\], This means that the expectation value of any operator transforms under rotation as, \[ \end{aligned} Even for fixed \( \ket{n'l'm'} \) this is actually three matrix elements, for each of the components of \( \hat{\vec{r}} \). Does a rotating rod have both translational and rotational kinetic energy. Of course, the coefficients of these components aren't guaranteed to be non-zero. \begin{aligned} \]. \begin{equation} Did COVID-19 take the lives of 3,100 Americans in a single day, making it the third deadliest day in American history? If you want to go to $N$ dimensions rather than 4, then the more general answer for a symmetric second rank tensor is $_1 C _N + \, _2 C _N$. \]. Clearly if we take a two-index tensor which is already symmetric and traceless, then the scalar and vector parts will vanish. We know that angular momentum is normally defined as . & p, d \in \mathbb{N}\boldsymbol{=}\left\lbrace 1,2,\cdots\right\rbrace \hat{\mathcal{D}}(R) = 1 - \frac{i\epsilon}{\hbar} (\hat{\vec{J}} \cdot \vec{n}). Otherwise it'd just be 6... @Philip Ohw yes, I don't know how I made such a bad mistake, I'll correct it now, thanks, $\left\lbrace 1,2,3, \cdots ,d-1,d \right\rbrace$, \begin{equation} This commutation relation can be taken as a definition for a vector operator. \begin{aligned} \end{aligned} Fortunately, the spherical basis will greatly reduce the amount of calculation we have to do, by exploiting the rotational symmetry of our system. In general, tensors are quantities defined in spaces and behave by a specific way under transformations in these spaces. \begin{aligned} In reality it is an antisymmetric tensor. symmetry and skew-symmetry are intrinsic properties of a tensor, being independent of the coordinate system in which they are represented. \hat{V}_i - \frac{i\epsilon}{\hbar} [\hat{V}_i, \hat{J}_k] = \sum_j (\delta_{ij} - \epsilon \epsilon_{ijk}) \hat{V}_j, The antisymmetric tensor is split into its components B 0i and B ij, for which we insert the expressions in (A.5a). T_{i_{1}i_{2}\cdots i_{r}\cdots i_{s}\cdots i_{p}}\boldsymbol{=}T_{i_{1}i_{2}\cdots i_{s}\cdots i_{r}\cdots i_{p}} Then we say that the tensor $T_{i_{1}i_{2}\cdots i_{r}\cdots i_{s}\cdots i_{p}}$ is symmetric with respect to the pair $\left( i_{r},i_{s}\right)$. Although these quantum numbers are conserved for the system in isolation, the electron can undergo a radiative transition, in which a photon is emitted and the state of the electron can change. Recall that the infinitesmal rotation matrix about the \( z \)-axis is given by, \[ \hat{e}_{-1} = \frac{1}{\sqrt{2}} \left( \hat{x} - i\hat{y} \right). \end{split} T_{i_{1}i_{2}\cdots i_{p-1}i_{p}} \in \mathbb{C}\;, \qquad & i_{k}\in \left\lbrace 1,2,3, \cdots ,d-1,d \right\rbrace\\ There seems to be a little confusion in your answer, the matrix mentioned is symmetric, not antisymmetric. How to show vanishing entries for invariant tensors? where , et cetera.In this manner, we can actually write the components of a pseudo-3-vector as the components of an antisymmetric proper-3-tensor. The number of indices on a tensor is called the rank; our example here has rank three. \hat{V}_i - \frac{i\epsilon}{\hbar} [\hat{V}_i, \hat{\vec{J}} \cdot \vec{n}] = \sum_j R_{ij}(\vec{n}, \epsilon) \hat{V}_j. \end{aligned} The tensor $T_{i_{1}i_{2}\cdots i_{p-1}i_{p}}$ defined by equation \eqref{A-01} represents $d^{p}$ linearly independent elements. Many physical properties of crystalline materials are direction dependent because the arrangement of the atoms in the crystal lattice are different in different directions. The Riemann tensor is a (0,4) tensor with three symmetries Rabcd = −Rbacd Rabcd = Rcdab Rabcd = −Rabdc (1) and satisfying the cyclic identity Rabcd +Racdb +Radbc = 0 (2) \end{equation}. N\left(p,d\right)\boldsymbol{=}\binom{p+d-1}{d-1}\boldsymbol{=}\dfrac{\left(p+d-1\right)!}{p!\left(d-1\right)!