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The de Broglie wavelength of a massive particle is redshifted in an expanding universe. The de Broglie wavelength is given by: $$\lambda = \frac{h}{p}$$ so a red shift of the de Broglie wavelength simply means that the momentum is decreasing, which for a massive particle means that its velocity relative to us is decreasing. And that is exactly what we ...

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The cosmologial redshift only requires expansion, not inflation. There is no way to distinguish just from observing the light of a single object how much of the shift is cosmological, gravitational or Doppler (peculiar motion). However, given that in all directions the red shift is increasing rather uniformly with distance, expansion seems like the only ...

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The cosmic background radiation has a wavelength about 2000 times longer than visible light. So you would need to be sitting deep in a gravitational well such that local time progresses 2000 times more slowly than distant time. Gravitational time dilation is given by the formula $\sqrt{1-r_\mathrm{S}/r}$ where $r$ is the distance from the center of the (...

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Well, the singularity does not concern the differentiable structure: Even around the tip of a cone (including the tip) you can define a smooth differentiable structure (obviously this smooth structure cannot be induced by the natural one in $R^3$ when the cone is viewed as embedded in $R^3$). Here the singularity is metrical however! Consider a $2D$ smooth ...

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Yes the effect is real, potentially at least, but no it's not measurable. As an aside, the redshift of light by its gravitational interaction with (homogeneous and isotropic) dust is exactly what the FLRW metric predicts, but this clearly isn't what the question means. The gravitational field of a beam of light is calculated in the paper On the ...

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The answer to your question is yes, but the gravitational redshift could not be confused with cosmological redshift because it is small. The basics of gravitational redshift can be grasped by the simple approximation that the redshift is given by $$z \simeq \Delta \Phi /c^2,$$ where $\Delta \Phi$ is the difference in gravitational potential between where ...

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Yes, you're quite correct that galaxies in a region of the universe with greater than average density would appear redshifted to observers outside that region. In fact this is the origin of the Sachs-Wolfe effect, which is an important way that we study various properties of the universe. Actually doing the calculation is hard because of course the ...

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A recent test (2018) using atomic clocks aboard two satellites found general relativity's gravitational redshift prediction to be accurate to $(+0.19 \pm 2.48)\times10^{-5}$. This was not a planned experiment and the satellites had accidentally been delivered on elliptic rather than circular orbits in 2014. They were useless for their original purpose of ...

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We know that some galaxies are moving away from us faster than the speed of light and we know it by measuring the redshift, but how's that possible? The following papers give good explanations: http://users.etown.edu/s/stuckeym/AJP1992a.pdf http://arxiv.org/pdf/astro-ph/0011070v2.pdf In summary, Hubble Law: $v = H(t)D$, where $v$ is recession velocity, $... 10 Stellar motion would imply we're in the place where the Big Bang occurred while everything else is speeding away from us. I can see how this would be a worthwile explanation for a creationist, but it doesn't really make much sense building just on physics. Doppler Effect has the same issue - it assumes that every star we can see moves in a speed ... 10 Update: According to this paper, "On the Interpretation of the Redshift in a Static Gravitational Field", the answer I give below is a common but misleading interpretation. The classical phenomenon of the redshift of light in a static gravitational potential, usually called the gravitational redshift, is described in the literature essentially in two ... 9 I posted more-or-less the same thing in response to another query about whether "tired light" still holds any water. I don't think there is any doubt that the redshifts we see in the spectra of distant objects are real doppler shifts that can be explained by an expanding universe. A crucial piece of evidence that seems to be ignored by almost everyone ... 8 SN 1987A is (was?) in the Large Magellanic Cloud, which is gravitationally bound to the Milky Way. This means that its motion relative to us is only minimally affected by cosmological expansion, and talking about it in terms of a$z$parameter is misleading at best. The best estimates of the distance to SN 1987A are about 168,000 light-years. If you ... 8 Gravitation clearly can change wavelength and frequency, and it does that for instance for the cosmological redshift. But with the speed of light being$c$locally, independent of frequency or anything else -- note: independence of frequency or wavelength$k$means that it is not dispersive, since dispersion is due to different wavelengths traveling at ... 7 It's true that stars can be very different but the spectral lines are the same in all of them, though the relative intensities of the lines will differ from star to star. So if for example you're measuring the Lyman$\alpha$line then every star in Andromeda will have this line at the same frequency. This allows you to measure an average position for this ... 7 My answer assumes that by "at the center", you mean orbiting the supermassive black hole (SMBH) at the centre. The amount of time dilation or gravitational redshift (as measured by an observer at infinity) depends on how deeply into a gravitational potential well the clock or light source is. From the point of view of various observers that are in the ... 7 The meaning of the coordinate t is much broader than that. In fact we might want to talk about events that never send or receive signals from infinity . After all signals to and from infinity will literally take infinite time to “get there”. The coordinate t in a stationary spacetimes is often chosen to be adapted to the ... 6 Gravitational redshift is extremely small for stars, it has no significant impact on cosmic redshift measurements of galaxies. The gravitational redshift of light emitted by a star is of the form $$z = \frac{1}{\sqrt{1 - \frac{GM}{Rc^2}}} - 1 \approx \frac{GM}{2Rc^2},$$ where$R$is (approximately) the radius of the star. For the Sun, this is of the order$...

