Justice Must Be Done.

The prior election must be acknowledged as fraudulent, and steps must be taken to prosecute the fraudsters and restore integrity to the system.

Nothing else matters at this point. Talking about trying again in 2022 or 2024 is hopeless otherwise. Which is not to say one must never talk about this, but rather that one must account for this in ones planning; if fixing the fraud is not part of the plan, you have no plan.

Lawyer Appeasement Section

OK now for the fine print.

This is the WQTH Daily Thread. You know the drill. There’s no Poltical correctness, but civility is a requirement. There are Important Guidelines,  here, with an addendum on 20191110.

We have a new board – called The U Tree – where people can take each other to the woodshed without fear of censorship or moderation.

And remember Wheatie’s Rules:

1. No food fights
2. No running with scissors.
3. If you bring snacks, bring enough for everyone.
4. Zeroth rule of gun safety: Don’t let the government get your guns.
5. Rule one of gun safety: The gun is always loaded.
5a. If you actually want the gun to be loaded, like because you’re checking out a bump in the night, then it’s empty.
6. Rule two of gun safety: Never point the gun at anything you’re not willing to destroy.
7. Rule three: Keep your finger off the trigger until ready to fire.
8. Rule the fourth: Be sure of your target and what is behind it.

(Hmm a few extras seem to have crept in.)

Spot (i.e., paper) Prices

Last week:

Gold $1880.70
Silver $27.63
Platinum $1172.00
Palladium $2834.00
Rhodium $25,500.00

This week, 3PM Mountain Time, markets have closed for the weekend.

Gold $1904.50
Silver $28.03
Platinum $1187.00
Palladium $2878.00
Rhodium $24,400.00

Unfortunately, when looking at the prices only on Friday, you lose some things. Rhodium took a hard hit during the week, dropping below $20,000. At least, according to Kitco. A different site I sometimes check never noticed that drop, so when rhodium came right back up again the downward plunge disappeared, for them. Gold definitely seems to be on an uptrend, and perhaps silver is too. Rhodium is off its all time high, but I am waiting to see if it will truly start to go down.

Torque and Angular Momentum (Part V of a Long Series)

Introduction

Having run out of precious metals to babble about, I’m going to change tacks. If you’ve been here a while, you might remember two postings I did on stars. These were independent posts, having nothing to do with politics (poly = many, ticks = blood sucking bugs) and at least some people enjoyed them. I wanted to go to the opposite end of the scale and talk about a certain sub-atomic particle, but then I realized that the best way to do that would be a very, very, very long post. (And yes, it’s a subatomic particle, but it has a lot to do with stars.) A huge part of it would be explaining where physics stood in 1895, and how three discoveries in the next four years basically overturned things, and eventually led to that subatomic particle, the real star (ahem) of the whole series.

So I decided to break this story up into pieces. And this is the fifth of those pieces.

And here is the caveat: I will be explaining, at first, what the scientific consensus was in 1895. So much of what I have to say is out of date, and I know it…but going past it would be a spoiler. So I’d appreciate not being “corrected” in the comments when I say things like “mass is conserved.” I know that that isn’t considered true any more, but the point is in 1895 we didn’t know that. I will get there in due time. (On the other hand, if I do misrepresent the state of understanding as it was in 1895, I do want to know it.)

Also, to avoid getting bogged down in Spockian numbers specified to nine decimal places, I’m going to round a lot of things off. I use 9.8, below, for a number that’s actually closer to 9.80665, for instance, similarly for the number 32.

Dimensions and Units

I have another go-back.

I’ve been re-educating myself on a lot of this stuff, and I find I’ve been glossing over some critical distinctions.

In particular I’ve been sloppy about the one between dimensions and units. I may very well have never misused the one word to refer to the other thing, but even if not, I haven’t clearly drawn the distinction, and it may have led to some confusion (or at least the sense that I am switching my terminology without any particular reason).

To try to make it clear, let’s take as an example something that, though we’ve not covered it here, is something quite visual. Area. We all know about the area of a room in a house, and we typically measure it in “square feet” which is to say, how many squares a foot on a side will fit into the area. (If it’s irregular, or has fractional measurements, obviously we can cut up our squares into pieces to fit them in.)

The units of area here are in square feet, or feet•feet or ft². But an area could just as easily be measured in square inches, or square yards, or square centimeters, or square meters. There is also the acre (which is 1/640th of a square mile), and the hectare (10,000 square meters or one square hectometer (100 meters on a side), roughly 2 1/2 acres). Those would all be different units for area. But the dimensions of area are length x length. Length is the thing a foot, or a meter, or a furlong, or a light year, actually measures, just as an acre measures area. When we go to talking about dimensions, we’ve divorced ourselves from any particular measuring system, and we’re talking in the abstract.

So, looking at work, in metric the unit of measurement is the joule, which is a newton-meter, which in turn is a kilogram-meter/second² • meter, or abbreviating, kg•m²/s². But the dimensions of work are mass•length²/time², or abbreviating, m•d²/t². (Unfortunately m can stand for the dimension mass, as well as the unit meter, adding to potential confusion.)

Physicists–and hard scientists in general, actually, engage in something called dimensional analysis from time to time. If they’re working on proving a relationship between two different things, the dimensions had better match up properly, or they’d know to go back to the begining and start again. For example, if Einstein, while writing that famous equation, E = mc², had not had the dimensions match up, he’d have wadded up the paper he was writing on and started over.

But there’s also unit analysis. And chemists do this a lot because they have all sorts of specialized units of convenience (like the calorie) and often have to convert from one unit to another. Of course we do this all the time as well, just rather informally, but there’s a way to lay things out so they come out right and we don’t just have to guess.

Let’s say you own a large warehouse, 330 x 660 feet. You want to brag about how big it is because you want to sell it to someone. How many acres does it cover?

As I mentioned above a square mile is 640 acres. We also know that there are 1760 yards to the mile and three feet to the yard. (Pretend you don’t know anything else, in particular forget the number 5,280.)

So you can get the right answer, guaranteed, by doing something like this:

Write: 640ac/mi², mi/1760yd, mi/1760yd, yd/3ft, yd/3ft, 330ft, 660ft.

The last two are the two numbers that go into your area, when you multiply them together you’re multiplying two lengths to get an area. So from a dimension analysis standpoint we’re good with those, but multiplying them together gives you 217,800 ft² and we don’t know how to relate that to acres. So for now let’s not multiply them!

That’s where the rest of the crap I told you to write comes in. Look at each one. They’re all fractions equal to one. In the first case both top and bottom are equal to a square mile (or they’re both equal to 640 acres), so that first term is equal to 1. That’s true for the other four terms too.

So you can take your 330ft x 660ft and multiply it by all five of these and not change it, since they’re all equal to 1.

Let’s combine things.

640ac • mile • mile • yard • yard • 330ft • 660ft
——————————————————————————————
mile²•1760 yd • 1760 yd • 3ft • 3ft

The first thing you can do is cancel out the units. Feet, for instance, appears on the top twice, and on the bottom twice. Remove them all! The same with yards (twice on top and bottom). And mile shows up twice too, so remove them all.

Now you’re left with nothing but “acre” in the numerator, and a bunch of numbers.

640 acre • 330 • 660
——————————————-
1760 • 1760 • 3 • 3

This means when you do that arithmetic, you will have your answer in acres.

You can do some cancelling, you can divide the 330 and 660 by 3 and get 110 and 220. Then it turns out that 1760 is 16•110 (and therefore 8•220) so you can do some more canceling and get 640 acres / ( 16•8 ). This should be readily digestible as 5 acres.

This sort of procedure can be used to convert from metric to US customary, too, provided you know a conversion factor somewhere. For example, I know the metric system weights fairly well, and I also know something about the US customary system weights, but the only conversion factor I can remember is that 31.1035 grams makes up a troy ounce. I know for regular grocery ounces it’s 28-point-something but can never remember. So if I have 500 grams of something, how many grocery ounces does that weigh?

OK, working “backwards” from the desired answer to what we have, start with 16 oz / 1 lb. Then get there from grains, and get to grains from troy ounces, and get to troy ounces from grams:

(16oz/1lb)•(1lb/7000gr)•(480gr/1ozt)•(1ozt/31.1035g)•500g.

When you go through and cancel out all the units, you’re left with oz as the sole standing unit. You can then multiply and divide all those numbers and get that 500 grams weighs 17.63696+ grocery ounces. I only have to remember ONE conversion facter from US customary to metric, so long as I know the conversion factors within the US customary system. (The internal metric ones are much easier to remember!)

OK, that’s out of the way. On we go.

Torque

Up until now we’ve been working with forces that go entirely into making the object move from one place to another.

That’s because we’ve implicitly assumed the force was directed through the center of mass of the object.

