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A scalar field is one which is unchanged under rotation. But how do we decide whether a given field (e.g., the temperature in a room) is unchanged under rotation or not? We need to measure the temperature at any point $P$ in two coordinates at some time $t$. In one $P$ has coordinates $(x,y,z)$ and in a rotated frame where $P$ has coordinates $(x^\prime,y^\prime,z^\prime)$. If we find $$T(x,y,z,t)=T^\prime(x^\prime,y^\prime,z^\prime,t)$$ for all points $P$, we will call it a scalar field.

To check whether it is unchanged under a Galilean boost or Lorentz boost, do we also need to perform experiments and decide? Can one exclude or establish whether the scalar-like behaviour holds under these transformations (even without measurements)? Stated differently, I mean

$\bullet$ Does one expect $$T(x,y,z,t)=T^\prime(x^\prime,y^\prime,z^\prime,t)$$ under Galilean boost where $\vec{r}^\prime=\vec{r}-\vec{V}t$ and $t^\prime=t$?

$\bullet$ Does one expect $$T(x,y,z,t)=T^\prime(x^\prime,y^\prime,z^\prime,t^\prime)$$ where the primed coordinates ${x^\prime}^\mu$ and unprimed coordinates $x^\mu$ are related by Lorentz boost?

Solidification
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1 Answers1

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Whether fields are scalar, vector, tensor or whatever else is not a question of experiment. Saying that temperature is a scalar field is not a claim about any concrete arrangement of temperature values in the real world, it is a claim about the mathematical nature of the temperature field in the formalism.

You decide the transformation behaviour of a field based on its definition in terms of other quantities, and their definitions, and so on, until you arrive at a list of quantities whose transformation behaviour is known axiomatically (e.g. coordinates). Then you figure out how the field transforms based on how these quantities are combined. Since we systematically label non-scalar quantities with indices, it is often enough to ascertain that the expression does not contain any free (i.e. non-summed-over) indices to know that it is scalar.

Temperature is an annoying example for a scalar field because its definition is $1/T = \partial_E S(E)$ but energy $E$ is not a scalar in relativity, but rather part of the 4-momentum vector. Depending on what exactly you think the energy apparing there is, and what an observer experiences as "temperature", you can come to the conclusion that it is a covariant, contravariant or scalar quantity, cf. this question and its answers

ACuriousMind
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  • "You decide the transformation behaviour of a field based on its definition" what is the definition of the density field $\rho(x,y,z,t)$? How does the Higgs field fit into your answer? – Solidification May 05 '19 at 12:26
  • @mithusengupta123 density is mass per unit volume, so if you know how mass and volumes transform, you also know how densities transform. The Higgs field is defined to be a scalar field, it is one of the things whose transformation behaviour is axiomatic. – ACuriousMind May 05 '19 at 16:46