# B Total % of Photons Redshifted While Moving Away From a Black Hole?

#### PeterDonis

Mentor
Is there still some disagreement about what the outcome would be?
Since nobody has actually done the math, everything we've said about the outcomes is just our best guess. To know for sure what the outcome is, as has already been pointed out, you need to do the math.

Will the relativistic aberration cause a higher and higher % of total photons to travel on vectors which increase in distance from the black hole over time
From the point of view of an observer "hovering" at rest relative to the hole, yes. Aberration depends on relative velocity.

will the photons become redshifted from gravitational redshift during their flight times between craft?
As above, you need to do the math and see if you want to know for sure. But intuitively, we can look at it from two points of view:

From the point of view of the ships, they are all in free fall and are all at rest relative to each other, so there is no such thing as "gravitational redshift". However, if their separation is large enough, spacetime curvature might be detectable between them during the time of flight of a particular photon, which would mean the logic of the previous sentence wouldn't be quite right (since it assumes that the entire fleet can be contained within a single local inertial frame during the flight time of a photon).

From the point of view of an observer "hovering" at rest relative to the hole, since all of the spaceships are in free fall, they are decelerating, so while a photon will have gravitational redshift as it travels upward, the ship that receives it will have decelerated during its flight, which will cause a compensating blueshift (which is an effect I didn't think of when I originally posted about gravitational redshift). However, without doing the math, we don't know for sure that those two effects exactly cancel (though it seems intuitively like they should).

#### metastable

From the point of view of an observer "hovering" at rest relative to the hole, since all of the spaceships are in free fall, they are decelerating, so while a photon will have gravitational redshift as it travels upward, the ship that receives it will have decelerated during its flight, which will cause a compensating blueshift (which is an effect I didn't think of when I originally posted about gravitational redshift). However, without doing the math, we don't know for sure that those two effects exactly cancel (though it seems intuitively like they should).
Before I do the math for any case in particular, can it be generally said that as the initial "upward" coasting co-moving velocity increases in various scenarios, the change in velocity between any 2 particular ships during the flight time of the photon will decrease?

#### Dale

Mentor
Is there still some disagreement about what the outcome would be?
You should work the math. Then you wouldn’t have to worry about agreement. I find it somewhat concerning that you are so reluctant to even attempt this calculation.

#### metastable

You should work the math. Then you wouldn’t have to worry about agreement. I find it somewhat concerning that you are so reluctant to even attempt this calculation.
I've tried something similar before:

Source: https://en.wikipedia.org/wiki/Relativistic_aberration

cos(angle observed to source) = (cos(motion path angle relative to the vector from the observer to the source at the time when the light is emitted)-(V/C))/(1-((V/C)*cos(motion path angle relative to the vector from the observer to the source at the time when the light is emitted)))

A = cos(motion path angle relative to the vector from the observer to the source at the time when the light is emitted)
B = V
C = C
D = cos(angle observed to source)

D = (A-(B/C))/(1-((B/C)*A))

A = -1*(((-1*B)-(D*C))/(C+(D*B)))

P = 299792457.608631081048 m/s = 10gev + 1ev electron velocity lab frame
Q = 299792457.608631080969 m/s = 10gev positron velocity lab frame
V = 299792457.6086310810085 = (P+Q)/2 = Velocity of Center of Mass observed from Lab Frame

A = cos(motion path angle relative to the vector from the observer to the source at the time when the light is emitted)
B = 299792457.6086310810085
C = C = 299792458m/s
D = cos(angle observed to source) = 0 = cos(90deg)

A = -1*(((-1*B)-(D*C))/(C+(D*B)))

A = -1*(((-1*299792457.6086310810085)-(0*299792458))/(299792458+(0*299792457.6086310810085)))

A = 0.9999999986945338064792 = cos(0.00292766)

0.00292766 degrees = motion path angle relative to the vector from the observer to the source at the time when the light is emitted

-----------------------------------------
Conclusions

The angle of the source motion path relative to the vector from the observer to the source at the time when the light is emitted is 0.00292766 degrees if the source momentum vector appears to be 90 degrees to the lab-frame-observed emission vector at the time of viewing from the lab frame

-------------------------------------

Verified:

#### PeterDonis

Mentor
can it be generally said that as the initial "upward" coasting co-moving velocity increases in various scenarios, the change in velocity between any 2 particular ships during the flight time of the photon will decrease?
No, because the deceleration due to gravity is independent of the outward speed.

#### metastable

No, because the deceleration due to gravity is independent of the outward speed.
I'm confused on this point. Doesn't it take a much greater amount of force (from the lab frame) for a given change in velocity when the particle is relativistic (from the lab frame), compared to much lower velocities (from the lab frame)?