However, you know, and have probably known since the first time you tried to push on an object as a baby, what happens when you don’t line up with the object’s center of mass: The object does some combination of turning and moving, and that motion isn’t in the direction you pushed!

Let us, for now, pretend we’re on a frictionless surface (or, perhaps, in orbit, freefall, which for complicated reasons is called “microgravity” by sticklers).

Figure 1A shows an object, and a vector representing a force applied to the object. The dot is the center of mass of the object. The force does not go through that center of mass.

Figure 1B shows the vector resolved into two components, a radial component (through the center of mass) and a transverse component, perpendicular to the radial component.

If your point of view is the center of mass, the transverse component is the one you see as a vector against the background. The radial component looks like a vector pointed right at you. (Figure 1C)

As it turns out the radial component goes into pushing the object, and the radial component’s direction, not the direction of the original force is the direction of the shove, the sort of shove we talked about way back in part 1 when we talked about mass and force.

The transverse component will set the object to turning around its center of mass. (Or, if the object is fixed to a pivot, the object turns about the pivot.) This action is called torque.

Not only that, the induced rotation will be around an axis that’s perpendicular to the radial component of the force. And it will also be perpendicular to the transverse component. (That sounds complex, it really isn’t. It’s definitely one of those picture-equals-a-kiloword things. See Figure 2.)

Different objects will, depending on not just their mass, but also their shape (sphere, donut, cylinder, cube, flat sheet), orientation (it makes a difference whether a cylinder is oriented so the ends are on the axis of rotation, versus whether the axis passes through the “wall” of the cylinder), and mass distribution (is the mass uniformly distributed or is, say, most of it near the center of the object), resist the torque trying to get it to turn. These factors would be multiplied by the mass of the object to come up with something called the “moment of inertia.”

(If you see the word “moment” in a physics term, it has to do with getting something to rotate, either something like this, or, say, a magnet wanting to swing to point towards/away from a magnetic pole somewhere–that’s a magnetic moment.)

The moment of inertia of a point mass, m, at a distance r from the center of rotation (the pivot point), is mr². You can determine the moment of inertia of actual, real shapes (not point masses) either around their own center of gravity or a pivot point elsewhere, by breaking the object up into infinitesimal (almost zero size) pieces, computing each piece’s moment of inertia, then adding them up again. This can be “automated” in large part by using calculus.

Note that moment of inertia seems to have dimensions mass • distance-squared.

For example, a solid, uniform sphere had a moment of inertia about its center of I = ²/₅mr², m being the mass of the sphere and r being its radius. And of course the center of gravity is the center of the sphere. But if it’s a spherical shell (where the thickness of the shell is very small compared to the radius), it’s ²/₃mr². If it’s a thin rod of length L, spinning around a perpendicular line through the center, it’s ¹/₁₂mL². And moment of inertia doesn’t just apply to objects completely free to move. If that rod is attached to a pivot at one end, like (say) the arm of a wrench, the moment of inertia is ¹/₃mL²—four times as much.

Tedious stuff and I had to memorize it then (of course) forget it.

OK, here’s an application. You’re driving down the road and have a blowout. You now have the task of loosening the lug nuts on the wheel so you can change to the spare. Out comes the lug wrench, and you push on it to loosen the nut.

The handle of the wrench is a radial (displacement) vector, and you know, intuitively, that you get the most leverage if you push on it at a right angle, as far out as possible. You’re trying to loosen the nut which not only has a (very small!) moment of inertia but a lot of friction.

If it doesn’t want to come loose, you need more torque. There are two ways to increase the torque: Apply more force to the end of the wrench (making sure it’s perpendicular), or get a longer wrench.

By now you’ll have noticed a pattern. Any time doubling some piece of the puzzle doubles the effect, the formula is going to involve multiplying by that factor. In this case this happens to both the the force and the distance (length of the wrench). We use the Greek letter tau, τ, to denote the torque. Then, if F_t is the transverse component of the force:

τ = F_tl

So you can imagine applying three newtons of force to the end of a two meter wrench (if you can imagine a wrench that long!) or twelve newtons to the end of a half meter wrench, and getting exactly the same torque either way.

Note this isn’t a vector…but really, it should be! Torque absolutely has direction! Not only righty-tighty, lefty-loosey, but you’d never try to remove the lug nut by pushing toward the car or away from it, even though that’s also perpendicular to the wrench. You’re applying a torque by doing this, but it’s not in a useful direction. (In fact if you manage to bend or snap the lug, it’s worse than useless.)

So we have two obvious vectors, a displacement (length) vector, and the direction of the force. We also know from our personal experience that a perpendicular force exerts the most torque because the entire force is transverse. Other angles exert less.

So our vector formula should depend on the angle between the vectors.

Well, we have the dot product. Is that what we want?

No, it’s not. First, the dot product does not give you a vector…and torque absolutely has direction, not just amount (magnitude).

But the second shortcoming is worse. A dot product is zero when the two vectors are perpendicular, and is maximized when the vectors are parallel (and minus that same maximum when the vectors point in exactly opposite directions). That’s the exact opposite of what we want.

The Cross Product

So pardon me for just a few seconds while I bust out in those peals of evil laughter once again. Bwahahahaha!!!

We need the other way to multiply vectors. We need the cross product.

The cross product, represented with ×, is maximized for perpendicular vectors, is zero when the vectors are parallel (or point in opposite directions) and gives you a vector answer. Perfect! It behaves exactly like torque with force vectors in various directions.

And now you know why the dot product is always written with a dot, never a “multiplication sign” like you saw in elementary school. Because when it comes to vectors, those two symbols do not mean the same thing.

First the pictorial description. Then the trig. Then how to compute it given some vectors.

First, a cross product only exists in 3D space. It won’t show up in the Donald Trump 64D Chess Open Championship. Nor will it show up in your 2D diagrams, unless you’re really showing a slice of 3D space, in which case it still won’t show up in your diagram because it will point straight up out of the diagram (or into it).

The result of a cross product between A and B, A×B, will be a vector that is perpendicular to both A and B. To visualize this, make a fat L with your right hand (your right hand, not your left hand), with the thumb sticking out from the hand at a 90 degree angle. Now point the fingers along vector A. Now bend the fingers in the direction of vector B. (Hopefully you don’t have to contort yourself too much for this part. Figure 3 is safe, as long as you follow directions and use your right hand.) Your thumb points in the direction of the cross product, so long as the angle you sweep through is less than 180 degrees. (And for that matter, more than zero–angles less than zero would bend your fingers backwards, anyways.) If it is more than that, the cross product points the opposite way.

This is called the right hand rule. You use your right hand to determine the direction of the cross product.

OK, now here’s a thing about cross products that will seem kind of odd. Do the same thing, only do B×A. Start with your fingers along B, turn your hand around so that you can sweep towards A. Now your thumb is pointed in the opposite direction from before.

Well, now, that’s odd! A×B actually is the opposite of B×A.

A×B = – B×A.

This GIF shows the cross product of two vectors in an animation, it pauses at 0, 90, 180, and 270 degrees but watch the cross product vector grow or shrink when it sweeps through the other angles.

Now for the trigonometric interpretation. The magnitude of the cross product AxB is equal to the sine of the angle between them, times the magnitude of both vectors.

∥A×B∥ = ∥A∥ ∥B∥ sinθ

If A and B are unit vectors, it reduces to just plain sinθ.

Here, if you look at the angle from B to A as the opposite of angle from A to B, you can see why AxB = –BxA, because the sine of a negative angle is the same as the sine of the positive angle—except for a minus sign.

OK, now, if you’re dealing with two “raw” vectors, triplets of numbers, how do you compute their cross product?

It’s quite a bit more complicated than a dot product. However, there are a number of gimmicks to help you remember, and I’ll share my personal favorite.

Let’s take the vectors A = [2,7,3] and B = [5, 4, 6]. Let’s also take three vectors, i, j, and k. These three vectors are a physicist’s best friends, they’re unit vectors along the X, Y and Z axes. In other words, i = [1, 0, 0], j = [0, 1, 0], and k = [0, 0, 1]. (Note: I know I should put hats on them…but those characters are unavailable, so I’m settling for just using lowercase to denote a unit vector.)

Arrange things like this:

i j k
2 7 3
5 4 6

In other words, our unit vector friends, then vector A, then vector B.

Now repeat the first two columns:

i j k i j
2 7 3 2 7
5 4 6 5 4

Now start with the first i and run down and to the right, multiplying: i•7•6, which is to say 42i. This is a vector of length 42 along the X axis.

Do the same for j and k, you should get 15j and 8k.

Add these together, and get 42i + 15j + 8k. But, if you think of it, that’s just the vector [42, 15, 8], isn’t it? OK, save that off, we have to go on to the next step.

