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The metric is ds^2=\frac{dx^2+dy^2}{y^2}. I have used the Euler-Lagrange equations to find the geodesics, and my equations are \dot{x}=Ay^2, \ddot{y}+\frac{\dot{x}^2-\dot{y}^2}{y}=0. I cannot seem to find the first integral for the second equation. I know it is \dot{y}=y\sqrt{1-Ay^2}, but I...

Does this only work because we are taking the average? Because if r>1, then dS(y) would be larger than dS(z), right? And so the left integral (without taking the average) would be larger than the right integral (without taking the average)...

\frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)}u(y)\,dS(y)=\frac{1}{|\partial B(0,1)|}\int_{\partial B(0,1)} u(x+rz)\,dS(z) Why does dS(y)\to dS(z) and not dS(y)\to dS(x+rz)? If you want more information, it comes from http://www.stanford.edu/class/math220b/handouts/laplace.pdf on page 8...

Is the answer to my question also because it would be physically irrelevant?

For this to be true, it must be assumed that the x derivative of the wave function is bounded near +\- infinity, or at least grows slower than the wave function goes to zero...how do we know this? Can we not have a wave function that goes to zero but oscillates quicker and quicker as x grows, or...

Wow that's a very interesting derivation, thanks! By the way, has it ever been the case that someone successfully found the laws of a theory based on minimizing the action, without knowing the laws first? Like just by guessing a Lagrangian? I mean when the least action principle came along, the...

But Newton's laws are invariant under Galilean boosts, but kinetic energy is not...does it still make sense to call kinetic energy a scalar if this is an allowable coordinate transformation? Thanks for the replies, good info.

So why wouldn't we have this problem when using Lorentz boosts?

So allowable coordinate transformations just means coordinate transformations that leave the equations the same? What about Galilean boosts? F=ma either way. I mean if Lorentz boosts are allowable in SR, why aren't Galilean boosts allowable boosts allowable in CM?

I've studied classical physics and never heard this before until recently...the allowable coordinate transformations for classical mechanics are rotations and translations. Could someone explain why this is so? What makes these "allowable" (I know they are orthogonal transformations).

I didn't mean it like that, I mean I've seen the derivation of the Lagrange equations without using the action, and that minimizing the action gives the same results.

Ok so you can't use L=T-V in SR, but why does that imply the action is invariant? The action being invariant is the only justification I've read for why the SR Lagrangian is what it is. But in classical mechanics, Galilean transformations are analogous to Lorentz transformations...so if the...

I'm confused. If you launch a ball in a train, and you are in the train, the kinetic energy of the ball will be less than if you were standing outside the train, so the calculated actions will be different, won't they? Which would mean the action isn't invariant. So why would we assume the...

In classical mechanics, isn't kinetic energy not a Galilean scalar? So the action isn't invariant under Galilean transformations, but we can still use it with Galilean transformations. So why must it be a scalar in special relativity? I think I'm missing something...

I know that, but those laws are derived based on the postulate that an object traveling at c in one reference frame travels at c in all reference frames. I want to know if you can get that based on the postulates: 1. All laws of physics are the same in all inertial frames. 2. There is a speed...

How I studied relativity, we postulated that a particle traveling at c in one inertial frame travels at c in all inertial frame. But now looking through a book, I see that they just postulate that all laws of physics are same in all inertial frames, and that there is a speed limit (c). However...

Ok so I get the motivation behind it, but are you saying that you can't actually represent a (2,0) or (0,2) as a matrix. Also I still don't quite understand what the contraction argument has to do with covariance and contravariance. Why does a matrix transform like a (1,1) tensor, ie...

I decided to take out a book and read about tensors on my own, but am having a bit of trouble, mainly with regards to the indexing (although I understand, at least superficially covariant and contravariant). Why is a matrix a tensor of type (1,1)? Obviously it is of order 2, but why not (2,0) or...

I think the bottle of ice will roll down faster. Water expands upon freezing, so the bottle will be bigger, so it will roll faster. Edit:Ok, I thought it was a drinking bottle of water being rolled, so what I said is void.

Ah. So to get hyperbolic geometry, could you just use the first four of Euclid's postulates, and for the fifth just define a different metric? (whatever a geodesic on an hyperbole is) and that would allow you to draw the shapes on a flat surface. Or for Euclidean geometry, replace the fifth...

But on a flat surface, can you prove that parallel lines are at a constant distance from each other, given that the surface is flat, and the first four postulates? If not, wouldn't that mean you can construct parallel lines on a flat surface that aren't at a constant distance from each other...

