# An Introduction…

A bit of background about myself: Since my sophomore year of high school I have been interested in astronomy, physics, and mathematics. I received my Bachelor’s degree in Earth and Planetary Sciences concentrating in astronomy/astrophysics from Western Connecticut State University. My coursework and independent readings ultimately led to minors in mathematics and physics.

My intention for this blog is to serve as a reference in astrophysics and related topics to myself as well as others. I aim to share my own research interests and consider selected problems that have fascinated me. I also hope to communicate recent news in the fields of physics and astronomy and discuss the implications of discoveries made.

DISCLAIMER: I am by no means an expert, and as such the posts that I create are of my opinion and my own logic. I may be wrong sometimes, and I hope that the people who see this (assuming that anyone sees this) will respect that.

That being said… Enjoy!

(ABOVE: An image of the moon taken with a lunar and planetary imaging camera mounted to a Newtonian 130mm reflecting telescope.)

# 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

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

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…

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

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

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

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

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

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.

# A Problem in Thermodynamics and Statistical Mechanics: Analytical and Numerical Study of an Einstein Solid

Every physics major at some point in their undergraduate career takes a course in thermodynamics and statistical mechanics. One of my problem sets included a problem that considers an Einstein solid with 50 oscillators and 100 units of energy and then increases the number of oscillators to 5000. I will be presenting my solution to the numerical side of the problem. An Einstein solid can be regarded as

“… a collection of microscopic systems which can store any number of energy ‘units’ of equal size which occur for any quantum-mechanical harmonic oscillator whose potential energy function has the form $\displaystyle \frac{1}{2}k_{s}x^{2}$…The model of a solid as a collection of identical oscillators with quantized energy units…”

described (defined) by Schroeder in his text Introduction to Thermal Physics. Figure 1 represents the Einstein solid as a whole (in a lattice) and Figure 2 depicts the quantum-mechanical harmonic oscillator interpretation of an Einstein solid.

The problem statement is:

“Use a computer to study the entropy, temperature, and heat capacity of an Einstein solid, as follows. Let the solid contain 50 oscillators (initially), and from 0 to 100 units of energy. Make a table, analogous to Table 3.2, in which each row represents a different value for the energy…Make a graph of entropy vs. energy, and a graph of the heat capacity vs. temperature. Then change the number of oscillators to 5000, and again make a graph of the heat capacity and temperature and entropy and energy, and discuss the predictions and compare it to the predictions to the data for lead, aluminum, and diamond. Estimate the numerical value of $\displaystyle \epsilon$ for each of those solids.”

This problem can be found in the aforementioned text.

Figure 1. Einstein Solid (Lattice); Image Credit/Obtained from https://mappingignorance.org/2015/12/17/einstein-and-quantum-solids/

Figure 2. Quantum-Mechanical Harmonic Oscillator interpretation of an Einstein solid as a collection of these oscillators. Image Credit: http://hyperphysics.phy-astr.gsu.edu/hbase/Therm/einsol.html

Part I: Let $q = 100$ units, and let $N = 50$. The corresponding data table for this Einstein solid follows. The following set of equations were used to determine the multiplicity and entropy.

$\displaystyle \Omega(N,q) = {q+N-1 \choose q} = \frac{(q+N-1)!}{q! (N-1)!}, (1)$

and

$\displaystyle S = Nk \ln{\Omega}, (2)$

where $\Omega$ is the multiplicity. The remaining quantities of temperature were obtained using a simplified form of the central difference equations for the first order derivative. The respective definitions of temperature and heat capacity are

$\displaystyle T = \frac{\partial U}{\partial S}, (3)$

and

$\displaystyle C_{V} = \frac{\partial U}{\partial T}, (4)$

where $U$ represents the internal energy of the Einstein solid, and $S$ is the entropy. The generalized from of the first order central difference approximation has the form

