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The Philadelphia Experiment: A Mathematical Analysis
editThe Fundamental Mathematics
editThe Metric Tensor and Field Tensor Problem
editLet us assume the most general possible metric tensor in (x,y,z,t) space:
![{\displaystyle [g]=[g_{\alpha \beta }]={\begin{bmatrix}g_{xx}&g_{xy}&g_{xz}&g_{xt}\\g_{yx}&g_{yy}&g_{yz}&g_{yt}\\g_{zx}&g_{zy}&g_{zz}&g_{zt}\\g_{tx}&g_{ty}&g_{tz}&g_{tt}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6699affc8526f6ad1ab9b53712820671173f80b7)
We will not assume any coordinate symmetries, nor any symmetries in the metric tensor; we will solve the problem in the full 16 metric tensor variables.
We will also need the inverse metric tensor, which in this case would be obtained by treating the tensor as a matrix and inverting the matrix.
![{\displaystyle [g]^{-1}=[g^{\alpha \beta }]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4708176a911c36f49c09131bceaf3e384e7962c3)
We note that the metric tensor times the inverse metric tensor yields the Kronecker-Delta tensor. This is equivalent to a matrix times its inverse being equal to the identity matrix.
![{\displaystyle [g]^{-1}[g]=[I]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1428c68a2acc2b61f0b9716d820be2be8e9ad66f)
or
Now that we have postulated the metric tensor, we have a lot of algebra to do in order to arrive at the Einstein tensor. The first thing to calculate are all of the connection symbols.
![{\displaystyle \Gamma _{\beta \gamma }^{\alpha }={\frac {1}{2}}g^{\mu \alpha }[{\frac {\partial g_{\gamma \mu }}{\partial x^{\beta }}}+{\frac {\partial g_{\beta \mu }}{\partial x^{\gamma }}}-{\frac {\partial g_{\gamma \beta }}{\partial x^{\mu }}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c52baf32fb7074f563c15ff582cfdac98ba0eb63)
Since there are 4 indices (x,y,z,t) that could be assigned to the connection symbol indices , then there are 4 x 4 x 4 = 64 different connection symbols that need to be calculated (since we are not assuming any coordinate symmetry, nor even any metric tensor symmetry). The index is an index that is repeated and it is an example of the Einstein notation, which specifies summation over all possible values of that index. For example, consider the product of the factor in front with the first term in brackets. What the repeated index really means is a sum of the form:
![{\displaystyle g^{\mu \alpha }[{\frac {\partial g_{\gamma \mu }}{\partial x^{\beta }}}]=g^{x\alpha }{\frac {\partial g_{\gamma x}}{\partial x^{\beta }}}+g^{y\alpha }{\frac {\partial g_{\gamma y}}{\partial x^{\beta }}}+g^{z\alpha }{\frac {\partial g_{\gamma z}}{\partial x^{\beta }}}+g^{t\alpha }{\frac {\partial g_{\gamma t}}{\partial x^{\beta }}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40f654767bf175d972ff223f2fc04c9a027e9945)
This is the sort of summation that is going to occur a lot in this article. Basically, whenever 1 repeated index occurs, it implies the sum of 4 terms, one for each of the indices (x,y,z,t). Whenever 2 repeated indices occurs, it implies the sum of 4 x 4 = 16 terms. In this fashion, the expressions that occur in the algebra of general relativity can quickly become millions of terms long.
Note that first derivatives occur in the formula for the connection symbol. These are expressions that are going to be replaced by finite differenced first derivative expressions. We will talk about the numerical aspects of this problem later.
Now that the connection symbols are all calculated, we move on to the Riemann tensor components. These are given by: <Wide image hidden> For this tensor, there are 4 x 4 x 4 x 4 = 256 components to calculate (since we are not assuming any symmetries whatsoever). Also, the product that occurs with the factor times the terms in the second set of brackets, involve a sum over 2 indices, and so this means the sum of 4 x 4 = 16 sets of such brackets, in each and every one of the Riemann tensor components.
Note that second derivatives occur in the first set of brackets. These are expressions that are going to be replaced by finite differenced second derivative expressions.
Next, we must calculate the Ricci tensor components. These are given by:
There will be 4 x 4 = 16 such tensor components. Each one will involve a sum of 4 x 4 = 16 terms.
