Undergrad Convert 2x2 Matrix to 1x1 Tensor

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To convert a 2x2 matrix representing a second-order tensor from M^{ab} to M_{a}{}^{b}, one must apply the metric tensor, which involves using the Einstein summation convention. This process does not yield a rank-1 tensor but results in a (1,1) tensor, which remains a rank-2 tensor. The correct operation for raising an index is M^{ab} g_{bc} = M^{a}{}_{c}, while taking the trace to obtain a scalar involves M_{ab} g^{ab}. Understanding the Einstein summation convention is crucial for performing these operations correctly. The discussion highlights the importance of notation and clarity in tensor operations.
Vitani1
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Trying to figure out how to transform tensors appropriately.
If I have a matrix representing a 2nd order tensor (2 2) and I want to convert this matrix from M$$\textsuperscript{ab}$$ to $$M\textsubscript{b}\textsuperscript{a}$$ what do I do? I'm given the matrix elements for the 2x2 tensor. When applying the metric tensor to this matrix I understand symbolically how to get this 1st tank tensor but I don't understand how to write it out.
 
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Sorry - these should be proper super/subscripts.
 
To write subscripts and superscripts use _ and ^. To get ##M^{ab}## use M^{ab}. To get ##M^a{}_b## do M^a{}_b (the empty {} produce a zero width blank which the subscript hangs off - missing it out gives ##M^a_b##).

Applying the metric tensor to a rank-2 tensor doesn't give you a rank-1 tensor - it gives you a (1,1) tensor, which is also a rank-2 tensor. If you are representing the components in a matrix, it's still a 4×4 matrix. If you know how to write the expression, do you understand the Einstein summation convention? If so, can you write out explicitly what the ##\mu\nu## component of the result is in terms of a sum of products of the components of ##M## and the metric?
 
@Vitani1 I'm confused about what you are trying to do. Are you trying to raise an index on the tensor ##M##? Or are you trying to compute its trace, which is a scalar?

Raising an index, which is what is implied by your saying you want to convert the matrix from ##M_{ab}## to ##M_a{}^b##, would look like this:

$$
M_{ab} \ g^{bc} = M_a{}^c
$$

Taking the trace, which is what is implied by your saying you want to convert the matrix to a 1 x 1 tensor, which is a scalar, would look like this:

$$
M_{ab} \ g^{ab}
$$
 
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The first thing you stated is correct. Say I have a tensor Mab (superscripts) and I apply this metric tensor to this matrix to lower one of its indices - how would I multiply this result out?
 
Vitani1 said:
Say I have a tensor Mab (superscripts) and I apply this metric tensor to this matrix to lower one of its indices - how would I multiply this result out?

So you have ##M^{ab} \ g_{bc} = M^a{}_c## [Edit: fixed] in the standard notation using the Einstein summation convention. Do you understand how that convention works?
 
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I take this to mean I multiply this matrix by a scalar given by the metric tensor g?
 
No.

Do you understand the Einstein summation convention?
 
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I guess not. I'll look it up. Thank you.
 
  • #10
Vitani1 said:
I'll look it up.
See the link in @PeterDonis' last post.
 
  • #11
PeterDonis said:
So you have

Shouldn't it be ##M^{a}{}_{c}##?
 
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  • #12
PeterDonis said:
So you have ##M^{ab} \ g_{bc} = M_{ac}## in the standard notation using the Einstein summation convention. Do you understand how that convention works?
It's ##M^{ab} g_{bc}={M^a}_c##, and whenever two indices are the same (where one must be an upper and the other necessarily a lower index) you sum over ##a## (in GR from 0 to 3).

The metric components ##g_{ab}## are used to convert an upper index (contravariant) to a lower index (covariant).
 
  • #13
Solved. Thanks! By the way there is a nice problem about this in A First Course In General Relativity by Schutz (Chapter 3, problem 24)
 
  • #14
weirdoguy said:
Shouldn't it be ##M^{a}{}_{c}##?

Yes, you're right, I'll fix the post.
 

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