Recent content by jk86

In central force problems, it is useful to instead use the coordinates: \begin{align} \vec{r} &= \vec{r}_2-\vec{r}_1\\ \vec{R} &= \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{M} \end{align} The reason is that with these coordinates the Lagrangian reduces to something that looks like a single body with a...

In order to equate coefficients, for example with the (2.4.11) expression, you can expand the commutator and collect all the coefficients together on the left hand side so that it looks like: -\frac{i}{2}\omega_{\mu\nu}\left[P^{\rho},J^{\mu\nu}\right] +...

You can say that the uncertainty Δx can certainly not be larger than the size of the well, but that doesn't imply that Δx=L. Remember that Δx is the standard deviation of the position given the wavefunction of the particle ψ: \Delta x \equiv \sqrt{\langle \psi | \hat{x}^2 |\psi\rangle...

Ah, OK I'm sorry I should have read your post more carefully. If you are calculating the dot product of \vec{a}\cdot\vec{b}, you can expand each in terms of its contravariant components. As an example, define a coordinate system (u,v,w) via the Cartesian coordinates (x,y,z) using some relations...

From dimensional analysis you can derive that it depends on [mass][velocity]2, but the factor of \frac{1}{2} comes from the work-energy theorem: W=\int_1^2 Fvdt = \int_1^2 \vec{F}\cdot d\vec{x} = \int_1^2\left(m\frac{dv}{dt}\right)vdt = \frac{1}{2}m(v_2^2-v_1^2) The dimensions of...

You could always use the \vec{u}\cdot\vec{v}=|\vec{u}|\ |\vec{v}|\cos\theta and \vec{u}\times\vec{v}=|\vec{u}|\ |\vec{v}|\sin\theta \ \hat{n} definitions. In general the dot and cross product are independent of coordinate system.

The problem stated that the voltage was 78.7 mV, but you have written that down as 0.00787=7.87mV. Also check your multiplication.

It sounds like you are assuming the uncertainty in the position of the particle Δx is set to the size of the well, but this is something that depends on the state of the particle. But yes, by the uncertainty principle, Δp has a lower bound. When you solve the Schrödinger equation and find...

You originally derived the conservation of current ∂μjμ= 0 from the equations of motion ∂μFμσ=jσ. The form that this current takes in terms of the Dirac fields ψ(x) was found by noting that the global U(1) transformations ψ(x)→e-ieΓψ(x), where Γ is a constant, leave the Lagrangian invariant...

You definitely can multiply an infinite series by a constant, but I am guessing you aren't cancelling the terms correctly. Consider just the sum going to N, and you can take the limit N→∞ afterwards: \sum_{i=0}^{N}2^i = (2-1)\sum_{i=0}^N 2^i = \left(2 + 4 + 8 + \cdots + \cdots 2\cdot...

No that does not mean it happens instantaneously. The "axes" on the Feynman diagrams should not be taken too literally. For example, many diagrams are drawn with exchanged particles following curved paths, but this does not mean that it actually does this.

If time is going from left to right, then the diagram on the left means an electron and a positron annihilate and become a photon, which then decays into an electron-positron pair. The diagram on the right shows an electron and positron exchanging a photon.

The second term in your original energy expression is missing a factor of 2, since it is coming from \mu r^2\dot{\phi}^2/2=L^2/(2\mu r^2), where L=\mu r^2 \dot{\phi}. I think the radius can also be calculated easier by just taking \dot{r}=0 in the radial equation of motion...