# Lorentz scalars - transformation of a scalar field

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1. Aug 2, 2015

### "Don't panic!"

When one considers a Lorentz transformation between two frames $S$ and $S'$, such that the coordinates in $S$ are given by $x^{\mu}$ and the coordinates in $S'$ are given by $x'^{\mu}$, with the two related by $$x'^{\mu}=\Lambda^{\mu}_{\;\;\nu}x^{\nu}$$ then a scalar field $\phi (x)$ relative to $S$ transforms as $$\phi '(x')=\phi (x)\qquad\text{or}\qquad\phi '(x)=\phi (\Lambda^{-1}x)$$ where $\phi '(x')$ is the form of the scalar field relative to the frame $S'$. Clearly, as it is a scalar it is a Lorentz invariant quantity.

My question (albeit a simple one, so apologies for that) is, when we transform the coordinates from those defined in $S$ to those defined in $S'$ (i.e. $x\rightarrow x'$), am I understanding it correctly that the field transforms as $\phi\rightarrow\phi '$, such that it has a different functional form in the two different frames, $S$ and $S'$, but in such a way that when evaluated at a given point $p$, then it has the same value in both frames? (That is, if a point $p$ has coordinates $x^{\mu}$ in $S$ and coordinates $x'^{\mu}$ in $S'$ then $\phi '(x')=\phi (x)$, however, the functional form of $\phi$ will be different to the functional form of $\phi '$)

2. Aug 2, 2015

### Staff: Mentor

Yes, this is how it works. Another way to put it is that the point $p$, physically, is identified by the value of physical quantities there, such as $\phi$; its coordinates may change but the physical values do not.

3. Aug 2, 2015

### "Don't panic!"

Ok great, thanks for confirming this. Is what I put about $\phi$ and $\phi '$ being different functional forms of the scalar field (relative to the two different reference frames $S$ and $S'$ respectively), but having the same numerical value at each point correct?

4. Aug 2, 2015

Yes.

5. Aug 2, 2015

### "Don't panic!"

OK, great. Thanks for your help! ☺