$\displaystyle \frac{dy_{j}}{dx}\approx \frac{y_{j+1}-y_{j-1}}{2h} + \mathcal{O}(h^{2}), (5)$

where $\mathcal{O}(h^{2})$ represents the higher order terms, in this case, the quadratic, cubic, quartic, and so on, and $h$ is the step size for each iteration. For the final iteration (when $q = 100$ units), instead of using the central difference approximation, a backward difference approximation was employed since there does not exist data for $q = 101$ units of energy. The backward difference approximation has the form

$\displaystyle \frac{dy_{j}}{dx}\approx \frac{y_{j}-y_{j-1}}{h}+\mathcal{O}(h). (6)$

Table I (Dimensionless Parameters):

 Energy q Ω S/k kT/ε C/Nk 0 1 0 0 N/A 1 50 3.912023005 0.27969284 0.121826198 2 1275 7.150701458 0.328336604 0.453606383 3 22100 10.00333289 0.367875021 0.536183525 4 292825 12.58733044 0.402937926 0.593741905 5 3162510 14.96687657 0.43524436 0.637773801 6 28989675 17.18245029 0.465656087 0.673043377 7 231917400 19.26189183 0.494675894 0.702124659 8 1652411475 21.22550156 0.522626028 0.72660015 9 10648873950 23.08871999 0.549726805 0.747522024 10 62828356305 24.86367234 0.576136157 0.765628174 11 3.427E+11 26.56012163 0.601971486 0.781456694 12 1.74206E+12 28.18608885 0.627322615 0.795411957 13 8.30828E+12 29.74827387 0.652259893 0.807805226 14 3.73873E+13 31.25235127 0.676839501 0.818880855 15 1.59519E+14 32.70318415 0.701107048 0.828833859 16 6.48046E+14 34.1049827 0.725100078 0.837822083 17 2.51594E+15 35.4614241 0.748849881 0.845974847 18 9.3649E+15 36.77574496 0.772382808 0.853399232 19 3.35165E+16 38.05081369 0.795721261 0.860184741 20 1.15632E+17 39.28918792 0.818884446 0.866406816 21 3.8544E+17 40.49316072 0.84188895 0.872129523 22 1.24392E+18 41.66479814 0.864749193 0.877407641 23 3.89401E+18 42.80597005 0.887477794 0.882288296 24 1.18443E+19 43.91837566 0.910085848 0.88681226 25 3.5059E+19 45.00356493 0.932583169 0.891014994 26 1.01132E+20 46.0629565 0.954978471 0.89492748 27 2.84667E+20 47.09785298 0.977279528 0.898576916 28 7.82835E+20 48.10945389 0.999493303 0.901987268 29 2.10556E+21 49.09886689 1.021626052 0.905179739 30 5.54463E+21 50.06711736 1.043683421 0.908173155 31 1.43087E+22 51.01515679 1.065670516 0.910984284 32 3.6219E+22 51.94387004 1.087591972 0.913628113 33 8.99987E+22 52.85408172 1.109452006 0.916118071 34 2.19703E+23 53.74656181 1.131254467 0.918466228 35 5.27286E+23 54.62203054 1.153002873 0.920683463 36 1.24498E+24 55.48116286 1.174700452 0.9227796 37 2.89374E+24 56.32459225 1.196350168 0.924763536 38 6.62514E+24 57.1529142 1.217954752 0.926643346 39 1.4949E+25 57.96668937 1.239516722 0.928426373 40 3.32616E+25 58.76644629 1.261038406 0.930119309 41 7.30133E+25 59.55268389 1.282521958 0.931728264 42 1.58195E+26 60.32587378 1.303969378 0.933258829 43 3.38465E+26 61.08646224 1.325382522 0.934716125 44 7.15391E+26 61.8348721 1.346763119 