Understanding Tensor Notation: What is the Difference?

[LagrangeEuler]

I am struggling with tensor notation. For instance sometimes teacher uses
\Lambda^{\nu}_{\hspace{0.2cm}\mu}
and sometimes
\Lambda^{\hspace{0.2cm}\nu}_{\mu}.
Can you explain to me the difference? These spacings I can not understand. What is the difference between $\Lambda^{\nu}_{\hspace{0.2cm}\mu}$ and $\Lambda^{\hspace{0.2cm}\nu}_{\mu}$?

 

[Ibix]

Generally, tensors can be contracted with different things to produce various other objects. For example, if you take a vector and parallel transport it around a small parallelogram (defined by two more vectors) then the vector will change slightly (in curved spacetime, anyway). This is characterised by the Riemann tensor, $R^a{}_{bcd}$. If you have a vector that you wish to parallel transport then you call it $V^b$, to associate it with the second index of the Riemann. If you know your parallelogram vectors you call them $p^c$ and $P^d$, to associate them with the third and fourth indices. Then $R^a{}_{bcd}V^ap^bP^c$ is a vector $(\delta V)^a$.

To put it more generally, tensors have different "slots" to which you can attach other vectors and tensors. In index notation, you keep track of which slot is which by the order of the indices, and remembering the significance of slot 1, 2, etc (which is just a convention).

The upper and lower positions relate to whether that part is covariant or contravariant. In relativity this isn't hugely significant, since you can always raise or lower an index by contracting it with the metric tensor (which sometimes makes maths easier, but doesn't otherwise change all that much). Do remember that if you wish to sum over a pair of indices, one must be upper and one lower.

So you need to write $R^a{}_{bcd}$ because if you wrote $R^a_{bcd}$ it wouldn't be clear if $a$ was the first index and $b$ the second, or the other way around after you'd raised one index and lowered the other.

Getting back to your example, $\Lambda$ (assuming you mean the Lorentz transform) is a coordinate transform, not a tensor. A lot of the same rules apply, though. As @etotheipi says, this is a standard way of notating forward and reverse transforms (I think that there are different conventions, though, so do be aware). Whenever you are transforming the components of a vector, match the vector's index (upper) to the lower index of the $\Lambda$. Whenever you are transforming a one-form, match its index (lower) to the upper index of the $\Lambda$, whichever way round they are.

 

[LagrangeEuler]

I thought that this is some fancy way to mark what is the column and what is the row
\Lambda^{\nu}_{\hspace{0.2cm}\mu}
For instance $a_{i,j}$ $i$ labels raw and $j$ column. So $a_{2,3}$ second row, third column. So maybe this is the way to say
\Lambda^{\nu}_{\hspace{0.2cm}\mu}
that $\nu$ is the row, but I am not sure.

 

[Ibix]

Tensors aren't matrices. They don't multiply the same way. It's often convenient to write them out on a grid the same as a matrix, but they are not the same thing.

For example, $g_{ij}V^iU^j$ is the inner product of two vectors $V$ and $U$. This expression is fine in index notation but would be illegal in matrix notation. You'd have to write $\vec V^T\mathbf{g}\vec U$, with a specific order and a transpose.

It's a conventional way to write the Lorentz transforms to distinguish forward from inverse transforms, that's all.

 

[vanhees71]

It's very important to keep both the vertical and horizontal spacing correct. It's also a bit sloppy physicists' slang not to clearly distinguish between tensors, which are invariant (geometrical) objects and their components with respect to a basis. The components of course transform under Lorentz transformations, which describe how to calculate the tensor components with respect to one inertial reference frame (with its Minkowski-orthonormal basis) to another such frame (with its different Minkowski-orthonormal basis).

