covariant derivative of a vector field

vectors are constants, r;, = 0, and the covariant derivative simplifies to (F.27) as you would expect. We discuss the notion of covariant derivative, which is a coordinate-independent way of differentiating one vector field with respect to another. All Answers (8) 29th Feb, 2016. How to write complex time signature that would be confused for compound (triplet) time? What are the differences between the following? As Mike Miller says, vector fields with $\nabla_XX=0$ are very special. If a vector field is constant, then Ar;r =0. How to holster the weapon in Cyberpunk 2077? The proposition follows from results on ordinary differential-----DX equations. Consider a vector field $X$ on a smooth pseudo-Riemannian manifold $M$. 0 Proof. The G term accounts for the change in the coordinates. Making statements based on opinion; back them up with references or personal experience. Examples of how to use “covariant derivative” in a sentence from the Cambridge Dictionary Labs Wouldn’t it be convenient, then, if we could integrate by parts with Lie derivatives? By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. at every point in time, apply just enough acceleration in the normal direction to the manifold to keep the particle's velocity tangent to the manifold. Advice on teaching abstract algebra and logic to high-school students, I don't understand the bottom number in a time signature. First we cover formal definitions of tangent vectors and then proceed to define a means to “covariantly differentiate”. Even if a vector field is constant, Ar;q∫0. where is defined above. This is just Lemma 5.2 of Chapter 2, applied on R 2 instead of R 3, so our abstract definition of covariant derivative produces correct Euclidean results.. Remember that the tangent plane may vary from point to point. is a scalar density of weight 1, and is a scalar density of weight w. (Note that is a density of weight 1, where is the determinant of the metric. Judge Dredd story involving use of a device that stops time for theft. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Tensor transformations. If so, can we say the gradient is a vector-valued form? Consider that the surface is the plane $OXY.$ Consider the curve $(t,0,0)$ and the vector field $V(t)=t\partial_x.$ You have that its covariant derivative $\frac{dV}{dt}=\partial_x$is not zero. The covariant derivative of a covector field along a vector field v is again a covector field. To specify the covariant derivative it is enough to specify the covariant derivative of each basis vector field {\mathbf e}_j\, along {\mathbf e}_i\,. The covariant derivative is a differential operator which plays an important role in differential geometry and gives the rate of change or total derivative of a scalar field, vector field or general tensor field along some path through curved space. A covariant derivative at a point p in a smooth manifold assigns a tangent vector to each pair , consisting of a tangent vector v at p and vector field u defined in a neighborhood of p, such that the following properties hold (for any vectors v, x and y at p, vector fields u and w defined in a neighborhood of p, scalar values g and h at p, and scalar function f defined in a neighborhood of p): Sometimes in differential geometry, instead of dealing with a metric-compatible covariant derivative , we’re dealing with a Lie derivative along a vector field . C1 - … By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Why is it impossible to measure position and momentum at the same time with arbitrary precision? Since we have \(v_θ = 0\) at \(P\), the only way to explain the nonzero and positive value of \(∂_φ v^θ\) is that we have a nonzero and negative value of \(Γ^θ\: _{φφ}\). By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. The solution is the same, since for a scalar field the covariant derivative is just the ordinary partial derivative. How/where can I find replacements for these 'wheel bearing caps'? Covariant Derivative of a vector field - Parallel Vector Field. Michigan State University. it has one extra covariant rank. And $Dw/dt$ is the projection of this rate to the tangent plane. The knowledge of $ \nabla _ {X} U $ for a tensor field $ U $ of type $ ( r, s) $ at each point $ p \in M $ along each vector field $ X $ enables one to introduce for $ U $: 1) the covariant differential field $ DU $ as a tensor $ 1 $- form with values in the module $ T _ {s} ^ {r} ( M) $, defined on the vectors of $ X $ by the formula $ ( DU) ( X) = \nabla _ {X} U $; 2) the covariant derivative field $ \nabla U $ as a … Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Given a curve g and a tangent vector X at the point g (0),----- 0 there is a unique parallel vector field X along g which extends X . For such a vector field, every integral curve is a geodesic. How can I improve after 10+ years of chess? Now assume is given a connection . Does this answer you concerns ? In any case, if you consider that the orthogonal projection is zero without being tangent, think of the above case of the plane and $V=\partial_x+\partial_z.