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### kstheking's blog

By kstheking, history, 10 months ago,

So we had an Interview Question at Codenation, This is the question Question Image. I understand that we can do the update node and sum of values through segment trees but what I can't understand is how to find the aith beautiful number, can someone help me out please! Also the constraints of X in the question are below 10^5

• +9

 » 10 months ago, # |   0 Auto comment: topic has been updated by kstheking (previous revision, new revision, compare).
 » 10 months ago, # |   +37 Hi!A simple bfs should be enough.Here's a code for creating series whose sum of square of digits <= x.The remaining is just classical euler tour and update using segment tree.I'll leave the complexity analysis to you.Thanks!
•  » » 10 months ago, # ^ |   0 Got it, Thanks for answering!
•  » » 10 months ago, # ^ | ← Rev. 2 →   0 Dsxv, could you please clear this for me. q.push({(cur*10 + i)%mod , sum + i*i}) ; Won't taking mod here change the number itself. It'll yield the wrong number when used further, right? For instance, for x=1, we get the first 15 numbers as this: Output for x=11 10 100 1000 10000 100000 1000000 10000000 100000000 1755647 17556470 175564700 757402647 586315999 871938225
•  » » » 10 months ago, # ^ |   +3 Hi!The deciding parameter for the series is actually the second number i.e, sum + i*i.As far as the numbers in series are concerned, notice that the question is asking us to take mod of sum of all the values (fi) in the subtree.Where the values (fi) in the subtree are nothing but the elements of our series.Now, property of modulus addition:(A+B+C...) % mod = (A%mod + B%mod ...) % mod so, taking mod of A, B ... beforehand doesn't affect our answer.Feel free to ask anything if you still have any doubt.Thanks!
•  » » » » 10 months ago, # ^ |   0 Yup, The deciding parameter for the series is actually the second number i.e, sum + i*i.I get it now. Thanks.
•  » » » » 10 months ago, # ^ |   +8 Hi, I have a little doubt, since we are beginning with 1, wouldn't this code leave out numbers which begin with 2, 3 etc.
•  » » » » » 10 months ago, # ^ |   +3 Yes, Updated. Thanks for pointing out!
•  » » » » 10 months ago, # ^ |   +3 Your logic won't work for x = 5, you push {1, 1} and then push {10*1 + i, 1 + i*i}, but the number {2, 4} also satisfies the constraints.....how will your algo detect that??
•  » » » » » 10 months ago, # ^ |   0 Updated. Thanks!
•  » » 10 months ago, # ^ | ← Rev. 2 →   0 Hi bro can you pls suggest some beginner level problems involving euler tours or similar concepts? would be really helpful
•  » » » 10 months ago, # ^ |   0 Hi!I suggest you to read LCA and learn the idea behind it as well as solve the questions mentioned at the bottom (First one is literally just template check).Thanks!
•  » » 10 months ago, # ^ |   0 why a segment tree is required?
•  » » » 10 months ago, # ^ | ← Rev. 2 →   0 It's required for finding the sum all nodes of subtree as well as updating the nodes.Edit : please read my comment below for more info
•  » » 10 months ago, # ^ |   0 How to update using segment tree, it is not given in the question that the tree is binary tree , how exactly will the indexing of segment tree will work????????
•  » » » 10 months ago, # ^ |   +3 Hi! We can't apply segment tree directly onto the tree for solving this problem. First we have to flatten the tree using Euler tour. After this we know the in and out time of every node. We know that all nodes in the subtree of (say X) also lie between the range (in and out of X) so we can just get sum of this subarray using segment tree and divide this by 2. for updating an element we just update the element at it's in and outindexes
•  » » » » 10 months ago, # ^ |   0 I think flattening the tree into an array of size n is more "natural". There, you only increase t when you enter a new node, so you still have t_in and t_out but you only need to update once and sum doesn't over-count.
 » 10 months ago, # |   -30 I thought companies ask to reverse a linked list.
 » 10 months ago, # | ← Rev. 2 →   0 how did you solve 2nd question? question 2 laser tagi thought of dp recurrence donno if its correct tho as hackerrank was very slow during the test(my test was completed before knowing the status of submission). Spoilerdp[i] = 1 + max(dp[i/2] , dp[i/2 +1] ) (if i is odd) dp[i] = 1 + dp[i/2]. (if i is even)dp[i] represents the number of games needed for i people. at last check if dp[i] <= 2*x
•  » » 10 months ago, # ^ |   +5 Just take ceil of log2(n). This will be your number of matches.
•  » » 10 months ago, # ^ |   +1 as one_last_chance mentioned, what we can do is first distribute them in teams of half, now consider one half and swap its half people to the other side,now you have a quarter of people who have been together, swap half of them to the other sideBy doing this log2(n) times you can generate configurations in which each person has fought with every other person
•  » » 10 months ago, # ^ |   0 I implemented the same solution. It passed all tescases.
•  » » 10 months ago, # ^ | ← Rev. 2 →   0 actually mine also should work! as it is same taking ceil(log2(i))but donno about the status of the submission thoughif(i%2==1){dp[i] = 1 + dp[i/2 +1];}else{dp[i] = 1 + dp[i/2];}
•  » » 10 months ago, # ^ |   0 Can u share the 3rd problem?
•  » » » 10 months ago, # ^ |   0 sure! link for 3rd problem
•  » » 10 months ago, # ^ |   0 just sharing a similar question if wanna try https://www.codechef.com/SEPT19A/problems/DOOFST
•  » » 10 months ago, # ^ |   0 it is correct
 » 10 months ago, # |   0 Did anyone solve the xor one?
•  » » 10 months ago, # ^ |   +4 i observed binomial coefficients pattern (pascal triangle) and solved using it. Spoilerlets say the numbers are arr = a b c d e f g h and every time arr is modifiedk = 1 ;arr = 1a xor 1b , 1b xor 1c , 1c xor 1d ... so onk=2 ; arr = 1a xor 2b xor 1c , 1b xor 2c xor 1d ... so onk=3; arr = 1a xor 3b xor 3c xor 1d , .. so onk=4; arr = 1a xor 4b xor 6c xor 4d xor 1eafter finding the binomial coefficients of level k = kgiven+1 (we can do that in O(K) time). initialise your ans as 0 ans start from idx = m (given) ans only xor array[idx] with ans if our current coefficients is odd. here array is initial array provided for us.but we can't calculate ncr (binomial coefficients list) for k = 1e5 level (as it overflows the integer). Actually we don't need to (think for a bit you'll get it).
•  » » » 10 months ago, # ^ |   0 I didn't know about Pascal Triangle, couldn't make sense of the pattern, thanks!
•  » » » » 10 months ago, # ^ |   0 the nth level is same asnC1 , nC2 , nC3 , nC4 ... .. nCnncr is n!/((n-r)! * r!)
•  » » » » » 10 months ago, # ^ |   0 how do u solve tournament question.i solved in o(n2) using dp stilll some cases are not passing.
•  » » » » 10 months ago, # ^ |   0 I knew about Pascal Triangle, still couldn't make sense of the pattern, thanks!
•  » » 10 months ago, # ^ | ← Rev. 4 →   +4 Lemme describe the question first for those who don't know. QuestionGiven an array A of length n, apply the following operation k times : Create a new array B of length n - 1 such that B_0 = A_0 XOR A_1, B_1 = A_1 XOR A_2,......,B_{n - 2} = A_{n - 2} XOR A_{n - 1}. Replace A with B i.e. A = B Now you have to tell m-th element of final array A Constraints1 <= n <= 1e51 <= k <= n - 10 <= m < n - k1 <= A_i <= 1e9 HintTry to find how A will look if k = power of 2. SpoilerLet n = 10 and k = 4A will look like (let final array is denoted by B)B_0 = A_0 XOR A_1 XOR A_2 XOR A_3B_1 = A_1 XOR A_2 XOR A_3 XOR A_4and so on. Code#include using namespace std; // main Observation vector solve(vector a, int p){ // p should be power of 2 vector ans; int x = 0; for(int i = 0 ; i < p ; i++) x = x ^ a[i]; ans.push_back(x); for(int i = p ; i < a.size() ; i++){ x = x ^ a[i - p] ^ a[i]; ans.push_back(x); } return ans; } // bruteforce soln int bf(int n, int k, int m, vector a){ while(k--){ for(int i = 1 ; i < n ; i++){ a[i - 1] = a[i] ^ a[i - 1]; } n--; } return a[m]; } int main() { srand (time(NULL)); int t; cin >> t; while(t--){ // prepare input int n, k, m ; vector a; #ifndef input n = rand() % 10000 + 2; k = rand() % (n - 1) + 1; // 0 to n - 1 -k m = rand() % (n - k); for(int i = 0 ; i < n ; i++) a.push_back(rand()%10000000); #else cin >> n >> k >> m; for(int i = 0 ; i < n ; i++){ int x; cin >> x; a.push_back(x); } #endif // duplicate input int d_n = n, d_k = k, d_m = m; vector d_a = a; // input preparation over int bf_ans = bf(n, k, m, a); while(1){ // calculate highest power of 2 less than k int p = 2; while(p * 2 < k) p *= 2; if(p - 1 <= k and k > 1){ k -= (p - 1); a = solve(a , p); continue; } //bruteforce; at this point value of k would be 1-digit so bruteforce will work n = a.size(); while(k--){ for(int i = 1 ; i < n ; i++) a[i - 1] = a[i] ^ a[i - 1]; n--; } break; } if(a[m] != bf_ans){ cout<<"failed"; break; } else cout<<"ok\n"; } return 0; } P.S. : I was not able to code it during the contest and just verified with bruteforce soln after the contest, so there might be some mistakes. I would appreciate if anyone points out any issue.
•  » » » 10 months ago, # ^ |   +2 How did u do tournamnet questions??i used dp for that like:dp[2]=1,dp[1]=1;thenfor(i=3;i<=n;i++){int t=INT_MAX;for(j=1;j=(dp[n]+1)/2))"1";else "0";what is wrong in this
•  » » » » 10 months ago, # ^ |   0 Could you please use good formatting?
•  » » » » » 10 months ago, # ^ | ← Rev. 2 →   +6 88179101 plz see at this dp[1]=0
•  » » » » 10 months ago, # ^ | ← Rev. 3 →   0 N was the total numbers of player . Assume N is a perfect power of 2. Divide them in teams of size N/2 and let them fight. Team A = N/2 Team B = N/2 players Now each player in Team A will fight against every player in Team B . Only the players of Team A & B haven't fight among themselves . So again divide them in two halves Team A ( N/2 ) -> Team C (N/4) Team D (N/4)Team B ( N/2 ) -> Team E (N/4) Team F (N/4)Team C & D & E & F will be of size N/4.Now if you let fight between C & D and E & F then +2 will get added in answer . But is this the optimal ?No , Because you can club Team C and E , Team D and F then let them fight . doing this +1 will get to the answer.So basically what is the terminating condition ? when size is equal to 1N -> N/2 -> N/4 — > N/8 -------------> 1 solve it to get the answer.So ans is log(N) base 2 if it is a perfect power of 2. if not then log(N) base 2 + 1 ( for the remaining )
•  » » » » » 10 months ago, # ^ |   +3 thnx.
•  » » 10 months ago, # ^ |   0 Just brute force with help of bitsets.
 » 10 months ago, # | ← Rev. 2 →   +4 For Question 3 Simple/well known fact is that for any given number $x$ we know $xor(x,x)=0$, so if we take xor of $x$ with itself an even number of times then $xor(x,x,x....)=0$, but if we take xor of $x$ with itself odd times the result is $x$.Notation Used- $xor(x,x,x...x)[k-times]=x^k$ Consider the simple array $a=1,2,3,4,5$For observing the pattern assume $m=0$ So we will look only at the value $a[0]$ Performing the operation 1 time we get — $a[0]=[1\cdot2]$ here $1\cdot2=xor(1,2)$ same notation is followed below 2nd time — $a[0]=[1\cdot2^2\cdot3]$ 3rd time — $a[0]=[1\cdot2^3\cdot3^3\cdot4^1]$ 4th time — $a[0]=[1\cdot2^4\cdot3^6\cdot4^4\cdot5^1]$ 5th time — $a[0]=[1\cdot2^5\cdot 3^{10} \cdot4^{10}\cdot5^5\cdot6^1]$ Now observe the powers of $1,2,3,4,5,6$ which are respectively $1,5,10,10,5,1$ which helps us notice the pattern as we can now see that powers obtained after 5th time are nothing but ${5 \choose 0},{5 \choose 1},{5 \choose 2},{5 \choose 3},{5 \choose 4},{5 \choose 5}$But how to find $4^{10}$ or $2^5$(xor-value) ? For finding the value we just need to know for any $k$, $i$ when ${k \choose i}$ is even and when it is odd. or we just need to find ${k \choose i} \% \space 2$ for doing this many methods are given on gfg you just need to select the appropriate method keeping in mind the time-complexity as here $k$ is the of the order of $10^5$. Some methods are given here, and here. Spoiler int main() { int parity[k + 1]; //compute parity of (k)C(j)for j=0 to k int ans = 0; int j = 0; for (i = m; i <= k + m; i++) { //if (k)C(j) is odd then xor it with //answer othwerwise not if (parity[j] == 1) ans = ans ^ a[i]; j++; } }
•  » » 10 months ago, # ^ |   +12 That's an interesting pattern. Why does this happen?
•  » » » 10 months ago, # ^ | ← Rev. 2 →   +15 (most of what is written below is based on this amazing video by kartik8800 I really recommend watching it link) Really Long comment ahead — We start off by looking at the really cool animation of pascal triangle on wikipedia, which offers us some satisfaction because something similar is happening there. Note how in the above animation the value of one row is calculated on the basis of previous row, which itself is based on the recurrence relation ${n \choose r}={n-1 \choose r}+{n-1 \choose r-1}$So for this particular question what we are essentially doing is sort of like pascal triangle. ImageIn the above example the value of a node is simply xor of the two nodes below it(like xor(1,2)=3). This makes sure that the size of the array is reduced by 1 after every step(as in our original question). Indeed another way of stating the above visual operation can be - given an array at each step $a[i]=xor(a[i],a[i+1])$ is done for a particular $k$ number of times.Let's now jump to a seemingly unrelated problem the famous dp problem. The question is about how to count the number of shortest path in a grid from (0,0) to (m,n) if only moving to the up or right cell is allowed at each step. This has a very short dp solution based on the recurrence - $f(m,n)=f(m,n-1)+f(m-1,n)$ where $f(m,0)=1$ and $f(0,n)=1$. However the number of such paths can be directly proven to be ${m+n \choose m}$. A very nice combinatorial/intutive proof is given at brilliant and stackoverflow. Jumping back to our question assume we have an array of size $k+1$ , since performing the operation each time reduces the size of our array by 1, so after $k$ such operations only one element will be left and we want to find that value. For finding the value all we need to find is for each index $i$ in the original array ( that is $i=0,1,2...,k-1,k$) what is its contribution in the final array. ImageNote that the number on top of each node is just it's index. The image also shows at which positons the index=3 is contributing. Now the real insight for linking the two problems is that the contribution of the index $i$ is nothing but the number of ways to reach the top element if we are only allowed to either move up-left or up-right(as shown in image) The reason for that is based on the idea that each time the element goes up-left or up-right it contributes it's current value to that index, this is even more clear by looking at the recurrence relation. Recurrence relationLets denote the contribution of the element $i$ in the original array to a position $j$ after performing $x$ operations be $f_i(j,x)$.Then $f_i(j,x)=f_i(j,x-1)+f_i(j+1,x-1)$(base case for above recurrence is left for the reader if any :P)So all that is left to calculate is to find the number of ways to reach the top-most element for the element at index $i$. It is easy to observe now that it takes exactly $i$ up-left moves and exactly $k-i$ up-right moves for it to reach the top(for example the element at index 3 requires 3 up-left move and 4-3=1 up-right move). So by the formula for number of such paths= ${m+n \choose m}$. For our question the number of such paths is given by $m=i,n=k-i$. ${i+(k-i) \choose i}={k \choose i}$Which explains the formula I got in my previous comment. Additional questionCan you observe something similar if instead of $xor$ , the operation $sum$ is used ? what about multiplication ? BonusYou may have come across a solution which uses bit-manipulation however that is also based on this pattern and uses lucas-theorem(without even knowing about it), it is essentially based on the fact that all we need is ${k \choose i} mod \space 2$, can you come up with that solution after seeing this pattern ?
•  » » 10 months ago, # ^ |   +1 That is a very neat and clever observation.
 » 10 months ago, # |   +22 Here is my O(N) solution to ZaurXOR: https://youtu.be/lccWXAd7enI I think the solution is very easy to comprehend and does not use any pattern observation etc. I have used number of paths in a graph to evaluate the answer. Basically I have developed a DP solution and optimized it using some very basic combinatorics!
 » 10 months ago, # |   0 Hey... How do you get to know about these tests?? I missed Facebook HackerCup and now this. Can anyone please tell me how can I keep track of these interview tests or competitions? Plz, it would be a great help!!!
•  » » 10 months ago, # ^ | ← Rev. 2 →   +13 use https://clist.by/ to know about ongoing contest. btw codenation test was only for college student for hiring and internship. Their college's training and placement cell float the test link.
 » 10 months ago, # | ← Rev. 2 →   0 The last Question zaurxor had weak test cases! I wrote a O(n) solution but some of my friends passed n.k solution which is not fair
•  » » 10 months ago, # ^ |   0 The n.k brute force indeed passed majority of the test cases but I think it failed about 4 of them
•  » » » 10 months ago, # ^ |   +5 Actually implemented O(n*k) soln and it failed just 1 testcase.
•  » » 10 months ago, # ^ |   0 Can u share your code and approach?


