# Coincidence of spacetime events & frame independence

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1. Jul 16, 2015

### "Don't panic!"

I have been reading these notes: http://isites.harvard.edu/fs/docs/icb.topic455971.files/l10.pdf
in which they claim that if two spacetime events are coincident in one frame of reference then they are coincident in all inertial frames of reference, thus spacetime events are absolute i.e. they exist independently of any reference frame.
Is this simply to do with the fact that if two spacetime events are coincident then their spacetime interval is null, i.e. $ds^{2}=0$, and as this is a frame independent quantity this must hold in any inertial reference frame, hence they are coincident in all inertial frames? Or does it simply follow from the fact that spacetime is modelled as a 4-D smooth manifold, and the points on a manifold exist independently of how you choose to represent them?
Is there anyway to prove the statement that if two spacetime events are coincident in one frame of reference then they are coincident in all inertial frames?

One of the reasons for me asking this is to try and understand why Lagrangian densities $\mathscr{L}$ are always expressed in terms of field values (and their derivatives) at a single spacetime point? Is this simply because the Lagrangian At a given instant in time is given by $$\mathcal{L}(t)=\int d^{4}x\;\mathscr{L}(t,\mathbf{x})$$ (I know that there is no explicit dependence on spacetime points, but this is just to highlight my question) and so fields can only interact with objects at a single spatial point (or infinitesimally close to that point), as otherwise they would be separated by a spacelike interval and their interaction would correspond to an instantaneous action at a distance (prohibited by the finite speed of light)?

2. Jul 16, 2015

### Orodruin

Staff Emeritus
This.

If you had terms which depended on different points in space-time, your theory would be non-local. Also note that the integral to obtain a Lagrangian is over space only, not over space-time. If you integrate over space-time, you obtain the action (which to some extent is more fundamental).

3. Jul 16, 2015

### Staff: Mentor

In SR just take the Lorentz transform of two events, and show that if $a=b$ then $a'=b'$

In GR it follows from the fact that coordinate charts are defined to be 1 to 1.

4. Jul 16, 2015

### "Don't panic!"

So when texts (such as the one that I linked to in my original post) state that the coincidence of spacetime events is frame independent (is this the statement that spacetime points are frame independent, i.e. if two spacetime events are coincident then they are labelling the same spacetime point?) then they are simply relying on the fact that spacetime is a 4-D manifold and nothing else? Is there any physical intuition to this as well?

I understand that interactions between two objects that are spatially separated at a given instant in time are non-local as they correspond to action at a distance (instantaneous influence of one object on another when they are spatially separated by some finite distance), but for instance, why couldn't one consider a Lagrangian density that involves fields evaluated at the same spatial point, but at different times? Or is it simply that the Lagrangian describes the dynamics of the state at a given instant in time and thus the Lagrangian density can only be local if it's value at each spatial point at a given instant only depends on the field values at that point (as otherwise this would correspond to instantaneous propagation of information which is prohibited due to the finiteness of light speed)?!

5. Jul 16, 2015

### "Don't panic!"

So would something like this be correct:

Let event $A$ be labelled by spacetime coordinates $x^{\mu}$ and event $B$ be labelled by spacetime coordinates $y^{\mu}$ in frame $S$. Furthermore, in this frame, let $x^{\mu}=y^{\mu}$. Now, in another inertial frame $S'$ we have that the coordinates of $A$ in this frame are related to its coordinates in $S$ in the following manner, $$x'^{\mu}=\Lambda^{\mu}_{\;\;\nu}x^{\nu}$$ Similarly, we have that the coordinates of $B$ in this frame are related to its coordinates in $S$ as, $$y'^{\mu}=\Lambda^{\mu}_{\;\;\nu}y^{\nu}$$ It follows then, that $$x'^{\mu}=\Lambda^{\mu}_{\;\;\nu}x^{\nu}=\Lambda^{\mu}_{\;\;\nu}y^{\nu}=y'^{\mu}$$ (as $x^{\mu}=y^{\mu}$). Therefore, if spacetime events are coincident in one frame they are coincident in all inertial frames (as the Lorentz transformation between frames was chosen arbitrarily).

6. Jul 16, 2015

### Staff: Mentor

Yes, exactly.

7. Jul 16, 2015

### "Don't panic!"

Is the reason why we require interactions to be localised in both space and time because intuitively (and philosophically so) it makes sense that objects (fields etc) should be in direct contact. If there is any finite separation between them, be it temporally, spatially, or both, then this would imply action at a distance (i.e. The two objects could immediately affect each other regardless of their temporal and spatial separation from each other without any intermediary medium in which the information can propagate between the two)?

Is it also the case that the only case in which an interaction is Lorentz invariant is when it occurs at a single point in spacetime - in all other cases Lorentz invariance is broken?!

(As an aside, in simplest terms, is locality simply the statement that we should be able to specify the location at which an interaction occurs. If it depends on two distinct points finitely separated, then it is not possible to specify an exact location at which the interaction occurs?)

