The post Lorentz transformation in-terms of hyperbolic rotation appeared first on Physics Log.

]]>, and

(say equation 1) where, is a Lorentz factor, and is the velocity of light.

You may aware or not that the matrix is another way of representing equations. So, in matrix form, we have our Lorentz transformation (say LT for short) as

.

This means, if you multiply right-hand side matrices then, you can get the equation 1. If we suppose velocity of light to be 1 then,

.

If you are curious why I set then, read this post. We want to see LT in terms of hyperbolic trigonometric functions so, we are interested in the matrix ( say)

.

So, if you do square of and (or, and ) and then subtract them so that you get one. i.e.

.

Similarly, .

This type of identity can be found in hyperbolic trigonometric function which I mean this

.

So, we set and . Now, we put these in our matrix and thus

.

This is a Lorentz transformation in terms of hyperbolic rotation.

Now, we may have two questions:

- Why do we make in terms of hyperbolic trigonometric functions? The answer is quite simple. Because we can have more mathematical tools (or, flexibility) when we working in hyperbolic trigonometric functions than in algebraic function so that we can quickly solve the problem.
- Why our is a hyperbolic rotation matrix? Let me make you clear, in the matrix , we only have which manipulate our transformation. This is the velocity of second observer measured by first observer . This means if second observer move in a velocity then, he will see all the points of spacetime which is seen by the first observer will shift (strictly speaking, linearly transform) towards the opposite direction of him (i.e. second observer). But, if we make our spacetime with hyperbolic coordinates then, all the points rotate on the hyperbola along by some angle. So, the mathematics for this case is Lorentz transformation in terms of hyperbolic rotation.

If you are interested to know the visualization of Lorentz transformation, and LT in terms of hyperbolic rotation then, see this, and this.

In the next post, there may be a guest post by my friend Bindesh Tripathi on the topic “Cosmic Concert”. I hope you will keep your patience alive.

Lastly, **“****Happy birthday to Prof. Einstein”**.

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]]>The post A lookup to Dirac’s Lightcone Coordinate System appeared first on Physics Log.

]]>Suppose, there is a bomb blast (strictly speaking, an event happen in the spacetime) at 1 kilometer from me. Believe or not, the only way you could get the information that there is a bomb blast is by hearing through your ears or by seeing through your eyes. Let’s say I’m blind, so the only way I could get information is through hearing. As soon as it blasts, the sound waves will take around 3 seconds to hit on my eardrum. Because in one second, sound waves can only travel 343 meters. i.e. velocity of sound is 343m/s. Now, I will know there is a blast then, I will note the time. Let’s change the scenario that I can see but cannot hear. If a bomb is blast, light waves will travel at that distance with seemingly takes no time. Because in one second, light waves can travel 299792458 meters which is very very large compared to the 1000 meters. Finally, I will see the difference in the time for the same event but from two different cases. In conclusion, the fastest way you could get the information is through light. Because most of the observers are us regardless there are signal detectors. So, in 1905 Einstein’s theory of special relativity (strictly speaking, in Lorentz transformation), he suggested the speed of light as a limit for a particle can travel, and the fastest carrier of information.

In 1907, Minkowski realized that his former student Einstein’s theory can be interpreted in a four-dimensional mathematical space (don’t be confused with physical space, simply space) which is known as the “Minkowski spacetime‘. After some years, Einstein used this framework to describe gravity in-terms of geometry, known as general relativity (GR) and realized Minkowski spacetime is a flat spacetime. In special relativity, we also need to deal with a skew coordinate system which uses contravariant/covariant vectors and it can be a tedious job to do. In 1949, Dirac when formulating the relativistic dynamics of atoms, he introduced a coordinate system (now known as lightcone coordinate system) which allows us to describe geometry when Lorentz transformation is applied. Thus, minimizes our task.

In this post, I’m interested to introduce the light cone coordinate system in 2D flat spacetime (for simplicity) and the distance associated with it. This type of coordinate system can be extended to curved spacetime (strictly speaking Pseudo-Riemannian manifold) of GR which might help us to understand the propagation of the gravitational waves [If you want to go further, read this]. But, this is out of the scope of this post.

Before we start, we need to define the velocity of light in a different way, I mean . This can be done if we define 1 second to be 299792458 meters then

.

We are doing this so that we can clearly separate the particle which has a velocity less than the velocity of light. This means, in the spacetime diagram, the line which is inclined by ° is a path (strictly speaking, called a null path) where light can only travel. This is because the slope in the spacetime diagram, and °. Thus we construct a visualization in the spacetime called light cone. Inside the lightcone, there are particles whose velocity is lesser than light and their paths are called timelike. Outside of it, there are particles whose velocity is greater than light and their paths are called spacelike. But, in nature, we have not seen any particle which can travel in a spacelike path. So, our spacetime diagram looks like this:

If we rotate a coordinate system with an angle of to then, the coordinate axes rotate as

and

.

In our case, the coordinate axes rotates with ° to and coincides with null lines. i.e.

and

.

This type of coordinate system is called Light cone coordinate system with coordinates .

Now, let’s move to our another proposal. In order to preserve the Lorentz invariance (which says the distance between two events is always same in whatever coordinate system, i.e. proper distance), we need to multiply and so that we can form a distance formula given by special relativity. i.e.

.

The proper distance for the two events when is with Minkowski signature (+ -). So,

.

Finally, the proper distance between two events for the Dirac’s lightcone coordinate system is . This type of coordinate system can be very helpful if you want to study causal relationship between the events (for example, in Causal Sets Theory [1, 2, 3], a candidate theory for Quantum Gravity), and maybe it help us to solve the problem very quickly.

