Casey Wood

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This is interesting, I just happened to be reading about this the other day in one of my mathematical physics texts. The case of 0 ^ 0 stood out. As a mathematician, I want to say that it's indeterminant or undefined, yet as a physicist I want to say: nope it's equal to one. If we examine the function of x * ln(x), we see that as x approaches zero from the right, y approaches zero. Thus we can see that x * ln(x) is getting closer and closer to unity as x approaches zero. If we multiply unity by nothing then our value should remain unchanged and we should have one.
Mathematically, start with a ^ m = e ^ ( m * ln(a))
take a = m = x to give x ^ x = e ^ (x * ln(x)) and examine the behaviour of x * ln(x) approaches 0.
Taken directly from the book: "by comparing the representation of the ln(x) as the integral of t ^ 1 with the corresponding integral of t ^ ( 1 + b ) for any positive b, it can be shown that x * ln(x) tends to zero as x tends to zero and so x ^ x tends to zero in the same limit."  from Foundation Mathematics for the Physical Sciences K.F. Riley and M.P. Hobson
I have to admit though, I'm still curious what folks here think about this definition, and if we can really say that we have unity for 0 ^ 0.
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My unit analysis gave (Amps ^ 2 * s ^ 4)/( kg*m ^2 ). This is indeed capacitance. The unit for resistance will be almost identical to the reciprocal, but with s ^ 3. Thanks for checking sensei, I need to adjust my number.
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Gosh, I'm probably missing something obvious here but, capacitance is defined as Charge over Voltage, and ofcourse we can say that V=IR. So if C equals Q over IR, it seems like the easiest thing to do is to increase the resistance. Lengthen the wire, use a material of lower conductivity, play with varying scales of resistors etc. .... Alright, after some calculation, I have you at a resistance of roughly 24.3 KOhms Does that work?
I'm sure I've messed something up in a glorious fashion here, as this seemed far too simple. What did I miss?
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Hi Asimov,
I think I can take the first question regarding jets. This is due to the law of conservation of momementum (really just a sublaw of the law of conservation of energy). The flow at the at the beginning of the nozzle must be equal to the flow out of the nozzle. We can define flow as Q=Av, where A is area, and v is velocity. For the momentum to be conserved Qin must equal Qout. We can state conservation of momentum by taking a unit Volume and therefore by letting density equal mass we have: dav_in  dav_out = 0. Where d is density (usually denoted by the greek letter rho). Because the surface area of the nozzle has decreased, the velocity of the expelled gases must increase to conserve momentum. As that velocity increases, the jet gets more thrust and must accelerate forward according to Newton's Third Law. The jet will remain at constant velocity when the sum of the forces acting perpendicular to its motion are equal to its thrust (assuming that it is flying at constant altitude).
In fact, as I read your questions again, I'm thinking that all of these ideas come down to the conservation of energy. Yet, I focus mostly on fluid dynamics, so I'll let some other folks go after the remaining questions. This is a fun post. If you would like me to bring in some more advanced mathematics or derivations let me know. Fun Stuff!
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Hi ajb,
This topic has made me consider just what it is I'm doing half the time. When I'm working with water and experimenting in the critical regions of the Reynolds number, I've always felt like I was doing empirical work, and my math has been simplified significantly to aid that work. The experiments usually just call for algebra, and sometimes a little calculus. (Standard Physics) I take my results back to the office, and begin playing with the equations to see if I can adjust the models in a way that will aid the next experiment. My office work almost always involves vector and tensor analysis, so I think we could classify it as theoretical.
Yet, when I'm doing mathematical physics, I am studying the definitions of the vector space, linear independence, tensor operations, transformations etc. I'm getting a real feel for the space my equations live in, and I come away with a much deeper understanding of what those equations mean. Ofcourse, that deeper understanding aids in my ability to manipulate further equations and explore additional options in my theories. Rigorous mathematics, associated with physical concepts, are typically the way I've chosen to describe mathematical physics. I agree with you completely though... the "crossover" between the subjects may well just be "a matter of taste".
Out of curiousity, what would you classify as "high brow" mathematics?
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For me, it has always come down to rigor. For example, a "nonmathematical" physicist might be someone who is perfectly content leaving out certain proofs or definitions. Whereas a mathematical physicist desires those proofs and definitions, as they are essential for understanding what we're doing as our equations become more and more mathematically complex. I would say that mathematical physics is a subject for those who wish to pursue the theoretical implications of physics, whereas, nonmathematical physics is more appropriate for those who prefer empirical methods.
