# Tag Info

4

Perhaps some additional information is in order to shed additional light... The whole discussion begs the question: If $\hbar$ is so convenient, why do we have $h$ around? As usual, "historical reasons". Planck originally invented $h$ as a proportionality constant. The problem he was solving was blackbody radiation, for which the experimental data came ...

7

To quote Stephen Gasciorowicz, Before evaluating these quantities to obtain an idea of their magnitude, we will introduce some notations that will be very useful. First, it is $h/2\pi$ rather than $h$ that appears in most formulas in quantum mechanics. We therefore define $$\hbar=\frac{h}{2\pi}=1.0546\times10^{-34}\,{\rm J\cdot s}$$ So basically it's ...

3

First, I want to say that different people use different notation and I welcome any comments. I also feel as if I am about to enter a minefield. Here the answer is made up with examples of use of $d$, $\partial$ and $\delta$. I would say for $d$ that $dV \over dx$ would be the total derivative in one dimension for $V(x)$ where the potential $V$ is a ...

19

Typically: $\rm d$ denotes the total derivative (sometimes called the exact differential):$$\frac{{\rm d}}{{\rm d}t}f(x,t)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}t}$$This is also sometimes denoted via $$\frac{Df}{Dt},\,D_tf$$ $\partial$ represents the partial derivative (derivative of $f(x,y)$ with respect to $x$ ...

0

In one dimensional motion, sign really does mean direction for vectors. You can say positive is upwards and negative is downwards or vice versa. Now, come to your case. You seem to have mathematically understood. Then start with making your intuition. As stated by other answer, suppose you are very,very far away where gravitational force of the earth ...

0

Here is a simple model as explanation: Imagine yourself far away from any gravitational field - your potential energy is zero. As soon as you are entering into a gravitational field, you are accelerating and winning kinetic energy. The origin of this energy is that you "borrowed" some energy which is the potential energy you are losing. When you want to ...

2

$$\delta^{[\mu_1\mu_2\ldots \mu_n]}_{\nu_1\nu_2\ldots \nu_n}~=~ \frac{1}{n!}\sum_{\pi\in S_n}{\rm sgn(\pi)} \prod_{i=1}^n \delta^{\mu_{\pi(i)}}_{\nu_i}.$$ More generally, $$T^{[\mu_1\mu_2\ldots \mu_n]}~=~ \frac{1}{n!}\sum_{\pi\in S_n}{\rm sgn(\pi)} T^{\mu_{\pi(1)}\mu_{\pi(2)}\ldots \mu_{\pi(n)}}.$$ Here $S_n$ is the symmetric group of permutations, ...

1

$\epsilon_{ab}$ are the components of a tensor, so you can raise and lower indices with the metric: \begin{align} \epsilon^a{}_b & = g^{ac} \epsilon_{cb} \\ \epsilon_b{}^a & = g^{ac} \epsilon_{bc}. \end{align} Note that order matters: $\epsilon^a{}_b = -\epsilon_b{}^a$. Since it seems you are just relying on heuristics for when signs get ...

0

Remember the definition of work: $\vec F.d\vec x=Fx\cos\theta$. In your case $\theta=180^o$, thus $\cos\theta=-1$. That is the minus sign you were missing.

0

In your formula you don't calculate the potential energy, but the difference between the potential energy of the body at the lower level $P_E(h_0)$, and that at the higher level $P_E(h_1)$. Indeed, since $P_E(h_0) < P_E(h_1)$ you get $P_E(h_0) - P_E(h_1) < 0$ as expected. Phenomenologically, starting from $h_0$, you have to invest some work to raise ...

6

Intensity has units of watts per area: $$\left[I\right]=\rm\frac{W}{ m^2}$$ where the area is the surface area of the emitting source (in this case, the sun). This tells you the total amount of radiation present (over all wavelengths). The extra factor of 1/nm in your plot gives the spectral irradiance: $$\left[\mathcal E\right]=\rm \frac{W}{m^2\,nm}$$ ...

2

$W/m^2$ would be the total energy emitted, regardless of wavelength. When you use $W/m^2/nm$ you are explicitly saying that it corresponds to a specific part of the spectrum (nm is a unit of wavelength). Which is what the graph you posted is showing. The first one is called "irradiance", the one plotted here is called "spectral irradiance". For more details ...

0

It may help: suppose we are close to the earth and at height $h$. So $$\Delta V=Gm_1m_E(\frac{1}{R}-\frac{1}{R+h})$$ where $R$ is the radius of the earth and $h \ll R$. Now we approximate this relation and it's turn out that $$\Delta V=Gm_1m_E(\frac1R-\frac1R+\frac{h}{R^2})$$ By calling $g=\dfrac{Gm_E}{R^2}$, we find $\Delta V=m_1 gh$. Even if we don't ...

2

Spin of an elementary particles is not necessarily the result of a movement of the particle around itself i.e. around some rotation axis that passes through the particle.If there were such an axis, the projection of the spin in the plane perpendicular to that axis were zero. But, this is not the case. So, along whatever axis we would measure the spin, we ...

1

Generally speaking, the choice of what the $z$-axis (equivalently $x,y$) is is arbitrary. You can choose any direction to be your $z$-axis, as long as you do the calculations consistently with this choice. If the system has a priviliged direction (like that imposed by the magnetic field in the Stern-Gerlach case) that is usually choosen to be the $z$-axis. ...

