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Messages - trfrm
1
« pada: Desember 04, 2019, 09:00:47 PM »
\section{Definisi dan Teorema Limit}
Andaikan $f, g\,:\,\mathbb{R}\to\mathbb{R}$ dan $c, \delta, \epsilon \in \mathbb{R}$. Definisi limit adalah
\[ \lim_{x\to c}f(x) = L \]
sedemikian rupa sehingga untuk setiap $\epsilon >0$, terdapat $\delta > 0$, sedemikian berlaku jika
\[ 0 < |x - c| < \delta \]
mengakibatkan
\[ |f(x) - L| < \epsilon. \]
Dari definisi ini, kita hendak mencari bentuk eksplisit dari nilai $L$.
Kita dapat menuliskan
\[ |x - c| = |r| \]
di mana $0 < |r| < \delta$ alias $r \in (-\delta, 0)\cup(0, \delta)$, sehingga
\[ x = c + r. \]
Kita dapat menuliskan pula
\[ |f(x) - L| = |R| \]
di mana $0 \leq |R| < \epsilon$ alias $R \in (-\epsilon, \epsilon)$, sehingga
\[ L = f(x) + R = f(c + r) + R. \]
Contoh kongkretnya adalah
\[ \lim_{x\to 0}x = (0 + r) + R. \]
Kita ambil $R = -r$, sehingga
\[ \lim_{x\to 0}x = r + (-r) = 0. \]
Selain itu,
\[ \lim_{x\to 0}\frac{x}{x} = \frac{0 + r}{0 + r} + R. \]
Kita ambil $R = 0$, sehingga
\[ \lim_{x\to 0}\frac{x}{x} = \frac{r}{r} = 1. \]
Selanjutnya kita hendak menurunkan beberapa teorema limit.
\[ \lim_{x\to c}(f(x) + g(x)) = (f(c + r) + g(c + r)) + R. \]
Apabila kita ambil $R = R_1 + R_2$, di mana $R_1, R_2 \in (-\epsilon, \epsilon)$, maka
\[ \lim_{x\to c}(f(x) + g(x)) = (f(c + r) + g(c + r)) + R = (f(c + r) + R_1) + (g(c + r) + R_2) = \lim_{x\to c}f(x) + \lim_{x\to c}g(x). \]
\[ \lim_{x\to c}(f(x)g(x)) = f(c + r)g(c + r) + R. \]
Apabila kita ambil $R = 0$, maka
\[ \lim_{x\to c}(f(x)g(x)) = (f(c + r) + R)(g(c + r) + R) = \left(\lim_{x\to c}f(x)\right)\left(\lim_{x\to c}g(x)\right). \]
\[ \lim_{x\to c}\frac{f(x)}{g(x)} = \frac{f(c + r)}{g(c + r)} + R. \]
Apabila kita ambil $R = 0$, maka
\[ \lim_{x\to c}\frac{f(x)}{g(x)} = \frac{f(c + r) + R}{g(c + r) + R} = \frac{\lim_{x\to c}f(x)}{\lim_{x\to c}g(x)} \]
di mana $\lim_{x\to c}g(x) \neq 0$.
2
« pada: Juni 09, 2019, 04:14:39 PM »
\section{Menalar Turunan Lie dengan Analisis Tensor Biasa}
Andaikan $T := {T^{\mu_1\cdots\mu_m}}_{\nu_1\cdots\nu_n}e_{\mu_1}\otimes\cdots\otimes e_{\mu_m}\otimes e^{\nu_1}\otimes\cdots\otimes e^{\nu_n}$ adalah sebuah tensor yang bergantung pada $p$ buah koordinat umum, yaitu $x^1,\cdots,x^p \in \mathbb{R}$ sebagai $p$ buah parameter bagi manifold $M$ yang berdimensi $p$, serta $X := X^\lambda e_\lambda$ adalah sebuah vektor yang bergantung pada $p$ buah koordinat umum tersebut juga. Selanjutnya, akan dicari turunan Lie dari $T$ sepanjang $X$, yang didefinisikan sebagai
\[ L_XT := \left(\lim_{\epsilon\to 0}\frac{T_x(x + \epsilon X) - T}{\epsilon}\right)_{\partial_\rho X = 0} \]
untuk semua $\rho\in\{1,\cdots,p\}$, di mana $\partial_\rho := \partial/\partial x^\rho$.