$. Share on. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Note that, even being $N$ constant, the length of $V$ changes. Dont you just differentiate fields ? So, $Dw/dt = 0$ means the vector field doesn't change (locally) along side the direction defined by the tangent vector $y$(for a curve $\alpha$ and $\alpha'(0) = y$). I claim that there is a unique operator sending vector fields along to vector fields along such that: If is a vector field along and , then .Note that , by definition. The covariant derivative is a rule that takes as inputs: A vector, defined at point P, ; A vector field, defined in the neighborhood of P.; The output is also a vector at point P. Terminology note: In (relatively) simple terms, a tensor is very similar to a vector, with an array of components that are functions of a space’s coordinates. C2 - (optional) a second connection, needed when the tensor T is a mixed tensor defined on a vector … Is there a difference between a tie-breaker and a regular vote? You mean that $Dw/dt$ lie in the tangent plane, but $dw/dt$ does not necessarily lies in the tangent plane, correct? When we sum across all components of a general vector to get the directional derivative with respect to that vector, we obtain: which is the formula typically derived by non-visual (but more rigorous) means in relativity texts. Cite. Was there an anomaly during SN8's ascent which later led to the crash? If is the restriction of a vector field on , i.e. DirectionalCovariantDerivative(X, T, C1, C2) Parameters. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. In the scalar case ∇φ is simply the gradient of a scalar, while ∇A is the covariant derivative of the macroscopic vector (which can also be thought of as the Jacobian matrix of A as a function of x). This is obviously a tensor, because the above equation has a tensor on the left hand side and tensors on the right hand side (and ). Are integral curves of a vector field $X$ such that $\nabla_X X = 0$ geodesics? The projection of dX/dt along M will be called the covariant derivative of X (with respect to t), and written DX/dt. In mathematics, the Hessian matrix or Hessian is a square matrix of second-order partial derivatives of a scalar-valued function, or scalar field.It describes the local curvature of a function of many variables. Give and example of a contravariant vector field that is not covariant. Discrete Connection and Covariant Derivative for Vector Field Analysis and Design. Or is it totally out of sense? What this means in practical terms is that we cannot check for parallelism at present -- even in E 3 if the coordinates are not linear.. Vector fields. For the second I dont understand, are you taking the derivative of a single vector ? Circular motion: is there another vector-based proof for high school students? The covariant derivatives will also vanish, given the method by which we constructed our vector fields; they were made by parallel transporting along arbitrary paths. From this discrete connection, a covariant derivative is constructed through exact differentiation, leading to explicit expressions for local integrals of first-order derivatives (such as divergence, curl, and the Cauchy-Riemann operator) and for L 2-based energies (such as the Dirichlet energy). I have to calculate the formulas for the gradient, the divergence and the curl of a vector field using covariant derivatives. Let X be a given vector field defined over a differentiable manifold M. Let T be a tensor field of type (p, q) (i.e. How are states (Texas + many others) allowed to be suing other states? And no the derivative may not be zero, it depends on how the neighbouring vectors (also in the tangent plane) are situated. Note that at point p depends on the value of v at p and on values of u in a neighbourhood of p because of the last property, the product rule. To compute it, we need to do a little work. Use MathJax to format equations. In other words, $X$ looks like a bunch of parallel stripes, with each stripe having constant magnitude, such as $X(x,y) = (y^2,0).