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Expansion of gauge potential on infinite dimensional manifold

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I'm studying geometrical approaches to locomotion at low Reynolds number by reading the article Geometry of self-propulsion at low Reynolds number by Alfred Shapere and Frank Wilczek and found a situation I'm not understanding from a mathematical point of view what is being done.

The idea itself is not that hard and I shall present it first to make things clear: we consider that a body inside a fluid is a three-dimensional subset of the fluid region and we consider the body's shape as a parametrization of its boundary by means of a function $S : U\subset \mathbb{R}^2\to \mathbb{R}^3$.

In that case, the idea as I understand is to consider a sequence of such shapes $t\mapsto S_t$ parametrized by time (that is a curve of shapes) which just contain deformations with the shape "held fixed" and try to find a curve of rigid motions resulting from the deformations.

If one considers $M$ the space of all shapes, this is a space of functions. A sequence of shapes will be one curve $\gamma : I\subset\mathbb{R}\to M$. Considering then the euclidean group in three-dimensions (that is, the group of rigid motions in $\mathbb{R}^3$) which we denote $E(3)$ there is a natural right action of $E(3)$ in $M$ in which a shape is translated and rotated by some element $g\in E(3)$. Indeed, if $S\in M$ then $S\cdot g$ will be the function

$$(S\cdot g)(u,v) = S(u,v) \cdot g$$

Where the right action on the right side is the natural action of $E(3)$ in $\mathbb{R}^3$. Given the space $M$ then we can also consider the quotient $M/E(3)$ where shapes related by rigid motions are considered equivalent.

If $\pi : M\to M/E(3)$ is the natural projection, $(M,\pi, M/E(3))$ has the structure of a principle bundle with structure group $E(3)$ and the problem at hand becomes the problem of finding a connection which allows the horizontal lift of sequence of unlocated shapes (just deformations) into sequences of deformations and movements.

In page 7 of the document, the author considers infinitesimal deformations, from which he will describe a way to obtain the connection (which he calls the gauge potential). He defines then one infinitesimal deformation by parametrizing a sequence of unlocated shapes by

$$S_0(t) = S_0 + s(t),$$

where $S_0$ is one initial shape and $s(t)$ is infinitesimal. He then says we can expand $s(t)$ as

$$s(t) = \sum_{i} \alpha_i w_i$$

where $w_i$ is a "fixed basis of vector fields on $S_0$". From one intuitive point of view I imagine that a tangent vector to $M$ is equivalent to a vector field defined on the shape, so this makes sense.

Now, he considers the gauge potential as a vector field $A$ with the property that denoting $A_{\dot{S}_0(t)}[S_0(t)]$ the component of $A$ at the point $S(t)$ in the direction of $\dot{S}_0(t)$ this object belongs to the lie algebra of $E(3)$, that is $\mathfrak{e}(3)$ and we can find the curve of rigid motions $\mathfrak{R}(t)$ by solving the equation

$$\dfrac{d\mathfrak{R}}{dt} = \mathfrak{R}A_{\dot{S}_0(t)}[S_0(t)].$$

My doubt is then in what he does next: he considers this infinitesimal deformations $S_0(t)$ I gave and then he "expands the gauge potential" to second order in this way

$$A_{\dot{S}_0(t)}[S(t)] = A_{v(t)}[S_0]+\sum_i \dfrac{\partial A_v}{\partial w_i} \dot{\alpha}_i\approx \sum_j A_{w_j}\dot{\alpha}_j + \sum_i \dfrac{\partial A_v}{\partial w_i}\alpha_i\dot{\alpha}_j$$

where this $v(t) = S_0'(t)$. Now what is going on here? I simply cannot understand where this equation came from and what does it means. What is really happening here?

This post imported from StackExchange Physics at 2015-04-09 12:10 (UTC), posted by SE-user user1620696

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# Social Encounters in a West Marches Campaign

I am a new DM with one session under my belt and am planning to run the campaign as a "West Marches" D&D 5e campaign. One of my players requested I have some social encounters next time and I am wondering how to have these in a WM game, especially as it progresses and players spend more time away from the main town.

• Can you narrow this down a bit? "how to have social encounters" is a really broad question. What kind of things are you worried about? Why are you afraid it'll get worse as people spend more time away from the main town? – Erik Sep 16 at 14:10
• I get the feeling that you’re thinking “social encounters can only happen when they’re in Town”, and you’re wondering how to do this when the party will be outside Town more often than not. Is that an accurate understanding of the question? – SevenSidedDie Sep 16 at 14:35
• @SevenSidedDie Pretty much yeah. Because if the wilderness is mostly untamed, who will the players interact with? – AuWiMo Sep 16 at 16:11
• Thanks, everyone! The original post warned against intelligent creatures, but as long as they are not savages it should be fine and add some nonviolent interaction. – AuWiMo Sep 16 at 16:19

## Every encounter with an intelligent creature can be a social encounter

In some cases, it is really smart for your PCs to first parley before deciding to get violent. You need to discuss with your players what their operating mind set is: when they meet a bunch of humans or humanoids (you don't need to say "you see seven bandits") what is their first reaction?
Fight, flight, or parley?
Let their choices and actions determine whether the encounter is social or not, at the beginning. An encounter can change in character as a consequence of the PC's actions and decisions.

### Social encounters can happen anywhere

In a sandbox campaign, in which category West Marches generally fits, the players are not guaranteed that their encounters are level appropriate. It is in their best interests to do some intelligence gathering on what/who they encounter before resorting to violence.

