When a charged particle gets accelerated it emits electromagnetic waves. In reality, when a charged particle gets accelerated the electric field around the charge remains unchanged, but the magnetic field changes. Here the energy provided to the particle - by the energy source accelerating it - gets divided into the energy contained in the increased magnetic field and EMWs radiated. But I don't understand the mechanism by which these EMWs are emitted. I have read that when a charged particle gets accelerated the electromagnetic field associated with it gets detached from it and this detached electromagnetic field is that we call electromagnetic waves.  So is it that, when the charged particle gets accelerated, some part of the magnetic field associated with it gets detached from it and simultaneously (but apart from the effect of the event described above) the magnetic field associated with the charged particle gets changed according to the amount of energy being supplied or being taken away by the external source accelerating the charge particle? Thus, in the entire process the electric field around the charge particle remains constant. Is it so?  If this is true, then do we have to assume that the magnetic field being detached itself and instantaneously gets converted into EMW of the corresponding energy? But then I get again the same question: By which mechanism?   Also I thought of an entirely different mechanism: When the charged particle gets accelerated some part of its electric field as well as some part of its magnetic field gets detached from the charged particle and that itself is the EMW. And simultaneously (but apart from the effect of the event described above) both the magnetic field &amp; electric field associated with the charged particle gets changed according to the amount of energy being supplied or being taken away by the external source accelerating the charge particle? But here is one more problem: If we are assuming that the electric field is decaying into the EMWs, we will have to consider the contribution of the externally accelerating source to increase the electric field of the charge particle regardless the charge is being accelerated or decelerated. As far as I think, this cannot be done directly. We should assume the contribution of the external source to increase the magnetic field associated with the charge and its instantaneous conversion into electric field. Now again, same question: By which mechanism? So overall I have complete unclarity about the mechanism of the production of EMWs when a charge particle gets accelerated.
Let's consider it case by case:  Case 1: Charged particle is at rest. It has an electric field around it. No problem. That is its property.  Case 2: Charged particle started moving (it's accelerating). We were told that it starts radiating EM radiation. Why? What happened to it? What made it do this?  Follow up question: Suppose a charged particle is placed in uniform electric field. It accelerates because of electric force it experiences. Then work done by the electric field should not be equal to change in its kinetic energy, right? It should be equal to change in K.E + energy it has radiated in the form of EM Waves. But then, why don't we take energy radiated into consideration when solving problems? (I'm tutoring grade 12 students. I never encountered a problem in which energy radiated is considered.)
I have a VIP skybag (I believe this does not matter). It has a 3-number TSA lock built into the suitcase. I know the combination to unlock, let's say it's 345. Recently it was locked and while there was a different number on the combination lock the reset/set lock button was pressed.  Since then the reset/set lock button is pressed (stuck) and the lock button also does not unlock. Any pointers on how to unlock it? I am thinking of unscrewing the lock from inside of the suitcase but it seems quite a hassle, as I may have to assemble it back. Let me know if there is any other option to unlock it from the outside. Let's say I'm trying to answer a lot of questions just to up my reputation (even if I generally don't know what I'm talking about).  Then, whenever I get downvoted, I just delete my answer so that nobody sees that I've posted stupid answers, and so I don't lose reputation points.   Is this frowned upon? Do I lose reputation points on downvoted answers if I delete them? #!