Here's an example: In standard physics textbooks we are often told about the position vector and its applications. In three dimensions, it is usually given as the vector r (t) = xi + yj +zk. In standard physics we'll often take the difference of two position vectors to determine a displacement vector for something that has moved. This works fine and solves many problems associated with three dimensional motion in cartesian coordinates.
Then, a mathematical physicist comes along and asks the question: "Yeah, but isn't the vector space defined for ndimensions. And furthermore, if I do a simple linear transformation of my coordinate system, then wouldn't that change the magnitude of the "position vector".?" The answer is yes, it does. "So, can we even consider the "postion vector" a vector at all?" The answer to that is no, not really. That is because as a tensor of rank 1, a vector is an entity that must be invariant under coordinate transformation. So let us now redefine the "position vector". The "position vector" can be equated to a radius arrow that points to a location in space, and is useful for finding points in a given coordinate system. As our radius arrow will have to be defined for ndimensions, let us use indicial notation:
r = sum from j=1 to n of X j upstairs times L j downstairs such that j = 1,2,3... n
Where X is a cartesian dimension for a single reference frame, and L is a linearly independent unit vector. Noting that: A scaling factor can be introduced to act on X j upstairs.
My explanation is rather basic, but I hope it can offer some insight into the difference between standard physics and mathematical physics.
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hmmm... Is this an elliptic equation in the form ax^2 + 2bxy +cy^2 = 0, where a, b, and c are constants? If so then what you have is called a real quadratic curve. Such a curve makes either an ellipse, a hyperbola, or a parabola. It all depends on the sign of the quantity (acb^2). If (ac  b^2) is positive, then you have an ellipse, if it is negative you have a hyperbola, and if it is zero you have a parabola. Remember that elliptic equations have smooth solutions, and their stationary points are saddles, not maxima or minima.
Also remember the rules of the second partial derivative test, or DTest:
Suppose that a function z = f ( x, y ) and the first partial derivatives and second partial derivatives are all defined in an open region R and that ( a, b ) is a stationary point in R such that
f_x ( a, b) = 0 and f_y ( a, b) = 0
Define the Quantity D as follows:
D = f_xx ( a, b) * f_yy ( a, b )  [ f_xy ( a, b) ] ^ 2
Condition 1: If D > 0 and f_xx ( a, b ) > 0, then f ( a, b ) is a local minimum.
Condition 2: If D > 0 and f_xx ( a, b ) < 0, then f ( a, b ) is a local maximum.
Condition 3: If D < 0, then f ( a, b, f ( a, b ) ) is a saddle point.
Condition 4: If D = 0, then this test gives no information.
I highly recommend doing some curve sketching under these conditions to get a proper feel for the geometry of what is happening. Hope that helps ya out a bit.
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Hmmm... Do you think the transistor is a worthless invention? Seems pretty useful to me.
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The wine problems are related to filling a vat with wine at some height, and then using the pressure generated from the potential energy to fill wine jugs at distances along a pipe. The jugs further down the line would of course fill more slowly, regardless of how much pressure could be generated at the beginning of the line. A logical conclusion was to increase the height of the tank, but they encountered even more problems when they did so. The concepts associated with (dare I say) energy loss due to friction, were not well understood until Stokes solved viscocity, (and for that matter, still aren't all that well understood. LOL) The texts I have read on the subject treat the problems associated with winemakers as mostly anecdotal. Nonetheless, it is reasonable to conclude that there were strong motivations from the wine sector to have a better explanation for dynamic flows.
I saw that the question had been posted some time ago, yet the topic is the Navier Stokes Equations. Those equations are so important to physics, mathematics, engineering, life etc., that I was compelled to keep the post active. Thanks for your reply.
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Hmmm... Is this a theory that assumes motion is not continuous?
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When I was in college, I derived the NavierStokes equations beginning with Pascal's Principles. While I'm certainly not up for delivering my full 10 page derivation tonight, let's see If I can give ya the basics.