Start with k. Run down and to the left and multiply. k•7•5 = 35k. Move to the second i and do the same thing, then the final j. You should get 12i and 12j. You can add these up and get 12i + 12j + 35k. But that’s just [12,12,35], right?

OK, last step. Take the second vector and subtract it from the first:

[42, 15, 8] – [12, 12, 35] = [30, 3, -27]

Now I can say that (unless I made a boo-boo), [30, 3, -27] is perpendicular to both [2, 7, 3] and [5, 4, 6]. Which also means it’s perpendicular to the plane those two vectors are in (two vectors that aren’t parallel or antiparallel define a plane, but if the two vectors are like that then the cross product is zero).

Of course I can check that statement, and so can you. I can dot [30, 3, -27] with each of those vectors, if they are perpendicular the dot product will be zero.

[2, 7, 3] • [30, 3, -27] = 60 + 21 – 81 = 0

and

[5, 4, 6 ] • [30, 3, -27 ] = 150 + 12 – 162 = 0.

(So I guess I didn’t make a boo-boo.)

OK, that’s a kind of lengthy process. If you don’t like that, please, just remember the right hand rule and remember the gif. Those will tell you the direction, and give you a qualitative understanding of what’s going on.

Torque as a Vector

Okay, with that out of the way, back to torque. It’s the cross product of the force and the displacement vector, r, from the center of mass to where the force is being applied.

τ = r × F

We no longer need to specify the transverse component of F, the cross-producting takes care of that.

Let’s re-use the picture from Figure 5-2 and show you the torque vector.

What about the units? F is a force and is in newtons, and r is a distance and is in meters. So torque is measured in newton-meters. Note this is the same units as work, but we don’t ever describe a torque in joules.

In the US customary system energy is sometimes measured in foot-pounds, and torque is quoted in pound-feet, just to keep them distinct.

And remember the direction of the torque is along the axis of rotation it’s trying to create. (In fact, it’s away from the car when loosening a lug nut, try the right hand rule to see.) If you find that counter-intuitive, you’re not alone. You might think the direction of torque ought to be the same direction as the force, or at least the transverse component of it. But on further reflection, that won’t work. In trying to loosen that rusted lug nut, I can be pushing down on a wrench sticking to the left, or I can be pulling up on the wrench while it sticks to the right. Those would be opposite torques if the direction of the torque were the same as the direction of the force. But they’re intuitively the same torque. If that isn’t so intuitive to you, then consider the case where you’re using one of those X shaped tire irons and you are pushing down on the left and pulling up on the right at the same time. The forces should cancel each other out, but clearly the torques they cause do not, they add together.

If you take both cross products left × down and right × up, both give you a vector pointing away from the car, and they add to each other.

Another way to look at is, the force, no matter where it is, is in a certain plane, the plane of the lug nuts. One way to specify a plane is to specify a line perpendicular to it. Of course, this is a vector, and can be anywhere so long as you don’t change its direction or magnitude, so a vector specifies any of a number of parallel planes (like pages in a closed book) depending on where you put it. A vector specifies a specific orientation of a plane, then,

There is actually another way to analyze a torque when you are forced to push on the lever at an angle that isn’t perpendicular. Extend the line of the force, either forward or backward. Find the point where it’s closest to the center of rotation. Use the full force at that point and the distance to that point, and simply multiply (the angle is 90 degrees, so the sine factor is 1). This works because the length reduces by the same factor as you lost by not applying the force perpendicularly; you can prove that geometrically.

Angular Displacement, Velocity, and Acceleration

Imagine a wheel, free to turn, frictionless. You push on the outer rim. That’s a torque. How much does the wheel speed up?

As you might guess, it won’t turn at a high RPM immediately but will speed up as you continue to apply the torque.

You can actually draw a useful analogy here. We talked in Part I about applying a force to am object with a certain mass, causing it to speed up and, given a certain amount of time, covering a certain distance or displacement.

For rotations, we can apply a torque to an object with a certain moment of inertia, causing it to speed up in angular velocity (RPM is a measure of angular velocity) and eventually turn through a certain angle.

It’s actually a pretty tight analogy, everything “works.”

Distance (displacement) is represented by d. Angle, as you’ve already seen, is represented by lower case Greek theta, θ. But here’s the schiff in the punchbowl: the angle isn’t measured in degrees, it’s measured in radians.

A radian is 57.295779513082320876798 degrees.

Approximately.

Where the heck did that number come from? Okay, imagine you’re at a Biden rally, there to heckle, and you’re standing on the edge of one of those silly-ass social distancing circles. And the circle has a radius of 6 feet.

Now walk along the arc of the circle exactly six feet. The angle you covered is one radian. If you were to walk completely around the circle (why not? It’s not as if Biden is worth listening to) you’d cover 6 × 2π feet (approximately 37.699111843 feet), because the circumference of a circle is 2πr. That’s 2π radians. In other words, if you’ve expressed an angle in radians, you’re giving the ratio between the distance along the arc and the radius of the circle. And for reasons I long ago forgot (if I ever truly understood them) this is the most “natural” way to measure an angle, from a mathematical standpoint. (If you take a trig class you will learn like Pavlov’s dogs to recognize, for example, π/6 as being 30 degrees [with a sine of 0.5 and a cosine of 0.866+].)

Since an angle measured in radians is distance along the arc divided by the radius, you’re dividing length by length and a radian is actually a dimensionless value.

Velocity in a straight line is represented by v, the dimensions are distance/time Angular velocity, measured in radians per second, is represented by lower case Greek omega, ω. The units are 1/s, because the angle is dimensionless. Physicists usually write it as s^-1, but I’ve avoided that so far and actually written fractions.

It’s possible to think of ω as a vector! It’s circular motion, though, so we cannot use the instantaneous regular velocity, just like we couldn’t define the torque vector as being in the same direction as the force producing the torque. You can define it as r × v or you can visualize it with a variation of the right hand rule. If the fingers of the right hand are curled in the direction of the circular motion, your thumb points in the direction of the vector. So if something is rotating counterclockwise (as seen by you), the angular velocity vector points towards you. [However, do not think of an angle as a vector; it doesn’t follow certain laws of vector addition. A long story…]

Mass is represented by m. Moment of inertia is represented by capital I: I.

Acceleration (in a line) is a. Angular acceleration is represented by lower case Greek letter α. And is in radians per second squared, i.e., 1/s² or s^-2.

And we’ve already seen F (linear force) and τ (torque).

You can follow through the analogy quite well. But I want to get to a specific destination, angular momentum.

But before we go there, if you’re really alert, you may have noticed one bit of the analogy doesn’t seem like the others.

Angular displacement, angular velocity, and angular acceleration are “sort of” like their linear counterparts, but in all cases, the displacement dimension disappears in the angular quantities.

But with torque, the displacement unit doesn’t disappear, it gets worse! Force is measured in newtons, kg m/s². Torque is measured in newton-meters, kg m²/s². There is a distance-squared in there, versus a distance, not the no-distance-at-all we’d expect from the analogy.

But in fact this is not a problem. A torque acts to accelerate an object with a moment of inertia at a certain angular acceleration. A torque, by analogy with F=ma, ought to be:

τ = Iα

I has units kg•m² and α has dimensions 1/s², combined they are kg m²/s². This turns out to be newton-meters. So the analogy actually continues to hold, thanks to the fact that the mass-analog includes d² in its dimensions.

And this is the case for momentum, and its analog, angular momentum as well.

Momentum is p = mv, yes, it’s a vector. Angular momentum is the same sort of thing, for a spinning object. It’s symbolized by L.

And you might expect angular momentum to be the mass-analog times the velocity analog. And indeed, it is:

L = Iω

This has dimension mass•distance²/time, md²/t or in MKS units, kg•m²/s.

You can rearrange this a tiny bit, and get L = md/t • d.

Notice, though, the first part of that has the same dimensions as momentum. And d of course is the distance.

It’s almost as if angular momentum is just regular momentum, times the distance from somewhere.

And indeed, the formal definition of angular momentum of a particle of mass m at a distance d from some point is:

L = r × p

It’s back!!! Here’s the cross product, again, and I could even just recycle some of my figures from earlier on by changing F to p and τ to L. In fact, what the heck, here’s figure five with the central mass removed.

People have a tendency to think of angular momentum as having to do with spinning objects only, or maybe their outlook is a little broader and they’ll give an angular momentum to one object running in rings around another.

But actually, angular momentum applies to everything. If you’re standing by a highway, and a car goes whizzing past, then from your standpoint the car has angular momentum, even on a dead-straight highway!

That definition above doesn’t say a single solitary thing about angular velocity. It does have linear velocity built into p, however! And the car certainly has a lot of that and a lot of mass so p is huge.

When the car was a mile away, it was headed almost directly at you. The radial component was almost as big as its total speed, and there was almost no transverse component. As it drives by, it’s closer, but all of the motion is transverse. This should sound familiar.