Ternyata, pengolahan lebih lanjut dari persamaan terakhir, menghasilkan
\[ L_XT = \left(X^\lambda\partial_\lambda T\right)_{\partial_\rho X = 0}. \]
\[ L_XT = (X^\lambda(\partial_\lambda {T^{\mu_1\cdots\mu_m}}_{\nu_1\cdots\nu_n}e_{\mu_1}\otimes\cdots\otimes e_{\mu_m}\otimes e^{\nu_1}\otimes\cdots\otimes e^{\nu_n} \]
\[ + {T^{\mu_1\cdots\mu_m}}_{\nu_1\cdots\nu_n}\sum_{s=1}^m e_{\mu_1}\otimes\cdots\otimes e_{\mu_{s-1}}\otimes\partial_\lambda e_{\mu_s}\otimes e_{\mu_{s+1}}\otimes\cdots\otimes e_{\mu_m}\otimes e^{\nu_1}\otimes\cdots\otimes e^{\nu_n} \]
\[ + {T^{\mu_1\cdots\mu_m}}_{\nu_1\cdots\nu_n}e_{\mu_1}\otimes\cdots\otimes e_{\mu_m}\otimes\sum_{r=1}^n e^{\nu_1}\otimes\cdots\otimes e^{\nu_{r-1}}\otimes\partial_\lambda e^{\nu_r}\otimes e^{\nu_{r+1}}\otimes\cdots\otimes e^{\nu_n}))_{\partial_\rho X = 0}. \]
Dengan memanfaatkan identitas $\partial_\alpha e_\beta = {\Gamma^\gamma}_{\alpha\beta}e_\gamma$ dan $\partial_\alpha e^\beta = -{\Gamma^\beta}_{\alpha\gamma}e^\gamma$, diperoleh
\[ L_XT = (X^\lambda\partial_\lambda {T^{\mu_1\cdots\mu_m}}_{\nu_1\cdots\nu_n}e_{\mu_1}\otimes\cdots\otimes e_{\mu_m}\otimes e^{\nu_1}\otimes\cdots\otimes e^{\nu_n} \]
\[ + {T^{\mu_1\cdots\mu_m}}_{\nu_1\cdots\nu_n}X^\lambda\sum_{s=1}^m {\Gamma^\sigma}_{\lambda\mu_s} e_{\mu_1}\otimes\cdots\otimes e_{\mu_{s-1}}\otimes e_\sigma\otimes e_{\mu_{s+1}}\otimes\cdots\otimes e_{\mu_m}\otimes e^{\nu_1}\otimes\cdots\otimes e^{\nu_n} \]
\[ - {T^{\mu_1\cdots\mu_m}}_{\nu_1\cdots\nu_n}X^\lambda e_{\mu_1}\otimes\cdots\otimes e_{\mu_m}\otimes\sum_{r=1}^n {\Gamma^{\nu_r}}_{\lambda\tau} e^{\nu_1}\otimes\cdots\otimes e^{\nu_{r-1}}\otimes e^\tau\otimes e^{\nu_{r+1}}\otimes\cdots\otimes e^{\nu_n})_{\partial_\rho X = 0} \]
di mana $\Gamma$ adalah lambang Christoffel.