In the next post, I will be presenting you the “Lorentz transformation in-terms of hyperbolic rotations”. I hope you will keep your patience alive.

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]]>The post A detailed procedure for the synthetic division when the divisor is a second-degree polynomial appeared first on Physics Log.

]]>In this post, I will try to explain a bit more detail about the method which will work for our proposed problem. Let me start by defining a general expression for the polynomial as . Lets say, our divisor will be and reminder be . I must say, when you divide a polynomial of degree by another polynomial with degree then, the quotient will have a degree of . This means, in our case, we have a divisor with degree . So, our quotient will look like this .

If we use the Euclidean division algorithm which says “if you divide integers then, it produces a quotient and a remainder smaller than the divisor”. For our case,

.

Now, we collect the terms so that we can form a polynomial. i.e.

It looks like we can compare with the polynomial as

and .

If we make supposition and then, above compare becomes

and .

At last, you will know why I made this supposition!

Have you notice there is no ? This means . This is because the degree of quotient is . So we can say, must equal to . i.e.

.

In most of the case, a remainder is simply a number (i.e. ) but, if the remainder is a polynomial then, we need to write . This happens if . I would be very happy if you could find it by yourself why we can write .

Now, we have all the ingredients to make a method that works for us. Thus, the method looks like this:

. The red line shows . This is our initial visualization.

In this figure, the blue line shows is multiplied by and the green line shows is multiplied by . This means, always multiply one term further than . All the elements of the fourth column are then added and equal to .

Similarly,

. The last figure is our final visual method for the synthetic division when the divisor is second-degree polynomial.

Before finishing this post, isn’t it great if we can implement this visual method in any one related example? So, our example goes like this: “Find the quotient and remainder when is divided by ?”

Our solution starts as and . Now, by using our visual method, we can have

. Hence, the quotient is . Since the sum of the second last column is not equal to zero so, the remainder becomes . I must say, when you multiply with second last column’s last row then, you should multiply it with , not .

**Quick Question for You**

Lastly, if you could write a program (in any programming language except maths ) to implement this method then, please feel free to share with all of us via comment.

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]]>The post An introduction to F-notation appeared first on Physics Log.

]]>In this post, I want to introduce a new type of notation which I named it as F-notation. You will know at last, why I need to introduce this, how the name came in my mind, and how it will help to prove one of the well-known theorem of Number theory.

So, the strategy goes like this: first, I will create the necessary ingredients for the given statement “The set is countably infinite”. This means the cartesian product of natural numbers is countably infinite, and countably infinite means you can able to find or count a natural number even at infinity. At the middle of the procedure, I will introduce F-notation. Let’s get started!

**Theorem. ***The set is countably infinite.*

*Proof. We know, *. So the cartesian graph looks like above figure [Caption: Cantor pairing function assigns one natural number to each pair of natural numbers]. I want to introduce two definitions.

**Definition 1. **The ordered pair can be written as where a is the first entry which lies on x co-ordinate and b is the second entry which lies on y co-ordinate. And, will give the solution as (i.e. ) such that and are greater than zero, and also not equal to zero.

First, we will pick out the diagonal’s points (i.e. ordered pairs) from the above figure. From the definition 1, we can write

where, the first diagonal () is a point, second () is a line and so on.

So, the pattern would be

.

Thus, we can write

.

This means,

when n =1 (first diagonal), gives ,

when n = 2 (second diagonal), gives and ,

when n = 3 (third diagonal), gives , and and so on.

This means you can generate the same as given by definition 1.

**Definition 2. ***Let us define a notation named as functional notation (i.e. F-notation) and given by . It states that, when different operation has same solution then, F-notation is .*

Since, satisfy the definition 2. So, . Now, it’s very easy to test whether this function gives countably infinite or not, by proving whether it is bijective or not.

**Test 1. ** Let,

.

Hence, the function is one to one.

**Test 2. **Let,

Therefore, the inverse function is .

Since, and .

Thus, i.e. .

The inverse of the given function where satisfy the domain elements. But as far we know, as . This means, range is equal to co-domain. Hence, the function is onto.

This means is bijective in nature.

Finally, we proved our statement using the new notation named F-notation. Let me know in the comment or via email, if you can apply this notation to solve any problems. For the pdf version of this post, download it.

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]]>The post Can you really measure distance between two points? appeared first on Physics Log.

]]>Of course! By using a ruler, right?

But in reality, you are actually coinciding given two points with the points of your ruler. Then you will say, these points are separated by a unit in length.

I want to mention now, in the above line consisting of points [see above figure], these points are equally separated. In order to make an equal separation, you do not need a standard ruler (i.e. with standard unit). Is that right? This means the line is a ruler but without unit.

Now, you can count all the points from a coinciding point with A to coinciding point with B including points lies between A and B in our new ruler then, you will say the distance between A and B is 6 (because in the ruler, counting starts from 0). But, where is its unit? Does 6 really mean the distance? No! This 6 really means the line coincided 7 points between A and B including themselves. Otherwise, nothing. We can even make more small spacing with more points in the line. So, how can we track different spacing and relationship between them? The solution is to assign “unit” to each and every spacing. You cannot simply measure spacing with itself. But, you can measure the ratio of the spacing to a standard ruler such that it gives one because of the coinciding concept.

Generally, if one measure a length, one actually measures the ratio of that length to a standard unit of length. i.e.

.

This is how AB get its unit.