Since for two four-vectors $V$ and $W$ the Minkowski product is a scalar (by definition) you have
$$\eta_{\rho \sigma} V^{\rho} U^{\sigma}=\eta_{\mu \nu} V^{\prime \mu} W^{\prime \mu}. \qquad (1)$$
The Lorentz transformation is given by a Matrix ${\Lambda^{\mu}}_{\nu}$ (which are no tensor components!)
$$V^{\prime \mu}={\Lambda^{\mu}}_{\rho} V^{\rho}, \quad W^{\prime \nu}={\Lambda^{\nu}}_{\sigma} W^{\sigma}. \qquad (2)$$
Plugging this into (1) and the fact that this must hold for all four-vectors $V$ and $W$, leads to
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} = \eta_{\rho \sigma}. \qquad (3)$$
Now let's see how covariant components of a vector transform. We have
$$V_{\rho}'=\eta_{\rho \sigma} V^{\prime \sigma}=\eta_{\rho \sigma} {\Lambda^{\sigma}}_{\nu} V^{\nu} = \eta_{\rho \sigma} {\Lambda^{\sigma}}_{\nu} \eta^{\mu \nu} V_{\mu}.$$
It's now convenient to apply the raising and lowering convention of tensor components also to the Lorentz-transformation matrix elements, i.e., to write
$$\eta_{\rho \sigma} {\Lambda^{\sigma}}_{\nu} \eta^{\mu \nu}={\Lambda_{\rho}}^{\mu}.$$
Then
$$V_{\rho}'={\Lambda_{\rho}}^{\mu} V_{\mu}. \qquad (4)$$
You can also contract (3) with $\eta^{\rho \alpha}$ leading to
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} \eta^{\rho \alpha} {\Lambda^{\nu}}_{\sigma}=\delta_{\sigma}^{\alpha}$$
but this can be written with our convention to raise and lower indices of the Lorentz-tranformation matrix as
$${\Lambda_{\nu}}^{\alpha} {\Lambda^{\nu}}_{\sigma}=\delta_{\sigma}^{\alpha}.$$
Thus
$${\Lambda_{\nu}}^{\alpha}={(\Lambda^{-1})^{\alpha}}_{\nu}.$$
Then (4) reads
$$V_{\rho}'={\Lambda_{\rho}}^{\mu} V_{\mu} = V_{\mu} {(\Lambda^{-1})^{\mu}}_{\rho},$$
as it should be.

 

[etotheipi]

I was just thinking that you could use the inverse satisfies ${(A^{-1})^{a}}_{b} {A^b}_c = \delta^{a}_c$ i.e. you have$$\begin{align*}

\eta_{\nu \rho} {\Lambda^{\nu}}_{\mu} {\Lambda^{\rho}}_{\sigma} &= \eta_{\mu \sigma} \\

\eta_{\gamma \rho} {\Lambda^{\rho}}_{\sigma} = \eta_{\nu \rho} \delta^{\nu}_{\gamma} {\Lambda^{\rho}}_{\sigma} = \eta_{\nu \rho} {\Lambda^{\nu}}_{\mu} {(\Lambda^{-1})^{\mu}}_{\gamma} {\Lambda^{\rho}}_{\sigma} &= \eta_{\mu \sigma} {(\Lambda^{-1})^{\mu}}_{\gamma} \\\eta^{\alpha \sigma} \eta_{\gamma \rho} {\Lambda^{\rho}}_{\sigma} &= \eta^{\alpha \sigma} \eta_{\mu \sigma} {(\Lambda^{-1})^{\mu}}_{\gamma} = \delta^{\alpha}_{\mu} {(\Lambda^{-1})^{\mu}}_{\gamma} = {(\Lambda^{-1})^{\alpha}}_{\gamma} \\

{\Lambda_{\gamma}}^{\alpha}&={(\Lambda^{-1})^{\alpha}}_{\gamma} \end{align*}$$but since $\Lambda$ isn't a tensor, and that's it's more notational convenience, are you guys saying that we can just define ${\Lambda_{\nu}}^{\alpha} {\Lambda^{\nu}}_{\sigma}=\delta_{\sigma}^{\alpha}$ straight away?

 

[vanhees71]

Sure, that's equivalent for Lorentz-transformation matrices to the raising and lowering of indices with the $\eta^{\mu \nu}$ or $\eta_{\mu \nu}$.

 

[etotheipi]

Sure, that's equivalent for Lorentz-transformation matrices to the raising and lowering of indices with the $\eta^{\mu \nu}$ or $\eta_{\mu \nu}$.

Thinking about it, I'm getting a bit confused myself now, since I thought the metric induced an isomorphism between a vector space and its dual space, and that's what let's you "raise and lower the indicies", i.e. $$\tilde{X}(\,\cdot \,) = g(\,\cdot \,, X) = g(\,\cdot \,, X^{\nu} e_{\nu})$$ $$\tilde{X}(e_{\mu}) = g(e_{\mu}, X) = g(e_{\mu}, X^{\nu} e_{\nu})$$ $$\tilde{X}_{\mu} = g_{\mu \nu} X^{\nu}$$i.e. there is a correspondence $\tilde{X} \in V^* \leftrightarrow X \in V$ but since this all involves tensors, why does the metric allow you to raise and lower the indices of the Lorentz transformations, which are not tensors? I wonder if that is the notational part, i.e. it's just convenience?

