# Tag Info

13

Let me first go through this without friction or air drag. You say $v_y$ along the $x$-axis and the train moves with $v_x$ along the $z$-axis. This is a little inconsistent. I will use the velocities, but not your description of the axes. So the train moves in the $x$-direction, the ball is thrown into the $y$-direction and it the $z$-direction is up-down. ...

11

Why does time stop in black holes? Time according to whom? The fact is that, in special and general relativity, there is no universal time. Indeed, time is a coordinate in relativity so one must be careful to specify the coordinate system when asking questions like this. Now, every entity also has an associated proper time which is not a coordinate ...

6

I think there are two answers to this, one emprical and one theoretical. First, the theoretical one: What you describe is essentially induction, the belief that we can generalize from a subset of a class events/situations to the whole class of events/situations. This belief is, by necessity, unprovable, only falsifiable, since proving it would require ...

5

Yes, that's ok! You've stumbled upon one of the basic strange phenomena of relativity.

5

If you stick to gases then things are relatively straightforward because the temperature is related to the relative velocity of the gas molecules, that is the velocity of the gas molecules relative to each other. If you put your canister of gas in a fast moving (but non-relativistic) rocket moving at some velocity $v$ then you add the same velocity $v$ to ...

5

Short answer: It doesn't stop. Slightly longer answer: The case of a non-rotating, non-charged black hole is described by the Schwarzschild solution. It is now the case that, if you draw the worldline of a particle falling into a black hole, you will find that the coordinate time in the Schwarzschild metric grows infinite as the particle approaches the ...

4

so when I arrive at B why would I be younger? I believe I addressed this in another question of yours. Once again, assume that when you pass planet A, your clock and planet A's clock both read $t = t_A =0$. Now, according to the inhabitants of planet A, planet B's clock is synchronized with their clock. However, in your inertial frame of reference, ...

4

Use \begin{align} \frac{a_{23}}{a_{33}} & = \frac{ -\cos b \sin a}{\cos a \cos b} = -\tan a \\ \frac{a_{13}}{\sqrt{ a_{23}^2 + a_{33}^2 }} & = \frac{\sin b}{\cos b \sqrt{\sin^2 a +\cos^2 a}} = \tan b \\ \frac{a_{12}}{a_{11}} & = \frac{-\cos b\sin c}{\cos b \cos c} = -\tan{c} \end{align} a = atan2(-a23,a33) b = atan2(a13, ...

4

There is no reason why physical laws should be absolute. But observation tells us they are. If you think about it, if the laws of the universe did change from place to place or if they were different at different times, there would be no laws, and there would be no science.

3

Let me try a more down to earth example: Let's say I formulate a law "I can kick with my leg in front of me without getting hurt." This law is indeed true in many cases, but in some cases it is not because there is a wall right in front of me and my leg kinda hurts after kicking. That is, the world is not everywhere the same. Say I come to the same place ...

3

You shouldn't use the "subjective/objective" distinction for a place where "relative/absolute" is much more appropriate, because they mean different things. For something to be subjective, it must be dependent on the knowledge or state of mind of an observer. As an example, suppose we define "depth" as "length along the direction an observer is facing". ...