In Quantum Gravity, we are interested to probe the spacetime even below the Planck length (order of m) but, we don’t know how to define spacing below that length. In reality, we don’t know does spacetime at that region is discrete or continuum? But, most of the candidate theory for Quantum Gravity believe that spacetime has a discrete nature, and even length and time has discreteness. So, if the length is discrete how to define a spacing between two points can be a major problem in Quantum Gravity. Because we define spacing AB supposing that it can include infinitely many points between A and B.

Lastly, this post is dedicated to Dr. Joshua Cooperman article’s arXiv:1604.01798.

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]]>The post Find the number of possible choices for x and y when A=100? appeared first on Physics Log.

]]>Our problem will go like this:

*Take any number* *, and then, take* *, where* *and* *. Find the number of possible choices for* * and* * when* *? Order doesn’t matter, e.g.* *and* * is same as* *and* .

To solve this problem first let’s tabulate the natural number and its sum as;

From the above pattern, we can say;

When is odd, the number of possibilities is

When is even, the number of possibilities is

And hence, the answer is 49 because is even.

If the pattern goes like this then, we can say these formulae are reliable. But if the pattern will break after some natural number then, it won’t work. *This means we need a theorem! If anyone could find(or invent) the theorem in regards to this problem. Please do comment below! And also, if you have another method to solve it. Please let me know! *

**Useful Resources**

One can find a wonderful discussion on my(this) problem at Mathematics Stack Exchange:

https://goo.gl/uYSzAS .

**Still, need to go further**

Perhaps one may find this problem very simple. But, I have another similar but extended problem for you. And the problem states like this:

*Take any number* *, and then, take where and . Find the number of possible choices for , and when ?*

* Order doesn’t matter, e.g. you can do: or or or or or . But you cannot do: or or or or or or or or soon.*

Looks like we also need general theorem for which we can solve any extended problem. I’m leaving these problems to you on behalf of willing to see its solution. Best of luck!

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]]>The post A full proof of Berry-Esseen inequality in the Central Limit Theorem appeared first on Physics Log.

]]>**[Note: I want to note here from the very beginning, this post is a bit technical. But I’m hoping that this will very helpful who is very needy of it. This proof is based on the book by W. Feller, “An Introduction to Probability and it’s application” and is only for identically independent distributed summands. Thus, I won’t prove the non-identical case because this post is a way longer. Please find it’s proof in Feller’s book.]**

Central limit theorem concern with the situation that the limit distribution of the normalized sum is normal as the sample size goes to infinity. But the question you may raise is, “*What is the rate of convergence of normalized sum distribution to the standard normal distribution?*“. Let’s answer this question by considering the case where the samples to be identical. To be more precise, let’s state like this:

*Let be independent variable with identical(or common) distribution such that, and, let stands for the distribution of the normalized sum . Then for all and , the supremum of convergence between and i.e. standard normal distribution is .*

Looks very boring! Right? Okay, let’s start with the history of the Central limit theorem(CLT).

The first proof of CLT was given by French mathematician Pierre-Simon Laplace in 1810. After fourteen years later, French mathematician Siméon-Denis Poisson improved it and provided us a more general form of proof. Laplace and his contemporaries were very interested in this theorem because they see the importance of it in repeated measurements of the same quantity. And thus they realized the individual measurements could be viewed as approximately independent and identically distributed, then their mean could be approximated by a normal distribution. Because this statistical plus probability theorem states that for a given sufficiently large sample size from a population with a finite level of variance, the mean of all samples from the same population will be approximately equal to the mean of the population regardless of the shape of the proposed distribution.

Then, the first convergence rate for CLT was estimated by Russian mathematician Aleksandr M. Lyapunov. But, the more refined version of proof is independently discovered by two mathematicians Andrew C. Berry (in 1941) and Carl-Gustav Esseen (in 1942), who then, continuously refined the convergence theorem of CLT and hence, given this theorem which is named as “Berry-Esseen theorem”. The best thing about this theorem is that it only considered first three moments.

Now, I think you are very eager to know about the proof of this theorem. Right? Let’s get started without any further delay!

From Feller’s book lemma 1 section 3.13 at Page 538, the upper bound between and is

where,

= characteristics function for which equals to ,

= characteristics function for which equals to ,

= maximum growth rate for such that .

The above expression can be found by starting with Fourier’s methods. And our proposed proof is based on smoothing inequality (refer Feller’s paper at section 3.5) such that,

.

The last inequality is the result of moment inequality. And, the normal density has maximum (I really don’t know why Feller chose this bound. I would be very happy if you guys could help me in this quest!).

So equation becomes,

Now, Let’s find ?

Isn’t it looks like the reverse triangle inequality with exponent ““? I mean this

if .

Thus, we can say

, if .

Again, let’s make our problem much simpler by proposing ?

First of all, let’s suppose so that . Thus, so that .

Look! How beautiful this looks like:

, putting the series of

, neglecting the higher order terms because for large then, .

The characteristics function for is

.

From the very first, I said as the sample size goes on increasing the shape of the curve of proposed distribution tends to match up with the normal curve. I mean this

Isn’t the smoothing concept looks like Taylor’s theorem? Exactly! Like Taylor theorem said, we can approximate any curve to a well-defined curve by a series expression. Likewise, we can estimate our proposed distribution with standard normal distribution by taking higher order terms. So, we will need to go like this

or, .

Now, multiply by and do integration both sides with the limit to i.e.

.

From characteristics property, the subtraction of two characteristics function gives another characteristics function and also we suppose, the result can be approximated by taking the higher order series. This is our trick:

, from equation .