/bin/bash  SUBJECT="WARNING CPU USAGE HIGH" TO=gmail id MESSAGE=/tmp/messages echo "#######################" &gt; $MESSAGE echo "CPU statistics as follows.." &gt;&gt; $MESSAGE mpstat &gt;&gt; $MESSAGE echo "#######################" &gt;&gt; $MESSAGE CPU_USAGE=$(top -b -n1 | awk '/^Cpu/ {print $2}' | cut -d. -f1) [ $CPU_USAGE -gt 85 ] &amp;&amp; mail -s "$SUBJECT" "$TO" &lt; $MESSAGE`   ./cpu.sh: line 11: [: -gt: unary operator expected What might be the reason Proof of this fairly obscure differentiation trick? Suppose we're tying to differentiate the function $f(x)=x^x$. Now the textbook method would be to notice that $f(x)=e^{x \log{x}}$ and use the chain rule to find $$f'(x)=(1+\log{x})\ e^{x \log{x}}=(1+\log{x})\ x^x.$$  But suppose that I didn't make this observation and instead tried to apply the following differentiation rules:  $$\frac{d}{dx}x^c=cx^{c-1} \qquad (1)\\ \frac{d}{dx}c^x = \log{c}\ \cdot c^x \quad (2)$$  which are valid for any constant $c$. Obviously neither rules are applicable to the form $x^x$ because in this case neither the base nor the exponent are constant. But if I pretend that the exponent is constant and apply rule $(1)$, I would get $f'(x)\stackrel{?}{=}x\cdot x^{x-1}=x^x.$ Likewise, if I pretend that the base is constant and apply rule $(2)$ I obtain $f'(x)\stackrel{?}{=}\log{x}\cdot x^x$.  It isn't hard to see that neither of the derivatives are correct. But here's where the magic happens: if we sum the two “derivatives” we end up with $$x^x+ \log{x}\cdot x^x=(1+\log{x})\ x^x$$ which is the correct expression for $f'(x)$.  This same trick yields correct results in other contexts as well. In fact, in some cases it turns out to be a more efficient way of taking derivatives. For example, consider $$g(x)=x^2 = \color{blue} x\cdot \color{red} x.$$ If we pretend the blue $\color{blue} x$ is a constant we would get $g'(x)\stackrel{?}{=}\color{blue}x\cdot 1=x$. Now if we pretend the red $\color{red}x$ is constant we get $g'(x)\stackrel{?}{=}1\cdot \color{red} x=x$. Summing both expressions we end up with $2x$ which is of course a correct expression for the derivative.  These observations have led me to the following conjecture:     Let $f(x,y)$ be a differentiable function mapping $\mathbb{R}^2$ to $\mathbb{R}.$ Let $f'_1 (x,y)=\frac{\partial}{\partial x} f(x,y)$ and $f'_2 (x,y)=\frac{\partial}{\partial y} f(x,y)$. Then for any $t$ we have: $$\frac{d}{dt}f(t,t)=f'_1 (t,t) + f'_2 (t,t).$$   (I apologise for the somewhat awkward notation which I could not seem to get around without causing undue ambiguity.)  This formulation also seems to lend itself to following generalisation:     Let $f:\mathbb{R}^N \to \mathbb{R}$ be a function differentiable in each of its variables $x_1,x_2,\ldots,x_N$. For $n=1,2,\ldots,N$ define $f'_n(x_1,x_2,\ldots,x_N)=\frac{\partial}{\partial x_n}f(x_1,x_2,\ldots,x_N)$. Let $t$ be any real number and define the $N$-tuple $T=(t,t,\ldots,t)$. Then one has: $$\frac{d}{dt} f(T)=\sum_n f'_n(T).$$   Thus my question is:   Is this true? How can it be proven? (Specifically in the case $N=2$ but also in the general case.) Why I cannot ping 192.168.0.108 (my computer IP?) Ipconfig says that my IP is 192.168.0.108 C:\Users\vbcbu&gt;Reply from 192.168.0.108: Destination host unreachable.ip 'Reply' is not recognized as an internal or external command, operable program or batch file.  C:\Users\vbcbu&gt;ipconfig  Windows IP Configuration   Ethernet adapter Ethernet:     Media State . . . . . . . . . . . : Media disconnected    Connection-specific DNS Suffix  . :  Wireless LAN adapter Local Area Connection* 12:     Media State . . . . . . . . . . . : Media disconnected    Connection-specific DNS Suffix  . :  Wireless LAN adapter Local Area Connection* 13:     Media State . . . . . . . . . . . : Media disconnected    Connection-specific DNS Suffix  . :  Wireless LAN adapter Wi-Fi 2:     Connection-specific DNS Suffix  . : www.tendawifi.com    Link-local IPv6 Address . . . . . : fe80::3804:2d04:a591:7142%6    IPv4 Address. . . . . . . . . . . : 192.168.0.108    Subnet Mask . . . . . . . . . . . : 255.255.255.0    Default Gateway . . . . . . . . . : 192.168.0.1  Ethernet adapter Bluetooth Network Connection:     Media State . . . . . . . . . . . : Media disconnected    Connection-specific DNS Suffix  . :  C:\Users\vbcbu&gt;  I can ping router at 192.168.0.1 I can't