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0

You are basically asking what happens if we use fractions of a second. Everything still appilies.


1

HyperLuminal asked: "Does that mean that electrons are infinitely stable?" Think about Dirac's model of an electron, which includes left and right handed contributions. Now add the (Nobel-worthy) Brout-Englert-Higgs idea, that the left-handed bit interacts with a condensate of weak hypercharge, while the right-handed bit does not. This suggests a ...


0

Special relativity itself makes it clear that absolutes cannot be detected, such as being at absolute rest in space, or the inability to detect absolute motion. This therefore prevents one from having an absolute understanding of special relativity, since that which special relativity reveals does not extend to the point of absolutes. Thus the absolute ...


0

Take a look at this pendulum periodicity calculator: http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html. You can compare the period of each swing of the pendulum under different conditions of gravitational acceleration (little g in your formula). Notice that as gravity grows stronger (as gravitational acceleration becomes greater), the period of the ...


1

No, the relationship between the period of a pendulum and $g$ is simple Newtonian mechanics and unrelated to special or general relativity. This is discussed in the answers to Time period related to acceleration due to gravity (though I hesitate to link this as that question was not well received). Time dilation was actually known before Einstein formulated ...


4

Yes, there's a very famous example: muons produced in the upper atmosphere can be detected on the surface of the Earth. Moving at nearly the speed of light, it takes them over 300 microseconds to get down to the Earth's surface, but the average muon decays after 2.2 microseconds. If it were not for time dilation, only a few in every $10^{60}$ muons (so, ...


0

The pendulum motion is caused by a restoring force whose whose tangential component is opposite to displacement in direction. The tension from the string, if any, would always be perpendicular to the path. When there is no such force to provide the restoring force, the type of oscillation you mentioned would not happen. BTW, a kind reminder- Mathematics is ...


0

In that place, where g=0. T goes to infity. What it means?, just that the pendulum will not move. Just that. The time will run normally in that place and in the earth.


2

Let me clear up a few misconceptions. The edge of our observable universe would contain information from the beginning of the universe, since it is a particle horizon. However, the edge of the observable universe is not currently visible to us. What we can currently see only goes as far back as the recombination era, when electrons first joined with nuclei ...


-3

the time period in a SHM is related to the force causing the motion, the force which is proportional to the negative o the displacement. In the case of gravitational force, one simple example is that of the simple pendulum.T=2πLg−−√


33

The statement is true for decays, where lifetimes can be measured. It is not true for interactions though. A suicidal electron meeting a positron has a good probability to disappear, together with the positron, into two gamma rays, at low energies. Electron-positron annihilation It is intriguing that this is not true for neutrinos. If an electron ...


12

This is not exactly true. It is believed that net charge is conserved, but there is a weak process called electron capture, where an electron is captured by a nucleus, (usually from an inner "orbital" so there is a spectroscopic signature), a neutrino is emitted and a proton changes to a neutron. So therefore your textbook is wrong!


167

Imagine you are an electron. You have decided you have lived long enough, and wish to decay. What are your options, here? Gell-Mann said that in particle physics, "whatever is not forbidden is mandatory," so if we can identify something you can decay to, you should do that. We'll go to your own rest frame--any decay you can do has to occur in all reference ...


3

The practice of dividing the degree used to measure angle into sixty minutes of arc, and that into sixty seconds of arc is over 2000 years old. The corresponding practice of dividing the hour used to measure time into sixty minutes, and that into sixty seconds, is over 1000 years old. Why sixty? That's over 5000 years old. The Sumerians and Babylonians used ...


2

It's historical. The second was originally defined so that $60\cdot 60\cdot 24\,\rm s$ added up to a solar day. But that's a little hairy to measure, because the length of the day varies through the year. The sunrise-to-sunrise time varies from winter to summer. The noon-to-noon time interval, which would be operationally defined as the interval between ...


