1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Scalars as Rank 0 Tensors?

  1. Mar 12, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm learning a bit about tensors on my own. I've been given a definition of a tensor as an object which transforms upon a change of coordinates in one of two ways (contravariantly or covariantly) with the usual partial derivatives of the new and old coordinates. (I apologize if I'm screwing up the terminology). I was trying to apply this definition to a scalar field, specifically to the field in Euclidean 2-Space where f = x²+2y² and trying to transform that to the coordinates (g,h) given by
    g = x+y; h=y-x (a simple rotation). It was easy to solve for f in the new coordinates:
    f' = (1/4)(3g² + 2gh + 3h²) but I've tried to transform f into f' using the definition and just run around in circles. Is the transformation covariant? contravariant? both? neither? Is this the wrong definition for a rank 0 tensor? If so, what is the correct definition? What is the simplest example of an object which transforms contravariantly and one which transforms covariantly? Is it a rank 1 tensor? (I'm following along with L. Susskind's Stanford video lectures on GR, and he shows how a 2-vector and the grad of it transform, are there any simpler examples or are they it?)

    2. Relevant equations
    f(x,y) = x²+2y²; f'(g,h) = (0.25)(3g²+2gh+3h²) where g = x+y and h=y-x

    3. The attempt at a solution
    I've tried dg = ∂x + ∂y and dh = ∂y - ∂x but then I get (using the definitions of co- and contra- variant tensors): (∂x/∂g)*f + (∂y/∂g)*f and (∂x/∂h)*f + (∂y/∂h)*f
    and just can't figure how these can be combined to get f'. I must be missing something basic.
  2. jcsd
  3. Mar 12, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper

    You aren't missing anything basic. You've already did the coordinate transformation on your scalar. If there were indices on the object then the tensor transformation laws would tell you additional factors you need. But a scalar is a rank 0 tensor and doesn't have any indices. You are all done.
  4. Mar 14, 2014 #3
    I'm still confused. I thought there was some #other# way to get from f=x²+2y² to f' = 0.25( etc.etc.) using the rules of tensor transformations (or is it the definitions of tensors?). Isn't it true that if f was the x-th component of a vector (say) then all I'd need to do is multiply x²+2y² by (dg/dx + dh/dx) to get f' ? (contraviariant rank 1 tensor transformation).
  5. Mar 14, 2014 #4


    User Avatar
    Science Advisor
    Homework Helper

    No, there's really not another way. Even in the case of tensors you still have to spell out the coordinate transformation. In the case of tensor components there needs be other factors applied depending on the indices. A scalar has no indices.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Scalars as Rank 0 Tensors?
  1. Rank/order of tensor (Replies: 1)