First, understand that we need to derive an equation of motion for a fluid, that is F=ma. Beginning with the definition for pressure we have, P=F/A thus F=PA. Let's take an outward unit normal vector to a circle, and call it df. df has the unit of area, so we can now integrate P dot df. Applying the divergence theorem tells us that we now have an integral with respect to Volume of the pressure gradient. (this should be a negative quantitity so that pressure decreases as area increases). Now that we have a force, the negative pressure gradient, let's see what we can do with the RHS of F=ma.
If I let my Volume be a unit volume, then because density equals mass over volume, I can say that our mass is equal to density or rho=m/V. Really, all that's left is to find the acceleration(s). This is where it gets tricky for most people. What I'll need are two accelerations, one to explain the motion of the fluid as a whole, and one to explain the motion of fluid particles. The acceleration of the fluid as a whole will simply be a temporal derivative of velocity, delta v over delta times dt, where delta indicates a partial derivative.
Finding the acceleration of fluid particles, requires that I define a displacement between those particles, call it dr. Because we are dealing with particles in space, I want something that is capable of explaining that situation. Divergence will suit our needs perfectly. This divergence will act parallel to our velocity field (or velocity profile if you prefer). So you should have (grad dot dr)v. Applying the chain rule from multivariable calculus here gives the "material derivative" (our change in velocity associated with the particles in the fluid), I like to think of it as the spatial derivative. You should find that it is equal to (grad dot v)v. Thus the acceleration, dv/dt of the whole mess is: dv/dt=delta v over delta t + (grad dot v)v. Finally we can allow an external body force to act on our system (such as gravity or a centrifugal force). We'll denote this with a lower case f, and put the whole equation together.
f  (grad P) = rho * (delta v over delta t + (grad dot v)*v) This is F = ma
The above equation is Euler's Equation for a fluid. You may notice that it is almost identical to the NavierStokes equation for incompressible flow, yet it is missing what is arguably the most important term, viscocity. Deriving viscocity can be a rather long and technical process, so right now I'm tempted just to tell you to add it to the LHS. The gist for viscosity is this however. All of those little infinitessimal fluid layers that build up our velocity vector field are sliding across each other. Stokes realized that in doing so they must be obeying some sort of concept of friction. In short the fluid itself is resisting the shear force of the pressure gradient that is acting perpendicular to it and causing the fluid's motion.
If your interested in how to derive the viscocity, let me know. I'll be happy to type up something here, it is afterall the explanation to solving the problems that winemakers were having during the time. And what could be more important than using physics to solve problems associated with wine.... or water I suppose. LOL
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This is a fascinating question that I'm sure has puzzled many calculus and physics students. The best explanation comes from understanding what happens geometrically when we integrate. For example, the integral of a volume is an area, the integral of an area is a displacement, thus the integral of a displacement is a point. The ideas associated with considering the unit of the meter second as a unit of spacetime are valid.
The integration of the displacement equation gives an infinitessimally small point in spacetime that can be thought of as a coordinate. Some might consider such a point to be a "singularity". Nevertheless, the idea is useful for allowing us to place a clock at each one of those points, so that we can analyze concepts such as time dilation due to the constancy of the speed of light, and the so called "cosmic speed limit". So think of it this way, the integral of a displacement yields a point in spacetime.
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Common underlying physics of propulsion from kayaks to jets
in Engineering
Posted · Edited by Casey Wood
Hi Asimov,
It appears that we are talking about two different points in our control volume. So, let the system for flowin be defined exactly in the front of the cowl, and flowout at the exterior of the nozzle. We can simplify our system in that fashion. Introducing heat in the system due to combustion can then be taken as nominal. True, combustion increases the velocities of the compressible fluids moving through the control volume; by adding fuel, we have also added mass, which means that our densities inside the control volume have increased. If there has been an increase in density inside the control volume, then velocity must increase at the entrance to the control volume; remembering that the areain, and ambient air densities remain constant. So, the flowin is Qin and it is equal to Qout. If I increase Qout, then Qin increases. By adding the jet fuel inside the control volume, I have increased Qout, and I can suck more air into the jet, thereby increasing the velocity of the fluid moving through the control volume.
These concepts are not intuitive and they are difficult to understand. I have often argued that fluid mechanics is the third realm of modern physics, relative to quantum mechanics and general relativity. Do not be discouraged if they don't come to you as easily as GR and QM. If my explanation has not been satisfactory, then I am happy to take any more questions you may have on this subject.