# Tag Info

1

If you really, really wanted to do a direct measurement of the separation energy (instead of doing precision mass spectroscopy), you could use the "shooting protons from a cyclotron at the $^{56}\mathrm{Fe}$ atom[s]" approach. But for practical reasons you would not look for the lowest energy at which you start getting neutrons out (which I would ...

0

It is perhaps interesting to think about Mach's paradox in this context. I'll get back to your question and the limits of the two-body way of discussing special relativity in the end. One form of the paradox is this: imagine a bucket of water standing on the floor. The surface is (almost) flat. Now start spinning it. The water's surface begins to form a ...

1

If there isn't, and both balls are absolutely identical, then how come one is still and the other one moving? Where does the difference of motion come from? I don't think this question is nearly as perplexing as you might think nor do I think it requires sophisticated physics like the best answer describes. Ask yourself, how do you show with a snapshot that ...

3

You are limiting your snapshot to a 3D picture. If you took a 2D snapshot, it would be impossible to tell how deep your tennis "balls" are (in addition to being unable to tell their motion). So, take a 4D "snapshot", and all'll be fine.

7

Cylinders Don't Exist If I show you a picture of two round objects and tell you that one is a sphere and the other is a cylinder you are looking at head-on, how can you tell whether I am telling the truth or lying? You can't, and therefore, I conclude that there is no difference between spheres and cylinders, because we lack the proper evidence for their ...

2

If we could take a snapshot of both tennis balls, would there be any evidence that could suggest that one is moving and the other one is still? Is there anything happening, at the atomic level or bigger, being responsible for the motion? If the balls are truly identical and you are at rest with respect to one of them, the light of the one moving will look ...

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According to classical physics: no. It is impossible to tell how fast something is moving from a snapshot. According to special relativity: yes. If we choose a frame of reference where one of the balls is at rest then only that ball will look normal. The other ball is moving in this frame so it will be length contracted. If its rest length is $L$ then its ...

1

The photos would look identical, but you would have to take each photo from a different inertial frame of reference. You have to be moving in a different speed in a different direction to take the photo. This shows that there is inherent differences between objects in motion.

2

Your question assumes one ball is moving and the other is still. That assumption is meaningless without specifying a frame of reference. All motion is relative. To each of the balls it would appear that the other was moving. The 'evidence' that they are moving includes the fact that they would appear smaller to each other, and that their separation was ...

26

If we could take a snapshot of both tennis balls, would there be any evidence that could suggest that one is moving and the other one is still? We can't. Problem solved. Well, almost problem solved. So in reality, we can take shorter and shorter exposures. I can take a 1 second exposure of the scene, where the moving tennis ball will be heavily blurred ...

10

It is about the frame of reference, in the frame of reference of the tennis ball pushed by the astronaut, it could be considered as standing still and the other ball, the astronaut, and everything else as moving. For the frame of reference of the other ball it could be considered as standing still, and the first ball as moving. If you were with either one, ...

1

If you look at the link you gave, the combined higgs mass is given as 125.18 +/-16. Differs from just Atlas. They have combined all possible seen channels and experiments to get at this accuracy. To see part of the complexity have a look at this talk at CERN. It is not simple.

0

This question and the answers really highlight the fundamental point that existing models don't really make a lot of sense for understanding the nucleus. To get a sense of things, you need to start looking at nucleons and electro-magnetic entities in close loop form. deBroglie did a lot of work on this in the 1920's and main stream physics has really missed ...

1

Does gravity affect a Fusion reactors ability to sustain a reaction? If so has there been any experiments done or planed for in space? This was one of the very first considerations noted in toroidal experiments. I seem to recall Sakharov'searly paper mentioned it. The gradient in pressure caused by gravity, small though it is in a plasma, causes a tiny ...

3

The limitations on SNO were not fundamentally technological, they were driven by simple energy concerns. The neutrinos generated by solar fusions events are low energy (just a few MeV), and that simply isn't enough energy to create heavy charged leptons (you need more than 100 MeV to create a muon and more for a tauon). That is why SNO could (and did) ...

