# Homework Help: 2 questions about tenzors

1. Jul 4, 2009

### ibc

1. The problem statement, all variables and given/known data
whether (Aijxi)/xk is a tenzor or not, if it is, what kind? (co-variant or contra-variant).

2. Relevant equations
given that Aij is a second degree tenzor and x are the coordinates.

3. The attempt at a solution
Aijxi)/ is clearly a first degree contra-variant tenzor, therefore the question reduces to whether
Bj/xk
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 dimention, 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 dimention 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?

1. The problem statement, all variables and given/known data
what kind of a tenzor (co-variant or contra-variant) is:
d2(phi)/dxpdxq

2. Relevant equations
phi is a scalar function
x are the coordinates
3. 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)/dxpdxq =
d/dxp(d(phi)/dxq), and d(phi)/dxq is a co-variant tenzor, so I do the transformation for it, and the take the derivative by xp, 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

2. Jul 4, 2009

### 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
$$\frac{1}{x_j'}= \frac{1}{x_k}\frac{\partial x_j'}{\partial x_k}$and so is also a first order contravariant tensor. The tensor product $A^{ij}x_i/x_j$ then transforms as [tex]A^{mn}'x_m'/x_j'= A^{ij}x_i/x_j\frac{\partial x^n'}{\partial x^j}\frac{\partial x_j'}{\partial x_k}$$ and so is a second order contravariant tensor. So, again unless you are working only with Euclidean tensors, for a scalar function, $\phi$, $$\frac{\partial \phi}{\partial x^j}$$ is a first order covariant tensor but [tex]\frac{\partial^2\phi}{\partial x^i \partial x^j}$
is NOT a tensor.

(Euclidean tensors allow only coordinate systems in which coordinate lines are straight lines and are orthogonal to one another.

In general, unless you are working in Euclidean tensors, in which the Christoffel symbols are all 0, derivatives of tensors are not tensors- you have to use the "covariant derivative".

3. Jul 4, 2009

### ibc

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

thanks

4. Jul 4, 2009

### ibc

oh I managed to see it now.

you said: 1/xj'=(1/xk)*(dxj'/dxk)
how do you know 1/x transforms that way?
by saying that you assume it's a tenzor, but my question is how do you know 1/x is a tenzor?