Ribeyre

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These are replies submitted by Ribeyre

@ecterrab 

Thank very much for your answer it is clear for me now.

I think I made a mistake I Mix two things: how to obtain g_[~mu,~nu] from g_[mu,nu] the metric and

how to obtain K[~mu,~nu] from K[mu,nu] a general Tensor (we need the metric to to that). I am sorry for this mistake.

Your explanations are very clear, Thank you for your time.

Xavier Ribeyre

@ecterrab 

Thanks for your fast answer.

However, I have additional questions:

1- Is there any new update of the Physics Package which gives the same result than the Maple 2019?

2-About the Tensor calculation:

I test Calculation on Shwarzchild metric:

This is the Maple code:

restart

Physics:-Version()

"/home/ribeyre/maple/toolbox/2018/Physics Updates/lib/Physics Updates.maple", `2019, August 28, 14:33 hours, version in the MapleCloud: 420, version installed in this computer: 329.`

with(Physics);

Define(K[mu, nu] = Matrix(4, {(1, 1) = exp(nu), (2, 2) = -exp(lambda), (3, 3) = -r^2, (4, 4) = -r^2*(sin(theta)^2)}, shape = symmetric));   

K[]

K[mu, nu] = Matrix(4, 4, [[exp(nu), 0, 0, 0], [0, -exp(lambda), 0, 0], [0, 0, -r^2, 0], [0, 0, 0, -r^2*(sin(theta)^2)]])

TensorArray(K[mu, ~nu])

Matrix(4, 4, [[-exp(nu), 0, 0, 0], [0, exp(lambda), 0, 0], [0, 0, r^2, 0], [0, 0, 0, -r^2*(sin(theta)^2)]])

K[determinant]

exp(nu)*exp(lambda)*(r^4)*(sin(theta)^2)

It seems that you are rigth K[determinant] calculate the determinant of K[mu,~nu] and not K|mu,nu]

K[~]

K[`~mu`, `~nu`] = Matrix(4, 4, [[exp(nu), 0, 0, 0], [0, -exp(lambda), 0, 0], [0, 0, -r^2, 0], [0, 0, 0, -r^2*(sin(theta)^2)]])

But For K[~] which is K[~mu,~nu], Maple found K[]=K[~] (There is a problem)

The Calculation given by Mapale 2018 is seems to be wrong because analytically it is:

K[~0,~0]=exp(-nu); K[~1,~1]= -exp(-lambda);K[~2,~2]=-1/r^2;K[~3,~3]=-1/(r^2 sin(theta)^2)

What Maple 2018 calculate really ?

Thanks for your answer.

Xavier Ribeyre

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