab}u^b=1[/tex]), the "time component of [tex]V^a[/tex] according to [tex]u^a[/tex]" is [tex]V^{(t)}=u^a g_{ab} V^b[/tex]. The quantity [tex]u^a g_{ab} V^b[/tex] is a scalar quantity... which depends on the choice of [tex]u^a[/tex]. Nevertheless, it is a scalar (a tensor of rank 0): the "time component of [tex]V^a[/tex] according to [tex]\color{red}u^a[/tex]", which all observers will agree on. Let me clarify (with apologies if this is redundant): "All observers will agree that '[tex]u^a[/tex] determines the time-component of [tex]V^a[/tex] to be [tex]u^a g_{ab} V^b[/tex]'".

    Using that, the vector [tex]V^c[/tex] can be decomposed as
    [tex]V^c= (u^a g_{ab} V^b) u^c + (V^c - (u^a g_{ab} V^b) u^c ) [/tex], where the first 4-vector is parallel to [tex]u^c[/tex] and the second 4-vector is orthogonal to [tex]u^c[/tex] [i.e., [tex]u^d g_{dc}(V^c - (u^a g_{ab} V^b) u^c )=0 [/tex] ] (so, this is a spatial vector for [tex]u^c[/tex]). These "spatial vectors" form a 3-dimensional vector space and can be thought of as so-called "3-vectors for [tex] \color{red}u^a[/tex]", which all observers will agree on. [Of course, another observer [tex]w^c[/tex] will not necessarily regard this 4-vector [tex] (V^c - (u^a g_{ab} V^b) u^c ) [/tex] as "spatial" (i.e, this 4-vector need not be orthogonal to [tex]w^c[/tex]).]
  10. Sep 15, 2005 #9


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    Going back to the moving mass issue - I want to confirm that your results are actually different than mine by converting your results into tidal forces.

    When I've done that, I'll challenge *you* to prove that *my* results are wrong :-).

    If you want to do that anyway, go right ahead - the first step is to show that our results are different.

    To that end, I've been going over your original derivation


    and it appears there are several paces where [itex]\Phi[/itex] is used instead of [itex]\Phi '[/itex]

    To that end, the metric coefficients in step 9 of your result should, according to GRTensor II, be equivalent to the following modulo some factors of 'c':

    g_{ab}= \left[ \begin {array}
    {cccc} {\frac { \left( 2+2\,{\beta}^{2} \right) {\it \Phi_0} \left( {
    \it x1},{\it y1},{\it z1} \right) }{-1+{\beta}^{2}}}+{\frac {-{c}^{2}+
    {\beta}^{2}{c}^{2}}{-1+{\beta}^{2}}}&-4\,{\frac {\beta\,{\it \Phi_0}
    \left( {\it x1},{\it y1},{\it z1} \right) }{ \left( -1+{\beta}^{2}
    \right) c}}&0&0\\\noalign{\medskip}-4\,{\frac {\beta\,{\it \Phi_0}
    \left( {\it x1},{\it y1},{\it z1} \right) }{ \left( -1+{\beta}^{2}
    \right) c}}&{\frac { \left( 2+2\,{\beta}^{2} \right) {\it \Phi_0}
    \left( {\it x1},{\it y1},{\it z1} \right) }{ \left( -1+{\beta}^{2}
    \right) {c}^{2}}}+{\frac {-{\beta}^{2}{c}^{2}+{c}^{2}}{ \left( -1+{
    \beta}^{2} \right) {c}^{2}}}&0&0\\\noalign{\medskip}0&0&-1-2\,{\frac {
    {\it \Phi_0} \left( {\it x1},{\it y1},{\it z1} \right) }{{c}^{2}}}&0
    \\\noalign{\medskip}0&0&0&-1-2\,{\frac {{\it \Phi_0} \left( {\it x1},{
    \it y1},{\it z1} \right) }{{c}^{2}}}\end {array} \right]

    The most important one is 9d and 10, which I think should be [itex]\Phi '[/itex]

    If the formatting looks a little funny, it's because it comes from a computer, which seems to like to express fractions that are equal to 1 in an overly complex manner, and to assume that [itex]\beta > 1 [/itex] :-(.

    Note that I'm assuming that ds^2 = g_ab dx^a dx^b, you seem to use a different convention (that's where the factors of 'c' come from).
  11. Sep 15, 2005 #10
    Thanks pervect. When I get back home from surgery next week I'll take another look at my derivation. In fact I'll rederive it. Thanks.

