# Tag Info

24

The picture isn't a "colour" picture - it is monochrome. i.e. It is obtained at a single microwave wavelength of 1.3 mm, and so not at any wavelength you could see (Akiyama et al. 2019). There isn't therefore any spectral information that would reveal the expected Doppler effect. Any difference of colour in the "false-colour picture" is ...

13

It's spread out in time. If a source emits a 1W pule of energy lasting 1 second and the receiver is receding so fast that it is Doppler shifted to a frequency which means the power is just 0.5W, then the pulse will take 2seconds to arrive (since the end of it had further to travel).

13

The colour of a star is due mostly to its temperature. Except at extreme velocities, the red-shift won't affect the light enough to change the visible colour. To detect a red or blue shift you need some marker in the spectrum that can be measured accurately. The spectrum of light from a star will contain dark lines, called Fraunhofer lines, which are due ...

12

Completely ignoring relativistic effects, it depends on which reference frame you are using; the "missing" energy is seen as kinetic energy in either the emitting or receiving atom as recoil depending on which one you view as moving. Energy is not conserved between reference frames. If I am travelling at a velocity away from you, and I emit a photon at you ...

9

After considering @benrg's comments, I realize that my first answer contained too strong statements about the relation between the two redshifts. I try here to moderate my answer, but you might want to accept their answer instead. It is common to think of the two redshifts as having nothing to do with each other. Doppler shifts arise when the observer and/...

9

If you had a simple slit spectroscope, and looked at an incandescent light, you'd see a smear of light with red on one end and blue on the other. This is because the filament is producing light by glowing from being heated.If you looked at one of those orange colored sodium vapor street lamps, instead of a smear of color, you'd see a group of lines. This ...

7

Without any other information, you cannot distinguish between the two effects. $$T = T_0 (1 + z)$$ A blackbody spectrum of temperature $T$ is identical to a blackbody spectrum of temperature $T_0$ with redshift $z$. For stellar/galactic radiation, we can use the fact that the radiation is not a perfect blackbody. For the CMB, we can use the fact that ...

6

So GN z-11 is the latest "furthest away" galaxy. It has a claimed redshift measurement of $z=11.1$, meaning that we are seeing the light it emitted about 400 Myr after the big bang (dependent on an assumed set of cosmological parameters). To get the redshift measurement, the discoverers used grism (relatively low resolution) spectroscopy in the near ...

6

Accounting for the transverse Doppler effect (and other relativistic effects) is essential in modelling the X-ray spectral emission lines from the accretion discs around black holes (e.g. Cadaz & Calvani 2005). In this case the transverse Doppler effect is "mixed up" with gravitational redshift and it is treated holistically in the Schwarzschild or Kerr ...

5

If stars were actually about to escape, they would do so on a so-called dynamical timescale: $$t_\mathrm{dyn} \sim \sqrt{\frac{R^3}{8GM}},$$ where $G$ is the gravitational constant, and $R$ and $M$ are the radius and the mass of the galaxy, respectively. For a Milky Way-sized galaxy, this is equates to roughly 30 million years$^\dagger$. But disk galaxies ...

5

Ultimately this is an application of Occam's razor In order for the light to be gravitationally redshifted, it would have to be coming out of a deep gravitational well. For the red shift observed in galaxies to be gravitational, you would have to suppose several things. First, that the stars in distant galaxies are somehow much denser: more than neutron ...

5

Conrad is almost right. It is true generally that if a Galaxy is close enough to take spectra of individual stars (e.g. luminous supergiants) then it is not far enough away to be regarded as part of the "Hubble flow" and so applying Hubble's law to this star, or its host galaxy, would not yield a reliable distance in any case, but would reflect the "peculiar ...

4

Before I start, let me just say that this topic is vastly more complicated than you've presented and what I will be showing. The trouble here is that ultimately, everything you've done and I will do (to a lesser extent) uses approximations and assumptions. Because of this, it can sometimes be hard to understand the true underlying physics when looking at ...

4

Yes, it is absolutely silly to predict an atmospheric/surface temperature without knowing it's composition or density. And we don't understand the formation process of atmospheres well enough to exclude any type of atmospheres around different stars. There is however one more or less objective quantity that you can calculate: The equilibrium temperature. ...

