Exploring the Effects of Black Hole Gravity: Blueshift

[Revolucien]

TL;DR Summary

gravity, blueshift, event horizon, photon, wavelength, frequency

Does the extreme gravity from a black hole cause a blueshift in light approaching the event horizon?

sorry for the newb question

 

[PeroK]

Summary:: gravity, blueshift, event horizon, photon, wavelength, frequency

Does the extreme gravity from a black hole cause a blueshift in light approaching the event horizon?

sorry for the newb question

Redshift and blueshift are a measure of the relationship between the source and the receiver. If you have a receiver near a black hole and a source further away, then the receiver will measure the light blue-shifted compared to the source.

 

[Ibix]

You have to be rather careful how you state questions like this. There are observers outside the horizon who see light blue shifted and observers who see it red shifted - it depends how they are moving. Typically of interest are observers hovering at constant altitude. These observers will see incoming light slightly blue shifted, yes.

However, you asked about the event horizon. Once you cross the horizon there are no hovering observers, and I'm not sure if there's any "obvious" group of observers to talk about. Some observers (probably most, to be honest) will see light from stars as blue shifted, but you can always find a state of motion in which the light would be redshifted.

 

[Revolucien]

I understand and thank you for the response, but was more curious about the if the intensity of the gravity would be enough to overcome any point of observation- source or receiver

 

[Ibix]

I understand and thank you for the response, but was more curious about the if the intensity of the gravity would be enough to overcome any point of observation- source or receiver

I have no idea what you are trying to ask. Can you restate the question?

 

[Revolucien]

does the gravity from the black hole cause a continued shortening wavelength so that even from the source it would appear to blueshift until it went beyond visible wavelengths

 

[PeroK]

I understand and thank you for the response, but was more curious about the if the intensity of the gravity would be enough to overcome any point of observation- source or receiver

Light itself does not change intrinsically. The only way you can talk about redshift and blueshift is in relation to sources and receivers. There is no other way to look at it.

If you are thinking of light being stretched or compressed, then that is a poor way to think about it.

Instead, the energy of the light is dependent on your reference frame. This creates the Doppler effect for relative motion between source and receiver. Likewise, the position and motion of source and observer in a gravitational field determines the relationship between their reference frames and whether the receiver measures a blue or red shift compared to the source.

 

[Ibix]

does the gravity from the black hole cause a continued shortening wavelength so that even from the source it would appear to blueshift until it went beyond visible wavelengths

How would the source see the light? It would have to get the light to bounce of a mirror or something and climb back away from the black hole, back to the source. And this would mean a gravitational red shift that exactly cancels the gravitational blue shift from the downward leg.

 

[PeroK]

does the gravity from the black hole cause a continued shortening wavelength so that even from the source it would appear to blueshift until it went beyond visible wavelengths

The source cannot measure light unless it is reflected back to it in some way. Measurement of the wavelegth of light is a local process. When we talk about the redshift from a distance galaxy, we actually only measure the light we receive. In this sense there is no such thing as redshifted light. There is only light of a measured wavelength. There is no way to measure that it has been redshifted.

But, if the light has a spectrum that looks like the spectrum of hydrogen, say, but redshifted by a certain amount then you may infer that the source of the light was hydrogen. But, you're not measuring this. You are inferring it from your single measurement of the light you receive.

 

[Revolucien]

Actually I just found this on Wikipedia and it kind of answers it "Matter waves (protons, electrons, photons, etc.) falling into a gravity well become more energetic and undergo observer-independent blueshifting "
I know your not supposed to trust Wikipedia but does that sound right that it would be an "observer-independent blueshifting"?

 

[PeroK]

Actually I just found this on Wikipedia and it kind of answers it "Matter waves (protons, electrons, photons, etc.) falling into a gravity well become more energetic and undergo observer-independent blueshifting "
I know your not supposed to trust Wikipedia but does that sound right that it would be an "observer-independent blueshifting"?

I can't find that phrase in either of the links you posted. Can you show us where it says that?

"Observer-independent energy" and "observer-independent blueshift" are a contradiction in terms.

 

[Revolucien]

https://en.wikipedia.org/wiki/Blueshift
The second header "Gravitational Blueshift" has an illustration of a Gravity Well to the right and that is the accompanying text.

 

[PeroK]

https://en.wikipedia.org/wiki/Blueshift
The second header "Gravitational Blueshift" has an illustration of a Gravity Well to the right and that is the accompanying text.

Maybe this is why Wikipedia is not a totally reliable source.

That's nonsense, I'm afraid. Energy is a measureable quantity. The only way you can give a number to the energy of a particle or the wavelength of light is to say how you'll measure it.

This is true for classical physics and relatively.

