Tag Info

New answers tagged

0

The notation is confusing. Typically, a $\Delta t$ and $\Delta t'$ denote the coordinate time difference of two events. For example, consider two events with coordinates in the unprimed system $(t_1, x_2)$ and $(t_2, x_2)$. The coordinate time difference for these two events is $$\Delta t = t_2 - t_1$$ For the same two events (this is crucial), the ...


0

So the clock which measures proper time is the one for which the two events occur at the same location. So lets say you are travelling on a train at constant velocity and you get hungry and eat a bar of chocolate. In your reference frame on the train you start eating the bar (event 1) and finish eating it (event 2) at the same spatial coordinate. So the time ...


1

Complex mass means gravitational mass + i.lambda.higgs mass. The weak coupling constants are all proportional to (higgs) mass with Higgs vacuum value (=246 GeV) as the proportionality constant. Thus mass necessarily becomes a complex number. The real part produces attractive gravity forces and the imaginary part produces repulsive "weak" forces. The factor ...


0

Well, an heuristic solution would be that it is actually impossible for anybody to cross the event horizon. As you noticed, Schwarzschild black hole is dual (at least locally, that is, if you don't make full turn) to Rindler's observer (the uniformly accelerated one). The stationnary observer in the Schwarzschild case is the uniformly accelerated one in ...


1

Yes, I do believe there is and that is the approach I outline in my answer to the Physics SE equation "What's so special about the speed of light" as well as my answer here. I like to think of SR as simply Galileo's basic idea but with the assumption of absolute time relaxed (not to devalue Einstein's bold step in making this relaxation). One begins with ...


2

From the reference frame of the space ship, your body is stationary. When you swing your arm back and forth, it has a non-zero speed in this frame, and thus its mass, or rather momentum, increases. The space ship's velocity of $v_\mathrm{ship} = 0.866c$ doesn't add to this. That means that the increase in momentum is exactly the same as if you swing your arm ...


0

On a vector space $V$ with metric $g$ - be that euclidean, lorentzian or whatever - the Orthogonal group $O(V,g)\subset GL(V)$ is defined to be the group of (linear) isometries on $V$. More precisely, for an element $\Lambda\in O(V,g)$, $$ g(\Lambda v,\Lambda u)=g(u,v)$$ holds for all $u,v\in V$. Orthorgonal trafos preserve lengths and angles. Expanding ...


0

I'll give this a shot. I think I follow what you're asking. I'm thinking of the to-be-fissioned-away material as a mass traveling at the speed of light in some sense, In a sense that's true, but it's probobly good to keep in mind that time isn't a dimension quite like the other 3 and traveling through time isn't exactly moving, so in a sense it's ...


2

Let us take a uranium nucleus being hit by a neutron .At the rest mass system of the two bodies there is an invariant mass m described by E=m*c^2. The CM system, seen as an excited U236 in the diagram below, is not moving in three dimensions nor in any other dimensions, velocity needs a dx/dt. An induced fission reaction. A neutron is absorbed by a ...


3

Relativists tend to use the proper time, $d\tau$, and the proper distance, $ds$, interchangably. If you're working with proper time you'd expect the equation for it to look like: $$ d\tau^2 = dt^2 + \text{other terms} $$ while if you're working with proper distance you expect: $$ ds^2 = dx^2 + dy^2 + dz^2 + \text{other terms} $$ The sign problem comes ...


1

$g$ denotes the metric. For Euclidean space the metric is just the unit matrix $I$. For Minkowksi space, which is of interest when talking about the Lorentz group it's the Minkowski metric $\eta_{\mu \nu}$. The lower right matrix inside the Minkowski metric is the 3-dimensional unit matrix and therefore for the space-like components of the Minkowski metric ...


0

E=mc2 applies equally to ordinary chemical bombs, except there is less "m" turned into "E", by around a factor of a million.


