Jaepheth wrote:Hey, Kuroneko. Most of what you (ever) said is over my head, but I was wondering if you could clarify this:
"inertial frames are only ever local: over non-infinitesimal regions, they are approximations correct only to first order."
Everything here is continuous. What I mean is that (a) in GTR, inertial frames are exact over infinitesimally small regions of spacetime, (b) over larger regions, they are only approximations.
The very rough intuitive meaning is basically the same as "if you zoom in close enough on curved surface, it looks flat". But it isn't actually flat, so if you treat it as such, you only get an approximation. I'll try to motivate it in a more precise way shortly.
Jaepheth wrote:I ask because I usually read "infinitesimal" as meaning small, which in this case I'm interpreting to mean something akin to "not continuous" likely related to the Planck constant.
I don't know how much calculus you're familiar with, but if you are, this is the same sense as used in calculus.
If you're not, let me illustrate it by explaining roughly what kind of thing I was talking about in referring to approximation 'to first order' or 'to second order', etc. Suppose you want to approximate some smooth function, say f(x) = e
x, near some known value, say at a = 0. I picked this because it's simple: e
0 = 1, but in any would do. You could say: if my argument is not too far away a, then f(a+Δx) is not too far away from f(a):
[0] f(a+Δx) ≈ f(x), so e
Δx ≈ 1, if Δx is small
Let's call this a
zeroth-order approximation. And because f is continuous, this works to some extent, e.g., e
0.01 ≈ 1.010 ≈ 1. But it's rather terrible if Δx is not small enough, e.g., e
0.25 ≈ 1.284. If you know the slope of the tangent line f'(a), and calculus will tell you how to find it, you could do better:
[1] f(a+Δx) ≈ f(a) + (Δx)f'(a), so e
Δx ≈ 1 + Δx, if Δx is small
Let's call this a
first-order approximation. Note that e
0.25 ≈ 1.284 ≈ 1+0.25, better than before.
You can see graphically how well it's doing. And calculus will also tell you how to do even better using the value of the second derivative at a:
[2] f(a+Δx) ≈ f(a) + (Δx)f'(a) + (Δx)²f"(a)/2
... and so forth. Also note that if Δx is very small in magnitude, than (Δx)² is much smaller, e.g. 0.001² = 0.000001, etc. If we're taking Δx to 0, then (Δx)² goes to zero "faster" than Δx does, because the limit of their ratio is also 0.
In calculus, the derivative is actually defined as the limit of [1] when Δx tends to zero:
[Def] f'(a) = lim
Δx→0[ ( f(a+Δx) - f(a) ) / ( Δx ) ].
We could write something like
[???] f(x + dx) = f(x) + f'(x) dx
with infinitesimal dx, as an exact equation. In standard calculus, this would only make sense as a roundabout way of talking about limits: what it would mean taking a real Δx, there may be some error |f(x+Δx)-f(x)|, but as Δx is taken to be of smaller and smaller magnitude, that error goes to zero faster than Δx itself does. In non-standard analysis, we can have infinitesimal numbers directly, but we're not obligated to go in that territory.
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The path of a particle in free-fall is described by the geodesic equation, which in arbitrary coordinates specifies the second derivatives of the trajectory to be what are called the "connection coefficients" (Einstein called them "components of the gravitational field"). In special relativity, one can build coordinates in which those second derivatives vanish everywhere in spacetime, and so in which the path of every particle in free-fall is given by a line, everywhere and everywhen.
In general relativity, this is no longer possible, although one could at any event (point in spacetime) choose coordinates that make both the connection coefficients and their first derivatives vanish
at that event. Which means that near that event, they are approximately zero, to first order in the much same sense as above. The same kind of statement applies to the metric tensor, which physically acts like the potential for the gravitational field: at any event, we can choose coordinates such that it matches the Minkowski metric (the one in special relativity) to first order.
There's also a mostly independent approach based on
frame fields, but as the name suggests, what it intuitively corresponds to is stitching together multiple frames in a certain way, and so it's still the case that we can't get a
single inertial frame that covers a non-infinitesimal region except as a certain kind of approximation.