Infinitesimal change of coordinates transformation law

Question

I have a question regarding coordinate transformations. When we have a vector, say $X^\mu\partial_\mu$, in some coordinates $\xi^\alpha$ and perform a change of coordinates $\xi^\alpha\to\eta^\alpha$, the components of the original vector written in the new coordinates can be retrieved using the vector transformation law:

$$\tilde{X}^\alpha=\dfrac{\partial\eta^\alpha}{\partial\xi^\beta}X^\beta$$

In certain cases we perform an infinitesimal change of coordinates $\xi^\mu \to \xi^\mu + \epsilon v^\mu(\xi)$. Applying the above formula, one would expect to get

$$\tilde{X}^\alpha=\dfrac{\partial(\xi^\alpha+\epsilon v^\alpha(\xi))}{\partial\xi^\beta}X^\beta=\delta^\alpha_\beta X^\beta+\epsilon\dfrac{\partial v^\alpha}{\partial \xi^\beta}X^\beta=X^\alpha+\epsilon\dfrac{\partial v^\alpha}{\partial \xi^\beta}X^\beta$$

However, in How does a vector field transform under an infinitesimal coordinate transformation? it is stated that the transformation requires the use of the Lie derivative. The result is similar if we perform a Taylor expansion:

$$\tilde{X}^\alpha=X^\alpha(\xi^\beta+\epsilon v^\beta(\xi))\simeq X^\alpha(\xi^\beta)+\epsilon v^\nu\dfrac{\partial X^\alpha}{\partial\xi^\nu}=X^\alpha+\epsilon v^\nu\dfrac{\partial X^\alpha}{\partial\xi^\nu}$$

However, $\epsilon\dfrac{\partial v^\alpha}{\partial \xi^\beta}X^\beta\neq \epsilon v^\nu\dfrac{\partial X^\alpha}{\partial\xi^\nu}$.

• Why does this happen? Which is the correct approach to this change of coordinates?

• If the correct approach is the Lie derivative, why is the formula for the change of coordinates used in Metric components transformation under change of coordinates instead of the Lie derivative?

Answer 1

The simple answer is that the two are different (but related) concepts and you have confused them. The Lie derivative of a vector $\mathbf{u}$ in the direction $\mathbf{v}$ (tangent to the flow $\phi_\epsilon$) is defined as $$\mathcal{L}_{\mathbf{v}} \mathbf{u}= \lim_{\epsilon \to 0} \frac{\phi_\epsilon^*[\mathbf{u}(\phi_\epsilon (x))]-\mathbf{u}(x)}{\epsilon}$$

The focus is the term $\phi_\epsilon^*[\mathbf{u}(\phi_\epsilon (x))]$. What this means is that we need to take the vector $\mathbf{u}$ at a nearby point $\phi_\epsilon (x)$ further down the flow, which is $\mathbf{u}(\phi_\epsilon (x))$, and then pull it back to our starting point $x$ using $\phi_\epsilon^*$.

To first order $\phi_\epsilon (x) \approx x+\epsilon\mathbf{v}$, so we obtain by Taylor expansion, in components, $$u^\alpha(\phi_\epsilon (x)) \approx u^\alpha(x+\epsilon\mathbf{v})= u^\alpha(x)+\epsilon v^\beta \frac{\partial u^\alpha (x)}{\partial x^\beta}$$

However, we are not done yet. We need to pull it back to our starting point $x$ and this is accomplished by performing an infinitesimal coordinate transformation $x^\mu \to x^\mu -\epsilon v^\mu$ on both of the above terms. Applying the standard tensor transformation law, the first term becomes $$u^\alpha \to u^\alpha-\epsilon u^\beta \frac{\partial v^\alpha}{\partial x^\beta}$$ while the second term gives an additional term of order $\epsilon^2$ which can be ignored. So we end up with $$\left(\phi_\epsilon^*[\mathbf{u}(\phi_\epsilon (x))]\right)^\alpha = u^\alpha + \epsilon v^\beta \frac{\partial u^\alpha}{\partial x^\beta}-\epsilon u^\beta \frac{\partial v^\alpha}{\partial x^\beta}$$

Inserting it into the definition gives us the Lie derivative $$\mathcal{L}_{\mathbf{v}} \mathbf{u}= [\mathbf{v},\mathbf{u}] = \left(v^\beta \frac{\partial u^\alpha}{\partial x^\beta}- u^\beta \frac{\partial v^\alpha}{\partial x^\beta}\right)\mathbf{e}_\alpha$$

The infinitesimal coordinate transformation is only part of the construction of the Lie derivative. If you simply perform an infinitesimal coordinate transformation (which is what you did in your first equation), you will, of course, not obtain the term arising from the Taylor expansion.