 

[LagrangeEuler]

The Lorentz transformation is given by a Matrix ${\Lambda^{\mu}}_{\nu}$ (which are no tensor components!)
$$V^{\prime \mu}={\Lambda^{\mu}}_{\rho} V^{\rho}, \quad W^{\prime \nu}={\Lambda^{\nu}}_{\sigma} W^{\sigma}. \qquad (2)$$
Plugging this into (1) and the fact that this must hold for all four-vectors $V$ and $W$, leads to
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} = \eta_{\rho \sigma}. \qquad (3)$$

Ok. Yes, how that matrix is orthogonal its inverse is just transposed matrix. Basically when this is a matrix could I just say when I write ${\Lambda^{\mu}}_{\nu}$ where $\mu$ is row end $\nu$ is column? If I want to use matrix notation, or to write down matrix in the basis.

 

[vanhees71]

In a coordinate free notation you never need Lorentz-transformation matrices, because you never need to introduce a reference frame ;-).

From the point of view of this coordinate-free notation you just introduce a pseudoorthonormal basis (tetrad) $\underline{e}_{\mu}$ and it's co-basis (or dual basis) $e^{\mu}$. You also canonically identify the covectors of the dual space (for which you don't need a pseudometric).

For the tetrad then you have
$$g(\underline{e}_{\mu},\underline{e}_{\nu})=\eta_{\mu \nu}, \quad g(\underline{e}^{\mu},\underline{e}^{\nu}), \quad g(\underline{e}_{\mu},\underline{e}^{\nu})=\delta_{\mu}^{\nu}.$$
The Lorentz-transformation matrices come into the game when you express this tetrad in terms of another tetrad $\underline{e}_{\mu}'$
$$\underline{e}_{\mu}={\Lambda^{\rho}}_{\mu} \underline{e}_{\rho}'.$$
Since vectors are invariant you have
$$\underline{V}=V^{\mu} \underline{e}_{\mu}= V^{\mu} {\Lambda^{\rho}}_{\mu} \underline{e}_{\rho} \; \Rightarrow \; V^{\prime \rho}=V^{\mu} {\Lambda^{\rho}}_{\mu}.$$
For the covariant vector components you get
$$V_{\mu}=g(\underline{e}_{\mu},\underline{V})={\Lambda^{\rho}}_{\mu} (\underline{e}_{\rho}',\underline{V}) = {\Lambda^{\rho}}_{\mu} V_{\rho}'.$$

 

[etotheipi]

By the way after a bit of screwing around with indices I think I finally settled on a few guidelines for converting between the component and matrix representations of these transformations, i.e.$$\begin{align*}

\Lambda &\rightarrow {\Lambda^{\mu}}_{\nu} \\
\eta &\rightarrow \eta_{\mu \nu}

\end{align*}$$then$$\begin{align*}

\Lambda^{-1} &\rightarrow {(\Lambda^{-1})^{\mu}}_{\nu} \equiv {\Lambda_{\nu}}^{\mu} \\
\Lambda^T &\rightarrow {(\Lambda^T)_{\mu}}^{\nu} \equiv {\Lambda^{\nu}}_{\mu}
\end{align*}$$I believe that's consistent with the usual rules of matrix algebra as well as the properties of the Lorentz transformation but taking the transpose can be a bit fiddly

 

[vanhees71]

Ok. Yes, how that matrix is orthogonal its inverse is just transposed matrix. Basically when this is a matrix could I just say when I write ${\Lambda^{\mu}}_{\nu}$ where $\mu$ is row end $\nu$ is column? If I want to use matrix notation, or to write down matrix in the basis.

Yes, that's right. The only problem with the matrix and (row and column) vector notation is that it's not immediately evident which components you are talking about, i.e., whether you refer to upper and/or lower indices, which tell you how the components behave under Lorentz transformations. The index notation tells you that objects with lower indices transform covariantly and those with upper indices contravariantly.

