New answers tagged

3

We exist along wordlines in 4D space. We like to measure three of those dimensions using one unit and the fourth using another, but we could use the same unit for all four and the constant $c$ would vanish from all calculations. We would then notice, for example, that nothing other than light can include both (0, 0, 0, 0) and (1, 0, 0, 1) in its wordline (in ...


5

First: From a practical standpoint, it is not exactly correct to say that we use the speed of light to define the value of the meter, or the value of a second. Rather, we should say that we used specific measurements of the speed of light, taken by specific scientists at specific places and times, to define these units. Suppose, for the sake of argument, ...


2

You actually describe, why the speed of light is necessarily assumed to be the fundamental constant with the definitions you mention. Although of course its actual numerical value in the end depends on OUR choice of definition (for example on which part of the distance traveled by light in one second defines a $m$.). Let's use an analogy: Let's assume the ...


8

The numerical value of the speed of light is not a universal constant. It's the quantity that is a universal constant. As an analogy, in Newtonian physics, the distance between any two points in space, or the time interval between any two events, is a universal constant. Pick any two points in space in a Newtonian universe. Say, one observer measures the ...


4

Proper time is the time experienced by a single clock. A proper time interval between two events is an interval measured by a clock that is present at both of them- the duration of the interval depends on the path taken by the clock, and is longest for a clock that can consider itself to have been stationary between the two events (ie a clock that has not ...


12

The question is very poorly formulated. Its answer depends a great deal on a subjective or context-dependent choice of terminology, not only on physical facts; it leaves some important points unspecified; and it confuses "being undefined" with "being unmeasurable". Let me address these points in turn. First, whether the term "proper ...


3

The right answers are either A, B, or C. The only one that's definitely wrong is D! Any inertial observer can measure proper time. It's only that it happens to be zero for trajectories of objects that move at the speed of light. Pending qualifications on proper time of what exactly?; the signals themselves?, then the answer is zero; of Ajax's and Hector's ...


2

Edit: I noticed one of your other questions so I'll preface this answer with an additional point, in special relativity it is better to think of reality as if everything that ever happened or ever will is fixed in stone. It turns out all the weirdness that arises from special relativity basically comes from looking at this from different angles, e.g. if you ...


3

From Britannica.com The proper time would be measured by any clock moving along the straight world line between the two events. In other words, the proper time is the time measured by that clock which is moving with a uniform velocity between the 2 events. Here, the 2 events are the event of the robot being at a certain point above the planet's surface and ...


11

According to Wikipedia, Proper time can only be defined for timelike paths through spacetime which allow for the construction of an accompanying set of physical rulers and clocks. For lightlike paths, there exists no concept of proper time and it is undefined as the spacetime interval is zero. The path connecting the 2 events you mentioned i.e. event of ...


0

The Cosmic Microwave Background (aka CMB) is light radiated from hot glowing matter as it cooled from the Big Bang. It took a few hundred thousand years for the universe to cool enough that electrons could stick to protons, forming neutral H atoms. Before this, the universe was full of free charges, which absorbed light. After this, the universe was ...


3

Classically, there would be no hard limit. In a perfectly silent, classical universe, you could imagine a device to detect the wave. But practically, the strength would eventually decay below the ability of your device to discriminate it from other sources of noise (including those from within the device). In QM, any detector has a lower and lower ...


3

Electromagnetic waves are made of photons. These will continue to travel along geodesics unless they scatter or are absorbed. Currently the main model of cosmology has an infinite Universe so if this is correct a photon could keep travelling indefinitely if it is not absorbed.


2

In your question you have a photograph of tracks in some kind of bubble chamber. This is an old technology, in which some kind of fluid is prepared in a supercritical chemical state, on the wrong side of a phase change. (In a "cloud chamber" the fluid is ready to condense liquid from vapor; in a "bubble chamber" the fluid is ready to ...


-1

Any background information about how, for example, in the Large Hadron Collider, these images are taken, would be helpful to understand what we can and can't conclude from these diagrams. Many things could be deduced from such particle chamber. I've repainted main directions here : Given, that main test particle enters chamber from outside world and moves ...


1

here is found in the list a "picture" of an antiproton: The dark lines in this picture are produced by charged particles as they force their way through liquid deuterium. The highlighted track is an antiproton , produced in the decay of an antilambda into an antiproton and a $π^-$. In the top left corner of the picture, this antiproton ...


5

There is a simple geometric interpretation of the quantity that is called the age of the universe. Spacetime has a large-scale shape. It looks something like this: Later times are at the top, earlier times at the bottom. You shouldn't take this too literally because it isn't a proper embedding (I flipped the sign of the metric to make it Euclidean, and ...