Here, I recycled figure 6, same substitutions. Instead of this being about the torque for a force applied anywhere on a straight line being the same, it’s the angular momentum that’s the same anywhere along a straight line, so long as the object is moving along with constant momentum.

I remember a story problem from a physics book (I cannot find it in my old college textbook, though). A child in a playground is running in a straight line, fixing to jump onto the edge of one of those rotating platforms that have probably been banned from playgrounds now because some idiot thinks they’re white heterosexual male. He has a constant angular momentum (seen from anywhere, but in particular the axis of the platform), then at the instant he jumps onto the platform, his motion is all transverse, and now that he’s revolving about the center of the platform, his motion will remain perfectly transverse. You can mentally relate angular momentum from rotation to angular momentum of an object moving in a straight line this way.

And, here is the freaky thing. You could pick any point on the diagram, and moving objects anywhere on the diagram would maintain the same angular momentum as they move along, relative to that point, as long as they don’t interfere with each other.

Conservation of Angular Momentum

You know, if momentum is conserved in a closed system, maybe angular momentum is also conserved. And indeed that turns out to be the case! Without exception, angular momentum in a closed system, relative to a point in that system, is conserved, and that includes objects in the system spinning about an axis. So even if objects interfere with each other by colliding, or whatever), the total angular momentum will remain the same.

The almost cliche illustration of the conservation of angular momentum is to watch a figure skater spin. When her arms are outstretched, she’s turning slowly, perhaps skating through a turn. Then she brings her arms in, raising them above her head, and suddenly she’s spinning, fast. Then she puts her arms out again and slows down. She’s reducing (and then increasing) the size of the displacement, so the rotation must increase (then decrease) so that the angular momentum will stay the same.

I also remember, but cannot find, a video of an astronaut on Skylab. He’s “standing” perfectly straight, perfectly still. His angular momentum is zero. He then kicks one leg forward, and one leg back, he then sweeps them around 90 degrees–which makes his body turn, but only while he is sweeping his feet around in arcs. Then he returns to standing. He’s managed to turn himself 90 degrees to the right, but he is again motionless. It’s a demo of the conservation of angular momentum because while his feet were moving in arcs, his body had to rotate in the opposite direction to keep his net angular momentum at zero.

And of course there is the gyroscope, but that one is complicated…and I’m going to skip it. Suffice it to say that the force pulling on the axis of the gyroscope is being crossed with the angular momentum vector (which is through the axis), and a vector in a totally different direction results. Optional homework, go find some youtube videos of gyroscopes and see what they have to say.

Applications

But now, let’s apply this to something a lot cooler than lug nuts and kids in a playground and an ice skater. How about an object in orbit around the Earth?

If it’s in a circular orbit, then it’s going to remain moving at the same speed and it’s a no-brainer, the angular momentum won’t change because neither the angular velocity nor the distance will change, and you don’t even need the vector form of the equation, because in a circle the two are at right angles, always. (Of course to verify that the direction doesn’t change, go ahead and take the cross product.)

But what about in an elliptical orbit? At one end of the ellipse, the satellite is closer than at the other end. At periapsis (closest point) and apoapsis (furthest point), furthermore, the motion at these two points is all transverse. So if angular momentum is conserved, the satellite must be moving slower at apoapsis than it does at periapsis. At any other place on the ellipse the satellite has some radial motion, it’s either climbing to its apoapsis or descending to periapsis. So those are harder to analyze.

Kepler’s second law, put forward in the late 1500s (!) describes the motion of a satellite in an elliptical orbit. But it doesn’t just say the satellite slows down the higher it goes, it goes further. It says if you draw a line from the primary through the satellite, and look at the area it sweeps out in some time interval, it’s constant! A fat wedge when the satellite is close in, a skinny one when the satellite is further out.

I always wondered how the heck Kepler figured that out.

I’ve seen how it’s done today; you do some calculus on the r and p vectors after setting their cross product to a constant (because angular momentum is conserved) and it pops out, very readily, in less than five minutes of chalkboard time. (And I don’t remember exactly how, just that I was surprised how readily it occured.)

But that’s not how Kepler did it. He didn’t know about the conservation of momentum, and he didn’t know calculus. No one did at that time, because Newton wasn’t even a gleam in his father’s eye.

So I’m still wondering how Kepler did it.

Another cool application of what we learned today to the orbiting satellite, is that it’s very easy to compute the orbital inclination. The orbit is in a plane. The primary is on that plane too, it’s at one focus of the ellipse. But the plane could have any arbitrary tilt. Maybe it sits right over the equator, and maybe it’s at some tilt (like the tilted circle on a globe that’s supposed to represent the ecliptic somehow–I always thought those were silly because as soon as the Earth rotates a tiny bit, that line is wrong).

If you have a measurement of the satellite’s position at a certain point in time and its velocity (including the direction!) at that same time, and they’re vectors in the right coordinate system (one where x and y point at two places over the equator and z points through the north pole), you can take the cross product. Both of those vectors are in the plane of the satellite’s orbit. so the cross product is perpendicular to that plane.

You can then turn that cross product into a unit vector. Take the dot product of it and the k unit vector (usually taken as pointing through the earth’s axis. (Actually you can save yourself some time. Just grab the third element of the unit vector). That’s the cosine of an angle, take the arccosine to get the angle. You now have the angle between a line perpendicular to the plane of the orbit and the earth’s axis, which is the same as the angle between the plane of the orbit, and the earth’s equatorial plane. Easy peasy, doable with almost no data.

This Week’s Mystery

We have a conservation law. I usually try to come up with an 1895 mystery too. Well we have one.

Consider the solar system. 99.9 percent of the mass is in the Sun, which is about 800,000 miles across, and rotates in about 28 days. That’s a certain amount of angular momentum.

The other 0.1 percent of the mass is in the planets (with a small fraction of that small fraction in asteroids, comets, etc). They’re light weight compared to the sun, but they are far out there, and remember there is an r² term in angular momentum. Mercury, the closest one out, is roughly 100 times as far out from the sun as the sun’s radius. Neptune is almost 100 times as far as that.

It turns out that the vast majority of the solar system’s angular momentum resides in the planets. The Sun is the “one percent” when it comes to mass, but the planets are the “one percent” when it comes to angular momentum.

The mystery is how that came about. And any theory of how the solar system was formed has to explain how the heck all the angular momentum ended up out there in the planets, because angular momentum is conserved. You can’t have the sun just shed angular momentum, it has to be transferred. So if your theory can’t explain that…it can’t explain Jack.

A number of different ideas were proposed as early as the late 1700s, perhaps the most prominent of them is called the nebular hypothesis. It suggests that the solar system formed from a shrinking nebula of dust and gas. The nebula, when initially all spread out, is going to have some very small net rotation (it’s a random melange of particles moving at random velocities, after all; the chance of them all cancelling out perfectly is close to zero). As the nebula shrinks it’s going to spin faster, a disk will end up being formed and the disk will be clumpy and the clumps will eventually form planets because the clumps will tend to attract more matter to them.

Fairly elegant, but it could not explain the distribution of angular momentum, so by the end of the 1800s it had fallen out of favor. I had a book on the planets as a kid (which was probably about ten years old when I was born) that still considered it a mystery, and contained some of the alternatives that had been proposed, including one that suggested the planets had been pulled out of the sun by another passing star’s gravity. (If that one is true, then solar systems ought to be rare, rare, rare.)

Just this once, I’ll give it away now. Unlike back then, today we can actually see some stars forming, and they are surrounded by disks of gas and dust, exactly like the nebular hypothesis. Some astronomers have done a lot of work to refine the nebular hypothesis to make it more detailed and try to address the angular momentum problem…but they still haven’t succeeded. Yet we now know it must be correct, because we can see it happening right now. So the answer to this one is, we still don’t really know. It’s conceivable (though not bloody likely) that the conservation of angular momentum is broken (even though it has been reliably true every. single. time. we have looked at it). More likely, there’s some process at work we don’t understand, perhaps even transfer via magnetic fields.

But we haven’t got to magnetic fields yet…

Obligatory PSAs and Reminders

China is Lower than Whale Shit

Remember Hong Kong!!!

中国是个混蛋 !!!
Zhōngguò shì gè hùndàn !!!
China is asshoe !!!

China is in the White House

Since Wednesday, January 20 at Noon EST, the bought-and-paid for His Fraudulency Joseph Biden has been in the White House. It’s as good as having China in the Oval Office.

Joe Biden is Asshoe

China is in the White House, because Joe Biden is in the White House, and Joe Biden is identically equal to China. China is Asshoe. Therefore, Joe Biden is Asshoe.