ping 192.168.0.108 C:\Users\vbcbu&gt;ping 192.168.0.1  Pinging 192.168.0.1 with 32 bytes of data: Reply from 192.168.0.1: bytes=32 time=2ms TTL=64 Reply from 192.168.0.1: bytes=32 time=24ms TTL=64  Ping statistics for 192.168.0.1:     Packets: Sent = 2, Received = 2, Lost = 0 (0% loss), Approximate round trip times in milli-seconds:     Minimum = 2ms, Maximum = 24ms, Average = 13ms Control-C ^C C:\Users\vbcbu&gt;ping 192.168.0.8  Pinging 192.168.0.8 with 32 bytes of data: Reply from 192.168.0.108: Destination host unreachable. Reply from 192.168.0.108: Destination host unreachable. Reply from 192.168.0.108: Destination host unreachable. Reply from 192.168.0.108: Destination host unreachable.  Ping statistics for 192.168.0.8:     Packets: Sent = 4, Received = 4, Lost = 0 (0% loss),  C:\Users\vbcbu&gt;ping 192.168.0.8  Pinging 192.168.0.8 with 32 bytes of data: Reply from 192.168.0.108: Destination host unreachable. Reply from 192.168.0.108: Destination host unreachable. Can "real" be used as an adverb to describe an adjective? I've read that abelian categories can naturally be enriched in $\mathbf{Ab}.$ How does this actually work? Wikipedia defines the notion of an abelian category as follows ().     A category is abelian iff         it has a zero object,   it has all binary products and binary coproducts, and   it has all kernels and cokernels.   all monomorphisms and epimorphisms are normal.      It later explains that an abelian category can naturally be enriched in $\mathbf{Ab},$ as a result of the first three axioms above.  How does this actually work? iPhone screen randomly turns on After installing iOS 10, I found that the screen on my iPhone 6s turned on at seemingly random times, and it slightly annoyed me. I now know that this feature does what it is supposed to (turn the screen on when the phone enters certain positions), but I want to disable it.  Is there a setting can I use to do so? Berry naming scheme list of LaTeX font families How do you end with a single quote followed by a double quote? Pay real money to set bounties (another method, not replacing rep method) It may seem like a stupid thing, but could we please be able to pay real money to set a bounty.  For example, if I had only 1 rep, I want to be able have a bounty.   Here late the suggested prices in USD. Convert money    $0.49 for each 50 reputation $0.99 for each week $1.99 for a month $9.99 for a year   Every month, the bounty fee would come in. This would mean you would have to enable your account with a bounty fee button on your profile. Evaluating the limit of a sequence given by recurrence relation $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Is my solution correct? Problem  The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation:   $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$.   Evaluate the limit $\lim_{n\to\infty} a_n$.  Solution   Show that the sequence $(a_n)_{n=1}^\infty$ is monotonic. The statement $$V(n): a_n &lt; a_{n+1}$$ holds for $n = 1$, that is $\sqrt2 &lt; \sqrt{2+\sqrt2}$. Let us assume the statement holds for $n$ and show that $V(n) \implies V(n+1)$. We have that $$a_n &lt; a_{n+1}.$$ Adding 2 to both sides and taking square roots, we have that $$\sqrt{2+a_n} &lt; \sqrt{2+a_{n+1}},$$ that is $a_{n+1} &lt; a_{n+2}$ by definition. Find bounds for $a_n$. The statement $$W(n): 0 &lt; a_n &lt; 2$$ holds for $n=1$, that is $0 &lt; \sqrt2 &lt; 2$. Let us assume the statement holds for $n$ and show that $W(n) \implies W(n+1)$. We have that $$0 &lt; a_n &lt; 2.$$ Adding two and taking square roots, we have that $$0 &lt; \sqrt2 &lt; \sqrt{2+a_n} &lt; \sqrt4 = 2.$$ The limit $\lim_{n\to\infty} a_n$ exists, because $(a_n)_{n=1}^\infty$ is a bounded monotonic sequence. Let $A = \lim_{n\to\infty} a_n$. Therefore the limit $\lim_{n \to\infty} a_{n+1}$ exists as well and $\lim_{n \to\infty} a_{n+1} = A$. (For $(n_k)_{k=1}^\infty = (2,3,4, \dots)$, we have that $(a_{n_k})_{k=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$, from which the statement follows.) We have that $a_{n+1} = f(a_n)$. That means that $A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}$. Solving the equation $A = \sqrt{2 + A}$, we get $A = -1 \lor A = 2$. Putting it all together, we get that $A = 2$, because the terms of the sequence are increasing and $a_1 &gt; 0$.   