0.936104855 45 1.49437E+27 62.57150439 1.368112778 0.937429341 46 3.08621E+27 63.29673989 1.389433002 0.938693566 47 6.30374E+27 64.01094048 1.410725193 0.9399012 48 1.27388E+28 64.71445045 1.431990666 0.941055635 49 2.54776E+28 65.40759763 1.45323065 0.942160007 50 5.04457E+28 66.09069447 1.4744463 0.943217222 51 9.89131E+28 66.76403902 1.495638697 0.944229971 52 1.9212E+29 67.42791582 1.516808861 0.945200757 53 3.6974E+29 68.08259672 1.537957749 0.946131903 54 7.05244E+29 68.72834166 1.559086264 0.947025573 55 1.33355E+30 69.36539938 1.580195257 0.947883783 56 2.50041E+30 69.99400804 1.60128553 0.948708411 57 4.64989E+30 70.61439586 1.622357843 0.949501213 58 8.57824E+30 71.22678169 1.643412912 0.95026383 59 1.57025E+31 71.83137547 1.664451416 0.950997796 60 2.85263E+31 72.42837879 1.685473998 0.951704548 61 5.14408E+31 73.01798529 1.706481267 0.952385432 62 9.20957E+31 73.60038111 1.7274738 0.953041713 63 1.63726E+32 74.17574525 1.748452147 0.953674575 64 2.89078E+32 74.74424999 1.769416829 0.954285135 65 5.06999E+32 75.30606117 1.79036834 0.95487444 66 8.83407E+32 75.86133855 1.811307153 0.955443478 67 1.52948E+33 76.41023613 1.832233716 0.955993177 68 2.63161E+33 76.95290236 1.853148456 0.956524415 69 4.50043E+33 77.48948048 1.874051781 0.957038017 70 7.65073E+33 78.02010873 1.894944079 0.957534764 71 1.29308E+34 78.54492059 1.915825721 0.958015394 72 2.17309E+34 79.06404502 1.936697061 0.958480604 73 3.63175E+34 79.57760662 1.957558438 0.958931052 74 6.03655E+34 80.08572588 1.978410175 0.959367363 75 9.98043E+34 80.58851934 1.999252581 0.959790129 76 1.64152E+35 81.08609973 2.020085953 0.960199908 77 2.68612E+35 81.57857622 2.040910573 0.960597234 78 4.37356E+35 82.06605448 2.061726714 0.960982609 79 7.08627E+35 82.54863689 2.082534635 0.961356514 80 1.14266E+36 83.02642266 2.103334587 0.961719401 81 1.8339E+36 83.49950796 2.124126809 0.962071705 82 2.92977E+36 83.96798603 2.14491153 0.962413835 83 4.65939E+36 84.43194735 2.165688971 0.962746183 84 7.37736E+36 84.89147968 2.186459344 0.963069122 85 1.16302E+37 85.34666822 2.207222854 0.963383006 86 1.82567E+37 85.7975957 2.227979695 0.963688173 87 2.85392E+37 86.24434247 2.248730056 0.963984945 88 4.44304E+37 86.68698658 2.269474119 0.964273631 89 6.8892E+37 87.12560389 2.290212057 0.964554522 90 1.064E+38 87.56026816 2.31094404 0.9648279 91 1.63692E+38 87.99105107 2.331670228 0.965094032 92 2.50876E+38 88.41802239 2.352390779 0.965353173 93 3.83058E+38 88.84124995 2.373105841 0.965605568 94 5.82737E+38 89.2607998 2.39381556 0.965851451 95 8.83307E+38 89.67673621 2.414520077 0.966091045 96 1.33416E+39 90.08912176 2.435219526 0.966324564 97 2.00812E+39 90.4980174 2.455914037 0.966552213 98 3.01218E+39 90.90348251 2.476603736 0.966774189 99 4.50306E+39 91.30557493 2.497288745 1.287457337 100 6.70955E+39 91.70435105 2.507672727 1.926043463

