The question Let $e_a$ be the coordinate basis vectors in a manifold described by coordinate system $x^a$. The vector displacement between two nearby points is given by \begin{equation} ds=dx^ae_a=dx'^ae'_a \end{equation} where a prime denotes the same quantity measured in a different reference frame.

Using the relationship $dx^a=\frac{\partial x^a}{\partial x'^b}dx'^b$, it is easy to see that \begin{equation} e'_a=\frac{\partial x^b}{\partial x'^a}e_b \end{equation} However, my textbook (General Relativity An Introduction for Physicists by Hobson Lasenby and Efsthathiou, pg 60-61) says that the following must obviously be true as well: \begin{equation} e'^a=\frac{\partial x'^a}{\partial x^b}e^b \end{equation} My question is how do I prove this?

Attempt at a solution

I can see that this would follow if we assume that $dx_a$ transforms as \begin{equation} dx'_a=\frac{\partial x^a}{\partial x'^b}dx_b \end{equation} And the proof goes as: \begin{eqnarray} dx_ae^a&=&dx'_ae'^a\\ &=&\frac{\partial x^a}{\partial x'^b}dx_b e'^a \end{eqnarray} Switching the dummy index to b on the left side we get: \begin{equation} dx_be^b=\frac{\partial x^a}{\partial x'^b}dx_b e'^a \end{equation} which gives: \begin{equation} e^b=\frac{\partial x^a}{\partial x'^b} e'^a \end{equation} Interchanging primed and unprimed variables as well as a and b we get \begin{equation} e'^a=\frac{\partial x'^b}{\partial x^a} e^b \end{equation}

This proof seems a bit clumsy to me, AND I am unable to prove my starting equation (my text uses the transformation rule for $e'^a$ to prove it later). Is there a better way of deriving this result WITHOUT using the equation I started with?

  • 1
    $\begingroup$ Which text? Which page? $\endgroup$ – Qmechanic May 14 at 8:32
  • $\begingroup$ The text is General Relativity An Introduction for Physicists by Hobson, Efsthathiou and Lasenby, page 60-61 $\endgroup$ – samgon May 14 at 8:57

I do not know what you mean by the "coordinate basis vectors" with their upstairs indices. The basis vectors of the tangent space ${\rm TM}_p$ at a point $p$ are usually written with downstairs indices as $$ {\bf e}_\mu= \frac{\partial}{\partial x^\mu}. $$
The basis vectors with upstairs indices are $$ {\bf e}^{*\mu}= dx^\mu $$ and they are the basis vectors of the dual space $({\rm TM}_p)^*$.

If we insert a displacement vector $$ \delta {\bf x}= \delta x^\mu \frac{\partial}{\partial x^\mu} $$ into the dual basis element $dx^\nu$ we use the definition of the dual basis $$ dx^\nu(\partial_\mu)= \delta^\nu_\mu $$ to get $$ dx^\nu( \delta {\bf x})= \delta x^\nu $$ Thus $dx^\nu$ is a machine into which we drop a vector a get back the numerical components of the vector.

If we change coordinates to $x'^\mu$ we have $$ {\bf e}'^{*\nu}= dx'^\mu= \frac{\partial x'^\mu}{\partial x^\nu} dx^\nu= \frac{\partial x'^\mu}{\partial x^\nu}{\bf e}^{*\nu} $$ and $$ {\bf e}'_\mu= \frac{\partial}{\partial x'^\mu}= \frac{\partial x^\nu }{\partial x'^\mu }\frac{\partial}{\partial x^\nu}= \frac{\partial x^\nu }{\partial x'^\mu }{\bf e}_\nu $$

  • $\begingroup$ This is really helpful. In the book I am following, the $e_{\mu}$ are defined as $\partial s/\partial x^{\mu}$ where $\delta s$ is the infinitesimal vector separation between two nearby points. Thinking of them as operators $\partial /\partial x^{\mu}$instead makes this a lot easier. Thanks so much! $\endgroup$ – samgon May 15 at 11:29
  • $\begingroup$ @samgon This is the usual approach by mathematicians. It is due to Elie Cartan. It looks a bit weird at first, but after a while one gets used to it and computationa advantages are substantial. $\endgroup$ – mike stone May 15 at 12:34

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