# 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.

• Can you give me an example of someone saying "It has to be X because it's the only Lorentz invariant object there exists"? I can't imagine that being a very satisfying argument for anything. – Abhimanyu Pallavi Sudhir Jun 14 '18 at 4:19
• @AbhimanyuPallaviSudhir Sure: ibb.co/ny5QvJ – Jonathan Tooker Jun 14 '18 at 6:09
• @JonathanTooker Could you clarify whether you want to know "how to check if an object is Lorentz invariant" or "how to find all Lorentz invariant objects"? – gj255 Jun 14 '18 at 11:49
• @gj255 I can't actually. I have asked about both in the two sentences that end with question marks. If you know the answer to either, please share it. – Jonathan Tooker Jun 16 '18 at 14:28

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.

• I like this answer, but it doesn't fully explain how to check whether something is Lorentz invariant, which seems to be the OP's question. – gj255 Jun 14 '18 at 11:48
• @gj255, you are right about that, unfortunately I can't do that. Hopefully this will point the asker in the right direction. – Ján Lalinský Jun 15 '18 at 19:20

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.

• Hi, I suppose that is good information, but I'm not quite sure how it answers my question. For instance, I don't know what you mean by Lorentz scalar, and I don't see how what you wrote sheds light on that highly common phrase of so many authors which can be roughly paraphrased as, "It has to be X because that is the only Lorentz invariant object that exists." Also, I don't see how it relates to my question about whether or not there is some expeditious way to quickly find all Lorentz invariant objects (of low rank and relating to relevant quantities.) Could you say a little more, or a lot? – Jonathan Tooker Jun 14 '18 at 4:03
• @JonathanTooker: A Lorentz scalar is the same thing as a Lorentz invariant. You find all possible Lorentz scalars by writing down all expressions that have the form of a scalar, according to the criteria I described. – Ben Crowell Jun 14 '18 at 11:48