Graphing the entropy vs. energy, and the heat capacity vs. temperature gives the following:

Graphs I & II

Part II: Let $q = 100$ units and let $N = 5000$. Using this in the calculation yields the following table for this Einstein solid. This “dilutes” the system and lowers the temperature:

Table II (Dimensionless  Parameters):

 Energy q Ω S/k kT/ε C/Nk 0 1 0 0 N/A 1 5000 8.517193 0.122388 0.003049 2 12502500 16.34144 0.131206 0.026553 3 2.08E+10 23.76042 0.137453 0.035575 4 2.61E+13 30.89192 0.14245 0.043342 5 2.61E+16 37.80047 0.146681 0.050387 6 2.18E+19 44.52691 0.150388 0.056922 7 1.56E+22 51.09939 0.153709 0.063064 8 9.74E+24 57.53854 0.156731 0.068885 9 5.42E+27 63.86011 0.159515 0.074436 10 2.72E+30 70.07651 0.162105 0.079755 11 1.24E+33 76.19781 0.164531 0.08487 12 5.16E+35 82.23229 0.166818 0.089804 13 1.99E+38 88.18693 0.168985 0.094575 14 7.13E+40 94.06767 0.171047 0.099198 15 2.38E+43 99.87961 0.173017 0.103687 16 7.47E+45 105.6272 0.174905 0.108052 17 2.2E+48 111.3144 0.176719 0.112303 18 6.14E+50 116.9446 0.178467 0.116448 19 1.62E+53 122.5209 0.180154 0.120494 20 4.07E+55 128.0462 0.181787 0.124447 21 9.73E+57 133.5229 0.183368 0.128314 22 2.22E+60 138.9532 0.184904 0.132098 23 4.85E+62 144.3393 0.186396 0.135806 24 1.02E+65 149.683 0.187849 0.13944 25 2.04E+67 154.9861 0.189265 0.143004 26 3.94E+69 160.2502 0.190646 0.146503 27 7.34E+71 165.4768 0.191995 0.149939 28 1.32E+74 170.6671 0.193314 0.153316 29 2.28E+76 175.8226 0.194604 0.156635 30 3.83E+78 180.9444 0.195868 0.159899 31 6.21E+80 186.0336 0.197106 0.16311 32 9.77E+82 191.0912 0.19832 0.166272 33 1.49E+85 196.1183 0.199512 0.169384 34 2.21E+87 201.1157 0.200682 0.172451 35 3.17E+89 206.0843 0.201831 0.175472 36 4.44E+91 211.025 0.202961 0.17845 37 6.04E+93 215.9384 0.204073 0.181387 38 8E+95 220.8254 0.205166 0.184283 39 1.03E+98 225.6866 0.206243 0.18714 40 1.3E+100 230.5227 0.207304 0.18996 41 1.6E+102 235.3343 0.208349 0.192743 42 1.9E+104 240.122 0.209379 0.195491 43 2.3E+106 244.8863 0.210395 0.198205 44 2.6E+108 249.6279 0.211397 0.200885 45 2.9E+110 254.3472 0.212386 0.203533 46 3.2E+112 259.0447 0.213363 0.20615 47 3.4E+114 263.7209 0.214327 0.208736 48 3.6E+116 268.3762 0.215279 0.211293 49 3.7E+118 273.0112 0.21622 0.213821 50 3.7E+120 277.6261 0.21715 0.21632 51 3.7E+122 282.2214 0.218069 0.218793 52 3.6E+124 286.7975 0.218978 0.221238 53 3.4E+126 291.3548 0.219877 0.223658 54 3.2E+128 295.8935 0.220766 0.226052 55 2.9E+130 300.4141 0.221646 0.228422 56 2.7E+132 304.9169 0.222517 0.230768 57 2.4E+134 309.4022 0.22338 0.23309 58 2.1E+136 313.8703 0.224233 0.235388 59 1.8E+138 318.3214 0.225079 0.237665 60 1.5E+140 322.756 0.225917 0.239919 61 1.2E+142 327.1743 0.226746 0.242152 62 1E+144 331.5765 0.227568 0.244363 63 8.1E+145 335.9628 0.228383 0.246555 64 6.4E+147 340.3337 0.229191 0.248725 65 5E+149 344.6892 0.229991 0.250876 66 3.8E+151 349.0296 0.230785 0.253008 67 2.9E+153 353.3553 0.231572 0.255121 68 2.2E+155 357.6663 0.232353 0.257215 69 1.6E+157 361.9629 0.233127 0.259291 70 1.1E+159 366.2453 0.233896 0.261349 71 8.2E+160 370.5137 0.234658 0.263389 72 5.8E+162 374.7683 0.235414 0.265413 73 4E+164 379.0093 0.236165 0.267419 74 2.7E+166 383.237 0.23691 0.269409 75 1.9E+168 387.4514 0.23765 0.271382 76 1.2E+170 391.6527 0.238384 0.273339 77 8.2E+171 395.8412 0.239113 0.275281 78 5.3E+173 400.0169 0.239837 0.277207 79 3.4E+175 404.1802 0.240556 0.279118 80 2.2E+177 408.331 0.24127 0.281015 81 1.4E+179 412.4696 0.24198 0.282896 82 8.4E+180 416.5962 0.242684 0.284763 83 5.2E+182 420.7108 0.243384 0.286616 84 3.1E+184 424.8136 0.24408 0.288455 85 1.9E+186 428.9048 0.244771 0.29028 86 1.1E+188 432.9845 0.245458 0.292092 87 6.5E+189 437.0529 0.24614 0.29389 88 3.7E+191 441.11 0.246819 0.295676 89 2.1E+193 445.156 0.247493 0.297448 90 1.2E+195 449.191 0.248164 0.299208 91 6.7E+196 453.2152 0.24883 0.300955 92 3.7E+198 457.2286 0.249493 0.30269 93 2E+200 461.2315 0.250152 0.304413 94 1.1E+202 465.2238 0.250807 0.306124 95 5.9E+203 469.2057 0.251458 0.307823 96 3.2E+205 473.1774 0.252106 0.30951 97 1.7E+207 477.1389 0.252751 0.311187 98 8.6E+208 481.0903 0.253392 0.312851 99 4.4E+210 485.0318 0.254029 0.418439 100 2.3E+212 488.9634 0.254345 0.628243

