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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?

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4  
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.

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There is a certain rule of thumb as to know when a quantity is a scalar or not: if the indices go through all space-time dimensions, i.e., $\alpha=0,1,2,3$, then the expression is probably a scalar. Your formula is not a scalar because $i=1,2,3$, so when performing a change of reference system, the zeroeth component of $A$ and $\partial$ will mess up your formula. On the other hand, something like $\partial_\alpha A^\alpha$, where $\alpha=0,1,2,3$ is in general a scalar, because the index can take four values.

But why does this work? what's the point of this?

Let's say you write, in Classical Mechanics, the following set of equations: \begin{aligned} ma_x&=F_x\\ma_y&=F_y\\ma_z&=\color{red}2F_z \end{aligned} This is clearly a "bad" set of equations, because it's not symmetric: the $z$ component of your vectors is treated very differently compared to the others. If you and me use this formula in order to get physical predictions, and we use different systems of reference, then we will get very different predictions. This formula can't work in general.

But let's say we by chance pick systems of reference such that my $y$ axis is your $x$ axis and vice versa, and the $z$ axes coincide; in this case our predictions will agree. This is similar to what happens in Newtonian Mechanics: the equations are not invariant, but our systems of reference are in general so similar that we can't detect any flaw in the equations. If you take a very fast-moving system of reference, then you will easily see that Newtonian mechanics predictions disagree between observers.

In Special Relativity, the time component of vectors is just as important as the others, so your formulas should be symmetric w.r.t the four components. If you have in Newtonian mechanics a formula that tells you something about the $x$ and the $y$ components of some vector, then there should be another analogous formula for the $z$ component; otherwise a rotation would make your formula nonsense. If you have in Special Relativity a formula that tells you something about the $x,y$ and $z$ components of some vector, then there should be another one that tells you the same thing about the time component; otherwise a boost would make your formula nonsense.

In this sense, you should be able to tell if this formulas are "good" formulas or not:

\begin{align} \partial_\alpha A^\alpha&=A^0\tag{1}\\ \partial_\alpha\partial^\alpha T^\beta&=j^\beta\tag{2}\\ j^0=|\phi|^2\ &\text{and}\ j^i=\phi^*\partial^i\phi-\phi\partial^i\phi^*\tag{3}\\ p_\alpha p^\alpha&=m^2\tag{4} \end{align}

and so on. Remember that a Greek index will always take four values and a Latin one only three.

So remember: invariant formulas are symmetric w.r.t. the four components of vectors, and everything should be written with Greek indices, not Latin ones. If a formula includes Latin indices but you don't see the 0th component around, this will mean that the formula is not invariant. This might be tricky, because you will sometimes find things like $\partial_t \rho+\partial_ij^i=0$, and this might look non-invariant, but it really is, provided you have $\rho=j^0$.

I don't know why you make any distinction between a tensor and a Lorentz-tensor: they are the same thing. When studying mathematics, we talk about general tensors because we don't care about physical theories, but when the vector space of interest is space-time, then the tensor is a Lorentz-tensor. In general we never say "Lorentz-tensor", but simply "tensor".

PS: only (2) and (4) are invariant. Can you tell why the others are not?

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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.

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1  
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.

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protected by Qmechanic Jul 21 '15 at 13:32

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