How can I transform L_V W^μ to another coordinate system and show that it transforms like a tensor, L is the lie derivative

Question

I have reached $L_V*W^μ=(dx^{ν'}/dx^ν)*V^{ν}*(dx^{ν}/dx^{ν'})*d_ν*(dx^{μ}/dx^{μ'}*W_μ)-dx^{ν}/dx^{ν'}*W_*dx^{μ}/dx^{μ'}*d_μ*dx^{ν'}/dx^ν*V^ν$ and I am stuck. I know i have to find that $L_V*W^μ= dx^{μ}/dx^{μ'}*(L_V*W_μ)$, which is the definition of a tensor. Any help would be really appreciated

Answer 1

We will begin by writing the Lie derivative acting on a tensor at a coordinate system $x^\mu$ given by $W_\mu$, as seen below

$ \mathcal{L}_V\cdot W_\mu=V^\nu\cdot \partial_\nu\cdot W_\mu + W_\nu\cdot\partial_\mu\cdot V^\nu $

We will change the coordinates from the $x^\mu$ coordinate system to a new system (primed), $x^{\mu'}$, transformed as seen below

$ W_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\cdot W_{\mu} $

$ V^{\nu'}=\frac{\partial x^{\nu'}}{\partial x^{\nu}}\cdot V^{\nu} $

We will now take the lie derivative acting on the new coordinate system tensor, $W_{\mu'}$

$ \mathcal{L}_V\cdot W_{\mu'}=V^{\nu'}\cdot \partial_{\nu'}\cdot W_{\mu'} + W_{\nu'}\cdot\partial_{\mu'}\cdot V^{\nu'} $

Furthermore, we can use the chain rule to apply the following transformations

$ \partial_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\cdot\partial_{\mu}, \partial_{\nu'}=\frac{\partial x^\nu}{\partial x^{\nu'}}\cdot\partial_{\nu} $

So using the above we get

$ \mathcal{L}_V\cdot W_{\mu'}=V^{\nu'}\cdot (\frac{\partial x^\nu}{\partial x^{\nu'}}\cdot\partial_{\nu})\cdot W_{\mu'} + W_{\nu'}\cdot(\frac{\partial x^\mu}{\partial x^{\mu'}}\cdot\partial_{\mu})\cdot V^{\nu'}=(\frac{\partial x^{\nu'}}{\partial x^{\nu}}\cdot V^{\nu})\cdot (\frac{\partial x^\nu}{\partial x^{\nu'}}\cdot\partial_{\nu})\cdot (\frac{\partial x^\mu}{\partial x^{\mu'}}\cdot W_{\mu}) + (\frac{\partial x^\nu}{\partial x^{\nu'}}\cdot W_{\nu})\cdot(\frac{\partial x^\mu}{\partial x^{\mu'}}\cdot\partial_{\mu})\cdot (\frac{\partial x^{\nu'}}{\partial x^{\nu}}\cdot V^{\nu})=V^\nu \cdot \partial\nu\cdot(\frac{\partial x^\nu}{\partial x^{\nu'}}\cdot W_{\nu}) + \frac{\partial x^\nu}{\partial x^{\nu'}}\cdot W_{\nu}\cdot\frac{\partial x^\mu}{\partial x^{\mu'}}\cdot\partial_{\mu}\cdot (\frac{\partial x^{\nu'}}{\partial x^{\nu}}\cdot V^{\nu})=...= \frac{\partial x^\mu}{\partial x^{\mu'}}\cdot(\mathcal{L}_V\cdot W_\mu)+V^\nu\cdot W_\mu\cdot \partial_\nu(\frac{\partial x^\mu}{\partial x^{\mu'}})+V^\nu\cdot\frac{\partial x^\nu}{\partial x^{\nu'}}\cdot W_{\nu}\cdot\frac{\partial x^\mu}{\partial x^{\mu'}}\cdot\partial_{\mu}(\frac{\partial x^{\nu'}}{\partial x^{\nu}}) $

As we can see, in order for the lie derivative of the new system tensor, we want the second term to be zero. This is true, as we will see below. We can start by using

$ \delta^{\nu'}_{\mu'}=\frac{\partial x^\nu}{\partial x^{\mu'}}\cdot\frac{\partial x^{\nu'}}{\partial x^{\nu}} $

Where

$ \partial_\mu(\delta^{\nu'}_{\mu'})=\partial_\mu(\frac{\partial x^\nu}{\partial x^{\mu'}}\cdot\frac{\partial x^{\nu'}}{\partial x^{\nu}})\Rightarrow\partial_\mu(\frac{\partial x^{\nu'}}{\partial x^{\nu}})\cdot\frac{\partial x^{\nu}}{\partial x^{\mu'}}=-\partial_\mu(\frac{\partial x^\nu}{\partial x^{\mu'}})\cdot\frac{\partial x^{\nu'}}{\partial x^{\nu}}\Rightarrow\partial_\mu(\frac{\partial x^{\nu'}}{\partial x^{\nu}})=-\partial_\mu(\frac{\partial x^\nu}{\partial x^{\mu'}})\cdot\frac{\partial x^{\nu'}}{\partial x^{\nu}}\cdot\frac{\partial x^{\mu'}}{\partial x^{\nu}} $

The second term becomes

$ V^\nu\cdot W_\mu\cdot \partial_\nu(\frac{\partial x^\mu}{\partial x^{\mu'}}) + V^\nu\cdot\frac{\partial x^\nu}{\partial x^{\nu'}}\cdot W_{\nu}\cdot\frac{\partial x^\mu}{\partial x^{\mu'}}\cdot\partial_{\mu}\cdot (\frac{\partial x^{\nu'}}{\partial x^{\nu}})= V^\nu\cdot W_\mu\cdot \partial_\nu(\frac{\partial x^\mu}{\partial x^{\mu'}}) - V^\nu\cdot\frac{\partial x^\nu}{\partial x^{\nu'}}\cdot W_{\nu}\cdot\frac{\partial x^\mu}{\partial x^{\mu'}}\cdot\partial_\mu(\frac{\partial x^\nu}{\partial x^{\mu'}})\cdot\frac{\partial x^{\nu'}}{\partial x^{\nu}}\cdot\frac{\partial x^{\mu'}}{\partial x^{\nu}}= V^\nu\cdot W_\mu\cdot \partial_\nu(\frac{\partial x^\mu}{\partial x^{\mu'}}) - V^\nu\cdot W_{\nu}\cdot\delta^\mu_\nu\cdot\partial_\mu(\frac{\partial x^\nu}{\partial x^{\mu'}})= V^\nu\cdot W_\mu\cdot(\frac{\partial x^\mu}{\partial x^{\nu}}) \cdot\partial_\mu(\frac{\partial x^\mu}{\partial x^{\mu'}}) - V^\nu\cdot W_{\nu}\cdot\delta^\mu_\nu\cdot\partial_\mu(\frac{\partial x^\nu}{\partial x^{\mu'}})=V^\mu\cdot W_\mu\cdot\partial_\mu(\frac{\partial x^\mu}{\partial x^{\mu'}}) - V^\mu\cdot W_{\mu}\cdot\partial_\mu(\frac{\partial x^\mu}{\partial x^{\mu'}})=0 $

Finally, we get

$ \mathcal{L}_V\cdot W_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\cdot(\mathcal{L}_V\cdot W_\mu) $

Which is the definition of a tensor, proving that the lie derivative in new coordinates transforms like a tensor.