0

I believe gravity's rainbow relates to gravitational lensing, which can be likened to refraction. In normal refraction we see a rainbow spectrum, because the degree of refraction depends on the photon E=hf energy. However in gravitational lensing, we do not. Gravity's rainbow is a speculation that we might. I'm not fond of it because it's akin to saying ...


0

Could you provide a simple reason for these two conventions? The reason behind the (-,+,+,+) convention (the "mostly plus metric") is that a positive length in 3 dimensional space (e.g., the distance from my head to my toes) should still be a positive length in 4 dimensional space-time. Why should the distance from my head to my toes all of a sudden ...


0

Given a certain four-current $J^\mu = (c \varrho, \vec{j})$, that is a charge density $\varrho$ and current density $\vec{j}$. the four-potential $A^\mu = (\Phi / c, \vec{A})$ is given by: $$ A^\mu(\vec{r},t) \propto \int \frac{j^\mu(\vec{r}\ ', t_r)}{|\vec{r}-\vec{r}\ '|} d^3r'$$ with $t_r = t - \frac{|\vec{r}-\vec{r}\ '|}{c}$ if one takes the retarded ...


0

Your question consists actually of two parts, I will answer them one-by-one: Why is the magnetic field circular? Any vector field $\vec F$ can be decomposed into a rotational part and a divergent part, according to the Helmholtz decomposition theorem $$\vec F = - \vec \nabla \Phi + \vec \nabla \times \vec A $$ This is a purely mathematical statement and ...


-1

Answer I feel strongly there is no way to prove that such thing as a "slowest possible non-zero speed" exists. However there are realistic limits regarding the degree to which slower and slower speeds can reliably be measured. Example (purely hypothetical) Consider a pair of neutrinos, ν1 and ν2, emitted from the sun, one right after the other, at times ...


1

Pressure is a single component of the rank-2 stress-energy tensor; it isn't a Lorentz scalar and observers will disagree about the pressure of a gas in a box. You say that the box might explode (perhaps the walls of the box break) if the pressure exceeds a critical value. Perhaps, then, observers disagree about whether an explosion occurs? I don't think ...


-1

What you are forgetting in this thought experiment is that the gas' volume would decrease too, allowing for pressure to stay the same.


13

We do. The LHC accelerates two protons, each with 3.5 TeV of energy, giving a total of 7 TeV in the CoM frame (The energies are from the previous LHC run which discovered the Higgs. The energies are doubling now for Run II). The main reason for this is as you mentioned, the energy involved. In any frame, we have the following invariant quantity, $s = (p_1 + ...


5

Many modern particle accelerators do accelerate both particles towards each other. LEP accelerated electrons and positrons in opposite directions in the same chamber, and the Tevatron did the same for protons and antiprotons. The LHC is a proton-proton collider, and so it has two stacked rings that accelerate protons in different directions. For the BaBar ...


0

which should not be possible. Indeed, uniform coordinate acceleration $a$ is inconsistent with special relativity however, uniform proper acceleration $\alpha$ is consistent. The proper acceleration is the acceleration of the object according to an attached accelerometer. For 1D motion, the relationship between $\alpha$ and $a$ is given by $$\alpha = ...


14

Have a look at the article by Phil Gibbs on the relativistic rocket. This describes the motion of a rocket that is accelerating with a constant acceleration. In this context constant acceleration means the crew of the rocket feel a constant acceleration. Technically the rocket has a constant four-acceleration. Anyhow, the velocity of the rocket as observed ...


1

Look at sparknotes.com/physics/specialrelativity/dynamics/…, you can see $dE/dx=F$ - if your force is constant, it is the energy that increases constantly. $E=\gamma(v)m_0c^2$, you can deduce the $v$. Beacause of laziness I used mathomatic, and it gives me something like this: $v=c\sqrt{1-\frac{m_0 c^2}{(F\cdot x + m_0 c^2)^2}}$ If you check it for x=0 and ...


