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Post by mondo on Jan 21, 2023 18:45:13 GMT
On page 242 in the paragraph before equation (32) author says that the differential equation governing the motion of a particle hanging at the end of the thread and moving on an elliptical trajectory is $\ddot{z} = -z$ I wonder how was it derived? There is no proof of it so I assume it is a known fact?
Additionally, why the solutions is $e^{\pm it}$? Finally I also don't understand the general solution $z = pe^{it} + qe^{it}$ what are $p,q$? Why the general solution is a sum?
Thank you.
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Post by Admin on Jan 21, 2023 21:18:42 GMT
On page 242 in the paragraph before equation (32) author says that the differential equation governing the motion of a particle hanging at the end of the thread and moving on an elliptical trajectory is $\ddot{z} = -z$ I wonder how was it derived? There is no proof of it so I assume it is a known fact? This is Newton's 2nd law of motion: the Force is proportional to the mass multiplied by the acceleration. For a full and rigorous explanation you need to read a textbook on solving differential equations, but if you substitute $e^{it}$ or $e^{-it}$ into the differential equation you will see that it satisfies the equation and the same thing with $pe^{it}$ and $qe^{-it}$. and then in the theory of differential equations you will see that the general solution is the sum of these two particular solutions. These are things which physicists and applied mathematicians have tattooed on the back of their hands.  Vasco |
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Post by mondo on Jan 22, 2023 22:19:36 GMT
This is Newton's 2nd law of motion: the Force is proportional to the mass multiplied by the acceleration. Yes I recognized it $F = ma$, and here $a = \ddot{z}$ (acceleration) is clear but why on the other side we have $-z$? Is that because the force towards the origin is $|z|$ and in the differential equation we need the opposite direction, plus a vector not it's modulus hence $z$ and not $|z|$? For a full and rigorous explanation you need to read a textbook on solving differential equations, but if you substitute $e^{it}$ or $e^{-it}$ into the differential equation you will see that it satisfies the equation and the same thing with $pe^{it}$ and $qe^{-it}$. and then in the theory of differential equations you will see that the general solution is the sum of these two particular solutions. These are things which physicists and applied mathematicians have tattooed on the back of their hands. Heh, ok thank you for a hint, I will check some supplement material to get that. Thank you. |
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Post by mondo on Aug 18, 2023 6:11:57 GMT
This is Newton's 2nd law of motion: the Force is proportional to the mass multiplied by the acceleration. Yes I recognized it $F = ma$, and here $a = \ddot{z}$ (acceleration) is clear but why on the other side we have $-z$? Is that because the force towards the origin is $|z|$ and in the differential equation we need the opposite direction, plus a vector not it's modulus hence $z$ and not $|z|$? Vasco, do you agree with what I said here? |
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Post by mondo on Aug 18, 2023 7:10:06 GMT
Beside this, I also wonder how on page 243 author can conclude that the formula for $E$ can be rewritten as $\frac{1}{2}(a^2 + b^2)$? So velocity is $a$ now? Why?
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Post by Admin on Aug 18, 2023 16:09:27 GMT
Yes I recognized it $F = ma$, and here $a = \ddot{z}$ (acceleration) is clear but why on the other side we have $-z$? Is that because the force towards the origin is $|z|$ and in the differential equation we need the opposite direction, plus a vector not it's modulus hence $z$ and not $|z|$? Vasco, do you agree with what I said here? Mondo Yes, note that on page 241 the first sentence in subsection 2 says that the force is proportional to $r$ the distance from the fixed point. Vasco |
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Post by Admin on Aug 18, 2023 16:19:58 GMT
Beside this, I also wonder how on page 243 author can conclude that the formula for $E$ can be rewritten as $\frac{1}{2}(a^2 + b^2)$? So velocity is $a$ now? Why? Mondo This doesn't mean that $v=a$, why do you say that? Vasco |
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Post by mondo on Aug 18, 2023 19:51:21 GMT
Beside this, I also wonder how on page 243 author can conclude that the formula for $E$ can be rewritten as $\frac{1}{2}(a^2 + b^2)$? So velocity is $a$ now? Why? Mondo This doesn't mean that $v=a$, why do you say that? Vasco On page 243 the energy is written as $E = \frac{1}{2}(v^2 + r^2)$ which is a sum of kinetic and potential energy. Next author says that as the particle travels around the ellipse in [22] we can rewrite it to $E = \frac{1}{2}(a^2 + b^2)$. This means either $v = a$ and $r = b$ or there is some other relation between them, how to see it? I think it would also help if I know why on figure [22] the speed at $a = b$? |
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Post by Admin on Aug 18, 2023 20:00:43 GMT
Mondo
You have an equation for $z$ so you can calculate the velocity $v$ and you can calculate $r$.
Vasco
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Post by mondo on Aug 18, 2023 21:29:35 GMT
Mondo You have an equation for $z$ so you can calculate the velocity $v$ and you can calculate $r$. Vasco Right and also notice on figure [22] that they form a right triangle so from Pythagoras theorem we can write $c^2 = a^2 + b^2 = v^2+r^2$. Clear, thank you. There are a few more things I don't get in this subsection: 1. The solution to the differential equation -> $\frac{d^2}{d^2t}z(t) = -z$ hence I try to solve for $z$ by $\int \frac{d^2}{d^2t}z(t) = \int -re^{it} = ire^{it}$ next we need to take the integral again to get $z = \int ire^{it} = re^{it}$. So two things here, why author neglects $r$? We are not on the unit circle right? Second why he got $\pm$ in the exponent? 2. Next he says the general solution is $z = pe^{it} + qe^{-it}$ - how was it obtained? Where are $p,q$ coming from? Why $p>q$? |
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Post by Admin on Aug 18, 2023 21:39:15 GMT
Mondo
What form a right triangle?
Vasco
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Post by mondo on Aug 18, 2023 21:44:54 GMT
Mondo What form a right triangle? Vasco On figure [22] $v$ and $r$ at a point $a$ can be visualized as sides of a right triangle right? |
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Post by Admin on Aug 18, 2023 21:47:56 GMT
Mondo
Yes but this is not true at all points of the orbit and so your argument doesn't work.
Vasco
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Post by Admin on Aug 18, 2023 21:56:14 GMT
Mondo
1. Can you please write your first integration properly so that I can understand it please? You have no differentials in the rhs of your argument.
Vasco
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Post by mondo on Aug 18, 2023 22:00:06 GMT
I agree; Author says that as the particle orbits round the elipse this is equal to to this formula so how is he sure about this equality?
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