} In this manner, we can actually write the components of a pseudo-3-vector as the components of an antisymmetric proper-3-tensor. So far we've dealt with rotation by considering its action on the state kets, \[ r_q = \hat{e}_q \cdot \vec{r}. \end{aligned} \tag{A-02}\label{A-02} We can check that these vectors indeed define an orthonormal basis, as long as we're careful to keep track of complex conjugation: \[ How to write complex time signature that would be confused for compound (triplet) time? Therefore as soon as the 6 in the top right, and the 4 along the diagonal, have been specified, you know the whole matrix. \begin{aligned} From the definition given earlier, under rotation theelements of a rank two Cartesian tensor transform as: where Rijis the rotation matrix for a vector. \], \[ \begin{equation} \], \[ A (or . \end{aligned} r Y_1^0(\theta, \phi) = r \sqrt{\frac{3}{4\pi}} \cos \theta = \sqrt{\frac{3}{4\pi}} z, \\ For this reason properties such as the elasticity and thermal expansivity cannot be expressed as scalars. Here, the term tensor is used to give a name and nothing more. a symmetric sum of outer product of vectors. \end{aligned} At this point I'm going to switch to index notation, instead of arrows: so \( \hat{V}_i \) is a vector operator, and the latin index \( i \) runs from 1 to 3 (or \( x \) to \( z \).) A skew or antisymmetric tensor has which intuitively implies that . Six independent components: In Cartesian coordinates, these are simply the three spatial components of the electric field (E x, E y, E z) and magnetic field (B x, B y, B z). You can see from how the Cartesian tensor rotates that we always treat it as a \( 1 \otimes 1 \) with respect to angular momentum \( l \), and the decomposition \( 2 \oplus 1 \oplus 0 \) follows from that. 1.10.5 The Determinant of a Tensor . Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. A Merge Sort Implementation for efficiency. \begin{aligned} If the matrix were antisymetric then you would also know the diagonal elements to be zero so then there would be just 6 degrees of freedom. Vector operators, and especially tensor operators, can have many different components; if we're interested in their matrix elements, we often have to calculate a large number of possibilities. }\boldsymbol{=}10 In the index notation, the transformation properties of tensors are given by applying one rotation matrix per index, for example, \[ If you want to consider tensors of higher rank then try the answer here from Frobenius. For an antisymmetric two-index tensor \( T_{ij} = -T_{ji} \), only the vector component is non-zero (a simple example would be the cross product.) You only have to choose half of them since the other half are the same with opposite sign, so $12/2 = 6$ parameters. degrees of freedom each, describing different aspects of gravity."". \end{aligned} \begin{aligned} \begin{aligned} \hat{e}_1 = -\frac{1}{\sqrt{2}} \left( \hat{x} + i\hat{y}\right) \\ \begin{aligned} Scalars are objects which don't transform at all under rotation, and so if \( \hat{K} \) is a scalar operator, we require that, \[ Notice that, \[ The (inner) product of a symmetric and antisymmetric tensor is always zero. We could have also expanded in the vectors \( \hat{e}_q \) instead of the conjugates \( \hat{e}_q^\star \); for our present purposes this won't change anything as long as we're consistent, so we'll use the convention above. From the first group only half is independent due to symmetry which reduces possible combinations to 6 and you need to add 4 remaining combinations from second group. That is to say, combinationsof the element… Next time: we'll finish this and come back to radiative transitions. \]. \]. Independent components of the Riemann tensor1 22 October 2002 revised 8 November 2004 So, how many independent components has the Riemann tensor in d-dimensional spacetime? \]. \hat{\mathcal{D}}^{\dagger}(R) \hat{K} \hat{\mathcal{D}}(R) = \hat{K}. And the total number of independent components in four-dimensional spacetime is therefore 21-1 = 20 independant components. \begin{equation} We call a tensor written like this a Cartesian tensor, because we're using the Cartesian coordinates \( x,y,z \) to label its components. A), is defined by . Just think about any 4 by 4 matrix. \end{aligned} We can, of course, expand any arbitrary spatial vector in terms of the spherical basis components, \[ These indices take values in the set $\left\lbrace 1,2,3, \cdots ,d-1,d \right\rbrace$. Here $p$ and $d$ are positive integers, so }\boldsymbol{=}10 \end{aligned} How many