6

Is there some other formula ... which ... does not allow the speed ... to surpass the speed of light? That would be the equations of special relativity mentioned by sahin in a comment. Image from Loodog? Another factor you have to take into account with classical mechanics is to work out how a constant force can be applied to your object over 11 ...


0

To answer my question #1: As Demosthene said, $Λ^μ_ν$ depends only on the relative velocity between frames $S$ and $S'$, but not the relative velocity between particles. This allows me to make the assumption, which I numbered 6 in my initial question, thus solving my problem.


2

We conclude that the lenght perceived by the observer in S is bigger than the lenght perceived by the observer in S. No, that would be an error. Why? The short answer is that the coordinate difference $(x_2 - x_1)$ is not a length in S. A length is the (spatial) coordinate difference at the same coordinate time. To be clear, there are two events - ...


0

I think that the fact that the clocks in the two ships appear synchronized to the initial observer in S distracts us and makes us think of the situation in non-relativistic terms. In general, most SR paradoxes can be eased by remembering the relativity of simultaneity. Let's say that the 2 spaceships have clocks at their centers, and the observer in S finds ...


0

EM waves are mass-less. The time dilation effect you mention only applies when a travelling body has mass. Note that for a body that has mass it would take an infinite amount of energy to travel at c.


2

According to relativity, If magnetic field is just an electric field viewed from a different frame of reference It is true that a pure electrostatic field in an inertial reference frame (IRF) will be observed as a mix of electric and magnetic fields in some relatively moving IRFs. However, in the general (time varying) case, it is not possible to ...


0

you are correct that the EM field propagates at c, but as the comment from Acuriousmind mentioned, there is no frame of reference traveling at that speed. What that means in simpler words, is that the equations break or diverge for anything traveling at the speed of light. Light is not bound to time dilation or length contraction for that same reason. Only ...


0

Electromagnetic waves rely upon oscillations to propagate Electromagnetic waves are propagating disturbances (or oscillations) in the electromagnetic field and the electromagnetic field exists everywhere and everywhen. But the electromagnetic field itself does not 'travel' at $c$ so your reasoning isn't clear to me.


5

Your statement is not really true, since if you only have a magnetic field in one frame of reference, then it can never be viewed as just an electric field in another frame of reference. And vice-versa. As described here, the magnetic field can be defined as (e.g. in Jackson's Classical Electrodynamics) the field that is responsible for the Lorentz force ...


-1

As I don't belong anymore to this site, my answer will be very brief. I apologize for not being interested to give explicit details, only certain hints. 1) Nature is local. 2) FTL signals are not only self-contradictory, as explained in my protocol, the nature doesn't use them. I repeat, they are not the way the nature works. (By the way, this is why we ...


0

Mathematically, since the coordinates $\alpha$ are functions of $\tau$, you can, assuming certain assumptions on the curve, reexpress the curve as a function from the time coordinate to the spacetime coordinates by writing $\alpha^\mu(t)=\alpha^\mu((\alpha^0)^{-1}(t))$, which is just a reparametrization of the curve from $\tau$ to $t$. The assumptions are in ...


0

It is incorrect to say that we can only differentiate $\alpha^i$ with respect to proper time. We know that proper time is the parameter along the curve, i.e. a point $x$ on the curve is a function of the proper time $\tau$. This relation is given by the vector $\alpha=\alpha(\tau)$. However, we also have the relation $$\tau(x)=\int^x \mathrm{d}\tau$$ which ...


1

The Hamiltonian $H$ generates time translations and the momentum $\mathbf p$ generates space translations. In a relativistic theory time and space can mix, so we should consider the 4-vector $$p^\mu = (H, \mathbf p).$$ Now whatever the $\mathbf p$ is in terms of mass, velocity and so on, certainly it is $\mathbf 0$ in a rest frame. Then so that there can be ...


0

First apply a transformation in the $x$ direction using the formulae you stated. This way you obtain $x'$ and $t'$. The next transformation you have to apply is in the opposite direction, so the velocity changes sign. Use again your formulae to calculate $x''$ and $t''$: $x''=\gamma( x' - (-v)t')=\gamma( x' +vt')$ ...