Ok, thanks. I was just confused because I didn't see him write anywhere that we consider a line to be the shortest distance between two points, which is what I assumed it would be. Can we replace the parallel postulate in Euclidean geometry with the postulate that the figures take place on flat...

I'm reading a book on an introduction to non-Euclidean geometry, and it starts off with the usual Euclidean geometry. I didn't really need a line to be defined in that case, since it's obvious, but now that the parallel postulate has been replaced and we are working with non-Euclidean geometry...

g(x) = -2x if x>0, 2x if x<0, -1 if x=0.

Given a bounded domain with the homogeneous Neumann boundary condition, show that the Laplacian has an eigenvalue equal to zero (show that there is a nonzero function u such that ∆u = 0, with the homogeneous Neumann B.C.). I said: ∇•(u∇u)=u∆u+∇u2, since ∆u = 0, we have ∇•(u∇u)=∇u2 ∫...

Or, 4x500=2000, so 400x500=100x(4x500)=200 000.

Hm, I think I got it, I`ll try to post it soon.

What do you mean? I thought I did write it down, u(x,t)=\frac{1}{2}[f(x-t)+f(x+t)]+\frac{1}{2}\int_{x-t}^{x+t} g(s)ds

anyone?

So, I do not think I did this properly, but if f(-x)=-f(x), then u(-x,0)=-u(x,0), and if g(-x)=-g(x), then ut(-x,0)=-ut(x,0). According to D`Alambert`s formula, u(x,t)=[f(x+t)+f(x-t)]/2 + 0.5∫g(s)ds (from x-t to x+t) so, u(0,t)=[f(t)+f(-t)]/2 + 0.5∫g(s)ds (from -t to t) f is odd, and so is...

I edited my post.

Thank you. You have been a great help. What do you think of my uniqueness theorem though? It does not seem like much of a theorem. Edit: I guess it means that since the integral is decreasing, the solution to the heat equation with a boundary condition is unique...

? My equation is the same as yours, I just used subscript notation. If the stuff in the square brackets vanishes, then as you change t, \int_{0}^{1}u^{2}dx decreases, so it will be smaller than \int_{0}^{1}f(x)^{2}dx since this is when t=0. Is that right? Edit: The term in the square...

I fixed my equation, any way that`s what I get, but I don`t see what to do with it.

\int_{0}^{1} \ (u^{2})_{t}dx = 2uu_{x}(1,t)-2uu_{x}(0,t)-2 \int_{0}^{1} \ (u_{x})^{2}dx This is what I get.

:cry: How do I do that?

So I multiplied the heat equation by 2u, and put the substitution into the heat equation, and get 2uut-2uuxx=(u2)t=2(uux)x+2(ux)2. I`m not sure where to go from there, I can integrate with respect to t, then I would have a u2 under the integral on the left side, but them I`m not sure where to...

I get u(r,t)=a(r)*r*t+r*b(r), where a,b are some functions of r. I don`t see how this means u vanished for the given condition though :S Edit: Oops I didn`t differentiate the v.

I wasn`t before :S, but now I get I get vtt=0, so I guess I can just integrate that and substitute u in.

So do you mean I should I write u=v/r, and insert that into the equation you wrote? Good night, thanks. If so, I get vtt=0, hope that is right!

I get v/r^3 = 0 :S...I edited my post if you didn`t see, can you check if that`s what I`m suppose to get? (but I still don`t yet see how this will show that u vanishes for the given condition, or why we can assume u satisfies laplace`s equation). Thanks for the help btw.

But u(x,y,t) is not in terms of r, can I write u(r,t) instead and use the laplacian on that? If that`s what you mean..or do you mean something else? I can expand it, and get (1/r)(du/dr)+(d/dr)(du/dr), but there is no u*r in this. If I let u=v/r, I get v/r^3.

In polar coordinates, it is (1/r)(d/dr)(rdu/dr)...Am I suppose to take the laplacian of u, and see something?

Yes that`s what it seems like.

We looked at the Cauchy problem in 1 spatial dimension. Most of what we have done is in two dimensions total.

No we`ve only done it for two variables :S...I`ll give it a shot, but can you let me know what substitutions to use please? Do i just introduce a third variable equal to like y-ct or what :S

Cauchy problems for the heat and wave equations. Poisson equation, laplace equation, characteristic curves, dirichlet problem, finite difference method, advection in 1d..

We haven`t learned that. There must be a more simple way. :S

Anyone?

Could someone tell me where to start? I tried separating variables, which got me no where (plus we haven`t technically learned it), and I tried putting it into a form of D^2U, but I couldn`t figure that out either. Please help. Thank you.