Also, another inequality we can suppose is this:

.

For n = 3,

.

So, the equation becomes

.

In the left part of this inequality, we’re going to apply the Cauchy-Schwarz inequality as

.

Then, this will turn into

where the third part of the inequality has higher value than others.

For our need, we will use

, if , and second part is from Cauchy-Schwarz inequality

, applying the properties of Riemann integral in second part

, applying Cauchy-Schwarz inequality.

such that

.

Returning back the value of . we get,

.

Now, we conclude to smooth our proposed PDF. So that we can use . So, the equation becomes

, converting into exponential form with .

We know, the assertion of the theorem is trivially true for and hence we may assume .

We taking exponent both side in equation . We can get,

Thus, for n = 10,

.

Let me remind you equation with maximum equality i.e.

if

So, the right part of equation may serve for the bound i.e.

or,

or,

Thus, we can have

.

Also, we need formulate one more inequality. Let’s start from this:

for

.

Oh! I almost forgot. We need to construct something very useful. i.e.

, I have added two terms and applied triangle inequality.

First term Second term

which means,

First term = , from equation .

and,

Second term = , from equation .

Returning the above results in equation . we get,

.

Since , the above inequality should follow the integrand which means

, from equation

.

Also, we know . So,

.

From equation (13), let’s consider for then, thus,

and,

.

These above integrations can be done by using by-parts rule and also from Gamma function. But, I found a difficulty when solving on . If you guys solve it, please comment your solution. Thanks in advance!

So, equation becomes

.

For simplicity, we use below as final form:

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]]>The post I want to be a future physicist. But how? appeared first on Physics Log.

]]>**[Note: Target audiences are bachelor students who are studying physics. ]**

Let’s get started!

You want to be a physicist in near future. Right? That’s why you are studying Physics in your bachelor. But you may encounter an interesting question that is, “But, how?”.

After that you will go on searching for the people who are very expert in their field of physics and then, you ask questions to them. They will give you the best possible answer they can have. Now, here is a time you need to be aware of! I mean, statistically, if you take those answers as grant then, your career is biased by them. So, what you need to do is: you’ll need to take a lot of sample answers from different people. Then, you will need to analyze by yourself on how you can have straight possible career paths in your interested field.

Stopped! Right?

Don’t worry. Listen I’m here! So that I can give you some strategies that I have followed in the duration of my Bachelor studies.

First thing first, *find your passion in physics!* Again, you may encounter the same question but with a different sense. Right?

So, what you need to do is “*Question yourself on what field of physics that you are very interested in?*“. Not just because of media or promoters but because with a concrete reason why you are really interested in this field.

To me, it was because I’m very interested in the field of the *concept of the field due to Sir. Micheal Faraday to explain Electro-magnetism. *I mean when I was in high school, I was very fascinated by the chapter Magnetism and I even try to give a theory about how repulsion and attraction work in magnets. And this trigger to me further and very eager to know, whether is it a mathematical trick to solve physics problem or is it a physical thing that we can see in magnets, gravity and soon.

Let’s not go there!

Okay, if you have this! I suggest you not to read until you arrive on the next section.

But, if you still don’t know? Okay! Let’s try this out!

*Find out what things makes you ask questions a lot to your teachers or your friends or even to you?*

For example, let’s start with this:

You may have a question like: why apple always fall towards the earth but not earth towards an apple? Then, you find that this is due to gravity. And you’ll go on reading a lot more about Newton’s law of gravitation. Now, you satisfied with the answer by Sir. Issac Newton. Because you learned how gravity works. And you know small bodies is pulled by heavenly bodies. But you still don’t know why gravity behave like this? Now, your quest for searching the answer may arrive at General Theory of Relativity that was formulated by Prof. Einstein. Now, you will understand gravity in-terms of the geometry of spacetime. And you know that you can derive Newton’s law of gravitation in a very weak gravitational field from General Relativity. Now, you may have a question like, if there is strong gravitational field i.e. Blackhole somewhere in our universe, will general relativity work? You may find the answer, obviously! Now, you will question what about in the singularity of that black hole where there is fluctuating strong gravitational field? Now, you may surprise that you need to account another theory that can explain the very small things i.e. Quantum theory. Then, you clearly see you need to be a quantum gravitist to understand this thing.

What I want to say is, from a very simple question you find yourself to be a physicist of Quantum gravity.

But I want to aware you, it takes a lot of time and effort to figure it out! And in the meanwhile you will get depressed, have frustrated hours, and at last, you will lose your hope. But I strongly recommend, you need to keep in mind that it’s common to every researcher and when you didn’t give up for a long time, everything you stuck at will be normal (Credit: Dr. Jonah M. Miller, for this motivating words). So, the inspiration you should force yourself is, never stop questioning, maybe one day, you will find your passion. And I suggest, your passion/field need not seem sounds great. But the main thing is, you need to be passionate about your interested field of Physics.

If you found your passion then, we are ready for the next section.

**Why not I built some basic skills from today which I surely needed when I’ll be a physicist in my interested field?**

Most of the people that I have seen ( or even me) entered in Physics is because they have some particular field of interest, they want to see themselves in near future. Okay! Everything is cool up to this point. But the trouble comes when they did not grow themselves in their interest fields. And it’s not because they cannot and not either their university did not furnish the syllabus. I think this is because they get desperate on what to do! And during my Bachelor, I understand what to do, and questioned on: *why not I built some basic skills from today which I surely needed when I’ll be a* physicist in my interested field?

