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1

In the underlying physics, c = 1 (Planck units). 1 is rational. But your unit system might not have a rational length. Speed of sound is rational in nature if macroscopic quantum mechanics holds (this is still open to debate that I will not enter). We should be able prove given macroscopic quantum that speed of sound is an integer multiple of Planck length ...


1

Well it's a tricky question in some way. You can for example consider the second as a rational number because its definition (a number of times the time needed for some atom to change state) is rational in nature (you can see it like this at least): you're technically just counting a number of occurrences of an event. Then if you consider the speed of ...


5

It depends on the unit you want to express it. If you choose c/100 as the speed unit, c will be expressed with a rational number. If you choose c/π, you'll have an irrational one. That depends on measure, not on nature.


36

Something I posted on reddit answers this question quite well, I think: "Rational" and "irrational" are properties of numbers. Quantities with units aren't numbers, so they're neither rational nor irrational. A quantity with units is the product of a number and something else (the unit) that isn't a number. By choosing the unit you use to express a ...


0

OK, just one more try to end this stupid question. There IS a way to formulate physics using light rays as your basis: a double null coordinate system${}^{1}$. If you have a ray moving in the $+x$ direction, define the two coordinates $$2\xi = t + x\;\;\quad\quad\quad2\eta = t - x$$ Then, the metric becomes $$ds^{2} = -4d\xi \,d\eta + dy^{2} + dz^{2}$$ ...


2

According to Einstein's theory of mass-energy equivalence, if the photon is a particle of pure energy, and if $E=mc^2$, then the photon is theoretically traveling at $c^2$; not $c$ You have neglected dimensional analysis: E $\to$ Joule = $\frac{\rm kg\,m^2}{\rm s^2}$ m $\to$ kilogram = $\rm kg$ which means that $c^2$ has units of m$^2$/s$^2$, not ...


9

Actually you're quite correct, though possibly not in the way you expected. Ordinary velocity isn't an invariant because obviously different observers moving at different speeds will measure different velocities. However there is an invariant form of velocity called the four velocity that is an invariant under special relativistic (i.e. Lorentz) ...


0

Well, obviously it would be the speed of light times two. However, this can be misleading. Equations are all fine and dandy, but if you do not understand them, they are not of complete help. Imagine that you have the following... 1) A 300,000 km long spaceship which is at rest in space. 2) A clock is located at each opposite end of the spaceship, and these ...


0

Your notion seems to be based on the thinking that light is a bunch of photons, and a photon is some kind of weird particle that travels at the speed of light, like some tiny spaceship. Then you ask, how can this tiny spaceship violate physical laws? What makes it so special? But a photon isn't a particle in any classical sense. It's not like a tiny ...


0

Special relativity states: ... I'll select and discuss the given statements in some particular order (which may be called "in order of simplicity of discussion") ... [...] The observer is (anything [...]) Right. Synonymous to "observer" or "anything", in the context of the theory of relativity, there are also the descriptions "material point" or ...


3

I will expand my comment above into an answer, but I will not comment further on it to avoid the usual very long discussions of your posts. In my opinion, you are trying to argue on a logical level, but it is not clear if you have enough knowledge of logical theories to do so on a mathematical/physical level. Without entering too much into details, a ...


10

To make progress we need to be clear what we mean by the laws of physics and observer. A law of physics is just some set of equations that we use to predict what happens. So if for example we're trying to describe how charges interact with light our set of equations, i.e. our law of physics, would be Maxwell's equations. But to write down Maxwell's ...


0

There is always a change of velocity after reflection. So in a sense 'yes'.


1

You vastly overestimate the meaning of frame. A frame is a (local) choice of coordinates on the spacetime manifold $\mathcal{M}$. All physical laws can be directly formulated on the manifold itself, without referring to frames at all. That is at the heart of relativity, and that is what Lorentz invariance means. Let's go through your numbered points one by ...


0

In response to the extended discussion in the comments I wrote a program to clarify things interactively. The program is here: https://www.khanacademy.org/cs/relativistic2/6050744190369792 Frame S: The three perfectly vertical lines are the points A, O, and B being stationary. The other three lines are A', O', and B'. In frame S, the flashes of light at ...


1

The spring would extend at a below light speed (but very close to it) this is because of it having some mass as you referred the spring to being "nearly mass-less" but it would extend on each side, therefore no physical law is broken however relative to an observer it would seem as if the object is expanding at faster than $c$ (Speed of Light) speed however ...