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# Capacitor questions

1. Apr 24, 2006

### Agent

Hey, just started this topic and I could use some help. Can someone explain the results? I got a bunch of equations and my head is starting to hurt. :grumpy:

First question.
Find the ammount of energy stored in a 1,000-microfarad capacitor that has been charged to 18 volts.

And
A 220-microfarad capacitor is being charged through a 330-ohm path from a 21 volt battery. Calculate the time required for the capacitor voltage to reach 14 volts after the final circuit connection is made.

2. Apr 24, 2006

### Ouabache

Welcome Agent to Physics Forums!! I see this is your 7th post. Feel free to browse around, it is a fun way to learn things.

Since this is a homework question, the best place to post would be: <Homework & Coursework Questions> .. your question can fit in either subtopic <Introductory Physics> or <Engineering, Computer Sci and Tech>

3. Apr 24, 2006

### Agent

4. Apr 24, 2006

### Staff: Mentor

And to get you started, Agent, can you tell us...

-1- What equation relates the charge on a capacitor to the capacitance and the voltage? What similar-looking equation relates the energy stored on the capacitor to the capacitance and the voltage?

-2- What can you tell us about how exponential decay (and exponential rise) equations are used in R-C circuits? What is the definition of the RC time constant? What is the equation that relates the voltage on a capacitor that is charging through a resistor from a constant voltage source? How do you solve that equation for a time, when you are given the other values?

5. Apr 24, 2006

### Staff: Mentor

Okay, this thread didn't get moved yet, so let's stop posting here, and move to the new thread started by Agent in the Homework area....

Last edited: Apr 24, 2006
6. Apr 24, 2006

### Staff: Mentor

Agent, you need to show *your* work, not the example work of your teacher. Can you answer some of these leading questions to get you started?

-1- What equation relates the charge on a capacitor to the capacitance and the voltage? What similar-looking equation relates the energy stored on the capacitor to the capacitance and the voltage?

-2- What can you tell us about how exponential decay (and exponential rise) equations are used in R-C circuits? What is the definition of the RC time constant? What is the equation that relates the voltage on a capacitor that is charging through a resistor from a constant voltage source? How do you solve that equation for a time, when you are given the other values?

7. Apr 24, 2006

### Agent

That was my work, I just didnt go through the steps because its hard to write it out here. Anyways, heres the first.

1.
C=Q/V

ES=
1
- CV^2
2

ES=energy stored
C=1,000mF
V=18
so then its 162? I'll be back with the 2nd question.

Last edited: Apr 24, 2006
8. Apr 24, 2006

### Ouabache

Good job Agent, on your equations.. Here is another reference you may want to keep handy for future questions like this.

Here is an example of how to write equations.
(select it and see the simple coding I used)
$$U = \frac{1}{2} CV^2$$

Be careful with the math (double check your decimal places). And what units is your answer in?

Last edited: Apr 25, 2006
9. Apr 26, 2006

### Ouabache

On your RC circuit question, tell us what you understand about this kind of circuit. Have you learned about the behavior of this kind of circuit with respect to time? After you give some of your thoughts we can help steer you in successful direction.

format: plain


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# Math Help - Calculating F value

1. ## Calculating F value

So I have the following information:

I have the one-factor model Y=W*m + e where e ~ N(0, (s^2)*I) and W = I(3) (x) J(6).

The (x) refers to the Kronecker product, I(3) refers to a 3-size identity matrix and J(6) refers to a 6-size vector of ones. Y and e are 18-by-1 vectors, W is a 18-by-3 matrix, and m is a 3-by-1 vector.

My given hypothesis is that m1 = m2 = m and m3 = d*m where d =/= 1 and m is some arbitrary value. Using this information, I am supposed to find an F value (test statistic) and its distribution. I can figure out the hypothesis matrix which is as follows:

H=
[1 -1 0]
[d 0 -1]
[0 d -1]

This is as far as I can go. Anyone know what I should do next?


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# Gates¶

Note

In Strawberry Fields we use the convention $$\hbar=2$$ by default, but other conventions can also be chosen by setting the global variable sf.hbar at the beginning of a session. In this document we keep $$\hbar$$ explicit.

## Displacement¶

Definition

$D(\alpha) = \exp( \alpha \ad -\alpha^* \a) = \exp(r (e^{i\phi}\ad -e^{-i\phi}\a)), \quad D^\dagger(\alpha) \a D(\alpha)=\a +\alpha\I$

where $$\alpha=r e^{i \phi}$$ with $$r \geq 0$$ and $$\phi \in [0,2 \pi)$$.

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.Dgate

We obtain for the position and momentum operators

$\begin{split}D^\dagger(\alpha) \x D(\alpha) = \x +\sqrt{2 \hbar } \re(\alpha) \I,\\ D^\dagger(\alpha) \p D(\alpha) = \p +\sqrt{2 \hbar } \im(\alpha) \I.\end{split}$

Definition

The pure position and momentum displacement operators are defined as

$\begin{split}X(x) &= D\left( x/\sqrt{2 \hbar}\right) = \exp(-i x \p /\hbar), \quad X^\dagger(x) \x X(x) = \x +x\I,\\ Z(p) &= D\left(i p/\sqrt{2 \hbar}\right) = \exp(i p \x /\hbar ), \quad Z^\dagger(p) \p Z(p) = \p +p\I.\end{split}$

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.Xgate

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.Zgate

The matrix elements of the displacement operator in the Fock basis were derived by Cahill and Glauber [1]:

$\bra{m}\hat D(\alpha) \ket{n} = \sqrt{\frac{n!}{m!}} \alpha^{m-n} e^{-|\alpha|^2/2} L_n^{m-n}\left( |\alpha|^2 \right)$

where $$L_n^{m}(x)$$ is a generalized Laguerre polynomial [2].