• For example, a party of four second level characters may run into an Ogre Mage / Oni (CR 7). If they try to take it on they are likely to either experience a TPK, or perhaps they all get captured and taken to that Oni's lair as prisoners or as food.

### Originally, most encounters with humanoids began with parley

When I first began playing OD&D (1975) and as I've found in some campaigns in 5e, and in the editions in between, encountering an unknown party of humanoids (or other intelligent creatures) is more often than not best begun with a parley. We have bargained with orcs and ogres, dervishes and brigands, dwarves and cut purses, and evil high priests. In dungeon crawls we made deals as often as we fought the encountered monsters/NPCs back when we first started.

• Our D&D 5e group tends to still do that unless we have a reason to jump right into combat with a given group - we don't normally default to combat (an exception being the wizard in my brother's campaign, who I now call Quickdraw). Black puddings? Not a lot of room for parley with them. Hungry owl bears rarely go in for small talk. :)

• Caveat: If a party encounters someone we've met before who are our enemies, trying to surprise / ambush them is usually a better choice, or evading them if they are far too powerful or too numerous.

The creatures / NPCs that your PCs meet may have clues and information; they may be potential allies; they may become foes if the social encounter does not go well. As a DM, it's good to introduce unknown friendly, neutral, and even hostile NPCs into encounters as sources of information for where a ruined old tower or lost temple is, where a dangerous swamp is, where a legend says a magic mace is housed in an Ogre's lair, etcetera. That way the world feeds the PCs information - not the DM.

DM Tip: Use the NPC reaction tables in the DMG to help you assess an initial attitude toward the PC's by the NPC's/monsters if the encounter begins as a social reaction - unless you have a good feel for how that creature already feels. My rule of thumb is "don't roll the dice unless you are at a loss for ideas" for the DM side of social encounters. OD&D had a 2d6 roll modified by Charisma that I still use1, but the 5e DMG's NPC reaction section fits into the d20 system a little better. (DMG, p. 245, "Conversation Reaction" table, part of the Social Interaction section; p. 244-246).

### Experience from a sand-boxy published adventure:Tomb of Annihilation

Our party of four 3rd level PCs ran into a party with a dozen soldiers, and a couple of Zhentarim (which we later found out was an Assassin NPC accompanied by a Knight NPC). Not realizing that if we had initiated combat we'd have been destroyed, we parleyed with them anyway and we all went our separate ways. My PC "in world" reason was numbers: there were too many of them for us to take and we had another mission / quest that we were on. Not worth starting a fight in the first place.

Later, when we got to 4th level, we ran into three frost giants who were escorted by pet winter wolves. (If you check the book that's way beyond a deadly encounter for four 4th level characters). We had a social encounter: as it worked out the parley helped further the mission we had been assigned by a merchant prince from Port Nyanzaru. We earned XP for getting that mission accomplished.

### How do I award XP for that?

I'd suggest reviewing the Dungeon Masters Guide (pages 260-261, and Chapter 3) for the options there. The XP award / level is up to you based on the risk / reward of the encounter, and how it moves the narrative forward.

### Some encounters can't be social encounters

When we ran into a T Rex in the jungle there was no parley. His alignment was hungry.
When we ran into a ghast leading a pack of ghouls, there was no parley. It was fight or flight.

1 The OD&D system is fast and easy to use (Men and Magic, p. 12, 1974, TSR, Gygax & Arneson)

The monster will react, with appropriate plusses or minuses, according to the offer, the referee rolling two six-sided dice and adjusting for Charisma:

$$\begin{array}{r|lll} \text{Dice Score} & \text{Reaction} \\ \hline 2 & \text{Attempts to attack} \\ 3-5 & \text{Hostile reaction} \\ 6-8 & \text{Uncertain} \\ 9-11 & \text{Accepts offer} \\ 12 & \text{Enthusiast, Loyalty} +3 \end{array}$$

An "Uncertain" reaction leaves the door open to additional reward offers, but scores under 6 do not.

While that table was based on hiring NPC's or convincing them to do something for the PC, using a 2d6 roll to gage the reaction of an NPC with a + or a - based on the charisma of the PCs works well enough - but it isn't strictly a d20 based outcome as the DMG's NPC reaction section is.

Based on my experience running a sandbox game in OSR, Pathfinder and D&D 5:

# Intelligent monsters

You place some intelligent, and not uncompromisingly bloodthirsty, creatures in your marches. Most intelligent creatures care for their life and are unlikely to attack on sight without provocation. Therefore, they can be talked to, in principle.

Then it is up to your players to approach these creatures in a social manner, or not, as they see fit. If the players decide to attack everything they meet, then there will not be many social encounters. If they choose to talk with everything they meet, there will be many more social encounters.

Note that "talking" does not have to use words. Gestures, warning shots, displays of power, etc., go a long way to saying that you do not wish others to enter your territory, while offering food and treasure acts as a sign of goodwill even without language.

# Consider reaction rolls

In old editions of D&D and most retroclones there is a rule called reaction rolls. These are not the reactions as defined in D&D 5. Rather, when a group of monsters meet the player characters and there is no particular reason for them to be friendly or hostile (they are not brain-eating zombies and do not have an established relationship with the player characters due to previous encounters, for example), the game master rolls to see if they are friendly, careful, hostile, or even outright attack (rare). The roll might be modified by charisma of some player character, or maybe not.

The idea here is that this makes encounters more interesting, as negotiating with monsters is fun, and reminds the game master that not everyone is always hostile, or friendly for that matter.

