B Particle Deceleration in Relativistic Jets?

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I have a question about this picture:

It shows matter traveling at relativistic velocities away from a black hole:

x313.jpg

"Figure 5.21: This VLA image of the radio-loud quasar 3C 175 shows the core, an apparently one-sided jet, and two radio lobes with hot spots of comparable flux densities. The jet is intrinsically two sided but relativistic, so Doppler boosting brightens the approaching jet and dims the receding jet. Both lobes and their hot spots are comparably bright and thus are not moving relativistically. Image credit: NRAO/AUI/NSF Investigators: Alan Bridle, David Hough, Colin Lonsdale, Jack Burns, & Robert Laing."

I'm confused by the image, because in a different thread it was stated:
the deceleration due to gravity is independent of the outward speed.
If the deceleration due to gravity is independent of the outward speed, how is it thought that the particles that have relativistic velocity with respect to the black hole retain a velocity which is so close to the speed of light after traveling many so many light years "upwards" away from the super massive black hole?
 

Ibix

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Qualitatively speaking, a small velocity less a certain amount of deceleration is a small velocity. A large velocity less that same amount of deceleration is a large velocity. The jets started with an enormous velocity.

There are subtleties around the behaviour of relativistic particles (velocity isn't well defined in the casual way you are using it), but that's the gist of it.
 
The jets started with an enormous velocity.

There are subtleties around the behaviour of relativistic particles (velocity isn't well defined in the casual way you are using it), but that's the gist of it.
10gev kinetic electron (hovering observer)-->

B = 0.5109989461MeV = electron rest mass

Z = 10000MeV = initial electron kinetic energy (lab frame) = 10gev

E = Electron Total Energy

E = B+Z

E = 10000.5109989461MeV

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*((0.5109989461^2-10000.5109989461^2)/10000.5109989461^2)

A = 0.999999997389067614404

A = V^2/C^2

can be rearranged to:

V = C*sqrt(A)

C = 299792458m/s

V = 299792458*sqrt(0.999999997389067614404)

V = 299792457.6086310809697

10gev Electron V=299792457.6086310809697m/s from hovering observer

-----

1000gev kinetic electron (hovering observer)-->

B = 0.5109989461MeV = electron rest mass

Z = 1000000MeV = initial electron kinetic energy (lab frame) = 1000gev

E = Electron Total Energy

E = B+Z

E = 1000000.5109989461MeV

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*((0.5109989461^2-1000000.5109989461^2)/1000000.5109989461^2)

A = 0.999999999999738880344

A = V^2/C^2

can be rearranged to:

V = C*sqrt(A)

C = 299792458m/s

V = 299792458*sqrt(0.999999999999738880344)

V = 299792457.9999608591483

1000gev Electron V=299792457.9999608591483m/s from hovering observer

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

Conclusions:

1000gev Electron V=299792457.9999608591483m/s from hovering observer
10gev Electron V=299792457.6086310809697m/s from hovering observer
Velocity difference between 1000gev and 10gev electron from hovering observer: 0.3913297781786m/s
(299792457.9999608591483/299792457.6086310809697)=100.0000001305335635527%
100.0000001305335635527%-100% = 0.00000013053...%

Can I say the 1000gev electron covers a given meter only 0.00000013...% faster, but has 10,000% as much energy compared to the 10gev electron, to a hovering observer?
 

Ibix

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What is the relevance of that to your question?
 
If the two particles spend 1 second (according to a hovering observer) transiting a region where the gravitational acceleration is fairly consistently 1m/s^2 (to a hovering observer), is it safe to assume the 0.00000013% difference in speed is effectively a rounding error, and both the 10gev and 1000gev electrons will have decelerated by almost exactly 1m/s in 1 second as recorded by the same hovering observer?
 

Ibix

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You didn't answer. How is this relevant to your original question?
 
It was stated in the OP:

re: "the deceleration due to gravity is independent of the outward speed."

I am asking for confirmation whether it is thought in this scenario that both particles (10gev and 1000gev electron) decelerate approximately 1m/s in 1 second to the hovering observer, while transiting a fairly consistently 1m/s^2 gravity region (as measured by a hovering observer).

