@nickbros123 no, you have to have the volume element that you are integrating over be an infinitesimal. Instead, the minimum volume you can integrate over is a unit cell. You can displace the integration volume smaller than the unit cell, but you integrate over a unit cell chunk centred upon the centre of the integration volume. Then you can see the smoothened variation of the charge density around the unit cell.
@naturallyInconsistent but aren't unit cells supposed to contain only a handful of molecules? We learn in solid chemistry that they contain at Max 4 atoms. Tell me one thing @naturallyInconsistent , the averaging of the charge density function, does it happen over the size of the unit cell( in Lorenz picture)
@nickbros123 It does not matter how many atoms there are in a primitive unit cell, as long as you have the full translational symmetry, then you just need to average any integer multiple of one primitive unit cell's worth and the averaging will be sufficient to be correct, so that any variation will really be macroscopic.
The averaging of the charge density function is not over the size of the unit cell in the Lorentz scheme because we tend to choose nice shapes for the hole in the Lorentz scheme. That is, the inner cavity is usually chosen to be a sphere, and then one unit cell is never spherically symmetric enough for this to work.
@naturallyInconsistent ok, I get what ur saying in that the integration volume must be of a particular size whilst the volume element is microscopically infinitesimal, which is fine. The interpretation works easily in this case as $\rho simply is not varying over a size of a unit cell$.
But when it comes to e field, the interpretation is lost and full subscription to math kicks in
At any rate I seem to have, based on what you are saying continually, an immense amount of misconceptions and misunderstandings of the entire averaging game
@nickbros123 I would also have to point out that this transition from micro to macro fields is considered so difficult that it is only ever taught properly in graduate level physics, and so your troubles are not just due to your fault
@naturallyInconsistent oh sorry, I'm on my phone, where I can't read in true latex fashion what I'm typing. In fact I've learnt to just read formulas in their naked form, like \rho \frac etc
@naturallyInconsistent but undergrads are taught about auxilary fields $D$ and $H$ , polarisation $P$ and all that. This stuff without knowing about macroscopic fields? Seems pretty wrong
What is happening is that $\rho_\text{micro}$ is varying wildly inside a unit cell, and that means that your E field is also wildly varying inside a unit cell. We really only want a smoothened function that can match with what Maxwell intuitively guessed that a smooth material would be doing, namely that $\rho_\text{macro}$ should be almost always zero and slowly smoothly varying over many unit cells, except maybe at boundaries.
@nickbros123 It is very wrong. The more I learn about electrodynamics, the less I think we should even be entertaining D and H and P and M, and I mean this even for graduates. These things are just incredibly badly defined and a continual source of headaches. We should still have macroscopic smoothed fields and so forth, but we should only have many different versions of E and B, not weird things flying around all the time.
@SillyGoose Yes, because $\mathrm{U}(N)\mapsto \mathrm{U}(1), A\mapsto \mathrm{det}(A)$ is a surjective group homomorphism with kernel $\mathrm{SU}(N)$ and so the isomorphism theorems yield $\mathrm{U}(N)/\mathrm{U}(1)\cong \mathrm{SU}(N)$
the $\mathbb{Z}_2$ is because in $\mathrm{SU}(2)\times \mathrm{U}(1)$, you have that $(g,\phi)$ and $(-g,-\phi)$ represent the same $\mathrm{U}(2)$ element
@SillyGoose unicode names all fun and games until you've had to debug a problem where someone accidentally added a zero-width space to a variable name so var and var were different objects and then you long to return to the purity of the blessed ASCII
@ACuriousMind This is usually only a problem for languages like Python, and not a problem for statically typed languages. Still, code environments really ought to keep unicode at bay. A way to show the hidden unicode will be helpful.
@nickbros123 I just flew a few books on the topic; However, my luggage was not big enough to bring the most important two that would be definitely of help. I did bring the T C Choy, though, and that is fun.
@naturallyInconsistent I mean, it can still take an annoyingly long time to figure out why you're getting a syntax error, but sure at least you don't only find this at runtime with static languages.
In any case most languages have tools to forbid non-ASCII in their source code even if they support Unicode so this is an easily solved problem
The way gamma matrices "transform" $S\gamma^\mu S^{-1}=\Lambda_\nu{}^{\mu}\gamma^\nu$ looks like the adjoint rep of the Lorentz group. Is there a way to connect these things or to explain the gamma matrices relation in terms of the adjoint rep?