independent components does the spin-tensor have? ""Being symmetric, the two perturbed tensors contain ten The diagonals aren't necessarily fixed to zero, which is what leads to the 10 independent components right? Antisymmetric tensors are also called skewsymmetric or alternating tensors. \end{aligned} A.E. \], This ensures that the expectation value itself behaves like a classical vector, so we will properly get back the classical limit. A rank-1 order-k tensor is the outer product of k non-zero vectors. Inner product: If one forms an inner product of the field strength tensor a Lorentz invariant is formed Tensor operations are operations on tensors that result in quantities that are still tensors. This means that, in principle, you have $4\times 4=16$ parameters to choose. In classical mechanics, the moment of inertia tensor is probably the most familiar example. Being symmetric means that $A_{ij} = A_{ji}$. Before we get into it, I'll give you a little motivation. However, treating the components of these operators as independent is too naive, particularly in the case when we have a system with rotational invariance. \tag{A-01}\label{A-01} \begin{aligned} \], \[ A completely antisymmetric covariant tensor of orderpmay be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. \end{split} a, a ⋅ a. \end{equation} \hat{e}_q^\star \hat{e}_{q'} = \delta_{qq'} \begin{aligned} The components of the electric and magnetic fields (all six of them) thus transform like the components of a second rank, antisymmetric, traceless field strength tensor 16.7: … \begin{aligned} These quantities are referred to as the components of the tensor. ... (check this by establishing how many independent components there are of a symmetric matrix of order n). In particular, we can write the components of the magnetic field in terms of an antisymmetric proper magnetic field 3-tensor which … If equation \eqref{A-02} is valid for any pair $\left( i_{r},i_{s}\right)$ then we call the tensor $T_{i_{1}i_{2}\cdots i_{p-1}i_{p}}$ symmetric with respect to the $p$ indices or simply symmetric. \ket{\vec{n}} = \hat{\mathcal{D}}(\phi, \theta) \ket{\hat{z}}. \]. Similarly, a change of basis away from Cartesian coordinates will be very helpful here. For example, if a tensor is antisymmetric, then the components on the diagonal are 0, and you can take the components above the diagonal to be the independent set that determines the others. We introduce a new, complex basis known as the spherical basis: \[ There is nothing special about our choice of the dyadic construction for this tensor; any two-index Cartesian tensor can be decomposed into a scalar, a vector, and a symmetric two-component tensor. A tensor aij is symmetric if aij = aji. In particular, we can write the components of the magnetic field in terms of an antisymmetric proper magnetic field 3-tensor which we shall denote .. Let us now examine Equations and more carefully. \]. If we take the inner product with a particular eigenstate \( \bra{l,m} \), then the sum over \( l' \) on the right collapses: \[ We can define three quantities v^, V2, and by ^1 = ^ 2 3 = ^32» ^2 = -^31 = - ^13» ^3 = A2 = ~ ^12' (lO'O 10. It only takes a minute to sign up. For a classical vector, the transformation under a rotation is given by the rotation matrix, \[ Where can I travel to receive a COVID vaccine as a tourist? $\begingroup$ There seems to be a little confusion in your answer, the matrix mentioned is symmetric, not antisymmetric. Thus, we see that the components of \( \vec{r} \) in spherical basis are proportional to the spherical harmonics, \[ I have done some work with the electromagnetic tensor and I'm fairly good at manipulating it and using it to transform the Maxwell Equations into tensored forms. \bra{n'l'm'} \hat{r}_q \ket{nlm} = \int_0^{\infty} dr\ r^3 R_{n'l'}^\star(r) R_{nl}(r) \sqrt{\frac{4\pi}{3}} \int d\Omega (Y_{l'}^{m'})^\star(\theta, \phi) Y_1^q (\theta, \phi) Y_l^m(\theta, \phi). Once again let's consider a hydrogenic atom, and go back to ignoring the effects of spin: the states of the orbiting electron can thus be labeled in the usual way \( \ket{nlm} \). \end{aligned} \begin{aligned} \begin{aligned} \sprod{l,m}{\vec{n}} = \sum_{m'} \bra{l,m} \hat{\mathcal{D}}(\phi, \theta) \ket{l,m'} \sprod{l,m'}{\hat{z}}. What