# Tag Info

2

There is absolutely a gravitational radiation reaction and solving for it is one of the very active fields in classical relativity theory at present. Basically, particles with nontrivial masses distort the spacetime around them; this causes them to not move on geodesics of the "background" spacetime (the spacetime that one would have found had the secondary ...

1

It depends what you mean by controlled. Without going into the details of each decay process you mention ($\alpha, \beta, \gamma$), the decay of an unstable nucleus is inherently random. At a quantum mechanical level, we can think of the system changing from one eigen-state to another and that is fundamentally unpredictable (this is the bit that Einstein ...

1

The color of an object says something about its emissivity/absorbance for visible radiation. But it has virtually no correlation to the behavior at IR wavelengths. A white object may reflect significantly more visible light than a black one, but may reflect identical amounts in IR. Since your roof will not be cooling via visible radiation, the color ...

1

I'm fairly certain that the energy difference between the two is negligible. Any heat as a result of light absorbed is localized on the roof, with layers of insulation between the inside of your house and the additional heat generated. Most of the additional energy absorbed would heat up the air rather than the inside of your house. Additionally, the optical ...

1

$p=\frac{1}{3}\rho$ is the well-known equation of state of a photon gas. It may be derived by looking at the ultra-relativistic limit of the energy momentum tensor for a bunch of particles.$^1$ $p=-\rho$ follows from the fact that the energy momentum tensor of $\Lambda$-style dark energy is proportional to the metric. Thus, at a point and in the proper ...

1

If we add or remove an election to the atoms of radioactive metals, they will become the isotopes of their adjacent chemical elements in the periodic table. You don't change isotopes by adding or removing electrons. That is called ionization, except that you really can't ADD electrons. You change isotopes by adding or removing protons or neutrons. The ...

2

There is no reason why you can't measure the rate frequently. However, in order to estimate the half life, you need to see a change in the rate of decay. How long you need to measure for, and how far apart you need to change your measurements, depends on the number of decays per second that you observe as well as the required accuracy. For example, if you ...

0

you should resolve the differential equation of N. If you do that you'll get that $$A = A_0 e^{\lambda (t-t_0)},$$ where $A_0$ is the activity at time $t_0$. From there you can obtain the value of $\lambda$ from two measurements of the activity whatever the time interval.

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Firstly, the activity formula is in fact: $$-\frac{dN}{dt}=A=λN,$$ because $\frac{dN}{dt}<0$. [...] is there any particular reason why our time interval for measuring the number of remaining Radionuclides should be close to the half-life of the substance? No and that's not how it's done in practice. $\lambda$ and the half-life are determined by ...

2

I've heard that the military has satellites orbiting earth with gamma detectors installed in them. The phenomenon of Gamma Ray Bursts was discovered when these satellites kept detecting gamma rays when no nuclear weapons were going off. See here: https://en.wikipedia.org/wiki/Vela_(satellite)

2

1) The relation $\frac{dF}{dS}=d\left(\frac{F}{S}\right)$ is certainly incorrect as @Floris has mentioned in the comment. As the simplest counter-example, consider a linear function, $F(S) = \alpha \, S$, with $\alpha \neq 0$ as a proportionality constant. Then, one could easily see that  \frac{dF}{dS}= \alpha \neq 0 = d\left(\frac{F}{S}\right) = ...

0

I've heard a tragic misconception that blackbody radiation is somehow related to the apparent color of an object. The uninformed reader might read these answers and get the idea that if you paint an object black, no infrared light will come off of it. The key here is that you may reduce the amount reflected - the object would be very cold indeed if it ...

-9

Why does a damped quantum harmonic oscillator have the same decay rate as the equivalent classical system? Because the latter is a complex macroscopic ensemble of the former. I presume you've read the answers to this question because you commented. There's some good stuff there. If S is excited, say into |1>, then coupling between S and the ...

1

I think most of you are confusing what contracts because of motion. Not the distance between the muon and the Earth, nor the distance between the point in which the muon began its motion and its destination. Length contraction deals with the contraction "of the moving object" (that is the muon's length, if we could talk of it). If you could see an airplane ...

2

Ionizing radiation is radiation that is strong enough so that, when it hits an atom or molecule, will knock off electrons. This happens even if the target object doesn't have freely mobile electrons, which leaves free radicals and broken bonds, both of which are harmful to complex biological processes. There's no selection based on electron binding energy; ...

1

Can a single particle be “heated” by radiation? Not really, because heat is an emergent macroscopic property of an ensemble of particles. But since you put the word "heated" in quotes, we can allow a yes of sorts. Take a look at the Wikipedia temperature page, where we can see this picture: CCASA image by Greg L, see Wikipedia It's to do with the ...

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A single particle can be a 'system' within itself having modes depending on the particle ' s structure. These modes may be in 'tune' with the incident radiation and thus capture the energy which can increase the particle ' s momentum and therefore its velocity. We never say the particle's temperature has increased but rather it's momentum. When a system of ...

0

Phase and random kinetic energy are macro. If you don't call the kinetic energy of a single molecule temperature then fine but it is still kinetic energy. In a single molecule you have rotation and vibration that can be effected by radiation. You can have an exited state where an electron is temporarily bumped to a higher orbit. Some reactions only take ...

3

Anna and Floris have given excellent descriptions of how the photon emission depends on the orientation of the nucleus, however there is another aspect of the question that I think is worth mentioning. If you conside an isolated nucleus, e.g. no magnetic field, then it will be in a suerposition of all possible orientations so overall the system is ...

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There are examples of nuclei whose emission is not isotropic in the presence of a magnetic field. Feynman gave an example of this during a lecture on symmetry in physical laws (vol 1-52). In particular in 52-7 he mentions an experiment in which the emission of an electron by a cobalt nucleus (Co-60) is asymmetrical with respect to the magnetic axis - more ...

13

1) To define an angle for a nuclear reaction one has to have an orientation for the nucleus. This could be a magnetic moment or a dipole moment. 2) a photon carries away spin 1, i.e. angular momentum, and will leave a nucleus minus that angular momentum. As the word states, angles are involved, and the probability distribution for the gamma ray will in ...

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