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1

If two events both have a spacetime interval of zero, can they both be said to be happening “now”? There is an interval associated with any two events but there is not an interval associated with an event. From the Wikipedia article "Spacetime": In spacetime, the separation between two events is measured by the invariant interval between the ...


0

I'm not sure this is a complete answer to your question, but thinking about special relativity that way will get you into trouble. Essentially, that way of interpreting special relativity attributes all of its weirdness to signal delay. Here's how I think you're interpreting the barn door experiment: The ladder is put stationary in the barn and is found to ...


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 ...


8

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

Instead of using existing spacecraft, let's use a photon rocket powered by the gamma rays emitted by matter anti-matter annihilations in its reacor. Where does the anti-mater come from? We will produce it using solar energy. We'll use giant solar panels that generate a huge voltage in vacuum which leads to Swinger pair production. Moving away from the Earth ...


0

From ScienceMuseum: Apollo 10 holds the record as the fastest manned vehicle, reaching speeds of almost 40,000 km per hour (11.08 km/s or 24,791 mph to be exact) during its return to Earth on 26th May 1969. Using the formula (as above). After traveling for 40 years, you would be a little over 0.86 seconds younger. Added: I did some calculating and ...


2

Almost none. Let's be much more generous than your idea of human-carrying craft. Let's just use the fastest probe. The Helios II craft, after nearing the sun, reached a heliocentric speed somewhere near 70 km/s. Obviously, its speed was more due to the gravitational influence of the sun than its engines. $$t = \frac{t_o}{\sqrt{1 - \frac{v^2}{c^2}}} $$ ...


2

So let's just say that the spacecraft can accelerate until it's moving away from the Earth at the speed of the fastest currently-existing spacecraft First, note that the fastest speed, relative to Earth, that a spacecraft has obtained is an exceedingly small fraction of the $c$ and, thus, one should not expect significant time dilation. For ...


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

Just to add to John Rennie's answer, the objects where we expect to see the largest frame dragging effects are spinning black holes. There, there is actually a surface called the ergosphere (outside of the event horizon), where it is impossible for observers to stay stationary with respect to observers far from the black hole. In a sense, their reference ...


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 ...


3

The spacetime outside a spinning mass is described by the Kerr metric. To explain how the Kerr metric produces frame dragging is hard, because it's not something for which there's an easy intuitive model. Frame dragging arises because the spacetime geometry links the angle measured around the spinning object to time, and this means the angle changes with ...


1

In special relativity, in the rest-frame of the proton, the moving magnet m appears as a magnet m’ and an electric dipole p’. The electrostatic E field created by the proton makes rotate this electric dipole, actually the magnet. And the E’ created by the electric dipole is the responsible for the force the proton experiences. Remark that you will not found ...


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 ...


1

This is my first problem, as the modulus of a vector shouldn't be negative. First, while there are many useful properties of introductory linear algebra you should keep in mind with GR, thinking in Cartesian terms with positive definite matrices simply has to go. Vectors in relativity can very much have negative norm. Even though it's not often done in ...


5

Let's start at the beginning: The setting for relativity - be it special or general - is that spacetime is a manifold $\mathcal{M}$, i.e. something that is locally homeomorphic to Cartesian space $\mathbb{R}^n$ ($n = 4$ in the case of relativity), but not globally. Such manifolds possess a tangent space $T_p\mathcal{M}$ at every point, which is where the ...


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 ...


2

First of all, light waves and matter waves may be treated together, using the same maths, because the waves associated with light and the waves associated with matter are fundamentally the same thing. Second, all the waves before they interfere and after they interfere may be written in terms of the probability current $j^\mu (x,y,z,t)$, and its ...


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 ...


6

Their race is ill-defined. You can't declare a winner if you can't agree on the ordering of events. If they failed to pick a reference frame for the race before starting it, then of course an argument over the winner may ensue. No laws of physics have been violated. If you try and extend this to "malfunctioning machines", then yes, a machine that was not ...


3

That way of thinking about this is a nice one. From “The Elegant Universe”? Anyway, mathematically, you can define this four-velocity like the normal velocity, the time derivative of the position. However, in special relativity, the position is space and time $(t, x, y, z)$. And the derivative has to be with respect to the proper time (or curve parameter) ...


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 ...