2

No, it doesn't mean that. One must distinguish two things: "laws of physics that apply to an object" and "laws of physics formulated from an object's viewpoint". These are two different things. Laws of physics apply to all objects. And the behavior of the objects may be described relatively to many coordinate systems or "frames of reference". The special ...


0

Your calculations are correct. The light does indeed take longer to reach the other end of the box when they are moving in the same direction, and vice-versa. How could this help, however, to convince the guy in the moving box that he was the one moving, and not me? When he compares the time that the light takes reaching the end of the box, and the time ...


3

I've said it before and I will say it again: There are no frames travelling at the speed of light As David Z says in the very link you give, it is meaningless to ask what you would perceive travelling at the speed of light. You cannot. And even though there are particles that can, there are no frames associated with them. Have a look at the Lorentz boost. ...


2

In this case which observer will find the events to be simultaneous and why? Briefly, the events are simultaneous in $S$ and the reason is that you've stipulated the events are simultaneous in $S$. In more detail... when two events occur at A and B in S and the corresponding points In S' are A' and B'. The wording here is puzzling. Events ...


3

We would know the difference. Here are two possible ways we could tell: One is that Maxwell's equations tell us exactly how fast light has to go, always. It doesn't matter how you're moving, theory predicts that every beam of light will always be moving at the same speed relative to you. There is only one speed for which that property can hold true, and ...


0

1. would it really be devastating for relativity that we find out in the future that photons have a tiny mass and move slower than ... uhh... the speed of light? As the phrase "... uhh ... " in your question anticipates: there is some devastation lurking in that question; namely a contradiction to the essential understanding of "light" as "any signal ...


2

A quick alternate perspective: Not really. Relativity only requires the existence of an invariant speed $c$, it doesn't require that anything actually travels at that speed. So if photons were massive, there would be no problem, although some results in cosmology might have to be modified a bit. Pretty much every time you see it, $c$ means the invariant ...


2

it appears that electromagnetism has some of preponderant role in the universe compared to other theories This is not true. It played an important historic role, but is in no way theoretically "unique" because the photon travels at speed c. Indeed, the gluon also travels at speed c. If photons were found to be slightly massive, it would change a lot of ...


1

what are we measuring when we measure light? When considering "(average) speed of light (in vacuum)", or rather primarily: "(average) speed of a signal front" this means the quotient of the distance of a signal source and a receiver between each other (i.e. under the condition that these two participants had been at rest with respect to each other), ...


-1

As far as I understand your experiment, A and B, as well as A' and B' should actually be considered sources of light. Now, if A and B are co-moving with O (and therefore, all three are stationary wrt. each other), the rays from A and B will reach the observer O at the same time. However, O and A' are moving toward each other, and therefore the observer O ...


0

In the first figure ... actually: in the bottom parts of both the first and the second figures ... A and B are two equidistant points from the observer O in S. In detail, therefore A, O, and B were and remained (pairwise) at rest to each other, with O having been and remained the "middle between" A and B, for each signal indication stated by ...


-1

According to Special Relativity: Given, at time t=0 AO=OB=A'O'=O'B'. To the observer O, since AO=OB, the rays from A and B will reach O at the same time and hence, O will find the events to be simultaneous. Furthermore, since, at t=0 (the instant when the events occurred) A coincides with A' and B with B' and the speed of light is absolute, the ...


1

If I (on the Earth) and you (on a very fast rocket) measure the time that takes you to travel a certain distance we would not agree. Still, nothing forbids me to compute your velocity using my measures of time and distance. The interesting point is that if your rocket is made of photons, even a third person trying to chasing you will agree with my measured ...


5

It's simpler than you think. When we measure the velocity of light we are simply measuring how much distance light travels per second. For example if we send a very short laser pulse to a reflector on the Moon and wait for it to return we find it covers the 768,934 kilometer round trip in about 2.56 seconds. This is measured in the Lunar Laser Ranging ...


5

Yet there are all kinds of reactions that cause light, seemingly without providing the infinite amount of energy needed to accelerate particles to light speed. If you're imagining that there are photons at rest within the flashlight when it's off and the flashlight accelerates photons to light speed when it's turned on, then I can see why you're ...


2

Photons have no mass, they always travel at the speed of light. Not so for massive particles. You could try building a collider with massless particles, but you would fail for several reasons: The only freely propagating massless particle is the photon, but it does not have a charge, so you cannot bend it to follow a circular collider. You could still build ...


4

It takes an infinite amount of energy to accelerate a particle with mass to the speed of light. A photon does not have mass, thus can move at the speed of light. Note that a photon does not accelerate; the moment it is created it moves at the speed of light.


