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I am studying QFT but I need to fill some gaps in my comprehension of special relativity (I didn't study it very well and I know I still misunderstand things in S.R).

In my book it is written:

" A system is Lorentz invariant if it is symmetric under the Lorentz group"

I would like to clarify a little more this sentence.

What do we exactly mean by "symmetric under the lorentz group" ?

Does that mean that if a given (but any) system is described by any equation $(1)$ in a referential $R_1$ (coordinates $x_1$), if I do an inertial change of frame to $R_2$, the new equation will be the same as $(1)$ but with $x_1$ replaced by $x_2$ ?

I insist on the "any" in my paragraphs.

Qmechanic
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StarBucK
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  • It means that the equation of motion (say) would be invariant under the transformation moving from inertial frame $R_1$ to $R_2$. The Lorentz invariance stems from the symmetry under Boosts and Rotations. –  Sep 28 '17 at 11:33
  • More general related questions: https://physics.stackexchange.com/q/98714/2451 https://physics.stackexchange.com/q/160596/2451 – Qmechanic Sep 28 '17 at 11:55

2 Answers2

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From the point of view of the lagrangian formulation of the theory, the system described by the lagrangian $L$ is lorentz-invariant if the lagrangian remains the same after the arbitrary Lorentz transformation, with the latter acting on the objects from which the lagrangian is composed (say, the coordinate $x$, the EM field $A_{\mu}$ and so on). This in general doesn't mean that the equations describing the system must be lorentz invariant. Rather they must be lorentz covariant. For example, Maxwell equations in vacuum (the pair of them), which have the form $$ \partial_{\mu}F^{\mu\nu} = 0, $$ under the Lorentz transformations $\Lambda^{\mu}_{\ \nu}$ are changed as $$ \Lambda^{\nu}_{\ \alpha}\partial_{\mu}F^{\mu\alpha} = 0 $$

From the point of view of the quantum theory, with the ray $|\Psi\rangle$ representing the one state of the system and the ray $|\Phi\rangle$ representing another state, the quantum system is lorentz invariant if the probability $|\langle \Psi|\Phi\rangle|^{2}$ is invariant under the Lorentz transformations. This again imply that the states are transformed non-trivially under the Lorentz transformation, but imposes restrictions on the corresponding transformation operator (Wigner theorem).

Name YYY
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  • Ok, so if I understand well, the postulate of special relativity is that the lagrangian of the system is unchanged under lorentz transformation. But not all the equation of the system. – StarBucK Sep 28 '17 at 15:20
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    @StarBucK : right. The equations of motion on the field (I quess that you're rather interested in field theory, so I talk in terms of fields) themselves are the variation of the action with respect to the given field (with requirement of this variation to be zero), and this variation is, of course, in general not lorentz invariant (as the field is, in general, not Lorentz invariant, but lorentz covariant). – Name YYY Sep 28 '17 at 15:24
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    In fact I have another question close to your answer. Do you agree with me if I say that all physical quantities must be lorentz invariant. What I want to say is that if I reason in $R_1$ or in $R_2$, it is finally just a change of variables on the quantity I work with. So the predictions I do in $R_1$ must be equivalent to the one I do in $R_2$. Thus all scalar quantities in physics must be independent of the frame. So in the case of lorentz transformation : they have to be lorentz invariant. The vectors also but not their components (because the axis can move from $R_1$ to $R_2$). – StarBucK Sep 29 '17 at 13:28
  • For example a transition amplitude (like an atom that is going to de-excite itself) must have the same value if I reason in $R_1$ or in $R_2$. The only things that can change from a frame to another is the quantities that depend on axes (like vector or tensors). In this case the quantity "in itself" doesn't change but the components of the quantity on the axes change. – StarBucK Sep 29 '17 at 13:32
  • @StarBucK : the transition amplitude between the states $|1\rangle, |2\rangle$ must obviously depend on the frame, since it is just the sandwiched S-operator, $M_{1\to 2} = \langle 2|S|1\rangle$, where the $S$-operator is lorentz invariant. The reason is that the states $|1\rangle, |2\rangle$ obviously depend on the frame (since are characterized by the spin, momentum, orbital momentum)...

    The same is true about any vectors, so I don't understand your passage in the previous comment.

     – Name YYY Sep 29 '17 at 16:54
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I will assume that spacetime is flat four dimensional manifold equipped with a Lorentzian metric and define,

Physical systems: any object that is capable of causing a response ( measurement) in the measuring apparatus(observer)

Observer: Devise that interact with a physical system and produce a real number and capable to communicate with the interpreters.

Interpreter: Intelligent device who is capable to communicate with the observers.

For me measurement should be Lorentz invariant, since Lorentz transformations relates inertial observers (observers free of interaction) and since measurement is an interaction between the observer and a physical system.

For example measurement of an accelerated observer differs from inertial observer because there is an extra interaction in the accelerated device (the interaction causing the acceleration ),so their measurement differs because they are measurement different things.

Now in most text books they define Lorentz transformation as a change of inertial coordinates between inertial interpreters, and then say that physics should be invariant under Lorentz transformation.

I think this point of view get people confused because,coordinates is how we label events is spacetime so is clearly that physics should be invariant under an arbitrary change of coordinates and even if they mean change of change of inertial interpreters,well how can an interpreter influence the results of a measurement?

To clarify my idea i will give an example.

Suppose we have to interpreters $A$ and $B$ with wave functions $\psi(x)$ and $\psi'(x')$. Suppose they want to know the probability that a particle is in point $p$ in the manifold. So they send an observer to the point $p$ and the observer measures the probability and the observer send light to inform them the result of the measurement. The result should be the same for both that is $|\psi'(x')|^2=|\psi(x)|^2$ independent if $A$ or $B$ are accelerated.

So in my point of view physics should be invariant for an arbitrary interpreter and invariant under inertial observers

amilton moreira
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