7
However if I do assume each photon gives up a bit of it's energy, thus changing it's wavelength, then the incoming momentum will not be equal to the outgoing momentum. But this would lead to a contradiction as we assumed the mirror was perfectly reflecting.
There is no contradiction. You should interpret "perfectly reflecting mirror" to mean that every ...
6
In the problem statement "perfect mirror" must mean that no light is absorbed and that the mirror is perfectly flat. However, there is a recoil effect that is measurable if the mirror is light enough. This is required by momentum conservation, as you state. Every photon will therefore be reflected with a slightly smaller energy, hence a slightly smaller ...
5
Perfect mirrors do not exist, but if you insist, let's discuss this case.
A perfect mirror is a mirror that reflects light (and electromagnetic radiation in general) perfectly, and does not transmit or absorb it.[1]
https://en.wikipedia.org/wiki/Perfect_mirror
This means, that as you see in the comments, each and every single photon that is incident on ...
4
Do not bother with converting the expression into coordinates; you will make your life much more difficult. Simply use the pertinent product rule for divergences: $$\mathbf{\nabla}\cdot(\phi\mathbf{A})=\phi(\mathbf{\nabla}\cdot\mathbf{A})+\mathbf{A}\cdot(\mathbf{\nabla}\phi)$$
where $\phi$ is a scalar function, and $\mathbf{A}$ a vector function. The desired ...
4
You are right to question the assumption that the impulse given to the mirror is twice the incoming photon's momentum, but this is not anything to do with the mirror being a perfect reflector. This is an approximation. You are correct that if the reflected photon's momentum is equal to the incoming photon's momentum, then the mass of the mirror must be ...
3
Treat it as an impulse (elastic collision) problem. You do not need force or power, you just need conservation of momentum when the photons hit.
You can reasonably assume equal and opposite momentum for the photons before and after collision (no change in wavelength, as the mirror is stationary at collision - you could include a doppler shift for the ...
3
You can use $E=\frac 12 \dot \theta^2+ V(\theta)$ being constant and separate variables to get something like
$$
\int_0^t dt +const. = \int_0^{\theta(t)} \frac{ d\theta}{\sqrt{2(V(\theta)-E)}},
$$
but, except for the case that the pendulum starts at rest at the topmost point, you are looking at an elliptic integral on the RHS. The start-at-top case is ...
2
It follows straightforwardly from $E=\gamma m$ and $p=\gamma\beta m$.
2
Performing the manipulation you suggest,
$$\langle x|x'\rangle=\langle x|\mathbb 1 |x'\rangle = \langle x|\left(\int dx'' |x'' \rangle\langle x''|\right)|x'\rangle $$
$$= \int dx '' \langle x|x''\rangle \langle x''|x'\rangle = \int dx'' \delta(x-x'')\delta(x''-x')$$
The defining property of the delta function is that $\int dx'' f(x'') \delta (x''-x') = f(...
2
Note that the question is about the work done on the weight, not the work done by the man. The work done by the man will be greater than the work done on the weight because the man's muscles are not 100% efficient and because energy is lost due to friction.
In case (i), if the weight is carried at a constant height about the ground, then the force required ...
2
Take the Sun(mass M) and the planet(mass m) to be point masses. Let the Sun be fixed at the origin and the planet be moving in the x-y plane, initial velocity of the planet be $v_o\hat{j}$ and initial position of the planet be $r_o\hat{i}$. At any given instant, let the planet's position, velocity and acceleration be $\vec{r}, \vec{v}, \vec{a}$ respectively. ...
2
I'm not totally sure, but this sounds like a problem concerning the frame of reference. In the center of mass frame there should be no energy transfer for an elastic collision. But as the mass of the mirror is not given, the center of mass frame may not even be approximately the lab frame where the mirror is at rest initially.
Differently phrased: One ...
2
The mass of the mirror is given to make you think this is a frame-of-reference problem. It is not. That, or it's required to derive the spring constant $k$ from the oscillator frequency $\Omega$. Let's suppose that is done.
Now the situation is in equilibrium with the mirror displaced by a distance $x$, which requires a force:
$$ F = kx $$
Whence does ...
2
Your equation (2) is wrong. To use $x$ as the integrating variable you need to change $x\,dm$ into $x\,\dfrac{dm}{dx}\,dx$. This means we need to define $m$ as a function of $x$, and the most reasonable way to do that while keeping the original meaning of $dm$ is to let $m(x)$ be the mass of the section that goes from $0$ to $x$.
2
The integral for the center of mass, using your notation, is correct:$$ \text{CM}=\frac{1}{M} \int_{a}^{b} xdm$$
Notice that $mdx \neq xdm$ in your equation 2. Thus, you are no longer calculating the center of mass.
An easier way to think about this is defining a "linear mass density variable" $\lambda$, which is given by $$\lambda = \frac{M}{X}$$ since ...
