"The age of the universe is 14 billion years"

Which observers' are these time intervals? According to whom 14 billion years?

Which observers' are these time intervals? According to whom 14 billion years?

Almintas2l
2022-07-21
Answered

Which observers' are these time intervals? According to whom 14 billion years?

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Tristan Pittman

Answered 2022-07-22
Author has **14** answers

An observer with zero comoving velocity (i.e. zero peculiar velocity). Such an observer can be defined at every point in space. They will all see the same Universe, and the Universe will look the same in all directions ("isotropic").

Note that here an "idealized" Universe described by the FLRW metric:

$\mathrm{d}{s}^{2}={a}^{2}(\tau )[\mathrm{d}{\tau}^{2}-\mathrm{d}{\chi}^{2}-{f}_{K}^{2}(\chi )(\mathrm{d}{\theta}^{2}+{\mathrm{sin}}^{2}\theta \phantom{\rule{thickmathspace}{0ex}}\mathrm{d}{\varphi}^{2})]$

where $a(\tau )$ is the "scale factor" and:

${f}_{K}(\chi )=\mathrm{sin}\chi \phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}(K=+1)$

${f}_{K}(\chi )=\chi \phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}(K=0)$

${f}_{K}(\chi )=\mathrm{sinh}\chi \phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}(K=-1)$

and $\tau $ is the conformal time:

$\tau (t)={\int}_{0}^{t}\frac{cd{t}^{\prime}}{a({t}^{\prime})}$

The peculiar velocity is defined:

${v}_{\mathrm{p}\mathrm{e}\mathrm{c}}=a(t)\dot{\chi}(t)$

so the condition of zero peculiar velocity can be expressed:

$\dot{\chi}(t)=0\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\phantom{\rule{thickmathspace}{0ex}}t$

The "age of the Universe" of about $14\phantom{\rule{thickmathspace}{0ex}}\mathrm{G}\mathrm{y}\mathrm{r}$ you frequently hear about is a good approximation for any observer whose peculiar velocity is non-relativistic at all times. In practice these are the only observers we're interested in, since peculiar velocities for any bulk object (like galaxies) tend to be non-relativistic. If you happened to be interested in the time experienced by a relativistic particle since the beginning of the Universe, it wouldn't be terribly hard to calculate.

Note that here an "idealized" Universe described by the FLRW metric:

$\mathrm{d}{s}^{2}={a}^{2}(\tau )[\mathrm{d}{\tau}^{2}-\mathrm{d}{\chi}^{2}-{f}_{K}^{2}(\chi )(\mathrm{d}{\theta}^{2}+{\mathrm{sin}}^{2}\theta \phantom{\rule{thickmathspace}{0ex}}\mathrm{d}{\varphi}^{2})]$

where $a(\tau )$ is the "scale factor" and:

${f}_{K}(\chi )=\mathrm{sin}\chi \phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}(K=+1)$

${f}_{K}(\chi )=\chi \phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}(K=0)$

${f}_{K}(\chi )=\mathrm{sinh}\chi \phantom{\rule{thickmathspace}{0ex}}\mathrm{i}\mathrm{f}\phantom{\rule{thickmathspace}{0ex}}(K=-1)$

and $\tau $ is the conformal time:

$\tau (t)={\int}_{0}^{t}\frac{cd{t}^{\prime}}{a({t}^{\prime})}$

The peculiar velocity is defined:

${v}_{\mathrm{p}\mathrm{e}\mathrm{c}}=a(t)\dot{\chi}(t)$

so the condition of zero peculiar velocity can be expressed:

$\dot{\chi}(t)=0\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\phantom{\rule{thickmathspace}{0ex}}t$

The "age of the Universe" of about $14\phantom{\rule{thickmathspace}{0ex}}\mathrm{G}\mathrm{y}\mathrm{r}$ you frequently hear about is a good approximation for any observer whose peculiar velocity is non-relativistic at all times. In practice these are the only observers we're interested in, since peculiar velocities for any bulk object (like galaxies) tend to be non-relativistic. If you happened to be interested in the time experienced by a relativistic particle since the beginning of the Universe, it wouldn't be terribly hard to calculate.

asked 2022-05-09

Formula for the Bekenstein bound

$S\le \frac{2\pi kRE}{\hslash c}$

where $E$ is the total mass-energy. That seems to imply that the presence of a black hole in the region is dependent on an observer's frame of reference. Yet, my understanding is that the Bekenstein bound is the maximum entropy that any area can withstand before collapsing into a black hole.

