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3.2: Vectors and Scalars

  • Page ID
    18058
  • 3.2.1 The position vector in a rotated coordinate system

    The position vector \(\vec{x}\) that labels a point in space (e.g., the blue vector in figure 3.1.1) is a physical entity that exists independently of any observer and of any coordinate frame used to measure it. If you use the basis vectors \(\hat{e}\) and I use \(\hat{e}^\prime\), then \(\vec{x}\) can be expressed as a linear combination of either set:

    \[\vec{x}=x_{i} \hat{e}^{(i)}=x_{i}^{\prime} \hat{e}^{\prime(i)}.\label{eq:1}\]

    Now suppose that our coordinate frames are related by a rotation matrix \(\underset{\sim}{C}\). If we know the components \(x_i\) that you measure and the rotation matrix \(\underset{\sim}{C}\), can we predict the components \(x^\prime_i\) that I will measure? We will do this by solving Equation \(\ref{eq:1}\) for the components of the vector \(\vec{x}^\prime\). Start with the second equality of Equation \(\ref{eq:1}\):

    \[x_{i}^{\prime} \hat{e}^{\prime(i)}=x_{i} \hat{e}^{(i)}.\]

    How do we solve this equality for \(x^\prime_i\)? If \(\hat{e}^{\prime(i)}\) were a scalar, we would simply divide it out of both sides, but division by a vector makes no sense. Instead, we dot both sides with \(\hat{e}^{\prime(j)}\):

    \[x_{i}^{\prime} \hat{e}^{\prime(i)} \cdot \hat{e}^{\prime(j)}=x_{i} \hat{e}^{(i)} \cdot \hat{e}^{\prime(j)}.\]

    Now use Equation 3.1.12 on the right-hand side,1

    \[\begin{aligned}
    x_{i}^{\prime} \underbrace{\hat{e}^{\prime(i)} \cdot \hat{e}^{\prime(j)}}_{=\delta_{i j}} &=x_{i} \underbrace{\hat{e}^{(i)} \cdot \hat{e}^{(k)}}_{=\delta_{i k}} C_{k j} \\
    x_{i}^{\prime} \delta_{i j} &=x_{i} \delta_{i k} C_{k j}
    \end{aligned}.\]

    Finally, multiplying out the deltas, we have

    \[x_{j}^{\prime}=x_{i} C_{i j}.\label{eq:2}\]

    This result is similar to the transformation rule Equation 3.1.12 for the basis vectors, though its meaning is entirely different. In each case, the dummy index is in the first position of the rotation matrix. The relation Equation \(\ref{eq:2}\) can also be written in vector notation:

    \[\vec{x}^{\prime}=\vec{x} \underset{\sim}{C}, \quad \text { or } \vec{x}=\underset{\sim}{C}^{T} \vec{x}\]

    Now consider the reverse transformation, i.e., my measurement into yours. Multiplying both sides of Equation \(\ref{eq:2}\) by \(C_{kj}\), we have

    \[x_{j}^{\prime} C_{k j}=x_{i} C_{i j} C_{k j}=x_{i} C_{i j} C_{j k}^{T}=x_{i} \delta_{i k}=x_{k}\]

    and therefore

    \[x_{j}=x_{i}^{\prime} C_{j i}.\label{eq:4}\]

    Compare this with Equation \(\ref{eq:2}\). The letters used to label the indices do not matter. The important distinction is that, for the reverse transformation, the dummy index is in the second position whereas for the forward transformation it is in the first. As with Equation \(\ref{eq:2}\), there is an equivalent expression in vector form:

    \[\vec{x}=\vec{x}^{\prime} \underset{\sim}{C}^{T}, \quad \text { or } \quad \vec{x}=\underset{\sim}{C} \vec{x}^{\prime}.\label{eq:5}\]

    3.2.2 Differentiating the position vector

    In a Cartesian coordinate system \({x,y}\), what is \(\partial y/\partial x\)? This is just the change in \(y\) as we move in the \(x\)-direction, i.e., zero. In contrast, \(\partial x /\partial x = 1\). In general, for a 3-dimensional system, we can write:

    \[\frac{\partial x_{j}}{\partial x_{i}}=\delta_{i j}.\label{eq:6}\]

