How to show something is a covector?

[Baggio]

Hi, think this question will be relatively simple for those familiar with tensor analysis

Need to show that (-x, -y, -z, t) is a covector. We're given a hint that we should apply a lorentz boost in the x direction but i don't see the point of this. If that is a covector then surely you can say

x
y
z
t

is a vector and just take the scalar product to find the interval?

There is also a samilar question where by I have to show that

(df/dx, df/dy,df/dz,df/dt) (f is a scalar field) is a covector

none of these questions are for credit or anything it's just to help with our understanding...but I don't :(

Thanks

 

[Hurkyl]

Well, there are two methods.

(1) You could, as you said, apply a coordiante transformation, and make sure the components transform appropriately.

(2) You could show that they are (coordiante-independent!) linear functionals on the space of vectors.


Your hint was to apply method (1). It has the advantage of being a mindless, straightforward algorithm.


But method (2) is often nicer, as you suggest. You know that \mathbf{x} := [x, y, z, t]^T is a vector, and left multiplying by [-x, -y, -z, t] is, as you say, the (coordinate independent!) function:

<br /> f(\mathbf{y}) = \langle \mathbf{x}, \mathbf{y} \rangle<br />

At least I think that's what you said -- I hope you didn't mean to simply take the product¹ of [-x, -y, -z, t] and [x, y, z, t]^T: sure, the result is an invariant, but you need to show it's invariant no matter what vector you use on the right.


For the second problem, you know a suitable geometric interpretation of the gradient, and what you can do with the gradient and a vector.



1: This is not a scalar product! The scalar product, at least in mathspeak, is a function of two vectors. When you have a covector and a vector, you're just plugging the vector into the covector. (Since a covector is a linear functional) In the coordinate representation, it's multiplying a 1xN matrix by an Nx1 matrix.