Dengan memanfaatkan identitas $\partial_\alpha X = D_\alpha X^\lambda e_\lambda$ di mana $D_\alpha X^\beta := \partial_\alpha X^\beta + X^\lambda{\Gamma^\beta}_{\alpha\lambda}$, maka diperoleh
\[ L_XT = X^\lambda\partial_\lambda {T^{\mu_1\cdots\mu_m}}_{\nu_1\cdots\nu_n}e_{\mu_1}\otimes\cdots\otimes e_{\mu_m}\otimes e^{\nu_1}\otimes\cdots\otimes e^{\nu_n} \]
\[ - {T^{\mu_1\cdots\mu_m}}_{\nu_1\cdots\nu_n}\sum_{s=1}^m \partial_{\mu_s}X^\sigma e_{\mu_1}\otimes\cdots\otimes e_{\mu_{s-1}}\otimes e_\sigma\otimes e_{\mu_{s+1}}\otimes\cdots\otimes e_{\mu_m}\otimes e^{\nu_1}\otimes\cdots\otimes e^{\nu_n} \]
\[ + {T^{\mu_1\cdots\mu_m}}_{\nu_1\cdots\nu_n}e_{\mu_1}\otimes\cdots\otimes e_{\mu_m}\otimes\sum_{r=1}^n \partial_\tau X^{\nu_r} e^{\nu_1}\otimes\cdots\otimes e^{\nu_{r-1}}\otimes e^\tau\otimes e^{\nu_{r+1}}\otimes\cdots\otimes e^{\nu_n}. \]
Dengan melakukan penukaran indeks boneka, akhirnya diperoleh bentuk eksplisit dari turunan Lie dari $T$ sepanjang $X$, yaitu
\[ L_XT = (X^\lambda\partial_\lambda {T^{\mu_1\cdots\mu_m}}_{\nu_1\cdots\nu_n} \]
\[ - \sum_{s=1}^m {T^{\mu_1\cdots\mu_{s-1}\sigma\mu_{s+1}\cdots\mu_m}}_{\nu_1\cdots\nu_n}\partial_\sigma X^{\mu_s} \]
\[ + \sum_{r=1}^n {T^{\mu_1\cdots\mu_m}}_{\nu_1\cdots\nu_{r-1}\tau\nu_{r+1}\cdots\nu_n}\partial_{\nu_r}X^\tau) \]
\[ e_{\mu_1}\otimes\cdots\otimes e_{\mu_m}\otimes e^{\nu_1}\otimes\cdots\otimes e^{\nu_n}. \]
3
« pada: Juli 20, 2018, 01:33:29 PM »
The limit $c\to\infty$ is non-relativistic limit . . .
5
« pada: Juli 20, 2018, 01:11:29 PM »
Excuse me ... .
As a real function on non-negative real line, the square root function is a non-negative single-valued-function ... .
As a complex function on complex plane, the square root function is a multi-valued-function ... .
6
« pada: Juli 20, 2018, 01:08:30 PM »
That is not true. The function $y= \frac{x}{x+1}$ maps the set of all rational numbers one-to-one onto the set of rational numbers between -1 and 1. Excuse me ... . The plot of f(x) = x/(x + 1) is in the following site ... . x/(x+1) - Wolfram|Alpha The domain of the function $f\,:\,\mathbb{Q}\to\mathbb{Q}~:~x\mapsto{x/(x+1)}$ is $\textrm{dom}(f)=(-\infty,-1)_\mathbb{Q}\cup(-1,+\infty)_\mathbb{Q}$ ... , and the range (or image) of the $f$ is $\textrm{im}(f)=(-\infty,-1)_\mathbb{Q}\cup(-1,+\infty)_\mathbb{Q}$ ... . Thus, the function is not bijection from $\mathbb{Q}$ to $(-1,1)_\mathbb{Q}$ ... . I'm sorry ... .  http://www.thescienceforum.com/mathematics/35211-rationalizing-irrational-number.html#post414277
7
« pada: Juli 20, 2018, 01:06:40 PM »
Is it all integers or only small ones ? What is small ? What is huge ? Oh, I'm sorry ... . The term "small" and "huge" are relative ... . I don't actually undrstand what you are trying to do. You define irrational numbers in a way that mean they are ... irrational. What do you mean by"rationalize"? (I can't follow from your notation)
I just tried to regard an irrational number as a rational number ... . Here is a simple proof that $ \sqrt{2}$ is irrational: Proof that the square root of 2 is irrational number
You can do this with most (all?) irrational numbers, for example: Proof that $\pi$ is irrational
Thank you very much for the explanation ... . Now, I understand why the square root of 2 and the pi are regarded as the irrational numbers ... . All rational numbers and all irrational number can form a set of real numbers ... . There are significant differences between character of $\mathbb{R}$ and character of $\mathbb{Q}$ ... . If in example we define the set $(-1,1)_\mathbb{Q}:=\{x\in\mathbb{Q}~|~-1<x<1\}\subset\mathbb{Q}$ and the set $(-1,1)_\mathbb{R}:=\{x\in\mathbb{R}~|~-1<x<1\}\subset\mathbb{R}$, then we can see the significant difference ... . We can find at least one bijective map $\mathbb{R}\to(-1,1)_\mathbb{R}$, in example $f\,:\,x\mapsto\tanh{kx}$ and $g\,:\,x\mapsto\frac{2}{\pi}\arctan{kx}$, where $k$ is a constant ... . But, however, we cannot find at least one bijective map $\mathbb{Q}\to(-1,1)_\mathbb{Q}$ ... . How can it ... ? Until now I have not understood ... .