 

[PeroK]

https://en.wikipedia.org/wiki/Blueshift
The second header "Gravitational Blueshift" has an illustration of a Gravity Well to the right and that is the accompanying text.

There is a good point here. Some people like to see "velocity-based" redshift and "gravity-based" redshift as different in some fundamental way. But, actually there is only the wavelength at the source and the wavelength at the receiver. That's it. Any breakdown of the redshift into this much is due to velocity and this much is due to gravity is arbitrary (coordinate dependent)

There are a few threads on here about this. Try here and look at the posts by @Orodruin and @kimbyd:

https://www.physicsforums.com/threads/cosmological-redshift.935943/

 

[Ibix]

Any breakdown of the redshift into this much is due to velocity and this much is due to gravity is arbitrary (coordinate dependent)

To be fair, in a stationary spacetime like that outside a black hole there is a way to do this breakdown in a way that has some kind of physical significance. You can pick hovering observers (formally, observers whose worldlines are integral curves of the timelike Killing vector field), who can make a reasonable claim to being stationary with respect to one another. Then any Doppler shift in signals between such observers must be due to gravitational red or blue shift.

Clearly that's not observer independent, though, since I invoked a particular family of obsevers in the definition.

 

[Revolucien]

Well I appreciate that and the reference threads, and I do not trust what's on Wiki either. Hence why I'm in here... I guess I'm a little curious because the way it seems to me (and I may be so far off it's not funny) is that the gravitational effect would increase the frequency of any visible light until it was no longer visible and I'm guessing that's what they mean by "no light escapes a black hole". So if radiation below the visible frequency were approaching the EH, would the gravitational effect cause frequency increases into the visible frequencies so that you could see flashes of visible light?

 

[PeroK]

Well I appreciate that and the reference threads, and I do not trust what's on Wiki either. Hence why I'm in here... I guess I'm a little curious because the way it seems to me (and I may be so far off it's not funny) is that the gravitational effect would increase the frequency of any visible light until it was no longer visible and I'm guessing that's what they mean by "no light escapes a black hole". So if radiation below the visible frequency were approaching the EH, would the gravitational effect cause frequency increases into the visible frequencies so that you could see flashes of visible light?

Light inside the EH simply doesn't get out. There are no light paths that cross the EH outwards.

Light emitted from outside the EH may eventually become undetectable but that's to an observer who is sufficiently far away. You can do the same with an accelerating observer moving away from a source. As you increase the starting distance the light gets fainter and more red shifted when it finally catches the receding observer. And, after a critical starting distance it never catches up.

 

[jartsa]

Let's say I start shining a laser pointer down towards a black hole. One meter above the event horizon there is a hovering mirror that reflects the light back. When I see the reflected light I turn the laser pointer off.

Now I calculate that the total number of produced waves was 10²⁰. And all those waves were in the space between me and the mirror when the first wave returned.

Next the experiment is repeated, but this time the mirror is one micrometer above the event horizon. This time 10²³ waves can be calculated having been between me and the mirror.

Well, it seems like in the latter experiment thousand times more waves were packed between me and the mirror compared to the first experiment.

(It took thousad times longer time for the light to return in the second experiment, because of Shapiro delay)

 

[Revolucien]

Thanks, I really appreciate the information from everyone!
I will stop back later because I have some questions about Photons (supposedly) having zero mass at rest and how we achieve an accurate rest frame when we can only base "rest" off relative motion.
Since we have no reference of a non-moving object in space, how can we get a actual true assessment of velocity and gravity effects? Those will be for a later time, I have to go do some more research first before broaching those topics. Thanks again.

 

[Orodruin]

To underline several issues that have been discussed in this thread:

Consider an observer falling radially into a Schwarzschild black hole from infinity. As that observer passes the event horizon, light signals sent to that observer from infinity will be red-shifted by a factor of two (i.e., not blue-shifted).

 

[PeroK]

Thanks, I really appreciate the information from everyone!
I will stop back later because I have some questions about Photons (supposedly) having zero mass at rest and how we achieve an accurate rest frame when we can only base "rest" off relative motion.
Since we have no reference of a non-moving object in space, how can we get a actual true assessment of velocity and gravity effects? Those will be for a later time, I have to go do some more research first before broaching those topics. Thanks again.

Strictly speaking it's better to say that the rest mass of a photon is undefined or that the photon is massless. Instead, you can define a quantity called the invariant mass, as follows. A key formula in SR is:
$$E^2 = p^2c^2 + m^2c^4$$
$E$ is the total energy of a particle and $p$ is the magnitude of its three-momentum. This formula holds for particles with mass $m$ and also for photons by setting $m = 0$, where the formula reduces to $E = pc$.

(Note that the above formula is a generalisation of the more famous $E = mc^2$, which is in fact not valid for massless particles.)

