Month: August 2018

Derivation of the Euler-Lagrange Equation for a Function of Several Dependent Variables

IMAGE CREDIT: NASA/JPL

SOURCE FOR CONTENT: Classical Dynamics of Particles and Systems. Thornton and Marion. 5th Edition. 

 

Consider a functional

\displaystyle \phi = \phi(y_{\mu},y_{\mu}^{\prime}; x), (1)

where \mu = 1,2,...,n. By the method used in a previous section of the aforementioned text, we may write

\displaystyle y_{\mu}(\alpha, x) = y_{\mu}(0,x) +\alpha \eta_{\mu}(x). (2)

Additionally, we will find it useful to define

\displaystyle y_{\mu}^{\prime}(\alpha,x) = y_{\mu}^{\prime}(0,x)+\alpha \eta_{\mu}^{\prime}(x). (3)

Further we may also define an integral functional by way of integrating Eq.(1) over the interval x_{1}\leq x \leq x_{2}, and introducing a variational parameter \alpha we have

\displaystyle J(\alpha) = \int_{x_{1}}^{x_{2}} \phi(y_{\mu},y_{\mu};x)dx. (4)

Two necessary conditions that are used to derive the Euler-Lagrange equation include

\displaystyle \frac{\partial J(\alpha)}{\partial \alpha}\bigg\|_{\alpha=0}=0, (5)

and

\displaystyle \eta_{\mu}(x_{1})=\eta_{\mu}(x_{2})=0. (6)

Let us take the derivative of J(\alpha) with respect to \alpha yielding

\displaystyle \frac{\partial J}{\partial \alpha}\bigg\|_{\alpha=0}=\frac{\partial}{\partial \alpha}\int_{x_{1}}^{x_{2}}\phi(y_{\mu},y_{\mu}^{\prime};x)dx. (7)

Carrying out the derivative operator on the right-hand-side of Eq.(7) we get

\displaystyle \frac{\partial J}{\partial \alpha}=\int_{x_{1}}^{x_{2}}\sum_{\mu}\bigg\{\partial_{y_{\mu}}\phi \partial_{\alpha}y_{\mu}+\partial_{y_{\mu}^{\prime}}\phi \partial_{\alpha}y_{\mu}^{\prime}\bigg\}dx. (8)

From Eqs.(2) and (3) we see that

\displaystyle \partial_{\alpha}y_{\mu}= \eta_{\mu}(x), (9)

and

\displaystyle \partial_{\alpha}y_{\mu}^{\prime} = \eta_{\mu}^{\prime}(x). (10)

Thus Eq.(8) becomes…

\displaystyle \frac{\partial J}{\partial \alpha}=\int_{x_{1}}^{x_{2}}\sum_{\mu}\bigg\{\partial_{y_{\mu}}\phi \eta_{\mu}(x)+\partial_{y_{\mu}^{\prime}}\phi \eta_{\mu}^{\prime}(x)\bigg\}dx. (11)

Consider the second term under the summation. We may make use of integration by parts to obtain the following

\displaystyle \frac{\partial J}{\partial \alpha}=\int_{x_{1}}^{x_{2}}\sum_{\mu}\bigg\{\partial_{y_{\mu}}\phi +\frac{d}{dx}(\partial_{y_{\mu}^{\prime}}\phi) \bigg\}\eta_{\mu}(x)dx. (12)

By the necessary condition (Eq.(5)), it follows that

\displaystyle 0 = \sum_{\mu}\bigg\{\partial_{y_{\mu}}\phi +\frac{d}{dx}(\partial_{y_{\mu}^{\prime}}\phi) \bigg\}. (13)

Additionally, in Eq.(13) above, we have also made use of the condition that \eta_{\mu}(x_{1})=\eta_{\mu}(x_{2})=0. Since \eta_{\mu}(x) \neq 0 for any x_{1}\leq x \leq x_{2}, then the terms in the brackets must vanish, yielding the Euler-Lagrange Equation for several dependent variables.

Update: the next posts will be those discussed on my Facebook page. Namely, I intend to continue with my Research Series and my series in Tensor Calculus and General Relativity with various ancillary posts in my Astrophysics Series.

 

Clear Skies!

Astrophysics Series: Derivation of the Total Energy of a Binary Orbit

SOURCE FOR CONTENT: An Introduction to Modern Astrophysics, Carroll & Ostlie, Cambridge University Press. Ch.2 Celestial Mechanics

Here is my solution to one of the problems in the aforementioned text. I derive the total energy of a binary system making use of center-of-mass coordinates. In order to conceptualize it I have used the binary Alpha Centauri A and Alpha Centauri B. While writing this I stumbled upon the Kepler problem, the two-body problem, and the N-body problem. Leave a comment if you would like me to consider that in another post.

Clear Skies!

Derivation of the Total Energy of a Binary Orbit:

Setup: Consider the nearest binary star system to our solar system: Alpha Centauri A and Alpha Centauri B. These two stars orbit each other about a common center of mass; a point called a barycenter. The orbital radius vector of Alpha Centauri A is \textbf{r}_{1} and the orbital radius vector of Alpha Centauri B is \textbf{r}_{2}. The masses of Alpha Centauri A and B are m_{1}, and m_{2}, respectively. The total mass of the binary orbit M is the sum of the individual masses of each component. In the context of this system, we encounter what is called the two-body problem of which there exists a special case known as the Kepler Problem (by the way let me know if that would be something that you guys would want to see…). We can simplify this two-body problem by making use of center-of-mass coordinates wherein we define the reduced mass \mu. Therefore, the derivation of the total energy of the binary system of Alpha Centauri A and B will be carried out in such a coordinate system.

To derive this energy equation, one would typically make use of center-of-mass coordinates in which

\displaystyle \textbf{r}_{1}=-\frac{\mu}{m_{1}}r,  (0.1)

and

\displaystyle \textbf{r}_{2}=\frac{\mu}{m_{2}}r, (0.2)

where \mu represents the reduced mass given by

\displaystyle \mu\equiv \frac{m_{1}m_{2}}{m_{1}+m_{2}}=\frac{m_{1}m_{2}}{M}. (0.3)

Recall from conservation of energy that

\displaystyle E = \frac{1}{2}m_{1}\dot{r}_{1}^{2}+\frac{1}{2}m_{2}\dot{r}_{2}^{2}-G\frac{m_{1}m_{2}}{|\mathcal{R}|}, (1)

where |\mathcal{R}| represents the separation distance between the two components. Let us take the derivative of Eqs.(0.1) and (0.2) to get

\displaystyle \dot{r}_{1}=-\frac{\mu}{m_{1}}v, (2.1)

and

\displaystyle \dot{r}_{2}= \frac{\mu}{m_{2}}v. (2.2)

Substitution yields

\displaystyle E = \frac{1}{2}\frac{\mu^{2}}{m_{1}}v^{2}+\frac{1}{2}\frac{\mu^{2}}{m_{2}}v^{2}-G\frac{m_{1}m_{2}}{|\mathcal{R}|}. (3)

Upon making use of the definition of the reduced mass (Eq. (0.3)) we arrive at

\displaystyle E = \frac{1}{2}\mu v^{2}-G\frac{M \mu}{|\mathcal{R}|}. (4)

If we solve for m_{1}m_{2} in Eq.(0.3) we get the total energy of the binary Alpha Centauri A and B. This is true for any binary system assuming center-of-mass coordinates.