8
« pada: Juli 20, 2018, 01:05:12 PM »
Excuse me ... .  A set of the rational numbers is denoted as $\mathbb{Q}:=\{m/n~|~m,n\in\mathbb{Z}~\textrm{and}~n\neq0\}$, where $\mathbb{Z}:=\{0,\pm{1},\pm{2},\pm{3},\dots\}$ is the set of integers ... . A set of the irrational numbers $\mathbb{R}-\mathbb{Q}$ consists of the real numbers which cannot be denoted as $m/n$ such that $m,n\in\mathbb{Z}$ and $n\neq0$ ... . The problem is in following ... . For example, in present time, here is $a\in\mathbb{R}-\mathbb{Q}$ which be agreed as an irrational number ... . Can the irrational number $a$ be rationalized to be a rational number if a person is able to denote the $a$ as $m/n$, where $m,n\in\mathbb{Z}$ and $n\neq0$, in the next time ... ? [Here, the absolute values of $m$ and $n$ may be the very very huge integers ... .] Thank you for the answer ... .
9
« pada: Juli 20, 2018, 12:58:14 PM »
I don't know why you insist on using the antiquate notion of relativistic mass, this is the source of confusion. Using the modern notion of invariant mass makes things a lot more straightforward. And yes, $\mathbf{a}=0$ is the only acceptable solution for the equation:
$0=m_0\gamma(\frac{\gamma^2\mathbf{v}\cdot\mathbf{a}}{c^2}\mathbf{v}+\mathbf{a})$
I’m sorry ... . There, $m_i=\gamma_im_0$ and $m=\gamma{m_0}$ are the initial and final relativistic mass, respectively ... .
10
« pada: Juli 20, 2018, 12:56:55 PM »
If $\mathbf{F}=0$ this means $\mathbf{p}=$const, i.e. $\mathbf{p}=\gamma(v_0)m_0 \mathbf{v_0}$ (i.e $\mathbf{v}=\mathbf{v_0}$) so $\mathbf{a}=0$. This is the correct chain of inference, what you posted so far, isn't. In this context (this relativistic case) ... , it is right ... . But, in general case, if $\mathbf{F}=\mathbf{0}$, then $\mathbf{p}=\mathbf{p}_i$ (where the index $i$ denotes the initial state) ... , then ... , $m\mathbf{v}=m_i\mathbf{v}_i$ alias $\mathbf{v}=m_i\mathbf{v}_i/m$ ... . Because in general, the mass is not constant, then $\mathbf{a}:=\dot{\mathbf{v}}=-m_i\mathbf{v}_i\dot{m}/m^2$ which must not to be zero ... . But, however, in the relativistic case, $\dot{m}=\gamma^2m(\mathbf{v}\cdot\mathbf{a})/c^2$, so the last equation becomes $\mathbf{a}=-m_i\mathbf{v}_i\gamma^2(\mathbf{v}\cdot\mathbf{a})/(mc^2)$ ... . The possible solution of the last equation is $\mathbf{a}=\mathbf{0}$ ... . I think it is important to distinguish pure mathematics from physics in this context. Mathematically it is certainly possible to come up with all sorts of relations affecting speed and acceleration even in the absence of a force, but that does not necessarily mean those relations are physically meaningful. In the context of physics, changing the state of motion of an object requires changing its momentum, and, at least classically, that will require a force.
