Here is an approach for lie derivative. And i would like to know how wrong is it.(adsbygoogle = window.adsbygoogle || []).push({});

Assuming lie derivative of a vector field measures change of a vector field along a vector field, take a coordinate system, x^{i}, and the vector field f^{i}along which T^{i}is being changed. I go this way, i take the vector 'T' at point x^{i}, transform the coordinate as- x^{i}-> x^{i}+ ef^{i}= x'^{i}, e nearly zero. Now T^{i}(x) is transformed to T'^{i}(x').

Now I subtract T'(x') with T(x).

We would do the same thing to find directional derivative of a scalar function- S( x + ef)- S(x)= S(x')-S(x)

Now back to vector T. T'^{j}(x')= T^{i}(x)dx'^{j}/dx^{i}.

Now take T^{i}(x') = T^{i}(x) + ef^{k}[tex]\nabla[/tex]_{k}T^{i}. Since T(x) is a vector field. But I don't ask for a definition of connection. We use arbitrary connection and try to see if T(x') is completely specified by T(x) by requiring specification of connection [tex]\nabla[/tex].

So T'^{j}(x')= T^{i}(x') - e( f^{k}[tex]\nabla[/tex]_{k}T^{i}(x) - T^{k}(x')df^{i}/dx^{k}

To first approximation.

This finally gives, considering first order in e,

T'^{j}(x')= T^{i}(x) - e( f^{k}[tex]\nabla[/tex]_{k}T^{i}(x) - T^{k}(x')[tex]\nabla[/tex]_{k}f^{i})

= T^{i}(x) + e(lie derivative required) + e^{2}(terms..)

As is seen here, the final form does not include the notion of a definite vector field for T. Even connection [tex]\nabla[/tex] can be removed from lie derivative for vector. I need to know if it is correct, the assumption that lie derivative can be considered as change in tensor, not necessarily a tensor field. Or is a precisely defined tensor field property already used here?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Lie derivative and vector field notion.

**Physics Forums | Science Articles, Homework Help, Discussion**