Last edited: Jul 16, 2015
8. Jul 17, 2015

### vanhees71

Indeed we use the field description of interactions in relativity as the most simple form to accomodate interactions with the relativistic notion of causality. "Action at a distance" means that there would be causally connected events that are space-like separated, but this would violate the fundamental notion of cause and effect. The cause should always be temporally before the effect, and this implies that the effect should be within (or at) the future lightcone of the cause. This is most simply achieved by describing, e.g., forces on point particles as mediated via a field, i.e., you have a local Lagrangian for the motion of the particle within a field. E.g., for a particle in an electromagnetic field you have
$$L=-m c^2 \sqrt{1-\dot{x} \cdot \dot{x}}-\frac{q}{c} F_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu},$$
where $$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$$ is the Faraday tensor of the electromagnetic fields (its components are the usual electric and magnetic field components wrt. to a given reference frame). The dot indicates the derivative with respect to an arbitary scalar world-line parameter, and $m$ is the invariant mass of the particle.

The equations of motion read
$$m \ddot{x}^{\mu}=\frac{q}{c} {F^{\mu}}_{\nu} \dot{x}^{\nu}.$$
From here on we use the proper time $\tau$ of the particle as the world-line parameter.

The equations of motion which are automatically compatible with the mass-shell condition
$$p \cdot p=m^2 \dot{x} \cdot \dot{x}=m^2 c^2=\text{const},$$
because of
$$\dot{p} \cdot p=m^2 \ddot{x} \cdot \dot{x}=m \frac{q}{c} F_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=0.$$
Thus the on-shell condition is fulfilled at all times, if it is fulfilled by the initial values of the momenta, and this one must obey for consistency with the classical particle picture. It implies that the tangent vectors on the the space-time trajectory of the particle are always time-like and thus the causality constraint is fulfilled by such manifestly covariant equations of motion automatically.

This building principle of manifestly covariant Lagrangians can be generalized to arbitrary types of fields.

The realization of Einstein causality in quantum theory is more complicated. So far the only successful realization is again in terms of local, microcausal quantum field theory. For a detailed explanation see my lecture notes on QFT:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

9. Jul 17, 2015

### "Don't panic!"

Would you be able to comment on whether my following summary is correct or not?

Pre-relativity, the notion of action-at-a-distance was simply that one object could affect another instantaneously, without any mediating agent, regardless of their physical (i.e. spatial) separation. This in itself is philosophically undesirable as why should one object be able to affect another without being in contact, or without the affect being transmitted from one to the other without a medium. When relativity is taken into account this situation becomes untenable as even if one introducs a medium to transmit the interactions between the two objects, the finiteness of light speed means that no two objects that are physically separated can interact with one another without violating causality. Thus we implement locality and require that objects can only be affected by physics in their immediate neighbourhood, i.e. an opoint located at a specific point can only directly interact with objects at adjoining points.

Enforcing locality requires that at a given instant in time, two objects can only interact if they are located at the same spatial point, right? Why is it though that when we construct a Lagrangian density to describe an interaction between two fields the interaction terms must be defined at the same spacetime point? Why couldn't we consider interaction terms that are evaluated at the same spatial point, but at different instants in time?! Is this simply because the Lagrangian density is constructed in order to describe the Lagrangian at a given instant in time (by integrating over a spatial volume), $$\mathcal{L}(t)=\int\;d^{3}x\mathscr{L}(t,\mathbf{x})$$ and thus all fields in the Lagrangian density are necessarily evaluated at the same instant, and this in turn forces them to be evaluated at the same spatial point to ensure locality (two fields can only interact instantaneously if they are located at the same spatial point)?! Or is there something else to it that I'm missing?

Also, is the above description correct at all?

Last edited: Jul 17, 2015
10. Jul 17, 2015

### Staff: Mentor

In a relativistic theory, the Lagrangian density is a function on spacetime, not space, and the action is the integral of the Lagrangian density over a spacetime volume, not a spatial volume:

$$S = \int d^4 x \mathscr{L} \left( x \right)$$

where $x$ labels a point in spacetime, not space. This form of the Lagrangian density automatically ensures locality. It is also required for Lorentz invariance, since in a Lorentz invariant theory there is no invariant notion of "an instant in time" for spatially separated points; or, to put it another way, there is no invariant way to split up spacetime into "space" and "time", so you can't separate out integration over "space" from integration over "time".

11. Jul 17, 2015

### "Don't panic!"

Is this simply because the Lagrangian density depends only on a single spacetime point and not on any finitely separated points (infinitesimally close points are ok though, right? They appear implicitly in the dependence of first-order spacetime derivatives of the fields?), and so the automatically ensures that spatially separated fields cannot interact instantaneously with one another?

Is it correct to say then that the only case in which an interaction can be local in all inertial reference frames (i.e. To ensure that the locality of the interaction is Lorentz invariant) is if the interaction (described by the Lagrangian density) occurs at a single spacetime point?

Why wouldn't it be ok to include interactions that are timelike separated in the Lagrangian density (as isn't the timelike nature a Lorentz invariant concept). For example, why would it be incorrect to have something like $$\mathscr{L}\sim\phi (t,\mathbf{x})\psi (t',\mathbf{x})$$ in which $(t,\mathbf{x})$ and $(t',\mathbf{x})$ are timelike separated?

