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In the Minkowski spacetime of special relativity it is apparent that along the path of a light ray $ds^2=0$ in any inertial reference frame given the Lorentz transformations and the invariance of $ds^2$. However this does not seem so apparent to me in the context of a curved spacetime given all reference frames, to my knowledge, are non-inertial. I have seen this taken to be fact in the context of curved spacetimes and have not came across justification for it yet.

QUESTION: Is there a rigorous mathematical proof that $ds^2=0$ along the path of a light ray, in any spacetime, in any reference frame, or is this an axiomatic fact?

Note: I assume this can be proven mathematically somehow and would be very interested in what the proof/logic is behind it. Also, the only curved spacetime I'm sufficiently familiar with is that described by the Schwarzschild metric.

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  • $\begingroup$ EdRich: "spacetime interval of light, in any spacetime" -- From Wikipedia page Spacetime intervals in flat space and from several discussions here I was given to understand that spacetime intervals $$s^2 : \mathcal S \times \mathcal S \rightarrow \mathbb R$$ are defined strictly only for flat spacetimes. (Some variant applicable in Lorentzian spacetimes of arbitrary curvature would be "Lorentzian distance".) Consequently your questions appears ill-posed. $\endgroup$
    – user12262
    Mar 15, 2017 at 22:56
  • $\begingroup$ @user12262 I believe you're right and I've made an edit to fix this. I was referring to the infinitesimal line element $ds^2$ along a ray of light and not the integer valued spacetime interval of Minkowski space. Thanks for pointing out the needed clarification. $\endgroup$
    – EdRich
    Mar 16, 2017 at 0:10
  • $\begingroup$ EdRich: "Thanks for pointing out the needed clarification." -- You're welcome. "I've made an edit to fix this." -- Thanks for having been responsive; +1. But perhaps you can appreciate that I'm not quite satisfied: "I was referring to the infinitesimal line element $ds^2$ along a ray of light" -- What does that even mean?? What exactly could you possibly mean by "$ds^2$" in settings in which $s^2$ is explicitly not defined? Why are you not asking an applicable question about Lorentzian distance $\ell$ instead?!? But, admittedly, my complaint goes beyond your specific question ... $\endgroup$
    – user12262
    Mar 16, 2017 at 5:56
  • $\begingroup$ @user12262 the metric - $ds^2$ physically describes the length between two infinitesimally separated points in a given spacetime, so that is the "distance" i am referring to. I believe then s is simply the spacetime interval between 2 points on a given spacetime and can be found by integrating a given metric. I looked at the link you provided however I'm not to sure what Lorentzian distance really is or how it fits into the context of which I'm asking the question in. However i could be wrong in my interpretations above. $\endgroup$
    – EdRich
    Mar 18, 2017 at 3:03
  • $\begingroup$ EdRich: "$ds^2$ physically describes the length between two infinitesimally separated points in a given spacetime" -- If there are indeed two distinct (separated) points (events) involved, case by case, we ought spell them out explicitly as two arguments to this (generalized) "lenght" (squared), and maybe drop that weird $d$ in front. Also: experimentalist like to know how to determine whether two distinct events under consideration are "infinitesimally separated", or just "separated". "what Lorentzian distance really is" -- Check google.com/#q=beem+ehrlich+easley+distance $\endgroup$
    – user12262
    Mar 18, 2017 at 6:12

2 Answers 2

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So, in general, a coordinate transformation involves giving each coordinate $x$ as a function of the new coordinates $x^{\prime}$:

$$x^{\prime\mu} = f^{\mu}(x^{\nu})$$

This gives a general relationship:

$$ds^{2} = g_{\mu\nu}dx^{\mu}dx^{\nu} = g_{\mu \nu}\left(\frac{dx^{\mu}}{dx^{\prime\alpha}}dx^{\prime \alpha}\right)\left(\frac{dx^{\nu}}{dx^{\prime\beta}}dx^{\prime \beta}\right) = g_{\alpha \beta}^{\prime}dx^{\prime \alpha} dx^{\prime \beta}$$

Where we can transform the old inner product defined by $g$ into a new inner product defined by $g^{\prime}$. So, if $ds^{2}=0$ in our old reference frame, then it is necessarily zero in our new reference frame. You can argue that the formalism is cooked to make this true, and you'd be right, but that's what the underlying mathematics of general and special relativity describe.