    Now imagine rotating the system using rotation matrix \(\underset{\sim}{C}\). How rapidly does the \(j^{th}\) coordinate in the rotated system change if we move along the \(i^{th}\) coordinate axis in the original system? First, expand Equation \(\ref{eq:2}\) as

    \[x_{j}^{\prime}=x_{1} C_{1 j}+x_{2} C_{2 j}+x_{3} C_{3 j}.\]

    We can easily differentiate \(x^\prime_j\) with respect to any of the un-primed coordinates. For example:

    \[\frac{\partial x_{j} \prime}{\partial x_{1}}=C_{1 j}.\]

    More generally:

    \[\frac{\partial x_{j} \prime}{\partial x_{i}}=C_{i j}.\label{eq:7}\]

    3.2.3 Defining Cartesian vectors and scalars

    We now establish a more precise definition of a vector using the position vector as a prototype.

    Definition

    A vector is a quantity possessed of direction and magnitude independent of the observer. A vector must therefore transform in the same way as the position vector:

    \[v_{j}^{\prime}=v_{i} C_{i j} ; \quad v_{j}=v_{i}^{\prime} C_{j i}.\label{eq:8}\]

    Before referring to any quantity as a vector, we should check to see that it satisfies this criterion.

    Examples:

    • Consider the velocity of a moving point:

    \[u_{i}=\frac{d}{d t} x_{i}.\label{eq:9}\]

    In rotated coordinates,

    \[u_{j}^{\prime}=\frac{d}{d t} x_{j}^{\prime}=\frac{d}{d t}\left(x_{i} C_{i j}\right)=\frac{d x_{i}}{d t} C_{i j}=u_{i} C_{i j}.\]

    This confirms that the velocity transforms in the same way as the position vector, and therefore that it qualifies as a vector.2

    • The wave vector of a plane wave transforms as a vector. Consider a surface gravity wave whose surface elevation is given by \(\eta=\eta^0\cos(\vec{k}\cdot\vec{x}-\omega t)\). The phase function \(\vec{k}\cdot\vec{x}-\omega t\) is a scalar; for example, the crest of a wave is a real entity that looks like the crest to every observer. Therefore \(\vec{k}\cdot\vec{x}\) must be a scalar. Defining that scalar as \(D\), we can write:

    \[D^{\prime}=k_{i}^{\prime} x_{i}^{\prime}=k_{i}^{\prime} x_{j} C_{j i}=\underbrace{k_{i}^{\prime} C_{j i}}_{k_{j}}x_{j}=D,\]

    which requires that \(k_j = k^\prime_iC_{ji}\).

    Counterexample: The phase velocity of a plane wave sounds like a vector. It has three components, and we usually use “velocity’’ rather than “speed’’ to differentiate the vector from the scalar. However, the definition of phase velocity is inextricably linked to a particular coordinate system. From the inverse relationship between phase velocity and wave vector, ci = ω/ki, it should be clear that ci does not transform according to Equation \(\ref{eq:2}\). If not, try it.

    Definition

    A scalar is a single number that is the same in every reference frame.

    Examples:

    Temperature is a scalar. You and I can look from different angles and we will still perceive the same temperature.

    The dot product of two vectors is a scalar. To see this, write the dot product of \(\vec{x}\) and \(\vec{y}\) as \(D=x_iy_i\). Then ask what the value of \(D\) would be in a rotated reference frame:

    \[\begin{aligned}
    D^{\prime}=x_{i}^{\prime} y_{i}^{\prime} &=\overbrace{x_{j} C_{j i}}^{x_{i}^{\prime}} \overbrace{y_{k} C_{k i}}^{y_{i}^{\prime}} \\
    &=x_{j} y_{k} C_{j i} C_{k i} \\
    &=x_{j} y_{k} C_{j i} C_{i k}^{T} \\
    &=x_{j} y_{k} \delta_{j k}=x_{j} y_{j}=D
    \end{aligned}.\]

    Counterexample: A single element of a vector, while consisting of a single number, is not a scalar. For example, the x-coordinate of a point in space differs, obviously, in different coordinate systems.

    1Compare with (3.1.9) and note that the dummy index i has been relabeled as k to avoid conflict with the dummy index i that is already in use. Also, the order of the basis vector and the rotation matrix has been reversed, as is always permissible.

    2The velocity components \({u_1,u_2,u_3}\) are commonly replaced by \({u,v,w}\).