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19

No, one doesn't need to measure the material for years - or even millions or billions of years. It's enough to watch it for a few minutes (for time $t$) and count the number of atoms $\Delta N$ (convention: a positive number) that have decayed. The lifetime $T$ is calculated from $$ \exp(-t/T) = \frac{N - \Delta N}{N}$$ where $N$ is the total number of atoms ...


11

I agree with @Ron Maimon that these ETS questions are problematic. But this is (i think) the reasoning they go with. Unlike @Mike's assumption you should not take the normal average, but as stated in the question the weighted average. A weighted average assigns to each measurement $x_i$ a weight $w_i$ and the average is then $$\frac{\sum_iw_ix_i}{\sum_i ...


9

Geant is a framework---which means that you use it to build applications that simulate the detector and physics you are interested in. The simulation can include all of physics and the complete detector including electronics and trigger (i.e. you can write your simulation so that it output a data file that looks just like the one you are going to get from ...


8

Typically one knows the functional form of something when they are interested in the FWHM. In this case, you can use least-squares fitting to fit the function to the data and extract the parameters. A very common functional form of a resonance is the Lorentzian: $$f(A,\gamma,x_0;x) = A\frac{1}{1 + \gamma^2\left(x-x_0\right)^2}$$ where $1/\gamma$ gives the ...


7

The formula you've specified $$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$ is the formula to obtain error of quantity $k$, as being dependent on $k_1$ and $k_2$ according to the following expression $$ k = k_1 + k_2.$$ Generally, to obtain experimental error of a dependent quantity (and the expression stated in your question), you start with ...


7

The fact that the quantities $\langle p^2\rangle$ and $\langle p\rangle^2$ are different is not something specific to quantum mechanics, but exists in any context where one can define an average value. Let's take a simple example and suppose that $p$ is a balanced binary random variable which can take the values $+1$ and $-1$. Since the variable is ...


7

You might like this 110-page paper by me and Alex Arkhipov, which is all about a quantum bosonic analogue of Galton's board (we even use the same graphic you did -- see Section 1.1!). In particular, we gave strong evidence that such a board (with an arbitrary configuration of "pegs," and with multiple entry points for the "balls") is exponentially hard even ...


7

Standard deviation adds uncertainties to the measured value: $23.3\pm 0.4\,{\rm m}$. One can quickly look at the error (which has units of ${\rm m}$ in my case) and think, The value could be as low as $22.9\,{\rm m}$ or as high as $23.7\,{\rm m}$ without much thinking. Modifying this to being a percentage of the value would be confusing. Plus it would be ...


7

No because none of them know the actual answer. The averaging process you describe only works if each estimate is of the exact answer plus noise. Otherwise it is known as the "Emperor's nose" problem. Nobody can see the Chinese emperor's face so they ask a million peasants how long his nose is, they average the results, and since they have such a large 'N' ...


6

I don't think there's any way you're going to do this in six months. I'll give a calculation below, but first an order of magnitude estimate. If you've detected a total of $N_{\rm events}$ events, your measurement of a modulation will have an error of order $N_{\rm events}^{-1/2}$ -- -- these things always do! -- so the number of events required is going ...


6

Errors in particle physics are of two kinds. Statistical, and systematic. Statistical is the usual standard deviation of gausian distributions, sqrt(n)/N for 1 sigma. It is the systematics that take a lot of effort, and often are not taken well into account. Systematic errors come from 1) the background to the signal expected. The background is calculated ...


6

The main problem is to determine what corresponds to zero mass of the harmonic oscillator. Remember that a fraction of the spring mass also participates in the motion. By introducing an intercept $\beta$, your friend takes into account that the true zero of the mass parameter $m$ may be shifted from what you think it is. So an affine model $T^2=Cm+\beta$ is ...


6

Well, the difference between the two expressions is exactly $(\Delta p)^2$, i.e. the squared uncertainty (variance) of the momentum, as the very question correctly says. To be sure, the real question is why it's not zero. It's not zero. Just write the function $\psi$ in the momentum representation. Then there is a probabilistic distribution $$ ...


6

The ultimate answer is the JCGM 100:2008 guide followed by most of the metrology institutes around the world. The specific chapter on combining uncertainties is Chapter 5. Specifically, for a two-variable function $f(t_1, t_2)$ of two random variables, Eq. (16) of Section 5.2.2. gives $$\Delta f^2= \left (\frac{\partial f}{\partial t_1} \right )^2 \Delta ...