Thus the graphs of the entropy vs. energy and heat capacity vs. temperature follow:

Figure 2. Graphs III and IV.

Figure 3. (Figure 1.14 of Schroeder’s Thermal Physics) Heat Capacity curves for Lead (Pb), Aluminum (Al), and Diamond, respectively as a function of temperature in Kelvin.

Graph II shows the prediction for heat capacity as a function of temperature of an Einstein solid for which there are 100 units of energy and 50 oscillators. The data exhibits a trend that appears to reach an asymptote quickly, then when the temperature reaches T ≈ 2.5, there is a sudden increase in the value of the heat capacity. The approach to determining the final data points was switched from a central difference approximation to a backward difference approximation of the last two entries corresponding to energies q = 99 and q = 100 units. If we ignore the last two, the curve approaches an asymptote at CV = 1. However, the graphs produced are of the dimensionless quantities involved. The overall curve appears to be logarithmic and resembles the heat capacity curve for lead. The initial increase is almost immediate and its slope appears to be slightly less than lead but greater than aluminum.

Graphs III and IV show the prediction for heat capacity in terms of temperature of an Einstein solid for which the energy is the same, but the number of oscillators is now 5000. The temperature has been reduced and the heat capacity vs. temperature yields a graph that shows a trendline that appears linear. Comparing to Figure 3(Fig. 1.14 in the text), this graph resembles the heat capacity curve for diamond. In Figure 3, the diamond curve is linear throughout. The only discrepancies among Graph IV and Figure 3 are the final two data points in Graph IV. Again, a backward difference approximation was used to determine the final data points for this Einstein solid as well.  The value for the constant ε was determined by finding the quotient of the entropy and temperature columns and taking the average value of ε for each energy.

This was the numerical analysis of an Einstein solid’s temperature, energy, entropy, and heat capacity. In the next post, I shall discuss the analytical version of this analysis.

# A “Proof” of the Sturm-Liouville Theorem/Problem

IMAGE CREDIT: NASA/JPL: This shows Jupiter’s Great Red Spot; a storm that has been occurring for over 300 years now. Quite recently, however, observations show that the Spot appears to be shrinking in size.

