# Tag Info

15

Your worry is not necessary. In the usual experiment the detector measures the distribution of the time between the muon stopping in the detector and the time of it's decay. Then an exponential curve is fit to the data and the lifetime taken from the fit parameters Muon decay is a processes analogous to radioactive decay, and (like all exponential ...

10

The general idea. Let's restrict the discussion to matrix Lie Groups for simplicity. Determining the generators of a given Lie group $G$ simply means (by definition) determining a basis for its Lie algebra $\mathfrak g$. Here's a standard method for finding such a basis: Recall that the Lie algebra $\mathfrak g$ of a matrix Lie group $G$ is defined as ...

10

On the actual Hilbert space of a consistent relativistic quantum mechanical system, the Lorentz transformations including boosts actually are unitary – which also means that the generators $J_{0i}$ are as Hermitian as the generators of rotations $J_{ij}$. We say that the Hilbert space forms a unitary representation of the Lorentz group. What the OP must be ...

8

Newton's third law is really a special case of the conservation of momentum. Suppose you have two rigid bodies with momenta $\mathbf{p}_1$ and $\mathbf{p}_2$. If they only interact with each other, then $\mathbf{p}_1 + \mathbf{p}_2$ is constant, since total momentum is conserved. Differentiating this gives $\frac{d\mathbf{p}_1}{dt} + \frac{d\mathbf{p}_2}{dt} ... 7 In the rest frame of the atom there are of course no changes, so I assume you're asking what the atom will look like to the stationary observer watching the moving atom. First note that electrons don't orbit the atom like planets orbiting a star. The electrons in atoms exist as a delocalised probability distribution. This distribution can have a non-zero ... 7 http://en.wikipedia.org/wiki/Light_cone It simply says that some parts of the space-time are not accessible to us. For example I assume :-) you are on (Earth, Now). No matter what you do (Moon, Now) is not accessible to you. (Moon, Now + 1 second) is also not accessible to you, because the Moon is 1.28 light seconds away from Earth. Some events from the ... 7 From special relativity we know that a Lorentz transformation: $$x'^\mu = \Lambda^\mu {}_\nu x^\nu$$ preserves the distance: $$g^{\mu \nu} \Delta x_\mu \Delta x_\nu = g^{\mu \nu} \Delta x_\mu' \Delta x_\nu'$$ The above two equations imply: g^{\mu \nu} = g^{\rho \sigma}\Lambda_\rho ... 7 Special relativity is used in the SM formulation. It is kinematics, so somehow more basic than interactions between bodies. A QFT derivation of General Relativity has been the Holy Grail of the field for many years. In the early times, Feynman, Dirac, and the others tackled this problem, but after decades of failures it was more or less considered ... 6 Your calculation is wrong because$E=mc^2$doesn't mean that the object has velocity$c$. To make my answer useful i will give a very brief overview of Dynamics at higher velocities which is a consequence of Special theory of relativity. At higher speeds(of order of$c\$) Newtonian mechanics is not valid. The linear momentum is defined as: $$\vec p=m_0\gamma ... 6 The mass m in the formula is NOT the rest mass m_0 and therefore dependent on the velocity:$$ m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} \equiv \gamma m_0 $$This means you cannot simply take the normal mass of a nitrogen atom and put it in there, if you assume a speed v \neq 0. The c in the formula doesn't mean that the particle travels at light ... 4 The situation you described gives evidence that, if we really believe nothing can travel faster than the speed of light, the usual view of an electron actually orbiting the nucleus (like a planet orbits a star) can't be correct. The model of the atom you described, with electrons moving around the nucleus, also has other problems. Physicists noticed these ... 4 The way to do problems like this is always to use the Lorentz transformations. Choose some sensible spacetime points in the rest frame S and use the transformations to see what those points look like in the moving frame S'. In this case this is what the points look like in S: The spacetime points are labelled as (t, x, y) - we'll ignore z since ... 4 However, I feel like having the denominator c+v is wrong, as in some sense I'm adding velocity v to the speed of light which is not allowed in SR. It is permissible, in SR, to have 'non-physical speeds' in excess of c. For example, you observe two trains speeding towards on another. You measure the speed of each train to be say, 0.9c relative to ... 4 As always, a spacetime diagram is crucial for insight. However, this can be dispensed with without one. What am I missing here? The event that Moe's wristwatch reads 1\mu s has coordinates, in Joe's frame, of (2.29\mu s, 618m). In other words, according to Joe, Moe's wristwatch reads 1\mu s when Joe's clock reads 2.29\mu s. According to Joe, ... 4 John has already provided a correct answer containing pretty much all the necessary information, but I'll just try to show the resolution of the paradox and visually point out where things go wrong for Bob. So let's draw a spacetime diagram of the situation. For simplicity we set c = 1 in what follows. The path Alice takes is simply a vertical line in ... 