But of course the much more important thing to realize:

Joe Biden Didn’t Win

乔*拜登没赢 !!!
Qiáo Bài dēng méi yíng !!!
Joe Biden didn’t win !!!

His Fraudulency

Joe Biteme, properly styled His Fraudulency, continues to infest the White House, we haven’t heard much from the person who should have been declared the victor, and hopium is still being dispensed even as our military appears to have joined the political establishment in knuckling under to the fraud.

One can hope that all is not as it seems.

I’d love to feast on that crow.

Justice Must Be Done.

The prior election must be acknowledged as fraudulent, and steps must be taken to prosecute the fraudsters and restore integrity to the system.

Nothing else matters at this point. Talking about trying again in 2022 or 2024 is hopeless otherwise. Which is not to say one must never talk about this, but rather that one must account for this in ones planning; if fixing the fraud is not part of the plan, you have no plan.

Lawyer Appeasement Section

OK now for the fine print.

This is the WQTH Daily Thread. You know the drill. There’s no Poltical correctness, but civility is a requirement. There are Important Guidelines,  here, with an addendum on 20191110.

We have a new board – called The U Tree – where people can take each other to the woodshed without fear of censorship or moderation.

And remember Wheatie’s Rules:

1. No food fights
2. No running with scissors.
3. If you bring snacks, bring enough for everyone.
4. Zeroth rule of gun safety: Don’t let the government get your guns.
5. Rule one of gun safety: The gun is always loaded.
5a. If you actually want the gun to be loaded, like because you’re checking out a bump in the night, then it’s empty.
6. Rule two of gun safety: Never point the gun at anything you’re not willing to destroy.
7. Rule three: Keep your finger off the trigger until ready to fire.
8. Rule the fourth: Be sure of your target and what is behind it.

(Hmm a few extras seem to have crept in.)

Spot Prices.

Kitco Ask. Last week:

Gold $1768.60
Silver $25.97
Platinum $1205.00
Palladium $2996.00
Rhodium $29,000.00

This week, markets closed as of 3PM MT.

Gold $1831.70
Silver $27.54
Platinum $1257.00
Palladium $2980.00
Rhodium $27,400.00

That is a big breakout for gold on the upside. It went up fifty dollars just since Wednesday. Platinum hasn’t done too badly either! It went up over $30 on Wednesday.

(Be advised that if you want to go buy some gold, you will have to pay at least $200 over these spot prices. They represent “paper” gold, not “physical” gold, a lump you can hold in your hand. Incidentally, if you do have a lump of some size, doesn’t it give you a nice warm feeling to heft it?)

Update: More info on Valcambi Combo Bars

Somebody asked me about the Valcambi combo bars for gold, silver and other precious metals. These are the ones that can readily be broken up into smaller pieces. I was at the Denver Coin Expo yesterday/Friday, and asked someone who had Valcambi bullion at his table if the bars were worth less broken apart, and he said yes, they are worth less.

Still, it might be worth holding a couple of them (and not breaking them apart) in case there’s a fiat money apocalypse. If that happens, you’ll have bigger problems than worrying about how much fiat you’ll get for a 1 gram bar as opposed to the full, 20 (or more) block chocolate bar, and it might be a good way to subdivide your gold holdings when you absolutely need to. Or, you can use silver for “small” change…but even there, it might make sense someday to be able to break down an ounce of silver.

Velocity and Momentum
(Part II of a Long Series)

Introduction

The general outline of this story is to start off by putting you “in touch” with the state of physics at the beginning of 1895. Physicists were feeling pretty confident that they understood most everything. Sure there were a few loose ends, but they were just loose ends.

1895 marks the year when people began tugging at the loose ends and things unraveled a bit. In the next three years, three major discoveries made it plain there was still a lot to learn at the fundamental level.

Once I’m there I will concentrate on a very, very small object…that ties in with stars, arguably the biggest objects there are (galaxies are basically collections of stars). And we would never have seen this but for those discoveries in the 1890s.

It’s such a long story I decided to break it down into pieces, and this is the second of those pieces.

And here is the caveat: I will be explaining, at first, what the scientific consensus was in 1895. So much of what I have to say is out of date, and I know it…but going past it would be a spoiler. So I’d appreciate not being “corrected” in the comments when I say things like “mass is conserved.” I know that that isn’t considered true any more, but the point is in 1895 we didn’t know that. I will get there in due time. (On the other hand, if I do misrepresent the state of understanding as it was in 1895, I do want to know it.)

Also, to avoid getting bogged down in Spockian numbers specified to nine decimal places, I’m going to round a lot of things off. I used 9.8 kg m/s² last time for a number that’s actually closer to 9.80665, for instance, similarly for the number 32.

A couple of Go Backs

Remembering my previous post on mass, one might wonder, “why bother with this sort of thing? Why should people investigate such things?”

We live in an orderly universe. This is a good thing. It gives us confidence that when we set the groceries on the counter, they won’t jump up and bite us. (Not even Darwin’s live food groceries.) It also means that when we drop a car battery on our feet, we know it!

The point of science, when it is being properly done, is to increase our understanding of this world we are in.

All that work on forces and masses and weight was at its core an exercise in breaking down phenomena we see every day into different effects and studying one of them. We removed friction and gravity from the picture and analyzed what was left. Then we took up gravity. What we didn’t get to was showing that friction is itself a force.

With that kind of understanding, you can predict what will happen in an environment where gravity is different, e.g., on the moon and in orbit. At first you won’t be used to it, but then you do get used to it. There are plenty of stories of astronauts who adapt so well to “microgravity” (i.e., the lack of any sort of sensation of “down”) and free fall that they will come home, put the toothpaste on the toothbrush, and drop the toothpaste tube, because they’re used to just letting it go and having it stay there until they’re ready to put things away. They’re applying Newton’s first law (objects at rest will stay at rest) since it is pretty much unmodified in orbit, no need to worry about gravity pulling things to the floor.

Another Go Back. Last time I rather casually used the concept of acceleration, without really going into it much. I assumed some knowledge that many might not have.

You can think of acceleration as change in speed, the faster the change, the greater the acceleration. And it can be a decrease as well as an increase; to a physicist it’s still an acceleration, albeit one which is in the reverse direction to the motion.

I know, too, I said things like “meters per second per second” a lot. That wasn’t a stammer. That’s truly how you measure acceleration. Think about a sports car, able to go from zero to 60 miles an hour in four seconds. That means, on average, every second of that four seconds, the car’s speed increases by 15 miles per hour. So it accelerates at 15 miles per hour every second, which is to say at 15 miles per hour per second. That can even be written as 15 miles / hour / second (the “per” functioning as a division). Or even, 15 miles • 1/hour • 1/sec.

Yes, when physicists do math, they will multiply and divide by units, not just the numbers. They can be divided and multiplied, and even follow the rules of cancelation. (Chemists often have to convert from one unit to another, like calories to Joules, so they really do this sort of thing a lot.)

Notice there are two time units (hours and seconds) in the denominator. It’s kind of funny that they don’t match. You could convert the hours into seconds, like this: Use the fact that there are 3600 seconds in an hour and do the following: 15 miles • 1/hour • 1/sec • 1 hour/3600 sec.

That last term is simply 1 hour divided by 3600 seconds, which is 1. You can multiply by 1 without changing anything. But now that it’s there, the two hours, one in the numerator and the other in the denominator, cancel

Do that cancellation and you have 15 miles • 1/hour • 1/sec • 1 hour/3600sec, and you can re-arrange to get 15/3600 • miles/sec/sec or 1/240 miles per second per second or 0.004167 miles / sec². You could then move on to convert miles to meters (1609 meters is one mile, roughly), and that works out to be 6.7 m/s², and it turns out that car is accelerating about 2/3 as much as gravity accelerates a falling object.

You may have noticed I seem to prefer metric, and that’s because just about every bit of my technical education was in metric; I’m used to it. When I got exposed to the occasional rocketry done in pounds force and pounds mass, it was like trying to use stone knives and bearskins (and you needed to know when you had to multiply or divide by that 32). Obviously it works (we put men on the moon after all), but it just seems cumbersome to me.

(And a note for the fussy, you’ll notice I’ve sometimes spelled out a unit, e.g., “second” and sometimes abbreviate it “sec” or “s.” That last one, just a plain s, is the “official” metric abbreviation for second. Likewise meters, m, and kilograms, kg. But sometimes I spell things out too, as a reminder.)

But the notion of acceleration (however you measure it) is itself depends on speed…actually, it is dependent on velocity.

Velocity

So my second “go back” actually leads us into today’s topics. Let’s flesh out velocity.

Velocity is both how fast something is moving, and in what direction. So it’s actually a more complex concept than the one we talked about last time, mass. The distinction between mass and weight was probably odd to many, but at least mass can be expressed as a number. Velocity? You need a number (your speed), and a direction. (Or maybe you can get by with a bit less…stay with me.)