0

If we estimate the size of the Earth to be a perfect sphere (which it isn't, but as a first approximation it will do), then you may apply the shell theorem. It states that the gravitational field of a spherically symmetric body appears as if it is concentrated in the center of mass of the body. Knowing how much Earth weighs ($5.97219\times10^{24}\,kg$), you ...

0

There is no need of a big theory to explain the direction of a dipole moment. It's just because the charge of an electron is negative (by convention).

3

Warning: This post contains a really stupid but important error. I will fix it in a couple of hours, but for now please don't read it. There's a bit more to this than the other answers are covering. As you already noted, in expressions like this $$(c_1^\dagger c_2^\dagger \ldots) |0\rangle,$$ it is important to keep a consistent ordering. The reason is ...

2

I would interpret $1$ and $2$ in the function arguments as shorthands for the coordinates belonging to particles 1 and 2; but the subscripts $1, 2, i, k$ of the single-particle wave functions $\phi$ / $\psi$ and corresponding creation/annihilation operators $a^+$ / $a$ as specifying a single-particle state. The first part is quite a common convention: ...

2

As OP pointed out, one should stick to a single convention. Hence the order of the creation operators should be reversed in one of the two cited equations. Reversing the order of operators in which of the equations is purely a matter of choice. However, to obtain the simplest relation between the Slater determinant and the occupation number representations, ...

3

Since the states $|\psi\rangle$ and $-|\psi\rangle$ differ only by a phase factor, they really describe the same electron configuration. So, while it is a bit sloppy to have inconsistent sign conventions in different places in a text, it's not surprising to see it, and it's not really a big deal. (Unless you have inconsistent sign conventions within a single ...

1

Lorentz spinors appear as irreducible representations of the group SL(2,C). Elements of the group are 2x2 matrices with complex entries and unity determinant. A Lorentz spinor is a two component vector $\psi^{A},\chi^{A}\in V_{2}$ with $A=1,2$. The Levi-Civita tensor $\epsilon_{AB}$ is an invariant tensor under SL(2,C). This means that if we have an irrep ...

0

I've clarified my doubt. The expression $(\bar{\psi} \chi )^{T }=\chi^{T }\bar{\psi} ^{T }$ is simply the definition of trasposition (the same argument for the adjoint), independently on the nature of the objects in the parenthesis. What is subtle is that, the following is true: $\bar{\psi} \chi =-\chi^{T }\bar{\psi} ^{T } =-(\bar{\psi} \chi )^{T }$ ...

2

Soliton is correct, there is a factor of two coming in when the indices are raised or lowered with the Levi-Civita tensors and then a contraction is made. Here is how it appears. Consider a four-vector as a Lorentz spinor, x^{\mu}\rightarrow X^{\dot{A}B}= \left[ \begin{array}{cc} x^{0}+x^{3} & x^{1}-ix^{2} \\ ...

0

What this refers to is the Rotation Reversal Theorem - rotating first about axis z with angle az , and then about the rotated y axis by angle ay , followed by rotation by the now twice rotated z axis by angle bz is the same as rotating first about the original z axis by bz, followed by rotation about the ORIGINAL y axis by ay and then finally about the ...

3

You could make an argument that the four-current is most naturally defined as a vector density. This is because a vector density uniquely defines a three-form (i.e. a totally antisymmetric tensor $J_{\alpha\beta\gamma}$), and three-forms can be integrated on hypersurfaces without any reference to a metric. So if you think of a current as an object which, ...

2

Clearly MTW's definition of $J^\alpha$ is a vector field due to the argument given. Note however that in the Wikipedia article you linked, under Summary, $J^\alpha$ is defined with an additional factor $\sqrt{-g}$, making it a vector density. You can easily see this by replacing the ordinary derivative in the definition of $J^\mu$ by a covariant one (which ...

0

The potential $V(r)=\int \frac{F(r)}{m} dr$ as integral of gravitational force $F(r)$ divided by mass $m$ is only determined up to a arbitrary constant. A satellite coming from infinity will get kinetic energy from falling to the gravity center. So it is a convenient convention to say the potential $V(r) = 0$ and therefore also the potential energy ...

1

What they are basically telling is that the gravitational potential at an infinite distance from the fixed mass is zero. The total energy of the satellite would be K.E. + P.E. Since, $$V(r)=-\int{E(r)}$$ $$V(r)=\frac{-GM}{r}$$ With M being the mass of the fixed object. As r tends to infinity, it is clear that the potential of the system rapidly decreases, ...

0

I think you can consider the anti-commutator, then use the linearity of the trace, as follows: $$\{\overline{\psi},\chi\} = 0$$ $$\{\overline{\psi},\chi\}^T = 0$$ $$(\overline{\psi}\chi)^T + (\chi\overline{\psi})^T = 0$$ $$(\overline{\psi}\chi)^T + \overline{\psi}^T\chi^T = 0$$ $$(\overline{\psi}\chi)^T - \chi^T\overline{\psi}^T = 0$$ $$... 0 Ok, tracked the 2005 paperback edition down and indeed those parts have been corrected. In front of Eq. (5.2.26) he now states \beta=i\gamma^0 and Eq. (5.2.26) has now changed to [2005 Paperback]$$D(L(p))\beta D^{-1}(L(p)) = i{L_\mu}^0(p)\gamma^0=-ip_\mu \gamma^\mu /m, with the sign change in the last step because of his $\eta^{00}=-1$ convention. ...

Top 50 recent answers are included