This allows us to define the invariant mass of any particle as:
$$m = \frac 1 {c^2}\sqrt{E^2 - p^2c^2}$$
Which gives the invariant mass of a photon as zero. And, for a single particle of rest mass $m$, the invariant mass is also $m$.

 

[Revolucien]

Strictly speaking it's better to say that the rest mass of a photon is undefined or that the photon is massless. Instead, you can define a quantity called the invariant mass, as follows. A key formula in SR is:
$$E^2 = p^2c^2 + m^2c^4$$
$E$ is the total energy of a particle and $p$ is the magnitude of its three-momentum. This formula holds for particles with mass $m$ and also for photons by setting $m = 0$, where the formula reduces to $E = pc$.

(Note that the above formula is a generalisation of the more famous $E = mc^2$, which is in fact not valid for massless particles.)

This allows us to define the invariant mass of any particle as:
$$m = \frac 1 {c^2}\sqrt{E^2 - p^2c^2}$$
Which gives the invariant mass of a photon as zero. And, for a single particle of rest mass $m$, the invariant mass is also $m$.

Yeah,
Something is just bugging me about the photon being considered massless, but I want to read a little more to get a better understanding about Planck Mass first. Thanks though.

 

[Revolucien]

To underline several issues that have been discussed in this thread:

Consider an observer falling radially into a Schwarzschild black hole from infinity. As that observer passes the event horizon, light signals sent to that observer from infinity will be red-shifted by a factor of two (i.e., not blue-shifted).

Orodruin,
Although originally I was only asking about the approach to EH, your statement is very interesting. I would definitely like to look at what happens after passing the EH, but I'm going to guess that we will need to figure out how to modulate a carrier that is unaffected by gravity to pass some data back from the opposite side of the EH to get any truth.

Let's say I start shining a laser pointer down towards a black hole. One meter above the event horizon there is a hovering mirror that reflects the light back. When I see the reflected light I turn the laser pointer off.

Now I calculate that the total number of produced waves was 10²⁰. And all those waves were in the space between me and the mirror when the first wave returned.

Next the experiment is repeated, but this time the mirror is one micrometer above the event horizon. This time 10²³ waves can be calculated having been between me and the mirror.

Well, it seems like in the latter experiment thousand times more waves were packed between me and the mirror compared to the first experiment.

(It took thousad times longer time for the light to return in the second experiment, because of Shapiro delay)

Jartsa,
I really like the way you explained that because it does shed light (pun intended) on what I was contemplating. I was trying to understand why a black hole was black- beyond the statement that I have heard so many times- "light cannot escape the gravity of a black hole".
If all photonic radiation approaching the event horizon at any frequency has that frequency increased through the immense gravity excitation, then two things are what I expect may be happening-
1. any photonic radiation is increased in frequency/decreased in wavelength beyond the visible spectrum by gravity excitation and that is a more fun explanation (at least for me to understand) why a black hole is black.

2. and if we were able to hover above the EH and facing outward any wavelength previously too large to be visible would decrease as through the visible spectrum and then beyond visible as it approached causing, however so brief, a possible flash of visible light.

Sorry if this is all rudimentary for you guys or if I have stepped on rules or laws to come to these thoughts.
I did post this a couple minutes ago but it somehow disappeared, apologies if this comes up twice.

 

[PeroK]

I was trying to understand why a black hole was black- beyond the statement that I have heard so many times- "light cannot escape the gravity of a black hole".

The simple answer is that all particle and light paths that start from inside the EH stay inside the EH. It's nothing to do with "gravity" as a force or an an acceleration.

Try this PBS video:

 

[PeroK]

And this one:

 

[PeterDonis]

we will need to figure out how to modulate a carrier that is unaffected by gravity

There is no such thing. "Gravity" is just spacetime geometry. Everything is affected by spacetime geometry.

 

[PeterDonis]

Try this PBS video

The title is a big red flag: "How Time Becomes Space Inside A Black Hole" is highly misleading. The video itself confirms the red flag, since it talks about the "switching" in question is an artifact of Schwarzschild coordinates and has nothing to do with the actual physics.

And this one

This one isn't much better; it talks about "density" of a black hole, which is meaningless. Also, it describes the standard GR viewpoint as "the official sanitized version". Then it mentions "black hole rotational growth", which makes no sense. The Penrose diagram, which the video goes into, is a valid tool, yes, but I think there are better ways of learning it.

I do not think either of these videos is a good source for learning.

 

[Orodruin]

The title is a big red flag: "How Time Becomes Space Inside A Black Hole" is highly misleading. The video itself confirms the red flag, since it talks about the "switching" in question is an artifact of Schwarzschild coordinates and has nothing to do with the actual physics.
This one isn't much better; it talks about "density" of a black hole, which is meaningless. Also, it describes the standard GR viewpoint as "the official sanitized version". Then it mentions "black hole rotational growth", which makes no sense. The Penrose diagram, which the video goes into, is a valid tool, yes, but I think there are better ways of learning it.

