Physicist's request for intuition on covariant derivatives and Lie derivatives - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:41:17Z http://mathoverflow.net/feeds/question/75220 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75220/physicists-request-for-intuition-on-covariant-derivatives-and-lie-derivatives Physicist's request for intuition on covariant derivatives and Lie derivatives Igor Rivin 2011-09-12T13:34:26Z 2012-01-03T19:59:04Z <p>A friend of mine is studying physics, and asks the following question which, I am sure, others could respond to better:</p> <p>What is the difference between the covariant derivative of $X$ along the curve $(t)$ and a Lie derivative of $X$ along $y(t)?$ I know the technical stuff about not needing to define a connection with a Lie derivative, needing to define the fields $X$ and $Y$ over a greater neighborhood, etc.</p> <p>I am looking for a more physical sense. If a Lie derivative gives the sense of the change of a vector field along the direction of another field, how does the covariant derivative differ?</p> http://mathoverflow.net/questions/75220/physicists-request-for-intuition-on-covariant-derivatives-and-lie-derivatives/75227#75227 Answer by Deane Yang for Physicist's request for intuition on covariant derivatives and Lie derivatives Deane Yang 2011-09-12T14:47:57Z 2011-09-12T14:47:57Z <p>The Lie derivative of a vector field $X$ with respect to another vector field $Y$ is just the Lie bracket of the two vector fields. It is well-defined given only the smooth structure and does not require any connection. In other words, it is independent of changes of co-ordinates and is preserved under any diffeomorphism. Given how flexible diffeomorphisms are, it can't be a pointwise or even curvewise concept, since you can basically map any pair of nonzero vectors to any other pair and even any nonvanishing transversal vector field along a curve to any other nonvanishing transversal vector field along another curve.</p> <p>But we know what the Lie derivative tells us. It tells us how "coherent" or "independent" the two vector fields are with respect to each other locally (on an open set and not just at a point). It measures to what extent the generated flows commute, i.e. what happens if you first travel along an integral curve of one and then along one of the other versus the opposite order.</p> <p>Another way to think about this is, discussed in control theory, to think about the set you get if you flow first along one vector field, then the other, then the first one again, etc. If the Lie bracket vanishes, then you stay inside a 2-dimensional surface. If it doesn't, then the value of the Lie bracket (and its iterates) tells you the dimension of the set that you stay inside.</p> <p>A connection allows you to define the concept of a "constant" vector along a curve, i.e. parallel translation along a curve. It is important to understand that defining parallel translation is an extra assumption or geometric structure added to the smooth manifold.</p> http://mathoverflow.net/questions/75220/physicists-request-for-intuition-on-covariant-derivatives-and-lie-derivatives/75230#75230 Answer by Qfwfq for Physicist's request for intuition on covariant derivatives and Lie derivatives Qfwfq 2011-09-12T15:44:12Z 2011-09-12T15:50:14Z <p>Let $T$ be a tensor field on the manifold $M$, $\nabla$ a connection, $v$ a tangent vector at $x\in M$, and $V$ a vector field such that $V(x)=v$.</p> <p>Then the intuition is as follows:</p> <blockquote> <p>The covariant derivative $\nabla_v T$ is the derivative of $T$ <strong>along a geodesic</strong> arc $\gamma$ for $\nabla$ which has direction $v$ at $x=\gamma(0)$. The derivation is computed in the finite dimensional tangent space $T_xM$, as nearby values $T(y)$, $y\in M$, are compared via parallel transport. </p> </blockquote> <p>(Remark: here "geodesic arc" should be made more precise, as geodesics emanating from $x$ are determined as <em>parametrized</em> curves and it may happen that the geodesic in the direction $v$ doesn't have velocity $v$)</p> <hr> <blockquote> <p>The (value at the point $x$ of the) Lie derivative $\mathcal{L}_VT$ is the derivative of $T$ <strong>along the flowline</strong> of $V$ (passing through $x$). The