About a week ago, I was looking through my notebooks and came across an unfinished problem posed by one of my professors. Unfortunately, I was not able to solve the problem during the semester. However, I thought it might be something interesting to consider. I did a quick search and found that the problem he gave us was to prove the Sturm-Liouville Theorem.

The main brute-force method to analytically solving a given partial differential equation is the separation of variables. This method is heavily used by physicists and in doing so transforms the initial boundary value problem (IBVP) into a Sturm-Liouville problem in which we have an ordinary differential equation and linear homogeneous boundary conditions.

# A Narrow, Technical Problem in Partial Differential Equations

While I was in school, one of my professors set this problem to me and my classmates and challenged us to solve it over the next few days. I found the challenge intriguing and it fascinated me, so I thought it was worth sharing. The problem was this:

Show that

$\displaystyle v(x,t) = \int_{-\infty}^{\infty} f(x-y,t)g(y)dy, (1.1)$

where $\displaystyle g(y)$ has finite support and also satisfies the PDE

$\displaystyle \frac{\partial v}{\partial t} = -\kappa \frac{\partial^{2}v}{\partial x^{2}}. (1.2)$

First off, what does finite support mean? Mathematically speaking, a function has support which is characterized by a subset of its domain whose members do not map to zero, and yet are finite. (Just as a quick note: much of the proper definitions require an understanding in mathematical analysis and measure theory, something which I have not studied in detail, so take that explanation with a grain of salt.)

As for the solution, we can rewrite the given PDE as

$\displaystyle \frac{\partial v}{\partial t} - \kappa \frac{\partial^{2}v}{\partial x^{2}} = 0. (2)$

The PDE requires a first-order time derivative and a second-order spatial derivative.

$\displaystyle \therefore \frac{\partial v}{\partial t} = \frac{\partial}{\partial t}\int_{-\infty}^{\infty} f(x-y,t)g(y)dy, (3.1)$

and

$\displaystyle \frac{\partial^{2} v}{\partial x^{2}} = \frac{\partial^{2}}{\partial x^{2}}\int_{-\infty}^{\infty} f(x-y,t)g(y)dy. (3.2)$

Next, we substitute Eqs. (3.1) and (3.2) into Eq.(2), yielding

$\displaystyle \frac{\partial}{\partial t}\int_{-\infty}^{\infty} f(x-y,t)g(y)dy -\kappa \frac{\partial^{2}}{\partial x^{2}}\int_{-\infty}^{\infty} f(x-y,t)g(y)dy = 0. (4)$

Note that taking the derivative of a function and then integrating that function is equivalent to integrating the function and differentiating the same function, in conjunction with the fact that the sum or difference of the integrals is the integral of the sum or difference (proofs of these facts are typically covered in a course in real analysis). Taking advantage of these gives

$\displaystyle \int_{-\infty}^{\infty} \bigg\{\frac{\partial}{\partial t}f(x-y,t)-\kappa\frac{\partial^{2}}{\partial x^{2}}f(x-y,t)\bigg\}g(y)dy = 0. (5)$

Notice that the terms contained in the brackets equate to $\displaystyle 0$. This means that

$\displaystyle \int_{-\infty}^{\infty} 0 \cdot g(y)dy = 0. (6)$

This implies that the function $\displaystyle v(x,t)$ does satisfy the given PDE (Eq.(2)).

References:

Definition of Support in Mathematics: https://en.wikipedia.org/wiki/Support_(mathematics)

# Observing the Variable Star W Ursae Majoris

While I was an undergraduate, one of my smaller research projects involved observing the variable star W Ursae Majoris.

In general, there are six types of binary star systems: Optical double, Visual binary, Astrometric binary, Eclipsing binary, Spectrum binary, and Spectroscopic binary.

