New answers tagged

1

I have never heard about an even-vs-odd rule for significant figures. It doesn't sound correct. Rather, simply round up the last digit that is to be shown if the next digit is in the upper half $(5,6,7,8,9)$. That's all. To reach two significant figures, you therefore round up the second digit only if the third digit is $5,6,7,8,9$. Note that only the next ...


0

The standard definition of the Killing form is, as a matter of fact (cf. ref 1) $$ \kappa(X,Y)=\frac{1}{2h^\vee}\operatorname{tr}(\operatorname{ad}X\operatorname{ad}Y)\tag{13.13} $$ which includes an explicit factor of $h^\vee$. (This definition is convenient because it makes $\kappa$ independent of conventions for how traces are normalized. The factor of 2 ...


2

It shifts the energy. In fact, it can be moved right away into the energy part of the equation. And just like in classical mechanics it has no significance. A bit more advanced view: in some cases this term will originate from the gap, e.g., when reducing the Dirac equation to the non-relativistic limit or when dealing with multiband crystals. However, this ...


9

It has no significance also in quantum mechanics. The solutions $\psi'$ of the Schrödinger equation with the new potential — viewed as an eigenvalue/eigenvector equation — are exactly the solutions $\psi$ of the Schrödinger equation with the old potential (obviously replacing $E$ for $E+k$). They are simply multiplied by a phase depending on $k$ and $t$ if ...


7

I don't think so. If $\gamma^\mu$ is a set of gamma matrices obeying the clifford algebra, then so is $S^{-1}\gamma^\mu S$ for any invertible $S$. But unless $S$ is unitary, $\gamma^\mu$ being hermitian does not imply that $S^{-1}\gamma^\mu S$ hermitian.


4

The answer is very simple: if $\vec{F}$ is a conservative vector field, then $\vec{G} = -\vec{F}$ is also a conservative vector field. So mathematically, the choice to include the minus sign or not doesn't change anything. But physically, the minus sign is important because it lets the physical quantities match our physical intuition. When we write $\vec{F} =...


0

Yes, in e.g. eq. (1) by convention all 4-momentum vectors are directed as incoming and the zero-component is positive (negative) if the particle is incoming (outgoing), respectively.$^1$ References: M. Srednicki, QFT, 2007; Chapter 16. A prepublication draft PDF file is available here. -- $^1$ Or vice-versa, depending on the author's conventions.


2

The reason for this is the clifford algebra, which the gamma matrices are required to obey: $$\{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu\nu}I_{4}$$ If you choose the $(+, -,-,-)$ signature, then $\gamma^{0}$ can be diagonal and real, and you can choose the standard "the blocks of the matrices are pauli spin matrices" representation for the $\gamma^{...


1

From the question it looks as though you are asking about the difference between adopting metric signature $(-1,+1,+1,+1)$ verses $(+1,-1,-1,-1)$. I found it more natural to do general relativity with $(-1,1,1,1)$. But there is now so much particle physics and quantum field theory that adopts $(+1,-1,-1,-1)$ that it is a pain to avoid that choice, because if ...


4

You are missing the physical information carried by the corresponding 4-vectors. The two versions just differ by "what remains positive". In QFT we often deal with collisions described in momentum space, where the zeroth component carries the interpretation of energy if you compute the norm of a four-momentum vector of a particle at rest, $$p=(E,\...


1

In the first reference you are quoting the chiral basis with $$\gamma^0=\begin{pmatrix} 0 & \mathbf{1}_{2\times 2} \\ \mathbf{1}_{2\times 2} & 0 \end{pmatrix} $$ is used while the second reference uses the (standard) Dirac basis $$\gamma^0=\begin{pmatrix} \mathbf{1}_{2\times 2} & 0 \\ 0 & -\mathbf{1}_{2\times 2} \end{pmatrix} $$ see for ...


0

This is almost correct, you have just made a mistake in the beginning in the index summation : $$\begin{align}1-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu} &= 1-\frac{i}{2}\omega_{0\nu}S^{0\nu} \color{red}{-} \frac{i}{2}\omega_{i\nu}S^{i\nu}=1 \color{red}{-} \frac{1}{2}\beta_i\begin{pmatrix}\sigma^i & 0 \\ 0 & -\sigma^i\end{pmatrix} + \frac{1}{2}\theta^...