## Squeezing¶

Definition

$\begin{split}& S(z) = \exp\left(\frac{1}{2}\left(z^* \a^2-z {\ad}^{2} \right) \right) = \exp\left(\frac{r}{2}\left(e^{-i\phi}\a^2 -e^{i\phi}{\ad}^{2} \right) \right)\\ & S^\dagger(z) \a S(z) = \a \cosh(r) -\ad e^{i \phi} \sinh (r)\\ & S^\dagger(z) \ad S(z) = \ad \cosh(r) -\a e^{-i \phi} \sinh(r)\end{split}$

where $$z=r e^{i \phi}$$ with $$r \geq 0$$ and $$\phi \in [0,2 \pi)$$.

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.Sgate

The squeeze gate affects the position and momentum operators as

$S^\dagger(z) \x_{\phi} S(z) = e^{-r}\x_{\phi}, ~~~ S^\dagger(z) \p_{\phi} S(z) = e^{r}\p_{\phi}$

The Fock basis decomposition of displacement and squeezing operations was analysed by Krall [3], and the following quantity was calculated,

$\begin{split}f_{n,m}(r,\phi,\beta)&=\bra{n}\exp\left(\frac{r}{2}\left(e^{i \phi} \a^2 -e^{-i \phi} \ad \right) \right) D(\beta) \ket{m} = \bra{n}S(z^*) D(\beta) \ket{m}\\ &=\sqrt{\frac{n!}{\mu m!}} e^{\frac{\beta ^2 \nu ^*}{2\mu }-\frac{\left| \beta \right| ^2}{2}} \sum_{i=0}^{\min(m,n)}\frac{\binom{m}{i} \left(\frac{1}{\mu \nu }\right)^{i/2}2^{\frac{i-m}{2}+\frac{i}{2}-\frac{n}{2}} \left(\frac{\nu }{\mu }\right)^{n/2} \left(-\frac{\nu ^*}{\mu }\right)^{\frac{m-i}{2}} H_{n-i}\left(\frac{\beta }{\sqrt{2} \sqrt{\mu \nu }}\right) H_{m-i}\left(-\frac{\alpha ^*}{\sqrt{2} \sqrt{-\mu \nu ^*}}\right)}{(n-i)!}\end{split}$

where $$\nu=e^{- i\phi} \sinh(r), \mu=\cosh(r), \alpha=\beta \mu - \beta^* \nu$$.

Two important special cases of the last formula are obtained when $$r \to 0$$ and when $$\beta \to 0$$:

• For $$r \to 0$$ we can take $$\nu \to 1, \mu \to r, \alpha \to \beta$$ and use the fact that for large $$x \gg 1$$ the leading order term of the Hermite polynomials is $$H_n(x) = 2^n x^n +O(x^{n-2})$$ to obtain

$f_{n,m}(0,\phi,\beta) = \bra{n}D(\beta) \ket{m}=\sqrt{\frac{n!}{ m!}} e^{-\frac{\left| \beta \right| ^2}{2}} \sum_{i=0}^{\min(m,n)} \frac{(-1)^{m-i}}{(n-i)!} \binom{m}{i} \beta^{n-i} (\beta^*)^{m-i}$
• On the other hand if we let $$\beta\to 0$$ we use the fact that

$\begin{split}H_n(0) =\begin{cases}0, & \mbox{if }n\mbox{ is odd} \$$-1)^{\tfrac{n}{2}} 2^{\tfrac{n}{2}} (n-1)!! , & \mbox{if }n\mbox{ is even} \end{cases}\end{split}$ to deduce that \(f_{n,m}(r,\phi,0)$$ is zero if $$n$$ is even and $$m$$ is odd or vice versa.

When writing the Bloch-Messiah reduction [4][5] of a Gaussian state in the Fock basis one often needs the following matrix element

$\bra{k} D(\alpha) R(\theta) S(r) \ket{l} = e^{i \theta l } \bra{k} D(\alpha) S(r e^{2i \theta}) \ket{l} = e^{i \theta l} f^*_{l,k}(-r,-2\theta,-\alpha)$

## Rotation¶

Note

We use the convention that a positive value of $$\phi$$ corresponds to an anticlockwise rotation in the phase space.

Definition

We write the phase space rotation operator as

$R(\phi) = \exp\left(i \phi \ad \a\right)=\exp\left(i \frac{\phi}{2} \left(\frac{\x^2+ \p^2}{\hbar}-\I\right)\right), \quad R^\dagger(\phi) \a R(\phi) = \a e^{i \phi}$

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.Rgate

It rotates the position and momentum quadratures to each other:

$\begin{split}R^\dagger(\phi)\x R(\phi) = \x \cos \phi -\p \sin \phi,\\ R^\dagger(\phi)\p R(\phi) = \p \cos \phi +\x \sin \phi.\end{split}$

Definition

A special case of the rotation operator is the case $$\phi=\pi/2$$; this corresponds to the Fourier gate,

$F = R(\pi/2) = e^{i (\pi/2) \ad \a},$

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.Fouriergate

The Fourier gate transforms the quadratures as follows:

$\begin{split}& F^\dagger\x F = -\p,\\ & F^\dagger\p F = \x.\end{split}$

## Quadratic phase¶

Definition

$P(s) = \exp\left(i \frac{s}{2 \hbar} \x^2\right), \quad P^\dagger(s) \a P(s) = \a +i\frac{s}{2}(\a +\ad)$

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.Pgate

It shears the phase space, preserving position:

$\begin{split}P^\dagger(s) \x P(s) &= \x,\\ P^\dagger(s) \p P(s) &= \p +s\x.\end{split}$

This gate can be decomposed as

$P(s) = R(\theta) S(r e^{i \phi})$

where $$\cosh(r) = \sqrt{1+(\frac{s}{2})^2}, \quad \tan(\theta) = \frac{s}{2}, \quad \phi = -\sign(s)\frac{\pi}{2} -\theta$$.

## Beamsplitter¶

Definition

For the annihilation and creation operators of two modes, denoted $$\a_1$$ and $$\a_2$$, the beamsplitter is defined by

$B(\theta,\phi) = \exp\left(\theta (e^{i \phi}\a_1 \ad_2 - e^{-i \phi} \ad_1 \a_2) \right)$

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.BSgate

Action on the creation and annihilation operators

They will transform the operators according to

$\begin{split}B^\dagger(\theta,\phi) \a_1 B(\theta,\phi) &= \a_1\cos \theta -\a_2 e^{-i \phi} \sin \theta = t \a_1 -r^* \a_2,\\ B^\dagger(\theta,\phi) \a_2 B(\theta,\phi) &= \a_2\cos \theta + \a_1 e^{i \phi} \sin \theta= t \a_2 +r \a_1.\end{split}$

where $$t = \cos \theta$$ and $$r = e^{i\phi} \sin \theta$$ are the transmittivity and reflectivity amplitudes of the beamsplitter respectively.

Therefore, the beamsplitter transforms two input coherent states to two output coherent states $$B(\theta, \phi) \ket{\alpha,\beta} = \ket{\alpha',\beta'}$$, where

$\begin{split}\alpha' &= \alpha\cos \theta-\beta e^{-i\phi}\sin\theta = t\alpha - r^*\beta\\ \beta' &= \beta\cos \theta+\alpha e^{i\phi}\sin\theta = t\beta + r\alpha\\\end{split}$

Action on the quadrature operators

By substituting in the definition of the creation and annihilation operators in terms of the position and momentum operators, it is possible to derive an expression for how the beamsplitter transforms the quadrature operators:

$\begin{split}&\begin{cases} B^\dagger(\theta,\phi) \x_1 B(\theta,\phi) = \x_1 \cos(\theta)-\sin(\theta) [\x_2\cos(\phi)+\p_2\sin(\phi)]\\ B^\dagger(\theta,\phi) \p_1 B(\theta,\phi) = \p_1 \cos(\theta)-\sin(\theta) [\p_2\cos(\phi)-\x_2\sin(\phi)]\\ \end{cases}\\[12pt] &\begin{cases} B^\dagger(\theta,\phi) \x_2 B(\theta,\phi) = \x_2 \cos(\theta)+\sin(\theta) [\x_1\cos(\phi)-\p_1\sin(\phi)]\\ B^\dagger(\theta,\phi) \p_2 B(\theta,\phi) = \p_2 \cos(\theta)+\sin(\theta) [\p_1\cos(\phi)+\x_1\sin(\phi)] \end{cases}\end{split}$

Action on the position and momentum eigenstates

A 50% or 50-50 beamsplitter has $$\theta=\pi/4$$ and $$\phi=0$$ or $$\phi=\pi$$; consequently $$|t|^2 = |r|^2 = \frac{1}{2}$$, and it acts as follows:

$\begin{split}& B(\pi/4,0)\xket{x_1}\xket{x_2} = \xket{\frac{1}{\sqrt{2}}(x_1-x_2)}\xket{\frac{1}{\sqrt{2}}(x_1+x_2)}\\ & B(\pi/4,0)\ket{p_1}_p\ket{p_2}_p = \xket{\frac{1}{\sqrt{2}}(p_1-p_2)}\xket{\frac{1}{\sqrt{2}}(p_1+p_2)}\end{split}$

and

$\begin{split}& B(\pi/4,\pi)\xket{x_1}\xket{x_2} = \xket{\frac{1}{\sqrt{2}}(x_1+x_2)}\xket{\frac{1}{\sqrt{2}}(x_2-x_1)}\\ & B(\pi/4,\pi)\ket{p_1}_p\ket{p_2}_p = \xket{\frac{1}{\sqrt{2}}(p_1+p_2)}\xket{\frac{1}{\sqrt{2}}(p_2-p_1)}\end{split}$

Alternatively, symmetric beamsplitter (one that does not distinguish between $$\a_1$$ and $$\a_2$$) is obtained by setting $$\phi=\pi/2$$.

## Two-mode squeezing¶

Definition

$S_2(z) = \exp\left(z \ad_1 \ad_2 - z^* \a_1\a_2 \right) = \exp\left(r (e^{i\phi} \ad_1\ad_2 -e^{-i\phi} \a_1 \a_2 \right)$

where $$z=r e^{i \phi}$$ with $$r \geq 0$$ and $$\phi \in [0,2 \pi)$$.

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.S2gate

It can be decomposed into two opposite local squeezers sandwiched between two 50% beamsplitters [6]:

$S_2(z) = B^\dagger(\pi/4,0) \: \left[ S(z) \otimes S(-z)\right] \: B(\pi/4,0)$

Two-mode squeezing will transform the operators according to

$\begin{split}S_2(z)^\dagger \a_1 S_2(z) &= \a_1 \cosh(r)-\ad_2 e^{i \phi} \sinh(r),\\ S_2(z)^\dagger \a_2 S_2(z) &= \a_2 \cosh(r) -\ad_1 e^{i \phi} \sinh(r),\\\end{split}$

where $$z=r e^{i \phi}$$ with $$r \geq 0$$ and $$\phi \in [0,2 \pi)$$.