In this project, my classmate and I were interested in the eclipsing binary (EW) W Ursae Majoris. An eclipsing binary is a binary system in which one of the stars will pass in front of its companion, effectively causing an eclipse. We are able to observe this by way of generating the light curves of the system. An example light curve is shown below:

(Image was obtained at the URL:  https://imagine.gsfc.nasa.gov/educators/hera_college/binary-model.html)

The graph shows a plot of intensity over time (which in this case is an orbital period). Observations of an EW should show dips in the intensity of the two stars. What is really fascinating to me is that we can gain valuable information from this graph. For example, the length of a dip can indicate the masses of the star. If we have a star of mass $m_{1}$ and the other is $m_{2}$ such that $m_{2}>m_{1}$, and if the duration of the decrease in intensity of the system is significant we can then infer that the mass passing in front of its companion is that of $m_{1}$. By default, the mass that is being “eclipsed” is $m_{2}$. Conversely, if the intensity decreases but only for a short while, the positions are reversed, with $m_{2}$ passing in front (relatively speaking) and $m_{1}$ is being “eclipsed”. (I am assuming that the barycenter (i.e. the system’s center of mass) is equidistant from the centers of the two stars.)

Another form of classification of binary stars is whether or not the binary system components are touching or not. More precisely, there are three kinds of close binaries: detached, semi-detached, and contact binary. There are sub-categories of contact binaries: near contact, contact, overcontact,  and double contact.

An equipotential surface map of a system (assuming that the binary system has a mass ratio of 2:1, which may be incorrect as most W UMa binaries have a mass ratio of 10:1) is shown below:

Image Credit: Fig.1 of Terrell, D., Eclipsing Binary Stars: Past, Present, and Future. JAAVSO Vol. 30, 2001.

To quickly elaborate, each type of contact binary will fill its inner Lagrangian surface (aka Roche lobes) to an extent. In the context of our project, W Ursae Majoris is an overcontact eclipsing binary system.  This type of binary will overfill its inner Lagrangian surface. As a result of this, processes such as mass transfer and accretion can occur. The diagram below shows the orbital evolution of a W UMa EW AC Bootis (in addition to being its own binary system, W UMa is also a class of close binaries)

Image Credit: Fig. 15 of Alton, K., A Unified Roche-Model Light Curve Solution for the W UMa Binary AC Bootis. JAAVSO. Vol. 38, 2010.

The objective of the project was to image the eclipsing binary, measure the apparent magnitude, to process the images, and to obtain a light curve. To observe this system, a classmate and I made use of the 20″ Ritchey-Chrétien telescope at the university observatory. We made use of the CCD camera attached and set a sequence of images to be taken every two minutes. W UMa has a period of approximately 8 hours, however, due to time constraints (and as much as I would have liked to, the weather was not conducive for observations exceeding two hours), we ended up only taking images for around two hours.

After the session was over, we ended up taking a total of roughly 40-50 images. Additionally, the software used to capture the images simultaneously measured the magnitude of W UMa at the time each image was taken. This allowed us to use Excel (and later on MATLAB) to obtain a partial light curve. However, since this is a partial light curve, we can say that an eclipse (and a short one at that) occurs, yet we cannot determine whether or not the local minimum depicted in the graph below is a primary or a secondary minimum–we simply do not have enough data.

In addition to the partial light curve above, we were able to process the images (using Registax v.6). Below is a stacked image of W UMa. The big blob near the center of the image is the binary. The binary is not able to be resolved by telescopes component-wise.

References:

Caroll, B.W., and Ostlie, D. A., Introduction to Modern Astrophysics. 2017. Cambridge University Press. 7.

Catalog and Atlas of Eclipsing Binaries (CALEB): Types of Binary Stars

http://www.daviddarling.info/encyclopedia/C/close_binary.html

American Association of Variable Star Observers (AAVSO) URL: https://www.aavso.org/vsots_wuma

Journal of American Association of Variable Star Observers: Figure References

# Coordinates and Transformations of Coordinates

SOURCES FOR CONTENT:  Neuenschwander, D.E., Tensor Calculus for Physics, 2015. Johns Hopkins University Press.

In this post, I continue the introduction of tensor calculus by discussing coordinates and coordinate transformations as applied to relativity theory. (A side note: I have acquired Misner, Thorne, and Wheeler’s Gravitation and will be using it sparingly given its reputation and weight of the material.)