0
$\begingroup$

What is the precise mathematical definition of an equation to be Lorentz invariant? Is it the same as being invariant under the maps $x \mapsto \Lambda x$, with $\Lambda$ being a given Lorentz transformation? I believe this is indeed the appropriate definition, but taking as an example the Klein-Gordon equation: $$\bigg{(}\frac{\partial^{2}}{\partial t^{2}} + \Delta + m^{2}\bigg{)} \psi(x) = 0$$ let us take $\Lambda = (\Lambda_{ij})$ a $4\times 4$ matrix and $x' = \Lambda x$. Then, if $x_{\mu}$ denotes any component of $x=(x_{0},x_{1},x_{2},x_{3})$: $$\frac{\partial}{\partial x_{\mu}}\psi(\Lambda x) = \sum_{i=1}^{4}\frac{\partial \psi}{\partial x_{i}'}(x')\frac{\partial x_{i}'}{\partial x_{\mu}} = \sum_{i=1}^{4}\frac{\partial \psi}{\partial x_{i}'}(x')\Lambda_{i\mu}$$ and, consequently: $$\frac{\partial^{2}}{\partial x_{\mu}^{2}}\psi(\Lambda x) = \sum_{i=1}^{4}\sum_{j=1}^{4}\frac{\partial^{2}\psi}{\partial x_{i}'\partial x_{j}'}(x')\Lambda_{i\mu}\Lambda_{j\mu}$$ How does this imply Lorentz invariance, i.e. how does one recover the original Klein-Gordon equation from this?

$\endgroup$
2
  • $\begingroup$ It is probably easier to prove that the action is invariant. But the same kind of math will be involved. see david tong . qft lectures. page 12 damtp.cam.ac.uk/user/tong/qft/one.pdf damtp.cam.ac.uk/user/tong/qft.html $\endgroup$ – user288901 Mar 12 at 2:06
  • $\begingroup$ transforming the lagrangian is much easier because you're only transforming contravariant vectors. And those are easy to transform. $\endgroup$ – user288901 Mar 12 at 2:10
1
$\begingroup$

The definition of a Lorentz invariant equation is that it should be written for a section of a vector bundle over a Lorentz manifold. Generally speaking, this would need to be soldered to the tangent bundle in order that there is a natural diffeomorphism of the bundle for every diffeomorphism of the base manifold.

Unpacking all this gives the usual definition of Lorentz scalars and Lorentz vectors. Unfortunately, although this language is widespread in mathematics, it's not so well known in physics, so it's difficult to point to an appropriate reference that demonstrates all this in detail.

Probably the simplest case of this is to take a section of the most natural vector bundle one can have on a manifold, the tangent bundle. After unpacking the definitions we get the usual definition of a contravariant tangent vector, for example, in Dirac's book on GR.

$\endgroup$
1
$\begingroup$

I think your equation has some minus signs wrong. Let $\eta^{\mu \nu} = \eta_{\mu \nu} = \mathrm{diag}(1,-1,-1,-1)$ be the Minkowski metric. Then $$ \partial_t^2 - \Delta = \eta^{\mu \nu} \partial_\mu \partial_\nu. $$ The Lorentz transformation matrices $\Lambda$ satisfy $$ \Lambda^T \eta \Lambda = \eta $$ by definition. In index notation, this is $$ \Lambda^\alpha_{\; \mu} \eta_{\alpha \beta} \Lambda^\beta_{\; \nu} = \eta_{\mu \nu}. $$ Then \begin{align} \eta^{\mu \nu} \partial_\mu \partial_\nu \psi(\Lambda x) &=( \eta^{\mu \nu} \frac{\partial (\Lambda x)^\alpha}{\partial x^\mu} \frac{\partial (\Lambda x)^\beta}{\partial x^\nu} \partial_\alpha \partial_\beta \psi)(\Lambda x) \\ &=( \eta^{\mu \nu} \Lambda^\alpha_{\; \mu} \Lambda^\beta_{\; \nu} \partial_\alpha \partial_\beta \psi)(\Lambda x) \\ &= (\eta^{\alpha \beta} \partial_\alpha \partial_\beta \psi)(\Lambda x). \end{align}

This means that if $\psi(x)$ satisfies the Klein Gordon equation, then $\psi(\Lambda x)$ does as well.

Edit: If we want to prove explicitly that $$ \eta^{\mu \nu} \Lambda^\alpha_{\; \mu} \Lambda^{\beta}_{\; \nu} = \eta^{\alpha \beta}. $$ perhaps it is easier to translate this to matrix notation, where it reads $$ \Lambda \eta \Lambda^T = \eta $$ which is clearly true.

There's also another whay to prove it, which involved using the formula $$ \eta \Lambda^T \eta = \Lambda^{-1} $$ which is true because $$ \Lambda( \eta \Lambda^T \eta ) = \eta \eta = I. $$

Edit: Let's do this another way. If we define $$ y^\mu = \Lambda^\mu_{\; \nu} x^\nu $$ where $$ \frac{\partial y^\mu}{\partial x^\nu} =\Lambda^\mu_{\; \nu} $$ then $$ \frac{\partial}{\partial x^\nu} = \frac{\partial y^\mu}{\partial x^\nu} \frac{\partial}{\partial y^\mu} = \Lambda^\mu_{\; \nu} \frac{\partial}{\partial y^\mu} $$ which means that $$ \eta^{\mu \nu} \frac{\partial}{\partial x^\mu} \frac{\partial}{\partial x^\nu} = \eta^{\mu \nu} \Lambda^\alpha_{\; \mu} \Lambda^\beta_{\; \nu} \frac{\partial}{\partial y^\alpha} \frac{\partial}{\partial y^\beta} = \eta^{\alpha \beta} \frac{\partial}{\partial y^\alpha} \frac{\partial}{\partial y^\beta} $$

$\endgroup$
6
  • $\begingroup$ You have assumed the invariance of $\eta^{\mu\nu} \partial_\mu \partial_\nu$ which is ok but I guess it was this invariance which the question was asking about. $\endgroup$ – Andrew Steane Mar 12 at 8:40
  • $\begingroup$ Please see my edit. I didn't assume the invariance of $\eta^{\mu \nu} \partial_\mu \partial_\nu$ anywhere. I proved it using the identity $\Lambda \eta \Lambda^T = \eta$. $\endgroup$ – user1379857 Mar 12 at 16:40
  • $\begingroup$ Let $\partial_a \equiv \partial/\partial x^a$ and $\partial'_a = \partial/\partial x'^a$ where $x'^a = \Lambda^a_{\; \nu} x^\nu$. Then one is needing to prove that $\eta^{\mu\nu} \partial_\mu \partial_\nu$ has the same effect on a scalar invariant function as does $\eta^{\mu\nu} \partial'_\mu \partial'_\nu$. But I don't see any mention of $\partial/\partial x'^a$ anywhere in this answer. If you prefer some other notation to the use of a prime, that's ok. The point is that one has to consider differential operators with respect to the coordinates in the new frame, not just $\partial_a$. $\endgroup$ – Andrew Steane Mar 12 at 16:46
  • $\begingroup$ Please see my latest edit $\endgroup$ – user1379857 Mar 12 at 16:58
  • $\begingroup$ Yes that's how I thought it would go. I am guessing the original version amounts to the same thing but I find that less intuitive so maybe this fuller version will be helpful; I hope overall this exchange has been useful rather than the reverse. I'll leave my comments for a little while and then delete them. $\endgroup$ – Andrew Steane Mar 12 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.