# Tag Info

22

Answering this requires a bit of a preamble, so bear with me ... Any observer can construct a coordinate system to locate points in spacetime. By construct a coordinate system I mean they choose a ruler for measuring distance and a clock for measuring time, then they choose the directions of their three spatial dimensions. Even before Einstein, coordinate ...

5

The E=mc^2 formula only applies to an object at rest, and light is never at rest. You want to use the more general formula: $E^2={m_0}^2c^4+p^2c^2$ Then you can set the mass to zero. $E=pc$ What this says is that light has momentum, which is related to its energy.

4

Einstein says this, not because your watch is some ancient artifact with power over time, but because in the theory of relativity, time is relative. We can not longer say that the time is blah blah o clock everywhere. The time is different at different points. Therefore, your time is the time on your clock, and this is the "correct" time for your reference ...

3

"Can we take the time interval between the two events 1 and 2 as observed by the mentioned observer as proper time interval?" No, if two events are simultaneous in any frame, the interval between them is a space-like one, not a time-like one. If you're not familiar with the idea of the three categories of intervals, time-like, space-like and light-like, see ...

2

This is really a comment on Hypnosifl's answer but it got a bit long to fit in the comment box. Be cautious about interpreting the word time in proper time too literally. We use the phrase proper time because it is the time measured by an unaccelerated observer travelling between the two points. However no such observer can exist in this case because that ...

2

It is an artificial distinction to say one is more fundamental than the other. The geometry of flat spacetime is given by the Minkowski metric: $$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$ and this is fundamental in the sense that all of special relativity is described by this equation. But it is also fundamental that the parameter $c$ in the equation (which ...

2

Which is more fundamental is probably a meaningless question, but you can think of the geometric notion of a manifold (the mathematical abstraction of spacetime) as being more general than Lorentzian manifolds, i.e. ones whose metric is locally like that given in John Rennie's Answer. So one could in principle conceive of universes that were described by ...

2

Mass-energy equivalence (Rest energy = rest mass * c^2) can be applied to a moving body: Total energy = relativistic mass * c^2 From the perspective of a stationary observer, the total energy in a moving body is greater than the rest energy in a stationary body. As velocity approaches c, it is much greater. The ensuing explosion would be much bigger than ...

2

The energy of a photon is given by the equation E = hf where h is Planck's constant and f is frequency. The energy would decrease, making the frequency decrease (since h is constant). So, if the photon was blue light, then it would get redder and redder as time when on. There is a point, however, when your system eventually stops working. This is because the ...

2

When we talk about the geometry of GR, it is understood that the manifold of spacetime is not a Riemannian one, but rather a Lorentzian manifold. This means that the metric is not positive definite. With this understanding, we call $g(.,.):=\langle.,.\rangle$ an inner product as usual. This lack of positive definiteness has many consequences. It is the ...

2

The local speed of light is always $c$. Local in this sense could mean that for each observer there exists a neigborhood of that observer such that, if we call $v_c$ the "observed speed of light", $|v_c - c| < a$ where $a$ is arbitrarily small. However $v_c$ is a slightly nebulous concept as beyond the inertial frames in special relativity, there ...

1

The reason why there is no contradiction is that the PEP states, essentially, that no two particles can be at the same place at the same time. Without going into difficult QFT concepts mentioned above, a simple answer to your question would be that "at the same place at the same time" is equivalent to saying "at the same point in space-time." While ...

1

This is because instead of $$\dfrac{1}{2}mv^2$$ or $$E = mc^2$$ the energy of light is given by $$E = hf$$ Where h is a number called Planck's constant and f is frequency (sometimes v is used) Here is an example, as requested: Imagine red light with $620. nm$ wavelength. The frequency of this light is $0.483$ x $10^{15}Hz$ This makes the energy of a ...

1

Let's say you have 3 systems. $B$ moving relative to $C$ with velocity $u$ and $A$ moving relative to $C$ with velocity $v$, all along one axis. $A$ will "measure" for the velocity of $B$: $$u' = \frac{u-v}{1-\frac{uv}{c^2}}$$ While $B$ will "measure" for the velocity of $A$: $$v' = \frac{v-u}{1-\frac{uv}{c^2}} = -u'$$ It holds true that the velocity ...

1

The velocity addition formula you cite does not quantify what you think it quantifies. I'm going to say see and when I do, I mean "computes relative to its own frame". The velocity addition formula describes the following setup. Frame 0 sees a particle (particle 1) moving in a direction at speed $v_1$. Frame 1 sees that particle 1 at rest and frame 1 ...

1

The expression for time dilation which we come across states the fact that the "time interval between two event is minimum when observed from a frame where two events occure at same place".. In your case the events did not occur at the same place.. so you can instead use the lorentz transformation equation to determine the time interval between event A and B ...

1

Well, a rocket traveling at close to the speed of light would be very hard to see at all cause it would go from a moon's distance in one direction to a moon's distance in the other direction in a little over 1 second, and seeing a rocket as far away as the moon would be difficult - but I'm thinking that's not what your asking, so lets pretend that we have a ...

1

vL/c^2 is the difference (for a stationary observer) between two clocks at the two ends of a rod of rest (or proper) length L traveling at velocity v. The clock at the forward-end (in the direction of v) of the rod trails by this amount (as perceived by the stationary observer). The two clocks are synchronized in the rest frame of the rod, but for the ...

1

I would like to bring the ladder paradox here to explain simultaneity of events.A ladder (an inertial frame) is moving horizontally with a relatively high constant speed with respect to a garage (another inertial frame). The garage has an open door where the ladder can not actually enter if the ladder was at rest in the garage's frame but that is not ...

1

Regarding your assertions: Events $\varepsilon_{AJ}$ and $\varepsilon_{BK}$ were simultaneous in the inertial frame of participants $A$, $B$, $M$. This is a perfectly reasonable statement and it is the sort of language used in everyday physics. Participant $M$ was the middle between $J$ and $K$, in the inertial frame of participants $A$, $B$, $M$. ...

1

"Coincident" is defined in the Google online dictionary as (1) "occurring together in space OR time" (emphasis mine), and (2) "in agreement or harmony". "Simultaneous" is defined in the same dictionary as "occurring, operating, or done at the same time". (This begs the question: "Whose time?") Unfortunately, this dictionary lists "coincident" as a synonym ...

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