- #1
romsofia
- 597
- 310
Hey, I am going to be a TA for an upper division classical mechanics class, and I'm going through the book, and making sure I can derive, or be able to justify equations used in the book. So far it has been okay, but one equation I can't justify. I will post the derivation here (or for those who have it, page 212 in Classical Dynamics by Thornton/Marion).The problem I'm sure we are all familiar with is given a particle in a constant force field, we need to find the path that the particle can take in the least amount of time.
So, let's begin: We assume our particle starts at rest at the origin ##x_1, y_1 ## and goes to some point ## x_2, y_2 ## We ignore friction in this problem, so our field in conservative thus T+U= Constant. We start measuring the particle at ## x = 0, U(0) = 0 ## and with the particle starting at rest, ## T+U = 0## which implies that ## T = - U ## thus, ##\frac{1}{2} mv^2 = mgx ##. This tells us that ## v = \sqrt{2gx} ##. So now let us set up our integral. We will see that ## t = \int^{x_2,y_2}_{x_1,y_2} \frac{ds}{v} ## where ds is the arc length, and v is the velocity. If we expand ds and substitute our velocity, we see the integral becomes: ## t = \int^{x_2,y_2}_{x_1,y_2} \frac{\sqrt{dx^2+dy^2}}{\sqrt{2gx}} ##. Now let us multiply the numerator by ## \frac{dx}{dx} ## to see it become ##\sqrt{(dx^2+dy^2)\frac{dx^2}{dx^2}} \rightarrow \sqrt{dx^2*\frac{dx^2}{dx^2}+dy^2*\frac{dx^2}{dx^2}} \rightarrow \sqrt{dx^2(1+y'^2)} \rightarrow dx* \sqrt{(1+y'^2)} ## (If anyone has an alternate way to justify this step, I'd appreciate it).So now our integral is: ## \int^{x_2}_{0} \frac{\sqrt{(1+y'^2)}}{\sqrt{2gx}} dx \rightarrow \int^{x_2}_{0} \sqrt{\frac{1+y'^2}{2gx}} dx ##Great, so we want a minimum for our time, and we can say our function is : ## \sqrt{\frac{1+y'^2}{x}} ## So let's use our euler lagrange equation. ## \frac{\partial f}{\partial y} = 0 ## because our function doesn't depend of y, so we're left with ## \frac{d}{dx} \frac{\partial f}{\partial y'} = 0 ##. Since the derivative of a constant is equal to 0, we can say that ## \frac{\partial f}{\partial y'} = CONSTANT ##. Now this is the step I can't justify very well, the book just says that ## \frac{\partial f}{\partial y'} = CONSTANT = \frac{1}{\sqrt{2a}} ## which is out of the blue, and would seem like magic to most students, which I don't like. Is there anyway to justify this other than it helps us finish the problem? I don't think any student would have any a priori reason to make the constant equal to this. Maybe I'm over looking something??Thanks for any help.
So, let's begin: We assume our particle starts at rest at the origin ##x_1, y_1 ## and goes to some point ## x_2, y_2 ## We ignore friction in this problem, so our field in conservative thus T+U= Constant. We start measuring the particle at ## x = 0, U(0) = 0 ## and with the particle starting at rest, ## T+U = 0## which implies that ## T = - U ## thus, ##\frac{1}{2} mv^2 = mgx ##. This tells us that ## v = \sqrt{2gx} ##. So now let us set up our integral. We will see that ## t = \int^{x_2,y_2}_{x_1,y_2} \frac{ds}{v} ## where ds is the arc length, and v is the velocity. If we expand ds and substitute our velocity, we see the integral becomes: ## t = \int^{x_2,y_2}_{x_1,y_2} \frac{\sqrt{dx^2+dy^2}}{\sqrt{2gx}} ##. Now let us multiply the numerator by ## \frac{dx}{dx} ## to see it become ##\sqrt{(dx^2+dy^2)\frac{dx^2}{dx^2}} \rightarrow \sqrt{dx^2*\frac{dx^2}{dx^2}+dy^2*\frac{dx^2}{dx^2}} \rightarrow \sqrt{dx^2(1+y'^2)} \rightarrow dx* \sqrt{(1+y'^2)} ## (If anyone has an alternate way to justify this step, I'd appreciate it).So now our integral is: ## \int^{x_2}_{0} \frac{\sqrt{(1+y'^2)}}{\sqrt{2gx}} dx \rightarrow \int^{x_2}_{0} \sqrt{\frac{1+y'^2}{2gx}} dx ##Great, so we want a minimum for our time, and we can say our function is : ## \sqrt{\frac{1+y'^2}{x}} ## So let's use our euler lagrange equation. ## \frac{\partial f}{\partial y} = 0 ## because our function doesn't depend of y, so we're left with ## \frac{d}{dx} \frac{\partial f}{\partial y'} = 0 ##. Since the derivative of a constant is equal to 0, we can say that ## \frac{\partial f}{\partial y'} = CONSTANT ##. Now this is the step I can't justify very well, the book just says that ## \frac{\partial f}{\partial y'} = CONSTANT = \frac{1}{\sqrt{2a}} ## which is out of the blue, and would seem like magic to most students, which I don't like. Is there anyway to justify this other than it helps us finish the problem? I don't think any student would have any a priori reason to make the constant equal to this. Maybe I'm over looking something??Thanks for any help.