1

If you suppose that simultaneity arises in the $S$ frame, this means that the emission of the rays correspond to the following space-time events , expressed in coordinates $(x,t)$ in the $S$ frame : $$A(-l,0)\quad B(l,0)$$ Simultaneity means that the time coordinates $t$ of $A$ and $B$ are equal. Now, these same space-time events, expressed in the $x'$, ...


1

Let us go the extreme case where o=o at t=0. If you follow the equations, there is only one observer who can perceive the events as simultaneous, the other one will perceive it as being not simultaneous. Just let's go to the classical example. A guy in the middle of a train turns on two lanterns in opposite directions. He will perceive the time when the ...


7

Since you only mention acceleration to 0.5c, we'll assume we're dealing with special relativity alone. In this case, your accelerating computer 'loses time' -- its clock moves slower. Computers ultimately work on clock cycles. Thus it is fair to say that, as its clocking is ticking slower -- from your point of view -- the computer on your desk will finish ...


3

You're thinking about gravitational time dilation. Time machines do exists. If you go in a space ship and travel around the supermassive blackhole in the center of Milky Way, close enough to not fall in it, and then come back to Earth, you just traveled to the future (relative to the space further from you). So in that thinking line, if you want to make a ...


14

1) No, because it's actually going slower from your perspective. In special relativity, "the fastest wristwatch is always your own". 2) Yes, but remember that it's farther away from us now, so it will take some time to get to us (if it was travelling at 0.5c it will take 50% longer to get to us). 3) Mostly in that as an observer the redshift effect would ...


2

The shapes of constellations (and there are several different depictions of any particular constellation) only depend on how they look at any specific time. Stars in any constellation are not necessarily close to each other in space. For example, the main stars in Ursa Major vary in distance from 58 to 124 light years. The boundaries of constellations are ...


2

Does that mean the speed of an individual photon is c even with respect to another photon? I mean, shouldn't the relative velocity be zero? When we write "the speed with respect to X" we mean precisely "the speed as observed from the inertial frame of reference in which X is at rest" Thus, if it is true that the speed of an individual photon ...


0

The idea behind special relativity is that the laws of physics do not change from one reference frame to another (this is key). So, if you're moving at velocity v, a photon traveling in that reference frame will travel at c (not c - v, which is the classical Galilean transformation). I believe this is what your book means when it says "with respect to ...


0

The whole point of special relativity is that imposing this postulate (on the constancy of the speed of light) forces a reunderstanding on the very geometry of space and time. This re-understanding can be summarized by the Lorentz transformations. These describe the coordinate transformations required when transferring measurement from one observer in ...


3

it doesn't move in time, so no time will have past when the light arrives at it's "destination". Right? A photon does 'move in time'. It just that, for a photon, the displacement in time, $c \Delta t$, equals the displacement in space, $\Delta x$. However, there is no proper time for a photon. Your proper time is, in words, the elapsed time ...


-1

When anything moves at the speed of light, all of our physical models break down. If you were to watch a spaceship speed up to the speed of light, you would see a clock on the ship slow down and come to a complete stop when it hit the speed of light (assuming you could even see it at this point). The ship would also contract so much in the direction of ...


4

To newcomers to relativity it seems to be based on the invariance of the speed of light. While this has some historic significance, these days we regard Lorentz invariance as the fundamental principle, and a constant speed of any massless particle is then just a consequence of Lorentz invariance. So your question could, and should, be written as the ...


0

The second postulate of special relativity is: "As measured in any inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body."(Wikipedia) Your question is not regarding the second postulate because no inertial observer can observe the constellation of ...


1

No, and the reason is quite simple. The proper time of a photon is zero (according to the principles of special relativity). That means that there is no time difference between the place of emission and absorption. By this, any hypothetical observation of a photon would be reduced to a time period of zero, and it would not be able to distinguish/ to ...


1

It is not hard to imagine a toy universe in which different fundamental forces propagate at different speeds. However, a necessary consequence of that would be violations of lorentzian symmetry, and the ability to triangulate a preferred rest frame. Although I don't see a theoretical reason why these speeds need be the same (I might be missing something ...


2

It's better to think of the deflection of the photon as an effect of its travel through curved spacetime. You can generally choose to analyze the problem in the rest frame of the massive object. In that case spacetime is curved in all directions around the object, and so the photon's path is deflected both as it approaches and as it recedes. If you want to ...


1

Any object that has mass just has a gravitational field around it and light (or particles, photons) just get refracted when it enters this field.



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