Direction is easy on a straight highway. You’re either going forward or backward. Since there’s only two choices, and they are opposite of each other, it’s natural to consider the forward direction positive and the backward direction negative. So driving 60 mph in fourth gear is +60 mi/hr, but switching to reverse (after first stopping, since we don’t want your engine to leap through your hood in an ugly mess), and going fairly fast in reverse might get you -10 mi/hr. The local constabulary, using a radar gun, will measure your speed and may depending on circumstances pull you over to inform you how fast you were going and how much you will have to pay for that.

You can even add and subtract your velocities, just like you do with masses. The usual example here is a railroad car moving along on tracks, also nice and tidy and one-dimensional. If the train is moving at 60 mi/hr and a pitcher is in a cattle car, playing catch with someone, and he throws the ball forward at 60 mi/hr, someone standing on the side of the tracks will see the ball moving at 120 mi/hr, because the speed of the train and the ball add together. If he and the catcher switch places, he’s now throwing the ball at -60 mi/hr and the ball is now stationary as far as the guy by the side of the tracks is concerned: +60 mi/hr plus -60 mi/hr = zero.

OK, I’ve used a very limited situation to make a couple of points, but it’s not very interesting in the real world. What about two or even three dimensions?

I’m going to do what everyone else does when explaining velocity in two dimensions: I’m going to use a pool table as my example. It’s the best choice I can think of, and I guess that was the best they could come up with too.

Let’s say the pool table points north-south along its length. A ball is moving directly north at 1 m/s. Another ball is moving directly east at 1 m/s. They have the same speed, but different velocities, because the direction of motion is different.

Now let’s consider a different ball, a red one moving at 1 m/s exactly to the north east. If you think about it, that ball is moving north at a certain rate, and at the same time it’s moving east at a certain rate. Or to put it another way, how fast would a ball (let’s make this one pale blue) moving straight east have to move so that it’s always directly south of the diagonally-moving ball? And how fast would a (purple this time) ball moving straight north have to move so that it’s always directly west of the diagonally-moving ball?

You can do this visually by drawing a diagram like this, then measuring the vertical and horizontal lines. You should get about .7 the length of the diagonal line. Since that diagonal line is 1 m/s, the horizontal and vertical lines should be .7 m/s. (The exact number is actually 1, divided by the square root of 2. That can be derived from the Pythagorean Theorem. To six places, it’s 0.707107, but you will never be able to measure quite that accurately off a drawing you made on a piece of paper.)

You can do this with any velocity, big or small, in any direction. You can break it down into a north-south component and an east-west component.

So any velocity on the pool table can be expressed with numbers, but by writing two numbers, not one. Our diagonal moving ball has a velocity of [ 0.707 north, 0.707 east ] meters per second.

That pair of numbers is enough to do the job of a speed and a direction.

The Vector

And this is what is called a vector in its mathematical form.

You can also represent a vector by picking a scale (1 inch equals 1 m/sec, for instance), and drawing an arrow with the appropriate length, pointed in the appropriate direction. We’ve already done that. You can’t compute things this way but it sure does help you visualize it. And you can get estimates by measuring off the diagram if you’re careful drawing it.

Vectors are considered equal if they have the same length (mathematicians call this the “magnitude” of the vector) and the same direction. There’s no notion built into a vector of “where it starts” and “where it ends.” We can move them around for convenience, especially on those diagrams, just so long as we don’t stretch them or rotate them.

If you think back to last time, I talked about force, mass and acceleration. F = ma. But it turns out the force is a vector. When you push on something, you’re pushing in a certain direction. Likewise, acceleration is a vector too, you’re speeding up in a certain direction. It’s customary to write vectors in bold face (or if on a blackboard, by drawing a line with a little arrowhead over the letter). So it’s actually F = ma.

Mass was not written in bold, because it takes a single number to express it; it doesn’t have direction. (Weight does. Why?) Such plain-old-number quantities are called scalars in distinction to vectors.

Returning to our current topic, velocity is abbreviated v, bold because it’s a vector. So in our diagonally moving ball example, v = [ 0.707 north, 0.707 east ] m/s.

When you take a vector and express it like this, you’ve broken it down into its north and east components. It actually doesn’t matter which two directions you use, so long as they’re perpendicular, but for now let’s stick with north and east.

Even a total distance moved can be a vector. The total distance is equal to the elapsed time, t, times the speed or velocity (depending on whether you want just the distance, or the distance and direction). d = vt.

What happens when you multiply a vector by a scalar, as shown here? What you do on a diagram, is make the arrow that much longer or shorter. Mathematically, you go to each component of the vector and multiply each one by the scalar. In the case of the diagonal moving ball, you have:

d = 5s [0.707, 0.707]m/s = [3.535, 3.535]m. This is how far the ball has gone, relative to where you first started watching it five seconds before. (And it would go right off the pool table, too, if not for the bumpers. More on that later.)

Mathematicians like to do things as generically as possible. So they will write vectors in terms of x and y, rather than north and south. That means they’re not really wedded to any particular orientation. Remember I said it didn’t matter which directions you used, so long as they were at right angles to each other. For convenience when they draw diagrams, the x direction is to the right, and the y direction is upward, the y axis being 90 degrees counterclockwise from the x axis.

You can do more to vectors than just multiply them by a scalar. They can be added together, provided they’re in the same units. (No fair adding speed to force!) This also means they can be subtracted.

Of course when dealing with pure mathematics (as opposed to mathematics applied to physics), generally units are not a concern. Like in the following example.

On a diagram, take your first vector, whatever it is, and then put your second vector so that its tail is right at the head of the first vector. Then draw a new vector from the tail of the first vector to the head of the second vector. That’s the sum of the two vectors. Mathematically, you add each individual element. So [ 3, 4 ] + [ -1, 6 ] = [ 3-1, 4+6 ] = [2, 10].

Conservation of Velocity?

So now let’s go back to the pool table, make the scenario slightly more complicated and see what we can use this whiz-bang vector thing to figure out.

This is pool, after all, balls are supposed to hit other balls. So, if we have a cue ball moving along in the x direction at, say, 1 m/s…or more rigorously [ 1.0, 0.0 ] m/s, and it hits another billiard ball head on, what happens? Well, the cue ball hits the other ball. Then the cue ball stops, and the second ball continues on along the x direction, also at 1 meters per second.

It’s as if the velocity transferred from the cue ball to the other ball, perfectly. So, is it possible we’re on the track of another conservation law, conservation of velocity?

Let’s do a little more investigation. For starters, consider a glancing blow. Let’s have the cue ball moving at 1 m/s in the x direction (ahem) v = [1.0, 0] m/s, and hit the other ball quite a bit off from head on, as shown below.

You’ve seen this happen often enough, you know the cue ball will, in this case, continue moving, up and to the right. And the second ball will move down and to the right. And perhaps one of the two balls moves at a steeper angle than the other. That doesn’t look very much like velocity was conserved, does it? A motion in the x direction turns into two sort-of-diagonal motions?

But actually, when you look at it a bit closer, it looks good. As you can see, we’ve broken the two vectors into their x and y components.

We started with the cue ball moving at [1, 0]m/s, and the other ball (not) moving at [0, 0]m/s. Afterwards, the cue ball is moving at [0.750, 0.433]m/s and the other ball is moving at [0.250, -0.433]m/s.

If velocity is conserved, the sum of the velocities before must equal the sum of the velocities afterwards. These are vectors, and I already told you how to add vectors. So let’s do some addition:

Before: Cue ball [1, 0]m/s + Other ball [0, 0]m/s = [1, 0]m/s.

After: [0.750, 0.433]m/sec + [0.250, -0.433] = [1, 0] m/s.

So it does look like velocity is conserved. Yes, here I could have just made up the numbers to make it work out, but the fact of the matter is in real life, these billiard ball examples really do work out like this.

(And, since I did contrive this scenario, the direction of the cue ball is 30 degrees “up” from the x axis, and its speed is 0.866 m/s. The other ball is moving “down” at a 60 degree angle, at a speed of 0.5 m/s. Those who took some trigonometry might remember there’s something special about 30 and 60 degree angles and the square root of 3, divided by 2.)

A pool player will have played so many games of pool that he knows this behavior in his gut; he knows exactly where to hit the other ball with the cue ball to get the angle he wants, to send that other ball into the corner pocket.

But if it’s a conservation law, it has to hold all of the time, not just in billiards scenarios. And this one doesn’t hold all of the time in billiards, much less in the “real world.”

Nope, No Conservation of Velocity

What happens when a ball hits the bumper? If it hits the bumper head on at 1 m/s, it bounces back at 1 m/s, in the opposite direction. In other words, whatever the vector was before, it’s now a vector in the opposite direction. That’s not conservation!! (And the pool player knows this one too, of course.)

