raapjeqp
2022-10-22
Answered

Dark energy physically can be interpreted as either a fluid with positive mass but pressure the negative of its density (pressure has units of energy/volume, and energy is mass), or a property of space. If it's a fluid, it should add to the mass of black holes like any form of energy (no hair), and the black hole should grow? However, if dark energy is a property of space, then this won't happen. Is my reasoning correct that we can differentiate (in theory) by looking at black hole's growth rate?

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fjaldangi

Answered 2022-10-23
Author has **9** answers

No, it is not possible to differentiate between these two interpretations because they're ultimately physically equivalent.

First, we should separate the discussions in cases where the total energy is conserved and where it isn't. The existence of conserved energy in general relativity (which must be ADM-like) actually requires vanishing or negative cosmological constant – the spacetime is Minkowski or AdS. In de Sitter space, there's no nonzero gauge-invariant definition of energy that would be conserved because the de Sitter space has no asymptotic region at infinity.

In the de Sitter space, masses of objects may therefore change in various general ways and by measuring them, you can't deduce pretty much anything.

In Minkowski or AdS space, the energy is conserved. Let's consider an anti de Sitter space with a negative cosmological constant. This means $\rho <0$, a negative energy density, with a positive pressure $p=-\rho >0$. The energy of the mass that ends up as the black hole is conserved, it's the total energy in the spacetime, assuming that everything collapses. However, the value of this total mass/energy is given before the black hole is formed – it stays the same by the conservation law – which means that we can't deduce anything new if we measure the same value at the end.

What you really want to do is to "attribute" or "divide" the total mass/energy of the black hole into different regions – either the generic black hole interior or the singularity. But this "attribution" or "localization" of matter is exactly what is impossible according to general relativity. The conserved total mass/energy cannot be written as an integral of a well-defined energy density. Such a thing may only be written in the "Newtonian" limit of weak gravitational fields and the existence of black hole is exactly the opposite situation in which the "weak fields" condition is dramatically violated.

So no, your verbal descriptions of the situations are just heuristic and to see what actually happens, you need to discuss things quantitatively, using the right concepts suggested by general relativity and using the right equations. The (Einstein's) equations say a very clear thing about the impact of cosmological constant in the absence of black holes much like in their presence and any idea about "two possibilities" (the cosmological constant is a property of space or a form of energy) is a mere illusion, an artifact of non-quantitative thinking about the problem.

First, we should separate the discussions in cases where the total energy is conserved and where it isn't. The existence of conserved energy in general relativity (which must be ADM-like) actually requires vanishing or negative cosmological constant – the spacetime is Minkowski or AdS. In de Sitter space, there's no nonzero gauge-invariant definition of energy that would be conserved because the de Sitter space has no asymptotic region at infinity.

In the de Sitter space, masses of objects may therefore change in various general ways and by measuring them, you can't deduce pretty much anything.

In Minkowski or AdS space, the energy is conserved. Let's consider an anti de Sitter space with a negative cosmological constant. This means $\rho <0$, a negative energy density, with a positive pressure $p=-\rho >0$. The energy of the mass that ends up as the black hole is conserved, it's the total energy in the spacetime, assuming that everything collapses. However, the value of this total mass/energy is given before the black hole is formed – it stays the same by the conservation law – which means that we can't deduce anything new if we measure the same value at the end.

What you really want to do is to "attribute" or "divide" the total mass/energy of the black hole into different regions – either the generic black hole interior or the singularity. But this "attribution" or "localization" of matter is exactly what is impossible according to general relativity. The conserved total mass/energy cannot be written as an integral of a well-defined energy density. Such a thing may only be written in the "Newtonian" limit of weak gravitational fields and the existence of black hole is exactly the opposite situation in which the "weak fields" condition is dramatically violated.