8

It has been done: D. Bouwmeester, J.-W. Pan, M. Daniell, H. Weinfurter and A. Zeilinger, Observation of three-photon Greenberger–Horne–Zeilinger entanglement, Phys. Rev. Lett. 82 (7), 1345–1349 (1999).

2

Is the above temperature calculations are correct or not? In some circumstances yes. In others no. The explanation follows. When in my younger working days I was conducting temperature tests on electrical products to make sure the components and materials used in the products did not exceed their temperature ratings, we had to deal with the issue of using ...

3

There are really two parts to this. The first is philosophical. The concern about the validity of using instruments to make measurements is an extension of the debate between realism and insturmentalism. That is quite a large topic, subject to its own SEP article. The actual question, however, is one for metrology. Metrology is the study of measurement, ...

0

I found two nice examples: The Particle Data Group provides a nice collection of various measurements of the total collision cross-section of $p\pi^\mp$. The $\Delta^0$ and $\Delta^{++}$ peaks are in good agreement with the expected ratio of $1/3$. The other example is the decay of the $K^+_1(1270)$ into $K^+ \omega$ and $K^+ \rho^0$. Again, the expected ...

0

It should be $\%=\frac{|x_t-x_e|}{\frac{1}{2}\left(x_t+x_e\right)}$. You dont have to worry about negative values as ${|x_t-x_e|}$ will take care of all possible values + or - Just plug the values as they are and the difference will be correct.

0

I realise this is an old question, but this is something that most newcomers to fibre optics struggle with, so I thought some more practical tips might be useful to others stumbling across this question. Coupling into a fibre is something that becomes a lot easier with experience/practice. The first couple of times it can be hugely time consuming. Here's a ...

0

There is plenty of evidence from nuclear structure. E1 transitions with $\Delta T=0$ are forbidden in $N=Z$ nuclei. We observe isospin multiplets of energy levels, with differences in energy that are small and can be explained from the Coulomb interaction. Direct tests have been made using nucleon-nucleon scattering: https://arxiv.org/abs/nucl-th/0011057

2

An even simpler variant of Ben's answer. Use the uncertainty principle. $\Delta p\ \Delta x \geq \hbar/2$. If a nucleon is trapped in a box of order $\Delta x \sim 1.2\times 10^{-15} A^{1/3}$ m, where $A$ is the atomic mass, then its momentum must be of order $10^{15} A^{-1/3}\hbar$. Inserting the mass of a nucleon gives a speed of order $0.2 A^{-1/3}c$. [...

0

Consider J/ψ ⟶ Λ $\bar Σ^0$ + cc , where the J/ψ and Λ are isosinglets and $Σ^0$ is centrally inside an isotriplet. It is squarely in the PDG listings, $\Gamma_{195}$, 28 parts per million. No external photons are involved, so a bona fide strong isospin violation. As suggesteded in the comments, from G-violation in 2nd-class weak decays, it is well-...

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There are a variety of ways of getting this kind of information. The simplest way to get a rough estimate is just to use the size of the nucleus and the de Broglie relation. A moderate-sized nucleus has a diameter of roughly 7 fm, so the lowest-energy standing wave would have a wavelength of 14 fm along any given axis $x$. Plugging in to the de Broglie ...

32

If you shoot an electron or a proton at a nucleus at moderate energies (a few hundred $\mathrm{MeV}$ to a few $\mathrm{GeV}$) it will usually either bounce off the whole nucleus or break up the nucleus. But every once in a while (and this gets rarer and rarer the harder you throw it in) it will actually bounce off of a single nucleon. At the right energies ...

1

Problem Statement Suppose that you have a set of $N$ values from measured data $(y_j,x_j)$ that are known from first principles to follow the function $y = mx$. The points in the data set may each have their own uncertainty values (errors) $\delta x_j$ and $\delta y_j$ You have two approaches to obtain $m$ from this data set. Method 1 - Point-by-Point ...