I do not think either of these videos is a good source for learning.

In the time it took me to look at the videos, somebody else already wrote my reply ...

 

[jartsa]

I was trying to understand why a black hole was black- beyond the statement that I have heard so many times- "light cannot escape the gravity of a black hole".
If all photonic radiation approaching the event horizon at any frequency has that frequency increased through the immense gravity excitation, then two things are what I expect may be happening-
1. any photonic radiation is increased in frequency/decreased in wavelength beyond the visible spectrum by gravity excitation and that is a more fun explanation (at least for me to understand) why a black hole is black.

2. and if we were able to hover above the EH and facing outward any wavelength previously too large to be visible would decrease as through the visible spectrum and then beyond visible as it approached causing, however so brief, a possible flash of visible light.

A normal human eye can't detect the EM-waves sent by a normal cell phone. But if the eye is far below the phone in a gravity well, then it can.

Or, you can't melt ton of steel in one second by the infrared radiation coming off of your hand, but if the chunk of steel is far below the hand, then you can.

Hmm ... the radiation from the hand or from the phone could drill a small hole on a block of steel too ... No, sorry I got confused, gravity may collimate an expanding wave, then if there is a lens, then the collimated wave may be focused on a small area. Are we interested about how much a smallish black hole between two radio antennas blocks transmissions when very long wavelengths are used?

 

[PeterDonis]

A normal human eye can't detect the EM-waves sent by a normal cell phone. But if the eye is far below the phone in a gravity well, then it can.

Or, you can't melt ton of steel in one second by the infrared radiation coming of your hand, but if the chunk of steel is far below the hand, then you can.

Note that "a gravity well" in general is not enough; it has to be the gravity well of a black hole. No other object allows you to get to altitudes that allow such large gravitational blueshifts will still remaining outside the object.

Also, to hover that close to a black hole's horizon, you need a rocket with an extremely powerful engine and lots and lots of fuel, since the rocket thrust required will be enormous. The effects you describe are best viewed as due to the enormous rocket thrust, not due to the radiation itself.

 

[jartsa]

Also, to hover that close to a black hole's horizon, you need a rocket with an extremely powerful engine and lots and lots of fuel, since the rocket thrust required will be enormous. The effects you describe are best viewed as due to the enormous rocket thrust, not due to the radiation itself.

Oh yes, if a spaceship is equipped with a ridiculously powerful engine, turning on a light bulb on the ceiling will cause a filament shaped hole to appear on the floor.

 

[PeterDonis]

if a spaceship is equipped with a ridiculously powerful engine, turning on a light bulb on the ceiling will cause a filament shaped hole to appear on the floor.

If the ship's acceleration is large enough relative to its height that the blueshift from ceiling to floor is that large, yes. But that is a much more stringent condition than what is required for the ship to hover close enough to a black hole that radiation coming in from far away is blueshifted to that extent. A ship in the latter situation could still have negligible blueshift over the range of its own height.

 

[Revolucien]

Let's say I start shining a laser pointer down towards a black hole. One meter above the event horizon there is a hovering mirror that reflects the light back. When I see the reflected light I turn the laser pointer off.

Now I calculate that the total number of produced waves was 10²⁰. And all those waves were in the space between me and the mirror when the first wave returned.

Next the experiment is repeated, but this time the mirror is one micrometer above the event horizon. This time 10²³ waves can be calculated having been between me and the mirror.

Well, it seems like in the latter experiment thousand times more waves were packed between me and the mirror compared to the first experiment.

(It took thousad times longer time for the light to return in the second experiment, because of Shapiro delay)

Jartsa,
Your experiment got me thinking and curious about what happens if the mirror is now 1 micro meter past the EH? Is the EH the critical point where space time dilation increases the path length beyond the photons ability to return?

 

[PeroK]

Jartsa,
Your experiment got me thinking and curious about what happens if the mirror is now 1 micro meter past the EH? Is the EH the critical point where space time dilation increases the path length beyond the photons ability to return?

You can't have a mirror hovering below the event horizon. Anything below the EH follows a path to the singularity.

Instead, you could think about someone falling through the EH and shining light in all directions. At the EH, light shone radially outward would theoretically hover at the EH indefinitely. Light in all other directions would follow a path inward towards the singularity. And, once below the EH, all light paths lead to the singularity. There are no light paths starting from inside the EH that cross the event horizon.

 

[timmdeeg]

I would like to add that a photon emitted inside the black hole upward "falls" towards the singularity. So even emitted upward it's r-coordinate decreases.

EDIT I just recognized @PeroK said the same: "And, once below the EH, all light paths lead to the singularity. "