Why is 1 metre defined as the distance travelled by light in 1/299792458 seconds rather than 1/300000000 seconds?
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Why is the speed of light defined as $299792458$ $m/s$? Why did they choose that number and no other number?  Or phrased differently:  Why is a metre $1/299792458$ of the distance light travels in a sec in a vacuum?
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I am currently trying to learn some basic quantum mechanics and I am a bit confused. Wikipedia defines a  as a quantum of light, which it further explains as some kind of a .     What exactly is a quantum of light?   More precisely, is a quantum of light meant to be just a certain number of wavelengths of light (something like "1 quantum = a single period of a sine wave" perhaps?), or is the concept completely unrelated to wavelengths? In other words, how much is a single quantum?
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I'm reading Nano: The Essentials by T. Pradeep and I came upon this statement in the section explaining the basics of scanning electron microscopy.     However, the equation breaks down when the electron velocity approaches the speed of light as mass increases. At such velocities, one needs to do relativistic correction to the mass so that it becomes[...]   We all know about the famous theory of relativity, but I couldn't quite grasp the "why" of its concepts yet. This might shed new light on what I already know about time slowing down for me if I move faster.  Why does the (relativistic) mass of an object increase when its speed approaches that of light?
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Let's suppose I have a spacetime manifold $M$. Let $p$ be a point on my manifold. Now I move from $p$ to some other point $p'$. Presumably I should have moved some "distance" right? How can I speak of notions of space and time if I have no conception of distance?   But now consider light moving through spacetime. Suppose my light starts at $p = (0,0,0,0)$ and travels to $p' = (1,1,0,0)$. By the definition of the spacetime interval $ds^2 = dt^2 - dx^2 - dy^2 - dz^2$, this should mean $ds^2 = (1)^2 - (1)^2 - 0 - 0 = 0$. So $ds^2=0$.   Yet I have moved from point $p$ to $p'$. So I clearly have moved along some path along the curve, but the length of this path is zero. Shouldn't that mean $p$ and $p'$ are the same point?   Note: I think I may be suffering from an overly Euclidean mindset and my brain hasn't adapted yet enough to the non-Euclidean logic of semi-Riemannian manifolds.   My Question   Can someone resolve this contradiction?
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Considering the definition of the SI unit of "length" [1] and [2 (" method a.")] I'm missing any requirements about the two "ends" of the required "path travelled by light" being "at rest to each other", or at least "rigid to each other".  Are such requirements perhaps presumed to be understood implicitly?  Accordingly, if some particular pair of "ends", $A$ and $B$, in some particular trial, were characterized as having been a certain "lenght of $x \, \text{m}$ apart from each other", where "$x$" is some particular positive real number and "$\text{m}$" denotes the SI base unit "metre", is it then understood:   that ends $A$ and $B$ had observed exchanging signal pings between each other; and not only once for each trial, but for any of their signal indications throughout a sufficiently extended trial?, that for any two (distinct) signal indications $A_J$ and $A_K$ of end $A$ during this trial the corresponding ping durations of end $A$ were equal to each other: $\tau_A[ \,_J, \,_{\circledR}^{B \circledR AJ} ] = \tau_A[ \,_K, \,_{\circledR}^{B \circledR AK} ]$ ?, that for any two (distinct) signal indications $B_P$ and $B_Q$ of end $B$ during this trial the corresponding ping durations of end $B$ were equal to each other: $\tau_B[ \,_P, \,_{\circledR}^{A \circledR BP} ] = \tau_B[ \,_Q, \,_{\circledR}^{A \circledR BQ} ]$ ?, and that for any signal indications $A_J$ of end $A$ and any signal indication $B_P$ of end $B$ during this trial the corresponding ping durations were equal to each other: $\tau_A[ \,_J, \,_{\circledR}^{B \circledR AJ} ] = \tau_B[ \,_P, \,_{\circledR}^{A \circledR BP} ] = \frac{2 x}{c} \text{m} $ ?.   References: [1] SI brochure (8th edition, 2006), Section 2.1.1.1;  ("The metre is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second.").  [2] "the mise en pratique of the definition of the metre";
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Meaning, why is it the exact number that it is? Why not 2x10^8 m/s instead of 3? Does it have something to do with the mass, size or behavior of a photon? To be clear, I'm not asking "how we determined the speed of light". I know there isn't a clear answer, I'm really looking for the prevailing theories.
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Galileo discovered that the distance fallen is proportional to the square of the time it has been falling.Why is it proportional to the square of the time and not just time? i.e $d \propto t^2$ why not just $d \propto t$  I know this question involves some common sense which i'am not able to use for some reason!