## Controlled-X gate¶

Definition

The controlled-X gate, also known as the addition gate or the sum gate, is a controlled displacement in position. It is given by

$\text{CX}(s) = \int dx \xket{x}\xbra{x} \otimes D\left(\frac{s x}{\sqrt{2\hbar}}\right) = \exp\left({-i \frac{s}{\hbar} \: \x_1 \otimes \p_2}\right).$

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.CXgate

It is called addition because in the position basis $$\text{CX}(s) \xket{x_1, x_2} = \xket{x_1, x_2+s x_1}$$.

We can also write the action of the addition gate on the canonical operators:

$\begin{split}\text{CX}(s)^\dagger \x_1 \text{CX}(s) &= \x_1\\ \text{CX}(s)^\dagger \p_1 \text{CX}(s) &= \p_1- s \ \p_2\\ \text{CX}(s)^\dagger \x_2 \text{CX}(s) &= \x_2+ s \ \x_1\\ \text{CX}(s)^\dagger \p_2 \text{CX}(s) &= \p_2 \\ \text{CX}(s)^\dagger \hat{a}_1 \text{CX}(s) &= \a_1+ \frac{s}{2} (\ad_2 - \a_2)\\ \text{CX}(s)^\dagger \hat{a}_2 \text{CX}(s) &= \a_2+ \frac{s}{2} (\ad_1 + \a_1)\\\end{split}$

The addition gate can be decomposed in terms of single mode squeezers and beamsplitter as follows $$\text{CX}(s) = B(\frac{\pi}{2}+\theta,0) \left(S(r,0) \otimes S(-r,0) \right) B(\theta,0)$$ where $$\sin(2 \theta) = \frac{-1}{\cosh r}, \ \cos(2 \theta)=-\tanh(r), \ \sinh(r) = -\frac{ s}{2}$$

## Controlled-phase¶

Definition

$\text{CZ}(s) = \iint dx dy \: e^{i s x_1 x_2/\hbar } \xket{x_1,x_2}\xbra{x_1,x_2} = \exp\left({i s \: \hat{x_1} \otimes \hat{x_2} /\hbar}\right).$

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.CZgate

It is related to the addition gate by a phase space rotation in the second mode: $$\text{CZ}(s) = R_{(2)}(\pi/2) \: \text{CX}(s) \: R_{(2)}^\dagger(\pi/2)$$.

In the position basis $$\text{CZ}(s) \xket{x_1, x_2} = e^{i s x_1 x_2/\hbar} \xket{x_1, x_2}$$.

We can also write the action of the controlled-phase gate on the canonical operators:

$\begin{split}\text{CZ}(s)^\dagger \x_1 \text{CZ}(s) &= \x_1\\ \text{CZ}(s)^\dagger \p_1 \text{CZ}(s) &= \p_1+ s \ \x_2\\ \text{CZ}(s)^\dagger \x_2 \text{CZ}(s) &= \x_2\\ \text{CZ}(s)^\dagger \p_2 \text{CZ}(s) &= \p_2+ s \ \x_1 \\ \text{CZ}(s)^\dagger \hat{a}_1 \text{CZ}(s) &= \a_1+ i\frac{s}{2} (\ad_2 + \a_2)\\ \text{CZ}(s)^\dagger \hat{a}_2 \text{CZ}(s) &= \a_2+ i\frac{s}{2} (\ad_1 + \a_1)\\\end{split}$

## Cubic phase¶

Warning

The cubic phase gate is non-Gaussian, and thus can only be used in the Fock backends, not the Gaussian backend.

Warning

The cubic phase gate can suffer heavily from numerical inaccuracies due to finite-dimensional cutoffs in the Fock basis. The gate implementation in Strawberry Fields is unitary, but it does not implement an exact cubic phase gate. The Kerr gate provides an alternative non-Gaussian gate.

Definition

$V(\gamma) = \exp\left(i \frac{\gamma}{3 \hbar} \x^3\right), \quad V^\dagger(\gamma) \a V(\gamma) = \a +i\frac{\gamma}{2\sqrt{2/\hbar}} (\a +\ad)^2$

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.Vgate

It transforms the phase space as follows:

$\begin{split}V^\dagger(\gamma) \x V(\gamma) &= \x,\\ V^\dagger(\gamma) \p V(\gamma) &= \p +\gamma \x^2.\end{split}$

## Kerr interaction¶

Warning

The Kerr gate is non-Gaussian, and thus can only be used in the Fock backends, not the Gaussian backend.

Definition

The Kerr interaction is given by the Hamiltonian

$H = (\hat{a}^\dagger\hat{a})^2=\hat{n}^2$

which is non-Gaussian and diagonal in the Fock basis.

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.Kgate

We can therefore define the Kerr gate, with parameter $$\kappa$$ as

$K(\kappa) = \exp{(i\kappa\hat{n}^2)}.$

## Cross-Kerr interaction¶

Warning

The cross-Kerr gate is non-Gaussian, and thus can only be used in the Fock backends, not the Gaussian backend.

Definition

The cross-Kerr interaction is given by the Hamiltonian

$H = \hat{n}_1\hat{n_2}$

which is non-Gaussian and diagonal in the Fock basis.

Tip

Implemented in Strawberry Fields as a quantum gate by strawberryfields.ops.CKgate

We can therefore define the cross-Kerr gate, with parameter $$\kappa$$ as

$CK(\kappa) = \exp{(i\kappa\hat{n}_1\hat{n_2})}.$

## References¶

1

K. E. Cahill and R. J. Glauber. Ordered expansions in boson amplitude operators. Physical Review, 177:1857–1881, Jan 1969. doi:10.1103/PhysRev.177.1857.

2

F. W. J. Olver, A. B. Olde Daalhuis, D. W. Lozier, B. I. Schneider, R. F. Boisvert, C. W. Clark, B. R. Miller, and B. V. Saunders (eds.). NIST digital library of mathematical functions. Release 1.0.16 of 2017-09-18. [Online; accessed 2017-10-25]. URL: http://dlmf.nist.gov/.

3

P Král. Displaced and squeezed Fock states. Journal of Modern Optics, 37(5):889–917, 1990. doi:10.1080/09500349014550941.

4

Gianfranco Cariolaro and Gianfranco Pierobon. Bloch-Messiah reduction of gaussian unitaries by Takagi factorization. Physical Review A, 94:062109, Dec 2016. doi:10.1103/PhysRevA.94.062109.

5

Gianfranco Cariolaro and Gianfranco Pierobon. Reexamination of Bloch-Messiah reduction. Physical Review A, 93:062115, Jun 2016. doi:10.1103/PhysRevA.93.062115.

6

M. S. Kim, W. Son, V. Bužek, and P. L. Knight. Entanglement by a beam splitter: nonclassicality as a prerequisite for entanglement. Physical Review A, 65:032323, Feb 2002. arXiv:quant-ph/0106136, doi:10.1103/PhysRevA.65.032323.


text:
My Math Forum Are we living in a @ light speed spinning black hole?

 Physics Physics Forum

 November 12th, 2014, 06:03 AM #21 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions Fine, okay. The paper you linked earlier is still trash though I would still advise anyone to read articles through arxiv or from MNRAS directly rather than through vixra. In fact, Nasa ADS is probably the best way of finding papers online. Thanks from topsquark Last edited by Benit13; November 12th, 2014 at 06:08 AM.
November 12th, 2014, 12:03 PM   #22
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Quote:
 Originally Posted by M_B_S American Journal of Astronomy and Astrophysics :: Science Publishing Group 1969MNRAS.142..129H Page 129 Title: On the rotation of the Universe Authors: Hawking, S. Journal: Monthly Notices of the Royal Astronomical Society, Vol. 142, p.129 Bibliographic Code: 1969MNRAS.142..129H
What point are you trying to make? You link to a highly disreputable journal, then to a famous paper in a different journal.

If you're using the paper to support the basic idea of this thread (but then why mention the journal?), then you have it backward. Hawking is showing that the Universe, if it is rotating at all, must rotate very slowly -- less than once per 600 trillion years, at a linear speed less than 0.00015c. Unless, perhaps, you interpret "near lightspeed" very charitably?

November 12th, 2014, 03:36 PM   #23
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Quote:
 Originally Posted by M_B_S A) Photons are massless particles because they move with Light Speed a) All massive partikels in a Black Hole are crashed to Photons ai) A spinning Black Hole is like a onion with orbits for every heavy partikel with mass. Postulate: All Black Holes are spinning If a massive object falls in it crashes threw its unique orbit and is destroyed untill it gets a photon. This process needs time. The Event Horizon is a Photon Corona
Actually you have it the wrong way around, though the difference is somewhat trivial: Photons move at the speed of light because they are massless, not the other way around.

$\displaystyle p^+ + p^+ \longrightarrow \pi ^+ + (deuteron)$

$\displaystyle \Sigma + p^+ \longrightarrow \Lambda + n^0$

These are two examples where massive particle interactions do not produce photons. (There are, of course, many others.) There is no reason to suspect that these interactions are different inside the Schwarzchild radius.

I have no problem with your description of a rotating black hole. Look up the Kerr black hole for more information on that. Rotating black holes engage in "frame dragging" which more or less corresponds to your statement of the onion.

Why should all black holes spin? It is true that the only way we know how to make black holes is via supernovas, which gives the black hole part of the angular momentum of the star that formed it. But why do all black holes have to be rotating? GR predicts that a non-rotating black hole is perfectly happy to exist. (If you really want to blow your mind we can talk about spinning black holes that have a net charge on them.)

Quote:
 Originally Posted by M_B_S B) Light (corona orbit) spinning in rotation direction NASA | Peer into a Simulated Stellar-mass Black Hole - YouTube NASA | Peer into a Simulated Stellar-mass Black Hole
"Corona orbit." Do you mean an accretion disk?

Quote:
 Originally Posted by M_B_S C) You need massive matter particles to get Newtons Gravitation Force Photons collide in a Big Bang and create massiv matter Hydrogen etc. =>Stars => New Black Holes are possible Mass is created in a Big Bang = exploding Black Hole = White Hole (only light) by photon collide.
Yes, Newtonian Physics states that we need a mass to have a gravitational field, but General Relativity says we don't. As GR is the more accurate theory I'm siding with GR on this one.

Here's a fact that many don't know: The event horizon is not a point of singularity. There is a metric (which is hideous) where no singularity exists until we get down to the center of the black hole. Since it is true in one metric it really has to be in all of them. The event horizon is just a location in space like any other. The problem comes in when we talk about what's inside. What we don't know is what states of matter exist inside...gravitation runs amok and nothing can stop the compression. On the other hand truly elementary particles cannot be "squeezed" so they must still exist as they are. But the scale of the gravitational force is immense and we don't yet have a theory of Quantum Gravity to say for sure what happens in there. As always we need experimental data and we can't get it in this case.

Look, you really need to learn some more Physics than you do. Your statements are almost completely speculation and based on "dodgy" sources. Remember: In Science a good theory is testable. I see no way to test many of your ideas.

Until you can talk about these matters with better base of information, I'm done with this.

-Dan

November 12th, 2014, 03:40 PM   #24
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Quote:
 Originally Posted by M_B_S E=mc² is Incomplete => E²=(mc²)²+(pc)² E_u with m=0 means only photons := pc the ultrarelativistic limit made possible in a Black Hole E = p c => The Big Bang
Just one more thing. $\displaystyle E^2 = p^2c^2 + m^2c^4$ is where we get $\displaystyle E = mc^2$ in the first place. It isn't that $\displaystyle E = mc^2$ is incomplete, it's that the popular media doesn't go into the matter deep enough. The full formula is used in Particle Physics all the time.

