Defining a tensor which take tensor as a prameter

[shubham kumar sinha]

Hi,

I want to define a tensor function which take some tensor as a parameter and give me a tensor.

for example if I want to define project tensor pT[v][a,b] which should give met[a,b] - p[a]p[b](p[c] p[-c])

where P[a] is a vector on manifold M.  PT i want to define such that i can call for different values of p.

DefManifold[M3, 3, {i1, i2, i3, i4, i5, i6, i7}];
DefMetric[-1, met[-i1, -i2], CD, {";", "\[Del]"}, PrintAs -> "g"];
DefTensor[p[i1], M3];
DefTensor[q[i2], M3];

DefTensor[PT[-i1, -i2], M3, Symmetric[{-i1, -i2}]];
PTDef = PT[p][-i1, -i2] ==
   met[-i1, -i2] - p[-i1] p[-i2] 1/(p[-i3] p[i3]);
PTRule = ToRule[PTDef, MetricOn -> All, ContractMetrics -> True];
AutomaticRules[PT, PTRule];

But this does not work. If anyone can give me suggestion or hint then it will be very helpful.

Best,
Shubham

[Jose]

Hi,

You don't need to define PT to construct a function that construct tensors. The Def commands are needed for objects that will appear in outputs of expressions.

Take this setup:

<< xAct`xTensor`

DefManifold[M3, 3, {i1, i2, i3, i4, i5, i6, i7}];
DefMetric[-1, met[-i1, -i2], CD, {";", "\[Del]"}, PrintAs -> "g"];

DefTensor[p[i1], M3];
DefTensor[q[i2], M3];

$PrePrint = ScreenDollarIndices;

Then I think you want a construction like this, where we use Scalar to wrap scalars that are dividing:

PT[p_][a_, b_] := met[a, b] - p[a] p[b]/Scalar[p[i1] p[-i1]]

Then you can check the projector property:

In[29]:= PT[p][i1, -i2] PT[p][i2, -i3] // PutScalar // Simplification
Out[29]= xAct`xTensor`delta[i1, -i3] - (p[i1]   p[-i3])/xAct`xTensor`Scalar[p[-i1] p[i1]]

Cheers,

Jose.

[shubham kumar sinha]

Thanks! I was wondering why there is reed bracket around scalar product p[i1]p[-i1] .

[Jose]

The red parentheses represent the Scalar head, which means that the indices inside are disconnected from the indices outside, and therefore can be repeated. You can see each Scalar[...] object as a fully independent tensorial expression, which happens to be a scalar.

Cheers,

Jose.