# Effects

In my experience, once the players know that you are not pulling punches or making things "balanced", they are more than happy to have their characters talk. Though sometimes they also decide that some creatures are to be murdered, even if they are trying to negotiate.

• Have you used these approaches? How have they gone? How did they interact with the WM world/theme? – NautArch Sep 16 at 14:56
• @Thanuir could you please link me to what you mean by reaction rolls or elaborate? Googling only turns up 5e reactions, even with other version keywords. – AuWiMo Sep 16 at 17:20
• @AuWiMo This blog post contains a clone of an early D&D Reaction Table that should work well for a West Marches game. (That table is from BFRPG, which matches the table from the old Basic/Expert D&D set.) It has a bit more explanatory text to help apply each result than the original. – SevenSidedDie Sep 17 at 6:47
• @AuWiMo My answer has the original reaction table for the original D&D game. Many of the OSR/retroclone games us that or a similar table. – KorvinStarmast Oct 8 at 22:08

## A Small Guide to Socializing and Domestic Encounters

It is common for D&D 5e to lead us down a path that everything in the Monster Manual is innately a Monster. The lore behind each monster entry is pared down from previous editions and the stats emphasized. But it is not the DM's job to make a bunch of combat encounters - it's the DM's job to make a narrative. Few narratives are a series of fight scenes connected by safe zone social encounters. With that in mind, let's look at some ways you can improve your narrative by having Domestic Encounters in a harsh wild land.

### Intelligence and Emotion

There's been some good advice already on Intelligent Monsters and Parleying. But just because something has or doesn't have intelligence doesn't detract from its emotions. Take this encounter for example:

Turning the corner you spot a giant bear, a grizzly, thrashing around. A simple glance tells you that it has been the target of a large trap. Despite its size and strength it can't seem to get free.

Having a Druid in your party or a Ranger (or anyone with Speak with Animals) can easily turn this into a Social encounter where you try to calm the beast so that you can free it. Of course, players can turn this into a combat encounter, but it creates a more interesting narrative as a social encounter. Set DCs for how persuasive you can be, let players be free to use skills like Nature or Medicine with Charisma (remember ability checks first then tag a skill to use). This creates a sense that your characters have a little bit of leeway in their decisions and creates good character moments.

In my personal experience, a good encounter that set up both Intelligent Monsters and Emotional Monsters was an encounter I did surrounding a Half-Ogre and his Hogs. In the scenario, the Half-Ogre was trying to calm his Dire Boar and removing an arrow from their haunch. Again, this could easily turn into a Combat Encounter but the players took that moment to look beyond the maddened boar and his Monster buddy. I'll admit, implementation wasn't the smoothest, but it presented an opportunity for the party to socialize with an unexpected source.

### The Town Folk Tale

Every Folk Tale about the woods begins with someone Jack/Jill going somewhere they don't belong. Bring the town folk out from the town and into the wilderness. That Tanner who made the studded leather armor needed materials from somewhere. And that Tanner almost certainly can't afford an Adventurer's salary. Or maybe you get two young bucks who are trying to strike it rich out in the wilderness from one good score. Take this encounter for example:

you hear a high scream from nearby. A little way through the woods you discover 2 humans, maybe 18 winters old. One of them is on the ground writhing in pain and the other is standing atop a rock. You discover the one in pain was stung by a vicious nettle that is strewn across the area here.

The easiest path here is to wander away. But the more interesting narrative is what you get for helping the NPCs. Did they acquire(steal) a map to hidden treasure? Did they hear about a rare resource nearby? The world's your oyster as to what they may know. The main point is that they were willing to go outside of town (like many of the townsfolk might).

In my experience, this is an easier buy into. Players will be more willing to accept talking with the common races than engage in a Parley with Monsters. And it's a good encounter to set up because players are notoriously loose tongued. Have them be followed by an Urchin or a couple of Noble kids. Personally, I went with the kid route because it's also an easy buy in for most 20-somethings to interact with their dumb kid selves. The purpose for doing this was to drive the adventure forward. It set up a trap in the form of stinging nettles, foreshadowing about something moving in the water (a water weird that I planned to jump a player with), and atmosphere.

### Adventurers in their Natural Habitat

Ever think it's funny that you only ever see your adventurers and no others. Sure you find corpses and skeletons in tombs and ruins. But it is a distinct rarity to find another group of adventuring sellswords or the like.

Much like intelligent monsters, the back of the Monster Manual gives you a lot of stats on civilized(used loosely) humanoid Monsters but literally nothing else. In Tomb of Annihilation, you can pretty easily run across some Flaming Fist Mercs among others. They have no reason to be hostile to you. So why would they just jump to guns or even require a lengthy Parley? Of course, some of the entries are innately more hostile than others (tribal warriors, scouts) but they don't just start as combat encounters.
However.
These encounters need a point. Socializing is nice but if it doesn't progress an adventure it can feel trite and contrived. Don't make a random encounter table and just use it without thinking about it. Take from DM Matt Mercer, who when asked about a scene he said "I rolled up a natural scene as a random encounter and thought about what would be best for the scene." (This is in reference to the Jellyfish Bloom during Season 2 Episode 36.)

### In Conclusion

You ask a broad question (without ever actually asking a question). It is a tenet of DMing that you need to separate the mechanical play of dice rolling with narrative elements. It can be hard to put in narrative elements deep in the woods. But as long as you put time and thought into your DM Planning you can make little story elements that help/challenge your players.

