# Infinitesimal change of coordinates transformation law

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

• The topmost equation is a tensor transformation, which you define at a given tangent space. Coordinate transformation is supposed to be linear, but that is not generally satisfied by the form of infinitesimal coordinate transformation you have considered. In your case, the vector transformation should look like $X(x)\to \Lambda X(\Lambda^{-1}x)$, where $\Lambda$ is the transformation matrix
– KP99
Nov 28, 2021 at 15:30

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.

• Thank you for your detailed answer! I think I have understood it, but just to make sure I'd like to confirm that if someone says "we perform an infinitesimal change of coordiantes" then I am supposed to use the standard transformation law, whereas if I am interested in constructing the Lie derivative I need to perdorm this computation you detailed. Moreover, it is not the same to perform the coordinate change than to evaluate the vector component at $x+\epsilon v$, since this can be taylored-expanded. Nov 29, 2021 at 23:46
• Also, why does the vector field change according to the Lie derivative under this infinitesimal coordinate change (see answer to physics.stackexchange.com/questions/310178/… )? Should it not change as the usual transformation law? As I understand it, the Llie derivative is another thing not related to coordinate changes Nov 29, 2021 at 23:52
• @Elementarium Yes, your first comment is correct. The linked post is unclear. If we are only considering a change of coordinates without any flow, we should indeed use the standard transformation law. Nov 30, 2021 at 7:35