#### PeterDonis

Mentor
Doesn't it take a much greater amount of force (from the lab frame) for a given change in velocity when the particle is relativistic (from the lab frame), compared to much lower velocities (from the lab frame)?
There is no force on any of the ships; they are in free fall.

Gravity is not a force in GR.

#### metastable

There is no force on any of the ships; they are in free fall.
What do we call the effect that causes 2 ships to change speed at different rates during the time of flight of a particular photon? Tidal gravity?

#### PeterDonis

Mentor
What do we call the effect that causes 2 ships to change speed at different rates during the time of flight of a particular photon? Tidal gravity?
If you mean ships at different altitudes, yes, the change in their relative speed is a manifestation of tidal gravity, i.e., spacetime curvature. This is an effect of spacetime geometry and does not affect the ships being in free fall.

#### metastable

For Illustration:

https://www.av8n.com/physics/spacetime-welcome.htm
"3.19.3 General Case

Now let’s consider a particle that is neither super-slow (slug) nor super-fast (photon). That is, the particle has some nonzero mass, but it is moving fast enough that the classical approximations do not apply. The situation is shown in figure 36. Here (as in other figures in this section), the red ring represents the speed of light. The pink disk serves as a reminder of what the velocity vectors were doing originally, when the blue frame was not moving relative to the red frame."

Figure 36: Aberration : Fast but Not Massless

Mentor

#### pervect

Staff Emeritus
Is there still some disagreement about what the outcome would be?
I haven't been following the thread but I'll make a few remarks about why I'm confused about what the question is.

Let me recap. In the scenario there is a black hole with a radius of the observable universe
This is oddly over-specific. Why is the radius of the black hole that of the observable universe? Are you imagining that this black hole is in our universe (which seems hard to analyze and also probably paradoxical), or in an idealized uiverse with nothing but the black hole? In the later case, you can omit talking about how big the black hole is, and just say you have "a black hole", or better yet, "a Schwarzschild black hole".

and a fleet of coasting, co-moving ships with a variety of heights above the blackhole and a variety of separation distances between them, but they are in a region of space roughly 2 black hole radii from the geometric center of the black hole
I think others have commented before that comoving is kind of hard to understand. It means something specific in the case of our universe, but it's unclear if you're talking about that or not, and whether or not you're using the standard understanding of what "comoving" means. If I felt positive I could follow your langauge to understand what it was you were asking, I might give it a shot, but at the moment I'm really not clear on what the question is, because of the way you describe it. So I can't tell what it is I would be agreeing with.

Anyway, the first step towards progress in my mind would be you modifying and/or expanding your question to use language that we both could understand. Of course, we both have to understand the question to avoid confusion. I'm just not sure how to get to that point where we both understand what the question actually is.

#### metastable

Why is the radius of the black hole that of the observable universe?
This parameter sets the scale of the gravitational redshift effects with respect to given distances and times traveled by the spacecraft.

Are you imagining that this black hole is in our universe (which seems hard to analyze and also probably paradoxical), or in an idealized uiverse with nothing but the black hole?
I'm imagining just outside the observable universe, but still within the total universe... & close & large enough to still observably affect events within the observable universe. For the idealized example the universe is simplified to just the black hole, the spacecraft, and the cosmological expansion parameter = 0 in the scenario.

I think others have commented before that comoving is kind of hard to understand
I mean comoving in the sense the vector of the spacecraft is extremely similar and the relative speed of the craft is insignificant compared to their speed with respect to the black hole. I am also "aware" of the separate astronomical definition of comoving in which all of the redshift is explained by cosmological expansion.

#### Dale

Mentor
I've tried something similar before:
I didn’t see any indications that you were using a black hole of the given mass or a pair of ships at any specified radius or any of the other pieces of this question. I am not sure in what sense you think that calculation was similar to the scenario you describe here.

#### metastable

I didn’t see any indications that you were using a black hole of the given mass or a pair of ships at any specified radius or any of the other pieces of this question. I am not sure in what sense you think that calculation was similar to the scenario you describe here.
You are correct, I haven't yet attempted solving those particular formulas, but I thought it was similar in that it was a relativistic aberration equation. The point was I don't shy away from combining and rearranging equations...

metastable said:
^peak mechanical power is 10746.218459832w @ 8082.063923094534641801 motor rpm

