How to prove an expression to be a tensor?How to prove that the expression φ , i j := ∂ 2 φ ∂ x i ∂ x j = ∇ ∇ φ is a tensor of second order where φ is a scalar? Furthermore, how to prove that q j 1 ⋯ j n is a vector?We can either prove it by definition or use the so-called "tensor recognition theorem" claiming that if q j 1 ⋯ j n , then p must be a tensor of order m+n, where q j 1 ⋯ j n is a tensor of order n and r i 1 ⋯ i m a tensor of order m.

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How to prove an expression to be a tensor?How to prove that the expression       φ          ,      i      j        :=                    ∂        2            φ              ∂              x        i            ∂              x        j              =  ∇  ∇  φ is a tensor of second order where φ is a scalar? Furthermore, how to prove that       q                  j        1            ⋯              j        n             is a vector?We can either prove it by definition or use the so-called &quot;tensor recognition theorem&quot; claiming that if       q                  j        1            ⋯              j        n            , then p must be a tensor of order m+n, where       q                  j        1            ⋯              j        n             is a tensor of order n and       r                  i        1            ⋯              i        m             a tensor of order m.

Lamar Casey · Accepted Answer

1 ) Suppose that   O      X    1        X    2        X    3   is the coordinate system corresponding to the given basis       e    1    ,      e    2    ,      e    3   on       E    3  . The gradient of a scalar function   φ  (      X    1    ,      X    2    ,      X    3    ), using this coordinate system, is defined by  O                    X        1            ~                          X        2            ~                          X        3            ~      Suppose that   O                    X        1            ~                          X        2            ~                          X        3            ~       is another orthonormal basis with corresponding cartesian coordinates   O                    X        1            ~                          X        2            ~                          X        3            ~       given by                     X        ~              i    =      q          i      j            X    j  , where   Q  =  (      q          i      j        )  ∈  S  O  (  3  )By the chain rule,                    X        ~              i    =      q          i      j            X    j  Using                     X        ~              i    =      q          i      j            X    j  , we have            ∂              X        k                    ∂                                    X            ~                          i              =      q          i      k        ,and hence  ∇  φThat is, by definition,   ∇  φ is a cartesian tensor of order 1. Doing this again, but with   ∇  φ in place of   φ, shows that             ∇              2        φ is a cartesian tensor of order 2.2) a and b are vectors so       a    i   and       b    j   are components of cartesian tensors of order 1, and the alternating tensor   ε  =      ε          k      l      m       is a cartesian tensor of order 3.Taking their product we get that       a    i        b    j        ε          i      j      m       are the components of a cartesian tensor of order 5.Contracting indices (that is, setting two indices equal and thus effecting a sum) gives that       a    i        b    j        ε          i      j      m       are the components of a cartesian tensor of order 4, and contracting again gives that       a    i        b    j        ε          i      j      m      are the components of a cartesian tensor of order 3.Hence,   a  ×  b is a cartesian tensor of order 3.

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