108
I work with an old toolmaker who also worked as a metrologist who goes on about this all day.
It seems to boil down to exploiting symmetries since the only way you can really check something is against itself.
Squareness:
For example, you can check a square by aligning one edge to the center of straight edge and tracing a right angle, then flip it over, re-...
58
The more you measure things and add or multiply those measurements, the greater your errors will become.
Not necessarily. If the errors in a series of measurements are independent and there is no systemic bias in the measurements then taking the average of all the measurements will give you a more accurate value than any individual measurement.
36
A second is a second long by definition, but if you measure any time in seconds, the number of seconds you infer will be subject to an error of at least $\mathcal O(10^{-15})$ because of the uncertainty of Caesium clocks as you correctly point out. This is true even if you make a higher precision measurement using new clocks like Quantum Clocks with ...
29
One thing I haven't seen mentioned is amplification.
Amplification: Imagine you have a lever that is 10 cm on one side of the pivot and 1 m on the other. Then any change in position on the short side is amplified 10 times on the long side. This new precision can be used to make new even more precise tools. Rinse and repeat.
Even just by iterating through ...
26
That's a really nice one! I'm not an expert on experiments and measurements but this is how I see it:
The ultimate calibration tool is always nature. We pick special phenomena which rely on certain parameters.
Take temperature, for example. The phenomenon here is the phase transition of water. You set boiling water to a $100^\circ$ C and freezing to $0^\circ$...
24
It basically comes from calculus (or more generally just the mathematics of change).
If you have a quantity that is a product $z=x\cdot y$, then the change in this value based on the change of $x$ and $y$ is$^*$ $\Delta z=x\Delta y+y\Delta x$. So then it is straightforward that
$$\frac{\Delta z}{z}=\frac{x\Delta y+y\Delta x}{xy}=\frac{\Delta x}{x}+\frac{\...
24
The calculations are done in
Schwartz (2019), Lecture 3: Equilibrium (https://scholar.harvard.edu/files/schwartz/files/3-equilibrium.pdf)
Here's a summary of the key equations, using the notation and equation numbers from those lecture notes.
Consider a classical system of spherical objects, each of radius $R$. Let $\ell$ be the average distance an object ...
20
Measurement errors can accumulate, yes.
But we are not talking about measurements here, we are talking about processes and tooling. That's another deal.
If you fling off a piece of flint from a flintstone by banging another stone into it, then the "precision" of that other stone doesn't matter. What matters is that you supply enough force - not ...
18
Accuracy can mean different things. While the question asks about the statistical accuracy, what immediately comes when talking about the Newton's laws is that they are non-relativistic, i.e., they are valid up to small corrections of order $v/c$.
Physics laws are based on empirical observations, the symmetries of the universe, and approximations appropriate ...
15
The way I understand it, errors only accumulate.
That is not always the case. Human ingenuity found and systematically selected processes which improved a particular quality.
Two examples:
You can hand polish a spherical mirror or even a plane mirror down below 100 nm surface roughness with easily available raw materials.
Zone melting of semiconductor or ...
14
I expect that worm drives are part of the story.
For every rotation of the driven axis the driving axis goes through multiple rotations. I expect that it is mechanically possible to capitalize on that ratio.
Other than that, forms of very high level consistency can be achieved with very low level means. A necessary tool for precision measurement is a large ...
13
Measurement errors or experimental errors can be reduced by, for example, using more accurate and more sensitive equipment; making multiple measurements and taking an average; thinking about possible sources of noise and trying to compensate for or reduce these. If you do all these things and there is still a difference between actual and expected results - ...
12
There are errors that come from measuring the quantities and errors that come from the inaccuracy of the laws themselves.
If we know only approximate values of parameters in the equation, then we can calculate how precise the result it using formulas for the propagation of uncertainty. For example, for the formula $F=ma$ we have
$$ \frac{\sigma_F}{F} = \sqrt{...
8
In my opinion, @gandalf61 has provided the correct answer. Let me just expand on it a little.
Measurement errors
The discrepancy between the law and the results of measurement may come from different sources, notably from the measurement errors (some of which cannot be controlled), but also possibly from sample rpeparation, different conditions at the time ...
8
The Higgs-discovery experiment is a particle-counting experiment. Lots of particles are produced by collisions in the accelerator, and appear in its various detectors. Information about those particles is stored for later: when they appeared, the direction they were traveling, their kinetic energy, their charge, what other particles appeared elsewhere in the ...
8
The key is that it's full width at half maximum, whereas you're thinking of $\pm 2.355\sigma$, which would have a total width of $4.710\sigma$.