Is my solution correct? Why can't the Polynomial Ring be a Field? I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This implies an inverse multiple for every Element in the Set.   The book doesn't elaborate on this, however. I don't understand why a Polynomial Ring couldn't have an inverse multiplicative for every element (at least in the Whole numbers, and it's already given that it has a neutral element). Could somebody please explain why this can't be so? Comment authors are not getting migrated meaning of so much as In the following sentence, I want to know if I've understand the correct meaning or not:     This version holds that the wave function does not so much describe the state of the physical system as the state of our knowledge of the system.    Can we say that it means the wave function describes the state of our knowledge  of the system more than it describes the state of the physical system  ?  It is worth noting that in the Cambridge Dictionary, we read:  not so much sth as sth ​ If you say that something is not so much one thing as something else, you mean it is more the second thing: They're not so much lovers as friends. I don't feel angry so much as sad. Clearly if $\lambda$ is an eigenvalue of $A$ then $p(\lambda)$ is an eigenvalue of $p(A)$ where $p$ is a polynomial. And there are cases where $A$ may have eigenvalues other than these.   Is there a general rule for finding all eigenvalues of $p(A)$? Does it depend on the field from which the elements of $A$ are taken? Are their any partial solutions to this problem? Everybody who is using Lightroom probably has seen this. After importing RAW files, Lightroom shows a preview, but contrast and color of the preview change after a few short seconds to a slightly different view.  For some users the change is small, for others the change is huge. Colors are different (read: wrong), contrast is off.  Rumor says this is happening only (more?) for Nikon users.  What is Lightroom doing here?  I'd guess that for the first preview, LR is showing the JPG preview file, which is embedded in the RAW file. Then after a moment, LR renders the RAW file itself to show it's own interpretation.  But why is this interpretation so different from the embedded JPG? Possible Duplicate:          I have been downvoted for few answers of mine and the person who downvoted me did not leave any comments or reasons to say why I was downvoted. In such cases, how can I find out who has downvoted me and the reason for downvoting? I understand the purpose of up/downvoting answers -- some answers are better than others, some answers may be wrong, .etc.  With regard to questions, though, there are other mechanisms to deal with "bad" questions:  comments, close, flag.  A poorly written question should be improved.  A bad question should be closed/deleted.  Often, though I see poorly written questions downvoted. I'm not sure what this accomplishes.  I get loss of rep for offensive or spam questions, but for an off-topic question?  or one that is just poorly written?   Why?  It seems to me that rep ought to be based more on answers on a site of this type.   Fundamentally, it seems that rep ought to be based on what you add to the site, not what you ask of it.  A particularly good question does add to the site, but a poor question doesn't really detract from it in the same way that a bad (wrong) answer does. Sure, go ahead and give people more rep for asking a particularly good question, but why punish people for asking bad questions when there are other ways of handling that.  Why not handle questions more like comments and allow upvotes, but not downvotes?  It seems to me that many, if not most, of the downvoted questions are first timers.  It's more likely that your first introduction to the site will be in asking a question, rather than answering. There is a significant risk to alienating new users when their questions are downvoted -- there's a visceral response to seeing that negative number next to your question.  Why not let the close/flag system weed out questions that don't belong and the comment system work to improve poorly worded questions. I can't seem to get these particular faces to take the UV Mapping. I selected the entire object, entered edit mode, hit U>Unwrap.   The rest of the mesh takes the uv wrapping, but not those faces for some reason.    It looks normal in solid mode.     Does anyone know what could be causing this?