Is (Aijxi)/xk a Tensor and What Kind?

[ibc]

Homework Statement

whether (A^ijx_i)/x_k is a tenzor or not, if it is, what kind? (co-variant or contra-variant).

Homework Equations

given that A^ij is a second degree tenzor and x are the coordinates.

The Attempt at a Solution

A^ijx_i)/ is clearly a first degree contra-variant tenzor, therefore the question reduces to whether
B^j/x^k
is a tenzor.
so I do the transformations, and get a normal tenzor transformation on the numerator (from B), and a normal transformation on the denominator (from x), so if we are working in 1 dimension, it's simply 1/(dx/dx')*x = (dx'/dx)*1/x, which means a co-variant tenzor in the denominator transforms as a contra-variant tenzor.
however, if the dimension is greater than 1, I get a sum of lots of partial derivetives, and 1 devided by that equals who knows what... so in that case I fail to determine whether it is a tenzor or not, it doesn't really look like one to me, but I'm really not sure, maybe there's some algebric work to do and make it look like a tenzor again?

Homework Statement

what kind of a tenzor (co-variant or contra-variant) is:
d²(phi)/dx^pdx^q

Homework Equations

phi is a scalar function
x are the coordinates

The Attempt at a Solution

the questions states that it is a tenzor, and asks what kind of a tenzor it is, yet I don't understand how come it is a tenzor.
the expression is
d²(phi)/dx^pdx^q =
d/dx^p(d(phi)/dx^q), and d(phi)/dx^q is a co-variant tenzor, so I do the transformation for it, and the take the derivative by x^p, I get 2 expresion, which one looks like a tenzor transformation, yet the other a second derivative of x by some x'i, x'j (x' being the new coordinates).
so I don't understand why is that expression a tenzor

thanks
ibc

 

[HallsofIvy]

As you say $A^{ij}x_i$ transforms as
$$A^{mn}'x_m'= A^{ij}x_i\frac{\partial x^n'}{\partial x^j}$$ and so is a first order contravariant tensor. $1/x_k$ transforms as
[tex]\frac{1}{x_j'}= \frac{1}{x_k}\frac{\partial x_j'}{\partial x_k}[itex] and so is also a first order contravariant tensor. The tensor product $A^{ij}x_i/x_j$ then transforms as <br /> $$A^{mn}'x_m'/x_j'= A^{ij}x_i/x_j\frac{\partial x^n'}{\partial x^j}\frac{\partial x_j'}{\partial x_k}$$<br /> and so is a second order contravariant tensor. So, again unless you are working only with Euclidean tensors, for a scalar function, $\phi$, <br /> $$\frac{\partial \phi}{\partial x^j}$$<br /> is a first order covariant tensor but <br /> $$\frac{\partial^2\phi}{\partial x^i \partial x^j}$$[/itex]$$is NOT a tensor.<br /> <br /> (Euclidean tensors allow only coordinate systems in which coordinate lines are straight lines and are orthogonal to one another.<br /> <br /> In general, unless you are working in Euclidean tensors, in which the Christoffel symbols are all 0, derivatives of tensors are <b>not</b> tensors- you have to use the "covariant derivative".$$[/tex]

 

[ibc]

hey, wherever you wrote an equation, I see it as a black stripe, do you know how I can see it normally?

thanks

 

[ibc]

As you say $A^{ij}x_i$ transforms as
$$A^{mn}'x_m'= A^{ij}x_i\frac{\partial x^n'}{\partial x^j}$$ and so is a first order contravariant tensor. $1/x_k$ transforms as
[tex]\frac{1}{x_j'}= \frac{1}{x_k}\frac{\partial x_j'}{\partial x_k}[itex] and so is also a first order contravariant tensor. The tensor product $A^{ij}x_i/x_j$ then transforms as <br /> $$A^{mn}'x_m'/x_j'= A^{ij}x_i/x_j\frac{\partial x^n'}{\partial x^j}\frac{\partial x_j'}{\partial x_k}$$<br /> and so is a second order contravariant tensor. So, again unless you are working only with Euclidean tensors, for a scalar function, $\phi$, <br /> $$\frac{\partial \phi}{\partial x^j}$$<br /> is a first order covariant tensor but <br /> $$\frac{\partial^2\phi}{\partial x^i \partial x^j}$$[/itex]$$is NOT a tensor.<br /> <br /> (Euclidean tensors allow only coordinate systems in which coordinate lines are straight lines and are orthogonal to one another.<br /> <br /> In general, unless you are working in Euclidean tensors, in which the Christoffel symbols are all 0, derivatives of tensors are <b>not</b> tensors- you have to use the "covariant derivative".$$[/tex]

[tex]$$<br /> oh I managed to see it now.<br /> <br /> you said: 1/xj'=(1/xk)*(dxj'/dxk)<br /> how do you know 1/x transforms that way?<br /> by saying that you assume it's a tenzor, but my question is how do you know 1/x is a tenzor?$$[/tex]