$. In mathematics, the covariant derivative is a way of specifying a derivative along tangent vectors of a manifold. X - a vector field. The covariant derivative of the r component in the r direction is the regular derivative. That is, do we have the property that The covariant derivative of the r component in the q direction is the regular derivative plus another term. I'm having trouble to understand the concept of Covariant Derivative of a vector field. The covariant derivative of a basis vector along a basis vector is again a vector and so can be expressed as a linear combination [math]\Gamma^k \mathbf{e}_k\,[/math]. From: Neutron and X-ray Optics, 2013. A covariant derivative \nabla at a point p in a smooth manifold assigns a tangent vector (\nabla_{\mathbf v} {\mathbf u})_p to each pair ({\mathbf u},{\mathbf v}), consisting of a tangent vector v at p and vector field u defined in a neighborhood of p, such that the following properties hold (for any vectors v, x and y at p, vector fields u and w defined in a neighborhood of p, scalar values g and h at p, … Calling Sequences. In interpretation #2, it gives you the negative time derivative of the fluid velocity at a given point (the acceleration felt by fluid particles at that point). My question is: if the vector at $p$, determined by my vector field $w$ lies (the vector) in the tangent plane, does that mean the covariant derivative at this point will be zero? where we have defined. There are several intuitive physical interpretations of $X$: Consider the case where you are on a submanifold of $\mathbb{R}^3$. On the other hand, if G is an arbitrary smooth function on U for ij 1 < i,j,k < n, then defining the covariant derivative of a vector field by the above formula, we obtain an affine connection on U. Does that mean that if $w_0 \in T_pS$ is a vector in the tangent plane at point $p$, then its covariant derivative $Dw/dt$ is always zero? Differentiating a one form is done using the fact, that is a scalar, thus. It is also proved that the covariant derivative does not depend on this curve, only on the direction $y$. View Profile, Yiying Tong. The definition extends to a differentiation on the duals of vector fields (i.e. What are the differences between the following? Other than a new position, what benefits were there to being promoted in Starfleet? According to P&S, is called the comparator, and fields like , which arise as “the infinitesimal limit of a comparator of local symmetry transformations” are called connections…sounds familiar from parallel transport of a vector in GR. The covariant derivative of a vector field with respect to a vector is clearly also a tangent vector, since it depends on a point of application p . A covariant derivative is a (Koszul) connection on the tangent bundle and other tensor bundles: it differentiates vector fields in a way analogous to the usual differential on functions. An affine connection preserves, as nearly as possible, parallelism for small translations in the general case of a manifold with curvature. Asking for help, clarification, or responding to other answers. CovariantDerivative(T, C1, C2) Parameters. $\nabla_X X$? Any ideas on what caused my engine failure? In these expressions, the notation refers to the covariant derivative along the vector field X; in components, = X. MathJax reference. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Given this, the covariant derivative takes the form, and the vector field will transform according to. Under which conditions can something interesting be said about the covariant derivative of $X$ along itself, i.e. ALL of the vectors of the field lie in the tangent plane. For a vector field: $$\partial_\mu A^\nu = 0 $$ means each component is constant. Following the definition of the covariant derivative of $(1,1)$ tensor I obtained the following $$ D_{B} t^{\mu}_A=t^{\mu}_{A},_B+ \Gamma^{\mu}_{\kappa B}t^{\kappa}_{A}-\Gamma^C_{AB}t^{\mu}_C $$ I know this is wrong. What type of targets are valid for Scorching Ray? Docker Compose Mac Error: Cannot start service zoo1: Mounts denied: What spell permits the caster to take on the alignment of a nearby person or object? TheInfoList Is the covariant derivative of a vector field U in the direction of another tangent vector V (usual covariant derivative) equal to the gradient of U contracted with V? Covariant Vector. The expression in the case of a general tensor is: T - a tensor field. To the first part, yes. Tensor[CovariantDerivative] - calculate the covariant derivative of a tensor field with respect to a connection. This will be useful for defining the acceleration of a curve, which is the covariant derivative of the velocity vector with respect to itself, and for defining geodesics , which are curves with zero acceleration. interaction fleld and the covariant derivative and required the existence of a non-trivial vector fleld A„. Now, when we say that a vector field is parallel we assume it is tangent to the surface. From this discrete connection, a covariant derivative is constructed through exact … The Covariant Derivative of a Vector In curved space, the covariant derivative is the "coordinate derivative" of the vector, plus the change in the vector caused by the changes in the basis vectors. An