Intelligent Monsters are great but don't forget Emotions they feel. Yes a T-Rex can be hungry and a Bear can be in pain, but a Goblin tribe can be in mourning of their lost leader. Townsfolk can flock to areas in between settlements or on the outskirts of society for a variety of reasons like fortune finding or resource gathering. And your players can even stumble upon other adventurers or other civilized yet armed groups.

Just remember, make it fun and make it fit the adventure.

• Have you utilized these ideas? Can you discuss your experience? – NautArch Sep 16 at 19:05
• I didn't know whether to discuss my experience as that can be deemed not answer material. But I'll edit it to reflect past experience with these methods. – GuidingOlive Sep 16 at 20:43
• What do you mean by domestic encounters? Can you please explain what that means? (+1 on this answer, by the way). Also, the Flaming Fist patrol has/had a reason to be hostile/rude to a party that did not, or may not, have the right 'adventurer's charter' in ToA. – KorvinStarmast Sep 16 at 21:23
• In West marches, the game master's job is not to make a narrative, but rather to offer a wilderness sandbox for the players to explore. – Thanuir Sep 17 at 5:54
• Domestic encounters fall under the "Town Folk Tale" section mostly. Bringing home out into harms way. Just because in my case it wasn't a character Contact in harms way doesn't mean you couldn't set that up. More narrowly, Domestic Encounters was the term I picked for Social Encounters involving the home base of the characters even when they were out in the wild. I also totally forgot that the Flaming Fist had a reason to be hostile (because I'm lawful good and made sure our party had their permits). And them being rude, eh, doesn't mean you have a lengthy parley if you have all your stuff. – GuidingOlive Sep 17 at 15:35

A great way to introduce social encounters, is

# Make a Random Encounters Table

The DMG (Dungeon Masters Guide) is a great resource for those. To add more, you can take monsters and make them more intelligent. It is a fantasy rpg, so if you wanted to, you could always add

• Talking animals
• Cannibalistic tribe
• Hermits (people who live in isolation)

to name a few.

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Now compare it with what I'd see if I connected the LED directly to the 220V wire. Too much resistance in the wire leads to less electrical current getting to the plug. If the component is left in the circuit, the readings could be affected by other components in parallel with the component to be tested. Zero Ohms indicates no resistance to the passage of electrons from one point to another. When I put a green LED into it, I saw it glow green - not even orange as would be in case of overcurrent. When working on a circuit board, it may be necessary to lift one of the leads of the resistor from the board to measure the correct resistance of the resistor. We used a transformer which had two long coils of copper wire inside, of different lengths (10m and 750m) and thicknesses. Why measure resistance? Is this correct, or is the multimeter only used to measure resistance of batteries and circuits and not a plain piece of wire? Other factors that can affect resistance readings: Foreign substances (dirt, solder flux, oil), body contact with the metal ends of the test leads, or parallel circuit paths. Breadboard (Radio Shack catalog # 276-174 or equivalent) 6. Since the coils of wire in the transformer are insulated, you can then place the coils in freezing/boiling water with the ends sticking out. How do I experimentally measure the resistance from using dielectric grease on battery posts? Enter the resistance … How would one justify public funding for non-STEM (or unprofitable) college majors to a non college educated taxpayer? I, chapter 8.10. Ion-ion interaction potential in Kohn-Sham DFT. Digital Multimeter Principles by Glen A. Mazur, American Technical Publishers, How to measure dc voltage with a digital multimeter, How to troubleshoot and validate repair of a pump control panel, How to Measure Duty Cycle with a Digital Multimeter, How to Verify Electrical Test Tool Operation with a Proving Unit, Fluke 117 Electrician's Multimeter with Non-Contact Voltage. For each wire plot a graph of resistance (y-axis) against length. Measure the resistance now, and you should notice a difference. 30 gage wire has a resistance of $0.1\: \Omega/\mathrm{ft}$ which is well below what a typical multimeter can read. Start by measuring the resistance of the length of wire. Touch each the end of each lead to the end of the wire you wish to measure. In general, resistance of any one component varies over time and from component to component. 2. With the 4-wire connection, the idea is to have separate wires to deliver the measurement current and to measure the voltage drop over the resistance. Press the RANGE button to select a specific fixed measurement range. The display should show OLΩ because, in Resistance mode, even before test leads are connected to a component, a digital multimeter automatically begins taking a resistance measurement. Such circuits can allow very accurate resistance measurements by comparing the resistance to be measured with resistors of known values. this one at amazon). The meter should display a … Is that not feasible at my income level? To measure true DC, resistance ohmmeters typically use 4 wire measurement. The larger the diameter of the wire, the smaller its gauge number will be. Best results will be achieved if the component to be tested is removed from the circuit. Set your multi meter to ohms and touch the contacts to either end of each of the coils (We got 2.2 ohms and 1500 phms respectively). Is there a remote desktop solution for GNU/Linux as performant as RDP for Microsoft Windows? Accept Cookies. Always check the circuit schematic for parallel paths. You'll likely end up measuring the resistance at the point of contact as much as the resistance of the wire. An Ohm (often symbolized by the Greek letter Omega) is a unit of resistance in electrical systems. Cirris.com will store cookies on your device for login, form submissions, and analytics. Regarding temperature, will a simple home thermometer (battery, not mercury) be sufficient? @DanNeely: yes, nichrome should work well. Turn power to circuit OFF. Talking about the relation between resistance of a wire and its resistivity, the resistivity of a wire is equal to is resistance multiplied by area of cross section overall divided by the Length or in other words you can say that resistance of a wire is equal to the resistivity multiplied by length overall divided by the area of cross section. Ohm’s law defines resistance, “R”, as the ratio of voltage “V” across a component, to the current “I” passing through it: R = V/I To measure resistance, we apply a test current to a wire and detect the voltage drop developed. A "candy thermometer" you can pick up at any grocery store will also go way above 110 F. It might not give the needed precision, but it's cheap and easy to get in a hurry. For example, as the resistance of a heating element rises, the current passing through the element decreases, and vice versa. We got our 1.5k ohm resistor up to 3k+ ohm just going from 25C to 100C. Why can a square wave (or digital signal) be transmitted directly through wired cable but not wireless? Cadmium Sulphide photocell (Radio Shack catalog # 276-1657) 5. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. This would allow you to do the measurement using only a single voltmeter. MathJax reference. 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text:
# Kinematic Equations of Motion

