# Tensor Homework For Kids

## At a glance

- Reading and problem solving are good activities for your child to do at home.
- Play is an important part of your child's learning.
- Any homework should reinforce what your child learns in class.
- Schools develop homework policies with the help of teachers and parents.
- If your child is having difficulties with any homework activities speak to their teacher.

Public schools in NSW don't expect children in Kindergarten to complete formal homework. They encourage families to read with their children and be involved in family activities that assist the development of their skills in reading, mathematics and problem solving to make the most of what they are learning.

**Learning through play**

Thea Eyles, an early childhood expert with the NSW Department of Education and Communities, says young children in the early years of school need the time to play.

Homework should help build good learning habits; it shouldn't exhaust or turn kids off. Ken OlahNSW Department of Education and Communities

"I think the most powerful message around the homework issue for parents of children who are in Kindergarten and Year 1 is that playing is learning," Thea says.

"Early childhood educators know the importance of learning through play – it is this philosophy that forms the basis of early childhood education, and all of the international research supports this approach."

**Formal homework**

More formal homework usually starts in Years 1 and 2 where children may be asked to complete some maths, simple writing tasks, or an activity sheet. The purpose of this homework is to reinforce what has been learnt in class.

**Homework helps:**

- to bridge the gap between learning at school and learning at home.
- parents to see what their child can do and to be involved in their learning.
- improve children's concentration and focus.
- children to retain and understand what they've been taught in class.
- prepare children for what they will be taught the next day.
- provide children with the challenges and stimulus they need to engage in their classwork.

**Perfect balance**

NSW Department of Education curriculum expert Ken Olah says the homework policy, which schools use as a guide, is based on common sense.

"The amount of homework depends on the age of the student, the school context, the subject, the purpose for which the homework is set, and so on," he says.

"Homework should help build good learning habits; it shouldn't exhaust or turn kids off."

Schools develop their own homework policy with input from staff and parents. Ask your school for a copy.

If you find homework is becoming too much or is too difficult for your child, or there is something specific going on for your family that makes getting homework done a real challenge, have a chat with your child's teacher.

## Tensors in physics

One of my professors at Cornell told me, possibly influenced by Anthony Zee, that the definition of a tensor in physics is

A tensor is anything which transforms like a tensor.

Our whole class laughed, which irked him, because, as he went on to point out: it's not quite circular. Once you know how one vector rotates under a coordinate transformation (e.g. the position vector!) you have these coordinate-change matrices, and a tensor of the appropriate rank is *anything whose components transform with the appropriate combination of coordinate-change matrices.* So "transforms like a tensor" is an external definition, not an internal one.

A scalar is a (0, 0)-tensor: it is any number (really, any assignment of numbers to points on the manifold -- generally we mean scalar "fields") which does not change under a coordinate transform.

In your case, since you're apparently "missing" the $i=0$ component, you should probably check for some simple $A^i$ and $\partial_i$ whether the $i=0$ component matters. You may find that $A^0 = \text{constant}$ with respect to time, and it will then emerge that your expression is indeed a scalar field whenever $A^\mu$ is a vector field.

## Tensors in geometry

As you can imagine, the above expression based upon coordinates is highly unsatisfactory in the mathematical profession of differential geometry. There is a great notation called abstract index notation which solves its coordinate-centric problems.

See, the physics definition is a "blacklist": it says "do whatever you want, and you'll figure out if it's a tensor or not afterwards by how it behaves when we change our coordinates." By contrast, geometric definitions are a "whitelist": they say "we're going to start with good objects and good operations, and then everything we create will be good."

Basically, we define a set of functions $\mathcal A \subseteq (\mathcal M \to \mathbb R)$ as "scalar fields", where $\mathcal M$ is whatever space we're interested in. For $\mathbb R^n$ a simple choice is the smooth fields $\mathcal A = C^\infty(\mathcal M, \mathbb R)$. Then the derivations on $\mathcal A$ form a vector space (you would normally write $v^\alpha \partial_\alpha$ for a derivation), and we postulate a metric, which makes the vector space isomorphic with its dual. With a metric tensor and an antisymmetric tensor we can then usually build up all of the other objects that we're interested in -- tensor fields, for example.

You can then insert these coordinates again, when you need them, by marking them differently (which could be by capitalization or boldface or underlining or primes/dots or switching to/from Greek letters...). So you use something like the covectors $c^a_\alpha$, $a \in \{0, 1, \dots n-1\}$, to turn some $v^\alpha$ (a vector) into its $n$ components $c^{a}_\alpha v^\alpha$.

In this sort of calculus, if you partially limit (in some coordinates) $$\phi = \sum_{a \in A} \partial_{a} A^{a}$$ and thus arrive at a smooth function from $\mathcal M \to \mathbb R$, then since that smooth function is in $\mathcal A$, it is **obviously a scalar field**, and everyone can agree on its existence and properties: however, someone else with different coordinates $v^{\bar a} = \bar c_\alpha^{\bar a} v^a$ will not necessarily agree that it can be represented as $\sum_{\bar a \in A'} \partial_{\bar a} A^{\bar a}$ for any set $A'$.

Or to take the reverse, in curved manifolds there are non-tensors called Christoffel symbols which are really useful in general relativity; you can use this approach to turn any reference frame's Christoffel symbol into a tensor. However: not all reference frames will agree that the resulting Christoffel tensor has any relationship to their own Christoffel symbols; it's not "the" Christoffel tensor, but rather "a" Christoffel tensor derived from this particular context. Similarly, in special relativity when we take the 4-velocity we unambiguously specify the reference frame that the $dt$ of time is being measured in to be the particle's own reference frame, so that it's a "proper time" $d\tau$. The resulting notion is indeed a (1,0)-tensor, because we *specified explicitly* the reference frame that we're stealing the time coordinate from.

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