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  • $\begingroup$ I understand the invariance of the metric equation and I understand the logic behind you're argument in answering my initial question. But given what you've said, I suppose my question is now, given the spacetime interval for light is always zero in an inertial frame in Minkowski space, can we always find coordinate functions between Minkowski space and any curved space, thereby concluding the interval must always be zero in any curved spacetime as well? $\endgroup$
    – EdRich
    Mar 15, 2017 at 22:04
  • $\begingroup$ @EdRich: no. You can compute tensors out of the metric that are coordinate-independent but which are nonzero in curved spaces, but zero in non-curved spaces. There is no coordinate transformation that will globally map the Schwarzschild metric to the Minkowski one. $\endgroup$ Mar 15, 2017 at 23:31
  • $\begingroup$ So, for a really blatant example, $R_{abcd}R^{abcd}$ is nonzero for every curved spaetime that I can think of, but will be zero for any coordinate transform of the Minkowski spacetime. $\endgroup$ Mar 15, 2017 at 23:39
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    $\begingroup$ So if I'm following what you're saying then, because $ds^2=0$ along the path of light ray in Minkowski spacetime, then this must be the case for any spacetime since as you've stated the metric equation is invariant? Is it that simple? $\endgroup$
    – EdRich
    Mar 15, 2017 at 23:58
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    $\begingroup$ @EdRich. I don't want to complicate this but in your last comment I think it is misstated. It is in fact true (I think contrary to what you are saying) that only in the same spacetime is ds invariant with respect to coordinate transformations. A different spacetime will have different form for ds, and if you somehow use the same coordinates as in the first spacetime you get a different value of ds. And yes ds is invariant for curved or flat spacetimes (but if you go from curved to flat something changes). $\endgroup$
    – Bob Bee
    Mar 16, 2017 at 1:20
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If you want a really rigorous proof, then the answer is that it is in fact not true that light rays have $ds^2 = 0$, because in a curved spacetime there is no general notion of a light ray.

What you would have to do is solve Maxwell's equations in a curved background; assuming that the typical length scale of the field (so the wavelength) is much smaller than the length scale of spacetime curvature you get something like the wave equation, and you can pretend your wave is a light ray. If you do this it turns out that the wave four-vector $k^\mu$ of this wave/ray must be null ($k_\mu k^\mu = 0$), which is the same as saying that $ds^2 = 0$ along the ray. For some insight into the subtleties of this, see for example this post.

However, there is another way of getting at this if you're willing to accept that in special relativity there is such a thing as a light ray. The argument rests on the fact that every manifold is locally flat; to put in physics words, in a curved spacetime you can choose coordinates near a point so that the metric is that of SR to first order, and is exactly the Minkowiski metric at your chosen point. You simply say that in these particular coordinates $ds^2 = 0$, and since $ds^2$ is a scalar, it will be zero no matter what system of coordinates.


There is a conceptual distinction which can be made here, which is that of speed of light vs. the speed of causality. The latter is the maximum speed at which things can travel, and almost by definition the maximum speed paths are those with $ds^2=0$; this is because the very structure of relativity enforces this speed limit. It just so happens that light travels at this maximum speed (particle physicists will mention the masslessness of the photon at this point), but there's no fundamental reason it must do so.

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    $\begingroup$ You have null geodesics in a curved spacetime, which is just as much a notion of a "light ray" as anything, at least in the geometric limit. $\endgroup$ Mar 16, 2017 at 5:32
  • $\begingroup$ @Jerry: that's what I tried to get at in my last paragraph, though maybe it didn't come out right. The point is that if you want to get really rigorous you have to show that light rays travel on null geodesics. $\endgroup$
    – Javier
    Mar 16, 2017 at 13:18
  • $\begingroup$ @Javier that makes more sense to me in how one would have to show that but seems awfully difficult. Also, as I mentioned in another comment I've looked up the approximate first order local flatness of curved spacetimes which I think serves as enough justification for me. I like the idea behind your final comment as well. Ill keep that in mind. $\endgroup$
    – EdRich
    Mar 18, 2017 at 3:13

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