Why should E = pc imply a Big Bang? It's nothing more than a formula for the high energy regime.

-Dan

 November 12th, 2014, 04:38 PM #25 Senior Member   Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory Dan Its the angluar momentum law who solves the problem with the singularity triggered by gravitation. Because its the simple reason why there is now physical Singularity in a spinning Black Hole Light corona : look the NASA video please. E=pc status in a Black Hole means no Newtonian Mass m = Gravity so no force against the centrifugal power => White Hole In physical terms, a spin-out Big Bang occurs when centrifugal force overpowers whatever is connecting a rotating object (Black Hole)to some central hub. The power acting to make the object independent is stronger than the power constraining it within the existing system. Objects would fly off the earth, for instance, if the earth were spinning fast enough to cancel the power of gravity. So in a Black Hole all Mass is transformed into massles particles (Light)=> Big Bang Last edited by M_B_S; November 12th, 2014 at 04:54 PM.
November 12th, 2014, 05:54 PM   #26
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Quote:
 Originally Posted by M_B_S Dan Its the angluar momentum law who solves the problem with the singularity triggered by gravitation. Because its the simple reason why there is now physical Singularity in a spinning Black Hole Light corona : look the NASA video please. E=pc status in a Black Hole means no Newtonian Mass m = Gravity so no force against the centrifugal power => White Hole
(sighs) I'm an idiot for responding to this and I'm done playing around.

The Schwarzchild metric for non-rotation black hole:
$\displaystyle d \tau ^2 = \left [ 1 - \frac{2MG}{r} \right ] dt^2 - \left [ 1 - \frac{2MG}{r} \right ]^{-1} dr^2 - r^2 d \theta ^2 - r^2~sin^2 ( \theta ) d \phi ^2$

Clearly there is a singularity here, when 2MG = r.

The Kerr metric for a rotating black hole:
$\displaystyle -d \tau ^2 = -dt^2 + d \textbf{x} ^2 + \frac{2MG \rho}{(\rho ^4 + ( \textbf{x} \cdot \textbf{a}^2 )) (\rho ^2 + \textbf{a} ^2 )^2} \cdot \left [ \rho ^2 \textbf{x} \cdot d \textbf{x} + \rho d \textbf{x} \cdot ( \textbf{a} \times \textbf{x} ) + ( \textbf{a} \cdot \textbf{x} )(\textbf{a} \cdot d \textbf{x} ) + ( \rho ^2 + \textbf{a} ^2 ) \rho dt \right ]^2$

where $\displaystyle \rho$ is defined by $\displaystyle \rho ^4 - (r^2 - \textbf{a}^2 ) \rho ^2 - ( \textbf{a} \cdot \textbf{x} )^2 = 0$
( $\displaystyle \textbf{a}$ is a constant and the dot and cross products are calculated using the standard Euclidean inner and cross products, and $\displaystyle r^2 = \textbf{x} ^2$.)

This does not appear to have a singularity but when we take the limit as $\displaystyle r \to \infty$ the $\displaystyle g_{00}$ component of the metric becomes $\displaystyle -1 + \frac{2MG}{r}$, so we have the same singularity as in the Schwarzchild solution.

I'm not going to sit down and work it all out for the Kerr black hole, but choosing the correct coordinates gives a metric for the non-rotating black hole:
$\displaystyle d \tau ^2 = \left ( \frac{32 G^3M^3}{r T^2} \right ) Exp \left [ \frac{-r}{2GM} \right ] (dt'^2 - dr'^2 ) - r^2 d \theta ^2 - r^2 sin^2 (\theta) d \phi ^2$

The coordinate system being used here is kind of wacky, but there is no longer a singularity at r = 2GM! So the event horizon really isn't a singularity at all.

That doesn't mean that we can simply ignore our usual coordinate system, which does have an apparent singularity. Even in a Kerr black hole the escape speed of an object is still equal or larger than the speed of light. And even when talking about Hawking radiation the black hole doesn't actually emit mass, or even light, from inside the event horizon. So no white hole.

Now, you have a serious problem with E = pc. I don't blame you for this as only someone who has studied the issue of mass/massless objects in Relativistic QM would likely know. You can't derive an expression in Relativity or QM with a mass m and smoothly take it to the m = 0 case. The reason is this: Relativistic QM is based on a number of basic symmetries, in this case the Poincare symmetry. When you work out the symmetry equations for a particle with mass Wigner's little group symmetry is essentially the rotation group in three dimensions, which has three independent degrees of freedom. On the other hand Wigner's little group symmetry for a massless particle is ISO(2), which only has two independent degrees of freedom. We cannot smoothly make a transformation from three degrees of freedom to two. So even though the high energy limit of $\displaystyle E^2 = p^2c^2 + m^2c^4 \approx p^2c^2$ the particle still has to have a mass that we can't just ignore. The approximation is good dynamically, but you have to be very careful of how you do Physics in this regime.

-Dan

 November 13th, 2014, 01:15 AM #27 Senior Member   Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory Yes Dan Its the math stupid there is a ring singularity in a Kerr Black Hole I think its the photon corona ?! Look Nasa Video simulation of a BH E²=p²C²+m²C²C² E = pc means here there is only light Photons => mC² = 0 Without mass no relativistic/newton gravitation => no force against pressure of light => Big Bang ( there will be light) Genesis Universe => Mass creation via Photon collide=> Newtonian/Relativistic Gravity => New Black Holes Excurs: Video: Pressure of Light Physics - Electromagnetic Radiation (6 of 6) Intro: Pressure of Light - YouTube Genesis of E=mc² Last edited by M_B_S; November 13th, 2014 at 01:30 AM.
 November 13th, 2014, 02:28 AM #28 Senior Member   Joined: Nov 2013 From: Germany Posts: 179 Thanks: 1 Math Focus: Number Theory Video Mass and Momentum of Photon Mass and Momentum of Photon - YouTube ************* Here you learn what i mean: In a Kerr Black Hole all mass is "transformed" to light in time. so that E = pC => Big Bang You could proof it in a computer simulation it will work. M_B_S Last edited by M_B_S; November 13th, 2014 at 02:30 AM.
November 13th, 2014, 03:15 AM   #29
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Quote:
 Originally Posted by M_B_S Yes Dan Its the math stupid there is a ring singularity in a Kerr Black Hole

Quote:
 I think its the photon corona ?! Look Nasa Video simulation of a BH
Don't search for truthfulness in anything but textbooks and published, peer-review papers. Online videos (regardless of the source) and other sources are unreliable and should be taken with a huge pinch of salt.

Also, remember that learning science is about learning truth, not making what you already believe sound like truth. In this regard, don't settle for anything less than the most professional, high quality work and be sceptical about everything. There are even papers published in peer review journals that are crappy, so don't just accept anything you see or read; question it and decide for yourself whether you think it is the truth or not.

Please... get a textbook on cosmology and follow through everything in it. By educating yourself with the state of the art, you'll be able to learn where there is wiggle room for new knowledge.

Quote:
 E²=p²C²+m²C²C² E = pc means here there is only light Photons => mC² = 0 Without mass no relativistic/newton gravitation => no force against pressure of light => Big Bang ( there will be light) Genesis Universe => Mass creation via Photon collide=> Newtonian/Relativistic Gravity => New Black Holes
This is poetry, not mathematics. Your logical steps don't link together at all. The formula that gives the energy carried by a particle based on its mass and momentum can be applied to many situations and is just a law that relates one parameter to others. It tells you nothing about black holes or the Big Bang or any of the conclusions you drew from the above. You might as well say "hair exists, unicorns supposedly have hair, so that means unicorns exist".

You are correct that there is something called "light pressure" because even though photons have zero mass, they carry momentum and can exchange that momentum in interactions with atoms. Poynting vectors tell you how much "force" is created when light shines on a surface and is the mechanism through which solar sails are based on. The magnitude of this pressure is very small however and it would take a solar sail with several km$\displaystyle ^2$ of surface area to enable motion with any relevance to space travel. Light pressure is relevant only in a few phenomena, such as novae and supernovae. It can safely be ignored even in standard stars because gas pressure and gravitational pressure exceed light pressure considerably.

As for black holes? No one knows the equation of state for matter inside a black hole and the usual laws don't apply anyway, so anyone claiming they know what is inside a black hole for sure is a conman. However, using general relativity, it is possible to generate some mathematical models to make predictions. This is what theorists do.

What about experimenters? Active Galactic Nuclei have supermassive black holes in the centre of them and there is lots of observational evidence gathered by observing the light given off by accretion disks surrounding the black hole. Properties of the black hole can be determined this way by comparing with theory. All the theory involves general relativity because Newtonian laws break down and simply don't work. There is also Cygnus X, which is an object that provides evidence of a stellar mass black hole. Cygnus X is still under scrutiny however.

Seriously... there is loads of high quality, interesting literature on black holes. Don't settle for anything other than the best papers. In fact, why don't you read that famous Stephen Hawking paper that you linked and tell us about it? I would genuinely be interested to hear more about his work on black holes. We can all read it and use this this opportunity to have a group literature review session!

Last edited by Benit13; November 13th, 2014 at 03:20 AM.

November 13th, 2014, 04:08 AM   #30
Senior Member

Joined: Nov 2013
From: Germany

Posts: 179
Thanks: 1

Math Focus: Number Theory
Quote:
 Originally Posted by Benit13 No... you need to learn more about general relativity. Don't search for truthfulness in anything but textbooks and published, peer-review papers. Online videos (regardless of the source) and other sources are unreliable and should be taken with a huge pinch of salt. Also, remember that learning science is about learning truth, not making what you already believe sound like truth. In this regard, don't settle for anything less than the most professional, high quality work and be sceptical about everything. There are even papers published in peer review journals that are crappy, so don't just accept anything you see or read; question it and decide for yourself whether you think it is the truth or not. Please... get a textbook on cosmology and follow through everything in it. By educating yourself with the state of the art, you'll be able to learn where there is wiggle room for new knowledge. This is poetry, not mathematics. Your logical steps don't link together at all. The formula that gives the energy carried by a particle based on its mass and momentum can be applied to many situations and is just a law that relates one parameter to others. It tells you nothing about black holes or the Big Bang or any of the conclusions you drew from the above. You might as well say "hair exists, unicorns supposedly have hair, so that means unicorns exist". You are correct that there is something called "light pressure" because even though photons have zero mass, they carry momentum and can exchange that momentum in interactions with atoms. Poynting vectors tell you how much "force" is created when light shines on a surface and is the mechanism through which solar sails are based on. The magnitude of this pressure is very small however and it would take a solar sail with several km$\displaystyle ^2$ of surface area to enable motion with any relevance to space travel. Light pressure is relevant only in a few phenomena, such as novae and supernovae. It can safely be ignored even in standard stars because gas pressure and gravitational pressure exceed light pressure considerably. As for black holes? No one knows the equation of state for matter inside a black hole and the usual laws don't apply anyway, so anyone claiming they know what is inside a black hole for sure is a conman. However, using general relativity, it is possible to generate some mathematical models to make predictions. This is what theorists do. What about experimenters? Active Galactic Nuclei have supermassive black holes in the centre of them and there is lots of observational evidence gathered by observing the light given off by accretion disks surrounding the black hole. Properties of the black hole can be determined this way by comparing with theory. All the theory involves general relativity because Newtonian laws break down and simply don't work. There is also Cygnus X, which is an object that provides evidence of a stellar mass black hole. Cygnus X is still under scrutiny however. Seriously... there is loads of high quality, interesting literature on black holes. Don't settle for anything other than the best papers. In fact, why don't you read that famous Stephen Hawking paper that you linked and tell us about it? I would genuinely be interested to hear more about his work on black holes. We can all read it and use this this opportunity to have a group literature review session!

