# Tag Info

## Hot answers tagged time

18

A physicist, me for example, identifies events by choosing a set of coordinates. For example I have a clock that I use to record time and a ruler that I can choose to measure distance. This allows me to set up some coordinates $(t, x, y, z)$ so I can assign every event to some point in my coordinate system. If I received a laser pulse from Mars at 16:05 ...

14

The way atomic clocks work is to produce a microwave signal with exactly the frequency of the atomic transition being used. So for caesium this would be 9,192,631,770Hz. Then we can count the oscillations of our microwave generator to measure the time. Practically you do this by tuning your microwave signal to maximise its absorption by the caesium atoms. ...

3

$\frac{dM}{dt} = \frac{\partial{M}}{\partial{t}}+\frac{\partial{M}}{\partial{x}}\frac{d{x}}{d{t}} = \frac{\partial{M}}{\partial{t}}+v\cdot\nabla{M}$ (with no assumption on what is M) . So if $v\cdot\nabla{M} \neq0$ you can have one of $\frac{dM}{dt}$ and $\frac{\partial{M}}{\partial{t}}$ that is zero when the other is not. ...

3

As ACuriousMind says in a comment, this isn't the approach Yukawa used in his 1935 paper (Yukawa H 1935 Proc. Phys. Math. Soc. Japan 17 48) though whether he did that calculation in the privacy of his own notebook only he knows. The calculation you describe is a rather arm waving sort of justification for the relationship between the mass of the mediating ...

2

They couldn't carry sufficently accurate time from London with portable clocks. But they were able to use clocks to time measure the time between the sun crossing and the transit of stars the night before and after. The absolute transit time of stars can be trivially obtained if you know the site's longitude. If you are on land and have an observatory ...

2

A space train leaves Mars at 14:00pm and arrives on Earth at 19:45. The train moves at 0.001C and has 40Km of length. How long will it take for the whole train to arrive on Earth? - Disregard re-entry and friction. Nobody on Earth will say the train is leaving mars now. Same thing with the light, just it moves faster and is smaller than the train above. ...

2

Addressing the 3rd question, the clocks' frequencies are the same when each is viewed in its own rest frame. When a clock is viewed in a moving frame, that's when its frequency is changed. You record the ticks and their locations (which are different since the clock is moving) and you discover that the time between ticks is now longer.

2

Just open any string text which has a discussion of the relativistic point particle. http://arxiv.org/abs/0908.0333 - Section 1 for example or Green, Schwartz, Witten Volume 1 Punchlines: 1) Time can be introduced as an operator but you need to introduce a 'proper time' parameter for which the system evolves with. In doing this you introduce a gauge ...

2

My name is Sarah Huggins, about three years ago, my dad wrote an article on Wikipedia to demonstrate how nothing on the site is reliable. That article was The Huggins Displacement Theory. A couple years later, it's in three books, multiple movie and book reviews, and on physics blogs like this. He took the article down once it got flagged, but I thought I ...

2

What you have done here is a Galilean transform, that is a non-relativistic transformation. Take your final result (which is quite correct): $$t' = \frac{\sqrt{\beta^2 + \alpha^2}}{\sqrt{\eta^2 + \mu^2}} \tag{1}$$ We know that the vertical velocity is $\eta$, so the vertical distance moved in our time $t$ is given by: $$\beta = \eta t$$ We also know ...

1

I don't know the "formal" proof, but here is my proof: Time dilation and length contractions are given to us by the Lorentz transformations by: t’ = t/(1-v2/C2)1/2  and d’ = d/(1-v2/C2)1/2 (in other words “same” or proportional to each other) where: t = distance/length traveled through the T dimension in observers own frame of ...

1

No. Cooling down the atom will not alter it's frequency. What the scientist meant by saying that the clocks will be better is that they will be much more accurate. This comes from the principle of Quantum mechanics, in fact one of its most beautiful consequence, Heisenberg's uncertainty principle. Which says (in one of it's variety) that the uncertainty in ...