7
In addition to excellent answer of Cream. Let us not confuse precision and accuracy.
Accuracy is the ability to obtain correct measurement in absolute scale. Following Cream's example, if you use a metal slab to define length of a meter, then indeed all your measurements are limited by how well you can produce and measure such slab. E.g. the reference ...
7
"Physical Laws" is a misleading expression. Physical theories and models expressed as mathematics are what we have. That does not mean the universe works that way "under the hood", it means those rules are good models to predict what will happen in given circumstances.
How Accurate a given theory is depends on three things :
How ...
7
I think this question may arise from a difference between somewhat rough layman's-terms presentations and the more careful statistics which goes on in the actual labs. But even after a given body of data has been analyzed to death, there is no formal way to capture in full the evidence underlying the way knowledge of physics grows. The evidence surrounding ...
7
The second itself does have an uncertainty. When we're using it without uncertainty, we're basically using the following trick:
The time span of $n$ seconds is defined as $2\times n\times 9\,192\,631\,770$ transitions between the two Cs hyperfine levels.
Making $n$ big, we can still count exactly how many transitions have occured, so the error on the total ...
6
I would also like to emphasize the importance of periodic waves which can magnify tiny errors. The most precise length measurements ever, by LIGO, are done with interferometry. A similar principle was used in the Michelson Morley experiment.
This also allows us to measure the speed of light for waves of different frequencies over long distances (gamma ray ...
6
In the uncertainty principle, $\Delta p$ is indeed the standard deviation, which is defined as
$$
(\Delta p)^2 = \left<(p-⟨p⟩)^2\right> = \left<p^2\right> - \left<p\right>^2
$$
(where $⟨\cdot⟩$ represents the average value of the quantity inside the brackets).
Now, in the calculation you mentioned, we expect $⟨p⟩$ to vanish, because ...
6
There's a shift from the old way of having standard examples of the units that everyone can compare against, to defining the units in terms of fundamental physics.
There was an uncertainty in measuring the speed of light. Then, the units were defined in terms of the speed of light, thus fixing the value exactly. Now there is still uncertainty in making the ...
6
The question is, a $10^{-12}\rm\,N$ force applied to what. A force of $10^{-12}\rm\,N$ applied to a hydrogen atom, with mass $10^{-27}\rm\,kg$, would produce an acceleration $F/m = 10^{+15}\rm\,m/s^2$.
A torsion pendulum is absolutely a way to allow very feeble forces to cause observable, macroscopic motion.
My favorite underrated classic paper is Beth’s ...
6
In the first place, I would write
$$ 5.868\,709\cdots×10^{−7} \pm 7.884\,31\cdots×10^{−12} $$
instead as
$$ (5.868\,709 \pm 0.000\,078\,8431 )×10^{−7} $$
with (a) grouped digits, (b) a common exponent, and (c) no ellipses. Next I would start removing “insignificant” digits.
The parenthesis notation gets rid of the leading zeros in the uncertainty. You ...
5
If an instrument involves counting, just count more
Some measurements are done by counting things. For example, the definition of the second is n = 9,192,631,770 cycles produced by Cs-133. The number of whole cycles that have elapsed on an atomic clock are exactly known, even if there is some uncertainty in the additional fraction of a cycle. No matter ...
5
Our partial differential equation (PDE) and boundary conditions (BCs) are:
$$T_t=\alpha T_{xx};$$
$$T(0,t)=30\text{ and }T_x(L,t)=0.$$
Let's use a generic initial condition (IC):
$$T(x,0)=f(x).$$
First, we transform the dependent variable $T(x,t)$:
$$u(x,t)=T(x,t)-30.$$
This means that:
$$\Rightarrow u(0,t)=30-30=0.$$
The derivative $T_x(L,t)$ isn't affected,...
5
The null hypothesis here is that the data was generated by physics which obeys the effective field theory describing all the Standard Model particles except the Higgs. This model doesn't usually have a name, but could reasonably be called the 'Standard Model without Higgs'. It's a perfectly good effective field theory. It's predictions are barely ...
5
It means 'of order' i.e. the approximate size
5
$A = O(1)$ means that $A$ has order of magnitude $10^1$. That means roughly
$$10^{0.5} \lt A \lt 10^{1.5} \\ 3 \lt A \lt 3 \times 10^{1}$$
$A = O(10^{-4})$ means that $A$ has order of magnitude $10^{-4}$. That means roughly
$$10^{-4.5} \lt A \lt 10^{-3.5} \\ 3 \times 10^{-5} \lt A \lt 3 \times 10^{-4}$$
The factor of $3$ here comes from $3 \approx 10^{0.5}$, ...
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