4 You can't calculate Bob's proper time using the Lorentz transformations because Bob does not travel in an inertial frame. The frame he sets out in is not the same as the frame he returns in because it's travelling in a different direction. You say Bob instantly turns around and travels back to Alice at −v, and what this means is that at the halfway stage Bob ... 4 Every Lie group has an adjoint representation. I'm not sure what definition you come at the adjoint representation from, but here's the fundamental one which I'm sure you'll see is always meaningful. Think of a C^1 path \sigma:[-1,1]\to\mathfrak{G} through the identity in a Lie group \mathfrak{G} with \sigma(0) =\mathrm{id} and with tangent X ... 4 Preliminary remarks. As Danu writes in his comment, the physics of the other four generators has to do with spacetime translations, one for each spatial direction, and one for time. But how do we see this explicitly in the math behind the somewhat odd-looking presentation of the Poincare group and its Lie algebra that Hall discusses. First, recall that ... 4 Isn't dt something that cannot change? In Special Relativity, time t is a coordinate rather than a (universal) parameter. To locate an event in spacetime in a particular reference frame, one must specify 4 coordinates, 3 spatial and 1 temporal. So, a quantity like \frac{dx}{dt} is a coordinate velocity; it is the rate of change of one coordinate ... 4 Causality is preserved, unless Tachyons exist. Part 1: STR doesn't assume causality. Causality is violated when you have a flow of information that goes back to the same place in space AND time, creating a contradiction. Both newtonian and STR guarantee causality. STR is more complex, but it still prevents anything from going back in time with respect to ... 4 First and foremost, draw the spacetime diagram. In the reference frame of the flash bulb, the other two clocks are always synchronized since those clocks have, at all times, the same speed as each other. If you trace the wordline of the two accelerated clocks (in the frame of the flash-bulb), you'll find they are congruent and so, the proper time along the ... 4 The problem with this sort of scheme is that Alice has no control over the results of her measurements, since those are random. This means that she can control which basis Bob's spin is projected on, but she cannot control which of the basis states gets chosen. Bob will then see a random mix of results which turns out to contain no trace of what Alice was ... 4 No, you do not want representations of the diffeomorphism group for the same reason that you do not want representations of the gauged Lie group in Yang-Mills. The diffeomorphisms are a gauge symmetry, not a real symmetry of the theory. Gauge transformations act trivially on physical states, they map one redundant description of a state onto another. They ... 3 Quantum field theory is a general framework. There are many different kinds of field theories, the most well-known of which are QED (quantum electrodynamics) and QCD (quantum chromodynamics). QED has been shown to agree incredibly well with experiment. The anomalous magnetic dipole moment of an electron, as computed using QED, agrees with experiment to 10 ... 3 In many contexts, we would like to determine how Lorentz transformations act on the mathematical objects that characterize a particular theory. In the case of classical, Lorentz-invariant field theories on Minkowski space for example, we need to specify how Lorentz transformations act on the fields of the theory. This leads naturally to determining how ... 3 It's the same way you know there are three parameters in SO(3). The equation \Lambda^T \eta \, \Lambda = \eta has (n^2+n)/2 independent scalar equations. To see this, write the equation in component form: \Lambda^{\mu\nu} \Lambda_\mu{}^\rho = \eta^{\nu\rho}. Now we see there are n^2 scalar equations equations, but because \eta is symmetric and ... 3 If I understand you correctly, your two points about apparent slowness of speeds is related to scale, and disappears when you quantify it using a common unit. ie: We think of 10m/s as relatively slow because the average human is 1.8 metres in height, and we can imagine that 10 metres per second, or 36 kilometer/hour as an achievable speed using a machine ... 3 An approach alternative to that discussed by David Bar Moshe is to start from a different coordinate system in the Rindler wedge W_R:$$ds^2 = e^{2y}(−g^2dt^2+dy^2)$$here t, y \in \mathbb R. The relation with the standard spatial coordinate in W_R is x=e^y, where x>0 is related with the alternate form of the (same) metric:$$ds^2 = -g^2 x^2 ...

3

Essentially by definition (due to Wigner), one-particle Hilbert spaces of elementary particles support unitary strongly continuous irreducible representations of Poincaré group. Conversely, any multi-particle Hilbert space, with either fixed or undefined number of particles either identical or distinguishable, cannot be irreducible under the action of the ...

2

No, the Lorentz transformations do not include the travel time of the photon from the transformed event to an observer. The Lorentz transformations do indeed derive from a geometrical property, but the property in question is the invariance of the spacetime interval: $$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$ I don't really understand your diagram, but it ...

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