So no, your verbal descriptions of the situations are just heuristic and to see what actually happens, you need to discuss things quantitatively, using the right concepts suggested by general relativity and using the right equations. The (Einstein's) equations say a very clear thing about the impact of cosmological constant in the absence of black holes much like in their presence and any idea about "two possibilities" (the cosmological constant is a property of space or a form of energy) is a mere illusion, an artifact of non-quantitative thinking about the problem.

asked 2022-07-31

At one poinn a pipeline the water's spced is 3.00 ms and the gange pressure is

Sx l0 Pa. Find the gauge peessure al a seoond point in the line, 11.0 m bower thas

the first, if the pipe dianeter at the seccond point is twice that at the first.

asked 2022-08-25

A proton accelerated with electric field gives off E.M. radiation and therefore should lose mass. Larmor's formula gives us a value for the power emitted (varies as acceleration squared). However, as the proton picks up speed, it also gains mass. Now, say I set up an immense electric field which provides an immense acceleration to the proton. In the initial moments of motion, even though its acceleration is extremely high, its velocity is low. In those moments, does the proton lose mass faster than it gains?

asked 2022-09-23

Trying to derive the infinitesimal time dilation relation $dt=\gamma d\tau $, where $\tau $ is the proper time, 𝑡 the coordinate time, and $\gamma =(1-v(t{)}^{2}/{c}^{2}{)}^{-1/2}$ the time dependent Lorentz factor. The derivation is trivial if one starts by considering the invariant interval $d{s}^{2}$, but it should be possible to obtain the result considering only Lorentz transformations. So, in my approach I am using two different reference frames $(t,x)$ will denote an intertial laboratory frame while $({t}^{\prime},{x}^{\prime})$ will be the set of all inertial frames momentarily coinciding with the observed particle, i.e. the rest frame of the particle. These frames are related by

${t}^{\prime}=\gamma (t-\frac{Vx}{c}),\phantom{\rule{1em}{0ex}}{x}^{\prime}=\gamma (x-Vt),$

where $V$ is some nonconstant (i.e. time dependent) parameter which is, hopefully, the velocity of the particle in the laboratory frame. Treating $x$, $t$ and $V$ as independent variables (for now) and taking the differential of the above relations,

$d{t}^{\prime}=\gamma (dt-\frac{Vdx}{c})-\frac{{\gamma}^{3}}{{c}^{2}}(x-Vt)dV,$ and

$d{x}^{\prime}=\gamma (dx-Vdt)-{\gamma}^{3}(t-\frac{Vx}{c})dV.$

Imposing either the definition of the rest frame $d{x}^{\prime}=0$ or (what should be equivalent) $dx=Vdt$, the only way in which i obtain $dt=\gamma d{t}^{\prime}$ is if $dV=0$. So, the derivation breaks badly at some point or I must be wrong in using some of the above equations. Which one is it?

${t}^{\prime}=\gamma (t-\frac{Vx}{c}),\phantom{\rule{1em}{0ex}}{x}^{\prime}=\gamma (x-Vt),$

where $V$ is some nonconstant (i.e. time dependent) parameter which is, hopefully, the velocity of the particle in the laboratory frame. Treating $x$, $t$ and $V$ as independent variables (for now) and taking the differential of the above relations,

$d{t}^{\prime}=\gamma (dt-\frac{Vdx}{c})-\frac{{\gamma}^{3}}{{c}^{2}}(x-Vt)dV,$ and

$d{x}^{\prime}=\gamma (dx-Vdt)-{\gamma}^{3}(t-\frac{Vx}{c})dV.$

Imposing either the definition of the rest frame $d{x}^{\prime}=0$ or (what should be equivalent) $dx=Vdt$, the only way in which i obtain $dt=\gamma d{t}^{\prime}$ is if $dV=0$. So, the derivation breaks badly at some point or I must be wrong in using some of the above equations. Which one is it?

asked 2022-07-16

Can one explain the relativistic energy transformation formula:

$E=\gamma \text{}{E}^{\prime},$

where the primed frame has a velocity $v$ relative to the unprimed frame, in terms of relativistic time dilation and the quantum relation $E=h\nu $?

Imagine a pair of observers, A and B, initially at rest, each with an identical quantum system with oscillation period $T$.

Now A stays at rest whereas B is boosted to velocity 𝑣.

Just as in the "twin paradox" the two observers are no longer identical: B has experienced a boost whereas A has not. Both observers should agree on the fact that B has more energy than A.

From A's perspective B has extra kinetic energy by virtue of his velocity $v$. Relativistically A should use the energy transformation formula above.

But we should also be able to argue that B has more energy from B's perspective as well.

From B's perspective he is stationary and A has velocity $-v$. Therefore, due to relativistic time dilation, B sees A's oscillation period $T$ increased to $\gamma \text{}T$.

Thus B finds that his quantum oscillator will perform a factor of $\gamma \text{}T/T=\gamma $ more oscillations in the same period as A's quantum system.