This seemly comes from the origin definition of the force as derivative of the linear momentum respect to time, that is the Newton’s second law of motion as a differential equation [Goldstein, Classical Mechanics, 1980, Chapter 1] ... . http://www.thescienceforum.com/physics/35197-acceleration-classical-particle-without-force-acting-particle.html#post413388
11
« pada: Juli 20, 2018, 12:52:48 PM »
I don't know where you are trying to get, you simply have:
$0=\mathbf{F}=m_0\gamma(\frac{\gamma^2\mathbf{v}\cdot\mathbf{a}}{c^2}\mathbf{v}+\mathbf{a})$
meaning that you get $\mathbf{a}=-\frac{\gamma^2 \mathbf{v}\cdot\mathbf{a}}{c^2}\mathbf{v}$
A little further algebraic manipulation shows that the above is impossible.
Here ... , incidentally, $\mathbf{a}=\mathbf{0}$ ... .
It's getting worse, you should really arrive to :
$\mathbf{F}=m_0\gamma(\frac{\gamma^2\mathbf{v}\cdot\mathbf{a}}{c^2}\mathbf{v}+\mathbf{a})$ It turns out that the equation $\mathbf{F}=m(\gamma^2(\mathbf{v}\cdot\mathbf{a})\mathbf{v}/c^2+\mathbf{a})$ is not different from the equation $\mathbf{F}=\gamma^2m(\mathbf{v}\times(\mathbf{v}\times\mathbf{a})/c^2+\mathbf{a})$ ... . http://www.thescienceforum.com/physics/35197-acceleration-classical-particle-without-force-acting-particle.html#post413250
12
« pada: Juli 20, 2018, 12:51:17 PM »
Oh O.K. ... . From the identity of vectors $\mathbf{A}\times(\mathbf{B}\times\mathbf{C})=(\mathbf{A}\cdot\mathbf{C})\mathbf{B}-(\mathbf{A}\cdot\mathbf{B})\mathbf{C}$, then $\mathbf{v}\times(\mathbf{v}\times\mathbf{a})=(\mathbf{v}\cdot\mathbf{a})\mathbf{v}-|\mathbf{v}|^2\mathbf{a}$ alias $(\mathbf{v}\cdot\mathbf{a})\mathbf{v}=\mathbf{v}\times(\mathbf{v}\times\mathbf{a})+|\mathbf{v}|^2\mathbf{a}$ ... .
Thus,
$\mathbf{F}=m(\gamma^2(\mathbf{v}\times(\mathbf{v}\times\mathbf{a})+|\mathbf{v}|^2\mathbf{a})/c^2+\mathbf{a})$ alias
$\mathbf{F}=m(\gamma^2\mathbf{v}\times(\mathbf{v}\times\mathbf{a})/c^2+(1+\gamma^2|\mathbf{v}|^2/c^2)\mathbf{a})$ ... .
Since $1+\gamma^2|\mathbf{v}|^2/c^2=\gamma^2$, then
$\mathbf{F}=\gamma^2m(\mathbf{v}\times(\mathbf{v}\times\mathbf{a})/c^2+\mathbf{a})$ ... .
In the last equation, we can see that there are transversal and longitudinal components of the force ... .
13
« pada: Juli 20, 2018, 12:48:42 PM »
Hence, if the force was zero, then the acceleration should not be zero ... . In other word (in the relativistic case without presence of force), $\mathbf{v}=m_i\mathbf{v}_i/m=\gamma_i\mathbf{v}_i/\gamma$, so $c^2|\mathbf{v}|^2={\gamma_i}^2|\mathbf{v}_i|^2(c^2-|\mathbf{v}|^2)$ alias $(c^2+{\gamma_i}^2|\mathbf{v}_i|^2)|\mathbf{v}|^2=c^2{\gamma_i}^2|\mathbf{v}_i|^2$ alias
$\displaystyle|\mathbf{v}|=\frac{c\gamma_i|\mathbf{v}_i|}{\sqrt{c^2+{\gamma_i}^2|\mathbf{v}_i|^2}}$ ... .
N.B. : The $\gamma_i$ and the $\mathbf{v}_i$, respectively, are the initial gamma and the initial velocity (at $t=0$) ... .