5

No. The age of the universe does not depend on any referential system. In order to measure time, you need some physical quantity that's changing to measure time against. In the case of cosmology, it's the time perceived by a typical observer based on the expansion parameter $a$ --see below. In a manner of speaking, it's the time that follows the galaxies in ...


2

How do I describe the interior? How can I even think about something that isn’t there until I go there? It seems to me that the future is always that way, whether it's a black hole interior or not. It's true that when there is a black hole, people can end up stuck in distinct "branches" of the universe in the sense that they can never meet or ...


4

"Do we have any evidence that flow of time is not relative in quantum scale?" No, we do not. In fact, quite the reverse; Wikipedia is not always reliable! Firstly, the basic quantum equations simply assume that time is absolute (Newtonian) and not relative, they make no claim that this accurately represents reality. This is what the Wikipedia ...


1

Imagine a totally perfect clock. It measures the time in the place where it is situated. Its own proper time. But due to general relativity the speed of time is affected by nearby objects, gravitational fields, spacetime curvature, speed of motion and gravitational waves. The ideally perfect clock would look not that perfect for anyone who is located in a ...


-1

It seems as though you are asking if a "clock" can exist that isn't impacted by relativity/time dilation. At the moment, nothing we can observe in the universe is free of time dilation. We would need to be able to observe or measure a property of the universe or particle that does not suffer any impact of time dilation. Unfortunately, there are ...


0

This is correct. The fact that they're inside a non-accelerating rocketship doesn't matter, Bob is simply in stronger gravity because he's closer to the surface of the earth. Gravity gets weaker as you go away from the earth's surface. Obviously the effect on these scales is miniscule.


4

If Bob says the time interval is smaller than Alice says, then time is going slower for Bob. For example, perhaps Bob says one second has passed while Alice says two second have passed. In this case Alice will think "come on Bob, your clock is ticking too slowly" and Bob will think "wow! Alice is aging rapidly."


0

The answer in your Edit seems right, You must have also needed to change one of the minus signs in $x_2(t) = v_ft - \frac{1}{2}(-9.81)t^2 \quad$ as if you are measuring positive to be upwards, just one of them will take account of gravity being downwards. Then you just put your first two formulae equal to each other to get $h=v_ft$ and since you now have $...


0

Starting with the basic equations for the kinematics, we have object 1 falling from h with zero starting velocity, and object 2 going up from the ground: $$ x_1(t) = h - \frac{gt^{2}}{2} \tag{1.1} $$ $$ x_2(t) = v_f t + \frac{gt^{2}}{2} \tag{1.2} $$ Now, recall that for the falling body 1: $$ v = gt \tag{1.3}$$ We have to compute the final velocity, taking ...


0

The answer uses the notation introduced in the OP and presents the complete analysis. The position kinematics of the first ball which falls down under gravity are given as $x_1(t) = h - \frac{1}{2}gt^2$ and that of the second which rises up due to the initial velocity as $x_2(t)=v_f\cdot t-\frac{1}{2}gt^2$, where $0<v_f$ is the speed of the first ball at ...


0

No they aren't the same thing. The following explanation is from a DoD technical document. https://apps.dtic.mil/sti/pdfs/ADA305743.pdf The idea here is that there are multiple transforms from the frequency domain to the time domain, and you need to have a compact notation for handling the transforms. The time convention is how that time term is rendered in ...


0

First, no theorem can definitively demonstrate anything about the real world. They can only show that their conclusions follow from their assumptions. Second, even when the assumptions of this particular theorem are met, the conclusion seems quite modest: [A] cosmological model which is inflating – or just expanding sufficiently fast – must be incomplete in ...


0

We can do it either by first taking the derivative and then taking the conjugate transpose of it. $(\partial U/\partial t)^\dagger$ = $(-{\frac\iota \hbar} \space U(t) \space H(t))^\dagger$ and we know $(AB)^\dagger$ = $ B^\dagger \space A^\dagger$ So we get, $\frac {∂U(t)^†} {∂t}=\frac iℏU(t)^†H(t)$ so we get the expression. We can also find it buy ...


1

$$[H,U(t)]=0\Rightarrow \text{Doesn't matter :)}$$ Note that $$\text{In general } \ \ U(t)\not=\exp\left(-\frac{iHt}{\hbar}\right)$$ You have used time dependent Hamiltonian and above is not valid for such case.


1

The terms are not exactly equivalent, although both likely hold in this context. To work in the time domain is (broadly) to consider a signal's strength as it changes as a function of varying time (as opposed to varying frequency, for example). Parametrizing a signal in terms of e^jwt means working in the time domain because one can plug in the time t (in ...


1

The 'why' of things is tricky in physics. Let me give a spin to this question by considering the alternative: why would space and time be completely uncorrelated to whatever happens in time and space? We would have two metaphysical dimensions in nature: one, similar to a container, the all and ever unchanging space/time background, like a stage where the ...