Next, we calculate the Ricci scalar. This is given by:
Finally, we form the Einstein tensor:
To get this far, in terms of writing out useful finite difference expressions, a tremendous amount of algebra is required! Probably, this is too much algebra to write down by hand, or even to code by hand. There will be an awful lot more before we are finished. It is suggested that the user code this expressions within a symbollic computation program such as Maple. One would start by coding the metric tensor exactly as we defined it at the start. One would then calculate the inverse metric tensor by simply calculating the inverse of the matrix, possibly using Kramer's Rule, or possibly using a built-in matrix inversion function. Note that calculating the inverse metric tensor alone is probably too much algebra to do by hand. Then, each set of tensor elements may be symbollically coded, until one finally has got symbollic expressions for all 4 x 4 = 16 components of the Einstein tensor. Next, all occurrences of first and second derivatives must be replaced with finite differenced expressions, which we will discuss toward the end of this article.
Eventually, we will end up with a residual tensor, and we will want to convert the symbollic expressions for all related elements to computer code, such as fortran 77. This is one good reason to use Maple. Maple is capable of automatically converting symbollic expressions to optimized code. The result will be code that is not human-readable. However, the Maple worksheet that has led to the auto-code will hopefully be somewhat human-readable, for error checking purposes.
We have got about half the material we need to formulate the Einstein equations, namely, the Einstein tensor. The second half of the material is the energy-momentum tensor for an electromagnetic field.
We must start with the electromagnetic field tensor:
The quantities and are the components of the electric and magnetic fields respectively, in our rectangular coordinate system (x,y,z,t). Since these are the field components that make up the field tensor, and hence which will make up the energy-momentum tensor, one might think that these were going to be the source variables. But this is not how we will formulate this problem! The field variables are going to be added to the list of dependent variables, and they are going to appear in the Maxwell equations, which will be added to the list of equations. And in the Maxwell equations, the field variables will be expressed in terms of the source current densities. The source current densities are going to be the sole source variables in the problem! This means that the entire problem is going to be solved based on the magnitude and direction of the source electric currents. These currents might possibly be an electric current running through a set of coils, or plasma running through some kind of plasma conduit, or whatever electric current source one might postulate was used in the actual Philadelphia Experiment.
In any event, we now need to calculate the energy-momentum tensor components. This components we require are given by:
Note that this formula uses the form of the field tensor. This is given by:
We finally have all the info we need to formulate the Einstein equations:
where G is the Universal Constant of Gravitation.
We need this equation to be in residual form. Therefore, let us define the residual tensor to be:
Therefore our gravitational field equations are simply:
We stated earlier that we needed the Maxwell equations. In their general relativistic form, the Maxwell equations are:
The covariant form of differentiation we have used here are expressions of the form:
Since both forms of the field tensor are antisymmetric, it turns out that the two Maxwell equations may be expressed in the simplified form:
In the first Maxwell equation, the scalar is the negative of the determinant of the metric tensor . Since a square root of the determinant occurs in the equation, we will not use this form; we will use the previous more complicated form. The reason for this is that we will be bending space by gradually increasing the magnitude of the source current . As this is occurring, there is no telling how the metric tensor will change. We will start it off as the Minkowski metric tensor for flat space. However, as the source current is increased, space will bend, and the metric tensor will change. If the metric tensor changes to something such that is imaginary, even just temporarily within the numerical code, then this will invalidate the computation. Therefore, it will be numerically simpler to use the previous more complicated version of the equation where does not appear and hence this problem will not occur.
![{\displaystyle [g]=[g_{\alpha \beta }]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/289aaa6f2d8d9f007efd343872288a802a665b43)
The second Maxwell equation, however, is considerably simpler and does not contain any such troublesome term. We will therefore use the latter simpler version of this equation.
At this point we assume a source current density of the form . These will be the only source terms appearing anywhere in the equations. Expanding the first Maxwell equation into 3 equations, one for each source current density component, we obtain:
Let us expand all the covariant derivatives in these equations to obtain: <Wide images hidden> In order for these equations to be useful in the numerical solution that will be described soon, they must be in residual form: <Wide images hidden> Expanding the second Maxwell equation into 3 equations:
However, we want equations with the form of the field tensor, and so the above equations become:
These equations are already in the residual form we require. We simply need to expand the derivatives with the product rule: <Wide images hidden> At this point we have got 22 equations in 22 unknowns. The 22 equations are the 16 Einstein equations:
plus the 6 Maxwell equations:
And the 22 unknowns are the 16 metric tensor components plus the 6 electromagnetic field components , .
Also, recall that the only source variables in this problem are the source current densities .