derivation is computed in the finite dimensional tangent space $T_xM$, as nearby values $T(y)$, $y\in M$, are compared via pullback along the local flow of $V$. </p> </blockquote> http://mathoverflow.net/questions/75220/physicists-request-for-intuition-on-covariant-derivatives-and-lie-derivatives/75237#75237 Answer by Xiao Xinli for Physicist's request for intuition on covariant derivatives and Lie derivatives Xiao Xinli 2011-09-12T17:43:35Z 2011-09-12T17:43:35Z <p>Lie derivative is based on a Lie group (or Lie algebra) which acts on the manifold. This derivative cannot be defined just at one point because the action cannot be defined at a point even if you give explicitly the direction at that point. On the other hand, using connection, covariant derivative can be defined pointwise. I think this is the main technical difference between them.</p> http://mathoverflow.net/questions/75220/physicists-request-for-intuition-on-covariant-derivatives-and-lie-derivatives/84821#84821 Answer by Liviu Nicolaescu for Physicist's request for intuition on covariant derivatives and Lie derivatives Liviu Nicolaescu 2012-01-03T18:33:42Z 2012-01-03T19:59:04Z <p>First let me say that what is intuitive to a physicist may be not be so to a geometer and vice-versa. To many physicists a connection is the potential of a field satisfying a gauge invariance. For this point of view I refer to vol. 1, Chap. 6 sect 41 of the three volume book by Dubrovin-Fomenko-Novikov: Modern Geometry-Methods and applications.</p> <p>I find this point of view less intuitive only because I was trained as a mathematician.</p> <p>The notion of covariant derivative appears naturally when one tries to solve the following problem. Suppose that $E\to M$ is a smooth vector bundle over a smooth manifold $M$. For example, $E$ could be the tangent bundle of $M$. We seek a notion of parallel transport. More precisely, this is a correspondence that associates to each smooth path</p> <p>$$\gamma: [a,b]\to M$$</p> <p>a linear map $T_\gamma$ from the fiber of $E$ at the initial point of $\gamma$ to the fiber of $E$ over the final point of $\gamma$</p> <p>$$T_\gamma: E_{\gamma(a)}\to E_{\gamma(b)}.$$</p> <p>The map $T_\gamma$ is called the <strong>parallel transport along the path $\gamma$</strong>.The assignment $\gamma\mapsto T_\gamma$ should satisfy two natural conditions .</p> <p>(a) $T_\gamma$ should depend smoothly on $\gamma$. (The precise meaning of this smoothness is a bit technical to formulate, but in the end it means what your intuition tells you it should mean.) </p> <p>(b) If $\gamma_0: [a,b]\to M$ and $\gamma_1:[b,c]\to M$ are two smooth paths such that the initial point of $\gamma_1$ coincides with the final point, then we obtain by concatenation a path $\gamma:[a,c]\to M$ and we require that</p> <p>$$T_\gamma= T_{\gamma_1}\circ T_{\gamma_0}.$$</p> <p>Suppose we have a concept of parallel transport. Given a smooth path $\gamma:[0,1]\to M$ and a section $\boldsymbol{u}(t)\in E_{\gamma(t)}$, $t\in [0,1]$ of $E$ over $\gamma$, then we can define a concept of derivative of $\boldsymbol{u}$ along $\gamma$. More precisely</p> <p>$$\nabla_{\dot{\gamma}} \boldsymbol{u}|_{t=t_0}=\lim_{\varepsilon \to 0} \frac{1}{\varepsilon} \left( T^{t_0,t_0+\varepsilon}_\gamma \boldsymbol{u}(t_0+\varepsilon)- \boldsymbol{u}(t_0)\right),$$</p> <p>where $T^{t_0,t_0+\varepsilon}_\gamma$ denotes the parallel transport along $\gamma$ from the fiber of $E$ over $\gamma(t_0+\varepsilon)$ to the fiber of $E$ over $\gamma(t_0)$. The left-hand-side of the above equality is called the covariant derivative of $\boldsymbol{u}$ along the vector field $\dot{\gamma}$ determined by the parallel transport. Thus, a choice of parallel transport leads to a concept of covariant derivative.</p> <p>Conversely, a covariant derivative $\nabla$ leads to a parallel transport. Given a smooth path $\gamma:[0,1]\to M$ the parallel transport</p> <p>$$T_{\gamma}: E_{\gamma(0)}\to E_{\gamma(1)}$$</p> <p>is defined as follows. Fix $u_0\in E_{\gamma(0)}$. Then there exists a unique section $\boldsymbol{u}(t)$ of $E$ over $\gamma$ satisfying</p> <p>$$\boldsymbol{u}(0)=u_0,\;\;\nabla_{\dot{\gamma}}\boldsymbol{u}(t)=0,\;\;\forall t\in [0,1].$$</p> <p>We then set</p> <p>$$T_\gamma u_0:= \boldsymbol{u}(1).$$</p>