弹性波动力学笔记（二）张量简介

Cartesian Tensors

In subsequent chapters the following three tensors will be introduced.

(1) The strain tensor \(\varepsilon_{ij}\) :

\[\varepsilon_{ij}=\frac{1}{2}(\frac{\partial u_{i}}{\partial x_j}+\frac{\partial u_j}{\partial x_i})\tag{1.4.1} \]

where the vector \(\mathbf{u}=(u_1,u_2,u_3)\) is the displacement suffered by a particle inside a body when it is deformed.

(2) The stress tensor \(\tau_{ij}\):

\[T_{i}=\tau_{ij}n_{j} \tag{1.4.2} \]

where \(T_i\) and \(n_j\) indicate the components of the stress vector and normal vector.

(3)The elastic tensor \(c_{ijkl}\) which relates stress to strain:

\[\tau_{ij}=c_{ijkl}\varepsilon_{kl} \tag{1.4.3} \]

Let us list some of the differences between vectors and tensors. Firstly, while a vector can be represented by a single symbol, such as \(\mathbf{u}\) , or by its components, such as \(u_j\), a tensor can only be represented by its components, such as \(\varepsilon_{ij}\) , although the introduction of dyadics will allow the representation of tensors by single symbols. Secondly, while vector components carry only one subindex, tensors carry two subindices or more. Thirdly, in the three-dimensional space we are considering, a vector has three components, while \(\varepsilon_{ij}\) and \(\tau_{ij}\) have \(3\times 3\), or nine components, and \(c_{ijkl}\) is a fourth-order tensor, with the order of the tensor being given by the number of free indices. There are also differences among the tensors shown above. For example， \(\varepsilon_{ij}\) is defined in terms of operations (derivatives) on the components of a single vector, while \(\tau_{ij}\) appears in a relation between two vectors. \(c_{ijkl}\), on the other hand, relates two tensors.

Clearly, tensors offer more variety than vectors, and because they are defined in terms of components, the comments made in connection with vector components and the rotation of axes also apply to tensors. to motivate the following definition of a second-order tensor consider the relation represented by equation (1.4.2). For this relation to be independent of the coordinate system, upon a rotation of axes we must have

\[T_l^{'}=\tau_{lk}^{'}n_{k}^{'} \tag{1.4.4} \]

In other words, the functional form of the relation must remain the same after a change of coordinates. We want to find the relation between \(\tau_{lk}^{'}\) and \(\tau_{ij}\) that satisfies (1.4.2) and（1.4.4). To do that multiply by \(a_{li}\) and sum over \(i\):

\[a_{li}T_i=a_{li}\tau_{ij}n_{j} \tag{1.4.5} \]

Before proceeding rewrite \(T_{l}^{'}\) and \(n_{j}\) using with \(\mathbf{v}\) replaced by \(\mathbf{T}\) and with \(\mathbf{v}\) replaced by \(\mathbf{n}\) . This gives

\[\mathbf{T}_l^{'}=a_{li}T_{i};\newline n_j=a_{kl}n_k^{'} \tag{1.4.6 a,b} \]

From (1.4.6 a,b), (1.4.2), and(1.4.5) we find

\[T_l^{'}=a_{li}T_{i}=a_{li}\tau_{ij}n_j=a_{li}\tau_{ij}a_{kj}n_k^{'}=(a_{li}a_{kj}\tau_{ij})n_k^{'}\tag{1.4.7} \]

Now subtracting (1.4.4) from (1.4.7) gives

\[(\tau_{lk}^{'}-a_{li}a_{kj}\tau_{ij})n_{k}^{'}=0 \tag{1.4.8} \]

As \(n_k\) is an arbitrary vector, the factor in parentheses in equation (1.4.8) must be equal to zero, so that

\[\tau_{lk}^{'}=a_{li}a_{kj}\tau_{ij}\tag{1.4.9} \]

Note that equation (1.4.9) does not depend on the physical nature of the quantities involved in equation (1.4.2). Only the functional relation matters. This result motivates the following definition.

---

Second-order tensor Given nine quantities \(t_{ij}\), they constitute the components of a second-order tensor if they transform according to

\[t_{ij}^{'}=a_{il}a_{jk}t_{lk} \tag{1.4.10} \]

under a change of coordinates \(v_i^{'}=a_{ij}v_j\).

To write the tensor components in the unprimed system in terms of the components in the primed system, multiply (1.4.10) by \(a_{im}a_{jn}\) and sum over \(i\) and \(j\) , and use the orthogonality relation:

\[a_{im}a_{jn}t_{ij}^{'}=a_{im}a_{jn}a_{il}a_{jk}t_{lk}=a_{im}a_{il}a_{jn}a_{jk}t_{lk}=\delta_{ml}\delta_{nk}t_{lk}=t_{nm} \tag{1.4.11} \]

Therefore,

\[ t_{nm}=a_{im}a_{jn}t_{ij}^{'} \tag{1.4.12} \]

As (1.4.10) and (1.4.12) are similar, it is important to make sure that arrangement of indices is strictly adhered to.

Equation (1.4.11) illustrates an important aspect of the Kronecker delta. For a given \(m\) and \(n\), the expression \(\delta_{lm}\delta_{kn}t_{lk}\) is a double sum over \(l\) and \(k\), so that there are nine terms. However, since \(\delta_{lm}\) and \(\delta_{kn}\) are equal to zero except when \(l=m\) and \(k=n\), in which case the deltas are equal to one, the only nonzero term in the sum is \(t_{mn}\). Therefore, the equality \(\delta_{lm}\delta_{kn}t_{lk}=t_{mn}\) can be derived formally by replacing \(l\) and \(k\) in \(t_{lk}\) by \(m\) and \(n\) and by dropping the deltas. The extension of (1.4.10) to higher-order tensors is straight-forward.

---

N-order Tensor. Given \(3^n\) quantities \(t_{i_{1}i_2…i_n}\) ，they constitute the components of a tensor of order \(n\) if they transform according to

\[ t_{i_1i_2…i_n}^{'}=a_{i1j1}a_{i2j2}…a_{i_nj_n}t_{j_1j_2…j_n} \tag{1.4.13} \]

under a change of coordinates \(v_{i}^{'}=a_{ij}v_j\). All the indices \(i_1,i_2,…\) and \(j_1,j_2,j_3,…\) can be \(1,2,3\).