When you enter in graduate school (most probably for Ph.D.) then, you’ll need to train yourself with the necessary skills before you start your own research. That means you will be investing more time in your training. So, why not you start to train yourself from bachelor after knowing what skills do you need. Then, you can focus more on your Ph.D. thesis proposal problem.

But the question is, how can I know what are the basic skills needed?

Don’t worry! There is one social site that may help us!

Guess what?

LinkedIn (I’m not advertising it!). Now, but how?

So, let’s sign up there and complete your profile. Now, we are ready to do some research for yourself.

Suppose you are very interested in Astrophysics. Let’s find out what are the basic skills for it?

After you created your LinkedIn account, you can see a search box in the topmost left part of its website. Now, let’s type astrophysicist. And hit enter. You can see organizations, people and etc related to this keyword. Then, click the “People” tab. This is the thing that we’ll need! This means we have a long list of astrophysicists. And at the right side, you can see filter option which you definitely like to use, if you knew which location has more experts in this field. But for our case, let’s make our proposal simple by ignoring this option.

Open every profile in a new tab. And ignore all their professional details (like organizations, achievements etc.) and directly jump into “Featured Skills & Endorsement” heading.

Let’s define what I mean about “basic skills” in this context.* If there are any skills common in between the astrophysicists then, we call these skills as basic skills. *Suppose, every people has spectroscopy as a common skill then, we need to know this belongs to the basic skills category. In the above image, the same color between three different people refers to the skill that is common.

Sometimes you see a skill that is not common to any profile that you are sampling. Then, you need to warn yourself it does not belong to the category. If you are very eager to know, search for it. If not, just ignore it. Again, you can see something like “Social media” and the same sort of skills then, you need to think I need to ignore it. If you put it in this category then, you will get into trouble.

And I want to note here, there is also an endorsement number that other people endorse them. One way you can neglect skills by seeing the endorsements for those skills on every profile.

Suppose 23 astrophysicists, know about “Image processing” then, you need to think this is a basic skill.

Hurray, you know how you can find what are the basic skills. Best of luck!

After you know what are the basic skills for your interested field then, why not you learn some of them. I don’t suggest you will need to learn all of them but strongly recommend know at least what they are! A quick tutorial is also a good start. But do not try to be an expert in the first learning stage rather go for searching some research projects that are finished with these skills which you have just learned its basics. You can find it on arXiv or researcher personal websites. And try to reproduce their projects. This is one way you can build up your skills level from basics to intermediate. If you got stuck then, email to the respective authors about your question.

Wow! You know what actually the research be like!

This is a time where you especially want to do your own research and you need to have a mentor to supervise/track your research. And it’s really the hardest part everyone will ever meet in their scientific career. So, here is the list of the points I would suggest:

- Find the list of people who are working in your field of interest.
- Email them mentioning that you have done this, …, this research works till now with your CV. And read their papers and give a concrete reason why you really wanted to work with him/her. Because they might think that there are others people why him/her!
- If they appreciate your effort then maybe they will happily accept your proposal.
- If yes! Best of luck on your research.
- But most of the time, they don’t! Because of that do not stop. Have faith and continue building up your skills.

Again, Best of luck! May your wishes be fulfilled soon!

From search box of LinkedIn to spreadsheet and sort out the basic skills is actually a tedious job. Right? Due to very busy in my research, I could not focus on this task. So, I’m sharing here a list of things that one will need if they want to build an app under my concept.

- Learn how to use LinkedIn API.
- Make a statistical model. We need it because how can you give 23 a threshold number to say that “Image processing” belongs to basic skills. So this job will be done by your model.
*If someone want to enroll me in this project then I’m very happy to work on making this model.* - Design the flowchart to in-cooperate all tasks from extracting data, analyzing it and give results.
- I suggest you mention me as “
*This app is developed under the concepts by Damodar Rajbhandari, connect him through dphysicslog@gmail.com*” in the documentation or about page of your app. - Compile it.
- Share it in app store.
- Notify me by email. So that I can give you some relevant feedbacks to your app.

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]]>**[ Note: Actually, this article represents the script of the embed video in here. The featured snapshot is one of the slides from my mid-term progress defense presentation.]**

Before I answer this question, let’s go back to the time of Sir. Issac Newton. Newton sees space, time and gravity are the three different things and are not connected to each other. He thought of our universe to be a space consisting of matters and matters does not play any role to define the structure of space. Because at that time, there was no concept that space has some structure. Newton says, our universe and space are the interchangeable terms. He said space does not change due to anything so, he called it as static space.

Now comes with the conception of time, if different observers are observing the same events happening somewhere in the space then, Newton says the time noted by different observers are same and he believed that there is one universal clock and it gives the information that events have happened in this specific time. He then called time to be static. This means you cannot control that clock. It is said to be static time.

With that concept of space and time, he developed an approach which explains the behavior of heavenly bodies in our universe which we now called Classical Mechanics.

Let’s comes with gravity, the conception of gravity was realized thousands of years ago, even to monkeys, they know how to jump from one tree to another tree in projectile motion. But there was no any formalism for gravity which we can be thought of as a force. Newton gave a description of gravity in the form of Newton laws of gravitation by observing an apple falling towards the earth and questioning, by how much force does the earth needed to pull the apple towards it!