1. ### randeezy

0
Can anyone help me familiarize myself with the Kinematic Equations of Motion?

I need to know them for this summer institute.

Thanks,

Randeezy

3. ### randeezy

0
Thank you, but is it possible that you could explain to me what each is specifically used for, and how to use them?

4. ### Feldoh

I'll take a shot...

The branch of kinematics deals with motion without focusing on the causes of the motion. Generally when we talk about motion we talk about a few different quantities.

When we talk about motion we are talking about quantities that have a direction and a magnitude associated with them. These quantities are called vectors. Quantities that just have magnitude are called scalars.

In kinematics we are primarily concerned with the vectors, position, velocity, and acceleration.

Position deals with the location of something in space. If we take cartesian coordinates and say you are located at (3,4). Well you've just defined a direction (3 units on the x axis, and 4 units on the y axis) and a magnitude (by the Pythagorean theorem the magnitude is 5). Usually position is denoted by x.

Velocity deals with a change in position. The scalar version of velocity is usually called speed. So with velocity we're concerned with a change in distance per unit time and the direction(s) in which this change occurs. Usually we denote velocity with v. The relationship between velocity and position can be seen by:

$$v_{ave} = \frac{\Delta x}{\Delta t}$$ where ave just denotes the average.

Acceleration is a change in velocity of a change of a change in position, if you'd like to think of it in that way. We denote acceleration usually as a.

$$a_{ave} = \frac{\Delta v}{\Delta t}$$.

In basic kinematics we usually only focus on problems involving constant acceleration, otherwise we have to discuss this with calculus. In fact all the equations I'm about to list are easily derived using differential and integral calculus but I'll try to explain them without all that.

Picture this example: You are walking in the positive x direction (still in Cartesian coordinates) with a constant velocity. Notice it's stated as constant velocity. It's constant in the sense that the velocity is not changing which means that there is no acceleration in this example. At some instant in time how do know where you are?

The relation $$\Delta x = vt$$, where $$\Delta x = x_{final}-x_{initial}$$ tells us where you are.

If you start walking at some x_i and want to know where you ended up we can rearrange this equation to give:

$$x_{final} = x_{initial} + v_{ave}t$$ (1)

By a similar argument it can be shown that if we have constant acceleration that the velocity is given by:

$$v_{final} = v_{initial} + at$$ (2)

Now if we have constant acceleration over some time interval then the velocity is changing so we can consider an average velocity over that time interval. This is given by:

$$v_{ave} = \frac{v_{final} + v_{initial}}{2}$$ (3)

Now if we sub equation (3) into equation (1) we get:

$$x_{final} = x_{initial} + v_{ave} t = x_{initial} + \frac{v_{final} + v_{initial}}{2}$$ (4)

Now if we sub equation (2) into equation (4) and some rearranging we get:

$$x_{final} = x_{initial} + v_{initial} t + (1/2) a_{ave} t^2$$ (5)

and

$$v_{final}^2 = v_{initial}^2 + 2 a \Delta x$$ (6)

This equations make up the basics of constant acceleration kinematics...

Equations 1-6 can be used to solve any problem that can be given to you. The reason for so many equations is because each equation is limited to certain quantities. For example it you wanted to find the final velocity knowing the initial velocity, acceleration, and the displacement you could use equation 6 to easily calculate it. Notice equation 6 is time independent. The other equations are listed because they are also independent of one of the quantities typically used.

6. ### randeezy

0
Thanks. I'll see if I can wrap my head around it....

Maybe if I slept with it under my pillow.... lol


text:
1. block 'n springs

A block of mass $m$ lies on a smooth plane $AB$ inclined at an angle $\alpha$ to the horizontal. The block is attached to $A$ and $B$ by a spring on either side (up the slope is positive).

The general solution for the equation of motion is:

$
x\left ( t \right ) = A~cos \left ( t \sqrt{\frac{5k}{m}} + \phi \right )
$

Given:

$
x\left ( 0 \right ) = l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha)
$

I have the particular solution as:

$
x\left ( t \right ) = l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha) cos \left ( t \sqrt{\frac{5k}{m}} \right )
$

The period as:

$
\frac{2\pi}{\sqrt{\frac{5k}{m}}}
$

and the amplitude as:

$
l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha)
$

seems weird... is all this correct?

Is $\phi = \frac{dx}{dt} = 0$ ?

If so, would the average position of the block be the equilibrium position?

