Show that an object is a tensor

**davesface** · March 23rd 2011, 07:12 PM

Suppose that an object with two indices has the property that $A(\mu,\nu)C^{\mu\nu}$ is a scalar for any arbitrary tensor $C^{\mu\nu}$ . Show that $A(\mu,\nu)$ is a tensor.

Hint: start with the relationship $A^{'\nu}_{\mu}x^{'\mu}x^{'\nu}=A^\beta_\alpha x^\alpha x^\beta$

**ojones** · March 26th 2011, 12:36 AM

Suppose that with respect to coordinate system $(x^i)$ we have that $ A(\mu, \nu)C^{\mu \nu}=k$ , where $k$ is a scalar. Now, in coordinate system $(\bar x^i)$ we have that $\bar A(\mu, \nu)\bar C^{\mu \nu}=\bar k =k.$

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

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

Hence,

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

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

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

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

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