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I'm watching (or trying to watch) this lecture from NPTEL on classical field theory. I've understood everything in the series up till this point, including the first half of the lecture on elementary group theory. However, at a certain point he begins talking about a "vector of $SO(d)$". He essentially defines such an object as an d-tuple $\begin{pmatrix}v^1\\v^2\\\dots\\v^d\end{pmatrix}$ such that it transforms as follows under matrices $R\in SO(d)$:

$$(v')^i=R^i_{\,j}v^j = \sum_j R_{ij}v_j$$

This is where I began to get confused. Is this not the definition of matrix multiplication by a vector where $v'=Rv$? Because of this I essentially could not make heads or tails of the rest of the lecture, which generalized this idea to a "tensor of $SO(d)$", definining it as an object $T^{i_1i_2\dots i_n}$ which transforms like

$$T'^{j_1\dots j_n} = R^{j_1}_{\,i_1}\dots R^{j_1}_{\,i_1}T^{i_1\dots i_n}$$

From here I am essentially lost. Googling relevant terms hasn't been much help - is this a standard notation that I'm not understanding or is it just the cryptic language of this particular lecturer?

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The equation you gave is indeed the definition of matrix multiplication, applied to a $d\times d$ matrix and a $d\times 1$ matrix. But the underlying concept is something more.

The thing about vectors is that they exist, in some sense, independent of the numbers used to represent them. For example, an ordinary 3D displacement vector represents a physical length and a physical direction. These things are not numbers, they are abstract ideas. You only get the numbers when you choose a coordinate system and then compare the vector to the coordinate axes. Different coordinate systems will give you different sets of numbers for the same vector.

Two coordinate systems can be related by transformations, such as rotation and reflection. In other words, given coordinate system A, you can identify some transformation that turns it into coordinate system B, and you can come up with a $d\times d$ matrix, $R_{d\times d}$, that represents that transformation. What makes a vector a vector is that the numbers that describe the vector in coordinate system A and the numbers that describe the vector in coordinate system B are related by the same matrix.

$$\begin{pmatrix}v_B^1 \\ \vdots \\ v_B^d\end{pmatrix} = R_{d\times d}\begin{pmatrix}v_A^1 \\ \vdots \\ v_A^d\end{pmatrix}\tag{1}$$

The group of all possible transformations has some name. For example, $SO(3)$ is the group of all rotations in 3D space. Accordingly, anything that behaves as a vector (i.e. it follows equation 1) when you rotate the 3D coordinate system is called a vector of $SO(3)$, or an $SO(3)$ vector.

In case this seems like it should be obvious, let me point out that there are sets of quantities which don't behave this way, especially when you start talking about other kinds of transformations besides 3D rotations. For example, all possible Lorentz transformations, including both rotations and boosts, form the group $SO(3,1)$. The energy and momentum (of a single particle) form a vector of $SO(3,1)$, because they change in accordance with equation (1) (with $R_{d\times d}$ being a Lorentz transformation matrix) when you change reference frames. But the electromagnetic field does not. You actually need two factors of $R_{d\times d}$ to account for how EM fields change between reference frames. That makes the EM field a rank-2 tensor of $SO(3,1)$.

I would also refer you to this question of mine on Math about the meaning of a "physical vector space", which touches on the difference between a mathematical vector and a physical vector. Only the latter is subject to the requirement of equation (1).

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  • $\begingroup$ Does your answer imply that we can not give a quantity the title of a vector, tensor or scalar until we haven't decided on the transformation and that these terms hold meaning only with respect to the transformations we do? Also can you give an example where a vector was a vector of some group and not a vector for some other group? $\endgroup$ – Naman Agarwal Dec 8 '17 at 8:45
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    $\begingroup$ @NamanAgarwal (1) Yeah, pretty much. You can only identify something as a scalar/vector/tensor with respect to a particular transformation group; it's meaningless to say "this is a vector" without it being specified which transformation group it is a vector with respect to. Of course, all this applies only to the meaning of vector used in physics, not the one used in math (cf. the math question I linked). (2) E.g. total energy of a particle is an $SO(3)$ scalar but a component of a vector in $SO(3,1)$. $\endgroup$ – David Z Dec 8 '17 at 9:01
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Yes, he defined the vector as behaving that way (a vector rotation is equivalent to a change of basis), otherwise it would not be a vector. A tensor is a different kind of object, it has at least two indexes and behaves different that a vector under transformation of coordinates (as defined in your book). You might probably read more about linear algebra up to tensors before reading a physics book (it will make your life easier), because in physics they introduce them as they were obvious things, but they are pretty well defined mathematical concepts. At the opposite end is a scalar, who does not change under a transformation of coordinates.

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In the context of e.g. a pseudo-orthogonal Lie group

$$\tag{1} O(p,q)~:=~ \{\Lambda\in {\rm Mat}_{n\times n}(\mathbb{R}) ~|~\Lambda^T\eta\Lambda= \eta \} $$

of pseudo-orthogonal matrices $\Lambda$ for the metric

$$\tag{2} \eta_{\mu\nu}~=~{\rm diag} (\underbrace{1,\ldots,1}_{p~\text{times}},\underbrace{-1,\ldots -1}_{q~\text{times}}), \qquad n~=~p+q,$$

a "vector of $O(p,q)$" is an element of the $n$-dimensional vector representation (aka. the defining representation or fundamental representation) of $O(p,q)$.

See also this related Phys.SE post.

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