I How did Hamilton derive the characteristic function V in his essay?

AI Thread Summary
Hamilton derives the characteristic function V in his essay by varying the function U and establishing relationships between kinetic energy T, potential energy U, and Hamiltonian H. He uses the principles of calculus of variations to manipulate integrals and derive the expression for δV. This involves integrating by parts and applying identities related to variations of the coordinates and velocities. The final result expresses δV in terms of variations of both initial and final conditions, along with the term involving δH. The discussion highlights the mathematical rigor in Hamilton's approach to dynamics and the foundational role of calculus of variations in his derivations.
selim
Messages
2
Reaction score
1
In Hamilton's "on a general method in dynamics", he starts with varying the function ##U## and writes the equation:
$$\delta U=\sum m(\ddot x\delta x+\ddot y\delta y+\ddot z\delta z)$$
Then he defines ##T## to be:
$$T=\frac{1}{2}\sum m (\dot x^2+\dot y^2+\dot z^2)$$
Then by ##dT=dU##, he writes:
$$T=U+H$$
Then he varies T and writes:
$$\delta T= \delta U+\delta H$$
note that he is also varying in the initial conditions, that's why he did not omit the term ##\delta H##.
Hamilton then multiplies this expression by dt and integrates and writes it as:
$$\int\sum m(dx \delta \dot x+dy \delta \dot y+dz \delta \dot z)=\int\sum m(d \dot x \delta x+d \dot y \delta y+d \dot z \delta z)+\int\delta H dt$$
Then comes the part where I got confused. He says "that is, by the principles of the calculus of variations" and writes:
$$\delta V=\sum m(\dot x \delta x+\dot y \delta y+\dot z \delta z)-\sum m(\dot a \delta a+\dot b \delta b+\dot c \delta c)+\delta H t$$
where (x,y,z) and (a,b,c) are final and initial conditions then he denotes V by the integral:
$$V=\int\sum m(\dot x \delta x+\dot y \delta y+\dot z \delta z)$$
My questions are as follows:
1-how did he get ##\delta V##, what "principle of the calculus of variations" did he use?
2-then how from that did he get the integral ##V##?
 
Last edited by a moderator:
Physics news on Phys.org
To my shame I must admit that I've never read the original writings by Hamilton on his action principle. Do you have a reference?
 
selim said:
2-then how from that did he get the integral ##V##?
Equation (B) of the paper gives the definition of ##V##: $$V \equiv \int \sum m(\dot x dx + \dot y dy + \dot z dz) = \int_0^t 2T dt \,\,\,\,\,\,\,\, (B)$$

1-how did he get ##\delta V##, what "principle of the calculus of variations" did he use?
With the definition (B) and some manipulations, you can derive the expression for ##\delta V## given in equation (A) of the paper. Start with equation (10) of the paper: $$\int \sum m(dx \delta \dot x + dy \delta \dot y + dz \delta \dot z) = \int \sum m(d \dot x \delta x + d \dot y \delta y + d \dot z \delta z) + \int \delta H dt \,\,\,\,\,\,\,\, (10)$$
The left-hand side of (10) is ##\int \delta T dt##. The last term on the right is just ##t \delta H## because ##H## is independent of time. So, (10) can be written as $$\int \delta T dt = \int \sum m(d \dot x \delta x + d \dot y \delta y + d \dot z \delta z) + t \delta H $$
The integral on the right can be manipulated using integration by parts. For example,
$$\int_0^t m d \dot x \delta x = \int_0^t m (\frac{d \dot x}{dt}) \delta x dt = m \dot x \delta x \bigg|_0^t - \int_0^t m\dot x \frac{d}{dt}(\delta x) dt$$ Since ##\frac{d}{dt}(\delta x) = \delta \dot x## and ##\dot x dt = dx##, we get $$\int_0^t m d \dot x \delta x = m \dot x \delta x \bigg|_0^t - \int_0^t m dx \delta \dot x = m \dot x \delta x - m\dot a \delta a - \int_0^t m dx \delta \dot x$$ The first term on the far right, ##m \dot x \delta x##, is to be evaluated at the time ##t## of the upper limit of the integration.
##\dot a## and ##\delta a## represent evaluation of ##\dot x## and ##\delta x## at the initial time ##t = 0##.

Doing the same thing for the ##\dot y \delta y## and ##\dot z \delta z## integrations in (10), you can see that equation (10) may be written as $$\int \delta T dt = \sum m ( \dot x \delta x + \dot y \delta y + \dot z \delta z ) - \sum m ( \dot a \delta a+ \dot b \delta b + \dot c \delta c) - \int \delta T dt - t \delta H$$ or $$2\int_0^t \delta T dt = \sum m ( \dot x \delta x + \dot y \delta y + \dot z \delta z ) - \sum m( \dot a \delta a+ \dot b \delta b + \dot c \delta c ) - t \delta H$$ According to (B), the left side is ## \delta V##. So, we finally get equation (A) $$\delta V= \Sigma m\left( \dot x \delta x + \dot y \delta y + \dot z \delta z \right) - \sum m ( \dot a \delta a+ \dot b \delta b + \dot c \delta c) - t \delta H$$

I'm unsure what Hamilton meant when he stated "by the principles of the calculus of variations". Here, we used integration by parts and the use of identities such as ##\frac{d}{dt}(\delta x) = \delta \dot x##. These are often used in derivations in the calculus of variations.
 
  • Like
  • Wow
  • Love
Likes SammyS, vanhees71 and selim
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...

Similar threads

Replies
5
Views
2K
Replies
11
Views
2K
Replies
1
Views
2K
Replies
3
Views
3K
Replies
21
Views
2K
Replies
2
Views
2K
Replies
10
Views
5K
Back
Top