# Why the topological term F\til{F} is scale independent?

 P: 4 why the topological term in gauge theory, $ε_{\mu\nu ρσ}F^{\mu \nu} F^{ρσ}$ ,is scale-independent?
P: 908
 Quote by sinc why the topological term in gauge theory, $ε_{\mu\nu ρσ}F^{\mu \nu} F^{ρσ}$ ,is scale-independent?
Do you know how the field tensor transforms under scale transformation?
P: 4
 Quote by samalkhaiat Do you know how the field tensor transforms under scale transformation?
if the spacetime coordinates scales as $x_{0}→S^{a}\bar{x_0}, x_{i}→S^{b}\bar{x_i}$,then the potential vector scales as follow: $A_{0}→S^{a+d}\bar{A_0}, A_{i}→S^{b+d}\bar{A_i}$

P: 908
Why the topological term F\til{F} is scale independent?

 Quote by sinc if the spacetime coordinates scales as $x_{0}→S^{a}\bar{x_0}, x_{i}→S^{b}\bar{x_i}$
Why do space and time scale differently? Can you tell me your background in physics?
 P: 4 We are considering the most general rescaling, so space and time scale differently. This is especially true for nonrelativistic case.
P: 908
 Quote by sinc We are considering the most general rescaling, so space and time scale differently. This is especially true for nonrelativistic case.
No, not “especially”. Time and space scale differently ONLY in non-relativistic theory. But, your original question is meaningless in the non-relativistic domain. This is why I asked you about your background in physics.
Any way, in relativistic field theories, the coordinates scale according to
$$x^{ \mu } \rightarrow \bar{ x }^{ \mu } = e^{ - \lambda } x^{ \mu } ,$$
and the field transforms as
$$F ( x ) \rightarrow \bar{ F } ( \bar{ x } ) = e^{ \lambda \Delta } F ( x )$$
where $\Delta$ is the scaling dimension of the field. In D-dimensional space-time:
$$\Delta = \frac{ D - 2 }{ 2 }, \ \ \mbox{ for } \ \ A_{ \mu } (x) ,$$
and
$$\Delta = \frac{ D }{ 2 } , \ \ \mbox{ for } \ \ F_{ \mu \nu } ( x ) .$$

See (for more detailed description) the link below