0

We need to first ask ourselves what is Space? What is Time? Then we can begin to answer your question after we define what these two are and the relationship between them. According to Geometrical Mathematics and based on Numerical Vector Space is nothing more then an empty construct and has no Dimensions until you give it a coordinate. We can define space ...


1

I feel like it is correct to call gravity a force. As you know, there are several models for how the universe works. The Newtonian model. The relativistic model. The quantum-mechanical model. Within certain different boundaries of scale, these each work very well at predicting things that will happen. However the language or terminology of each ...


0

The problem with the question is the lack of definition for "same." In the "real/'practical" world, nothing is just the same, you have to specify the measurement and the precision (accuracy) associated/required for the measurement! Because of the higher precision/accuracy required for our modern scientific experiments, a need for higher precision standards ...


0

Can anyone describe, (or point me to a paper that describes) time without referring to something else. The principal relevant description is surely the one which Einstein put at the foundation of the (special) theory of relativity: "[... that instead] of ``time´´ we substitute ``the position of the little hand of my watch´´." [Punctuation marks as in ...


1

Can anyone describe, (or point me to a paper that describes) time without referring to something else. No. Nobody can. However I can point you to Presentism, and to A World Without Time: The Forgotten Legacy of Godel and Einstein. I can also point out that relativity accurately models our world using spacetime and worldlines and geometry, but that the map ...


2

It's worth drawing a diagram: The equations of motion are: $$x_1 = x_0 + \frac12 a_1 t^2$$ for $t_1 < \frac{v_1}{ a_1}$ And $$x_1 = x_0 + \frac{v_1^2}{2 a_1} + v_1 (t - \frac{v_1}{a_1})$$ for the steady state. The same equations, with different suffixes, hold for the tiger. To solve, you start by considering all different orders of $t_1, t_2, ...


1

I'm going to first address some misconceptions you seem to have and then I will get to answering your question. Now, as stated in the comments, @Pulsar did a very thorough job of answering this question in another post. But I read through that answer and it's a bit technical. I already knew the stuff, so it made sense to me, but I can see how someone ...


0

I am not sure that the alien can "aim" at our past, nor that we can "aim" at the future because I see time as a construct. We see two or more events occur in 3-space and we measure some interval between them. That is true. But, I have yet to hear anyone describe "naked time" (that is just time without allusion to some other event that defines it.) As such, ...


2

Ok, so here are is my solution. I'll be happy is someone can provide something simpler. $a_1, v_1, t_1 - acceleratin, terminal\ velocity,\ time\ to\ reach\ it\ (for\ Tiger)$ $a_2, v_2, t_2 - acceleratin, terminal\ velocity,\ time\ to\ reach\ it\ (for\ You)$ $s_0 - starting\ distance$ Not let's consider three cases: 1) $a_1 < a_2 \wedge v_1 < v_2$ ...


3

But what experiment showed that cesium-beam's period was so terribly consistent? They compared it with other clocks. That frequency is terribly consistent, because it isn't actually constant, because it varies with gravitational potential. Moreover this frequency is quoted in Hertz, which is cycles per second. And it's used to define the second. Spot the ...


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The frequency is determined by the energy spacing between two configurations of the caesium atom. Caesium has a single electron in the outermost $6s$ orbital, and this electron can be aligned with or against the nuclear spin. These two configurations differ in energy by about 0.000038 eV, and transitions between them produce/absorb light with a ...


5

Is time an illusion? No. I think it's best to think of it as something like heat. You know what heat is, especially if you put your hand on a stove: szzz aaargh! Heat is definitely not some illusion. However it is an "emergent property". Think about the kinetic theory of gases. The temperature of a hot gas is something like a measure of the average kinetic ...


0

An answer you might find satisfactory is that our model of spacetime is "larger" than what we observe. Yes, there is a preferred direction which gives us a (1,3) signature, but actual, real objects in spacetime must travel on timelike curves (we normalize the length of their geodesics to -1). These timelike curves only transverse part of the entire ...


