In the answer to the question: Coordinate Transformation of Scalar Fields in QFT by joshphysics a very nice mathematical explanation (using manifolds and charts) is given for the transformation of the scalar field. In physics we usually write this simply as:

$$ \phi(x) \rightarrow \phi(x)' = \phi(\Lambda^{-1}x), $$

where $\Lambda$ is e.g. a representation of the Lorentz transformation and $\phi(x)$ is a scalar field.

Transformation of vector field $V^{\mu}(x)$ is then given as:

$$V^\mu(x) \rightarrow V^\mu(x)' = \Lambda^{\mu}_{\nu}\,V^{\nu}(\Lambda^{-1}x),$$

and analogously for the tensor field we have:

$$T^{\mu\nu}(x) \rightarrow T^{\mu\nu}(x)' = \Lambda^{\mu}_{\alpha} \Lambda^{\nu}_{\beta}\,T^{\alpha\beta}(\Lambda^{-1}x).$$

How does one see/get this transformation rules in the same "mathematical" picture (using manifolds and charts picture) as done by joshphysics for the scalar field?

  • $\begingroup$ The reason for the argument $\Lambda^{-1}x$ is the same as the scalar case. So this question really boils down to asking for the reasoning behind the tensor transformation rule (which does of course have an answer involving charts and frames etc). $\endgroup$
    – jacob1729
    Commented Feb 4, 2020 at 17:03

1 Answer 1


Absolutely! Take vector field under Lorentz transformation as an example. What is on manifold is a vector field $A$, suppose we have two charts $\{x^\mu\}$ and $\{x'^\mu\}$, and there is a coordinate transformation $x'^\mu={\Lambda^{\mu}}_\nu x^\nu$ . The vector fields can be expanded in the two charts as $A=A^\mu(x)\partial_\mu=A'^\nu(x')\partial'_\nu$, note that $x$ and $x'$ are two coordinates of the same point. $ \\$Then there is a relation: $\partial_\mu=\frac{\partial x'^\nu}{\partial x^\mu}\partial_\nu'$, take it in and we get :$A^\mu(x)\frac{\partial x'^\nu}{\partial x^\mu}=A'^\nu(x')$, which is just $A'^\nu(\Lambda x)={\Lambda^\mu}_\nu A^\mu(x)$. Take a replacement of $x$ by $\Lambda x$, we get : $A'^\nu(x)={\Lambda^\mu}_\nu A^\mu(\Lambda^{-1}x)$

  • $\begingroup$ This is valid for any type of tensor fields. And for more general fields like spinor fields, I don't know how to argue, but it's natural to accept we need the representation of the group element, like $\phi'_a(x')=\pi(g)_{ab}\phi_b(x)$ $\endgroup$
    – Daniel YUE
    Commented May 27, 2021 at 8:39
  • $\begingroup$ Also, do remember that the fields we care about are the irreducible representation of Lorentz group. If you do a dilation transformation $x'=\lambda x$ for a scalar field $\phi$, the answer is not $\phi'(\lambda x)=\phi(x)$ but $\phi'(\lambda x)=\lambda^{-\Delta}\phi(x)$, we need more algebra structure to get the answer. $\endgroup$
    – Daniel YUE
    Commented May 27, 2021 at 8:46

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