# Tag Info

-3

Is photon fixed in space time? Photon is not fixed to a point in space time, it moves as light at speed c relative to us. It covers 0 distance in in zero time relative to itself. In Last question I said, "They don't experience time, that is they are stationary with respect to time axis". The time axis is a virtual axis. It has no physical significance. ...

2

Photon experience infinite time dilation and hence, time is stationary for it. Does photon experience time  photon travels at c through the three spatial dimensions. All of its velocity is directed through the three spatial dimensions. Thus Brian and Einstein are stating that a photon must be stationary in the fourth dimension. For if the photon had any ...

1

A black hole in the Schwarzschild case has a radius $$r=\frac{2GM}{c^2}$$ Let us solve for the mass with a radius of about Fermi or $10^{-15}$m, or the radius of a baryon such as a neutron, $$M = \frac{rc^2}{2G}=\frac{10^{-15}m\times 9\times 10^{16}m^2/s^2}{2\times 6.67\times 10^{-11}Nm^2/kg^2}$$ $$=6.7\times 10^{11}kg.$$ This is close to a billion ...

2

The general relativity model for an electron is the Kerr-Newman solution, that is a charged rotating black holes. Unfortunately the radius of the exterior horizon in geometric units is given by: $$r_{+}=M + \sqrt{M^2-Q^2-P^2-\left(\frac{J}{M}\right)^2}$$ Here P is the magnetic charge, zero in this case. If you convert this relation to standard units ( for ...

3

If you believe classical parameters for black holes down to a neutron's scale, then it turns out that a neutron has too much angular momentum to form a black hole, and would be best interpreted as a naked singularity. The reason why is that the radius of a black hole's event horizon is predicted to be: $$M \pm \sqrt{M^{2} - a^{2}}$$ where $M$ is the mass ...

1

Invariants are useful, in general, because they represent something that all observers can agree upon. Relativity showed us that the concept of time-intervals, spatial-distances, and even sequences of events can be drastically different from different observers. So how can one observer 'relate' to another? I.e. how could I, standing still, figure out what ...

1

The spacetime interval invariance property allows us to, for example, compare the rate of time passing for two observers moving at relative velocities to each other. Although no observer in the universe is at complete rest, the interval is a benchmark for comparison of the physical effects of differences in velocity, or indeed location. Say one observer is ...

0

Could the time for light to go that distance (4734 feet) be the same as the time difference caused by [relativity gravitational time dilation]? No, for two reasons. The units are not the same, one is nanoseconds per day. The other is nanoseconds. light travels 4734 feet in about 4813 nanoseconds not 20.

0

I believe $F^{\mu\nu}$ is better known as the Faraday tensor; Maxwell tensor is only the spatial part of it, 3x3 matrix. The argument goes like this: $F$ is a four-tensor, so if any quantity is constructed from it by multiplication of its components, this new quantity also has to transform as a tensor. The construct $$... 4 How does a scalar quantity transforms under a Lorentz transformation? I will show you here how this works for a vector (contravariant) and a covector. The transformation rule for these objects are$$u^a = { \partial x^a \over \partial x' ^b} u' ^b u_a = { \partial x'^b \over \partial x^a}u' _b $$Multiplying the two:$$ u^{a} u_{a} = { \partial ...

1

A body with finite dimensions will always contract in the direction of motion. A square sheet will contract in either direction by a factor of $L = L_0/\gamma$ where $\gamma = 1/\sqrt{1-\frac{v^2}{c^2}}$. A sheet moving in both directions will experience contraction in both dimensions. Length contraction is motivated, as you say, by a need to keep the speed ...

0

I think you misunderstood something.Length contraction in a moving body,does not occur in the perpendicular direction of its velocity (not length as you mentioned). Length contraction occurs in the direction of its motion.

-3

If a black hole travels from point A to point B, long distance at the speed of light you can say it's a wormhole.

3

This is a subtle and somewhat complicated question, but I think the basic answer is no''. 1) The relativistic Boltzmann equation is $$p^\mu\partial_\mu f = C[f]$$ which has the same structure as the non-relativistic Boltzmann equation. This equation can be used to derive relativistic Fokker-Planck equations. One example is the Landau collision term, ...

1

As you described the book you are searching for I thought of a book that fit that description except it may not be the same book. I have a text book for a course given at Cal Berkeley by Professor Richard A. Muller. It is called "Physics and Technology for Future Presidents" with a subtitle of "An Introduction to the Essential Physics Every World Leader ...

