How to show something transforms as a covector?

[Baggio]

Considering a boost in the x direction, how do you show that

(d/dx,d/dy,d/dz,d/dt)

transforms as a covector?


thanks

 

[Meir Achuz]

I assume you mean partial derivs.
Use the calculus rule relating \partial_x' to \partial_x.

 

[pmb_phy]

Considering a boost in the x direction, how do you show that

(d/dx,d/dy,d/dz,d/dt)

transforms as a covector?


thanks

I don't see that this is a covariant vector. If it was the -gradient then I'd see that (you used total derivatives rather than partial derivative. I've never seen such an object). If it was then I'd use the chain rule for partial differentiation.

Pete

 

[Baggio]

yes sorry they are partials... why is the chain rule used.. I don't see how that leads to the solution

thanks

 

[pmb_phy]

yes sorry they are partials... why is the chain rule used.. I don't see how that leads to the solution

thanks

\frac{\partial}{\partial x'} = \frac{\partial x}{\partial x'}\frac{\partial}{\partial x} + \frac{\partial y}{\partial y'}\frac{\partial}{\partial y} + \frac{\partial z}{\partial z'}\frac{\partial}{\partial z} + \frac{\partial t}{\partial t'}\frac{\partial}{\partial t}

Now change coordinates (x, y, z, t) - > (x¹, x², x³, x⁴)

Then the above sum can be written as a sum which is identical to the law of covariant transformation.

Pete