3

The quantity that tells you what time an observer travelling along a path $\gamma : [t_0,t_1] \rightarrow \mathbb{R}^4$ experiences is the proper time $$\tau = \int_\gamma \sqrt{\mathrm{d}x_\mu\mathrm{d}x^\mu}$$ Assuming flat spacetime, i.e Minkowski metric/special relativity, this reduces to $$\tau = \int_\gamma\sqrt{\mathrm{d}t^2 - ... 3 No. Time-dilation is the slowing of time as experienced by the fast moving craft, not the 'stationary' observer. Remember that light moves at c, and we see it move at c, not some slower or stationary speed. As the craft approaches c, it appears to accelerate increasingly slowly; from 0.99999c to 0.999999c is only a difference of 2.7 km/s, but it is still ... 3 it doesn't move in time, so no time will have past when the light arrives at it's "destination". Right? A photon does 'move in time'. It just that, for a photon, the displacement in time, c \Delta t, equals the displacement in space, \Delta x. However, there is no proper time for a photon. Your proper time is, in words, the elapsed time ... 3 Let \Delta_S and \Delta_G be the time dilation effects due to General Relativity (gravity) and Special Relativity (motion) respectively (i.e. the clock rate on the satellite due to SR and GR is 1 - \Delta_S + \Delta_G, signs chosen for simplicity). If these are small, they can be approximated as : \begin{eqnarray} \Delta_S &=& 1 - \sqrt{1 - ... 2 No, it doesn't mean that. One must distinguish two things: "laws of physics that apply to an object" and "laws of physics formulated from an object's viewpoint". These are two different things. Laws of physics apply to all objects. And the behavior of the objects may be described relatively to many coordinate systems or "frames of reference". The special ... 2 The time dilation due to motion in a circle, relative to an observer at the centre, is just the usual Lorentz time dilation due to the velocity of the motion. If you're interested, in my answer to Is gravitational time dilation fundamentally different than other forms of time dilation? I showed how this is derived from the metric. Anyhow, as you say, the ... 2 Almost none. Let's be much more generous than your idea of human-carrying craft. Let's just use the fastest probe. The Helios II craft, after nearing the sun, reached a heliocentric speed somewhere near 70 km/s. Obviously, its speed was more due to the gravitational influence of the sun than its engines.$$t = \frac{t_o}{\sqrt{1 - \frac{v^2}{c^2}}} $$... 2 So let's just say that the spacecraft can accelerate until it's moving away from the Earth at the speed of the fastest currently-existing spacecraft First, note that the fastest speed, relative to Earth, that a spacecraft has obtained is an exceedingly small fraction of the c and, thus, one should not expect significant time dilation. For ... 2 Inertial forces are considered non-assigned forces and thus are regarded as fictitious forces (even though they are real and the observer experiences these forces). They are mostly related to relative motion between (non-inertial, accelerated) frames of reference and transformation(s) thereof (eg centrifugal, Coriolis force, etc.. ). Apart from that, ... 2 There are three different effects to be distinguished here: (1) The universe is expanding. Let object A be at rest with respect to the average motion of the matter in its neighborhood of the universe. We describe A as being at rest relative to the "Hubble flow." Let object B also be at rest relative to the Hubble flow, at some cosmological distance x from ... 2 Ok, before we fill up the comment section with this, I will write this as an answer: Proper time \tau along a path \gamma is$$ \tau := \int_\gamma \sqrt{\mathrm{d}x^\mu\mathrm{d}x_\mu}$$and a clock moving along \gamma will have \tau as its elapsed time at the end of the path. Yet, the definition of proper time \tau involves such clocks not ... 2 It means that time is no longer an absolute concept, yes. The time a specific observer experiences in a specific frame of reference, i.e. his proper time depends on the path (worldline) he takes through spacetime. In other words, it depends on his state of motion, the way he accelerates. This is the reason for the famous twin paradoxon: the resolution is ... 2 The answer is that not everything is relative. Indeed, you've just shown that it is possible to detect rotation in an absolute sense. More generally, the principle of relativity says that all inertial frames are equivalent. In other words, it is impossible to detect (or even to define) absolute motion at a uniform velocity; it doesn't make sense to say that ... 2 As the others have said, it's simply true because it is a fundamental axiom upon which we build the the entirety of our physical laws. You can also approach an answer from a different perspective, though: relativity. In studying special and general relativity, one of the most important (and most difficult) concepts to grasp is that of 4-dimensional ... 2 that to an external observer it would appear to be moving very much slower? I don't understand the reasoning here. When you write assume that if a craft was travelling at a speed very close to the speed of light I take that to mean that the craft is travelling very close to the speed of light according to an external observer. Keep in mind ... 2 Does that mean the speed of an individual photon is c even with respect to another photon? I mean, shouldn't the relative velocity be zero? When we write "the speed with respect to X" we mean precisely "the speed as observed from the inertial frame of reference in which X is at rest" Thus, if it is true that the speed of an individual photon ... 2 Events which lie within each other's light cones are called "timelike separated." All observers agree on the ordering of these events. Events which lie on each other's light cones are separated by "lightlike" or "null" interval. All observers also agree about the time ordering of these events. Events which lie outside of each other's light cones are called ... 1 This is the question of the universal laws of gravitation. I think that may be because we are governed by a set of laws. And going against the nature you cannot set predefined restrictions to the dimensions of space and time. These are in the untraceable aspects of science. Maybe someday you could get a possible mathematical custom principle applied to it. 1 Now if I consider a particular set of rotation (say X first, then Y , then Z), with the corresponding Tait-Bryan angles --- a,b and c. My rotation matrix will be the following ... Look at that array. It's of the form$$\begin{array}{ccc} \cos(b) \cos(c) & -\cos(b) \sin(c) & \sin(b) \\ \cdots & \cdots & -\cos(b) \sin (a) \\ \cdots & ...

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