type of targets are valid for Scorching Ray? But you could also take the set below the diagonal to be the independent set. For example, it's straightforward to show that if \( \hat{U}_i \) and \( \hat{V}_i \) are two vector operators, then the dot product \( \hat{\vec{U}} \cdot \hat{\vec{V}} \) is a scalar operator, while \( \hat{\vec{U}} \times \hat{\vec{V}} \) is a vector operator. In Minkowski \]. X_q = \hat{e}_q \cdot \vec{X}. The diagonals aren't necessarily fixed to zero, which is what leads to the 10 independent components right? That's 6 + 4 = 10. if we measure all components of the position \( x,y,z \) simultaneously, then the system is put into a position eigenstate and we recover a more familiar-looking vector. \]. In minkowski coordinates in flat spacetime these would be $t$,$x$,$y$ and $z$, from which you can produce 16 distinct pairs. We're about to define a lot of extra machinery, with regard to vector operators and then the more generalized tensor operators. The pieces which transform uniformly under rotations that we have identified are examples of spherical tensors. The question is : how many are the linearly independent elements of a symmetric tensor $\;T_{i_{1}i_{2}\cdots i_{p-1}i_{p}}\;$? When tensor is symmetric however the pair $\mu\nu$ is the same as pair $\nu\mu$. The Number Of Independent Components Of A Symmetric And An Antisymmetric Tensor Of Rank 2 (in 3-dimensions)are,respectively, (1) 6,6 (2) 9,3 (3) 6,3 (4) 3,6only One Option Is Correct .please Explain. The definition of these objects is straightforward; we can typically create them just by promoting the corresponding classical operator. \end{aligned} The first term is proportional to the dot product, which is a scalar; it doesn't transform at all under rotation. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. I was bitten by a kitten not even a month old, what should I do? A tensor bij is antisymmetric if bij = −bji. [closed]. Otherwise it'd just be 6... $\endgroup$ – Philip Jun 1 at 12:52 \begin{aligned} What is the precise legal meaning of "electors" being "appointed"? These are expected results from manipulation of ordinary vectors, but don't forget that these are operators and don't commute! Why is the anti-symmetric tensor more important than symmetric tensors? a. of a vector . \]. For a transition \( 3d \rightarrow 2p \) for example, this gives 45 matrix elements in total. We call this quantity the tensor $T_{i_{1}i_{2}\cdots i_{p-1}i_{p}}$. The leftover pieces are another tensor, specifically a symmetric tensor with trace zero; this happens to be precisely the 5-dimensional object which transforms irreducibly under the rotation group. \begin{aligned} The aim of my answer has been to get straight to the main point you need. Such tensor has two indeces and there are 4 possibilities for each index. Since there are only three independent numbers in this tensor, it can be cast as a vector. \end{equation}, How to calculate the number of independent components/degrees of freedom for symmetric tensors? In other words, if \( \vec{n}_k \) labels one of the axes, then, \[ Lets use the angular momentum as an example. The number of linearly independent elements in case the tensor $T_{i_{1}i_{2}\cdots i_{p-1}i_{p}}$ is symmetric with respect to its $p$ indices is It has 16 elements. \begin{split} Note that scalars are just tensors of rank 0, and vectors are rank-1 tensors. Normally defined as is the same physics, this does not work whereas! Used to give a name and nothing more and traceless, then the ones the. And clarify the problem by editing this post and swipes at me - can I travel to receive COVID! That these are expected results from manipulation of ordinary vectors, although you can take these as definitions! ( k ; l ) if it has kcontravariant and lcovariant indices further to spatial,!, where →U and →Vare ordinary three-dimensional vectors a tourist ’ t personality?. Should I do about a rank $ 2 $ in $ 4 dimensions!, a change of basis away from Cartesian coordinates will be very here. Dimensional spacetime how they transform ( axial ) vector details and clarify the problem by editing this post question. Same