# Tag Info

7

There is a recoil when each photon leaves, but they radiate in all directions at once, as ACuriousMind intimates in his comment, so there is no collimated beam to concentrate the recoil. Even if the total recoil were concentrated, its effect is so small that the screw-in base of the bulb is more than sufficient to hold the bulb steady. Theoretically, it ...

4

No. There are loads of conserved quantities in decay processes like the one you are talking about. Lepton number conservation and charge conservation are the most notable ones for the case of a positron. Also, the sum of the masses of decay products should always be lesser than the mass of the initial particle. (This naturally follows from energy ...

3

Provided the bat delivers exactly horizontal momentum impulsively to the second ball, it will not travel as far due to its initial downward velocity, as you say. Dissipating the downward momentum doesn't make much sense in the scenario you described.

3

Even if it did, the energy-to-momentum ratio of light is $c$, a fantastically huge number. For example, even if you house the light bulb in one big perfect mirror with a tiny collimated (all in one-direction) aperture that all of the light goes out of, and even if the light bulb is a perfectly efficient radiator which is using 100W of power, the ...

3

Given you've read only QED, this is a highly astute question. Conservation laws in the quantum world work a little differently from classical conservation grounded on Noether's theorem (there is a kind of quantum analogue in the Ward-Takahashi identity). If a quantum entity has state $|\psi\rangle$, then conserved quantities are measurement means defined ...

3

First of all, my opinion is that the paper on your link is full of notational inconsistencies and therefore causes a great amount of confusion for someone who struggles to understand Noether’s theorem. So, allow me to formulate the field-theoretic version of Noether’s theorem in a more, according to my taste, charming way. Preliminaries 1: Lie Groups and ...

3

This is just a property of Fourier transformations. If the correlation function is translational invariant then, by definition, the position space representation $D(x,y)$ transforms as $D(x+a,y+a) = D(x,y)$ for any constant $a$. Thus $D(x,y) = D(x-y,0)$ and so the correlator depends only on the difference $x-y$. For simplicity, we'll define $D(x-y) = ... 1 I'll risk moderatorial opprobium with a partial answer because you have come so close. You correctly use the SUVAT equation$v^2 = u^2 + 2as$to find that the velocity of the ball just before it strikes the ground is$v_i = -7$m/s (using the sign convention that upwards is positive). So far so good. Now you know the ball rises back up to a height of 1.8m, ... 1 The charge associated to the$U(1)$symmetry you mention is called weak hypercharge. The relation with the electric charge is the following$Q= T_3+Y/2$where$T_3$is the third generator of the$U(2)$symmetry representing the weak isospin and$Q$the electric charge. This relation holds for all leptons. Neutrinos have weak isospin$+1/2$and weak ... 1 I guess what is to be noted here is the fact that, the Correlation function (operator), commutes with the momentum operator, since $$[D,\text e^{ixp}] = 0 \implies [D,p] = 0$$ Having that to be the case, one can recollect that any operator represented in its own eigenspace is diagonal should answer your question. PS : I am not completely sure about the ... 1 There is no general algorithm for doing so, and even figuring out how many conserved quantities a system has can be difficult. A famous example is the Toda lattice, a system which was originally proposed by Toda in 1967 and was believed to be chaotic, but was in fact proven to be integrable (to have too many conserved quantities to be chaotic) in 1974 by ... 1 Yes a positron can decay without encountering an electron. But it must encounter another particle because as it is said in another answer, the positron is a stable particle (in the vacuum), so it cannot decay on its own. An example of "decay" not involving an electron: $$e^+ + \mu^- \to \bar{\nu_e} +\nu_{\mu}$$ this decay proceeds via the weak interaction (a ... 1 There is a transient tension$\Delta T\$ in the string that will give rise to a change in momentum of the counterweight, the pan, and the mass on the pan. At the ceiling, the pulley transfers twice this force to the support: The change in momentum of the three components is (with + direction up): $$\Delta p = m v' + m(v-v') - mv' = m(v-v')$$ This change ...

Only top voted, non community-wiki answers of a minimum length are eligible