2. Originally Posted by billym
A block of mass $m$ lies on a smooth plane $AB$ inclined at an angle $\alpha$ to the horizontal. The block is attached to $A$ and $B$ by a spring on either side (up the slope is positive).

The general solution for the equation of motion is:

$
x\left ( t \right ) = A~cos \left ( t \sqrt{\frac{5k}{m}} + \phi \right )
$

Given:

$
x\left ( 0 \right ) = l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha)
$

I have the particular solution as:

$
x\left ( t \right ) = l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha) cos \left ( t \sqrt{\frac{5k}{m}} \right )
$
These last two lines are confusing me.

If
$x(t) = A~cos \left ( t \sqrt{\frac{5k}{m}} + \phi \right )$

the only way that you can possibly satisfy your initial condition with the solution
$
x\left ( t \right ) = l_0 - \left ( \frac{mg}{2k} \right ) sin (\alpha) cos \left ( t \sqrt{\frac{5k}{m}} \right )
$

is if you aren't using parenthesis.

-Dan

3. Ahhhh! Success!

I want to state first that I do not understand the Physics of this situation. I have tried to "reverse engineer" the problem and determine spring constants, initial positions etc, from the x(t) equation and the initial conditions and have come up with conflicting facts. Perhaps there is just something here I'm not seeing, but I would double check to make sure your general solution here is valid.

I see what has happened here. You didn't list all of your iniitial conditions. You needed to state the condition:
$v(0) = 0$

So what we have here is a general solution
$x(t) = A~cos \left ( t \sqrt{\frac{5k}{m}} + \phi \right )$

with
$x(0) = l_0 - \left ( \frac{mg}{2k} \right ) \cdot sin( \alpha )$
and
$v(0) = 0$

For convenience, I'm going to define
$\omega = \sqrt{\frac{5k}{m}}$
and
$L = \left ( \frac{mg}{2k} \right ) \cdot sin( \alpha )$

$x(t) = A~cos( \omega t + \phi )$
with
$x(0) = l_0 - L$
and
$v(0) = 0$

Let's work with the velocity condition first:
$v(t) = \frac{dx}{dt} = -A \omega ~sin( \omega t + \phi ) = 0$

For t = 0:
$-A \omega ~sin(\phi) = 0$

Thus $\phi = 0, \pi$. Which one we choose is going to depend on which spring is exerting the greater force. You have not given us that information. I will show you how the equation shows this factor.

So the solution will take the form of:
$x(t) = A~cos( \omega t )$ or $x(t) = A~cos( \omega t + \pi )$
with
$x(0) = l_0 - L$

Now, $cos( \omega t + \pi ) = cos( \omega t )~cos( \pi ) - sin( \omega t)~sin(\pi) = -cos(\omega t)$

Thus the solution is
$x(t) = \pm A~cos( \omega t )$
with
$x(0) = l_0 - L$
Which one of the + or - depends on the phase angle.

So to continue:
$x(0) = \pm A~cos(0) = \pm A = l_0 - L$

The size of L is going to determine if A is positive or negative. I leave it to you to decide if $l_0 - L$ can be negative or not. (The general solution for x(t) seems to imply one thing and the initial condition seems to imply another so I cannot reliably predict the size of L.)

I will simply leave this as
$A = \pm (l_0 - L)$

As a curious consequence of the initial condition notice that
$x(t) = \pm ( \pm (l_0 - L)~cos( \omega t)$
The $\pm$ signs are taken to be in the same sense, so they cancel out, leaving
$x(t) = (l_0 - L)~cos( \omega t)$
which, when put into the form of the original variables leads to

$x(t) = \left [ l_0 - \left ( \frac{mg}{2k} \right ) \cdot sin( \alpha ) \right ] ~ cos \left ( t \sqrt{ \frac{5k}{m}} \right )$

(We also see that I was right: You didn't use parenthesis in your original post, as this is obviously what you had meant to write.)

As to the rest, yes, the amplitude is indeed
$l_0 - \left ( \frac{mg}{2k} \right ) \cdot sin( \alpha )$

But $\phi$ has nothing at all to do with dx/dt = v. I don't know why you would say that. This is not true, if for no other line of reasoning, because the units are wrong. And besides
$v(t) = \frac{dx}{dt} \neq 0$
in general as you can see by taking the derivative of x(t).

-Dan

4. Whoops, sorry...

Background:

The first spring attached from $A$ to the block has stiffness $2k$, and natural length $l_0$.

At time $t$, the displacement of the block from $A$, measured along the slope is $x$.

The second spring attached from $B$ to the block has stiffness $3k$ and natural length $2l_0$.

The distance between $A$ and $B$ is $5l_0$.


text:
# How do you find the derivative of 5x^2(2-x^3)^2?

Jul 6, 2015

I found: $20 x \left(2 - {x}^{3}\right) \left(1 - 2 {x}^{3}\right)$
I would use the Product Rule and the Chain Rule to deal with ${\left(\right)}^{2}$, and get:
$y ' = 10 x {\left(2 - {x}^{3}\right)}^{2} + 5 {x}^{2} \cdot 2 \left(2 - {x}^{3}\right) \cdot \left(- 3 {x}^{2}\right) =$
$= 10 x {\left(2 - {x}^{3}\right)}^{2} - 30 {x}^{4} \left(2 - {x}^{3}\right) =$
$= 10 x \left(2 - {x}^{3}\right) \left[2 - {x}^{3} - 3 {x}^{3}\right] =$
$= 10 x \left(2 - {x}^{3}\right) \left[2 - 4 {x}^{3}\right] = 20 x \left(2 - {x}^{3}\right) \left(1 - 2 {x}^{3}\right)$


text:
# Kris's Research Notes

## April 10, 2012

### Nucleation Regime. Part 2

Filed under: GaAs Simulations — Kris Reyes @ 11:41 pm

This note is a follow-up to this post. In that note, we developed an expression for when nucleation is possible by expressing the rate of change in concentration $c_r$ as the net effect of atoms impinging upon the nucleus (given by an As flux and Laplace’s equation) and atoms leaving the nucleus (given by the Gibbs-Thomson relation). In this note, we discuss the problems with the discussion in that post and describe alternative models for surface nucleation.