I mean, it has to be related somehow through spin/pin groups
Or the fact that the commutators of gamma matrices furnish a representation of the generators of LG
@ACuriousMind That's the Lorentz algebra, yes. Sorry, I was comparing (not saying it is) it to the adjoint rep of the Poincaré algebra of which we've discussed in the past i.e. how $P^\mu$ tranforms under such rep
@Mr.Feynman Im one of those rare unicorns with a totally different way of looking at physics. If you look at the Dirac spinor from the point of view of GA, then it kinda makes sense that, since the spinor would be a rotor, it would rotate the gamma matrices (which in this case is to be interpreted as an axis frame of the observer). Then their Lorentz transform would drop out.
@Slereah When you unify spheres with planes, circles with lines, so that collision algorithms become deterministic as if reduced to quadratic formula, it is no wonder that game engines would want to play with them.
Using some � emojis for the Schiehallion experiment
To do the Schiehallion experiment they apparenly straight up split the mountain into ten pages of polyhedrons and then just summed the gravitational contribution of all of them
i am wondering why he is keeping the a constant while varying b
its about the angular momentum and the eigenvalkues of j^2 and j_z
my assumption is it that the ket $ j_+ \vert a , b > $ is an eigenket of j^2 with a as an eigenvalue and an erigenket of j_z with an eigenvalue of b + h
thus justifying writing it as a new ket, that keeps a constant
If I throw an object horizontally from the building it has constant horizontal velocity $V_{ox}$ throughout his journey. Its initial vertical velocity $V_oy = 0$ which increases continuously and after 1 second it becomes $V_y = 9.8 \frac{m}{s}$.
i have another q about gr hehe. so in carroll he says "if a coordinate system exists in which components of the metric are constant, the reimann tensor will vanish" this seems reasonable to me since the RCT encodes curvature but he claims that $\partial_{\sigma}\Gamma^{\rho}_{\mu \nu}$ vanishes. then in a separate context, he says "lets consider the components of the tensor $R_{\rho \sigma \mu \nu}$ in locally inertial coordinates the christoffels will vanish but their derivatives will not."
i feel like im missing smth here because arent these both saying different things ab whether or not the derivatives of christoffels will vanish in locally inertial coords?
@Amit Yes i understand in projectile force is not always perpendicular to the direction of motion. But i want to draw your attention the instant when i throw an object horizontal. At that instant force is perpendicular to the direction of motion.
locally inertial coordinates (mathematically also: Riemann normal coordinates) always exist, but coordinates in which the components of the metric are constant only exist when the Riemann tensor vanishes as you just said
@Amit What is understand direction change process is only instantaneous if force is perpendicular to the direction of motion. It repeat instantly only when force is always perpendicular. Am i correct?
Note that even if you throw the object at an upwards angle, there will come a point where $v_y=0$ so for an instant anyway the velocity will be purely horizontal. This is no different than the statement that if you derive $y(t)$ you will find $y'(t)=0$ at precisely that instant. The height function is "stationary" at that point
okay what is the proper qualifier to the statement "if a coordinate system exists in which the components of the metric are constant, the RCT will vanish" -- is it that this statement is valid in a neighborhood? @ACuriousMind
in normal coordinates at $x=0$ you have: $g_{\mu\nu}(x) = \eta_{\mu\nu} + \frac{1}{3}R_{\mu\nu\rho\sigma}x^\rho x^\sigma + \mathcal{O}(x^3)$, i.e. the measure of how much locally inertial coordinates are not "coordinates in which the metric is constant" is exactly the curvature!
@Relativisticcucumber I mean what does "constant" mean if not "in a neighbourhood"?
stuff can't be "constant at a point" - everything's just a value at a point
In general: $\mathbf{v}(t) = \mathbf{v}(0)+\mathbf{a}t$ , now put $t=\Delta{t}$ and rearrange: $\mathbf{v}(\Delta{t})-\mathbf{v}(0)=\mathbf{a}\Delta{t}$ , that's it, call the LHS $\mathbf{\Delta{v}}$ to indicate it's the change of the velocity vector in a very small time. Now for any finite amount of time in your projectile motion problem, not only the direction will change. Only for $\Delta{t} \rightarrow 0$ there is no change of direction...
