Is B^{ij}_i a Contravariant Vector?

[cristina89]

Homework Statement

B is a third order tensor. Show that B^{ij}_{i} is a contravariant vector.

The Attempt at a Solution

Well... I just thought about a simple solution but I don't think I'm right. But anyways.
Considering B^{ij}_{i}. If I raise the index i: g^{ij}B^{ij}_{i} = B^{ijj}
And so I can say that B is a first order tensor = vector. And this is a contravariant vector.
Is this right?

 

[HallsofIvy]

Homework Statement

B is a third order tensor. Show that B^{ij}_{i} is a contravariant vector.

The Attempt at a Solution

Well... I just thought about a simple solution but I don't think I'm right. But anyways.
Considering B^{ij}_{i}. If I raise the index i: g^{ij}B^{ij}_{i} = B^{ijj}

You can't have the same index three times! Also, using the "summation convention" you sum over one "upper" and one "lower" index so even if you wrote g^{kj}B^{ij}_k, you would not be summing. Perhaps what you mean is g^k_jB^{ij}_k= B^i

And so I can say that B is a first order tensor = vector. And this is a contravariant vector.
Is this right?

Well, what is the definition of a contravariant vector? Can you show that this satisfies that definition? Remember that it is NOT enough just to show that something can be written with a single index- you can write an array, in a given coordinate system, indexed with a single index- that does not make it a "vector". A vector must change coordinates correctly as the coordinate system changes. If I remember correctly (it's been a while!) one test for a contravariant vector, written as v^i, is that the combination g_{ij}v^iv^j (essentially the vector length) is a scalar. Which, again, does not just mean "a number" in a given coordinate system. A scalar (0 order tensor) must not change when you change coordinate systems.

 

[clamtrox]

Equally well you can show that B^{ij}_i transforms as a vector, ie. if I do a coordinate transform x^{i} \rightarrow x^{i'}, B should transform as
B^{i' j'}_{i'} = \frac{\partial x^{j'}}{\partial x^{j}} B^{ij}_i