12. Jul 17, 2015

### Staff: Mentor

Yes.

Because there will be multiple timelike paths between those two events, and the action might depend on which path is taken. (In fact, this is exactly what happens in the path integral formulation of quantum mechanics.) In other words, timelike separated events are still not "local" (unless they are infinitesimally close). Another way to put it might be that the Lagrangian density you wrote down leaves out intermediate steps: what happens at events between the two you included?

13. Jul 17, 2015

### "Don't panic!"

So is the point that we require the Lagrangian density to be local in time as well?

Also, is it correct to say then that the only case in which an interaction can be local in all inertial reference frames (i.e. To ensure that the locality of the interaction is Lorentz invariant) is if the interaction (described by the Lagrangian density) occurs at a single spacetime point?

(As an aside, in simplest terms, is locality simply the statement that we should be able to specify the location at which an interaction occurs. If it depends on two distinct points finitely separated, then it is not possible to specify an exact location at which the interaction occurs?)

14. Jul 17, 2015

### Staff: Mentor

The point is that, in a relativistic (Lorentz invariant) theory, there is no such thing as "local in space" vs. "local in time"; there is only "local in spacetime".

Yes. If you think about it, you will see that this is the same thing I said just above, just in different words.

This is a reasonable way of putting it, yes.

15. Jul 17, 2015

### "Don't panic!"

Is this because in a relativistic framework it is simply not possible to separate time and space into separate entities and thus something can only be local if it is local in spacetime (as you said)?

16. Jul 17, 2015

### Staff: Mentor

Yes.

17. Jul 18, 2015

### "Don't panic!"

Does the same kind of idea apply in the classical mechanics case, i.e. Before constructing a Lagrangian I'm I correct in saying that we require it to be local in time, such that the dynamics of a physical system at a given instant in time are only affected by the local behaviour of the environment at that given point in time (and infinitesimally close to it)?

18. Jul 18, 2015

### Staff: Mentor

Sort of. You can formulate Newtonian mechanics using a Lagrangian that is apparently local, but it still doesn't work quite the same as in the relativistic case.

For example, Newton's law of gravity in its usual formulation is explicitly nonlocal; it describes a force that acts instantaneously at a distance. We can also formulate the law using an apparently "local" Lagrangian that includes a potential energy term due to the "gravitational field". But all that really does is hide the nonlocality in the potential energy term.

19. Jul 18, 2015

### "Don't panic!"

How does one motivate the Lagrangian depending on only one instant in time and it being dependent on position and velocity at that time then? I thought it was simply a requirement of locality again, i.e. that the dynamics of a physical system should depend only on the physics in its immediate neighbourhood? Or, in the classical mechanics case, is it's functional dependence purely based on empirical evidence (that we can determine the dynamic evolution of a system given its initial position and velocity), and the fact that Newton's laws hold?

In this case am I correct in saying that the force is nonlocal because it acts instantaneously (as you said) without any discernible contact between to massive bodies and without anything to mediate the interaction across the separation?!

How does the gravitational potential "hide" the nonlocality? Is it simply because it's value a a given point depends on masses that are finitely separated from it, i.e. it's of the form $$V\sim \frac{GM}{\vert\mathbf{r}-\mathbf{r}'\vert}$$

Last edited: Jul 18, 2015
20. Jul 18, 2015

### Staff: Mentor

Some history might help here. At the time Newton formulated his law of gravity, the behavior of gravity appeared (given the accuracy of measurement at the time) to be instantaneous action at a distance, so that's how Newton described it. He was uncomfortable with this, and so were others, because instantaneous action at a distance didn't seem physically reasonable; but at the time nobody knew any other way of formulating the theory.

In the 19th century, the Lagrangian method for formulating theories was discovered, and Newtonian gravity was reformulated that way, and people were happy because it appeared that now gravity could be accounted for as a local response to the "gravitational field". They didn't realize (at least, it appears that they didn't until relativity was discovered) that the theory was still nonlocal; the nonlocality had just been hidden inside the dynamics of the "gravitational field" (see further comments below).

Once Einstein discovered special relativity, and began work on trying to come up with a theory of gravity that was consistent with it, it became clear that the nonlocality in the classical theory, even in its Lagrangian formulation, was still there. The only way to fix it was to come up with a fully relativistic theory, based on a fully relativistic Lagrangian (in other words, one that doesn't have even an implicit, "hidden" nonlocality). Once he had this theory (General Relativity), it was easy to show that Newtonian gravity is a very good approximation to it for weak fields and slow motion (all speeds much slower than the speed of light), which is the regime in which Newtonian gravity was discovered. So gravity can "look like" instantaneous action at a distance (to a good approximation) under certain conditions, even though it really isn't.

So I would say it was the discovery of relativity that really made physicists adopt locality as a fundamental requirement.

Yes.

Because the potential energy itself depends on the instantaneous distance to other gravitating bodies; the "action at a distance" is just reformulated to determine the potential energy directly instead of the force. The force (or "gravitational field") is the gradient of the potential energy, so it is still being indirectly determined by action at a distance.