This logic is not disproofable:

When a Black Hole spins with light speed.

What is "in" the Black Hole?

Answer only Energy = light with momentum! => E_u=pC the Big Bang

This is what E²=p²C²+m²C²C² is telling us.

You need good New papers on Black Hole Universe theory?

Nikodem Poplawski's articles on arXiv

Nikodem Poplawski's articles on arXiv

Your marks on light pressure are a bit to soft.

The Black Hole transformation into a Big Bang

black files := matter => red files := light

The onion of a Black Hole (Star) := parts different wavelenghts of light

inner blue => outer red

Last edited by M_B_S; November 13th, 2014 at 04:38 AM.

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# Math Help - Orthogonalization of polynomial

1. ## Orthogonalization of polynomial

Hi guys.
I have the following base for $\mathbb{R}_2[x]$: $\left \{ 1,x,x^2 \right \}$
I want to orthogonalize this base with respect to the inner product:
$=\int_{0}^{1}p(x)q(x)dx$

somthing doesn't add up, and I suspect that I have made some calculation error, even though I checked several times.

so let's say the new orthogonal base will be $\left \{ \tilde{v}_1, \tilde{v}_2, \tilde{v}_3 \right \}$
I first set $\tilde{v}_1=v_1=1$.
then:
$\tilde{v}_2=v_2-\frac{v_2\cdot v_1}{v_1\cdot v_1}v_1=x-\frac{1}{2}$ (I checked it - it is correct, since it's orthogonal to $v_1$, which is 1)

but when I try to find $\tilde{v}_3$, I get:

$\tilde{v}_3=v_3-\frac{v_3\cdot v_1}{v_1\cdot v_1}v_1-\frac{v_3\cdot v_2}{v_2\cdot v_2}v_2$

$\tilde{v}_3=x^2-\frac{\int_{0}^{1}x^2}{\int_{0}^{1}1}1-\frac{\int_{0}^{1}x^3}{\int_{0}^{1}x^2}x$

$\tilde{v}_3=x^2-\frac{1}{3}-\frac{3}{4}x$

so this is what I get, but $\tilde{v}_3$ is NOT orthogonal to neither $\tilde{v}_1$ or $\tilde{v}_2$!

any help would be greatly appreciated!

2. ## Re: Orthogonalization of polynomial

In your expression for $\tilde{v}_{3},$ the $v_{1}$ and $v_{2}$ (wherever they appear), on the RHS should be $\tilde{v}_{1}$ and $\tilde{v}_{2}$ shouldn't they ?

(You got lucky with $\tilde{v}_2},$ because $\tilde{v}_{1} = v_{1}.)$

3. ## Re: Orthogonalization of polynomial

Hi BobP, and thanks for the help.

I don't see why does it matter if I take $v_1$, $v_2$ or $\tilde{v}_1$, $\tilde{v}_2$.
both $v_1$, $v_2$ and $\tilde{v}_1$, $\tilde{v}_2$ spans the same vector space, so if a vector is orthogonal to $v_1$, $v_2$, it will be orthogonal to $\tilde{v}_1$, $\tilde{v}_2$ as well.
nevertheless, I tried to take $\tilde{v}_1,\tilde{v}_2$ instead, and this is what I get:

$\tilde{v}_3=v_3-\frac{v_3\cdot \tilde{v}_1}{\tilde{v}_1\cdot \tilde{v}_1}\tilde{v}_1-\frac{v_3\cdot \tilde{v}_2}{\tilde{v}_2\cdot \tilde{v}_2}\tilde{v}_2$

$\tilde{v}_3=v_3-\frac{1}{3}-\frac{\int_{0}^{1}x_2(x-\frac{1}{2})}{\int_{0}^{1}(x-\frac{1}{2})^2}(x-\frac{1}{2})$

$\tilde{v}_3=x^2+\frac{1}{5}x-\frac{7}{30}$

which is still not orthogonal to $v_1$/ $\tilde{v}_1$/ $v_2$/ $\tilde{v}_2$

4. ## Re: Orthogonalization of polynomial

The formula that you have for $\tilde{v}_{3}$ comes from the assumption that $\tilde{v}_{3}=v_{1}+\alpha \tilde{v}_{1}+\beta\tilde{v}_{2}.$
To calculate the coefficient $\beta$ (for example) you take the inner product of this equation with $\tilde{v}_{2}.$
That gets you
$(\tilde{v}_{3},\tilde{v}_{2})=(v_{1},\tilde{v}_{2} )+\alpha(\tilde{v}_{1},\tilde{v}_{2})+\beta(\tilde {v}_{2},\tilde{v}_{2}),$
from which
$\beta=-\frac{(v_{1},\tilde{v}_{2})}{(\tilde{v}_{2},\tilde {v}_{2})}.$
That doesn't happen if you use $v_{1} \text{ and } v_{2}$ rather than $\tilde{v}_{1} \text{ and } \tilde{v}_{2},$ since $v_{1} \text{ and } v_{2}$ are not orthogonal.

For the last bit, check your integration.
I get
$\tilde{v}_{3}= x^{2}-x+1/6.$

5. ## Re: Orthogonalization of polynomial

Great! Thank you so much.

6. ## Re: Orthogonalization of polynomial

Actually, I made a mistake !

I should have put $\tilde{v}_{3}=v_{3}+\dots .$

That changes the next two lines. You get the drift though ?

7. ## Re: Orthogonalization of polynomial

yeah, I'll get it from here. thanks.


text:
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$...\text{are parallel if } a=2$
RECALL: (i) Non-vertical lines are parallel if they have equal slopes. (ii) Non-vertical and non-horizontal lines are perpendicular if their slopes have a product of $-1$ Thus, for the given lines to be parallel, then $a$ must be $2$.


text:
# Find A Counter-Example to This

by Atran
Tags: counterexample
 P: 84 Hi! Say p(x) is the product of the first x odd prime numbers (e.g. p(4)=3*5*7*11) and i is at least one. Then consider: 1 < p(x) ± 2*i < (p(x+1)/p(x))2 My hypothesis is that the above formula, obeying the restrictions, always produces a prime number. For example if x=3 and i=13, then p(3)-2*13=79. 79 is bigger than 1 and smaller than 112, therefore it's a prime number. I'm not a python programmer, but this code yields prime numbers using the above formula: from math import ceil def is_prime(number): if(number < 2): return False i = 2 while i <= ceil(number**0.5): if(number%i == 0): return False i = i + 1 return True def f(number_of_primes): flag = True prime_product = 1 prime_array = [] n, i = 3, 0 while i < number_of_primes: if(is_prime(n) == True): prime_product = prime_product * n prime_array.append(n) i = i + 1 n = n + 1 while True: if(is_prime(n) == True): largest_prime = n break n = n + 1 i = 1 while True: for x in range(len(prime_array)): if(i%prime_array[x] == 0): flag = False break if flag == False: flag = True i = i + 1 continue n = prime_product - 2*i if n <= 1: break if n < largest_prime**2: if is_prime(n) == False: print("NOT : ", n, " : ", i) break else: print("NMB : ", n, " : ", i) i = i + 1 def g(number_of_primes): flag = True prime_product = 1 prime_array = [] n, i = 3, 0 while i < number_of_primes: if(is_prime(n) == True): prime_product = prime_product * n prime_array.append(n) i = i + 1 n = n + 1 while True: if(is_prime(n) == True): largest_prime = n break n = n + 1 i = 1 while True: for x in range(len(prime_array)): if(i%prime_array[x] == 0): flag = False break if flag == False: flag = True i = i + 1 continue n = prime_product + 2*i if n < largest_prime**2: if is_prime(n) == False: print("NOT : ", n, " : ", i) break else: print("NMB : ", n, " : ", i) else: break i = i + 1 x = 4 # Enter a non-negative integer f(x) print("- - - - -") g(x) Scroll down and assign a non-negative integer to x in the code and run it. If the program detects a non-prime then it should output "NOT" followed by the generated number. If the generated number is a prime, then it outputs "NMB". The second number after the second colon is the value of i. So the above example would be displayed as: "NMB : 79 : 13" (excluding the double quotes) Thanks for help.
 P: 2,969 What is your question? to find a counter example? or to prove your conjecture?
 P: 84 I want somebody to prove this wrong, by finding a counter example. And I'm really sorry, I forgot to mention that i must not be divisible by any of the primes found in the product p(x). So again, 1 < p(x) ± 2*i < (p(x+1)/p(x))2 where p(x) is the product of the first x odd prime numbers, and i is a positive integer not divisible by any prime in the product p(x). Example: p(4) - 2*514 = 127. 514 is positive and not divisible by 3, 5, 7, or 11. And 127 is less than 169, so according to the conjecture 127 should be a prime, and it is.
 P: 2,969 Find A Counter-Example to This Your best strategy here would be to let the program find the counter example by finding a list of primes for input and then running it thru its paces. Also if you could find a related program that can check for primeness of a number you could incorporate it in to the mix.
Mentor
P: 18,277
 Quote by Atran I want somebody to prove this wrong, by finding a counter example. And I'm really sorry, I forgot to mention that i must not be divisible by any of the primes found in the product p(x). So again, 1 < p(x) ± 2*i < (p(x+1)/p(x))2 where p(x) is the product of the first x odd prime numbers, and i is a positive integer not divisible by any prime in the product p(x). Example: p(4) - 2*514 = 127. 514 is positive and not divisible by 3, 5, 7, or 11. And 127 is less than 169, so according to the conjecture 127 should be a prime, and it is.
It's a true conjecture. Here's a proof:

First, we note that since ##p(x)## is not divisible by ##2## and ##2i## is divisible by ##2##, then ##p(x) - 2i## is not divisible by ##2##.
Second, if ##p## is an odd prime occuring in the product of ##p(x)##, then ##p## divides ##p(x)## and ##p## does not divide ##2i##, thus ##p## does not divide ##p(x) - 2i##.

So we deduce that if ##p## is any prime dividing ##p(x)-2i##, then ##p## cannot occur in ##p(x)##. Thus ##p\geq p(x+1)/p(x)##.

So, assume that ##p(x) - 2i## is composite, then there are there are two prime numbers ##p## and ##q## that divide ##p(x) - 2i##. Thus we have ##pq\leq p(x)-2i##. But we also have that ##p\geq p(x+1)/p(x)## and ##q\geq p(x+1)/p(x)##. Thus ##pq\geq (p(x+1)/p(x))^2##. So we see that
$$(p(x+1)/p(x))^2\leq pq\leq p(x)-2i < (p(x+1)/p(x))^2$$


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text:
## Elementary Geometry for College Students (6th Edition)

$\angle U = 69^{\circ}$
Since $\overline{VU} = \overline{VT}$ This means that $\angle T = \angle U = 69^{\circ}$ Therefore $\angle U = 69^{\circ}$


text:
## Non-positively curved metric spaces

This post is based on parts of Martin Bridson’s 2006 ICM article.

In the next post we will give an overview of non-positively curved groups. In preparation, here is an introduction to non-positively curved metric spaces.