Because the final velocity has same direction with the initial velocity, then
$\displaystyle\mathbf{v}=|\mathbf{v}|\frac{\mathbf{v}_i}{|\mathbf{v}_i|}=\frac{c\gamma_i\mathbf{v}_i}{\sqrt{c^2+{\gamma_i}^2|\mathbf{v}_i|^2}}=\frac{c\mathbf{v}_i}{\sqrt{c^2/{\gamma_i}^2+|\mathbf{v}_i|^2}}=\mathbf{v}_i=\textrm{constant}$
Thus, in this case, $\mathbf{a}=\dot{\mathbf{v}}=\mathbf{0}$ if $\mathbf{F}=\mathbf{0}$ ... .
14
« pada: Juli 20, 2018, 12:47:15 PM »
This stuff is really bad. Care to try again, given the information that $\mathbf{p}=\gamma m_0 \mathbf{v}$ . May I suggest that you forget about the nonsense $\dot{m}:=dm/dt$. This doesn't exist.
If you want to do "classical domain", just set $\gamma=1$
O.K. ... . Thanks ... . We have known that the relativistic momentum is $\mathbf{p}=m\mathbf{v}=\gamma{m_0}\mathbf{v}$, where $m=\gamma{m_0}$ and $\gamma:=1/\sqrt{1-|\mathbf{v}|^2/c^2}$ ... . [Here, the $m_0$ is not the initial mass, but is the rest mass ... .] Then, $\mathbf{F}=\dot{\mathbf{p}}=\dot{m}\mathbf{v}+m\mathbf{a}$ where $\dot{m}=\dot{\gamma}m_0$ and $\dot\gamma=\gamma^3(\mathbf{v}\cdot\mathbf{a})/c^2$ ... , so $\dot{m}=m\gamma^2(\mathbf{v}\cdot\mathbf{a})/c^2$ ... . Thus ... (by subtituting), this yields $\mathbf{F}=m(\gamma^2(\mathbf{v}\cdot\mathbf{a})\mathbf{v}/c^2+\mathbf{a})$ which be known as the relativistic force ... . [Note that the special relativity (without quantum aspect) is sometimes classified into the classical mechanics subject ... . I saw it in the Classical Mechanics textbook by Hebert Goldstein, 1980 ... .]
15
« pada: Juli 20, 2018, 12:44:43 PM »
Excuse me ... . In sight, it is impossible that a classical particle can be accelerated without a force acting on it ... . In other word ... , the velocity of the particle cannot change if the force acting on it is zero, sightly ... . But ... , in fact, this opinion is not always true generally ... . Really, the Newton’s second law of motion is a differential equation $\displaystyle\mathbf{F}=\frac{d\mathbf{p}}{dt}$ where the $\mathbf{F}$ and the $\mathbf{p}$, respectively, are the force and the momentum of the particle as function of the time $t$ ... . If the velocity and mass of the particle, respectively, are $\mathbf{v}$ and $m$ as function of time $t$, then $\mathbf{p}=m\mathbf{v}$ ... . Thus, in general, $\mathbf{F}=\dot{m}\mathbf{v}+m\mathbf{a}$ where $\dot{m}:=dm/dt$ is rate of the mass, and $\mathbf{a}:=d\mathbf{v}/dt$ is its acceleration ... . The last expression doesn’t only obey in non-relativistic cases, but also in relativistic cases ... . If we set $\mathbf{F}=\mathbf{0}$, then $\mathbf{a}=-\dot{m}\mathbf{v}/m$ ... . In this case, $d\mathbf{p}/dt=\mathbf{0}$ alias $\mathbf{p}=\mathbf{p}_0$ alias $m\mathbf{v}=m_0\mathbf{v}_0$ alias $\mathbf{v}=m_0\mathbf{v}_0/m$ ... . Thus (by subtituting or by differentiating), this yields $\mathbf{a}=-\dot{m}m_0\mathbf{v}_0/m^2$ ... . [Here, the $m_0$ and the $\mathbf{v}_0$ denote the initial mass and the initial velocity, respectively ... .] In another case, if $\mathbf{F}=\dot{m}\mathbf{v}$, then $\mathbf{a}=\mathbf{0}$ ... . Therefore, really, the force merely changes the momentum rather than the velocity ... .
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