# Show that an object is a tensor

• Mar 23rd 2011, 07:12 PM
davesface
Show that an object is a tensor
Suppose that an object with two indices has the property that $\displaystyle A(\mu,\nu)C^{\mu\nu}$ is a scalar for any arbitrary tensor $\displaystyle C^{\mu\nu}$. Show that $\displaystyle A(\mu,\nu)$ is a tensor.

Hint: start with the relationship $\displaystyle A^{'\nu}_{\mu}x^{'\mu}x^{'\nu}=A^\beta_\alpha x^\alpha x^\beta$
• Mar 26th 2011, 12:36 AM
ojones
Suppose that with respect to coordinate system $\displaystyle (x^i)$ we have that $\displaystyle A(\mu, \nu)C^{\mu \nu}=k$, where $\displaystyle k$ is a scalar. Now, in coordinate system $\displaystyle (\bar x^i)$ we have that $\displaystyle \bar A(\mu, \nu)\bar C^{\mu \nu}=\bar k =k.$

Since $\displaystyle C^{\mu \nu}$ are the components of a contravariant tensor of order two, we have that

$\displaystyle \displaystyle{\bar C^{\mu \nu}=\frac{\partial \bar x^\mu}{\partial x^i}\frac{\partial \bar x^\nu}{\partial x^j}C^{i j}}.$

Hence,

$\displaystyle \displaystyle{ [\bar A(\mu, \nu)\frac{\partial \bar x^\mu}{\partial x^i}\frac{\partial \bar x^\nu}{\partial x^j}-A(i, j)]C^{i j}=0.$

Since this holds for arbitrary components $\displaystyle C^{i j}$, we have that

$\displaystyle \displaystyle{ \bar A(\mu, \nu)\frac{\partial \bar x^\mu}{\partial x^i}\frac{\partial \bar x^\nu}{\partial x^j}-A(i, j)=0}.$

Multiplying both sides by $\displaystyle \displaystyle{\frac{\partial x^i}{\partial \bar x^r}\frac{\partial x^j}{\partial \bar x^s}}$ we get

$\displaystyle \displaystyle{ \bar A(\mu, \nu)\delta_r^\mu\delta_s^\nu=A(i,j)\frac{\partial x^i}{\partial \bar x^r}\frac{\partial x^j}{\partial \bar x^s}}.$

Hence,

$\displaystyle \displaystyle{ \bar A(r,s)=A(i,j)\frac{\partial x^i}{\partial \bar x^r}\frac{\partial x^j}{\partial \bar x^s}}.$

That is, $\displaystyle A(\mu ,\nu)$ are the components of a covariant tensor of order two.