7

What I would like to know is how can one motivate that the correct choice for p is the γ(v)mv In Newtonian mechanics, the momentum of a particle of mass $m$ is given by $$\mathbf p = m\frac{d {\mathbf r}}{dt} = m \mathbf v$$ where $\mathbf r$ is the position vector and $t$ is a universal parameter. However, in relativistic mechanics, $t$ is a ...


7

Special relativty is about Minkowski spacetime. A line element is given by $$ ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$ A free particle will move on a straight line, that is, it will minimize the path length $$ L = \int ds = \int \sqrt{c^2 \left(\frac{dt}{d\lambda}\right)^2 - \left(\frac{dx}{d\lambda}\right)^2 - \left(\frac{dy}{d\lambda}\right)^2 - ...


0

the quantity $ds^2 = g_{\mu,\nu} \ dx^\mu \ dx^\nu $ is a measure for distance on a manifold. It is indeed invariant under coordinate transformations, since $g_{\mu,\nu}$ transforms like $$ \tilde{g}_{\mu,\nu} = \frac{\partial x^\alpha}{\partial \tilde{x}^\mu} \frac{\partial x^\beta}{\partial \tilde{x}^\nu} \ g_{\alpha,\beta} $$ and $dx^\mu$ transforms like ...


-1

First of all to achieve speed of light the energy requirement tends to infinite, which is only possible by entire mass conversion, as if there will be not much to percieve as you wouldnt be alive to achieve speed of light, as in case of zero rest mass particles there is no mass, hence they achieve light of speed that is zero rest mass particle are not ...


0

One of the main effect that would be affecting how this man would see is the contraction of length that are not in its referential(travelling with him). If a referential is moving at the speed v close to c in the +x direction. Then every object that are not in its referential will be shorten along the x axis by a factor $\gamma$ such that : ...


0

Based on the comments from @ACuriousMind, the question itself is poorly formed. To calculate an interval between two events on a general manifold, one needs both a metric and a connection. Using the process of parallel transport, one moves one of the points next to the other. Then one can subtract the two points. The metric at that point is then used to ...


5

Although the Lorentz boosts have "tensor indices", they are not defined as tensors on the space, but as elements of the isometry group. Let $M,M'$ be two (pseudo-)Riemannian manifolds with metric tensors $g,g'$. Given a diffeomorphism $\phi : M\to M'$, you could let the isometry $\Lambda : M\to M$ (which, on $\mathbb{R}^{1,3}$ would be given by ...


1

As @Hypnosifl points out, a single event cannot have $\Delta t$. You must define two events such as a light flashing or the beginning and end of a process or a particle moving from point A to point B. You must the define the pair of events (call them $A$ and $B$ with both a time coordinate and a spacial coordinate, e.g. $$ \pmatrix{ct_A \\ x_A} \text{ and } ...


2

The basic confusion is in your comment that "During $\Delta t_3$ an event occurs in $O_3$'s frame that is measured as $\Delta t_2 = \gamma^\prime \Delta t_3$". In SR a single "event" occurs at a single point in spacetime--an interval like $\Delta t$ can only describe the time between a pair of events. What's more, the time dilation formula $\Delta t_B = ...


2

The second approach is wrong. You cannot use the relation $E_{total} = h\dfrac{c}{\lambda}$ because this is equivalent to saying that $E_{total} = pc$ which is true only for massless particles. In this case it should be $E_{total} = \sqrt{p^2c^2+m_0^2c^4}$.


1

Let me first say that I do not work in quantum foundations, really, so I might have a few misconceptions myself. I beg anyone to correct me, where I err and I will try to provide more references upon request. After the question seems to have cleared up in chat, let me rewrite my answer: You basically seem to ask: What if entanglement would allow ...



Top 50 recent answers are included