0

You have quoted that photomultiplier tubes have a higher gain and lower dark current than avalanche photodiodes and considered that noise and signal are both amplified by gain, but what is important is the lower dark current of photomultiplier tubes compared with avalanche photodiodes. This means that there is less inherent noise associated with ...

1

Have you heard of Monte Carlo calculations? This talk may enlighten you on how it is used in high energy physics experiments. In this Higgs discovery plot Note in the fine detail the various lines beneath the Higgs. The are Monet Carlo calculations which use the Standard model of particle physics and various assumtions given in the caption, which have in ...

0

Normally, a linear regression assumes no uncertainty in the independent variable (e.g., x). If this assumption is valid and the relationship is actually linear, a linear regression drives a line through the data such that the errors in the dependent variable normally distribute themselves around the regression line. From a statistical point of view, this ...

0

One possible reason is the time it takes the particle to dump energy in the scintillator. This will affect the rise time of the pulse. The neutrons must scatter off a nucleus to be detected while a gamma-ray interacts with the atoms in the scintillator which are much larger than nuclei of course.

0

A theory of everything will always have an explanation for why the observed world fits into it's explanation. So cries of "it's not falsifiable!" don't really add anything to the discourse. I suspect that this might be what you're thinking: A theory-of-everything would correctly explain everything, and therefore always be true. If it's always true, it can'...

0

So how do we do science on theories of everything? You mean that research into physics will stop if we have a theory of everything? Think a bit. As far as solid state physics goes, there exists a theory of everything: electromagnetic and quantum theory. Solid state research has been a vibrant branch of research and continues to be, even though the theory ...

1

You are wrong about this, and about what a theory of everything (TOE) actually is. First of all, as it is currently used in the physics community, "TOE" means a single mathematical framework that merges general relativity and quantum mechanics. How do you "do science" with TOE's? Here are the three initial steps in the process: 1) For a mathematical model ...

1

A 'theory of everything' is not what you think it is. It's not trying to somehow handwavy-highschooly explain everything, but unifiying all known forces of nature, to predict how the universe should look according to that theory. A 'theory of everything' is still a scientific construct that starts with axioms, held to be true, and derives further statements ...

3

At present we are at the following level of mathematically modeling interactions of matter: 1) the underlying framework for all matter is quantum mechanical, all other theories will be emergent ( gravity has a question mark as it is only effectively quantized). This framework is evident in the microworld of particle physics and is important as long as ...

0

an adiabatic wall is one which prohibits heat transfer. A 1" thick wall of styrofoam is an adequate approximation for an elementary lab experiment.

1

I have changed the cable (connecting between signal generator and the laser driver) from to Now the output is expected, and does not depend on orientation orposition of the photodiode.

2

The response of a platinum resistance thermometer (PRT) is better approximated with the so-called Callendar-Van Dusen equation (this is defined in the standard IEC EN 60751): $R(t) = \begin{cases} R_0\left[1+At+Bt^2+C(t-100\,{}^\circ\mathrm{C})t^3\right]& t<0 \\[5mm] R_0(1+At+Bt^2)& t\geqslant 0 \end{cases}$ where $t$ is the Celsius ...

2

I was wondering if they are suitable for a spinthariscope as well. The long and short of it is $\text{NO}$. The 'glowing' of certain $\text{U}$ salts (mainly diuranates) is simply (chemical) fluorescence and has nothing to do with $\text{U}$'s feeble radioactivity, which is in any case many orders of magnitude lower than that of $^{241}\text{Am}$.

1

There is probably an easier way to do this. Use a cylindrical plastic bucket as a reservoir. Put a hole in the bucket, and run a short piece of tubing from the bucket to a small diameter glass tube (below the bucket) which will be your Poiseuille device. Use another piece of tubing connected to a faucet to run water into the bucket, and adjust the flow ...

0

Thanks all for kind replies to my question.Regarding the problem carefully, I convinced myself that the anomaly in the Young interference pattern could be due to a return of marginal rays back from the double slit (in glass) to the detector.To verify such assumption I planned to change my setup with a plastic not reflecting double slit end a better quality (...

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