4

The two effects are observationally indistinguishable for an individual galaxy. The redshift associated with universal expansion is often interpreted as a recession velocity and there is nothing in a spectrum that tells you what fraction of the redshift is contributed by which effect. If a distant galaxy is associated with a cluster of galaxies, then you can ...

4

Yes, there is a transverse relativistic Doppler shift. You can think of it as being caused by time dilation. https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect There can be a redshift or a blueshift depending on when, where and who does the measurement. e.g. a receiver with a source going around it in a circular orbit. The receiver sees a lower ...

3

If the object you're observing is something good and hot, like a star, it's pretty easy to identify a set of Ballmer lines by the relative spacing between the lines. Then compare the absolute wavelengths to the known stationary values and viola you've got the Doppler shift value. So this would "technically" be using the blackbody radiation from the ...

3

This seems unintuitive, but the Universe doesn't appear to have a center for galaxies to move away from. This is because the Big Bang seems not to be a single event at a particular place, but instead an event at every place. In other words, the Big Bang seems to have happened at every point in the Universe, possibly infinitely in all directions (note that "...

3

There are lots of unknowns however, you have a star that is more massive than the sun. More massive stars (on the main sequence) are brighter than smaller ones. So the star is brighter than the sun. You also have a planet that is closer to the star than the Earth is to the sun. Both these effects would suggest that the planet would be warmer than the Earth. ...

2

We measure the aggregated red-shift of galaxies not single stars, which averages out the motion of the stars within the galaxy. This leaves us with the proper motion of the galaxy to deal with, which can be estimated from the red-shifts of other galaxies in the same group/cluster/super-cluster and their distribution. But this is all moot at high red-shift ...

2

Your understanding is correct. The doppler shift observed from a galaxy is the sum of its peculiar velocity with respect to the "Hubble flow" and the redshift due to the Hubble flow, which is caused by the expansion of the universe. There is no direct way from a spectrum to separate these two components - they have the same qualitative result. In principle,...

2

In the past, the universe was very hot and dense, and we are seeing the light from this time period as the Cosmic Microwave Background. This statement is (partially) correct. However.... Since gravity apparently travels at the speed of light, we should be experiencing a gravitational effect from this shell around us as well. As I understand, the ...

2

You can start by working out how fast a galaxy must be receding if its peculiar velocity (given as about 600 km/s) is less than or equal to 10% of its expansion velocity. I'm not going to spell out the next step as you said you don't know where to start.

2

There's only one kind of redshift in general relativity. The cosmological redshift, gravitational redshift, and special-relativistic redshift formulas are special cases of it, which apply to spacetimes with certain symmetries. If you put approximate Minkowski coordinates on a patch of spacetime that's small enough to be approximately flat, you'll find that ...

2

Since an object moving perpendicular to a given "line of sight" has a constantly changing range , there is a Doppler shift, blue when approaching and red when leaving. The shift drops to zero at the point of crossing the line of sight because at that instant the radial speed is zero, as you suggested. So, the general magnitude is calculated using ...

1

Your intuition is correct - a moving source emitting wavefronts periodically will be closer to the previously emitted wave in the direction of motion, and farther from the previously emitted wave in the opposite direction - see the simulation here. You are also correct that the size of the effect depends on the speed of the observer relative to the speed of ...

1

The term "Hubble flow" refers to the homologous expansion of space and the resulting recession of all galaxies from each other (if they're not close enough to be gravitationally bound). This effect causes the "cosmological redshift", i.e. the redshift that light from distant galaxies attain as it travels through space. In addition to this ...

1

To use the Doppler shift, you need to know by how much the light is blue/red-shifted. Suppose you have a light curve where the wavelength varies between $\lambda_1$ and $\lambda_2$. Using these wavelengths, the Doppler shift will give you an interval of radial velocities. This gives you a value for K that you can plug into your last equation to derive $M_p$...

1

You don't need to collect data for one cycle in order to determine the period. From Newtonian's physics, we know precisely how a trajectory of one particle orbitting around another one would be. Since the instantaneous velocity (v_inst) at any point on the trajectory is tangential to the point on the trajectory, we can decompose the v_inst into the radial ...

1

Conservation of energy doesn't apply to this situation because the energy you measure when at rest with respect to the source and the energy you measure when moving with respect to the source are in different reference frames. Energy is not conserved between different reference frames; in other words, if you're going to use conservation of energy, you have ...

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