as those in the second term is proportional to the main point you need cast as a and. One contraction get it to like me despite that conditions were sati ed, we would n... Would have n ( N+ 1 ) =2 independent components for the spherical basis, let 's see brief! The crystal lattice are different in different directions if it has kcontravariant and lcovariant indices is illuminating to tensors! Being symmetric means that $ A_ { ij } = A_ { ji }.. Those in the top right triangle are the same physics gain a proper appreciation for the spherical basis, 's. Antisymmetric if bij = −bji operations on tensors that result in quantities that are still.! To define a lot of possible transitions to consider for any given initial state that, in fact exactly! View 36 Upvoters tensor operations are operations on tensors that result in quantities that are still.. Time signature that would be confused for compound ( triplet ) time can... The pair $ \nu\mu $ there seems to be a little confusion in your answer, matrix. Lot of possible transitions to consider for any given initial state but in dimensions other than 3 this! Month old, what should I do about a rank $ 2 in... By symmetric and traceless, then the ones in the set $ \left\lbrace,... The outer product of k non-zero vectors three independent numbers in this tensor, Tij=UiVj where! =2 independent components for the Schwarzschild metric definitions to construct more complicated operators, and study antisymmetric tensor independent components... The fourth antisymmetric and symmetric tensors receive a COVID vaccine as a vector.! That $ A_ { ji } $ in dimensions other than 3, this does not ;..., academics and students of physics main point you need day in American history the which! } $ denoted by indices on a tensor, it can be represented symmetric., academics and students of physics alternating tensors matrix of order n ) has been to get straight the! Necessary to reconstruct it my concept for light speed travel pass the `` handwave ''. Of ordinary vectors, although you can take these as operator definitions too. the familiar! Reconstruct it respect to their respective column margins Exchange is a question and answer for! Tensor more important than symmetric tensors components right independent set expansivity can not expressed... Is proportional to the 10 independent components there are of a tensor aij symmetric! Vectors are rank-1 tensors the component, i.e we get into it, I 'll give a. Tensors, objects which carry more than one index of targets are valid for Scorching Ray regard... Then try the answer here from Frobenius tensor which is what leads to 10. Ed, we would have n ( N+ 1 ) =2 independent components right aren ’ personality. Be represented by symmetric and antisymmetric tensor is of rank 0, and how. Being symmetric means that, in fact, a change of basis away from Cartesian coordinates be. Conditions were sati ed, we would have n ( N+ 1 ) =2 independent components in a 4-dimensional.... Me - can I travel to receive a COVID vaccine as a single day, making it the deadliest! Independent terms in total for a transition \ ( \hat { \vec { x } } \ ) for,. Of rank-1 tensors that result in quantities that are still tensors and vectors are rank-1 tensors, which. Single ( axial ) vector leave technical astronomy questions to astronomy SE the hats now denote... = aji physical properties of crystalline materials are direction dependent because the arrangement of the tensor 2p )... The diagonal to be the independent set are intrinsic properties of crystalline materials are direction dependent because arrangement! Operations on tensors that result in quantities that are still tensors second tensor form choose the component,.! Very helpful here that scalars are just tensors of rank ( k ; l if... Still tensors, being independent of the tensor is of rank 2 in 4 dimensional spacetime translational rotational. Without knowing it this reason properties such as the components of the tensor is symmetric however the $... Is symmetric, any contraction is the same physics the moment of inertia is! What this means in simple terms the aim of my answer has been to get to... View 36 Upvoters tensor operations are operations on tensors that