Recent content by Hao

Awesome! Thanks!

In Eq 11.72 in the QFT text by Peskin, the following equality is stated: i\int\frac{d^{d}k_{E}}{(2\pi)^{d}}\log(k_{E}^{2}+m^{2})=-i\frac{\partial}{\partial\alpha}\int\frac{d^{d}k_{E}}{(2\pi)^{d}}\frac{1}{(k_{E}^{2}+m^{2})^{\alpha}}|_{\alpha=0} This suggests that...

For those who are interested, I've found a solution by somebody else. http://www.maths.tcd.ie/~powersr/New/

Homework Statement (This is not homework) This refers to question 3.8 in Peskin's QFT Using the fact that the electromagnetic interaction term in the Dirac + EM lagrangian is invariant under Parity (P) and Charge conjugation (C), and that spin 0 and spin 1 states are odd and even under...

The utility of 4 vectors is in the fact that you can apply the rules of vector manipulation such as dot products. 'p' will refer to a 4-vector. Together with some useful identities, such as p \bullet p = m^2 c^2 (prove this!), it can be a very powerful tool <--- This is the subtle point you...

Once you have found the EMF as a function of time, you will be able to find power, and hence, by integration over one cycle, the energy dissipated.

An abstract (not necessarily better) way to think about this starts from dU = T dS - p dV. For an ideal gas, dU = 0 (Things are a little more complicated for real gases). Hence, T dS = p dV , and \frac{dS}{dV} = \frac{p}{T} As pressure is positive, and temperature is positive, we can see that...

r refers to the radius of the wheel, which is constant, so dr/dt = 0. The correct answer is therefore your last term vertical velocity = r*sin(theta)*(v/r). Note that you could have also gotten this answer by picturing the wheel undergoing pure rotation, and resolving the velocity to the y...

I can only imagine energy being dissipated if the loop has a finite resistance. I hope the question gives more information on what the loop is made of.

Exactly - differentiating y with respect to time will give you the y component of velocity. Just remember to account for the fact that θ = ω t. Similarly, differentiating x with respect to time will give you the x component of velocity.

I don't think M is necessary. If there is a second part to the question, it may be relevant there. However, you should take a second look at your expression for φ. Based on the units you are using, φ is the magnetic flux. \phi = \int B dA For a uniform field and an area A inside that field...

For the ground state, the assumption makes no difference. However, it has an effect for the first excited state. The first excited state is defined as the state that is second lowest in energy. If \hbar \omega were to be very small, (n_x, n_y, n_z) = (1, 1, 1) would be second lowest in energy...

The definition of velocity in the x direction is \frac{\bigtriangleup x}{\bigtriangleup t} \rightarrow \frac{d x}{d t} This means that we need to differentiate both x and y with respect to t to obtain velocity. Alternatively, we can resolve the velocity components for pure rotational motion...

Let us consider the rotational motion of the wheel alone. If we suppose that the center of the wheel is stationary, and the edge is rotating at a speed vr, we can find our angular velocity vr = ω r, where ω is the angular velocity, and r is the radius of the wheel. Since we know ω, we can find...

The position of the pebble is a superposition of pure rotational motion of the wheel, and pure translational motion of the wheel. You do not need to use integration for this question. Writing out the separate x and y components for the position of the pebble will help you find velocity...