Understanding the Equivalence Principle

[Sagittarius A-Star]

TL;DR Summary

Is a violation of the Equivalence Principle proven by the following calculation?

Alan Macdonald claims in "4 Appendix: The Equivalence Principle" of his text "Special and General Relativity based on the Physical Meaning of the Spacetime Interval", that his calculation regarding a 2D-surface of a sphere proves, that the equivalence principle is violated.

He defines the local measurement of

$M = ({3 \over \pi}){2\pi r - C(r) \over r^3}$

with $C(r) = 2 \pi R \sin{ \Phi } = 2 \pi R \sin{ (r/R) } = 2 \pi R [r/R - {(r/R)}^3 /6 + ...]$.

He argues, that the equivalence principle is violated because of $\lim_{Area \rightarrow 0} M = 1/R^2$ instead of zero.

The calculation of the limit is correct. But can his argument regarding violation of the equivalence principle be correct?

Source (see under "4 Appendix: The Equivalence Principle"):
http://www.faculty.luther.edu/~macdonal/Interval/Interval.html

 

[PeterDonis]

He argues, that the equivalence principle is violated because of $\lim_{Area \rightarrow 0} M = 1/R^2$ instead of zero.

The equivalence principle does not say what he claims it says. He is claiming that the equivalence principle, or more precisely a generalized version of it that he is implicitly talking about, says that the curvature of a curved manifold has to vanish as the size of the patch of the manifold we are looking at goes to zero. But the EP does not say that. The Riemann curvature tensor of a curved manifold is a tensor in the tangent space at a point of the manifold, which is the correct technical version of "taking the limit as the size of the patch we are looking at goes to zero", and it is nonzero in a curved manifold.

To put it another way, he is claiming that the EP says that no observables can couple to the curvature of the manifold. But that is obviously false; if it were true, we could not measure the curvature of the manifold! But we obviously can.

The paper references other arguments in the literature along the same lines, but AFAIK none of those arguments are generally accepted by physicists.

 

[A.T.]

TL;DR Summary: Is a violation of the Equivalence Principle proven by the following calculation?

He argues, that the equivalence principle is violated because of limArea→0M=1/R2 instead of zero.

He seems to have the common confusion I described here:

This confusion is not specific to relativity, but is a more general issue about "curvature" vs. "effect of curvature".

When you consider a region of a spherical surface that gets smaller and smaller, the effects of curvature (deviation from the Euclidean circumference/radius-ratio or triangle angle sum) tend towards zero.

But the curvature stays constant in this process, because of the way curvature is defined: The "effect of curvature" is divided by the size of the considered region.

"Locally flat" doesn't mean that curvature goes to zero, but that, for a fixed curvature, deviations from (pseudo)Euclidean geometry go to zero, due to locality.

 

[Sagittarius A-Star]

He seems to have the common confusion I described here:

But the curvature stays constant in this process, because of the way curvature is defined: The "effect of curvature" is divided by the size of the considered region.

Is it possible to locally measure curvature, as described by Alan Macdonald for the sphere?

According to Sean Carroll, 4D-curvature cannot be measured locally.

Einstein Equivalence Principle, or EEP: “In small enough regions of spacetime, the laws of physics reduce to
those of special relativity; it is impossible to detect the existence of a gravitational field.”

Source (page 99):
https://arxiv.org/abs/gr-qc/9712019

 

[Ibix]

Is it possible to locally measure curvature, as described by Alan Macdonald?

Consider his $\Delta a/\Delta r$. What you would do is release two test masses $\Delta r$ apart, wait $\delta t$, then measure the new separation $\Delta r+\delta r$. Then his parameter is $2\delta r/\delta t^2\Delta r$. Note that all these things are independently measured, so the fractional error on the result is going to look like $$\sqrt{\left(\frac{\sigma_{\delta r}}{\delta r}\right)^2+\left(\frac{\sigma_{\delta t}}{\delta t}\right)^2+\left(\frac{\sigma_{\Delta r}}{\Delta r}\right)^2}$$where $\sigma_x$ is the standard deviation of your measurement of $x$ and I've neglected a shed load of constants I would need to care about if I were actually doing this. Note that his region in spacetime is defined by $\delta t\times\Delta r$, so both of those need to go to zero for this to be a pointlike measure. But the $\sigma$s won't go to zero.

So yeah, you can measure whatever you like, but only in a region of spacetime big enough that your errors don't swamp your measures. Macdonald's maths says that $\Delta a/\Delta r$ tends to a finite value; mine says the errors on the measure tend to infinity. So over a small region it is indistinguishable from zero, in agreement with Carroll. I expect that to be always true - I can't see how you can measure anything in zero time and zero space, even if you idealise away the non-continuous nature of matter.