After 200 years, Prof. Einstein was there to oppose Newton concept of static time by giving an example. Suppose I have two light bulbs in my left and right hand then I set up these light bulbs in such a way that they light up simultaneously, this means I will see these two events happened in the same time. But if the observer who is driving and observing these events from the left side from me. He says the left light bulb lighten up first then the right one. And Einstein answered the “why” by saying: light also takes some time to take the information from the events happened somewhere in the universe to the observer’s retina. This causes him to give a value for the velocity of light and is the highest velocity you can ever imagine. So, time is not same for different observers and suggested that time should be relative and different from different observers. Since space consists of matters and Einstein said, matters are the source of creating events like dead/birth of stars. And events depends upon the observer’s observations! i.e. the variable for observations is “relative time’. And he said these two things are not different but the same i.e. from space and time, he gave us the term spacetime. And he gave the suggestion that spacetime and our universe are interchangeable terms, not space! With this concept, he developed the approach called Special Relativity which can explain the behavior of particle which has high velocity compared to the speed of light. But, he hasn’t touched on gravity. At the same time, there was also a revolution in the quantum theory that can explain the behavior of sub-atomic particles.

In 10 years, Einstein again put a question like this: Okay! An apple is falling towards the earth because I’m on the earth observing the motion of that apple. But what if I somehow live on the surface of that apple and observing the motion of earth then, will I see the earth is falling towards the apple? His answered was YES! Because of his Special Theory of Relativity, he said everything in the universe is in the motion relative to each other. To have a conclusion like this is seems doubt to see gravity as a force. This means Einstein understands that spacetime and gravity are the same things. Einstein realized that spacetime is a container that store events and matters, and this matters defined the structure of spacetime/our universe. This means our universe has also some structure and from the observations, our universe is rapidly expanding. With this concept, Einstein formulated the gravity in-terms of geometry and which is presented in General Theory of Relativity. On the other side, there was the development of the different approaches for quantum theory i.e. Matrix mechanics, Schrodinger picture of wave functions, Feynman path integral and now we have Quantum field theory. One must also need to be aware that General relativity is also a field theory.

Now, what is remaining?

If we are probing our universe on a very small scale i.e. nearly to the Planck scale or even below, then we cannot neglect the quantum effects. And even if we assume our universe is empty but due to vacuum fluctuation, the universe can get enormous fluctuating energies and this is the source of strong gravity. This means we have our universe to be “superposition of curved spacetimes”. Superposition means you cannot predict the geometry of our universe and curved spacetimes mean, our universe will have strong gravitational field due to vacuum fluctuations. And such universe is our an experimental lab for quantum gravity.

*So, we need an approach to give a better understanding the behavior of such quantum spacetimes.*

*I hope this script covers this snapshot!*

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**Special Thanks**

I would like to thank my supervisors (Dr. Jonah Miller & Prof. Udayaraj Khanal) & dearest Swastika Stha for always encouraging me when I’m feeling low.

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]]>**[Note: This article is somewhat technical.]**

Many many years ago, there was a problem which created a mind-boggling puzzle to the eminent mathematician named Leonard Euler. But this isn’t the main starting of our story. There was a historical city of Königsberg in Prussia but after World War II, it has been named as Kaliningrad of Russia. Well, I’m not trying to make this story by creating some sad mood but it is the starting point where the problem was created. In that city, there is an island, called the kneiphof; the river (Pregel) which surrounds it is divided into two branches, and these branches are crossed by seven bridges [1]. Usually, in the evening time, the people who lived in this city used to gather near to the bridges and entertain themselves. These bridges act as a junction for them. The most entertaining things to them were the seven bridges of that city. Can you guess, how? *They were trying to devise a route around the city which would cross each of the seven Königsberg bridges just once and only once.* Every time they went to the junction and returned with no answer to that problem. Their ego was hurt badly. Some of them said, there is no any path to cross all bridges at once but they don’t have any concrete ideas to support it. And, many of them believed that the task was impossible. But it was not until the 1730s that the problem was treated from a mathematical point of view and the impossibility of finding such a route was proved [2].

Finally, the problem was given to Euler. As he was trying to give some thoughts in that problem, he saw it has a little link with mathematics and he added, its discovery doesn’t depend on any mathematical principle [3]. Again, he was trying to catch which known mathematical principle can be applied. His good luck!* He got some clue*. *The clue was very simple; he stated, this is a geometry problem with no relationship with distances or angles.* How can it be a geometry problem which does not have any relationship with distances or angles? As he was trying to understand the problem, he realized there’s the only relationship with bridges, lands, and river. He went through the primitive definition of Geometry. Have you ever stated the primitive definition of Geometry? I’ll say, it is a pure branch of mathematics which deals with points, lines, curves, and surfaces. The geometry which we were taught in elementary classes was Geometry of magnitude. He begins to thought bridges as curves or lines and lands as points. And, this is an elementary concept to start with Graph Theory. So, he laid the foundations of Graph Theory. He created the notion that you can start from any of the given bridges. So, he thought that the point where you starts acting like the position where you situate. Because you can get the same result from the different position. So he claimed that it should be the problem related to the Geometry of Position. He was aware that,* he was dealing with a different kind of geometry where it doesn’t involve measurements and calculations*. This kind of geometry was defined by Leibniz and it was in the primitive phase in Euler’s time. Though Leibniz initiated this concept, there wasn’t such a problem which can be solved by this concept. But, Euler found the problem that can be solved by this concept. *Today, we call it Topology*. This concept suggested that how things connect one to another. It is especially concerned with space and it also studies how properties of a space change and don’t change when space is distorted. In this problem, it doesn’t depend upon either the bridges are long or short, either the position where you start is farther or nearer from the bridges. The points which are concern with land is actually a position of an experimenter. And the point can be specified in any place within the specific land.