Recall in our previous note, we expressed the nucleation condition by a function $\phi(F, T, r)$ where $r$ was the radius of the droplet so that $\phi > 0$ implied growth of a nucleus while $\phi = 0$ meant decay. For a fixed flux and temperature, the radius $r_c$ such that

$\phi(F, T, r_c) = 0$

was called the critical radius. We modeled the droplet size as a function of temperature

$\displaystyle r(T) = \left(D_G/F_G\right)^\alpha$,

where $D_G$ was the diffusion coefficient for Ga-on-Ga diffusion and $F_G$ was the deposition rate of Ga during droplet growth. The parameter value $\alpha \approx 1/4$ was found empirically from the simulations. From this, we obtained a necessary condition of $F, T$ for growth: $\phi(F, T, r(T)) > 0$. This is the condition for the droplet to be sufficiently large in order to contain a nucleus of critical size (it is equivalent to $r(T) > r_c$ since $\phi_r > 0$).

From this condition, we considered the critical curve $\phi(F, T, r(T)) = 0$ in the $(\log(F), \frac{1}{T})$ plane and argued (correctly) that it is essentially linear. The slope of the line was given by the expression

$\displaystyle \frac{\alpha E_{Ga} - E_{As}}{k_B} \approx -3\times 10^3$,

for choices of the energy barrier for Ga-Ga diffusion $E_{Ga} = 0.9$ eV and energy barrier for As diffusion in liquid Ga $E_{As} = 0.7$ eV. While the slope is steep, it is not sufficiently steep to match our data. Indeed, we had considered instead (without proper justification) an alternative barrier $E_A = 1.7$ in order to match the data more closely.

Moreover, the condition $r(T) > r_c$, while necessary for nucleation, is not very strong as it is not directly a statement about the onset of nucleation. That is, asking for a domain to be sufficiently large to host a nucleus of critical size is not the same as asking whether (through random events) a nucleus of critical size is favored to form. In addition, the critical radius for GaAs is suspected to be small, on the order of 1 to 2 atoms. Hence, this condition is not that useful to determine a nucleation regime.

## A Simplified Domain

As a simplification, we consider instead a semi-infinite system consisting of a slab of liquid Ga on top of a GaAs substrate:

Let $L$ be the thickness of the liquid Ga. We assume uniformity along the horizontal direction so that the concentration $c(y)$ of As atoms depends only on the height $y$. Within the domain, we assume Laplace’s equation holds:

$\displaystyle c_{yy} = 0$,

and on the boundaries:

$\displaystyle \ell D_{As} c_y(L) = F_{As}$,

$\displaystyle c(0) = c_0$,

where $c_0$ is the equilibrium concentration of As atoms above the solid/liquid interface

The solution

$\displaystyle c(y) = \frac{F_{As}}{\ell D_{As}}y + c_0$

means the concentration gradient is linear within the liquid.

## Equilibrium Concentration

In equilibrium (i.e. when $F_{As} = 0$), the concentration $c_0$ is governed by the solubility of As from the substrate in liquid Ga. In equilibrium, the rate of detachment from the substrate equals that of attachment onto the substrate, and hence $c_0$ satisfies

$\displaystyle R_{D} - c_0 R_A = 0,$

where $R_D, R_A$ are the rates of detachment, attachment respectively, which are in Arrhenius form

$\displaystyle R_D = R_0 \exp\left[- \frac{E_D}{k_B T}\right], R_A = R_0 \exp\left[- \frac{E_A}{k_B T}\right],$

where $E_D, E_A$ are the energy barriers for detachment and attachment respectively and $R_0 \approx e^{26}$ is a constant prefactor. Hence, the equilibrium concentration may be written as

$\displaystyle c_0 = \exp\left[\frac{E_A - E_D}{k_B T}\right].$

We may obtain the energy barriers empirically from simulation results. We consider the configuration depicted above, with $L = 32$ and set the flux $F_{As} = 0$. Allowing the simulation to run for 600 seconds, we measure the frequency and associated rates of attachment and detachment events. The (harmonic) average of these rates $\bar{R}_D, \bar R_A$ are used as empirical estimates for the attachment, detachment rates above. From this, we may solve for the energy barriers, yielding empirical values:

$E_A = 0.33$ eV,

$E_D = 1.29$ eV.

These are very close to the computed energy barriers for the following transitions:

Indeed, most of the attachment/detachment events in the simulations are of this form (I say this because of the way attachment/detachment events are “measured” in the simulations, events other than attachment/detachment from a perfectly flat interface could be counted.)

These barriers yield an equilibrium concentration of $c_0 = 5.5 \times 10^{-10}$ at 523K. We may alternatively measure the attachment/detachment rates in the presence of an As flux when nucleation is first present. These rates and corresponding barriers describe the equilibrium concentration in the liquid when surface nucleation occurs. In this case, the barriers are

$\displaystyle E^{nuc}_A = 0.55$ eV,

$\displaystyle E^{nuc}_D = 1.15$ eV.

