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Yes, of course they are acting on different spaces in most cases. Was trying to keep things very generalized in an attempt to "rescue" invariance, but I think that may be overkill. And, Psi is indeed a dirac spinor. My question still remains on what to do about T being non-unitary. As ##A_\mu...

You see in the literature that the vector potentials in a gauge covariant derivative transform like: A_\mu \rightarrow T A_\mu T^{-1} + i(\partial_\mu T) T^{-1} Where T is not necessarily unitary. (In the case that it is ##T^{-1} = T^\dagger##) My question is then if T is not unitary, how is...

You’re right. I am just trying to figure out *how* this could be zero at this point, as in what conditions. Otherwise I’m stumped.

ok i think i have solid reasoning here: Suppose ##C^{ij} = M^{ij} + N^{ij}## From symmetry and antisymmetry we have: ##\epsilon_{ijkl} C^{ij}C^{kl} = 0## Also if you foil the CC product in terms of M and N you get ##C^{ij}C^{kl} = M^{ij}M^{kl} + N^{ij}N^{kl} + M^{ij}N^{kl} + N^{ij}M^{kl}##...

ep_{ijkl} M^{ij} N^{kl} + ep_{ijkl}N^{ij} M^{kl} The second term can be rewritten with indices swapped ep_{klij} N^{kl}M^{ij} Shuffle indices around in epsilon ep{klij} = ep{ijkl} Therefore the expression becomes 2ep_{ijkl}M^{ij}N^{kl} Not zero. What is wrong here?

Okay so the Lagrangian behavior is straightforward then. What about the Lagrangian density? Where rho is the mass density of a particle cloud. $$ \mathcal{L} = -\rho(y) \sqrt{\dot{y}_\mu \dot{y}^\mu} - A_\mu J^\mu -\frac{1}{4} F_{\rho\sigma} F^{\rho\sigma}$$ $$ \frac{\partial...

The last paragraph is basically asking, how do I write the full Lagrangian of a massive charged particle in an electromagnetic field? From what you've said, I gathered that it would be written like: $$ L = -m\sqrt{\dot{y}_\mu \dot{y}^\mu} - q A_\mu (y) \dot{y}^\mu - \frac{1}{4} \int d^3 x...

How would you unify the two Lagrangians you see in electrodynamics? Namely the field Lagrangian: Lem = -1/4 Fμν Fμν - Aμ Jμ and the particle Lagrangian: Lp = -m/γ - q Aμ vμ The latter here gives you the Lorentz force equation. fμ = q Fμν vν It seems the terms - q Aμ vμ and - Aμ Jμ account for...

Amazing. Thank you.

For example, can you predict the permittivity and permeability of a substance if you know what the atomic composition is? Is it a stat mech problem?

Maybe I'm not explaining it right. I swear I have seen this done before. Here is the process: Start with this generic form: R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} \pm \Lambda g_{\mu\nu} = \frac{8 \pi G}{c^4} T_{\mu\nu} Take the trace to get: -R \pm 4\Lambda = \frac{8 \pi G}{c^4} T Take the...

This is what I thought at first. But if you take the 00 element of the equations, and the trace of the equations, and solve for Lambda, you get different results depending on your choice of sign and metric. Would just like some clarity on the sign conventions for this.

My questions is: Depending on which metric you choose "east coast" or "west coast", do you have to also mind the sign on the cosmological constant in the Einstein field equations? R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} \pm \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu} For example, if you...

so could one say SU(1) = U(1)? If not, why not.

If you have a U(1) generator, can it just be normalized to SU(1)?

It seems a gravitational field does not alter the electromagnetic field strength. Is this correct? My reasoning: With no gravity, field strength is: F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu Introduce gravity: \partial_\mu A_\nu \rightarrow \nabla_\mu A_\nu = \partial_\mu A_\nu +...

The general uncertainty principle is derived to be: \sigma_A^2 \sigma_B^2 \geq \left(\frac{1}{2} \langle \{A,B\} \rangle -\langle A \rangle \langle B \rangle \right)^2 + \left(\frac{1}{2i} \langle [A,B] \rangle \right)^2 Then it is often "simplified" to be: \sigma_A^2 \sigma_B^2 \geq...

Right. I get that but I’m not sure where I go wrong here.

Is the following true if the momentum operator changes the direction in which it acts? \langle \phi | p_\mu | \psi \rangle = -\langle \phi |\overleftarrow{p}_\mu| \psi \rangle My reasoning: \langle \phi | p_\mu | \psi \rangle = -i\hbar \langle \phi | \partial_\mu | \psi \rangle \langle...

Is it possible to use separation of variables on this equation? au_{xx} + bu_{yy} + c u_{xy} = u + k Where u is a function of x and y, abck are constant. I tried the u(x,y) = X(x)Y(y) type of separation but I think something more clever is needed. Thank you.

It is easy to find that the equation for an ellipse is: $$1 = x^2/a^2 + y^2/b^2$$ Then according to Kepler's equation: $$x = a(\cos(E)-e)$$ $$y = b\sin(E)$$ where E is the eccentric anomaly and e is the eccentricity. If you plug the Kepler's equations' x and y into the equation for the ellipse...