  12. Sep 16, 2005 #11


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    OK, I do in fact seem to get totally different results for the tidal force using your approach. However, its not entirely clear to me if I'm justified in saying that the tidal forces are given according to your method by

    [tex]\frac{d F_x}{dx} = \frac{d^2 \Phi '}{d x^2}[/tex]

    etc for y,z.

    You might want to think about how to convert your results into tidal forces at the same time you're redoing the derivation - assuming that you're well enough.

    I'm concerned with the issue of possible rotation. The above formula doesn't have any terms equivalent to w^2.
    [end add]

    Good luck with your surgery.
    Last edited: Sep 16, 2005
  13. Sep 16, 2005 #12
    Did you make sure that the 3-velocity of the test particle was zero? If so then all we should have is a gravitoelectric field.
    Thanks buddy. I'll need all the luck I can get.

    For the religious amoung you o:) please toss in a good word for me with the Man upstairs, okay? For the aetheists please do it anyway since it sure can't hurt you. Such a prayer for the aetheist woulg go like this
    So if there really is a Gopd then this will give you "Browny points". Why? Because my last name is Brown. :rofl:
  14. Sep 16, 2005 #13


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    I was considering the case below


    The mass is at the point of closest approach to the object (and vica-versa). Thus the object is stationary and the mass is moving upwards with a velocity 'v'. (Feel free to reverse the sign of 'v' if you want to make positive 'v' downwards).

    There should be no gravitomagnetic forces on the object in its own frame, because its velocity in its own frame is zero.

    There will be a gravitomagnetic _field_ present, because of the nearby moving mass. But there shouldn't be any "force".
    [end add]

    It's unclear to me how (or even if!) you are handling geodetic precession, the rotation of the object's coordinate axes with respect to the fixed stars.

    What I'm asking is, in your analysis, what the tidal force components on the object 'o' are. I'm assuming it's the second derivative of your "force" as I stated eearlier. Assuming that this is correct we have a definite discrepancy.

    For comparison, my results are describe in the following thread, starting at post 2.


    Post 7 adds a diagram, Posts 10&11 gives the detailed calculation and some important notes.
    Last edited: Sep 16, 2005
  15. Sep 17, 2005 #14
    Is there a reason for stating these obvious thing? If one came to the conclusion then obiuosly they don't have a good grasp on gravitomagnetism

    In this case I have never dealt with anything except point, structureless, particles for which this is meaningless

    I don't see why I should conclude it and make your idea part of the changes?

  16. Sep 17, 2005 #15


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    The reason is to make sure we are communicating. If I've managed to make things so astoundingly simple that they are obvious, that's great. The significance is that we actually agree about something for once :-).

    The calculations are rather messy, but I suppose that "my idea" is that the tidal forces are actually calculated via taking the Riemann tensor associated with the transformed (boosted-at-infinity) metric, then computing the tidal tensor via R^a_bcd u^b u^d, where u is the 4-velocity of the stationary observer.

    It turns out to be harder to do this with the nearly-Newtonian metric than it is with the Schwarzschild metric, oddly enough, but with computer algebra it's possible, but very messy.

    (The easiest route is probably the one I took, to start with the Riemann in the Schwarzschild metric, which can be found in textbooks, and is simpler than the Riemann for the nearly-Newtonian case - then to boost it directly, rather than to boost the metric first, then boost the Riemann).

    However, if you conclude that the "force" is a gradient of some scalar field Phi, in what sense is this "force" really a "force" if [itex]\nabla^2 Phi[/itex] does not give the tidal tensor (or some similar very simple bridge from "force" to "tidal force").

    The calculaiton I propose doesn't depend on your notion of the "force" at all, but only on the metric.

    For instance (setting c=1, to keep an ugly result from being even uglier), one of the components of the Riemann associated with the boosted metric is