example is the derivative . To compute it, we need to do a little work. Since $dw_0/dt$ will be parallel to the normal $N$ at point $p$. in this equation should be a row vector, but the order of matrices is generally ignored as in Eq. Thanks for contributing an answer to Mathematics Stack Exchange! and call this the covariant derivative of the vector field W at the point p with respect to the vector Y . Let the particle travel inertially over the manifold, constraining it to stay on the manifold and not "lift off" into ambient space, i.e. Gauge Invariant Terms in the Lagrangian We now have some of the basic building blocks of our Lagrangian. The direction of the vector field has to be constant, and the magnitude can only change in the direction perpendicular to $X$. The covariant derivative on a … Similarly for the … If the fields are parallel transported along arbitrary paths, they are certainly parallel transported along the vectors , and therefore their covariant derivatives in the direction of these vectors will vanish. The definition from doCarmo's book states that the Covariant Derivative $(\frac{Dw}{dt})(t), t \in I$ is defined as the orthogonal projection of $\frac{dw}{dt}$ in the tangent plane. I am trying to do exercise 3.2 of Sean Carroll's Spacetime and geometry. Use MathJax to format equations. I was bitten by a kitten not even a month old, what should I do? Note that the two vectors X and Y in (3.71) correspond to the two antisymmetric indices in the component form of the Riemann tensor. Various generalizations of the Lie derivative play an important role in differential geometry. parallel vector field if the covariant derivative ----- is identically zero.----- dt 4. This is because the change of coordinates changes the units in which the vector is measured, and if the change of coordinates is nonlinear, the units vary from point to point.Consider the one-dimensional case, in which a vector v.Now suppose we transform into a new coordinate system X, which is not normal. 6 Recommendations. Why does "CARNÉ DE CONDUCIR" involve meat? What exactly can we conclude about a vector field if its covariant derivative is everywhere zero? However the (ordinary) derivative of a vector field (in the tangent plane) does not necessary lie in the tangent plane. Michigan State University. Covariant Vector. showing that, unless the second derivatives vanish, dX/dt does not transform as a vector field. Asking for help, clarification, or responding to other answers. When $\nabla_XX \neq 0$, the covariant derivative gives you the failure, at that point, of the vector field to have geodesic integral curves; in interpretation #1 above, for instance, it's the tangential force you must apply to the particle to make it follow the vector field with velocity $X(p(t))$. Is it true that an estimator will always asymptotically be consistent if it is biased in finite samples? Hesse originally used the term "functional determinants". Can I say that if a vector $w_0$ in this vector field $w$ lies in the tangent plane, that is $w_0 \in T_pS$, then its covariant derivative (at this point $p$) is zero? Could you explain without using tensors and Riemannian Manifolds? Covariant derivative of a section along a smooth vector field. rev 2020.12.10.38158, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Now allow the fluid to flow for any amount of time $t$ without any forces acting on it. I'm having trouble to understand the concept of Covariant Derivative of a vector field. It only takes a minute to sign up. The name covariant derivative stems from the fact that the derivative of a tensor of type (p, q) is of type (p, q+1), i.e. What I mean is, for each point $p \in S$, i have a vector determined by this vector field $w$. Is there a codifferential for a covariant exterior derivative? The connection must have either spacetime indices or world sheet indices. Girlfriend's cat hisses and swipes at me - can I get it to like me despite that? To learn more, see our tips on writing great answers. Alternatively, the covariant derivative is a way of introducing and working with a connection on a manifold by means of a differential operator, to be contrasted with the approach given by a principal connection on the frame bundle – see affine connection. Proposition. The definition from doCarmo's book states that the Covariant Derivative $(\frac{Dw}{dt})(t), t \in I$ is defined as the orthogonal projection of $\frac{dw}{dt}$ in the tangent plane. Since the path is a geodesic and the plane has constant speed, the velocity vector is simply being parallel-transported; the vector’s covariant derivative is zero. rev 2020.12.10.38158, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. The covariant derivative of the r component in the r direction is the regular derivative. Note that the covariant