Thank you!

Yes. I get something non-analytic.

It is fairly trivial to do this with a circular orbit: $$(x,y) = (cos(\omega t),sin(\omega t))$$ where t is time, and $$\omega = \sqrt{GM/r^3}$$ How this parametric equation look for an elliptical orbit?

$$t = \pi \sqrt{\frac{(r_1+r_2)^3}{8GM}}$$ $$r_1 + r_2 = 4e8 m$$ $$M = 6e24 kg$$ $$t=4.9 days $$

I see. Free return trajectories are faster then? Since they have enough energy to return a craft with minimal/no burns? I’m just trying to account for the discrepancy in time between the Hohmann transfer time of 5 days and the actual 3 day time.

Thanks. Are Bi-elliptic transfer orbits quicker than Hohmann transfers as well? Are they more akin to what the apollo missions used?

It's common knowledge that it takes about 3 days to get to the moon. With a Hohmann transfer, I get a transit time of 5 days, not 3. I see NASA used something called "trans-lunar injection". Is this distinct from a Hohmann transfer, and more time efficient? What makes this trajectory different...

In collisions that are inelastic or partially elastic, how can we predict how much of the energy lost to the surroundings becomes heat, and how much becomes sound? What determines that fraction?

Hi everyone, I am looking at a paper on compact dimensions. Equation 65 makes sense except for the term of 4*pi*n*R in the denominator. Why is it 4*pi and not 2*pi? I cannot rationalize this. Please help. Thank you. https://arxiv.org/ftp/hep-ph/papers/0609/0609260.pdf

So it is not yet known/understood?

So the universe is expanding, and galaxies are getting farther apart from one another on average. Does this motion count the same as ordinary motion, in that if a galaxy is being expanded away from us at 0.5c, that clocks in that galaxy would appear to tick slower at 0.866 the rate of clocks here?

None. I took a class that covered SR. That's why I am asking here.

Ok. What is it I am computing? How do I compute the velocity, momentum and energy of the photon in this scenario?

Some follow-up questions. If there's a photo traveling toward the center of a gravitational body, we have: m^2 = 0 = g_{\mu\nu} p^\mu p^\nu If we simplify by saying motion is along the x-axis: g_{00} E^2 + g_{11} p^2 = 0 Plug in the Schwartzchild metric, and we get \frac{\left(1 -...

Thanks for clarifying this. It was what I was thinking but did the math wrong.

I see now. So it should be written: 0 = g_{\mu\nu}p^\mu p^\nu - \eta_{\mu\nu} p'^\mu p'^\nu

In Special Relativity, you learn that invariant mass is computed by taking the difference between energy squared and momentum squared. (For simplicity, I'm saying c = 1). m^2 = E^2 - \vec{p}^2 This can also be written with the Minkowski metric as: m^2 = \eta_{\mu\nu} p^\mu p^\nu More...

That term comes from the chain rule of δL I have seen the least action principle shown as 0 = δS = ∫δL dt, which I guess is misleading. I have seen the form you have, and that makes more sense. You are explicitly minimizing with respect to epsilon.

Can you show why? How is the term ∂L/∂t δt handled in Least Action?

If a Lagrangian has explicit time dependence due to the potential changing, or thrust being applied to the object in question, how does calculus of variations handle this? It's easy to get the Lagrange equations from: δL = ∂L/∂x δx + ∂L/∂ẋ δẋ What is not clear is how this works when t is an...

Silly me, yes you can just forget about it being angles. Uniform distribution sample - uniform distribution sample = non-uniform sample. Still not sure why this is.

The vectors aren't *really* vectors computationally. I'm just generating angles using a uniform random number generator. Then taking the differences between them.

I am simulating random angles from 0 to 2π with a uniform distribution. However, if I take the differences between random angles, I get a non-uniform (monotonically decreasing) distribution of angles. In math speek: Ai = uniform(0,2π) dA = Ai - Aj dA is not uniform. Here is a rough image of...

Is the following true? ψ*∇^2 ψ = ∇ψ*⋅∇ψ It seems like it should be since you can change the direction of operators.

This cannot be solved analytically. You have to use an approximation. For example, you can take the approximation (taylor series) of e^y = 1 + y + y^2/2 Then you will have 0 = y^2 - 2y - 6, and can solve for y. You will get -1.6458 which is approaching the correct answer. To get more correct...

Thanks. I was trying to think of a counter example. This is very obvious.

For example, does this always hold true? M_ab = v_a × w_b If not, where does it break down?

It could become a big number of possibilities, but remember that if k > 0, then k' = 0, since n and n+1 do not share prime factors. So the number of combinations to evaluate is 2^a, where a is the number of primes below 7 (or whatever prime you are considering).

Yes, that is correct, 7 is the largest prime factor present in the pairs (6,7),(7,8),(14,15),(20,21),(27,28),(35,36),(48,49),(49,50). But where does this stop? Is there some integer n such that factoring the pair (n,n+1) gives you the last instance of 7 being the largest prime factor?