1

The problem is that you are equating too many things to $\dot{q_k}$. Usually $\dot{q_k} = \frac{dq_k}{dt}$, a total derivative, as opposed to a partial derivative. If $q_k$ has no explicit time-dependence, i.e. it does not depend directly on $t$ itself, then $\frac{\partial q_k}{\partial t} = 0.$ In this case, the Poisson bracket reduces to: $... 1 yes, time dilation still occurs. The reason is that the mirrors are there to provide an intuitive view of how/why time dilation occurs, not to to create it. This time dilations still occurs regardless of the existence of the mirrors or the direction of motion. But the drawing will no longer have explanatory power. 1 In that case, time dilation still occur, of course. In order to show this using t=d/v, you'd have to take into account the space contraction in the direction of motion. Mathematically, if d is the height of the clock, then the time taken from a photon at the bottom to reach the top of the clock isn't$\frac{d+vt}{c}$but$ \frac{d/\gamma+vt}{ c}$. When you ... 1 I think most of you are confusing what contracts because of motion. Not the distance between the muon and the Earth, nor the distance between the point in which the muon began its motion and its destination. Length contraction deals with the contraction "of the moving object" (that is the muon's length, if we could talk of it). If you could see an airplane ... 1 We could try this one: People down on that planet got few hours older while people on the ship got twenty years older. So, lets do this. Put a telescope on the surface on that planet and observe motion of people and such in the ship. If you could somehow do this, you should be able to see everything on the ship happening faster. Also, if people from the ... 1 You can use hook's law$F=-kx$. If$\tau$is the torque on board , then force at the spring,$F=\frac{\tau}{L}$and$\tau=I\alpha$, where$I=\frac{mL^2}{3}$, M.I of rod about one of its end . U can replace x by$x=L\theta$and appliying to hook's law yield$I\dfrac{d^2\theta}{dt^2}=-L^2k\theta$this gives$\omega=\sqrt{\frac{L^2k}{I}}$. Therefore, time ... 1 The state is "measured" in the sense you are imagining - that is, it becomes definite - at whatever time it becomes possible in principle to infer its having a particular measurement outcome. In your case, if the probe provides unambiguous information about the measurement result, the time of measurement will be found to have been delta-t back in time. This ... 1 Do you know Bernard Schutz's book: A First Course in General Relativity? Check out the first chapter of that book. There is a derivation of invariance of proper time using first principles in section 1.6. Basically, the idea is to start from expressing$\Delta \bar s ^2$(interval in the barred frame) as a linear combination of$\Delta x_i$'s (vector ... 1 I assume you mean by "the effects of time dilation...due to gravity vs. ... due to velocity" you mean can we tell the difference between the relativistic time dilation $$t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ and the gravitational time dilation $$t=\frac{t_0}{\sqrt{1-\frac{2GM}{rc^2}}}$$ where$t_0$is the proper time in both cases. The answer is ... 1 As you may know, it takes infinite time to charge a capacitor. So, the time when the capacitor is 100% charged never comes. Thus, we require a Time Constant to help us understand the time when the capacitor has got a decent amount of charge and after which the rate of charging becomes really slow and thus charging further is not of much use. You may also ... 1 Time dilation: linear or exponential or other? Other $$\Delta t' = \gamma\Delta t = \frac{\Delta t}{\sqrt{1 - \frac {v^2}{c^2}}}$$ Lorentz factor$\gamma$as a function of speed (in natural units where$c=1\$) - Image by Zayani CC BY-SA 3.0

1

One thing you have to note is that speed is relative, Clock A would see clock B moving from A's point of reference, and B would see A moving in B's reference, so you shouldn't be using the word "stationary" in this context. Both the clocks would see the other clock tick slower, B would see A's future only if it returns back to A, this makes it obvious to A ...

1

Well, the "standard" answer is the one given below (that's why we use telescopes to look at the past universe). However, I'd like to qualify this. The Einstein twin paradox (time contraction) implies the following thought experiment: With a very good telescope, you look at a planet 10 light-years away. And there you see an ET entering his spaceship and ...

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