1

The Einstein summation convention posits that if an index label appears both in an "up" and a "down" position, it should be summed over. This part of the convention is more or less unambiguous. It does happen sometimes that you repeat indices and you do not want to sum, but then you should put a note next to your equations. Now to the ...


0

The key point is that in practice we are only ever interested in the difference in potential energy between one position and another. Indeed, we can never measure absolute potential energy, only its difference between two points; or rate of change in a given direction etc. When calculating this difference, the constant of integration cancels out. So we can ...


0

In most cases, the potential implicitly uses a constant of integration such that the potential at infinity is $0$. That is where you get the formula $V = kq/r$ due to a point charge, which you utilize in your example; in general, it would be $V = kq/r + C$.


-1

The integration constant should always be supplied when performing an indefinite integral. This is because the purpose of an indefinite integral is to find the function whose derivative is the integrand. That is, if we are looking to compute $\int f(x)dx$ as an indefinite integral, this is the instruction to find any function $g(x)$ such that $\frac{d g}{dx}=...


2

For both signatures, the terms in the EFE (except the cosmological constant) have the same sign (see the answers here for why Einstein GR and metric signature). So without more information, no, it is not possible to tell just based on the EFE. The reason for different signs in the EFE is based on other conventions, e.g. for defining the Riemann tensor and ...


0

The distinction between degrees Celsius (or Fahrenheit) and Celsius degrees is nuanced but not unimportant. Consider a temperature specification of 900C +/- 14C. Were this spec converted to Fahrenheit, the correct interpretation of +/- 14C would be +/- 25F, not +/- 57F, even though 14C = 57F.


2

The infinitesimal generators for $\mathrm{SO}(3)$ are a basis for the Lie algebra $\mathfrak{so}(3)$, which is the vector space of $3\times 3$ antisymmetric matrices with real entries. As with any vector space, this basis is not unique - any linearly independent spanning set will do. $(\mathcal K_l)_{mn}=\epsilon_{lmn}$ is obviously a valid choice for such ...


0

Since no one has indicated an error in my mathematical argument, I will assume that part is correct. The question thus remains: is this a mistake in MTW, or is it simply adherence to a nonstandard parity? We might ask Kip Thorne. Since the other parts of Gravitation adhere to the conventions found in, for example Rotations in 3D, so(3), and su(2). version ...


1

you have been provided with theory also you can try this applet


0

Think about how you might measure the speed of a water wave. You might well follow a crest and measure how far the crest travels $\Delta x$ in a time interval $\Delta t$. From those readings you can obtain the speed of the wave $\frac {\Delta x}{\Delta t}$. For the function that you have chosen a crest means that $\omega t\pm kx = \frac {\pi}{2}+2n\pi $ ...


2

Here is what you should do: Select any fixed phase, e.g. $0 = \phi = \omega t \pm kx$, Determine the position $x$ of this phase for $t=0$, Increase the time by a "small number" $t=dt$ and check at which position $x$ the selected phase has moved. Hence, in which direction does the wave move? This can be done either using the formulas or by ...


1

The Schwinger parameter itself is manifestly positive. In particular, it is not Wick-rotated, so there are not different versions of it. Rather it is OP's $A$ operator that is Wick-rotated. OP lists a few references in above comments. Ref. 1 works in Euclidean signature, so it's well-defined. Ref. 2 & 3 only use the Schwinger parameter to derive the ...


0

As with most things in physics, the answers to your questions depend on your considerations in the system. Generally, potential energy is related to the forces that an object is under influence by. If the object has no forces acting on it, you can't really say it possesses any potential energy. The potential energy indicates the ability for the force to do ...


1

No. Only systems of objects which interact with forces which do path-independent work have potential energy. If your system is simply a book it doesn't have gravitational potential energy (taking a guess at what you're thinking.) If your system is the book and the earth, that system has gravitational potential energy. Individual objects don't have potential ...