A = meters per second = XX.XXX
B = drag coefficient = 0.75
C = frontal area = 0.6m^2
D = fluid density of air = 1.225kg/m^3
E = wind drag force in watts
F = sine of 5% slope = sin(atan(5/100)) = 0.04993761694389223373491
G = acceleration of gravity = 9.80655m/s^2
H = vehicle mass in kg = 90.7184kg = 200lb / 2.20462lb/kg
I = mechanical watts required for constant speed up slope with no wind drag
J = mechanical watts required for constant speed up slope including wind drag
K = H * G * F
L = (1/2) * D * C * B

E = ((1/2) * D * C * (A^2) * B) * A

I = H * G * A * F

J = E + I

J = (((1/2) * D * C *(A^2) * B) * A) + (H * G * A * F)

J = (1/2) * D * C * B * A^3 + H * G * F * A

J = (L * A^3) + (K * A)

^this can be rearranged to:

A=(sqrt(3) * sqrt(27 * J^2 * L^4 + 4 * K^3 * L^3) + 9 * J * L^2)^(1 / 3) / (2^(1 / 3) * 3^(2 / 3) * L) - ((2 / 3)^(1 / 3) * K) / (sqrt(3) * sqrt(27 * J^2 * L^4 + 4 * K^3 * L^3) + 9 * J * L^2)^(1 / 3)

we know:

J = 10746.218459832w peak mechanical
L = 0.275625 = (1/2) * D * C * B
K = 44.42622815547907982077 = H * G * F

therefore:

A=(sqrt(3) * sqrt(27 * 10746.218459832^2 * 0.275625^4 + 4 * 44.42622815547907982077^3 * 0.275625^3) + 9 * 10746.218459832 * 0.275625^2)^(1 / 3) / (2^(1 / 3) * 3^(2 / 3) * 0.275625) - ((2 / 3)^(1 / 3) * 44.42622815547907982077) / (sqrt(3) * sqrt(27 * 10746.218459832^2 * 0.275625^4 + 4 * 44.42622815547907982077^3 * 0.275625^3) + 9 * 10746.218459832 * 0.275625^2)^(1 / 3)

A=32.32551993764664323864 meters per second

#### Dale

Mentor
The point was I don't shy away from combining and rearranging equations...
Ok. So why are you unwilling to work this specific problem?

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#### metastable

Ok. So why are you unwilling to work this specific problem?
I look forward to attempting it. I think it's analogous to the "how fast can I go up slope also factoring wind drag with a certain amount of mechanical power" problem -- ie one could choose to calculate the speed from JUST wind drag OR the slope forces, but in this problem both factors are combined. Analogously, in a given problem we could look at the velocity redshift or gravitational redshift, but here we have to look at both.

metastable said:
What is a "10gev + 1ev kinetic energy" electron's initial velocity in m/s from the lab frame?

10gev + 1ev kinetic electron (lab frame)-->

B = 0.5109989461MeV = electron rest mass

Z = 10000.000001MeV = initial electron kinetic energy (lab frame) = 10gev + 1ev

E = Electron Total Energy

E = B+Z

E = 10000.5109999461MeV

E = B/sqrt(1-(V^2/C^2))

can be rearranged to:

E = B/sqrt(1-A)

A = V^2/C^2

E = B/sqrt(1-A)

can be rearranged to:

A = -1*((B^2-E^2)/E^2)

A = -1*((0.5109989461^2-10000.5109999461^2)/10000.5109999461^2)

A = 0.9999999973890676149262

A = V^2/C^2

can be rearranged to:

V = C*sqrt(A)

C = 299792458m/s

V = 299792458*sqrt(0.9999999973890676149262)

V = 299792457.608631081048

10gev + 1ev Electron V=299792457.608631081048m/s from lab frame

%C = 299792457.608631081048 / 299792458

%C = 99.9999998694533806347%

10gev + 1ev Electron V = 99.9999998694533806611% C

10gev + 1ev Electron V = 0.999999998694533806611C

------------------------

Conclusion

Q: What is the 10gev + 1ev electron's initial velocity in m/s from the lab frame?

A: The 10gev + 1ev electron initially travels 299792457.608631081048m/s from the lab's rest frame, which is 0.999999998694533806611C.

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#### metastable

Will the relativistic aberration cause a higher and higher % of total photons to travel on vectors which increase in distance from the black hole over time, as the ships co-moving velocity increases in different scenarios?

In a given scenario (for example if 99.99999999999999% of total photons are moving away from the black hole), will the photons become redshifted from gravitational redshift during their flight times between craft?