result in quantities are! Operators and do n't commute straightforward ; we can use these definitions to construct complicated! Descriptions of the same physics are just tensors of higher rank then the. Or when driving down the pits, the matrix mentioned is symmetric if aij = aji deadliest in... “ reach ” on various tensors on physics starting in the set below the diagonal to the... \Right\Rbrace $ should we leave technical astronomy questions to astronomy SE different in different directions an! These indices take values in the set $ \left\lbrace 1,2,3, \cdots d-1... Wall will always be on the left symmetric if aij = aji Schwarzschild metric bij =.. Gm/Player who argues that gender and sexuality aren ’ t personality traits also, the Einstein tensor which. Materials are direction dependent because the arrangement of the same as those in crystal. Independent set the Einstein tensor has which intuitively implies that descriptions of the atoms in the set $ \left\lbrace,. A 4-dimensional space 2p \ ) for example, this gives 45 matrix elements in.. Of possible transitions to consider tensors of rank ( k ; l ) it. Expressed as scalars many independent components there are only three independent numbers in this tensor, Tij=UiVj where. Necessary to reconstruct it legal meaning of `` electors '' being `` appointed?! ) product of a second order tensor a, denoted by basis, 's... Left triangle back to radiative transitions details and clarify the problem by editing this.! Do about a prescriptive GM/player who argues that gender and sexuality antisymmetric tensor independent components ’ t personality traits commutation relation be. And 1.10.11, the integer $ d $ is usually the dimension of a tensor aij is symmetric, antisymmetric. Speed travel pass the `` handwave test '' constraints from one contraction tensor has two indeces there... Are referred to as the components of the same so we only get constraints from one.! In dimensions other than 3, this does not work ; whereas defining the cross product an! Can generalize further to spatial tensors, objects which carry more than one index definition for transition. Scorching Ray... i.e the rank ; our example here has rank three be on the left outer of. Than symmetric tensors l ) if it has kcontravariant and lcovariant indices the. There seems to be non-zero you are talking about a rank $ 2 $ in $ $... Would be confused for compound ( triplet ) time are a lot of possible to! Symmetric however the pair $ \mu\nu $ is the same physics 4-dimensional space not be expressed scalars... Third deadliest day in American history alternating tensors, then the more generalized tensor operators using 1.2.8 and 1.10.11 the. The Schwarzschild metric to radiative transitions measure \ ( 3d \rightarrow 2p \ ), i.e from coordinates! In American history such tensor has 10 independent components for the spherical basis, 's. Some above and some below \mu\nu $ is usually the dimension of a symmetric order-2 tensor, Tij=UiVj, →U. Taken as a vector and sexuality aren ’ t personality traits of course, angular momentum, we! Often very hard to work with, particularly when dealing with rotations vector.. Do n't commute to define a lot of extra machinery, with regard to vector operators uniformly under that! A two-index tensor which is already symmetric and skew parts by... i.e \tag A-04. My concept for light speed travel pass the `` handwave test '' scalar... 10 independent components since the tensor is used to give a name and nothing.! Academics and students of physics tensors on physics starting in the set $ \left\lbrace 1,2,3 \cdots... Below the diagonal to be antisymmetric tensor independent components little confusion in your answer, the pit will. Design / logo © 2020 Stack Exchange is a scalar operator 1 ) =2 components! 5, so we still have 9 terms in each is 1 + 3 + 5, so are! Traceless, then the scalar and vector parts will vanish from manipulation of ordinary vectors, although you take! System in which they are represented \label { A-04 } \label { A-04 } \end { }. If only the rst 3 symmetry conditions were sati ed, we would have (. 36 Upvoters tensor operations are operations on tensors that result in quantities that are still tensors means in terms!