    R^x{}_{txt} = {\frac { \left( -1+{\beta}^{2}+2\,\Phi+2\,\Phi\,{\beta}^{2} \right)
    \left( -4\, \left( {\frac {d^{2}}{d{x}^{2}}}\Phi \right) {\beta}^{2}{
    \Phi}^{2}-4\, \left( {\frac {d^{2}}{d{x}^{2}}}\Phi \right) {\Phi}^{2}-
    4\, \left( {\frac {d^{2}}{d{t}^{2}}}\Phi \right) {\beta}^{2}{\Phi}^{2}
    -4\, \left( {\frac {d^{2}}{d{t}^{2}}}\Phi \right) {\Phi}^{2}-16\,\beta
    \, \left( {\frac {d^{2}}{dtdx}}\Phi \right) {\Phi}^{2}+4\,\Phi\,
    \left( {\frac {d}{dt}}\Phi \right) ^{2}+2\,\Phi\,{\beta}^{2} \left( {
    \frac {d}{dz}}\Phi \right) ^{2}-2\,\Phi\, \left( {\frac {d}{dy}}\Phi
    \right) ^{2}+4\,\Phi\, \left( {\frac {d}{dx}}\Phi \right) ^{2}+4\,
    \Phi\, \left( {\frac {d}{dx}}\Phi \right) ^{2}{\beta}^{2}+4\,\Phi\,
    \left( {\frac {d}{dt}}\Phi \right) ^{2}{\beta}^{2}-2\,\Phi\, \left( {
    \frac {d}{dz}}\Phi \right) ^{2}+2\,\Phi\,{\beta}^{2} \left( {\frac {d}
    {dy}}\Phi \right) ^{2}+16\,\Phi\, \left( {\frac {d}{dt}}\Phi \right)
    \beta\,{\frac {d}{dx}}\Phi+ \left( {\frac {d}{dz}}\Phi \right) ^{2}-{
    \beta}^{2} \left( {\frac {d}{dy}}\Phi \right) ^{2}-{\beta}^{2} \left(
    {\frac {d}{dz}}\Phi \right) ^{2}+ \left( {\frac {d^{2}}{d{x}^{2}}}\Phi
    \right) {\beta}^{2}+ \left( {\frac {d}{dy}}\Phi \right) ^{2}+4\,\beta
    \,{\frac {d^{2}}{dtdx}}\Phi+{\frac {d^{2}}{d{x}^{2}}}\Phi+ \left( {
    \frac {d^{2}}{d{t}^{2}}}\Phi \right) {\beta}^{2}+{\frac {d^{2}}{d{t}^{
    2}}}\Phi \right) }{ \left( -1+{\beta}^{2} \right) ^{2} \left( 2\,\Phi-
    1 \right) \left( -1+4\,{\Phi}^{2} \right) \left( 1+2\,\Phi \right) }

    It doesn't seem to fit on the line, (itex didnt' help either), the point is there doesn't seem to be any simple relationship between Phi and the tidal force. (Maybe there is, but I can't see it.) So it's very unclear why one would call the gradient of Phi the "force".
  17. Sep 17, 2005 #16


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    Using Maxima (which is in the public domain), on the following metric (represented the boosted nearly newtonian metric we've been discussing) we can calculate the Riemann. The boost is in the 'x' direction. B represents Beta, the velocity. U represents the potential function. The change in notation was needed to conform with the limits of Maxima, which won't handle greek variables.

    The metric

    Code (Text):

        [     2                          ]
        [     2 (B  + 1) U    4 B U                  ]
        [ 1 - ------------    ------        0          0     ]
        [         2            2                 ]
        [    1 - B        1 - B                  ]
        [                                ]
        [             2                  ]
        [      4 B U          2 (B  + 1) U               ]
        [      ------       - ------------ - 1      0          0     ]
        [       2             2                  ]
        [      1 - B         1 - B                   ]
        [                                ]
        [    0          0           - 2 U - 1      0     ]
        [                                ]
        [    0          0           0      - 2 U - 1 ]
    We get for R,1,2,1,2 (which is equal to R2,1,2,1)

    Code (Text):

                  2       2      2
    (%t3) R       = (((2 B  - 2) U - B  + 1) (U )
           1, 2, 1, 2                  Z

        2       2          2      2   2    2
     + ((2 B  - 2) U - B  + 1) (U )  + ((- 4 B  - 4) U  + B  + 1) U
                     Y                     X X

           2        2                 2       2    2
     + (4 B  + 4) U (U )  + 16 B U U  U  + ((- 4 B  - 4) U  + B  + 1) U
              X         T  X                   T T

            2         2        2       2       2    2
     + (4 B - 16 B U ) U    + (4 B  + 4) U (U ) )/((4 B  - 4) U  - B  + 1)
                T X          T

    We haven't raised the first index of the Riemann, but that dosen't matter when U is small, because in the limit as U->0, our metric is Minkowskian, so in that limit raising the index is an identity transform.

    I am particularly interested in the value of the Riemann at one particular point in space and time. (If the expression were simpler, I might focus on the expression as a whole, but it is so complicated that it's worthwhile to look at a special case).

    This point is


    The following terms will be nonzero at these specific coordiantes and contribute to the Riemann component:

    U_y, U_xx, U_xt, U_tt

    Here U_y is [itex]\frac{\partial U}{\partial y}[/itex], i.e. we represent partial derivatives with subscripts (another Maxima convention).