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Another line of thought is that there are other observations that support the notion of an expanding universe - it does not all hinge on redshift. For instance, I believe numbers 3 onwards would struggle to explain why the light curves of otherwise similar supernovae become stretched (at a variety of wavelengths) at larger distances from us and by an amount ...

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There are two effects that will change the frequency of the light: the gravitational time dilation the Doppler shift due to the velocity of the infalling observer To a first approximation we can simply multiply these together to work out what the falling observer sees. The gravitational time dilation for a freely falling observer is calculated in my ...

5

Here are some ideas re your question: Let us consider the path taken by the torch in the presence of a black hole, and let us assume the observer is outside the horizon. For the sake of simplicity let us assume the light source (the torch) is falling in a rectilinear way. Some algebra and quite a bit of physics, combining the principle of equivalence and ...

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The answer is yes they can but the effect is trivially small. Observed gravitational redshift just depends on the difference in gravitational potential between source and observer. Roughly, $$z \sim \frac{\Delta \Phi}{c^2}.$$ Density is is only indirectly involved in the sense that high density regions can produce large gravitational potentials. In ...

5

I'm not sure I would agree the S2 observations are the best test of GR, or even the most rigorous test. However they are a test in a regime that we have not been able to directly probe before. The easy tests of GR are all in the weak field limit. For example we can measure gravitational red shift and time dilation on Earth, and we can measure perihelion ...

4

Gravitational red shift is due to the energy of a photon in a gravitational field. So the photon needs energy to escape the gravitational force and therefore the wavelength increases (redshift). The cosmological redshift is due to the expansion of the universe and thus is not connected to a gravtiational field. The wavelength of the photon stretches like ...

4

Bremsstrahlung (yes, that's the correct english term) is only emitted if a charged particle is deflected - as a photon has no charge it won't ever emit bremsstrahlung. Another way to see that a photon cannot simply emit another photon is by conservation of energy-momentum. If the process $\gamma \rightarrow \gamma +\gamma$ were possible, one should be able ...

4

Light from beyond the Hubble sphere (the place where recession velocity equals the speed of light) reaches us daily. I'm not good enough a physicist to come up with a nice layman's explanation for this fact, but it might help to think in comoving coordinates: This is a special coordinate system where the coordinate grid expands with space, ie even though ...

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I had a quick look at the paper - it's mostly nonsense. The intrinsic light from a quasar is completely dominated by its emission line spectrum and a mostly featureless continuum. The observed wavelength of the emission lines compared to the rest wavelength give the true redshift of the quasar. Absorption lines in quasar spectra are predominantly due to ...

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