Download dAlembertian_(reviewed).mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Hi Inacio,

Below is your worksheet with the answer to your two questions italized.

Download TensorEnergiaMomento_(reviewed).mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Hi Escorpsy

Within the Physics package, there is a command, Fundiff, to compute functional differentiation, and it does work with a tensor function as functional differentiation variable; check its help page ?Physics,Fundiff.

So the answer to your question is: yes, this can be done with existing functionality, you can derive "Einstein's equations" for any particular form of the Lagrangian density in the Action for the problem.

Having said that, you will need to be sufficiently explicit at the time of entering the Lagrangian density: Fundiff will differentiate F(x) with respect to F(y) returning a Dirac function (that dissapears upon integration, and Fundiff takes care of that automatically) only if it sees 'F'. It will not guess, for instance, that the Christoffel symbols can be rewritten in terms of derivatives of the metric, or for the case that the Riemann tensor can be written as a commutator involving covariant derivatives, etc.

I also don't discard having to do some adjustments here and there in order to have the computation performed, so if you have trouble please feel free to post the concrete Action for your problem (within a worksheet) and we move ahead from there.

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Hi Shillis

This kind of operation is simpler to perform using the Physics package. Suppose the metric is given by some matrix M that is 4 x 4, or by the square of a line element ds2. Then enter Setup(metric = M) or Setup(metric = ds2) to set the metric. You could also set the coordinates you are using together, for instance as in Setup(coordinates = [x,y,z,t], metric = M).

Then just use Christoffel[1,1,1] to get the all covariant 1,1,1 component you want to pick out, or Christoffel[~1,1,1] to get the component where the 1st index is contravariant (so prefixed with ~). You see that the somehow artificial distinction between 1st and 2nd kind dissapears, you just indicate whether the 1st index is covariant or contravariant.  Check the help page ?Christoffel for examples, it is pretty straightforward. To see for instance the matrix corresponding to Christoffel[3, mu, nu] enter Christoffel[3, mu,nu, matrix], to see the definition (only in Maple 2015) enter Christoffel[definition], or if you have a previous version of Maple check the definition directly in the help page for Physics:-Christoffel. For example:

Download PickOutChristoffelComponents.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Hi zia

I gave a look at your worksheet, where you use RootFinding:-NextZero. There is one limitation in that command: if, as the second argument, you send to it something containing 0.0*I, it will refuse to work, while if you simplify that second argument, this spurious imaginary part dissapears. So, first thing: instead of 

R[k] := RootFinding:-NextZero(K, R[k-1], maxdistance = 10000, guardDigits = 3)

use

R[k] := RootFinding:-NextZero(K, simplify(R[k-1]), maxdistance = 10000, guardDigits = 3)

The second limitation is that RootFinding:-NextZero can return FAIL, and then your conditional 

for k to 20 while R[k-1]<10000 do 

won't work, because the system cannot determine wheter FAIL < 10000. Use instead 

for k to 20 while evalb(R[k-1]<10000) = true do 

Putting altogether, use the following and you do get the results you are looking for:

for k to 20 while evalb(R[k-1] < 10000) = true do
    R[k] := RootFinding:-NextZero(K, simplify(R[k-1]), maxdistance = 10000, guardDigits = 3) e
od

=============================================

Edit: originally evalb(R[k-1] < 10000), corrected to evalb(R[k-1] < 10000) = true

=============================================

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Hi Inacio

Note that in d_[mu](A[nu]) you are applying an operator, i.e. in computer language A[nu] is the argument of a 'function' called d_[mu]. On the other hand, in A[nu] d_[mu] you are multiplying two objects, i.e. in computer language this is the`*` (product) operator being used. These two operations are different, of course, and the Commutator, actually computes the `.` (dot operator `.`) product, that depending on the case (arguments passed to Commutator) is transformed into the `*` or Vectors:-`.` (scalar vector product) operator.

To have Commutator behaving as you were expecting we would need to decide, on the fly, when to apply or multiply two objects, which may make Commutator a bit unpredictable, although in the example you are posted (one of the arguments is a recognized differential operator) this can be implemented. I will think about this and post again here in this thread.

Meantime, the answer from all the above is: no, you cannot use Commutator with the purpose you are mentioning.

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

@Michael_Watson 

Please note that I cannot compare an image (that is what you posted) with a Maple expression (as in a worksheet, that is what I asked), unless I read your image and try to retype everything, which is time consumming and prone to errors. In any case attached is the worksheet you posted originally, without changing anything, just executed via menus (Edit -> Execute -> Worksheet). In this reviewed worksheet you see that (19) is not 0 = 0 as you posted originally. From this reviewed worksheet you can also copy the result (19) and paste it into the worksheet you used to obtain the image you presented in your reply, then try simplifying one expression minus the other one to see if if you get 0 or what. Hope it helps.

Question_algsubs_3.27.15_(reviewed).mw

Edgardo S. Cheb-Terrab 
Physics, Differential Equations and Mathematical Functions, Maplesoft

Hi,

In the eqdN you show there are parenthesis that are open and not closed (e.g. in eqd3 and eqd4), or you open with `(` but close with `}`. You need to correct that first, I suggest you entering one equation at a time and only entering the next one after you are sure the previous one is correctly entered.

Then, do not enclose each equation within {}; In Maple, {} is used represent sets. Copying from the error message you posted (it only contains three equations, not four), correct input for your system of equations is:

> sys := {diff(v(x), x) = V(x), 1/250*x*(diff(U(x), x))+(x-8*x*V(x)+3/500)*U(x)+(3/2-12*V(x)-8*x*(diff(V(x), x)))*u(x) = 0, 1/125*x*V(x)^2+2*(3/5+x)*v(x)*V(x)+v(x)*u(x)+v(x)^2 = 0}
and for this system you get:

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Download Fundiff_Intc.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Download PhysicsDiffBug_(reviewed).mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Download connection_between_functions_and_their_derivatives.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Enter your expression, highlight the input, right click and: 2D-math -> convert to -> atomic identifier, and that produces what you want, that is an arbitrary expression as a singlie variable name. This image illustrates the final result:

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Hi Trace

There is a problem with the input syntax, you are: missing a closing parenthesis, have extra spaces after d in 'd theta' and 'd phi' (should be dtheta and dphi), and you missed a multiplication sign after r in the last term of the square of the line element. Please give a look at the revised worksheet attached, where all this is adjusted, and the nonzero components of the Ricci tensor shown.

non_zero_components_of_ricci_tensor_(reviewed).mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Hi Michael
Unfortunately, there seems to be a GUI (Graphical User Interface) issue around, related to using the default document mode. Concretely, if instead of expand((13)) you open a prompt and execute > expand((13)); the result is not 0 = 0. See the attached worksheet.

I will give a further look tomorrow to see if there is a programatic (library) way to workaround this GUI issue until it is fixed.

Question_R12_(reviewed).mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Hi Codge

You know, the sections plotted are projections up to 3 dimension while the phase space has more than 3 dimensions; there is no programatic way to restrict the value of one of the variables (dimensions) out of sight - in your example, p1. I didn't investigate your problem further but it springs to my mind: why don't you analyze the problem trying to determine the range of t such that p1 > 0? Then you can specify that range instead of using t = -5000..5000 as you are using. For the purpose of this investigation, you can write Hamilton equations (you can use DEtools[hamilton_eqs]), then a numerical or exact solution of the system of equations will let you visually or algebraically determine the range of t for which p1 > 0.

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft