Series Expansion of the Infinitesimal Spacetime Interval

[Sciencemaster]

TL;DR Summary

I found a derivation of the spacetime interval's invariance that says if you expand a function of infinitesimals (i.e. dx), you can ignore terms of second order and above. Does this work?

I was watching an explanation of why the spacetime interval is invariant in all inertial frames (even when it's not lightlike) and the author made the assertion that if we have the relationship ds'=f(ds), we can expand the function as A+B*ds+C*ds^2+... (where C is not the speed of light). That's all well and good, but then he says that since we are working with infinitesimals, we can set all terms of O(ds^2) and greater to 0. I'm having trouble understanding why. Do infinitesimal series expansions really work like this?
Also, while I'm at it, the reason we can treat ds as a single quantity in the function is because each individual coordinate (i.e. dx) must have the relative weighting whether the interval is lightlike or not, lest the coefficient be coordinate-dependent (a big no-no), right?

The Video
Time: 9:58

Thank you for your help and patience!

 

[PeterDonis]

From what I can see, this video is arguing in a circle. It assumes that $ds^2 = c dt^2 - | d\vec{x} |^2$ is invariant, and then claims to "prove" that this is invariant.

Also, the video's claim to "prove" this invariance without using Lorentz transformations is obvious nonsense, since only under Lorentz transformations is that particular form of $ds^2$ preserved. Transform to a non-inertial frame and $ds^2$ takes a different form.

So this video does not look to me like a reliable source. Which would not be expected in any case for a pop science video.

 

[Sagittarius A-Star]

I was watching an explanation of why the spacetime interval is invariant in all inertial frames

For a better explanation it might be helpful to multiply equations (1) and (2) in combination with the derivation of $\gamma(v)$ behind equation (4) in
https://arxiv.org/abs/physics/0606046

 

[Ibix]

Is there something wrong with just writing $ds'^2=c^2dt'^2-dx'^2-dy'^2-dz'^2$, plugging in $dt'=\gamma(dt-vdx/c^2)$ etc and noting that this reduces to $ds'^2=ds^2$? As Peter noted, the chosen form of the metric is only invariant under this class of transforms. For example, if you pick anisotropic speed of light coordinates you get cross terms coming in, although of course the $ds^2$ remains invariant - only the relationship to coordinates changes.

 

[Sciencemaster]

First of all, thanks for the help. I especially appreciate Sagittarius A-Star's resource.
I would like to preface this next bit by saying that I'm not trying to disprove or argue with anyone--you've been helpful. I only say this because I don't want my next paragraphs to be interpreted as such.
However, many of the Lorentz Transformations' derivations I know of begin with the spacetime interval and derive from there. Even the original version of Sagittarius's resource does this. In addition, there are other more reputable resources that similarly derive the invariance of the spacetime interval. (In the latter resource, it's on pages 3-5). The point being, I think there's at least some value in deriving the spacetime interval.
The main differences are that the video uses a series to explain why the relationship is linear and that it has a definitive + sign at the end. I wanted to see if there was merit to either of these since the other sources seem to just assume a linear form for the relationship. As you said, it is a pop science video, so I could see the series bit being wrong. What do you think of this?

 

[Sagittarius A-Star]

Even the original version of Sagittarius's resource does this.

It is another text of the same author.
I like chapter "2 The Interval in Special Relativity".

In chapter "4 Appendix",
Alan Macdonald wrote (see cited link above)

However, Synge¹¹ and Ohanian¹² have argued, for me persuasively, that the equivalence principle, as usually understood, is false. I am not aware of any attempt to refute their arguments. I consider here the equivalence principle in the context of this paper.

Their arguments are refuted at mathpages.com

For example, Ohanian and Rufinni (1994) emphatically assert that “gravitational effects are not equivalent to the effects arising from an observer's acceleration...", even limited to sufficiently small regions. In support of this assertion they describe how accelerometers “of arbitrarily small size” can detect tidal variations in a non-homogeneous gravitational field based on “local” measurements. Unfortunately they overlook the significance of their own comment regarding gradiometers, i.e., “the sensitivity attained depends on the integration time… with a typical integration time of 10 seconds the sensitivity demonstrated in a recent test was about the same as that of the Eotvos balance…”. Needless to say, the “locality” restriction refers to sufficiently small regions of spacetime, not just to small regions of space. The gradiometer may be only a fraction of a meter in spatial extent, but 10 seconds of temporal extent corresponds to three billion meters, which somewhat undermines the claim that the detection can be performed with such accuracy in an arbitrarily small region of spacetime.