Also, not quite within the realm of billiards, what if the balls are of different weights…er, masses? You already know from your own personal life what will happen. Hit a pool ball with a cannonball and the pool ball will go rocketing away, much faster than the cannonball was moving, and the cannon ball will slow down the tiniest but not stop moving. Reverse the process, hitting the cannonball with the cue ball, and it will barely budge, but the cue ball will bounce back the way it came.

If you want to mess with a pool player, randomize the masses of the balls. Because normally all of the balls have exactly the same mass, at least as close as the manufacturer can make it. In real life very few objects have the same masses. As soon as the masses are different the tidy behavior we illustrated above goes right out the window and the player can’t predict what will happen.

So if you do some experimenting, it seems like what might be getting conserved is not velocity, but something that is the product of mass times velocity. You have to add the mass times velocity, before and after, and that will be conserved. A heavy object will move less under the same impetus from some other object, than a light one would. If mass goes up, velocity goes down to compensate, and vice versa.

Momentum

That product of mass times velocity is known as momentum. And it’s a scalar times a vector, so it’s a vector, too. And for some reason, they chose to symbolize it with p. (They didn’t use m because m is mass, but why did they pick p instead of q or u or…?). p = mv. And if m is in kilograms, and velocity is in meters per second, we can define the momentum as being in kilogram meters per second, kg•m/sec. That way we can avoid the use of a fudge factor, since the units are already consistent with each other. There is, unfortunately, no named unit of momentum like there is with force (the Newton), so “kilogram meter per second” it is.

OK, that takes care of the unequal masses behaving oddly, but what about a ball rebounding off one of the bumpers?

Actually, what’s happening there is that the ball is striking a much more massive object–the pool table. And the pool table is firmly fixed to the entire planet, if nothing else by friction.

So the entire Earth, it turns out, is reacting to that ball hitting the bumper, and picking up motion in that direction, but the earth is so massive that the motion is very, very small. In fact, in order to make the ball rebound, the momentum of the ball is changing by twice its prior value. If the mass of the ball is b, and it was moving at 1 m/s in the x direction before, its momentum was [ b, 0 ]kg•m/s before, and afterwards its moving in the opposite direction with a momentum of [ -b, 0 ]kg•m/s. Net change in momentum is [ -2b, 0 ]kg•m/s. The earth has to make up this change by gaining [ 2b, 0] kg•m/s. But the earth’s mass is much, much, much more than b, so the velocity imparted by the ball to the earth is microscopic. If one goes up the other has to come down to compensate.

One could complain that since we can’t measure the earth’s “rebound” in this case, maybe it isn’t rebounding. But the absence of evidence (i.e., the failure to be able to measure it) isn’t the same as the evidence of absence (i.e., evidence the earth doesn’t actually rebound when the ball hits the bumper). If we had a way of measuring the earth’s rebound that was sensitive enough to show what we expect based on theory, and it didn’t show that change, then we’d have evidence that momentum isn’t conserved. But if we know our measuring is inaccurate enough that we can’t see it even if it’s there, then not seeing it doesn’t mean anything, one way or the other.

Conservation of Momentum

Since this is a part of the story of where physics was in 1895, I’ll put it out, here, that as of that time, no exception was known. Every time we could measure things, momentum was conserved. It was considered a solid part of physics.

Because a vector consists of two components, and the addition rules keep the two components separately, you could treat the conservation of momentum as if it were two separate laws, conservation of momentum in the x direction, and conservation of momentum in the y direction. No one actually does this, but from a bookkeeping standpoint it’s definitely twice as much time with the ledger as conservation of mass is.

And, Oh By The Way…vectors can be three dimensional, too! It’s then a triple number, and the new axis is the z axis, perpendicular to both the x and y axes. The three edges of a cube that meet at the corner are a good representation of this.

Rockets and Guns

Now for an application. How does a rocket work? It works entirely through momentum. Let’s say the rocket’s mass is a thousand kilograms (one metric ton or “tonne”), including the fuel it has on board. And let’s furthermore imagine that it’s out in space somewhere.

A rocket engine works by shooting matter–burnt rocket fuel, to be specific–out the nozzle at very high velocity.

So if the rocket burns one kilogram of fuel plus oxidizer, and shoots the combustion product out the nozzle at 4000 m/s, what happens?

Let’s do this in one dimension for simplicity. The direction the rocket is pointed is positive. And we’re moving along with it, so it looks stationary to us. The rocket, including the fuel, has a momentum of zero.

The momentum of the rocket fuel after it has been burned is 1kg • -4000m/sec. (Negative because the rocket is blowing the exhaust out behind it, the nose points in the positive direction, the nozzle points in the negative direction.

If momentum is conserved, the rocket must now also have a momentumm, this time of +4000 kg•m/s. The rocket has a mass of 1000 kg, so that works out to the rocket now moving at 4 m/s in the forward direction.

So if we want another 4 m/s, burn another kilogram of fuel and oxidizer, right?

Good logic, but there’s a complication here. Because the rocket burned 1 kg of its own mass to get to this point, and now it masses 999 kg, So another kilogram of fuel, adding 2000 kg•m/s to the rocket’s momentum, will actually add slighly more than 4 m/s, precisely 4000/999 m/s, in fact.

For that matter, if you think about it, the mass of the rocket was declining while we did that first burn, so we must have gained a tiny bit more than 4 m/s even the first time around.

That’s quite true, actually, and the real formula for how much velocity a rocket gains by burning some amount of fuel is a bit more complex. But the takeaway is that even in following the other formula, the rocket and its burnt fuel are abiding by the conservation of momentum; in fact it relies on it to operate.

(If you’ve ever heard astronauts, or NASA types, talking about “delta vee”, that’s a reference to the total change in velocity given how much fuel is left, or alternatively, they’re talking about the total change in velocity for a specific maneuver, because that will be equivalent to a cost in fuel for that rocket, with its current mass.)

How about firing a gun? It’s sort of the flip side of a rocket. With a rocket the goal is to make the big thing move, and flinging the fuel out as fast as possible is a means to that end. With a gun, the goal is to make the little thing (the bullet) move, and the gun kicking in the opposite direction is the price paid.

Why does the muzzle flip up on a handgun? Shouldn’t it go straight back, instead of up? It would, except that the line of the barrel does not go through the gun’s center of mass, so there’s a bit of torque there, that causes the whole gun to rotate. If you grip it solidly enough, it kicks your arm up too. Tense up your arm and the entire weight of your body resists the torque and you don’t move much. (Torque, by the way, is another concept that beginning physics studies…)

These particular scenarios are also vivid illustrations of Newton’s third law: for every action, there is an equal but opposite reaction.

This is actually just another way of stating the conservation of momentum. And the first person to put forward the observation that momentum seemed to be conserved was John Wallis in 1671. Newton put forward his three laws in 1687.

Vector fun.

OK, here’s another application of stuff we’ve learned today. You have two identical cannonballs. You drop one. (Hopefully not on your foot.) At the same instant, the other is fired out of a cannon, perfectly horizontally. (Hopefully at a deserving target.) Oh, and you do your drop at the same height as the cannon’s muzzle.

If you’re on perfectly flat ground, which cannonball hits the ground first?

They hit at the same time.

Look at it from a vector standpoint. X is the direction the cannon fires. Y is straight up.

The dropped cannonball starts the experiment with v=[0,0]m/s. The fired cannonball, on the other hand, starts out with v=[200, 0]m/s. I just made that x number up; it doesn’t matter what it is as you’ll see in a moment. (Well, you’ll see it if if I did my job right, today.)

The force of gravity imparts an acceleration of –9.8 m/s² in the y direction, i.e., straight down.

This acceleration can only affect the y component of the velocity vectors, since it’s purely in that direction. And in the y direction, both cannonballs are stationary and in the same place when the experiment starts.

Thus, they both have the same fall, and they will both hit at the same time. It doesn’t matter how fast one of them is moving sideways! And in fact they don’t even need to be the same weight.

I’ve seen demos of this principle done where steel ball bearings are used, in a special little gizmo that drops one the same time a spring shoots the other one out horizontally. You only hear one clack as both balls hit at the same time.

Can’t get the drift.

Last time around, I highlighted what was, in 1895, a standing mystery. Gravitation seemed to work, except they couldn’t figure out what was going on with Mercury’s orbit about the sun. A similar problem with Uranus had led to the discovery of Neptune, so it seemed as if there must be some planet closer to the sun than Mercury, lost in the Sun’s glare, perturbing its orbit. Despite the best efforts of astronomers, that planet (already pre-named Vulcan) had never been found.

This time I’m going to highlight a different little issue.

I mentioned before that velocities were additive, right? A ball thrown by a pitcher on board a moving train ends up moving, relative to the outside observer, faster or slower than the pitcher threw it, by the speed of the train, depending on the direction of the throw.