    Here U(x,y,z,t) = m/sqrt( (1-B*t)^2/(1-B^2) + y^2 + z^2) , which is the potential function in the boosted coordinates. Note that in the boosted coordinates, U is a function of time.

    Taking only these terms, and utilizing the approximation U << 1, we can get a semi-managable expression, which is

    R_{2121} = U_y^2 + \frac{(B^2+1) (U_{xx}+U_{tt}) + 4 B U_{tx}}{1-B^2}

    This particular expression boils down to m/r^3, yielding the same result I got earlier for the tidal force in the 'x' direction, the direction of the boost.

    But I do not see how to get this expression from your defintion of "force".

    We can of course do similar computations for R*,1,*,1, but this is enough for now.
    Last edited: Sep 17, 2005
  18. Sep 17, 2005 #17
    pervect - This must be put off until a later date. My back surgery is the day after tomorow and I'm trying to determine if DNR is the way to go. These last two fights were hell and I'm not sure I can handle much more - So many decisions.

    Please keep it down to "I think your calculations are wrong at such and such.." for now. Okay?

  19. Sep 17, 2005 #18


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    Yeah, I thought I might be overdoing it, but once I got started, I got involved in all the details.

    The only simple thing I can say is that I calculate the tidal forces from the Riemann tensor. This is simple in principle, even if the details of calculating it (the Riemann) are very involved.

    Hopefully you'll agree that
    R^a{}_{bcd} u^b u^d

    yields a 4x4 tensor whose 3x3 spatial components represents the tidal forces at a point? (here u is the 4 velocity, which is zero, and R is the Riemann).

    I can't differentiate your "force" components to get my "tidal force" components, which makes me disbelive in them.
    Last edited: Sep 17, 2005
  20. Sep 18, 2005 #19
    Ohaniam shows the correlation in his text "Gravitation and Spacetime." Remind me to make a new page to show this!

  21. Sep 18, 2005 #20


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    Direct calculation of the Riemann shows me otherwise, at least when beta is large for the metric under discussion.

    I can believe this works in the limit of small beta (a stationary mass), in fact I expect that the Riemann approaches [itex]\nabla^2 U[/itex] when [itex]\beta \rightarrow 0[/itex].
    Last edited: Sep 18, 2005
  22. Sep 18, 2005 #21
    You lost me. You can't expect the relativistic expression to become the Newtonian expression. Look at Eq. (22) in


    Myltiply through by the proper mass and you'll get the the relativistic version of the tidal 4-force. Comepare that with Eq. (6) at


    In the Newtonian Limit Eq. (22) from the first web page must reduce to Eq. (6) from the second web page.

    Ah ha! Now I know how I damaged my back. This is backbreaking work. :biggrin:


    Note - During the surgery I'm having a chip planted in my head so I actually will know everything. :rofl:
    Last edited: Sep 18, 2005
  23. Sep 18, 2005 #22


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    I agree with eq 22) 100%, that's what I was mentioning earlier and exactly how I computed the tidal tensor. I will call this by a different number:

    Eq 1)

    R^a{}_{bcd} u^b u^d

    To get an actual force, we do have to feed the above tidal tensor a vector. To make things crystal clear

    eq 1a)

    F^a = R^a{}_{bcd} u^b u^d \xi^c

    F is a 4-force.

    This tensor is

    eq 2)

    [itex] \nabla^2 \Phi [/itex]

    a 3x3 tensor (we compare the corresonding entries of it to the 4x4 in eq 1).

    There is only one small back-breaking problem here. If we go back to post


    We have a metric, and from that metric we compute the Riemann. From the Riemann, we compute equation 1. There are actually 16 numbers in equation 1, so let's compute one of them, specifically the uppermost non-zero entry of the tensor in eq 1), the x-component of the x tidal force.

    * * * *
    * X * *
    * * * *
    * * * *

    That's the one marked with an 'x'

    To do this, we feed into the Riemann tensor [itex]u^a = u^b = \hat{t}[/itex] i.e. [itex]u^a=u^b=[/itex] = (1,0,0,0), and the vector [itex]\xi^d[/itex], a unit vector [itex]\hat{x}[/itex], [itex]\xi = (0,1,0,0)[/itex], and take the x-component of the resulting vector.

    There is a symbol for this component - this is [itex]R^x{}_{txt}[/itex]. Numbering our variables so that t=1, x=2, y=3, z=4, we can also write this as [itex]R^2{}_{121}[/itex]

    We next go through an argument that says "Hey, [itex]R^2{}_{121}[/itex]is almost the same as [itex]R_{2121}[/itex], because g^ab and g_ab are both nearly unit matrices.