These barriers were for $T = 498$K and $F_{As} = 0.1$ monolayers/second and were selected because they minimized the difference

$\displaystyle E_{nuc} = (E^{nuc}_A - E_A) - (E^{nuc}_D - E_D),$

among all conditions which facilitated surface nucleation, and hence described the minimal concentration observed to allow nucleation. Here, $E_{nuc} = 0.36$ eV. This critical concentration $c_{nuc}$ present when nucleation occurred may be written as

$\displaystyle c_{nuc} = c_0 \exp\left[ \frac{(E^{nuc}_A - E_A) - (E^{nuc}_D - E_D)}{k_BT}\right] = c_0 \exp\left[\frac{E_{nuc}}{k_B T}\right].$

## Nucleation Condition

From this, we get a condition for nucleation by comparing the concentration at the surface $c(L)$ with that of the critical concentration $c_{nuc}$. If $c(L) > c_{nuc}$, then surface nucleation will occur. From above, we know

$\displaystyle c(L) = \frac{F_{As} L}{\ell D_{As}} + c_0$,

and

$\displaystyle c_{nuc} = c_0 \exp\left[\frac{E_{nuc}}{k_B T}\right]$,

so that the necessary condition for surface nucleation is

$\displaystyle \frac{F_{As} L}{\ell D_{As}} + c_0 > c_0 \exp\left[\frac{E_{nuc}}{k_B T}\right]$

We may write

$\ell = 1$,

$c_0 = \exp\left[\frac{E_A - E_D}{k_B T}\right]$,

$\displaystyle D_{As} = R_0 \exp\left[-\frac{E_{As}}{k_B T}\right]$,

where $E_{As} = 0.7$ is the energy barrier for As diffusion thorough liquid Ga.

For the growth conditions we consider, $\frac{F_{As} L}{\ell D_{As}} \gg c_0$, so that the conditions above is well approximated (after performing the proper substitutions):

$\displaystyle F_{As} > \frac{R_0}{L} \exp\left[\frac{E_A - E_D - E_{As} + E_{nuc}}{k_B T}\right]$,

or alternatively:

$\displaystyle \log F_{As} > \frac{E_A - E_D - E_{As} + E_{nuc}}{k_B }\frac{1}{T} + \log R_0 - \log L$.

Plugging in all parameter values, we obtain:

$\displaystyle \log F_{As} > -\frac{1.3}{k_B}\frac{1}{T} + 24.5$.

## Simulation Results

We simulated the system discussed at the beginning on a domain of width 128 atoms, a GaAs substrate 16 monolayers thick, and a liquid Ga slab $L = 32$ monolayers thick. We deposited As atoms with flux $F_{As}$ varying between 0.1 and 4.0 monolayers/second at a temperature $T$ varying between 423 and 623 K. We deposited material for 30 seconds, and for each choice of $(T, F_{As})$, we indicate whether surface nucleation takes place. This results in a phase diagram in the $(1./T, \log F_{As})$ plane which we display below, along with the critical curve derived above (shown in green).

We see that all nucleation occurs above and to the right of the critical line.

Recall the radius $R$ of a Ga droplet formed from the deposition of Ga at a rate of $F_{G} = 0.1$ monolayers/second at temperature $T$ is given by

$\displaystyle R = r_0 \left( \frac{D_G}{F_G} \right)^\alpha,$

where $r_0$ is some constant of proportionality and $D_G$ is the diffusion coefficient

$D_{G} = R_0 \exp\left[ \frac{-E_G}{k_B T}\right]$,

for the energy barrier $E_G = 0.9$ eV of Ga-on-Ga diffusion. The power $\alpha = 0.25$. Substituting $R = L$ into the above expression for the nucleation condition yields:

$\displaystyle \log F_{As} > \frac{E_A - E_D - E_{As} + E_{nuc} + \alpha E_{G}}{k_B }\frac{1}{T} + \log \frac{R_0^{1-\alpha}F_G^\alpha}{r_0}$.

Plotting this line against our droplet crystallization simulation results (counting the number of anti-site defects to measure the presence of nucleation), we observe a good fit (fixing $r_0 = 0.5$)

Here, blue crosses correspond to growth conditions favorable to nucleation, while black dots indicate no nucleation present.


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## Timelike Hilbert geometry of the spherical simplex

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#### Abstract

We prove the following result on the timelike spherical Hilbert geometry of simplices: Let $\Delta_2$ be a simplex on the 2-sphere and $\tilde{\Delta}_2$ the antipodal simplex. We show that the timelike spherical Hilbert geometry associated with the pair $\Delta_2, \tilde{\Delta}_2$ is isometric to a union of six copies of vector spaces equipped with a timelike norm, isometrically and transitively acted upon by the group $\mathbb{R}_{>0}^2 \times \mathbb{Z}_3\times \mathbb{Z}_2$. This is a timelike spherical analogue of a well-known result (due to Busemann) stating that the Hilbert metric of a Euclidean simplex is isometric to a metric induced by a normed vector space. At the same time, this gives a new example of timelike space.

### Dates and versions

hal-04015534 , version 1 (07-03-2023)

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• HAL Id : hal-04015534 , version 1
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### Cite

Athanase Papadopoulos, Sumio Yamada. Timelike Hilbert geometry of the spherical simplex. 2023. ⟨hal-04015534⟩

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