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This is not a contradiction and I know it is impossible but still consider a thought experiment by me and point out if something is wrong. See the following picture and then the explanation follows.    Rest frames are easy to understand. They are just for clarification. Lets move on to the moving frames. The velocity of the box is $0.1\ c$. Now a photon is emitted (I am not taking a light ray to avoid complication in discussion). After emission the box also moves a certain distance ahead. So the photon takes more than one second to reach the wall. But even the source moves that much distance ahead. When the light is reflected back, the source and the wall again move forward as the box is moving. But after reflection the wall has no role to play. We are concerned about the source. The source moves a certain distance ahead and therefore light takes less than one second to reach back. But the difference of the first case and the second case is not the same. To explain, I will give some equations.  Moving frame 1:  Velocity of the box: $0.1\ c = 30,000km/s$  Time taken by light to reach the wall $=\frac{330,000\ \text{km}}{300,000\ \text{kms}^{-1}}= 1.1s$  Distance moved ahead by the box from original position: $30,000\ \text{km}$  Moving frame 2:  distance moved ahead by the box from original position  $&gt;30,000\ \text{km}$ as the box also moves ahead after reflection.  let $d$ be the distance moved ahead by the box after reflection.  Time taken by light to reach the source: $\frac{270,000\ \text{km}\ -\ d}{300,000\ \text{kms}^{-1}} &lt;0.9s$  Therefore we see that this could possibly determine the absolute motion.  So in total we see that the light will take lesser time to reach the source. Please correct me if I am wrong and give your opinion why is it so.  EDIT 2 for clarification  The box is the compartment in space in which you are moving and I am calculating the time the light takes to reach back the source. I am not even adding velocities. Please correct me and tell me where did I add velocities. For clarification just let me give a simple example. Suppose person A is standing still and B is running towards a light source. Obviously he will see the light before person A even though the speed would be the same. So, what I am doing over here is not increasing the speed of light, but instead decreasing the distance to be covered by it.
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Why does light have the speed it does? why is it not considerably faster or slower than it is?  I can't imagine science, being what it is, not pursuing a rational scientific explanation for the speed of light.  Just saying "it is what it is" or being satisfied saying it is 1 ($c=1$), does not sound like science.
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I'm preparing for my exam, but I have difficulties in perceiving why there is a $\frac{1}{2}$ in the distance formula $d=\frac{1}{2}at^2$?
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The metre is the length of the path travelled by light in vacuum during a time interval of 1  ⁄ 299792458 of a second. – 17th CGPM (1983, Resolution 1, CR, 97),    The meter (or metre) is considered one of the seven base units in SI, but since it depends on what is “a second”, shouldn’t it be a derived unit? Why is the meter considered a base SI unit, even though it depends on the definition of a second?  I am aware that other base units such as the ampere also depend on other base units (namely meter, kilogram and second), and this just increases my confusion even more. Perhaps my question boils down to: What property must a unit have so that it is called a base SI unit?
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Regarding the paper, what does Einstein means when he says:     "If we call the velocity of light at the origin of co-ordinates $c_0$, then the velocity of light $c$ at a location with the gravitation potential $\Phi$ will be given by the relation: $c = c_0\cdot\left(1+\frac{\Phi}{c^2}\right).$ The principle of the constancy of the velocity of light holds good according to this theory in a different form from that which usually underlies the ordinary theory of"   relativity"   Does the velocity of light is constant only into a space where the gravitation potential is constant either?
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Since electromagnetic energy is carried by photons and moves in forms of waves, does it mean that a single photon when propagating through space doesn't follow the straight path but instead always moves up and down, up and down like a wave. If so another question arises the speed of propagation of light in vacuum is fixed meaning that it will always take the same amount of time for it to travel from point A to point B, but if a photon always moves up and down it will also mean that it travels longer distance than the distance between A and B and so it ill travel faster than light propagates, is it even possible, could you please clarify these concepts for me?
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I've been thinking about this for quite some time, and from looking online I haven't found a satisfying answer.  Lots of photons, such as visible-light photons have very small wavelength (which from my understanding of basic physics is the distance between two crests/troughs), but I also know that some EM waves have wavelengths a few metres or even kilometres long e.g radio waves.   What keeps me up at night is the question "How can a photon have a wavelength of a few kilometres and yet still be thought of as a particle?"  Does this mean that one individual photon is several kilometres long? If so, wouldn't it be subject to so many variations between the beginning of the wave and its end?  I realise that matter is also wave-like, where it's uncertainty in position is given by its De Broglie wavelength. Does this apply to the photon?  In other words, is the wavelength of a photon simply the uncertainty in its position?
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$(\mathbf60\times\mathbf60)$ Seconds = $3600$ Seconds or $60$ Minutes  while  $(1\times 1)$ Minute = 1 Minute  What am I missing? and sorry about formatting i am new here on math network
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During a mock Cambridge interview, one of the questions was about how small a computer could theoretically be.The way I approached it was in terms of what something must be to be considered a computer.  I eventually made some progress, but I made a point that I am quite doubtful about. I said that one can use the photoelectric effect as a timer.  When you shine light on an atom and promote an electron to a higher energy level, there is a time delay before it returns to ground state (or at least there is a time delay between light being absorbed and light being re-radiated), which ought to be constant. I know this point may not really relate to the interview question.  Nevertheless, is what I said true? Is there a (measurable) time delay between the absorption of a photon and the re-radiation of a photon?
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When light is emitted (or received) one photon at a time, what is the duration of each pulse? Is it related to the wavelength? Is it possible to create a laser pulse (or a radio wave) shorter than the wavelength (period of oscillations)?