In 1736, Euler published a paper on the solution of the Königsberg bridges problem entitled “Solutio problematis ad geometriam situs pertinentis” which translates into English as “The solution of a problem relating to the geometry of position” [4]. His original paper was first published in Commentarii academiae scientiarum Petropolitanae 8, 1741, pp. 128-140 [7] and it was the first journal series published by the St. Petersburg Academy [8]. In this paper, he proved the problem has no solution. The difficulty was the development of a technique of analysis [5] and to establish the general method to compute not only that problem but others too.

Now, we are dealing with “How Euler solved the Königsberg bridge problem?” and some basics of Graph theory.

In the Euler’s original paper, He was asked as “*Whether anyone could arrange a route in such a way that he would cross each *bridge* once and only once?*” [2]. Here, I’ll be discussing only two methods out of three, if you want to read the last method then, you can prefer reference 2 that is mentioned in the section “References”. Also, *I’ll be pointing each and every steps, analysis, results and conclusions that he made and, make you simple to understand.*

**Step I:**

First, we have to give notation to the seven bridges and four land areas. Here, we gave notation to seven bridges as a, b, c, d, e, f, g and four land areas separated by rivers as A, B, C, D and also, replacing the map of the city into simpler diagram showing only river, bridges and lands which is exactly done by Euler.

Howdy, I’m Damodar. I love to introduce myself as a curious & focalize person. If you want to know more about me and my blog head over to my “About the Author” and “About the Blog” pages.

**Analysis I:**

He formulated the general problem as, “Whatever be the arrangement and division of the river into branches, and however many bridges there be, can one find out whether or not it is possible to cross each bridge exactly once?” [2].

**Step II:**

He said, If you go from A to B over bridges a or b, you can write as AB where the first letter says you are leaving from this place and the second letter says you are entering in this place.

Similarly, if you leave B and crosses into D over bridge f then, it can be represented by BD. If you would like to cross AB and BD, it can be represented by the three letters ABD where the middle letter B refers to both the area which is entered in the first crossing and to the one which is left in the second crossing. Then, if you go from D to C over the bridges g, it can be represented by four letters ABDC which says you started from A, crossed to B, went on to D, and finally arrived in C.

**Result I:**

Hence, the crossing of three bridges gave us four letters and, the crossing of four bridges would give five letters. In general, how many bridges the traveler crosses, *his journey is denoted by a number of letters one greater than the number of bridges*. So, the crossing of seven bridges requires eight letters to represent it.

**Conclusion I:**

Thus, we have to use four letters (i.e. ABCD) in order to represent the possible paths with eight letters.

**Analysis II:**

We are now trying to find such a rule whether or not such an arrangement can exist. Also, we are trying to find what is the number of repetition of a letter A that will appear when we cross an odd number of bridges (i.e. five bridges in our case) from the A or into the land A?

**Step III:**

Let’s make our diagram simple, we will make a river with no branches in it and, which separate the land areas A and B. We put six bridges over the river.

Now, consider bridges an only then, we have two cases;

Case I: You must be in A before crossing the bridges a. (i.e. AB)

Case II: You have reached A after crossing the bridges a. (i.e. BA)

Hence, A appears exactly once.

Again, consider three bridges a, b and c then, we will found A will occur twice whether you begin from A or not. (i.e. ABAB or BABA)

Similarly, consider five bridges a, b, c, d and e then, we will found A will occur thrice whether you begin from A or not. (i.e. ABABAB or BABABA)

**Result II:**

In general, *if the number of bridges is odd, first increase it by one and, take half the sum; the quotient represents the number of times the letter A appears*. Hence, in the given problem we can say, A occurs thrice, B occurs twice, D and C must occur twice. So, the total number of letters will be nine.

**Conclusion II:**

It says, we have to use four letters (i.e. ABCD) in order to represent the possible paths with nine letters.

**Comparison between the Results I and II:**

From the result I, the possible number of letters should be eight but from the result II, *we got nine letters within the similar condition* (exactly cross the bridges once and only once). So, we arrived at a contradiction. Hence, the crossing of the seven bridges of Königsberg at once is not possible.

**Analysis III:**

What is the number of repetition of a letter A that will appear when we cross even number of bridges from the land A or into the land A?

**Step IV:**

If the number of brides leading to A is even, then in describing our journey; we have to consider whether or not we start our journey from A. But in the former case, we don’t need to consider this situation. Now, consider two bridges then, we have two cases;

Case I: If we start from A then, the letter A must occur twice. (i.e. ABA)

Case II: If we start from B then, the letter A must occur once. (i.e. BAB)

Again, consider four bridges then, we have two cases;

Case I: If we start from A then, the letter a must occur thrice. (i.e. ABABA)

Case II: If we start from B then, the letter A must occur twice (i.e. BABAB)

Similarly, consider six bridges then, we have two cases;

Case I: If we start from A then, the letter A must occur four times (i.e. ABABABA)

Case II: If we start from B then, the letter A must occur three times. (i.e. BABABAB)

**Result III:**

In general, if the number of bridges is even, then the number of times the letter A appears will be half of this number if our journey is not started from A, and if our journey does start at A, it will be one greater than half the number of bridges.

**Conclusion II:**

Hence, we can say the number of times the letters appears denoting that area depends on whether the number of bridges leading to each area is even or odd.