2

To think about this, first we need to think about what we mean by time passing. A good way is consider some regularly repeating process, such as an object undergoing simple harmonic motion. We count the number of oscillations of the object, between two events nearby that are of interest to us, and this count gives a measure of the amount of time that has ...


2

You've asked a challenging question that can have several interpretations, but I'll try to give you one that you can understand. But you should realize that you've set off on a journey that could take you years. You ask "why does time change when gravity increases?" This is not a good interpretation. Many scientists believe that the gradient of ...


1

Yes, gravitational time dilation is due to the warping of space-time by matter. The pop-science picture of this warping as being like a heavy mass sitting on a rubber sheet is misleading, because that shows only the spatial warping. But in fact in ordinary circumstances by far the largest component of the warping is in the time dimension; that is, while ...


0

Fact: Yes, space is curved around a body. Newtonian gravity takes this into account. The surprise was that time is also curved, and that curve is defined by $\sqrt{1 - R_s / r}$. Don’t forget that, with all that speed as one falls in, that there is also kinematic time dilation, defined by $\sqrt{1 - v^2/c^2}$. The integration of the precession of the ...


5

A way to understand why that must happen uses the equivalence principle. A free fall observer can be considered as a inertial frame. If he compares the rate of his clock with another clock fixed somewhere at height $= h_0$, according to the special relativity the fixed clock (that is moving with respect to the free fall observer) shows a smaller tick rate. ...


-1

When gravity increases, speed increases more quickly, and time decreases in the same proportion. So when gravity increases, time decreases. That's the change that you'll get - a decrease in time.


-4

If the speed of light falling in gravity varies as per Newton, as suggested in the texts below, then there is no gravitational time dilation (general relativity is nonsense): University of Illinois at Urbana-Champaign: "Consider a falling object. ITS SPEED INCREASES AS IT IS FALLING. Hence, if we were to associate a frequency with that object the ...


2

It's recommended that you learn about energy to start with, looking up the definition of the unit of energy, the Joule and learning about mechanical energy or 'work done'. You could then move on to find out about other types of energy. Time is different and you probably won't find a simple explanation of what it actually is. The physics unit of time is ...


4

Time is no more than a coordinate. It could seem as it is special regarding the physical universe in a classical sense, dissociating time and space, but long years of physics develompent and the theory of relativity have proved this point of view wrong. You can't untangle time from space, so it must be regarded as just another coordinate labelling an event, ...


-1

It depends on the speed of the airplane. If the plane has speed zero, then they will return older. If they travel around the Earth at the speed for which time goes equally slow as the time on Earth, then they will return with the same age as the people who stayed on the ground. If they travel faster than this speed they will return younger.


0

The proper time is the time experienced by the particle itself. When you thus choose your reference frame as that of the particle itself, you fix your spatial coordinates at the particle's location. So $dx^i=0$.


2

In SI units, no, but in geometrized units like we often use in general relativity there is a geometric interpretation of acceleration. Specifically, in geometrized units $L/T^2 \rightarrow L/L^2 = 1/L$. Geometrically this gives a curvature which has units of inverse length. So the acceleration of an object, interpreted geometrically, is the curvature of that ...


1

The unit $\mathrm{m/s^2}$ represents metres-per-second added/removed per second, so $$\mathrm{\left[\frac{m}{s^2}\right]=\left[\frac{m}{s}/s\right]}.$$ Now, in classical non-relativistic mechanics space is three-dimensional while time is one-dimensional. With space being three-dimensional, we can see 1D, 2D og 3D space-property combinations. Area, such as $A=...


0

A good reference for such things is Wald's General Relativity. In that book, you'll find the following result: Theorem. Let $(M,g)$ (a manifold with a metric) be globally hyperbolic. Then There exists a global time function, that is a map $t:M\to \mathbb{R}$ such that $-dt$ is future-directed and timelike; Surfaces of constant $t$ are Cauchy surfaces, and ...


0

This argument isn't correct. A photon does change phase as it travels. "Objects traveling at the speed of light do not experience time" is a common idea, but there is no such thing as a frame of reference traveling at the speed of light, so you can't consistently talk about what something traveling at the speed of light experiences. See How does a ...


2

A lot of technology is facilitated and enabled by the ability to better measure things. And you never know what you might need it for until you need it. You could ask the same thing about a lot of research: "What's it good for right now?" People try to design around limitations if they can, but if they can't and are forced to deal with it head on, ...


9

I don't know the intended, specific purpose of those better clocks, but I can speak to the question on a more general front. "Mo betta accuracy" has generally been a plus in the past, especially when we thought we had things figured out. It's historically been one of our best ways to find new phenomena, right alongside being able to measure new ...


Top 50 recent answers are included