CAT(0) and CAT(-1) spaces

Named by Gromov in honour of Cartan, Alexandrov and Toponogov, a CAT(${\kappa}$) condition is a notion of curvature for a geodesic space — a metric space in which every pair of points ${a}$, ${b}$ is connected by a path of length ${d(a,b)}$.

The CAT(${\kappa}$) condition compares geodesic triangles in ${X}$ with triangles in a model space of constant curvature ${\kappa}$. It is a generalisation of a classical notion of curvature: assuming they are sufficiently smooth (such as ${C^3}$), then for Riemannian manifolds, locally the CAT(${\kappa}$) condition is equivalent to all the sectional curvatures being at most ${\kappa}$ — see the Appendix to Chapter II.1 of Part II of Bridson and Haefliger’s book. We will limit our discussion to CAT(0) and CAT(-1) spaces; in these cases the model spaces are the Euclidean plane ${\mathbb{E}^2}$ and the hyperbolic plane ${\mathbb{H}^2}$, respectively.

Here is what the CAT(0) and CAT(-1) conditions demand. Suppose ${\Delta}$ is a geodesic triangle in ${X}$. Let ${\bar{\Delta}}$ be a triangle (a comparison triangle) in ${\mathbb{E}^2}$ (for the CAT(0) condition) or ${\mathbb{H}^2}$ (for the stronger CAT(-1) condition) with the same side–lengths as ${\Delta}$. Suppose ${p}$ and ${q}$ are points on ${\Delta}$. Let ${\bar{p}}$ and ${\bar{q}}$ be the corresponding points on ${\bar{\Delta}}$ — that is, the sides they lie on are those that correspond to the sides ${p}$ and ${q}$ lie on, and their locations on those side agree with those of ${p}$ and ${q}$ on theirs. Then ${d(p,q) \leq d(\bar{p}, \bar{q})}$.

The usefulness of the CAT(0) and CAT(-1) conditions is greatly enhanced by the fact that they can readily be checked for suitable spaces (certain complexes) using Gromov’s link condition — more on which in some future post (probably).

We will now give two salient geometric features of a CAT(0) or CAT(-1) space ${X}$.

I. Triangles in CAT(-1) spaces are uniformly thin — that is, there exists ${\delta \geq 0}$ such that every geodesic triangle in ${X}$ is ${\delta}$–thin, meaning that each side is in the ${\delta}$-neighbourhood of the other two sides. This is true of CAT(-1) spaces because it holds for ${\mathbb{H}^2}$, where in fact ${\delta}$ can be taken to be any number greater than ${\ln(1+ \sqrt{2})}$. As explained in Gersten’s Banff Notes, one needs only consider ideal triangles since a triangle’s vertices can be pushed out to the boundary whilst only making it less thin. Then, as ideal triangles in ${\mathbb{H}^2}$ are equivalent, we can pick our favourite one; calculating ${\delta}$ for that triangle is a straightforward calculus problem.

II. The metric in a CAT(0) space is convex — that is, if ${c_1, c_2 : [0,1] \rightarrow X}$ are geodesics, parametrized proportional to arc–length, then

$\displaystyle D(t) \ \leq \ (1-t) D(0) + tD(1)$

for all ${t \in [0,1]}$, where ${D(t) := d(c_1(t), c_2(t))}$. Again, to see why this is true in ${X}$, we look at the corresponding claim in the model space — this time, ${\mathbb{E}^2}$. Suppose ${\bar{c}_1, \bar{c}_2 : [0,1] \rightarrow \mathbb{E}^2}$ follow opposite sides of a quadrilateral and ${\bar{c}_3 : [0,1] \rightarrow \mathbb{E}^2}$ follows the diagonal from ${\bar{c}_2(0)}$ to ${\bar{c}_1(1)}$ (all parametrized proportional to arc length). Then ${\bar{D}(t) \leq (1-t) \bar{D}(0) + t\bar{D}(1)}$ for all ${t \in [0,1]}$, where ${\bar{D}(t) := d(\bar{c}_1(t), \bar{c}_2(t))}$, since ${(1-t) \bar{D}(0)}$ is the distance from ${\bar{c}_1(t)}$ to ${\bar{c}_3(t)}$ and ${t\bar{D}(1)}$ is that from ${\bar{c}_3(t)}$ to ${\bar{c}_2(t)}$. So if ${c_3}$ is a geodesic from ${c_2(0)}$ to ${c_1(1)}$, then ${d(c_1(t), c_3(t)) \leq d(\bar{c}_1(t), \bar{c}_3(t))}$ and ${d(c_3(t), c_2(t)) \leq d(\bar{c}_3(t), \bar{c}_2(t))}$ by the CAT(0) condition, and therefore ${D(t) \leq \bar{D}(t)}$. So

$\displaystyle D(t) \ \leq \ \bar{D}(t) \ \leq \ (1-t) \bar{D}(0) + t\bar{D}(1) \ = \ (1-t) D(0) + t D(1).$

Convexity has two important consequences.

1. Uniqueness of geodesics. For each pairs of points in ${X}$, there is a unique geodesic from one to the other.
2. Contractibility. The space can be null–homotoped to a basepoint ${x_0 \in X}$ by ${h_t : X \rightarrow X}$ which takes ${x}$ to the point on the geodesics from ${x_0}$ to ${x}$ a distance ${(1-t)d(x_0,x)}$ from ${x_0}$. (Thus CAT(0) spaces have the potential to be universal covers of classifying spaces for discrete groups.)


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# 07 Gravitation

## Escape Velocity

Using the GeoGebra app above, I intend to demonstrate the relationship between total energy, kinetic energy and gravitational potential energy in a rocket trying to escape a planet’s gravitational field.

By changing the total energy of the rocket, you will increase the initial kinetic energy, thus allowing it to fly further from the surface of the planet. The furthest point to which the rocket can fly can be observed by moving the slider for “distance”. You will notice that the furthest point is where kinetic energy would have depleted.

Gravitational potential energy of an object is taken as zero at an infinite distance away from the source of the gravitational field. This means gravitational potential energy anywhere else takes on a negative value of $\dfrac{-GMm}{r}$. Therefore, the total energy of the object may be negative, even after taking into account its positive kinetic energy as total energy = kinetic energy + gravitational potential energy.

The minimum total energy needed for the rocket to leave the planet’s gravitational field is zero, as that will mean that the minimum initial kinetic energy will be equal to the increase in gravitational potential energy needed, according to the equation $\Delta U = 0 – (-\dfrac{GMm}{R_P})$, where $R_P$ is the radius of the planet.

Since $\dfrac{1}{2}mv^2 = \dfrac{GMm}{R_P}$, escape velocity, $v = \sqrt{\dfrac{2GM}{R_P}}$.

## How to Understand the Image of the Black Hole

Here’s a good explanation by Veritasium on why the image of the black hole looks the way it does.

## Simulation for Gravitational Field Strength and Potential

This simulation allows students to observe the variation of gravitational field strength and potential between two masses. Field strength is shown as vectors whereas potential is shown as scalar values on a plot. The resultant field strength and potential are shown in red.

## Getting Tides Right

I was reading up on tides this morning as I have to relieve a colleague’s IP2 class on forces, when I stumbled upon this video. It explains tides in a way that differs from most textbooks that clarified for me why the oceans bulge at both ends along the Moon-Earth line at the 2’30” mark onwards.

## Using Socrative for Pop Quizzes

Since almost every student owns a smartphone in SIngapore now, it has become possible for us to gather instant feedback or conduct an on-the-spot formative assessment to identify common misconceptions during class time.

A simple true-and-false quiz conducted during my recent lesson on Gravitation, after the lecture series was completed, showed that students are confused about the different treatment between vectors (force and field strength) and scalars (potential and potential energy).

1. If you were sitting half as far from your nearest classmate, the gravitational force of attraction between the two of you would be four times as strong. (81.0% correct)
2. At the midpoint of the distance from the center of the earth to the center of the moon, the gravitational field strength is zero. (90.5% correct)
3. At the midpoint of the distance between two identical masses, the gravitational potential is zero. (23.8% correct)
4. Work done by gravitational force in bringing an object from infinity to the surface of a planet is negative. (38.1% correct)
5. Satellites in a geostationary orbit must move directly above the equator. (52.4% correct)

This quiz was effective in helping students realise that potential cannot cancel out.

I also took the opportunity to explain to the class why the definition of gravitational potential (as well as potential energy) was based on the work done by an external force, rather than work done by gravitational force.

The tool that I used was Socrative and if anyone would like to use my quiz, you can follow the steps below:

1. Create a teacher account (not a student account) at socrative.com and log in.

2. Click on Manage Quizzes.

3. Select “Import Quiz”

4. Key in the SOC number 12289667

Since you have imported the quiz, you have effectively duplicated the quiz for yourself and are at liberty to edit the questions or even modify the quiz altogether as it does not affect mine.

To conduct the quiz, all you need to do is to go back to the dashboard and click on “Start a Quiz”, selecting the quiz you have saved. Follow the rest of the instructions to select the type of quiz you want (whether at students’ own pace or at a pace determined by yourself).

Once you have started the quiz, get your students to log in at m.socrative.com on their smartphone browsers and key in your Room Number (shown at the top of the screen). You will be able to see the number of students logging in at the top left corner and the attempts of your students in the main area.

If you have any questions regarding the use of Socrative in teaching, feel free to leave a comment below and I’ll try to answer it. Have fun!


text:
# What makes an equation equivalent to another?

1. Jun 14, 2013

### Atran

I know that this is basic, but I don't get it. Say we have the system,
$x + y = 32$
$3x + 2y = 70$
Each equation represents a graph, and the solution is the point where the two lines intersect.

Here is where I am confused:
$x + y = 32$
$2x + 2y = 64$
What is the proof that the two equations are equivalent / represent the same graph?

I read that two equations are equivalent if they share the same implication:
Thanks for help.

2. Jun 14, 2013

### Staff: Mentor

Start by saying that the two equations represent two distinct lines.

What can you conclude from that?

3. Jun 14, 2013

### Staff: Mentor

Equivalent equations have the same solution set.

Your first two equations are not equivalent, since their graphs are different. They do intersect at a point, though. Assuming that they intersect at (2, 3), the equations x = 2 and y = 3 would be equivalent to the first pair of equations you have above. (Note that I just made up a solution, so my equations x = 2 and y = 3 aren't actually equivalent to yours.)

For your second system of equations, both equations represent exactly the same line. Each solution of the first equation is also a solution of the second equation. In different words, if (x0, y0) is a solution (i.e., makes the equation a true statement) of the first equation, it will also be a solution of the second equation. For this example, the two equations are equivalent.

4. Jun 15, 2013

### Atran

So there are two ways of thinking:

1) If the two equations have one solution (which is one pair (xi, yi)) then I treat x and y in the equations as two (different) values.
$x+y=32$
$3x+2y=70$

$2x+2y=64$
$3x+2y=70$

$(3x+2y) - (2x+2y) = 70 - 64$
$x = 6$ and $y = 26$

2) (x+y=32) is equivalent to (2x+2y=64), since they share the same set of coordinates, therefore the same graph. Lets say set A has the coordinates of (x+y=32), and set B of (3x+2y=70). The solution is the set, $A \cap B$.

