Question

Show that \(T^{\alpha \beta \gamma \delta \rho} S_{\beta \rho}\) is a tensor. What is its rank?

Short answer

The object \(T^{\alpha \beta \gamma \delta \rho} S_{\beta \rho}\) behaves as a rank-3 tensor.

Step-by-step solution

Understanding the tensor notation and contraction

In the expression \(T^{\alpha \beta \gamma \delta \rho} S_{\beta \rho}\), \(T\) is a five-index tensor and \(S\) is a two-index tensor. The indices \(\beta\) and \(\rho\) appear as both upper indices in \(T\) and lower indices in \(S\), indicating that we're contracting over these indices.

Transforming the tensor product

Under a general coordinate transformation \(x^{\mu} \rightarrow x'^{\mu}\), the tensors transform as follows: \(T^{\alpha \beta \gamma \delta \rho} \rightarrow T'^{\alpha \beta \gamma \delta \rho} = \frac{\partial x'^{\alpha}}{\partial x^{\mu}}\frac{\partial x'^{\beta}}{\partial x^{\nu}}\frac{\partial x'^{\gamma}}{\partial x^{\sigma}}\frac{\partial x'^{\delta}}{\partial x^{\tau}}\frac{\partial x'^{\rho}}{\partial x^{\lambda}} T^{\mu \nu \sigma \tau \lambda}\) and \(S_{\beta \rho} \rightarrow S'_{\beta \rho} = \frac{\partial x^{\mu}}{\partial x'^{\beta}}\frac{\partial x^{\nu}}{\partial x'^{\rho}} S_{\mu \nu}\) Now, the tensor product \(T^{\alpha \beta \gamma \delta \rho} S_{\beta \rho}\) should transform according to \(T'^{\alpha \beta \gamma \delta \rho} S'_{\beta \rho} = \frac{\partial x'^{\alpha}}{\partial x^{\mu}}\frac{\partial x'^{\beta}}{\partial x^{\nu}}\frac{\partial x'^{\gamma}}{\partial x^{\sigma}}\frac{\partial x'^{\delta}}{\partial x^{\tau}}\frac{\partial x'^{\rho}}{\partial x^{\lambda}} T^{\mu \nu \sigma \tau \lambda} \frac{\partial x^{\mu}}{\partial x'^{\beta}}\frac{\partial x^{\nu}}{\partial x'^{\rho}} S_{\mu \nu}\). This simplifies to \(T'^{\alpha \mu \gamma \delta \nu} S'_{\mu \nu}\) via the Kronecker delta \(\delta_{\mu}^{\nu}\), showing that the contraction behaves as a tensor under the coordinate transformation.

Finding the rank of the tensor

The rank of a tensor is given by the number of its indices. Here, the contracted tensor \(T^{\alpha \beta \gamma \delta \rho} S_{\beta \rho}\) now has three upper indices (\(\alpha, \gamma\) and \(\delta\)) after the contraction, so it's a rank-3 tensor.

Key concepts

Tensor Transformation

The process of tensor transformation explains how the components of a tensor change when we move from one coordinate system to another. For any given tensor, the rule of transformation maintains the tensor's intrinsic properties, ensuring that it remains the same physical object despite changes in the way it is represented.
This is analogous to changing the language you use to describe a physical phenomenon; the phenomenon doesn't change just because you describe it in French instead of English. In mathematical terms, the transformation of a tensor of rank n can be expressed with the partial derivatives of the new coordinates with respect to the old ones. These derivatives form transformation matrices that multiply with the components of the original tensor to yield the components in the new coordinate system.

Rank of a Tensor

The rank of a tensor provides information about its structure in terms of the number of components it possesses, signifying its complexity and the types of geometric objects it can represent. A tensor of rank 0 is a scalar, possessing only magnitude; a rank 1 tensor is a vector, with magnitude and direction; and higher ranks correspond to more complex structures.
For instance, the contracted tensor from our exercise, after combining the five-index tensor with the two-index tensor and summing over the common indices \beta and \(\rho\), results in a new tensor that has three indices left, making it a rank-3 tensor. Explaining the complexity of this tensor to students can be done by visualizing it as a 3-dimensional grid or array, where each point in the array corresponds to a different component of the tensor.

Coordinate Transformation

When we dive into the realm of coordinate transformations, we are essentially talking about changing the viewing angles or the 'glasses' through which we observe our mathematical world. It is vital in physics and engineering as it allows us to reconcile measurements seen from different perspectives.
Coordinate transformations can be rotations, shifts, or more complex mappings, but the core principle remains the same: translating the description of a tensor from one set of coordinates to another while maintaining its intrinsic physical characteristics. These transformations are foundational to understanding the concepts in fields like relativity, where how we measure space and time can change dramatically based on our point of view.

Kronecker Delta

The Kronecker delta is a function that plays a starring role in simplifying tensor equations during contractions and transformations. Think of it as the identity element for tensor indices. It equals 1 if the indices are the same and 0 otherwise.

What Makes the Kronecker Delta Special?

It enables the cancellation of terms when used in tensor contractions, as seen in the given exercise. When the Kronecker delta interacts with a tensor, indices that are summed over collapse, reducing the rank of the tensor. This behavior is akin to how in arithmetic, multiplying any number by 1 yields the original number. Such a simple and elegant property is supremely powerful in the complex dance of tensor algebra, enabling us to confirm that certain operations, like contracting tensors, result in new tensors.