Testing the Lorentz invariance of tensors Often one encounters statements like, "We know $X$ has to be $Y$ because it is the only Lorentz invariant object that exists."  What is the most expeditious way to demonstrate that a tensor object is, or is not, invariant under Lorentz transformations?  Right now, I feel like I have to plug and chug through a hefty matrix algebra problem to show something is, or is not, Lorentz invariant.  Is there a better way that I can quickly get to "the only Lorentz invariant object that exists?," as so many authors often do?  Thanks.
 A: 
What is the most expeditious way to demonstrate that a tensor object is, or is not, invariant under Lorentz transformations?

Being Lorentz invariant has at least two different meanings. One is invariance of value, the other is invariance of form.
Lorentz invariance of value means we have a quantity whose components have the same value in all inertial frames, like:


*

*electric charge of isolated body $q$(single component);

*Kronecker tensor $\delta_{\mu\nu}$ (zeroes excepts units on diagonal);

*anti-symmetric Levi-Civita tensor $\epsilon_{\mu\nu\rho\sigma}$, with values $(0,1,-1)$ depending on the values of the indices). 

*"dot-product" of any four-vector with itself: for example radius 4-vector  $x^\mu x_\mu$

*contraction of a tensor with respect to all of its indices


If the tensor expression is a combination of the above, it has invariant value.
Lorentz invariance of form means that some physical law or quantity has the same expression (in terms of physical quantities and operations such as differentiation) in all inertial frames, but value of the quantities does not have to be the same. In relativity, any time you see equation or expression involving only usual operations with four-tensors, it has invariant form, or "is Lorentz covariant". For example:


*

*The equation of motion of point particle in external EM field
$$
qF^{\mu\nu}u_\nu = mdu^{\mu}/d\tau
$$
(where all quantities are four-tensors) has invariant form and is Lorentz covariant.



Right now, I feel like I have to plug and chug through a hefty matrix algebra problem to show something is, or is not, Lorentz invariant. Is there a better way that I can quickly get to "the only Lorentz invariant object that exists?," as so many authors often do? Thanks.

The authors have more experience, so get that too. However, sometimes these kinds of claims are simply not true. In the example you gave, the author claims that the only Lorentz-covariant constraint on $A^\mu$ that is linear in $A$ is the condition
$$
\partial_\mu A^\mu = 0.
$$
But there are other possible constraints, for example,
$$
X^\mu A_\mu = 0
$$
where $X^\mu$ is some fourvector field. Another example:
$$
\partial_\nu\partial^\nu A_\mu = 0.
$$
The author is implicitly assuming much more than he states: he wants a constraint that is first-order in derivatives, and also that it is not too restrictive, such as
$$
\partial_\mu A_\nu = \delta_{\mu\nu}.
$$
In other words, there is no "single possible Lorentz covariant expression" here. He uses standard gauge condition because it is customary in that part of theory.
A: Tensor notation is designed so that we can write expressions that manifestly have the transformation properties of tensors. In abstract or concrete index notation, anything you build out of tensors using the allowed operations (addition, contraction, covariant differentiation) is also a tensor, and you can tell its rank and valence based on the indices that have not been contracted over. For example, suppose that $P$ and $Q$ are known to be tensors. Then the expression
$$ Q^b{}_a{}^c \nabla_b P^a{}_{c}  $$
is a tensor because it's constructed in a legal way from tensors. It has rank 0 because it has no free indices -- all indices are "dummy" indices that have been contracted over. Therefore it's a Lorentz scalar.