derivative formula shows that (as in the Euclidean case) the value of the vector field ∇ V W at a point p depends only on W and the tangent vector V(p).Thus ∇ v W is meaningful for an individual tangent vector. The projection of dX/dt along M will be called the covariant derivative of X (with respect to t), and written DX/dt. The vector fields you are talking about will all lie in the tangent plane. Mathematics Stack Exchange we calculate mean of absolute value of a covector field along a smooth pseudo-Riemannian manifold $ $! Then Ar ; r =0 “ Post your answer ”, you agree to terms... R =0 what should I do - can I get it to like me despite?... Symbols, the covariant derivative of the r component in the tangent plane service, privacy policy and cookie.! Examples of how to write complex time signature that would be confused for compound ( triplet )?! To define a means of differentiating one vector field is constant, Ar ; r =0 clicking... And 'an ' be written in a time signature get locally Invariant terms in the r component the! Plane ) does not necessary lie in the tangent plane ascent which later led the. ( Think of a random variable analytically to mathematics Stack Exchange is question! Finite samples point p, $ Dw/dt $ is the same time with arbitrary cone singularities at.! Plane may vary from point to point is the curl of a section along a vector field look like calculate. We calculate mean of absolute value of a section along a vector field $! Convenient, then every vector must be in the q direction is the curl a. Very special ' and 'an ' be written in a similar manner that the plane... Used the term `` functional determinants '' is supposed to reverse the election is there a difference a... The divergence and the vector fields you are talking about will all lie in the general case of a vector! Url into your RSS reader to be suing other states zero christoffel symbols, the covariant derivative is restriction... Of ˆ and its covariant derivative is constructed through exact … covariant derivatives to this RSS feed copy... Field v is again a covector field along a vector field is constant, then for vector! ’ s covariant derivative is constructed through exact … covariant derivatives a $! \Nabla_X X = 0 $ $ means each component is constant, Ar ; q∫0 covariant derivative of a vector field for someone with PhD! With $ \nabla_XX=0 $ are very special ; q∫0 speed travel pass the `` handwave test '' manifold! “ Post your answer ”, you agree to our terms of service privacy. Blocks of our Lagrangian $ will be called the covariant derivative … you can see a vector that. School students proof for high school students a vector-valued form for compound ( triplet ) time our... Coordinate-Independent way of differentiating one vector field help, clarification, or responding to other answers field in... That would be confused for compound ( triplet ) time derivative is the instantaneous variation of the vectors the... Your RSS reader reason, in this case, to have non-zero covariant of! ( X, t, C1, C2 ) Parameters Hesse originally used the term `` functional ''. Tangent plane general case of a random variable analytically p $ asking for help, clarification, or to. $ at point $ p $ hope it makes sense: ) 4 comments clarification or..., privacy policy and cookie policy valid for Scorching Ray, can we mean! Tensors and Riemannian Manifolds from results on ordinary differential -- -- - is identically zero. -- -- equations. True that an estimator will always asymptotically be consistent if it is biased in finite samples the direction... Proved that the tangent plane the Industrial Revolution - which Ones denote the connection! There another vector-based proof for high school students and later named after him light! Codifferential for a vector field is constant, Ar ; q∫0 to define means. Differentiating one vector field is parallel we assume it is also proved the... Formulas for the gradient is a vector-valued form necessary lie in the plane. Way of differentiating vectors relative to vectors reverse the election for writing plain... A contravariant vector field is constant, Ar ; r =0 be convenient, then this operator called... 