6

think about this with an example: the sine and cosine functions. They both average individually to zero over an interval. You can multiply those averages and still obtain zero. But if you multiply sin by itself and then average, you get a very distinct non-zero result. When the functions are arbitrary, the average of the product quantifies statistical ...


6

I believe you are thinking of the Central Limit Theorem. The mean and variance of the averages of many measurements are better estimates of the precision of your measuring rule, but don't tell you anything about the accuracy of your measuring rule. Your measuring rule may be biased. The Central Limit Theorem is a part of mathematics. IMO you should also ...


5

I think you're exercising an incorrect picture of statistics here - mixing the inputs and outputs. You are recording the result of a measurement, and the spread of these measurement values (we'll say they're normally distributed) is theoretically a consequence of all of the variation from all different sources. That is, every time you do it, the length of ...


5

Yes, the only sensible formula for the total error is the sum in quadrature, $$ \Delta X_{\rm total} = \sqrt { \Delta X_{\rm syst}^2 + \Delta X_{\rm stat}^2 } $$ The key assumption behind the validity of the formula is that the two sources of error are independent i.e. uncorrelated. $$ \langle \Delta X_{\rm syst} \Delta X_{\rm stat} \rangle = 0$$ Because of ...


5

Maybe for the same reason that experimental physics groups do not have a theoretician as a group member. One could think of experimental groups as ruled by "control freaks", they need as members experimetnalists who have mastered enough theory to set up the experiments and enough theory to interpret the results. Within this "theory" one could count ...


5

On the deepest level, particles are indistinguishable if and only if they have the same quantum numbers (mass, spin, and charges). However, in statistical mechanics one ofte studies effective theories where there are additional means of distinguishing particles. Two important examples: In modeling molecular fluids, two atoms on the same molecule are ...


5

Statistics are used in physics to provide a conceptual link between the 'macroscopic view' and the 'microscopic view'. For example, when studying gases, we can examine the statistical distribution of particle velocities and energies to gain an understanding of the relationship between the macroscopically observable quantities (pressure, volume & ...


5

Depends what you are measuring. If you are measuring the temperature of a glass of water to 20C +- 0.1C what would the percentage be? Now what would it be if you measured it as 293Kelvin +- 0.1K


5

No, of course not. Yes, some people will overestimate and others will underestimate. Averaging would cancel out the bias to some extent, but there's no reason to expect it to cancel out the bias perfectly. We all have similar eyes and brains. We are all deceived by the same optical illusions, in the same way. We all have a shared cultural understanding of ...


5

$10\%$ of $10\text{ cm}$ is $1\text{ cm}$. $10\text{ cm}\times\frac{10}{100}=1\text{ cm}$. I really don't see how this is a 'nit-pick', or 'relative' vs 'absolute', as suggested in the comments. Are you absolutely, $100\%$, or at least relatively, sure you got the quote right? (There are very close variations of it, where indeed it would be rather stupid ...


5

If the 10 cm was a measured value with a corresponding error estimate, then the results would be different. Suppose the bar was $10 \text{ cm} \pm 0.10 \text{ cm}$. Adding exactly $1 \text{ cm}$ to the length of the bar would yield $11 \text{ cm} \pm 0.10 \text{ cm}$, whereas increasing the length of the bar by 10% yields $11 \text{ cm} \pm 0.11 \text{ cm}$. ...


4

Yes, it is simple to prove using moment generating functions. And yes, the mathematics is very closely related to that of quantum field theory. You compute $G(j) = <exp(\sum j_i x_i)>$ where each $j_i$ is a "source" for the corresponding $x_i$. This is easily shown to be something like $G(j) = exp(\sum j_i \mu_{ij}^{-1} j_j)$ To get expectation ...


4

For statistical analysis Gnumeric works very well, as it has passed a lot of statistical test. This report explains why it is a much better choice than Excel. Of course there is also R, which is the largest free statistical package and is used in a lot of research areas. Personally I think going the plain python route is also not a bad idea, as there is ...


4

That depends entirely on what you consider to be "expected range of values." When you see a value like $3.43\pm 0.04$ (I will omit units for brevity), in many cases, it actually represents a normal probability distribution with a mean of $3.43$ and a standard deviation of $0.04$. If the $3.43\pm 0.04$ is the result of an experiment, for example, then the ...



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