# Is this a Lorentz-scalar? How do I tell?

I'm struggling to identify whether a scalar is a Lorentz-scalar. E.g:

$$\partial_i A^i \quad i \in {1,2,3}.$$

How do I determine if this is a Lorentz-scalar or not?

If got the same problem with tensors. How do I differentiate between a tensor and a Lorentz-tensor?

• Well, how did you define "Lorentz-tensor" and "tensor" in the first place? – ACuriousMind Jul 20 '15 at 17:56

# 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.

This is not that difficult: A scalar is something that is just a number and a Lorentz scalar is a scalar that is invariant under Lorentz transformations. For example distance is a scalar but not a Lorentz scalar and a proper time interval is a scalar and a Lorentz scalar.

To find out if something is a Lorentz scalar we simple check how it transforms under a Lorentz transformation. For your example:

$$\partial_i A^{i} \rightarrow \partial'_i A'^{i} = \rightarrow \Lambda_i{}^j \partial_j \Lambda^{i}{}_k A^{k} = \Lambda_i{}^j \Lambda^{i}{}_k \partial_j A^{k} = \delta^j{}_k \partial_j A^{k} = \partial_j A^{j}$$

and in fact all scalar products of a covariant and a contravariant vector are Lorentz invariant. To see that this is a scalar just write $\partial_i A^{i}$ in it's components.

For an explanation of what a tensor is look here. A Lorentz tensor is then a tensor that transforms like a tensor under Lorentz transformations:

$$T^{\mu_1\dots\mu_n}{}_{\nu_1\dots\nu_m}\rightarrow T'^{\mu_1\dots\mu_n}{}_{\nu_1\dots\nu_m} = \Lambda^{\mu_1}{}_{\lambda_1}\dots\Lambda^{\mu_n}{}_{\lambda_n} \Lambda_{\nu_1}{}^{\sigma_1}\dots\Lambda_{\nu_m}{}^{\sigma_m} T^{\lambda_1\dots\lambda_n}{}_{\sigma_1\dots\sigma_m}$$

hope this helps.

• The OP said $i = 1,2,3$ so it wouldn't be a Lorentz scalar. – FenderLesPaul Jul 20 '15 at 23:45
• Ah ok, well in this case $\partial_i^{i} = \nabla A$, now it should be obvious that this is not a Lorentz scalar, as the electric field and the magnetic field are not Lorentz invariant and $A$ is directly related to $B$. But I can't think of a good way to prove it. – john Jul 21 '15 at 0:22
• @FenderLesPaul But from the way the question was asked I assumed this was just a typo. – john Jul 21 '15 at 0:52

I'm struggling to identify whether a scalar is a Lorentz-scalar. E.g: ∂iAii∈1,2,3. How do I determine if this is a Lorentz-scalar or not? If got the same problem with tensors. How do I differentiate between a tensor and a Lorentz-tensor?

By "tensor" do you mean a tensor in three space dimensions? If so, then a tensor is never a Lorentz tensor. A Lorentz tensor needs the fourth components to be included in any sum to produce a lower rank tensor.

## protected by Qmechanic♦Jul 21 '15 at 13:32

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