 

[etotheipi]

Whoops, actually I made a bit of an oopsie! I should have said$${(\Lambda^T)_{\mu}}^{\nu} \equiv {\Lambda_{\nu}}^{\mu} \quad \text{(correct)}$$and not$${(\Lambda^T)_{\mu}}^{\nu} \equiv {\Lambda^{\nu}}_{\mu} \quad \text{(wrong)}$$and furthermore the first is only really correct for fixed indices. See relevant postings here:
https://www.physicsforums.com/threads/index-notation-for-lorentz-transformation.909570/post-5730603

 

[vanhees71]

Hm. I thought your first version were right. Because of these confusions I try to avoid the matrix notation.

What you labelled as correct cannot be correct, because a lower index on the left-hand side of an equation must also be a lower index on the right side. Transposition means to write the columns of the original matrix as the rows of the transposed matrix. That means you exchange the left with the right index but keeping their "vertical position". So I think, what you labelled as "wrong" is in fact "correct". The problem seems to be the "transposition symbol".

Let's see. In the vector notation you write the contravariant components $V^{\mu}$ in a column vector and the matrix ${\Lambda^{\mu}}_{\nu}$ in the usual way labelling the rows with $\mu$ and with $\nu$ the columns. Then you can write the transformation of the contravariant components
$$x^{\prime \mu} ={\Lambda^{\mu}}_{\nu} x^{\nu} \; \Leftrightarrow \; x'=\hat{\Lambda} x.$$
The Lorentz property reads (writing $\hat{\eta}$ for both $\eta_{\mu \nu}$ and $\eta^{\mu \nu}$)
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=\eta_{\rho \sigma} \; \Leftrightarrow \; \hat{\Lambda}^{\text{T}} \hat{\eta} \hat{\Lambda}=\hat{\eta},$$
where
$${(\hat{\Lambda}^{\text{T}})_{\rho}}^{\mu}={\Lambda^{\mu}}_{\rho}.$$
You also get from the matrix equation
$$\hat{\Lambda}^{-1}=\hat{\eta} \hat{\Lambda}^{\text{T}} \hat{\eta},$$
which in index notation translates to
$${(\hat{\Lambda}^{-1})^{\mu}}_{\nu} =\eta^{\mu \rho} {(\hat{\Lambda}^{\text{T}})_{\rho}}^{\sigma} \eta_{\sigma \nu} = \eta^{\mu \rho} {\Lambda^{\sigma}}_{\rho} \eta_{\sigma \nu} = {\Lambda_{\nu}}^{\mu}.$$
This example shows that the trouble with the matrix notation is that you have to carefully think about where the indices have to placed (up or down as well as left or right), while the index (Ricci) calculus let's you do this right automatically.

I think it's safer to stay completely in the index notation (with properly taking into account both the vertical and the horizontal order of the indices).

 

[etotheipi]

What you wrote does make sense, and it's the only way that the conventional index placement is retained.

It's just a little confusing, because @strangerep made a good point in the other thread something along the lines of that if $L$ is a linear map between two vector spaces, then $L^T$ is a linear map between their respective dual spaces, and it would thus make sense to swap the vertical placement of the indices.

Maybe then the best thing is just to avoid using indices when using the transpose of the Lorentz matrix!

 

[vanhees71]

Well, since you always need the tensor calculus in SR, I'd say, it's better to avoid the matrix notation (though it's sometimes shorter than to handle carfully all the indices as long as you only deal with scalars and vectors and not tensors of higher rank).

 

[etotheipi]

What you labelled as correct cannot be correct, because a lower index on the left-hand side of an equation must also be a lower index on the right side. Transposition means to write the columns of the original matrix as the rows of the transposed matrix. That means you exchange the left with the right index but keeping their "vertical position". So I think, what you labelled as "wrong" is in fact "correct". The problem seems to be the "transposition symbol".

Yes, you are right! I also found this post explaining some details:
https://physics.stackexchange.com/a/329969/84729

It says that if $e_{\mu}$ is a basis of $V$ and $\theta^{\nu}$ is a basis of $V^*$, it gives the forms$$\Lambda = e_{\mu} {\Lambda^{\mu}}_{\nu} \otimes \theta^{\nu}$$which acts on vectors, and the transposed map$$\Lambda^T = \theta^{\nu} {(\Lambda^T)_{\nu}}^{\mu} \otimes e_{\mu}$$which acts on dual vectors. So we do indeed have$${\Lambda^{\mu}}_{\nu} \equiv {(\Lambda^T)_{\nu}}^{\mu}$$as is suggested by the Ricci calculus.