# Tag Info

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 ...

9

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 ...

7

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) ...

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 ...

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 ...

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 ...

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.

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 ...

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 ...

3

If you measure the speed of light using the Schwarzschild coordinates then you will calculate a speed that is less than $c$ when the light passes near a massive body. In this sense gravity does slow light. There are already several questions hereabouts that discuss this. The issue tends to be a bit controversial because there are differing interpretations ...

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. ...

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 ...

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 ...

2

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 ...

2

The speed of light in vacuum is invariant in any inertial reference frame. In your reference frame (if you could of course ride along the car) the light beam leaves the car at speed $c$. From the point of view of your friend who is watching you sitting on the pavement he will see the light leave at a speed which is given by the relativistic addition formula: ...

2

Your reasoning seems to be that because $\frac{t}{\sqrt{1-v^2}}\to \infty$ when $v\to 1$, it must be the case that $\frac{t}{\sqrt{1-v^2}}\to 0$ as $v\to 0$. But in fact when $v=0$ we have $$\frac{t}{\sqrt{1-v^2}} = \frac{t}{\sqrt{1-0^2}} = \frac{t}{\sqrt{1}} = \frac{t}{{1}} = t.$$ So time does not pass infinitely rapidly but instead passes at exactly its ...

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 ...

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 ...

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 ...

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

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

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

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 ...

1

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 ...

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 ...

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 ...

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