5

One piece of physics that you've missed is the most pulsars spin down due to the emission of magnetic dipole radiation. For instance, the crab pulsar has a period of 33.5028 (plus a few more sig figs) milliseconds, but slows down by 38 nanoseconds per day. Furthermore, the size of several more increasing order derivatives is known accurately. So in ...


3

Yes, in a sense, clocks are always wrong, namely in the sense that finding two that "agree" by continuously always showing the exact same time would require a very peculiar arrangement, crucially involving the observer's speed and position relative to the clocks! However, physicists have adopted another sense of what being simultaneous (or clocks agreeing) ...


3

The experiment is real. If you take an atomic clock that is accurate and stable enough and you fly it around the world, it will disagree with an identical clock that has remained on the ground. This is because of the combination of two effects: relativistic time dilation, which means that any clock that moves will run more slowly compared to one that's ...


0

Surely therefore our locally measured time is not the cosmological time t but rather the conformal time T? I don't think so. Our locally measured time is our locally measured time. If you had a clock that started ticking when the big bang occurred, the clock reading would be 13.8 billion years. If it displayed the conformal time, the clock reading would be ...


0

You've mostly only stumbled upon one fact with your list of paradoxes: a lot of the things presented in the movie Interstellar are not parts of a physical theory. This includes: "The bulk", whatever that's supposed to mean, and especially Arms reaching through "the bulk" (???) True back-in-time communication/time travel, "Tessaracts" (the word "tessaract" ...


0

A flaw in Wilczek's argument has been pointed out in PRL 110, 118901 (2013). In short, the rotating soliton exhibited by Wilczek is not the correct ground state of the system. The true ground state can be shown to be stationary, irrespective of the flux. Impossibility of a quantum system with time-crystal-like ground-state has been further addressed in PRL ...


0

Actually here $\rho$ and ${\bf v}$ are function of $(t,\vec{x}) \in \mathbb R \times \mathbb R^3$, as it is usual in the so-called Eulerian description of a continuous body. There is no reference to the curves describing the evolutions of the particles of the fluid $\vec{x}_{\vec{x_0}}= \vec{x}_{\vec{x_0}}(t)$ as instead, it is the standard in the so-called ...


0

From the chain rule we have, $\frac{d \rho}{dt} = \frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial x}\frac{dx}{dt} + \frac{\partial \rho}{\partial y}\frac{dy}{dt} + \frac{\partial \rho}{\partial z}\frac{dz}{dt}$ $\therefore \frac{d \rho}{dt} = \frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial x} v_x + \frac{\partial ...


0

Notice that one of the observers has to stop and turn back in order to compare watches and hence feel an acceleration. Thus they are not symmetrical in this sense. That's why the observer, who experiences the acceleration would be younger. As Feynman puts this fact in his famous lectures: This [twin paradox] is called a “paradox” only by the people who ...


-1

Special relativity is define has either points A or B being motionless to a third point, even if neither observer at point A or B can define that third point. Only after comparing stopwatches will the observers know which of them was at relative rest and which of them was in relative motion to the third point. Can't define absolute rest...but the classic ...


3

The link Qmechanic has suggested is a duplicate and does discuss the question you ask. However there is another point that is worth making here. In general relativity we describe the universe as a manifold equipped with a metric, and the metric is the FLRW metric that desciribes expanding spacetime. However the FLRW metric does not include the point(s) at ...


1

Yes, both special relativity and general relativity have to be taken into account. The total time dilation is given by $$ \frac{\text{d}\tau}{\text{d}t} = \sqrt{ \left(1 - \frac{2GM}{rc^2}\right) - \left(1 - \frac{2GM}{rc^2}\right)^{-1}\frac{v^2}{c^2}}, $$ where $\text{d}\tau$ is the time measured by a moving clock at radius $r$, and $\text{d}t$ is the ...



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