Can we do this with other things? Sound, for instance, travels at a specific speed (one which varies depending on temperature, humidity, pressure, whether His Fraudulency is on or off his meds, and a host of other factors, but still, a speed that will remain constant until one of these factors changes). Trains have a nice source of sound on them, the whistle (or today the horn). So how fast is the sound travelling in front of the train, and how fast is it travelling behind the train? Measurements from the ground show that they are travelling at the same speed, not different speeds. (They also show that in front of the train the pitch is higher, but that effect is a different rabbit hole. Some other time, perhaps. No, some other time, definitely.) What’s going on here?

It turns out that sound is a wave that travels through a medium, air. It’s going to move at a certain speed relative to the air.

The train is moving, the air is not (unless it’s Wyoming). Thus the sound wave travels the same speed in all direction from the train’s whistle (horn), as seen from someone on the ground on a breezeless day. If some bored passenger on the train were to measure the speed of sound (assuming they’d let him climb around on top of the train in the first place), he’d see the sound move slower, relative to the train, when measured from in front of the whistle, at a normal speed to the side of the whistle, and faster behind the whistle. He could even figure out how fast the train was going by taking the difference between his “in back of” reading and his “in front of” reading and dividing by two. If he were really ignorant of how trains move, he could even prove it wasn’t moving sideways, but rather forward, this way.

Looking at light, we had originally thought it was instantaneous, it was so doggone fast. But then…well, remember Jupiter’s moons from last time? We could predict their motions once we knew Newton’s law of gravitation, actually we could do so from Kepler’s laws of planetary motion, known earlier. (We didn’t even really need to know the mass or distance to Jupiter to be able to do that.) Well, there was one little anomaly. We could predict the motions all right, but the motions were about eight minutes and twenty seconds early when we were closest to Jupiter, and eight minutes and twenty seconds late when we were farthest from Jupiter.

A little thought and someone realized, that the difference was due to the speed of light not being infinite. What we see now going on around Jupiter actually happened at some time in the past, when the light left Jupiter. It then took some amount of time for the light to get here.

The eight minutes and twenty seconds, really in total a 16 minutes and 40 seconds difference, reflect the amount of time it took light to span the entire width of the earth’s orbit about the sun, because it has to cover that much additional distance when we are farther away, versus closer, to Jupiter. (This works out to about a thousand seconds, by the way, a neat coincidence.)

We didn’t know how big the earth’s orbit was, and wouldn’t until the 1760s. Before that we just knew however big it was, light took a thousand seconds to cross it, you could even call it a distance of one thousand light seconds. But once we discovered that the earth’s orbital diameter is roughly 300,000,000 kilometers, we now knew light moved at about 300,000 kilometers per second (186,000 miles per second). This was another product of all that work measuring the solar system that I totally forgot about when writing that article (which is OK, because it fits better here anyway).

Light was, and is, believed to be a wave. So, presumably it goes through a medium, just like sound does. But it must be an otherwise intangible medium, or planet earth would be suffering drag plowing through it. Only light could “feel” that medium. We knew it had to be there, and so we gave it a name: It was the ether.

We might not be able to feel the ether, but we sure as heck ought to be able to measure the Earth’s velocity through it, the same way as the man on the train: by measuring the speed of light in different directions here on Earth.

Measuring the speed of light in a laboratory was difficult to do accurately in the mid 1800s, but we could be much more precise by comparing two different beams of light in two different directions, and seeing what the difference in their speed is.

Michelson and Morley tried this in 1887. They found no difference in the speed of light no matter which way they measured.

Well, it’s possible that at that point in our orbit, we just happened to be stationary with respect to the ether. But that couldn’t be true a couple of months later, because the earth would at the very least be orbiting in a different direction, so they kept trying.

Others have tried too, with much better equipment.

No difference. Ever. No one has ever “got the drift.”

What’s going on here? Well, that, like Vulcan, was a mystery as 1895 dawned.

Obligatory PSAs and Reminders

China is Lower than Whale Shit

Remember Hong Kong!!!

中国是个混蛋 !!!
Zhōngguò shì gè hùndàn !!!
China is asshoe !!!

China is in the White House

Since Wednesday, January 20 at Noon EST, the bought-and-paid for His Fraudulency Joseph Biden has been in the White House. It’s as good as having China in the Oval Office.

Joe Biden is Asshoe

China is in the White House, because Joe Biden is in the White House, and Joe Biden is identically equal to China. China is Asshoe. Therefore, Joe Biden is Asshoe.

But of course the much more important thing to realize:

Joe Biden Didn’t Win

乔*拜登没赢 !!!
Qiáo Bài dēng méi yíng !!!
Joe Biden didn’t win !!!

His Fraudulency

Joe Biteme, properly styled His Fraudulency, continues to infest the White House, we haven’t heard much from the person who should have been declared the victor, and hopium is still being dispensed even as our military appears to have joined the political establishment in knuckling under to the fraud.

One can hope that all is not as it seems.

I’d love to feast on that crow.

Physics?

Part 1 was last week. I intended to do Part 2 this week, but both of my drafts are an utter mess. I’m going to keep plugging away at it, though, so hopefully next week.

Justice Must Be Done.

The prior election must be acknowledged as fraudulent, and steps must be taken to prosecute the fraudsters and restore integrity to the system.

Nothing else matters at this point. Talking about trying again in 2022 or 2024 is hopeless otherwise. Which is not to say one must never talk about this, but rather that one must account for this in ones planning; if fixing the fraud is not part of the plan, you have no plan.

Lawyer Appeasement Section

OK now for the fine print.

This is the WQTH Daily Thread. You know the drill. There’s no Poltical correctness, but civility is a requirement. There are Important Guidelines,  here, with an addendum on 20191110.

We have a new board – called The U Tree – where people can take each other to the woodshed without fear of censorship or moderation.

And remember Wheatie’s Rules:

1. No food fights
2. No running with scissors.
3. If you bring snacks, bring enough for everyone.
4. Zeroth rule of gun safety: Don’t let the government get your guns.
5. Rule one of gun safety: The gun is always loaded.
5a. If you actually want the gun to be loaded, like because you’re checking out a bump in the night, then it’s empty.
6. Rule two of gun safety: Never point the gun at anything you’re not willing to destroy.
7. Rule three: Keep your finger off the trigger until ready to fire.
8. Rule the fourth: Be sure of your target and what is behind it.

(Hmm a few extras seem to have crept in.)

(Paper) Spot Prices

Last week:

Gold $1778.20
Silver $26.07
Platinum $1236.00
Palladium $2914.00
Rhodium $28,000.00

This week, 3PM Mountain Time, markets have closed for the weekend.

Gold $1768.60
Silver $25.97
Platinum $1205.00
Palladium $2996.00
Rhodium $29,000.00

(Be advised that if you want to go buy some gold, you will have to pay at least $200 over these spot prices. They represent “paper” gold, not “physical” gold, a lump you can hold in your hand. Incidentally, if you do have a lump of some size, doesn’t it give you a nice warm feeling to heft it?)

Of these, palladium continues its steady climb. In fact, on Thursday it had closed at $3012, above three thousand dollars for at least the second time in the last week.

I did, long, long ago, buy some palladium, not a huge amount. (But I can drop it on my foot. I won’t hurt if I do so, because, like I said, not a huge amount.) I even told the guy I was buying it because it had gone over 900 bucks earlier (which at the time was higher than gold had ever been), maybe it would do so again. He scoffed, and explained that was due to a temporary supply interruption from Russia, unlikely to ever happen again. But now, it’s worth well over ten times what I paid for it. In fairness, I had to hold onto it for over 20 years for it to finally do what I wanted it to do, so this isn’t the short of thing that is a “short term” play.

I knew someone who, about the same time, wanted to find a way to get into rhodium, but back then there was no such thing as a rhodium one ounce bar. I don’t know if he managed to find a way to do it, but if he did, he has done very, very well, even if it took a couple of decades for it to finally move.

Obligatory PSAs and Reminders

China is Lower than Whale Shit

Remember Hong Kong!!!

中国是个混蛋 !!!
Zhōngguò shì gè hùndàn !!!
China is asshoe !!!

China is in the White House

Since Wednesday, January 20 at Noon EST, the bought-and-paid for His Fraudulency Joseph Biden has been in the White House. It’s as good as having China in the Oval Office.

Joe Biden is Asshoe

China is in the White House, because Joe Biden is in the White House, and Joe Biden is identically equal to China. China is Asshoe. Therefore, Joe Biden is Asshoe.

But of course the much more important thing to realize:

Joe Biden Didn’t Win

乔*拜登没赢 !!!
Qiáo Bài dēng méi yíng !!!
Joe Biden didn’t win !!!