2
From OP one has (the $\kappa^2$ goes away)
$$\eta_{\mu\nu} \Omega^2 = (1+2\kappa)\eta_{\mu\nu}\,,\tag{5}$$
and
$$\frac{\partial(x^\sigma +\xi^\sigma)}{\partial x^\mu} \frac{\partial(x^\rho + \xi^\rho)}{\partial x^\nu} \eta_{\sigma \rho}\,.\tag{6}$$
Now let's use $\partial_\mu x^\nu = \delta^\nu_\mu$ and keep only the linear order in $\xi$. So
$$
\begin{...
1
While @G. Smith's answer is correct and straightforward (with $\gamma = 1/\sqrt{(1 - \beta^2)}$), it might be of interest to interpret the form of this equation.
Using the substitutions in terms of hyperbolic-trigonometric functions of rapidity $\theta$:
$\beta=\tanh\theta$, $\gamma=\cosh\theta$, and $\cosh^2\theta - \sinh^2\theta\equiv 1$ (so $E=m\cosh\...
1
We know that net force can be expressed as:
$$F_{net}=m\dfrac{dv}{dx}$$
Therefore we have:
$$m\dfrac{dv}{dx}=mg-kv$$
$$\dfrac{dv}{dx}=g-\dfrac{kv}{m}$$
$$\dfrac{dv}{dx}+\dfrac{kv}{m}=g$$
This is a linear ODE. We have the integrating factor: $e^{kt/m}$
Therefore the solution to the ODE is:
$$v\cdot e^{kt/m}=\frac{mg}{k}e^{kt/m}+c$$
Assuming velocity at time ...
1
By the way you wrote the momentum of fuel, it can be assumed that all the $\varepsilon m_0N$ of fuel is moving as one chunk with the velocity $-u$. However, in reality only the first chunk will move with $-u$, the next chunk will move with a different velocity, because the rocket wasn't stationary when it dumbed this chunk.
1
In particle physics people refer to the masses and states of the physical particles as the spectrum. So when asked for the spectrum of this theory, they are asking to find the physical states and their masses.
I will proceed with only one field $\phi$ as the other one is analogous. Your parametrization suggests that your scalar fields have non-vanishing ...
1
You can use Newtown's equations of motion with constant acceleration (Here, the acceleration is only due to gravity neglecting the air resistance). Try to solve for the the displacement (height) for each ball separately as a function in time. Once you get both displacements as functions in time, you can equate them and solve for the time and consequently, ...
1
Rather than thinking of the force applied by the man and the distance moved by the block, think of the energy difference.
You will notice that in the first case, the body is just shifted parallel to the horizontal surface i.e. its P.E does not increase. But in the second case, the potential energy increases, which invariably means that he has done work.
...
1
The particle-hole transformation you are looking for is a little more complicated than exchanging creation and annihilation operators - as you found this has the undesirable effect of changing the sign of the hopping. Instead, let's introduce the operators:
$$d_{j \sigma}^\dagger = (-1)^j c_{j \sigma} \ .$$
On a 1D chain, clearly the factor $(-1)^j$ takes ...
1
The derivation you refer to is done locally in a set of coordinates in which $g_{ik,l} = 0$, which can be also characterized as related to a set of Riemann normal coordinates by a constant linear transform. This means, that you can make operations such as this one
$$g^{ij} (g_{kl,mn}) = (g^{ij}g_{kl,m})_{,n}$$
Another useful identity is $g^{ij}g_{jk,l} = -g^{...
1
From the diagram below, how does the newton's 2nd law equation change from top to bottom of the motion? (with mv^2, r and T). Whilst the slotted mass's weight will indicate the tension, the overall resultant force changes because of the mass of the object you are rotating is either added or taken away from the tension.
You should have the time period for a ...
1
Yes, your final result is correct, assuming you're treating everything as a scalar.
As you noticed, Griffiths' rules are a little bit clunky. They were chosen to be as explicit as possible, so people have a chance to see everything working. If you go to a "proper" quantum field theory textbook, its Feynman rules for the same theory will be more streamlined:
...
1
First, this has nothing to do with the capacitor. Regardless of what load you connected there (capacitor, inductor, some complicated transistor circuit, or whatever), you'd find the Thevenin equivalent of the source the same way.
Second, when you ask for the Thevenin equivalent resistance of the circuit, you're asking what's $\frac{dV_o}{dI_o}$, where $V_o$ ...
1
What class is this "homework" for? Or is this just something you find interesting. I will assume the latter.
You will not get very far modeling an entire instrument with a single wave function. But here is a start.
The equation you posted has two free parameters, amplitude and frequency. The amplitude is related to the volume of sound and the frequency ...
1
When you calculate the orbital eccentricity in terms of energy and angular momentum, the same formula applies for circular orbits, elliptical orbits, parabolic trajectories, and hyperbolic trajectories. That formula is
$$e=\sqrt{1+\frac{2\epsilon h^2}{\mu^2}}.\tag1$$
Here $\epsilon$ is the specific orbital energy (the energy -- kinetic plus potential -- ...
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