Source:
https://www.mathpages.com/rr/s5-06/5-06.htm

 

[Ibix]

I think the linear form of the relationship comes from the definition of an inertial frame, surely. If you are moving inertially then in an inertial frame your coordinates are $x=vt+x_0$, so a transform between inertial frames must map such a straight line to another one. It must also preserve parallel lines, so objects at relative rest are so in both frames. The most general transform that does that is the affine transform:$$\Lambda=\left(\begin{array}{cc}A&B\\C&D\end{array}\right)$$where $A$, $B$, $C$ and $D$ cannot depend on the coordinates. If you write the metric tensor as $\eta$, insisting $\eta=\Lambda\eta\Lambda^T$ gets you three of the constants. Insisting that $\Lambda$ should map $(t,vt)^T$ to $(t',0)^T$ gets you the last one.

 

[PeterDonis]

I think there's at least some value in deriving the spacetime interval.

It depends on what you want to take as given. If you take the Lorentz transformations as given, you can derive the standard SR form of the interval as the form which is left invariant by Lorentz transformations. Or if you take that form of the interval as given (for example by considering the behavior of expanding spheres of light from a point source, as IIRC Einstein did in one of his 1905 papers), you can derive the Lorentz transformations as those that leave that form of the interval as invariant. The relationship between the two is logical equivalence, so you can run the logic either way.

 

[Sciencemaster]

I think the linear form of the relationship comes from the definition of an inertial frame, surely. If you are moving inertially then in an inertial frame your coordinates are $x=vt+x_0$, so a transform between inertial frames must map such a straight line to another one. It must also preserve parallel lines, so objects at relative rest are so in both frames. The most general transform that does that is the affine transform:$$\Lambda=\left(\begin{array}{cc}A&B\\C&D\end{array}\right)$$where $A$, $B$, $C$ and $D$ cannot depend on the coordinates. If you write the metric tensor as $\eta$, insisting $\eta=\Lambda\eta\Lambda^T$ gets you three of the constants. Insisting that $\Lambda$ should map $(t,vt)^T$ to $(t',0)^T$ gets you the last one.

That's helpful, but there's something that confuses me here. Doesn't this formulation require you to know what $\eta$ is for $\eta=\Lambda\eta\Lambda^T$? Since $\eta$'s norm is the spacetime interval, it seems to me that you need to know the spacetime interval first to use this derivation anyway. Am I missing something?

 

[Sciencemaster]

It depends on what you want to take as given. If you take the Lorentz transformations as given, you can derive the standard SR form of the interval as the form which is left invariant by Lorentz transformations. Or if you take that form of the interval as given (for example by considering the behavior of expanding spheres of light from a point source, as IIRC Einstein did in one of his 1905 papers), you can derive the Lorentz transformations as those that leave that form of the interval as invariant. The relationship between the two is logical equivalence, so you can run the logic either way.

That's a good way of thinking of it. I think of the spacetime interval as a more mathematical way of stating that the speed of light is invariant (ds'=ds=0 when x=ct from Michaelson Morley, Maxwell's equations, etc.). As such, to me, it makes more intuitive sense to derive the Lorentz Transforms from them than the other way around (not that working in the other direction is 'wrong'). This being the case, starting with the spacetime interval seems somewhat in line with other derivations of the LT's which use the geometry of light rays. The last 'piece' of this invariance is showing that ds'=ds when ds/=0, which is what I'm trying to do here.

 

[Ibix]

That's helpful, but there's something that confuses me here. Doesn't this formulation require you to know what $\eta$ is for $\eta=\Lambda\eta\Lambda^T$? Since $\eta$'s norm is the spacetime interval, it seems to me that you need to know the spacetime interval first to use this derivation anyway. Am I missing something?

Sure. That's a derivation of the Lorentz transforms from the interval, which is what I thought you were after. Perhaps I misunderstood.

 

[Sciencemaster]

Sure. That's a derivation of the Lorentz transforms from the interval, which is what I thought you were after. Perhaps I misunderstood.