Thus B sees that the frequency of his quantum system has increased by a factor of $\gamma $ over the frequency of A's system.

As we have the quantum relation, $E=h\nu $, this implies that B observes that the energy of his quantum system is a factor of $\gamma $ larger than the energy of A's stationary system.

Thus observer B too, using his frame of reference, can confirm that his system has more energy than observer A's system.

Is this reasoning correct?

$E=\gamma \text{}{E}^{\prime},$

where the primed frame has a velocity $v$ relative to the unprimed frame, in terms of relativistic time dilation and the quantum relation $E=h\nu $?

Imagine a pair of observers, A and B, initially at rest, each with an identical quantum system with oscillation period $T$.

Now A stays at rest whereas B is boosted to velocity 𝑣.

Just as in the "twin paradox" the two observers are no longer identical: B has experienced a boost whereas A has not. Both observers should agree on the fact that B has more energy than A.

From A's perspective B has extra kinetic energy by virtue of his velocity $v$. Relativistically A should use the energy transformation formula above.

But we should also be able to argue that B has more energy from B's perspective as well.

From B's perspective he is stationary and A has velocity $-v$. Therefore, due to relativistic time dilation, B sees A's oscillation period $T$ increased to $\gamma \text{}T$.

Thus B finds that his quantum oscillator will perform a factor of $\gamma \text{}T/T=\gamma $ more oscillations in the same period as A's quantum system.

Thus B sees that the frequency of his quantum system has increased by a factor of $\gamma $ over the frequency of A's system.

As we have the quantum relation, $E=h\nu $, this implies that B observes that the energy of his quantum system is a factor of $\gamma $ larger than the energy of A's stationary system.

Thus observer B too, using his frame of reference, can confirm that his system has more energy than observer A's system.

Is this reasoning correct?

asked 2022-07-14

If you are traveling at $x$ speed the time will pass for you slower than to an observer that is relatively stopped. That's all just because a photon released at the $x$ speed can't travel faster than the $c$ limit.

What happens if you have two bodies, $A$ and $B$ moving towards each other. If $A$ releases a light beam, and $B$ measures it (the speed of the photons), the speed measured is still the same? The only difference will be the wave length? And if we have the opposite case, $A$ and $B$ are moving away from each other, we get the red shift, but the speed measured will be still the same?

What happens if you have two bodies, $A$ and $B$ moving towards each other. If $A$ releases a light beam, and $B$ measures it (the speed of the photons), the speed measured is still the same? The only difference will be the wave length? And if we have the opposite case, $A$ and $B$ are moving away from each other, we get the red shift, but the speed measured will be still the same?

asked 2022-07-16

Special relativity says that anything moving (almost) at the speed of light will look like its internal clock has (almost) stopped from the perspective of a stationary observer. How do we see light as alternating electric and magnetic fields?

asked 2022-10-09

The distance between interference lines in the double-slit experiment is:

$\text{}w=z\lambda /d$

Where:

$w$: Distance between fringes

$z$: Distance from slits to screen.

$\lambda $: Wave length of light

$d$: Distance between slits.

Now, looking at the setup from a reference frame traveling in the direction of the distance 𝑧 between slit and screen and moving away from the slit in direction of the screen, the light gets redshifted and the distance 𝑧 gets Lorentz contracted:

$\text{}{w}^{\prime}=\frac{z}{\gamma}\gamma (1+\frac{v}{c})\lambda /d=z(1+\frac{v}{c})\lambda /d$

This only holds for the first fringe or so, that aren't too far from the center, because the red-shift formula changes farther out. However, it's obvious that the distance between fringes has increased!

Where is my mistake?

$\text{}w=z\lambda /d$

Where:

$w$: Distance between fringes

$z$: Distance from slits to screen.

$\lambda $: Wave length of light

$d$: Distance between slits.

Now, looking at the setup from a reference frame traveling in the direction of the distance 𝑧 between slit and screen and moving away from the slit in direction of the screen, the light gets redshifted and the distance 𝑧 gets Lorentz contracted:

$\text{}{w}^{\prime}=\frac{z}{\gamma}\gamma (1+\frac{v}{c})\lambda /d=z(1+\frac{v}{c})\lambda /d$

This only holds for the first fringe or so, that aren't too far from the center, because the red-shift formula changes farther out. However, it's obvious that the distance between fringes has increased!

Where is my mistake?