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In order to derive the conservation of four-momentum of a field, $P^\mu$, it is assumed that the total change in the field, defined as $$\delta\phi_a(x)\equiv {\phi^{\prime}}_a(x^{\prime})-\phi_a(x)$$ is zero for any field, under a spacetime translation $x^\mu\to x^{\prime\mu}=x^\mu+\varepsilon^\mu.$ Here, $\varepsilon^\mu$ is an arbitrary constant 4-vector.

Why should the fields not change under translation (but may change under Lorentz transformations)?

ACuriousMind
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Solidification
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  • Why should they? Do you have an example for a field on $\mathbb{R}^4$ for which $\delta \phi_a \neq 0$? – ACuriousMind May 16 '23 at 09:31
  • Is it saying that there are no four vectors that change under translation? Well, I understand that under $x^{\prime\mu}=x^\mu+\varepsilon^\mu$, four-vectors such as four-velocity $U^\mu=dx^\mu/d\tau$, four-acceleration $\alpha^\mu=d^2x^\mu/d\tau^2$ etc are unchanged. But I am not sure why $A^\mu$, the electromagnetic four-potential, must be invariant. – Solidification May 16 '23 at 09:49
  • ?? look here https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity for the A – anna v May 16 '23 at 10:26
  • Could you add the source you are working with? – AlmostClueless May 16 '23 at 11:10

3 Answers3

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A tensor field on $\mathbb{R}^n$ changes under a diffeomorphism $f : \mathbb{R}^n\to\mathbb{R}^n$ by the Jacobian of $f$ (pullback by diffeomorphism), e.g. a vector field has $A(x)\mapsto Df\cdot A(f(x))$, where $Df$ is the Jacobian matrix. The Jacobian of a translation $x\mapsto x+a$ is the identity matrix, while the Jacobian of a Lorentz transformation $x\mapsto \Lambda x$ is $\Lambda$.

In more common physics notation, this is just the claim that a vector field obeys $$ V^{\mu} = \frac{\partial y^\mu}{\partial x^\nu}V^\nu $$ for a coordinate transformation $x\mapsto y(x)$, which is more or less the definition of a vector field.

ACuriousMind
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Relativity teaches us that the physics should be independent of our choice of coordinate system, so the (physical) fields don't actually change under (mathematical) coordinate transformations. What the transformations laws for tensor fields are really describing is not how the fields themselves change, but rather how the components of a field expressed in a particular basis change according to how you change the basis you're using to represent the field. In other words, the value of a field at a given point is going to be the same regardless of whether you call that point $x$ or $x'$, since your still talking about the same point.

Sandejo
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  • While this is correct for coordinate changes, the transformations we consider in the context of Noether's theorem are not coordinate changes but actual "active transformations" on the manifold. While the behaviour under these is closely related to the behaviour under coordinate changes, these notions are not exactly identical, see also my answer at https://physics.stackexchange.com/a/759904/50583 – ACuriousMind May 17 '23 at 16:01
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I recently learned that the transformation matrices of fields under translations and Lorentz transformations form representations of the Poincaré group, whose properties can be obtained by studying the corresponding Lie algebra. Different irreducible representation results into different field, such as scalar, spinor, vector fields and so on.

An infinitesemal translation can be written as $$ \phi_i'(x')=[\exp(-\mbox i a^\mu P_\mu)]_{ij} \phi_j(x), $$ where $P_\mu$ is the generator for translation. In representation for scalar fields, $P_\mu=0$. So actually $\phi'(x')=\phi(x)$. But I don't know what it is for other fields.

Luessiaw
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