 

[Sagittarius A-Star]

Macdonald's maths says that $\Delta a/\Delta r$ tends to a finite value; mine says the errors on the measure tend to infinity.

Does the same kind of argument also apply to Macdonalds local measurement of $M$ in his sphere example?

 

[Ibix]

Does the same kind of argument also apply to Macdonalds local measurement of $M$ in his sphere example?

I would think so. You'd be measuring $C$ and $r$ at the very least, and you'd need to show that the measurement errors were anti-correlated so that the error on the combination didn't go to infinity.

I can imagine circumstances where this is the case. For example, if I have a fan but no aircon then I'm sure the temperature is the same everywhere during each measurement, but measurements on different days scale due to thermal expansion. In that case I expect my larger $r$ values to coincide with my larger $C$ values and I'll get some cancellation. But as you let $r\rightarrow 0$ the errors will be dominated by the precision of the rulers, which won't get better as the measures get smaller.

 

[Sagittarius A-Star]

I would think so. You'd be measuring $C$ and $r$ at the very least, and you'd need to show that the measurement errors were anti-correlated so that the error on the combination didn't go to infinity.

I can imagine circumstances where this is the case. For example, if I have a fan but no aircon then I'm sure the temperature is the same everywhere during each measurement, but measurements on different days scale due to thermal expansion. In that case I expect my larger $r$ values to coincide with my larger $C$ values and I'll get some cancellation. But as you let $r\rightarrow 0$ the errors will be dominated by the precision of the rulers, which won't get better as the measures get smaller.

I made some calculations with Wolfram Alpha. Critical is, that the definition of the measurement value $M$ contains a division by $r^3$. If I change this into a division by $r^2$, then the limit of $M$ becomes zero and the curvature cannot be measured. The $r^3$ singles out the 3rd order term of the Taylor series of the sine function.
Does this division by $r^3$ increase the error of the measurement?

https://www.wolframalpha.com/input?...ivide[r,R]]\(41),π\(40)Power[r,3]\(41)],r->0]

compare to:
https://www.wolframalpha.com/input?...ivide[r,R]]\(41),π\(40)Power[r,2]\(41)],r->0]

 

[Ibix]

Does this division by $r^3$ increase the error of the measurement?

As long as you are measuring two or more things and combining them to get what you want to measure you have to worry about the errors in what you are directly measuring. And as long as the things you are directly measuring go to zero and the errors in them don't, the fractional errors diverge. If the thing you are calculating goes to zero you may be able to argue for a finite absolute error, depending on whether the errors diverge faster than the value goes to zero, but if it goes to a finite value you're going to get error bars that grow without bound.

I mean, imagine that the best ruler money can buy has graduations of 1mm. Once $r$ is of order 0.1mm, your best estimate of $M$ is going to be $0/0$ anyway, no? Substitute smaller units for mm as much as you like.

 

[JimWhoKnew]

Consider his $\Delta a/\Delta r$. What you would do is release two test masses $\Delta r$ apart, wait $\delta t$, then measure the new separation $\Delta r+\delta r$. Then his parameter is $2\delta r/\delta t^2\Delta r$. Note that all these things are independently measured, so the fractional error on the result is going to look like $$\sqrt{\left(\frac{\sigma_{\delta r}}{\delta r}\right)^2+\left(\frac{\sigma_{\delta t}}{\delta t}\right)^2+\left(\frac{\sigma_{\Delta r}}{\Delta r}\right)^2}$$where $\sigma_x$ is the standard deviation of your measurement of $x$ and I've neglected a shed load of constants I would need to care about if I were actually doing this. Note that his region in spacetime is defined by $\delta t\times\Delta r$, so both of those need to go to zero for this to be a pointlike measure. But the $\sigma$s won't go to zero.

So yeah, you can measure whatever you like, but only in a region of spacetime big enough that your errors don't swamp your measures. Macdonald's maths says that $\Delta a/\Delta r$ tends to a finite value; mine says the errors on the measure tend to infinity. So over a small region it is indistinguishable from zero, in agreement with Carroll. I expect that to be always true - I can't see how you can measure anything in zero time and zero space, even if you idealise away the non-continuous nature of matter.

I think that an additional way to look at it, Despite Macdonald's footnote 15, is that if $\Delta a\approx 0$ , then it takes a long time to measure, which makes the measurement nonlocal.

 

[PeterDonis]

it takes a long time to measure, which makes the measurement nonlocal.

Exactly. The EP is about locality in spacetime, not just space.