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

  • #1
selim
2
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
  • #2
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?
 
  • #3
  • Love
Likes vanhees71
  • #4
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

Similar threads

  • Classical Physics
Replies
5
Views
1K
Replies
17
Views
2K
Replies
3
Views
642
  • Classical Physics
Replies
3
Views
2K
  • Classical Physics
Replies
1
Views
1K
Replies
1
Views
875
Replies
8
Views
1K
  • Special and General Relativity
Replies
7
Views
565
  • Advanced Physics Homework Help
Replies
3
Views
528
Replies
27
Views
2K
Back
Top