5. Jun 15, 2013

### HallsofIvy

Two equations are equivalent if they have exactly the same solution set. If (x, y) is in the solution set of the first equation then x+ y= 32. Multiplying both sides of the equation by 2, 2x+ 2y= 64 so (x, y) is in the solution set for that equation. That proves that the solution set for x+ y= 32 is a subset of the solution set for 2x+ 2y= 64. Now we have to do it the other way: if (x, y) is in the solution set for 2x+ 2y= 64, then they satisfy that equation. Dividing both sides by 2, x+ y= 32 so (x, y) is also in the solution set for that equaton and the solution set for 2x+ 2y= 64 is a subset of the solution set for x+ y= 32. Since each is a subset of the other, they are equal and the two equations are equivalent.


text:
Journal of Formalized Mathematics
Volume 7, 1995
University of Bialystok
Copyright (c) 1995 Association of Mizar Users

## Introduction to Circuits, II

Yatsuka Nakamura
Shinshu University, Nagano
Piotr Rudnicki
University of Alberta, Edmonton
Andrzej Trybulec
Warsaw University, Bialystok
Pauline N. Kawamoto
Shinshu University, Nagano

### Summary.

This article is the last in a series of four articles (preceded by [22], [23], [21]) about modelling circuits by many sorted algebras.\par The notion of a circuit computation is defined as a sequence of circuit states. For a state of a circuit the next state is given by executing operations at circuit vertices in the current state, according to denotations of the operations. The values at input vertices at each state of a computation are provided by an external sequence of input values. The process of how input values propagate through a circuit is described in terms of a homomorphism of the free envelope algebra of the circuit into itself. We prove that every computation of a circuit over a finite monotonic signature and with constant input values stabilizes after executing the number of steps equal to the depth of the circuit.

Partial funding for this work has been provided by: Shinshu Endowment Fund for Information Science, NSERC Grant OGP9207, JSTF award 651-93-S009.

#### MML Identifier: CIRCUIT2

The terminology and notation used in this paper have been introduced in the following articles [25] [13] [32] [29] [33] [28] [11] [12] [18] [14] [2] [8] [16] [5] [6] [7] [30] [1] [3] [31] [4] [15] [9] [26] [19] [27] [10] [20] [17] [24] [22] [23] [21]

#### Contents (PDF format)

1. Circuit Inputs
2. Circuit Computations

#### Bibliography

[1] Grzegorz Bancerek. Cardinal numbers. Journal of Formalized Mathematics, 1, 1989.
[2] Grzegorz Bancerek. The fundamental properties of natural numbers. Journal of Formalized Mathematics, 1, 1989.
[3] Grzegorz Bancerek. K\"onig's theorem. Journal of Formalized Mathematics, 2, 1990.
[4] Grzegorz Bancerek. Cartesian product of functions. Journal of Formalized Mathematics, 3, 1991.
[5] Grzegorz Bancerek. K\"onig's Lemma. Journal of Formalized Mathematics, 3, 1991.
[6] Grzegorz Bancerek. Sets and functions of trees and joining operations of trees. Journal of Formalized Mathematics, 4, 1992.
[7] Grzegorz Bancerek. Joining of decorated trees. Journal of Formalized Mathematics, 5, 1993.
[8] Grzegorz Bancerek and Krzysztof Hryniewiecki. Segments of natural numbers and finite sequences. Journal of Formalized Mathematics, 1, 1989.
[9] Grzegorz Bancerek and Piotr Rudnicki. On defining functions on trees. Journal of Formalized Mathematics, 5, 1993.
[10] Ewa Burakowska. Subalgebras of many sorted algebra. Lattice of subalgebras. Journal of Formalized Mathematics, 6, 1994.
[11] Czeslaw Bylinski. Functions and their basic properties. Journal of Formalized Mathematics, 1, 1989.
[12] Czeslaw Bylinski. Functions from a set to a set. Journal of Formalized Mathematics, 1, 1989.
[13] Czeslaw Bylinski. Some basic properties of sets. Journal of Formalized Mathematics, 1, 1989.
[14] Czeslaw Bylinski. The modification of a function by a function and the iteration of the composition of a function. Journal of Formalized Mathematics, 2, 1990.
[15] Patricia L. Carlson and Grzegorz Bancerek. Context-free grammar --- part I. Journal of Formalized Mathematics, 4, 1992.
[16] Agata Darmochwal. Finite sets. Journal of Formalized Mathematics, 1, 1989.
[17] Malgorzata Korolkiewicz. Homomorphisms of many sorted algebras. Journal of Formalized Mathematics, 6, 1994.
[18] Jaroslaw Kotowicz. Monotone real sequences. Subsequences. Journal of Formalized Mathematics, 1, 1989.
[19] Beata Madras. Product of family of universal algebras. Journal of Formalized Mathematics, 5, 1993.
[20] Beata Madras. Products of many sorted algebras. Journal of Formalized Mathematics, 6, 1994.
[21] Yatsuka Nakamura, Piotr Rudnicki, Andrzej Trybulec, and Pauline N. Kawamoto. Introduction to circuits, I. Journal of Formalized Mathematics, 6, 1994.
[22] Yatsuka Nakamura, Piotr Rudnicki, Andrzej Trybulec, and Pauline N. Kawamoto. Preliminaries to circuits, I. Journal of Formalized Mathematics, 6, 1994.
[23] Yatsuka Nakamura, Piotr Rudnicki, Andrzej Trybulec, and Pauline N. Kawamoto. Preliminaries to circuits, II. Journal of Formalized Mathematics, 6, 1994.
[24] Beata Perkowska. Free many sorted universal algebra. Journal of Formalized Mathematics, 6, 1994.
[25] Andrzej Trybulec. Tarski Grothendieck set theory. Journal of Formalized Mathematics, Axiomatics, 1989.
[26] Andrzej Trybulec. Many-sorted sets. Journal of Formalized Mathematics, 5, 1993.
[27] Andrzej Trybulec. Many sorted algebras. Journal of Formalized Mathematics, 6, 1994.
[28] Andrzej Trybulec. On the sets inhabited by numbers. Journal of Formalized Mathematics, 15, 2003.
[29] Andrzej Trybulec. Subsets of real numbers. Journal of Formalized Mathematics, Addenda, 2003.
[30] Wojciech A. Trybulec. Groups. Journal of Formalized Mathematics, 2, 1990.
[31] Wojciech A. Trybulec. Pigeon hole principle. Journal of Formalized Mathematics, 2, 1990.
[32] Zinaida Trybulec. Properties of subsets. Journal of Formalized Mathematics, 1, 1989.
[33] Edmund Woronowicz. Relations and their basic properties. Journal of Formalized Mathematics, 1, 1989.


text: Thank you for taking the time to write this - people are unlikely to present a more somber take on higher mathematics. Do you know where can I find these Mumford-Lang lecture notes? Their algorithm is based on algebraic geometry methods, speciﬁcally cylindrical algebraic decomposition A road map for learning Algebraic Geometry as an undergraduate. The notes are missing a few chapters (in fact, over half the book according to the table of contents). So when you consider that algebraic local ring, you can think that the actual neighbourhood where each function is defined is the complement of some divisor, just like polynomials are defined in the coplement of the divisor at infinity. Even so, I like to have a path to follow before I begin to deviate. New comments cannot be posted and votes cannot be cast, Press J to jump to the feed. Yes, it's a slightly better theorem. I disagree that analysis is necessary, you need the intuition behind it all if you want to understand basic topology and whatnot but you definitely dont need much of the standard techniques associated to analysis to have this intuition. particular that of number theory, the best reference by far is a long typescript by Mumford and Lang which was meant to be a successor to “The Red Book” (Springer Lecture Notes 1358) but which was never finished. 2) Fulton's "Toric Varieties" is also very nice and readable, and will give access to some nice examples (lots of beginners don't seem to know enough explicit examples to work with). A roadmap for a semi-algebraic set S is a curve which has a non-empty and connected intersection with all connected components of S. Fine. There's a huge variety of stuff. I find both accessible and motivated. And now I wish I could edit my last comment, to respond to your edit: Kollar's book is great. MathJax reference. Is there ultimately an "algebraic geometry sucks" phase for every aspiring algebraic geometer, as Harrison suggested on these forums for pure algebra, that only (enormous) persistence can overcome? The books on phase 2 help with perspective but are not strictly prerequisites. An example of a topic that lends itself to this kind of independent study is abelian schemes, where some of the main topics are (with references in parentheses): You may amuse yourself by working out the first topics above over an arbitrary base. What is in some sense wrong with your list is that algebraic geometry includes things like the notion of a local ring. Underlying étale-ish things is a pretty vast generalization of Galois theory. The following seems very relevant to the OP from a historical point of view: a pre-Tohoku roadmap to algebraic topology, presenting itself as a "How to" for "most people", written by someone who thought deeply about classical mathematics as a whole. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. I left my PhD program early out of boredom. DF is also good, but it wasn't fun to learn from. Finally, I wrap things up, and provide a few references and a roadmap on how to continue a study of geometric algebra.. 1.3 Acknowledgements I would suggest adding in Garrity et al's excellent introductory problem book, Algebraic Geometry: A Problem-solving Approach. proof that abelian schemes assemble into an algebraic stack (Mumford. This is is, of course, an enormous topic, but I think it’s an exciting application of the theory, and one worth discussing a bit. real analytic geometry, and R[X] to algebraic geometry. However, I feel it is necessary to precede the reproduction I give below of this 'roadmap' with a modern, cautionary remark, taken literally from http://math.stanford.edu/~conrad/: It is interesting, and indicative of how much knowledge is required in algebraic geometry, that Snapper recommends Weil's 'Foundations' at the end of this "How to get started"-section. Right now, I'm trying to feel my way in the dark for topics that might interest me, that much I admit. ... learning roadmap for algebraic curves. It is interesting, and indicative of how much knowledge is required in algebraic geometry, that Snapper recommends Weil's 'Foundations' at the … Unfortunately I saw no scan on the web. We ﬁrst ﬁx some notation. Algebraic Geometry seemed like a good bet given its vastness and diversity. Here is the current plan I've laid out: (note, I have only taken some calculus and a little linear algebra, but study some number theory and topology while being mentored by a faculty member), Axler's Linear Algebra Done Right (for a rigorous and formal treatment of linear algebra), Artin's Algebra and Allan Clark's Elements of Abstract Algebra (I may pick up D&F as a reference at a later stage), Rudin's Principles of Mathematical Analysis (/u/GenericMadScientist), Ideals, Varieties and Algorithms by Cox, Little, and O'Shea (thanks /u/crystal__math for the advice to move it to phase, Garrity et al, Algebraic Geometry: A Problem-solving Approach. And it can be an extremely isolating and boring subject. Take some time to develop an organic view of the subject. For a small sample of topics (concrete descent, group schemes, algebraic spaces and bunch of other odd ones) somewhere in between SGA and EGA (in both style and subject), I definitely found the book 'Néron Models' by Bosch, Lütkebohmert and Raynaud a nice read, with lots and lots of references too. Then they remove the hypothesis that the derivative is continuous, and still prove that there is a number x so that g'(x) = (g(b)-g(a))/(b-a). Authors: Saugata Basu, Marie-Francoise Roy (Submitted on 14 May 2013 , last revised 8 Oct 2016 (this version, v6)) Abstract: Let $\mathrm{R}$ be a real closed field, and $\mathrm{D} \subset \mathrm{R}$ an ordered domain. Seemed like a good bet given its vastness and diversity is based on algebraic seemed... People are unlikely to present a more algebraic geometry roadmap take on higher mathematics not be cast, Press J jump. Your edit: Kollar 's book is great and it can be an isolating... 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