In Graph theory, we call bridges as edges, arcs, or lines and area as vertices, nodes, or points. This will result in a graph! Never confuse with term graph that uses co-ordinates. This is different, it doesn’t depend upon magnitude. Now we want to define graph; it is an ordered pair with vertices and edges. It can be denoted as G(V, E). The number of edges connected to a vertex is known as degree or valency of a vertex. An edge starts from a vertex and end on the same vertex is known as a loop and it can be counted as twice. Also, there is a kind of graph named by several authors as multigraph. If there are parallel edges and both ends within the same node then, the graph is said to be a multigraph [6]. Also, the path that crosses all the given edges at once is said to be a Eulerian path. In the given problem, let’s make the graph of it;

How many nodes, edges, a degree of a vertex and the total number of degree of vertices are there? Is there is any loop in the graph? From the above comparison between the results I and II, Is there is any Eulerian Path? Is it simply a graph or is it a multigraph? Answer the above question to yourself! It will make you understand some basics of Graph theory.

To be noted: The edges shouldn’t intersect each other. If you made it intersected mistakenly then, the point of intersection shouldn’t be counted as a vertex.

**Analysis IV:**

Since we can start our journey from any one land either it may have odd or even number of bridges leading to it. Euler defines, corresponding to the number of bridges leading to each area, the number of occurrences of the letter denoting that area to be half the number of bridges plus one, if the number of bridges is odd, and if the number of bridges is even, to be half of it [2].

**Step V:**

First, take a total of all the occurrences. Then, if the total of all the occurrences is equal to the number of bridges plus one, the required journey will be possible and will have to start from an area with an odd number of bridges leading to it. However, if the total number of letters is one less than the number of bridges plus one, then our journey is possible starting from an area with an even number of bridges leading to it. Thus, the number of letters will be increased by one. One less is because, we use the result III of not starting from the land, and we don’t know whether we have to start our journey from the land with an even number of bridges leading to it or not start from that land. Increased by one is because we have to use the result III of start from the land with an even number of bridges leading to it.

**General Steps:**

1. For various land areas which are separated from one another by the river, denote it by the letters A, B, C,….

2. For a possible path, the total number of letters occur = 1 + total number of bridges.

3. Write the letters A, B, C,… in a column, and write next to each one the number of bridges leading to it.

4. Indicate with an asterisk those letters which have an even number of bridges leading to it.

5. Next to each even bridges, write half the number and, next to each odd, write half the number increased by one.

6. Add all the third column elements.

7. If this sum is one less than or equal to, the number of bridges plus one then, we can conclude that the journey is possible. So, there are two cases:

Case I: If we get the sum to be one less than the number of bridges plus one then, our journey must begin from one of the areas marked with an asterisk.

Case II: If it equals to the number of bridges plus one then, our journey must begin from an unmarked one.

**The final solution to our Königsberg bridge problem:**

We now are using the above general steps to work out the given problem as:

The number of bridges = 7, which yields 8 letters.

Land | Leading bridges to it | Using Step 5 |

A |
5 | 3 |

B |
3 | 2 |

C |
3 | 2 |

D |
3 | 2 |

**Result IV:**

Since we got more than 8 (i.e. 9). So, such a journey can never be made.

**Conclusion IV:**

Hence, from the comparison between the result I and II, and result IV; we don’t have to be afraid to say, this problem has no solution.

**A problem for you:**

**Condition:** You can cross each bridge once and only once.

1. How many letters that we need to represent our possible path?

2. Can we make our journey possible or not?

3. If yes then, can you find such a route in which you cross each and every bridge once and only once?

4. Can you write all the path for the possible route?

5. Draw a graph of this problem?

6. How many nodes, edges, a degree of a vertex and the total number of degree of vertices are there?

7. Is there is any loop in the graph?

8. Is there is any Eulerian path?

9. Is it simply a graph or is it a multigraph?

Apply the general steps in this problem and answer the above questions to yourself, also convince your answer is right and whether you have understood the Euler’s method or not?

**For curious people**

*How many possible ways to cross the bridges at once, if we know the problem has a *solution*? (Problem relating to Combinatorics)*

In a nutshell, Euler’s work on this problem showed us some understanding of the primitive phase of Topology and Graph Theory.

**Good resources**

[1] Barnett, J. H. (n.d.). Early Writings on Graph Theory: Euler Circuits and The Königsberg Bridge Problem. Resources for Teaching Discrete Mathematics Classroom Projects, History Modules, and Articles, 197-208. doi:10.5948/upto9780883859742.026

[2] Biggs, N. L. (1976). Graph theory, 1736-1936: Norman L. Biggs, E. Keith Lloyd Robin J. Wilson.

[3] Alexanderson, G. L. (2006). About the cover: Euler and Königsberg’s Bridges: A historical view. Bull. Amer. Math. Soc. Bulletin of the American Mathematical Society, 43(04), 567-574. doi:10.1090/s0273-0979-06-01130-x

[4] A history of Topology. (n.d.). Retrieved August 16, 2016, from http://www-gap.dcs.st-and.ac.uk

[5] Seven Bridges of Königsberg. (2016, July 29). In Wikipedia, The Free Encyclopedia. Retrieved 06:22, August 16, 2016, from https://en.wikipedia.org/

[6] Multigraph. (2016, March 24). In Wikipedia, The Free Encyclopedia. Retrieved 08:29, August 19, 2016, from https://en.wikipedia.org

[7] E53 – Solutio problematis ad geometriam situs pertinentis. (n.d.). Retrieved August 19, 2016, from http://eulerarchive.maa.org/pages/E053.html

[8] Commentarii. (n.d.). Retrieved August 19, 2016, from http://eulerarchive.maa.org/

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