@Amit A is stationary and at origin, where B and C is moving with same velocity in same direction. From A frame of reference B & C are moving with the same speed. And from B & C frame of reference both are stationary. B is above at C.
at moment instantaneous perpendicular force is applied to the B. From C frame of reference B initially at rest then gain some velocity and coming toward C with that velocity
@Amit But from A frame of reference B initial moved along x-axis with $V_x = 3m/s$. Due to application of perpendicular force on B it should change only direction not magnitude. But the problem is that it look B maintain x-comp velocity and due to force it also gain some y-com velocity. What is the problem? Where i am wrong.
@ACuriousMind well i see in my notes that $g_{\mu \nu} \vert _p = \eta_{\mu \nu}$ and $g_{\mu \nu} \approx \eta_{\mu \nu} + O(x^2)$ so i think what i was getting confused on might be related to the first being applicable at a single point and the second being not just a point, and im trying to understand this. so it seems we can describe the metric as exactly minkowski at one point but then the second statement i wrote, what area of the manifold does this describe? the entire thing?
B is moving with the same velocity same direction as C. Both are stationary from their frame of reference. So B is stationary from C frame of reference. When instantaneous force is applied to B it gain some constant velocity and it move linearly toward C.
@ACuriousMind Okay i see. i think im trying to picture how this argument works like i see we can visualize the metric at any arbitrary point as mink but its only for that point so im trying to understand how we obtain curvature then. i guess this is because we can show that the derivatives of the metric vanish at a point but the derivatives of the christoffel symbols do not so as we move around the neighborhood, we curve? hm this argument is a bit strange.
But from A frame of reference same instantaneous force is applied to B but it is now perpendicular from A reference. So perpendicular force change only direction not magnitude. But in this example i calculated B maintains its x-comp velocity it gain y-comp velocity. By rule it does not change magnitude of velocity
@123 Idk what you mean. Both In A's and C's frame only a vertical force is applied on B. That's because Newton's second law looks the same when you move to a frame with constant velocity
But by the rule this from A frame of reference when instantaneous perpendicular force is applied to B. it change direction but due to change in velocity y-comp. Because x-comp of velocity is still the same. If i add x-com and y-comp velocity. It increase velocity. How perpendicular force increases velocity?
@Amit Oooh okay, you are saying in my example if i look carefully and see it component wise. So x-comp decreases and y-increases so that inertia remains the same?
@Amit I know it. I want to understand the physical process. how direction change happened? And what is the process of knowing how much direction change occur?
You can try the following... take: $\mathbf{v}=(v_x,v_y)$. Now you can find the perpendicular direction to be: $(-v_y,v_x)$ right? Now you can write a simple program, one that applies acceleration $(-v_y,v_x)\times \Delta{t}$ say 10 times with some small value for $\Delta{t}$, then do the same 20 times but with half the value of $\Delta{t}$. Then compare the length of the initial vector and see how much it changed. It will change less the smaller you make the time interval...
@Amit Okay. I think i am mixing the concept. I can understand you are getting me closer to the integration of velocity over time to calculate an acceleration and find vector how much direction change. Am i correct?
@Relativisticcucumber well, it's the unique ID of the chat message you're replying to. If you put the ID of one of your own messages, you can reply to yourself. You get the reply from the permalink of the message: The message I replied to has https://chat.stackexchange.com/transcript/message/63928283#63928283, so it's ID is 63928283. So I type : 63928283 and:
@123 The thing is, in both loops the force is applied for the same amount of time! 1 second in total. But in smaller chunks, it is closer to having an "instantaneous" force applied at every instant.
That the direction changes is not your question. Your question is how can it not affect the velocity's magnitude. You just proved algebraically that it changes magnitude only to the extent that the force doesn't change for a finite $\Delta{t}$. But we require that it changes smoothly, so that no matter how small $\Delta{t}$ is it is not the same force.
@Amit I can understand derivative and integral. But still direction change thinking is confusing. Without changing x and y-comp of velocity how can direction change. There must be slight change in both components. One slight decrease and another increase to preserve magnitude in rotation (direction change).
@Amit It is more that Java always gave meow the impression of incredible bloat, and that Java's default maths capabilities do IEEE floating point wrongly, so that it can be a source of hard to trace bugs if you are not insanely careful.