The Geodesic Problem
editThe solution will amount to the metric tensor and the electromagnetic field components being specified at all points in 4D space. In order for us to be able to visualize this space, and the bending of space that has occurred, we would need to map out the space by solving the geodesic equation for the paths of rays of light. The geodesic equation is:
This equation is in terms of proper time . The proper time is different from the coordinate time .
At this point we have covered all of the fundamental mathematics that is required. Next, we move on to describe how this problem would be solved numerically.
The Numerical Mathematics
editThe fundamental mathematics that has been covered so far is straight out of any of the accepted texts in general relativity.
Now, we discuss how the problem may be solved numerically. The 22 equations contain first derivatives of the metric tensor components and electromagnetic field components. They also contain second derivatives of the metric tensor components. We have carefully written all expressions in a form such that these first and second derivatives appear explicitly. Therefore, in a symbollic computation program such as Maple, these derivatives may be replaced with finite difference expressions; there should be no need for long partial derivative expansions.
Let us suppose that the quantity is either one of the 16 metric tensor components, or one of the 6 field components. Let us consider first and second derivative expressions with respect to coordinates and in the rectangular (x,y,z,t) space in which the finite difference grid is defined.
Suitable central first derivative expressions are:
Suitable central second derivative expressions are: <Wide images hidden> Substituting finite difference expressions like these into all the first and second derivative expressions in our 22 residual equations, amounts to a set of nonlinear algebraic residual equations that must be solved. This problem is going to be very large. Even with only a modest 100 grid points per dimension, the problem ends up being in terms of 2.2 billion variables.
Boundary Conditions
editThe obvious boundary condition to impose on all the grid boundary surfaces, is flat space (Minkowski space) and zero fields. Flat space is described by the metric tensor:
![{\displaystyle [g]=[g_{\alpha \beta }]={\begin{bmatrix}+1&&&0\\&+1&&\\&&+1&\\0&&&-1\\\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2d3d5e2f44ab34ba71ea77bbfd3ea0234670e98)
and zero fields are simply:
Initial Conditions
editThe obvious initial conditions are flat space and zero fields, the same as those given above, but imposed initially at all grid points.
Problem Geometry
editThe only source terms are the source current densities:
or in terms of a magnitude and a unit vector:
The source current geometry is most likely the only geometry that needs to be specified. An interesting source current geometry might be a hoop current, or two hoop currents, separate by a given distance, or possibly a helix current.
The source current direction vectors should be assigned values that describe any one of these possible current geometries. In a given time frame, a hoop current, say, will have the geometry of an ordinary hoop in ordinary 3D space. We will assume that this same hoop geometry will exist at all time frames within the simulation. Therefore, this same hoop geometry should be replicated for all time points in the 4D grid. In this way, we will have specified the source current direction vectors over 4D space. The source current direction values will have appropriate values where-ever the hoop resides in 4D space, and will be zero everywhere else.
The source current magnitude could be any function in time. It could be a non-zero constant value for all time. It could ramp up over an initial time interval and then stay constant at a non-zero value for all time there-on. There might be many interesting source current magnitude patterns that could be studied. In any event, at every point in 4D space where a non-zero source current direction vector resides, an appropriate source current magnitude should be specified.
We might want the hoop current, or the helix current, or whatever, to encircle a metal military hardware structure like a ship or an airplane. However, we may most likely ignore the geometry and the composition of such a metal structure, for purposes of our simulation. In order to bend space, we will need very very strong source current magnitudes. The very very strong electromagnetic fields that will arise as a result, will most likely take the magnetization of any metal structure way past magnetic saturation, so that for purposes of computation, the metal structure may be neglected.
Numerical Method of Solution
editOur system of nonlinear algebraic equations could very well contain 2.2 billion variables. For a smaller problem, we might attempt a numerical solution using a Newton nonlinear method. However, in order to use this class of methods, we would need to formulate a Jacobian matrix, and we would need to solve it often. However, for a problem size on the order of 2 billion, the Jacobian matrix is likely to contain on the order of a trillion non-zero entries. Such a matrix would most likely be too large for early 21st century computers.
A simpler method that might work would be to let the problem variables relax to whatever steady-state values they will take. In order to make this happen, we would invent a hypothetical time paramter . We would change our 22 residual equations to a system of coupled first order equations in time
The general idea is to pick a time-marching scheme that will advance these equations through our hypothetical time , until all the right-hand-sides have relaxed to zero. When this occurs, the first derivatives on the left-hand-sides are also zero, meaning that the original residual equations are satisfied, and that all problem variables have relaxed to the solution values.