Oh, I see. I was looking for a way to derive the interval *in order* to derive LT's (like you have done). That's still helpful, though!
This isn't what I was originally after, but while I'm here, you say that that matrix formulation is the *simplest* way to satisfy the given requirements. Doesn't this leave a lot of room for alternative solutions?

 

[PeterDonis]

I was looking for a way to derive the interval *in order* to derive LT's

What would you derive it from in this case?

 

[Ibix]

Oh, I see. I was looking for a way to derive the interval *in order* to derive LT's (like you have done). That's still helpful, though!

Use the invariance of light speed to state the invariance of the interval for null separations. Derive the Lorentz transforms as above. Wonder if they also preserve the invariance of the interval for non-null separations, and plug them back in to the interval to find out.

you say that that matrix formulation is the *simplest* way to satisfy the given requirements. Doesn't this leave a lot of room for alternative solutions?

I think it's the simplest. There isn't an objective measure of mathematical simplicity, though, so feel free to disagree and look for something else.

 

[Sagittarius A-Star]

more reputable resources that similarly derive the invariance of the spacetime interval. (In the latter resource, it's on pages 3-5).

You find the proof for $n$ time-dimensions and $p$ space-dimensions in the chapter "Rigorous Statement and Proof of Proportionality of $ds^2$ and ${ds'}^2$ " under:
https://en.wikipedia.org/wiki/Deriv..._and_Proof_of_Proportionality_of_ds2_and_ds′2

 

[SiennaTheGr8]

I think the approach taken in Chapter III.2 ("Einstein's Clock and Lorentz's Transformation") of Anthony Zee's Einstein Gravity in a Nutshell might be more or less what the OP is looking for. Zee derives the LT from the interval's invariance, and his derivation of the interval's invariance isn't surreptitiously limited to lightlike intervals (he even includes an endnote about this).

 

[Sciencemaster]

You find the proof for $n$ time-dimensions and $p$ space-dimensions in the chapter "Rigorous Statement and Proof of Proportionality of $ds^2$ and ${ds'}^2$ " under:
https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations#Rigorous_Statement_and_Proof_of_Proportionality_of_ds2_and_ds′2

Yes, this is effectively what I was asking about. I especially appreciate this derivation's treatment of the coefficient (where it says that while #B(|v_2|)/B(|v_1|)=B(|v_{21}|)#'s left side doesn't depend on angle, the right one would, and so there can't be any v-dependence), it's more solid than most of the explanations I've seen for that. The issue I was originally having is with the linearity of the relationship between ds and ds', and this certainly addresses it better than the original source: the interval in both frames are both of the same order, so they must be proportional makes more sense than that series stuff.

 

[Sagittarius A-Star]

The issue I was originally having is with the linearity of the relationship between ds and ds', and this certainly addresses it better than the original source: the interval in both frames are both of the same order, so they must be proportional makes more sense than that series stuff.

In the 1960 book "Special Relativity" of Wolfgang Rindler, he wrote in §8, page 16:

For from a (by no means trivial) theorem of algebra we know that two irreducible polynomials of the same degree, which share all their zeros, can differ by at most a constant factor.

Source:
https://www.amazon.com/-/de/dp/101342879X/?tag=pfamazon01-20

 

[WWGD]

In the 1960 book "Special Relativity" of Wolfgang Rindler, he wrote in §8, page 16:

Source:
https://www.amazon.com/-/de/dp/101342879X/?tag=pfamazon01-20

View attachment 349345​

I'm a bit confused, though. When you say they're irreducible, but share their zeros, do you mean in some field extension?

 

[Sagittarius A-Star]

I'm a bit confused, though. When you say they're irreducible, but share their zeros, do you mean in some field extension?

I didn't find via Google the related theorem of algebra, which Rindler mentioned. To my understanding, he meant irreducible over real numbers.

 

[Sagittarius A-Star]

In the 1960 book "Special Relativity" of Wolfgang Rindler, he wrote in §8, page 16:

Source:
https://www.amazon.com/-/de/dp/101342879X/?tag=pfamazon01-20

View attachment 349345​

In his sold-out 1st edition (1982) of the book "Introduction to Special Relativity", Rindler described this in more detail. Unfortunately, this was removed from the 2nd edition of the book, in which the LT is derived differently.
Sold-out:
https://www.amazon.com/Introduction-to-Special-Relativity/dp/0198531818?tag=pfamazon01-20