2

Assume that $g$ is a function in the space of rapidly decreasing functions with all their derivatives of any order, the so-called Schwartz space ${\cal S}(\mathbb R^4)$. Define $$\hat{g}(k) = \frac{1}{(2\pi)^2} \int_{\mathbb R^4} e^{-i k_\mu x^\mu} g(x) d^4x\::$$ As a consequence, as is well known, $\hat{g} \in {\cal S}(\mathbb R^4)$ and $$g(x) = ... 2 PART 1 : Taylor Expansion ( see @Prahar comment ). In the following :$$ \mathbf{x}=x^{\mu}, \quad \mathbf{k}=k_{\mu}, \quad \partial_{\mu}=\dfrac{\partial}{\partial x^{\mu}} , \quad \mu=0,1,2,3 \tag{1-01}  \mathbf{k}\cdot\mathbf{x}=k_{\mu}x^{\mu}=k^{\mu}x_{\mu}, \quad \text{Einstein's convention on}\: \mu \tag{1-02} $$Suppose now that the ... 3 Yes, both the internal potential energy and the internal kinetic energy of a bound system (in the rest frame of its center of mass) contribute to the bound system's inertial mass according to E=mc^2. For a paper discussing the evidence that this is true for internal kinetic energy in particular, see Kinetic Energy and the Equivalence Principle. 2 Theoretically, the answer is yes. However, looking at the practicality of the situation, the answer is no. The photons can not be contained inside the box unless they are either 1. Created inside the box itself, or 2. They are trapped beforehand, and then brought inside the box. In 1., they do not add to inertia because, they are created using energy ... 6 Yes, mass and energy are equivalent. A more competent relativist might be able to give you the complete description, but to first order you can say that the mass of an object is simply the total energy in its volume divided by c^2. That mass is equivalent to the inertial mass by the weak equivalence principle, which is a cornerstone of GR. That is to say, ... 23 Yes! In fact, this kind of phenomenon is very common. For example, the mass of a proton is much greater than the sum of the masses of the constituent quarks. Much of the extra mass comes from the gluons that bind the quarks together; like photons, gluons are massless, but they contribute to the inertia. 3 The short answer is special relativity. classically According to classical mechanics, there is nothing to prevent an object from accelerating faster than the speed of light. By conservation of energy you can calculate the speed of a dropped object by looking at the change in gravitational potential. As you note, the equation PE=mgh needs to be modified ... 4 You have slightly misunderstood time dilation. Suppose you are moving at some speed v relative to me, then in my coordinates the time t I measure for you to move a distance d is:$$ t = \frac{d}{v} $$Andd even if you are a ray of light this still applies so the time I measure for a ray of light to move a distance d is:$$ t = \frac{d}{c} $$So ... 4 A better way to think of it is "speed of causality". That's the fastest any cause-and-effect will spread over space. With nothing to cause it to go slower, changes to electric and magnetic fields will occur at that speed. No coincidence that changes to spacetime (causing gravity) propigate at the same speed. You really need to show how Minkowski spacetime ... 2 Nonsense. Maxwell derived his electromagnetic equations, with \epsilon_0 and \mu_0, and those quantities were known. The fact that his equations led to the speed of electromagnetic waves to be, in terms of \epsilon_0 and \mu_0, equal to the approximately then known speed of light is a big part of what led Maxwell to conclude that light is ... 2 Interesting question. Taking a stab at it - not absolutely sure this is correct, but let the comments begin. In the frame of reference of Earth, the light travels straight out to the reflector, and straight back. You are asking about the case where an observer is in a reference frame that is moving with respect to Earth/moon, and the picture would have to ... 12 Not really. The "speed of light" has very little to do with light; it is built into the actual geometry of spacetime independent of what matter fills it. In particular, \epsilon_0 and \mu_0 don't tell us anything physical about the vacuum; looking at the (simplified) expressions$$E = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2}, \quad B = ...

2

This question raises several concerns. From a purely signal processing point-of-view, where the aforementioned pulse is not a photon, but rather a time-domain pulse of a certain duration (here less than 10^-100 seconds)--then NO: you cannot have pulses shorter than the period of wave in question--as the wave an be defined. Consider sound: when a 440Hz A note ...

2

I am adding this answer to respond to your last followup comment on L.Levrel's answer. Point One: Alice and Bob have to agree on how many times Alice's clock ticks per rotation, and they have to agree on how many times Bob's clock ticks per rotation. Therefore, if Alice says Bob's clock is slow, Bob must say Alice's clock is fast, and vice versa. Point ...

2

$kT$ is related to the kinetic translation energy by the equipartition theorem. You are saying that the mean kinetic energy, is much greater than the rest energy. The particle has a large or relativistic velocity. The limit $kT>> mc^2$ is called ultrarelativistic limit. It means you can approximate the energy momentum relation $E^2=(pc)^2+(mc^2)^2$ by ...

3

Think about this from the perspective of a person in the elevator. No windows, they can't look outside. As far as they are concerned, they live on a small box-like planet where the acceleration due to gravity is 9.8 + 1.2 = 11 m/s$^2$. In a system where the acceleration due to gravity appears to be 11 m/s$^2$, a bolt drops 2.7 m. How long does it take to ...

0

I dont think it's a problem, that the satellite is accelerating. Locally the time will nevertheless run slower. If the camera is directed onto a watch, you would see it ticking more slowly. If the webcam is directed on the outside, you will see the normal speed of happenings, e.g. it would show you the same number of rounds as you have seen by looking at ...

Top 50 recent answers are included