Index notation - I never know when to introduce a new symbol?

In summary, introducing a new index can help to simplify expressions involving tensors and make them easier to work with.
  • #1
jeebs
325
4
This isn't strictly a homework problem but anyway...
I'm reading through a QFT textbook that is using index notation, and sometimes a new index symbol will be introduced during some mathematics and it always throws me off. I'll give a simple example, take the Minkowski metric:

[tex]g^{\mu\nu} = \left(\begin{array}{cccc}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}\right) [/tex] and its inverse: [tex]g_{\mu\nu} = \left(\begin{array}{cccc}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}\right) [/tex]

We can multiply these 2 matrices together, ie. we could take [itex]g^{\mu\nu}g_{\mu\nu}[/itex] to get the identity matrix. However - and this confuses me - we could also take [itex]g^{\mu\nu}g_{\mu\nu}[/itex] to mean just the sum of the products of the matrix elements over both indices, as both are repeated:

[tex] g^{\mu\nu}g_{\mu\nu} = g^{00}g_{00} + g^{01}g_{01} + g^{02}g_{02} + g^{03}g_{03} + g^{10}g_{10} + g^{11}g_{11} + g^{12}g_{13} + g^{20}g_{20} + g^{21}g_{21} + g^{22}g_{22} + g^{23}g_{23} + g^{30}g_{30} + g^{31}g_{31} + g^{32}g_{32} + g^{33}g_{33} [/tex]

So, the first thing that confuses me is, how come we use indices when we refer to the full matrix [itex]g^{\mu\nu}[/itex], when normally we would just call a matrix (for example) [itex]A[/itex], and only mention indices [itex]i, j[/itex] when we want to refer to the [itex]i^{th}, j^{th}[/itex] element of the matrix, [itex]A^{ij}[/itex] ?
It seems to me that there is ambiguity here, when is [itex]g^{\mu\nu}g_{\mu\nu}[/itex] a matrix and when is it just a number?

Also, to get to the main part of my question, my book makes the statement that [itex]g^{\mu\nu}g_{\nu\rho} = \delta^{\nu}_{\rho}[/itex], the kronecker delta.
Here it has introduced a new index [itex]\rho[/itex]. I can see that this is true if I do the summation over [itex]\nu[/itex]:
[itex]g^{\mu\nu}g_{\nu\rho} = g^{\mu 0}g_{0\rho} + g^{\mu 1}g_{1\rho} + g^{\mu 2}g_{2\rho} + g^{\mu 3}g_{3\rho}[/itex]

then if we set, say, [itex]\mu = 0, \rho = 0[/itex], we get
[itex]g^{\mu\nu}g_{\nu\rho} = g^{0 0}g_{00} + g^{0 1}g_{10} + g^{0 2}g_{20} + g^{0 3}g_{30} = (1)(1) + (0)(0) + (0)(0) + (0)(0) = 1[/itex]

or if we set, say, [itex]\mu = 0, \rho = 1[/itex], we get
[itex]g^{\mu\nu}g_{\nu\rho} = g^{0 0}g_{01} + g^{0 1}g_{11} + g^{0 2}g_{21} + g^{0 3}g_{31} = (1)(0) + (0)(-1) + (0)(0) + (0)(0) = 0[/itex]

So clearly the Kronecker delta condition is satisfied, so the statement [itex]g^{\mu\nu}g_{\nu\rho} = \delta^{\nu}_{\rho}[/itex] is true. However, if I was writing out my own solution to a problem that involved index notation, I would never know to introduce a new index symbol myself. It's just lucky that the textbook told me and I could verify it with an explicit calculation.

Can anyone explain to me how to know when a new index symbol should be introduced?
 
Last edited:
Physics news on Phys.org
  • #2
Thanks in advance! A:The idea of introducing a new index is to make the statement of the problem more concise. The simple example you have provided is actually just a special case of a much more general idea.In general, when dealing with tensors (which are objects that can be indexed by multiple indices) it is often useful to introduce a new index in order to avoid having to write out the same expression multiple times. For example, if we have a four-dimensional tensor $T_{\mu \nu \rho \sigma}$ and want to contract it with its inverse $T^{\mu \nu \rho \sigma}$ we could write out the full expression as$$T_{\mu \nu \rho \sigma} T^{\mu \nu \rho \sigma} = T_{0000} T^{0000} + T_{0001} T^{0001} + \dots + T_{3333} T^{3333}.$$This looks unwieldy, so it is often more convenient to introduce a new index $\lambda$ and rewrite the expression as$$T_{\mu \nu \rho \sigma} T^{\mu \nu \rho \lambda} = T_{0000 \lambda} + T_{0001 \lambda} + \dots + T_{3333 \lambda}.$$The idea here is that we are saying that the expression $T_{\mu \nu \rho \sigma} T^{\mu \nu \rho \lambda}$ should be evaluated for all possible values of $\lambda$, and then the result should be summed over all possible values of $\lambda$. This is equivalent to the original expression, but it is much more concise.
 

1. What is index notation?

Index notation is a mathematical notation system used to represent and manipulate vectors, matrices, and tensors. It uses subscripts to indicate the position of a component within a vector or matrix.

2. How do I know when to introduce a new symbol in index notation?

In index notation, a new symbol is usually introduced when there is a need to represent a different quantity or when the existing symbol has already been used. It is important to use clear and concise notation to avoid confusion and to make the equations easier to understand.

3. Can I use the same symbol for different indices?

In general, it is not recommended to use the same symbol for different indices as it can lead to confusion. However, in some cases, it may be necessary to use the same symbol with different indices, as long as it is clearly defined and does not cause any ambiguity in the equations.

4. How do I keep track of the indices in complicated equations?

One way to keep track of indices in complicated equations is to use consistent and logical labeling. It is also helpful to write out the indices explicitly when first introducing them and then use the shorthand notation in subsequent equations.

5. Are there any rules for raising and lowering indices?

In general, there are specific rules for raising and lowering indices, depending on the type of tensor being used. For example, in metric tensors, the index can be raised or lowered by using the metric tensor itself. It is important to follow these rules carefully to ensure the accuracy of the equations.

Similar threads

  • Advanced Physics Homework Help
Replies
2
Views
479
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Advanced Physics Homework Help
Replies
0
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
339
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
11
Views
2K
Back
Top