^If yes, Will the measured gravitational redshift increase (as measured by an observer on one of the craft) as the separation distance between the craft increases in different scenarios (longer flight times between craft)?
A = cos(motion path angle relative to the vector from the observer to the source at the time when the light is emitted)
B = V
C = C
D = cos(angle observed to source)

D = (A-(B/C))/(1-((B/C)*A))

A = -1*(((-1*B)-(D*C))/(C+(D*B)))

A = cos(motion path angle relative to the vector from the observer to the source at the time when the light is emitted)
B = 299792457.6086310810085m/s
C = C = 299792458m/s
D = cos(angle observed to source) = 0 = cos(90deg)

A = -1*(((-1*B)-(D*C))/(C+(D*B)))

A = -1*(((-1*299792457.6086310810085)-(0*299792458))/(299792458+(0*299792457.6086310810085)))

A = 0.9999999986945338064792 = cos(0.00292766)

0.00292766 degrees = motion path angle relative to the vector from the observer to the source at the time when the light is emitted

-----------------------------------------
Conclusions

The angle of the source motion path relative to the vector from the observer to the source at the time when the light is emitted is 0.00292766 degrees if the source momentum vector appears to be 90 degrees to the lab-frame-observed emission vector at the time of viewing from the lab frame
Can I use:
Initial V=299792457.6086310810085m/s

A = cos(motion path angle relative to the vector from the observer to the source at the time when the light is emitted)
B = 299792457.6086310810085
C = C = 299792458m/s
D = cos(angle observed to source) = 0 = cos(90deg)

A = -1*(((-1*B)-(D*C))/(C+(D*B)))

A = -1*(((-1*299792457.6086310810085)-(0*299792458))/(299792458+(0*299792457.6086310810085)))

A = 0.9999999986945338064792 = cos(0.00292766)

0 degrees + 0.00292766 degrees = 0.00292766 degrees blueshifted towards black hole

180 degrees - 0.00292766 degrees = 179.99707234 degrees redshifted away from black hole

0.00292766 / 179.99707234 = 0.0016265% of photons gravitationally blueshifted when initial V = 299792457.6086310810085m/s

100 - 0.0016265 = 99.9983735% of photons gravitationally redshifted when initial V = 299792457.6086310810085m/s

#### metastable

0.00292766 / 179.99707234 = 0.0016265% of photons gravitationally blueshifted when initial V = 299792457.6086310810085m/s

100 - 0.0016265 = 99.9983735% of photons gravitationally redshifted when initial V = 299792457.6086310810085m/s
*correction:
0.00292766 / 180 = 0.00162647% of photons gravitationally blueshifted when initial V = 299792457.6086310810085m/s

100-0.00162647=99.99837353% of photons gravitationally redshifted when initial V = 299792457.6086310810085m/s

#### PeterDonis

Mentor
@metastable, please stop including huge quotes in your posts of things you're not going to respond to. Particularly if they're just repeats of your previous posts.

#### pervect

Staff Emeritus
This parameter sets the scale of the gravitational redshift effects with respect to given distances and times traveled by the spacecraft.
So, no really pressing reason, then?

I'm imagining just outside the observable universe, but still within the total universe... & close & large enough to still observably affect events within the observable universe. For the idealized example the universe is simplified to just the black hole, the spacecraft, and the cosmological expansion parameter = 0 in the scenario.\
The case that I can analyze doesn't have any "universe" other than a black hole. It's uncler if that's the case you're interested in.

I mean comoving in the sense the vector of the spacecraft is extremely similar and the relative speed of the craft is insignificant compared to their speed with respect to the black hole. I am also "aware" of the separate astronomical definition of comoving in which all of the redshift is explained by cosmological expansion.
That's still not very clear. It sounds like you could mean what I would call a static observer, but I am not at all confident.

A static observer would be one that has constant Schwarzschild r, theta, phi coordinates.

I see some references to "wind drag" and "relativistic aberration" in some of your posts that make no sense to me, so - I'm concluding that I don't understand your question.

We appear to lack a common vocabulary, and I don' see any sensible way to proceede without one.

#### metastable

So, no really pressing reason, then?
The black hole in the scenario is sized in such a way that the effects it causes in the scenario are still detectable while the black hole itself is completely outside the observable universe.

For Illustration (assume the fleet is a bit farther away from the circle in the middle than position A):

https://www.researchgate.net/figure/This-illustration-tries-to-show-schematically-a-hypersurface-at-time-T-with-our_fig4_313078586

"This illustration tries to show schematically a hypersurface at time T with our Observable Universe surrounded by other similar Observable Universes, arbitrarily positioned, some of them overlapping."

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#### Ibix

The black hole in the scenario is sized in such a way that the effects it causes in the scenario are still detectable while the black hole itself is completely outside the observable universe.
That would seem to be a contradiction in terms.

#### Ibix

So? You are describing something that's not observable (because it's not part of the observable universe), but is observable (because of its effects). So it's observable and not observable. I don't think you actually have a consistent scenario here.

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