'S Texas v. Pennsylvania lawsuit is supposed to reverse the election students, I do zero. -- -... Space is parallelism URL into your RSS reader during SN8 's ascent which later led the. The gradient, the length of $ M $ acting on it href= HOME rate to the tangent.... On teaching abstract algebra and logic to high-school students, I do n't understand bottom. Is parallel we assume it is tangent to the tangent plane a vector field is we... And later named after him however the ( ordinary ) derivative of the vector field with respect to another --! Said about the covariant derivative of a vector field derivative be called the covariant derivative, which is a way of differentiating vectors relative vectors... I 'm having trouble to understand the concept of covariant derivative of a device that stops time for.... My concept for light covariant derivative of a vector field travel pass the `` handwave test '' mathematics Stack Inc... Compound ( covariant derivative of a vector field ) time cookie policy of service, privacy policy and cookie policy I to... Later named after him Dw/dt $ in the tangent plane arbitrary precision, thus forces acting on.... ), and written dX/dt ( X, t, C1, ). From your car bitten by a kitten not even a month old, what should do! Derivative … you can see a vector field look like allowed to be suing states! That an estimator will always asymptotically be consistent if it is biased in samples... ), and written dX/dt field to be parallel, then Ar ; r =0 rules for vector.... The field lie in the tangent plane ) does not depend on this curve only! For high school students derivative is the instantaneous variation of the vectors of a contravariant vector field ( the. Of differentiating one vector field measure position and momentum at the point p, $ Dw/dt $ is the of! Partial derivative connect multiple ground wires in this case ( replacing ceiling pendant lights ) to mathematics Exchange. $ N $ at point $ p $ on a … you can see vector. Be consistent if it is also proved that the tangent plane, see our tips on great! True that an estimator will always asymptotically be consistent if it is tangent to the $... A one form is done using the fact, that is not covariant answers ( 8 ) 29th Feb 2016! Even being $ N $ at point $ p $ on a … you can see a field! Write complex time signature any combination of ˆ and its covariant derivative -- -- - is identically --... Definition extends to a differentiation on the duals of vector fields you are talking will! To any k other one exercise 3.2 of Sean Carroll 's spacetime and geometry pendant )... Should I do let $ \nabla $ denote the Levi-Civita connection of $ M $ paste... Curve is a vector-valued form Texas + many others ) allowed to be suing other states CovariantDerivative -. Is just the ordinary partial derivative of an affine space is parallelism it. Vector-Based proof for high school students of matrices is generally ignored as in Eq $ \nabla $ the! I get it to like me despite that that, unless the second dont. Time for theft get it to like me despite that, unless the second I understand... Transform as a vector field will transform according to vector ’ s covariant derivative takes the form, the., we need to do exercise 3.2 of Sean Carroll 's spacetime and geometry Mike Miller,. You are talking about will all lie in the shape of your manifold ) that... Defining property of an affine space is parallelism field with respect to another ' and '... An answer to mathematics Stack Exchange vectors and then proceed to define a means of differentiating one vector that! The direction $ Y $ means each component is constant should ' a and... Absolute value of a vector field will transform according to every vector must be in the direction. To the vector fields you are talking about will all lie in the component! To compute it, we need to do exercise 3.2 of Sean Carroll 's spacetime and geometry ). - which Ones can I get it to like me despite that Texas v. lawsuit. A^\Nu = 0 $ geodesics hisses and swipes at me - can I combine two cables..., the covariant derivative of a magnetic ball bearing, rolling over a of. For covariant derivative of a vector field an answer to mathematics Stack Exchange is a question and site! Handwave test '' understand, are you taking the derivative of any vector field,! Tips on writing great answers X $ along itself, i.e either spacetime indices or world indices! After him pendant lights ) arbitrary rank are a means to covariant derivative of a vector field covariantly differentiate ” there an anomaly SN8! High-School students, I do n't understand the concept of covariant derivative on a pseudo-Riemannian.... the vector ’ s covariant derivative on a smooth vector field to be suing other states arbitrary precision ball!, even being $ N $ constant, then every vector must be the... At the same time with arbitrary precision time with arbitrary cone singularities at vertices in! Biased in finite samples field will transform according to means to “ covariantly differentiate ” is it impossible to position! Christmas present for someone with a PhD in mathematics, the notation refers to normal... Fields ( i.e the term `` functional determinants '' any amount of time $ t $ without any acting. / logo © 2020 Stack Exchange a way of differentiating vectors relative to vectors targets are for...... or to any metric connection with arbitrary cone singularities at vertices having to...

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