Perhaps a time-marching scheme as simple as the Runge Kutta Method would work here. Perhaps an adaptive Runge Kutta algorithm would be required, though perhaps a simple dumb Runge Kutta algorithm would work.
If the set of equations turned out to be a stiff set of equations, then a solution might be impossible for early 21st century computers. A time-marching scheme for stiff systems requires a Jacobian, and then we are right back to the practical limitations that made the Newton nonlinear class of methods impractical.
Mapping the Solution Space
editAssuming that we do end up with a numerical solution to this problem, we will need to map the warped space in order to be able to view a 3D projection of it. A suitable 3D projection could be the bent paths that light rays take through the 3D space, at each and every time frame. In fact, the mappings for each time frame could be strung together into a movie so that we could watch how the space warps as time advances. This could be particularly interest if we were using a source current magnitude that ramped up from zero to a very high value strong enough to warp space.
In order to map the bent paths that light rays would take through the system, we would need to take our metric tensor solution, and use it calculate all 4 x 4 x 4 = 64 connection symbols. Then, we would use the connection symbols to formulate the geodesic equation:
The derivatives in the geodesic equation are taken with respect to the proper time . The proper time is a parameter that expresses the distance travelled along a given geodesic.
The geodesic equation is second order, and hence requires an initial coordinate and an initial velocity, for whatever geodesic we wish to calculate. For a light ray, we could pick an initial coordinate to be a point on one of the boundary surfaces. We could pick the initial velocity direction to be pointing inwards from the boundary, into the problem space, and perpendicular to the boundary surface. We would pick the velocity magnitude to be that of the speed of light. Within this unitless system, the speed of light would be of unit magnitude:
The geodesic equation would perhaps also be solved using a Runge Kutta Method.
Some Final Thoughts
editThe body of math covered in this article would be laborious, but also straightforward, to use to construct a numerical solution of the general electromagnetic field problem arising from an arbitrary source current density distribution. The math would be useful for a very general class of applied physics problems.
However, it is hoped that this body of math will be used one day soon to clear up the mystery that has surrounded the conjectures that have been made for decades now on the Philadelphia Experiment. The math covered in this article is the genuine science behind the science fiction. A numerical solution of this problem would clarify for the world, once and for all, what would have really happened over the course of the experiment.
My own conjecture is that the math will demonstrate that the source current magnitude that it takes to bend space significantly enough for the unusual effects of general relativity to occur, would be so large that they could not be generated by man-made machinery on Earth. Quite possibly, it is only current magnitudes on the astrophysical scale, such as spinning charged stars or black holes, that will prove to be large enough.
This, however, does not make this body of math useless for man-made applications. it is quite possible that the very small general relativistic effects arising from small current source magnitudes, do or will carry some importance in the development of novel technologies.
On a practical note, in order for a computer to hold 2.2 billion variables in memory, with each variable being a 32 bit real number, the computer would have to have 70.4 Gigs of RAM. The time-marching schemes like the Runge Kutta methods, typically require that 5 or so copies of the solution vector be held in memory at any one time. For 5 copies of the solution vector, the computer would have to have 352 Gigs of RAM, meaning that for good measure this amount of memory should be doubled, so that a computer having on the order of 700 Gigs of RAM would be necessary to solve this problem.
That is one powerful and mighty computer that is required. But this is, after-all, the 21st century, and these are the sorts of applied physics problems that it is time to attempt.
Sources
editSteven Weinberg, Book: Gravitation and Cosmology: Principles and Applications of the General Theory of Relativity, John Wiley and Sons, Inc. (1972), ISBN 0-471-92567-5
216.168.125.113 20:01, 8 January 2006 (UTC)
- Do not create. No original research . Can we snip this to save space? Tearlach 21:03, 8 January 2006 (UTC)
- Declined. The proposed article is not suitable for Wikipedia. -- kenb215 talk 20:23, 26 November 2006 (UTC)
Little Wee Wee Head
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graeme tolmie
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| This is an archived discussion. Please do not modify it. |
Melendez Valdes=
editIN THE ARTICLE IN ENGLISH ON MELENDEZ VALDES YOU SAY THERE IS NO ARTICLE ON MANUEL GODOY. NOT TRUE. BOTH ENGLISH AND SPANISH WIKIS HAVE GODOY.
- Mended (assuming you mean Juan Meléndez Valdés and Manuel de Godoy). You're allowed to make such edits yourself. Tearlach 22:06, 8 January 2006 (UTC)
Luke Hemming
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