4

If you look at the Earth's geoid, you can see that there isn't a particular "band" of gravitational variation along the equator. So while time would move slower / faster at varying locations around the globe, it is not correlated with the equator. Here is, I believe, the latest model, in 2D: And there is also a very nice 3D animation here. Regarding the ...


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. ...


12

The current answers by Luboš and David do a good job of explaining why it is essential to include general relativity in the picture. In fact, this is even more of an issue because the irregularities in the shape of the Earth do matter. It's fairly easy to understand why this is the case: it's been known since 2010 that atomic clocks are sensitive to height ...


0

Things are much simpler here if one thinks in terms of spacetime events and their coordinates in the relatively moving reference frames. As best as I can tell, there are three events of interest: Event A: two photons are emitted from station A Event B: one of the photons is received at station B Event C: the other photon is received at station C ...


9

Is there a flaw in my reasoning or have I simply not been reading the right journals? Yes. The flaw is that you are ignoring general relativity. The poles are closer to the center of the Earth and are thus deeper in the Earth's gravity well than is the equator. The combined effects of gravitational and special relativistic time mean that clocks at sea ...


25

The difference would indeed be measurable with state-of-the-art atomic clocks but it's not there: it cancels. The reasons actually boil down to the very first thought experiments that Einstein went through when he realized the importance of the equivalence principle for general relativity – it was in Prague around 1911-1912. See e.g. the end of ...


2

The time dilation due to motion in a circle, relative to an observer at the centre, is just the usual Lorentz time dilation due to the velocity of the motion. If you're interested, in my answer to Is gravitational time dilation fundamentally different than other forms of time dilation? I showed how this is derived from the metric. Anyhow, as you say, the ...


-1

Your calculation refers to time intervals between two events inside the ship, as seen from a different reference frame (Earth?), and in Special Relativity this is equivalent to what would be seen from the ship if in the other reference frame the same experiment was being performed. However this is not the correct way to solving the paradox, this paradox ...


1

I do not find easy to understand your calculations, but can give you an explanation which is not based on specific distances. It is easy to see why the observer inside the ship will perceive the events as simultaneous and the one outside the ship will not. First, notice that for every observer the speed of light is the same, c. So the observer on the ships ...


1

If you do not know anything more about $F$, the only way to do this is to simply plug the transform result $\Lambda^\mu_\nu x^\nu$ into $F$ and see what happens. General functions into $\mathbb{R}$ cannot be expected to behave nicely in any way. However, I cannot really fathom a situation where you would have to use a function in $\mathbb{R}$ where you ...


0

So how much has the exposure to light darkened the photographic plate? As much as expected for 301 ns or 33 ns ? For 33 ns. That is the time that the clock inside the ship-muon indicates. We measure 301 ns of exposure, but a much shorter time was experienced by the moving photographic plate. This question is tricky also because the light will be ...


1

Trying to defend the application of SR in my question, I discovered my mistake. From the observation framework of the photographic plate, it's a simple race between a muon speed space ship and light. Not so to the other observer. from the muon speed space ship's observation, the space ship is stationary and points m and e are approaching at 0.994c. When ...


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 ...


1

Because of length contraction predicted by special relativity, the distance between points m and e is 1637 m. The light from point m takes 5.49540 μs to reach point e. Point e reaches the ("stationary") space ship after 5.46254 μs. So the photographic plate is exposed to light for 32.9 ns 1637m is after length contraction in the reference frame ...


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 ...


0

The Transactional Interpretation of QM suggests that Maxwell's equations work backwards in time, carrying the information of one test back to the point of entanglement, where it can affect the entangled particle. This explanation bypasses any issue of violation of SR.


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 ...


0

There is what appears to be a very good lecture that deals with many of the issues. It is a Google Techtalk from Ron Garret entitled "The Quantum Conspiracy:What the Popularizers of QM don't want you to know". A rather tongue-in-cheek title that should not mislead you into thinking it is not a serious and worthwhile talk. www.youtube.com/watch?v=HQIJgheuYNU ...


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 ...


0

The ladder won't be trapped in both reference frames. If an observer with the garage closes the doors at the same time, so as to "trap" the ladder inside the garage, the events corresponding to the two doors closing will be simultaneous in the garage frame, but not in the ladder frame. An observer with the ladder will see the front door closing first and the ...


-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 ...



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