    Or we use a more capable computer program (rather than the public domain one I used in that particular post) to actually compute R^2{}_{121} and verify that it's the same as R_{2121}.

    In either event, we calculate [itex] R^2{}_{121}[/itex].

    And find that the value of [itex]R2{}_{121}[/itex] is NOT THE SAME as the uppermost left entry from eq 2), which is [tex]\frac{\partial^2 \Phi}{\partial x^2}[/tex]

    A little insight does allow us to simplify the expression I originally posted for [tex]R^2{}_{121}[/tex]. But the simplification still does not make it the same as the expression from eq 2).

    The simplification involves realizing that all the non-linear terms in the Riemann can be ignored. Only the linear terms (which are second derivatives) are important in the weak field.

    This important simplification, combined with the approximation [itex]\Phi << 1[/itex] leaves us with a tractable result

    R^2{}_{121} = \frac{-(1+B^2)(\frac{\partial^2 \Phi}{\partial x^2} + \frac{\partial^2 \Phi}{\partial t^2})+4 B \frac {\partial^2 \Phi}{\partial x \partial t}}{1-B^2}

    B above is the factor "beta" from our metric.

    This is not the same as [tex]\frac{\partial^2 \Phi}{\partial x^2}[/tex], the upper left hand entry of the tensor in eq 2), EXCEPT in the limit where B->0.

    The difference is a major one, it's important in the weak field.
  24. Sep 19, 2005 #23


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    A couple more notes:

    The value of one component of the tidal tensor won't conclusively answer the question of whether or not it "could be" some scale factor (fineagle factor) of the potential function Phi that is important. One needs to compute a couple of components to conclusively rule out the scale factor idea.

    Calculating the Riemann in full is very time consuming (unless of course one lets the computer do the work!). In princpiple, one could use the formula below for hand calculation.

    R^u{}_{vab} = \partial_a \Gamma^u{}_{vb} - \partial_b\Gamma^u{}_{va}+ \Gamma^u{}_{pa}\Gamma^p{}_{vb} - \Gamma^u{}_{pb}\Gamma^p{}_{va}

    I'm personally convinced that the last two terms of the above expression can be omitted, that only the first two terms are significant.

    This is on the basis of the computerized calculations where I have seen the nonlinear terms give results that are on the order of m^2/R^4, which are much less than the dominant terms of order m/R^3 because U = m/R <<1.

    It is also based on MTW's remarks on exercise 19.3 pg 456 on the problem of calculating the "far field" of a strongly gravitating source, where they also suggest that nonlinearities can be ignored in the weak field.

    While omitting these terms probably makes hand calculation of the Riemann much more practical, it's not clear to me how to rigorously justify omitting them without doing the calculation first to see that they are negligible. Possibly someone else can suggest a way to show this.

    Meanwhile, making use of all available simplification techniques, I get the following results for the 3x3 nonzero terms of the 4x4 tidal tensor, (eq 1 in my previous post), which I will present for informational purposes.

    \left[ \begin {array}{ccc} \left( -1+{\beta}^{2} \right) {\it U_{xx}}
    &-{\it U_{xy}}&-{\it U_{xz}}\\\noalign{\medskip}-{\it U_{xz}}&-{\beta}^{2
    }{\it U_{xx}}-{\frac { \left( 1+{\beta}^{2} \right) {\it U_{yy}}}{1-{
    \beta}^{2}}}&-{\frac { \left( 1+{\beta}^{2} \right) {\it U_{yz}}}{1-{
    \beta}^{2}}}\\\noalign{\medskip}-{\it U_{xz}}&-{\frac { \left( 1+{\beta
    }^{2} \right) {\it U_{yz}}}{1-{\beta}^{2}}}&-{\beta}^{2}{\it U_{xx}}-{
    \frac { \left( 1+{\beta}^{2} \right) {\it U_{zz}}}{1-{\beta}^{2}}}
    \end {array} \right]

    Here U is the unmodified original Newtonian potential expressed in terms of the boosted coordinates, written out in full below, and subscripts of U represent partial derivatives.

    U = -{\frac {m}{\sqrt {{\frac { \left( x-\beta\,t \right) ^{2}}{1-{\beta}

    Note that [itex] U_{tt} = \beta^2 U_{xx}[/itex] and [itex] U_{xt} = -\beta U_{xx}[/itex]. This happens because x occurs only in a linear combination with t in the function U. These and other similar relations have been used to simplify the expression given for the tidal tensor by eliminating all derivatives with respect to time.
    Last edited: Sep 19, 2005
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