0

What you're referring to is "gravitational" potential energy.. which is one of many forms of potential energy (gravitational, electrical, chemical, nuclear, etc.) The idea behind potential energy is that it is energy that hasn't been yet released in an "obvious" form like movement, heat or radiation. Think of it like money in the bank.. ...


0

Yes, chemists like J couplings in Hz. The NMR J or scalar coupling and the Heisenberg model occur in very different contexts. In NMR, the scalar coupling is typically more than 1000 times smaller than the Zeeman term in the Hamiltonian. Moreover, the temperature is more than 1000 times larger than the Zeeman term. The NMR J coupling is a residual magnetic ...


0

As others have pointed out, the magnitude of the force, $ |F| $ is: $$ G\frac{m_1 m_2}{r^2} $$ The sign is just related to the vector direction and your choice of axis. In your picture, you can choose a coordinate axis denoted by $\widehat{r} $ and, let's say, $ \widehat{s} $ in the following way (please, ignore the small angles between your force lines and ...


1

This is just according to different definitions of $\hat r$. In the first pair of formulas $\hat r$ is the unit vector from one object to the other, so an attractive force is positive. In the second pair of formulas $\hat r$ is the spherical coordinate basis vector in the (outward) radial direction, so an attractive force is negative. There is nothing more ...


1

The decimal system is conveniant in conversion of magnitudes. In a positional system it is easy to multiply and divide in factors of ten and ten itself is easy enough to conceptualise. This fact is used in the SI system where there is a system of prefixes that indicate these scale factors. For example, $k$ for kilo and which means multiplying by a factor of ...


3

This is a classic misuse of the Einstein summation convention. The correct expression in your final equation is $$ \gamma^{\mu} \partial_{\mu} = \sum_{\mu = 0}^3 \gamma^{\mu} \partial_{\mu} = \gamma^0 \partial_0 + \vec{\gamma} \cdot \vec{\nabla} $$ The reason for your confusion is because you're used to taking the Minkowski inner product between two ...


4

As a warning, there are no absolute rules about notation, and you should always consult the particular source you are reading and make sure you understand their conventions. If their conventions are not spelled out clearly, you should find another source. Having said that, generally you can expect: The 3-dimensional (scalar) volume element is $dV = d^3 \...


0

If this is an exercise for path integration then it's Mathematics. But as Physics problem it's important to note first that the given electric field is irrotational \begin{equation} \boldsymbol{\nabla\times}\mathbf{E}\boldsymbol{=}\left(\dfrac{\partial E_y}{\partial x}\boldsymbol{-}\dfrac{\partial E_x}{\partial y}\right)\mathbf{k} \boldsymbol{=0} \tag{1}\...


1

My guess would be that originally the degree was the nightly shift in the position of the stars in the sky, but as with the original defnition of the meter, they didn't get it quite right. Or, mabe someone later decided to round it so that number of degrees in a circle could be factored.


1

Apart from the temperature, the units measured in degrees are referring to something periodic (for example, the phase of oscillations), which can then be reduced to angle, and consequently measured in degrees ($1/360$ of the full circle), grads ($1/400$ of the full circle) or radians. Angles are obviously dimensionless.


1

In some ways this is not a question about Physics rather one needs to look at the etymology of the word degree. Latin *degradus "a step", and so the notion of "one of a number of subdivisions of something". As to the angle measure it is a legacy of counting to base $60$, sexagesimal, by the ancient Sumerians in the 3rd millennium BC ...


1

As the term "quintal" is inherited from French language, the 'official' symbol would be the one given by the French 'Imprimerie Nationale' in its book about typographic rules and conventions (available here, beware it's quite a big book and it's all in French obviously...). Anyway, that symbol is q for the quintal. Nevertheless, as the quintal is ...


0

I don't know why nobody pointed about this, but a silvered plano-convex lens acts as a concave mirror. Silvering a lens makes it a mirror, then it depends which lens is mirrored to find what kind of mirror is formed. Also, Focal length is half the radius of curvature of a sphere whose mirror is made, so whichever side the rays are meeting (actually or ...


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