Hot answers tagged reference-frame
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In special relativity there is no way you can see someone elses time going faster. This is because in SR all motion is relative. There is no notion of an absolute state of rest. In dmckee's example of the muon experiment, we see time moving more slowly for the muons. However the muons (if they were sentient) would see time moving more slowly for us. This ...
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Moving away from earth at some speed can mean moving faster in space
OR moving slower in space and thus also mean slower or faster time
flow than the one on the earth.
There is no speed relative to space, there is just relative motion between objects.
But since the earth already travels at some always varying speed and
thus is not standing ...
3
In special relativity, everything is relative. This includes time dilation. As mentioned, it does not allow for a "privileged" reference frame because special relativity requires that all physics appears the same from every inertial frame.
You could travel at what you believe to be the exact opposite direction as Earth and at the same speed, but that would ...
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In principle there is no limit to the ratio of observed time between two frame of reference.
The highest artificially generated time dilation factor is around 200,000 but because it involved electrons there isn't a simple clock to point to.
The muon g-2 experiment1 used muons with a time dilation factor a bit over 29 and was able to measure the dilation ...
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I think this is an answer to Van's comment rather than the original question, but this is how you unambiguously measure acceleration in SR. Simply arrange a sphere of test particles around yourself then watch them. As long as the particles remain in a sphere centred on you know you aren't accelerating. If you are accelerating the sphere of particles will ...
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You can define a product in any way you want, so your definition is perfectly fine. The question is whether that product is useful. And one of the things we usually require is that a product be invariant under rotations, for example. What this means is that if you rotate each of the vectors that goes into the product, you should get the same result, but ...
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I don't think you can do much better than getting your head around the identity
$$\frac{d}{dt} \rightarrow \frac{d}{dt}+\vec \omega \times,$$
which holds when the former is applied to vectors. The essential point of the identity is that even if a vector is stationary in one reference frame, it will have some rotational motion in the rotating frame.
It may ...
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When you are moving from A to B, the distance between A and B shrinks relative to you. This is known as length contraction.
The equation is as follows:
$$L' = L \sqrt{1 - v^2/c^2}$$
where L' is the length you will see at move. L is the length at the resting reference frame. v is your speed, and c the speed of light. In your case:
$$L' = L \sqrt{1 - ...
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The time stands still for light indeed, so it will never age.
You can think of the photon as a sine wave shaped electric field fragment traveling at $c$, and you can measure it's amplitude and frequency as it flys past your instrument. The photon itself does not oscillate.
(Bit oversimplified but probably you get the point.)
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As far as I can tell, you're confusing the coordinate time and proper time of the light ray.
$$\tau^2 = t^2 - x^2$$
The proper time of a light ray is $0$ since $t = x$, meaning that, in its own reference frame, it is not 'moving' through time.
Its coordinate time, though (as should be clear from the proper time definition), or time measured by an outside ...
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The relative velocity between two points depends on the frame.
In general, $A$ will see $B$ going at a different velocity than $B$ sees $A$.
Imagine that, in your example, $\mathbf{V_B}=\mathbf{V_A}$ (zero relative velocity) in the inertial frame. In the rotating frame, $B$ will lag a bit behind, because it is further than $A$ from the rotation center. ...
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