Does this mean that the existence of black holes is observer dependent? Or that even if an observer does not report a black hole in their frame, one is guaranteed to form there in the future?

$S\le \frac{2\pi kRE}{\hslash c}$

where $E$ is the total mass-energy. That seems to imply that the presence of a black hole in the region is dependent on an observer's frame of reference. Yet, my understanding is that the Bekenstein bound is the maximum entropy that any area can withstand before collapsing into a black hole.

Does this mean that the existence of black holes is observer dependent? Or that even if an observer does not report a black hole in their frame, one is guaranteed to form there in the future?

asked 2022-08-14

According to the equivalence principle, gravity and inertial forces are similar. And according to general Relativity, If there’s a large object in space-time, it warps the space time’s geometry and causes the light to bend ( or to be seen as if it was coming from a different source). Now, if there’s an observer in an accelerating frame of reference and there’s no mass, will the observer still see the light to be coming from a different source considering the frame’s acceleration to have the same effect as that of a mass according to the principle of equivalence?

asked 2022-04-07

I was studying about potential energy, and I suddenly thought that is there any relation between potential energy and frame of reference?

For example, we say for an object raised to a height $h$, potential energy is $mgh$. But this is so when we are talking about its distance from Earth. If a person is holding a suitcase then for the person shouldn't be the P.E. of suitcase be zero?

For example, we say for an object raised to a height $h$, potential energy is $mgh$. But this is so when we are talking about its distance from Earth. If a person is holding a suitcase then for the person shouldn't be the P.E. of suitcase be zero?

asked 2022-09-27

How to solve this:

A ${\pi}^{+}$ decays into a muon and neutrino. Find the pion's energy if

$maxE\nu /minE\nu =100/1;$

${m}_{\nu}=0$

${m}_{\pi}\ast {c}^{2}=140\text{peta-eV}$

A ${\pi}^{+}$ decays into a muon and neutrino. Find the pion's energy if

$maxE\nu /minE\nu =100/1;$

${m}_{\nu}=0$

${m}_{\pi}\ast {c}^{2}=140\text{peta-eV}$

asked 2022-07-21

What reference frame is used to measure the velocity, $v=\frac{distance}{time}$?

$\mathbf{p}=\frac{mv}{\sqrt[2]{1-\frac{{v}^{2}}{{c}^{2}}}}$

Suppose the momentum of an electron is being observed in a lab. It could be observed in a lab that as the speed of electron increases, it becomes harder to accelerate it. To measure the velocity of electron, we need to measure distance or displacement and time.

$\mathbf{p}=\frac{mv}{\sqrt[2]{1-\frac{{v}^{2}}{{c}^{2}}}}$

Suppose the momentum of an electron is being observed in a lab. It could be observed in a lab that as the speed of electron increases, it becomes harder to accelerate it. To measure the velocity of electron, we need to measure distance or displacement and time.

asked 2022-08-16

Relativistic mass = ${m}_{0}\gamma $ where $\gamma $ is the lorentz factor.

So, if a mass that is $.5$ at rest then it is safe to say that the relativistic mass will be $1$ if it goes at $\frac{\sqrt{3}}{2}c$.

What happens if that $.5$ is actually a radioactive isotope and is decaying while speeding up? Then at what speed will it approximately equal $1$?

So, if a mass that is $.5$ at rest then it is safe to say that the relativistic mass will be $1$ if it goes at $\frac{\sqrt{3}}{2}c$.

What happens if that $.5$ is actually a radioactive isotope and is decaying while speeding up? Then at what speed will it approximately equal $1$?

asked 2022-10-08

Consider two identical charges moving with uniform velocity. There will be a magnetic force of attraction between them as two currents in the same direction attract each other. If I sit on one of the charges, according to me the other charge is not moving. So there wont be any magnetic attraction. How does changing the frame of reference change the outcome of the interaction?