Based on the statement "the deceleration due to gravity is independent of the outward speed" I'll assume the correct answer is both will decelerate by approximately 1m/s.
 
1000gev Electron V=299792457.9999608591483m/s from hovering observer
V=299792457.9999608591483m/s - 1m/s = 299792456.9999608591483m/s

299792456.9999608591483m/s

E = Electron Total Energy

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

E = 0.5109989461/sqrt(1-(299792456.9999608591483^2/299792458^2))

E = 0.5109989461/sqrt(1-(299792456.9999608591483^2/299792458^2))

E = 6256.145768696381MeV

B = Electron Rest Energy = 0.5109989461MeV

E-B = 6256.145768696381MeV - 0.5109989461MeV = 6255.634769750281MeV

Z = E-B = Final Kinetic Energy of 1000gev electron after 1 second in 1m/s^2 gravity = 6.25GeV

1000-6.25 = 993.75

993.75/1000 = 99.375%

Can I say after 1 second traversing a fairly uniform 1m/s^2 gravity region, a 1000gev kinetic electron (to a hovering observer) loses 99.375% of its kinetic energy (according to the same hovering observer)?
 
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Based on the statement "the deceleration due to gravity is independent of the outward speed" I'll assume the correct answer is both will decelerate by approximately 1m/s.
If we assume for the sake of argument that you are correct that electrons traveling at almost the speed of light will take 1 second to cover a region over which the deceleration due to gravity is, to a good enough approximation, a constant 1 m/s^2, then yes, obviously this statement is correct.

However, you just pulled those numbers out of thin air. You have no idea whether they're anywhere even close to being relevant for the jets you are asking about. You need to actually do the math to see what the relevant numbers are.
 

kimbyd

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I have a question about this picture:

It shows matter traveling at relativistic velocities away from a black hole:

View attachment 245386
"Figure 5.21: This VLA image of the radio-loud quasar 3C 175 shows the core, an apparently one-sided jet, and two radio lobes with hot spots of comparable flux densities. The jet is intrinsically two sided but relativistic, so Doppler boosting brightens the approaching jet and dims the receding jet. Both lobes and their hot spots are comparably bright and thus are not moving relativistically. Image credit: NRAO/AUI/NSF Investigators: Alan Bridle, David Hough, Colin Lonsdale, Jack Burns, & Robert Laing."

I'm confused by the image, because in a different thread it was stated:


If the deceleration due to gravity is independent of the outward speed, how is it thought that the particles that have relativistic velocity with respect to the black hole retain a velocity which is so close to the speed of light after traveling many so many light years "upwards" away from the super massive black hole?
In special relativity, acceleration due to gravity the particle's momentum rather than its speed.

Classically, subtracting from momentum and speed are the same thing: the momentum ##p = mv## is proportional to speed. In fact, you can rewrite Newton's second law in terms of momentum:

$$\sum\vec{F} = {d\vec{p} \over dt}$$

This equation is also correct in special relativity, provided we use the relativistic momentum. The relativistic momentum is:

$$\vec{p} = \gamma m \vec{v}$$
$$\gamma = {1 \over \sqrt{1 - {|\vec{v}|^2 \over c^2}}}$$

The "acceleration due to gravity" in relativistic terms is then the derivative of ##\gamma \vec{v}## rather than just the derivative of ##\vec{v}##.
 
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Vanadium 50

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First, the exact same thing happens on earth with Newtonian gravity. The acceleration (down, so it's negative) due to gravity is independent of velocity. That is exactly the same as "the deceleration due to gravity is independent of the outward speed ".

Second, the wall-O-numbers doesn't clarify anything. Especially when you refuse to clarify when asked.

Third, writing 16 significant digits is innumerate. And hard to read.
 
If we assume for the sake of argument that you are correct that electrons traveling at almost the speed of light will take 1 second to cover a region over which the deceleration due to gravity is, to a good enough approximation, a constant 1 m/s^2, then yes, obviously this statement is correct.
Assuming what you say is all correct, can I then say that the more kinetic energy an electron initially has to the hovering observer, the more it will lose in the first second in the region which is to a good approximation uniform 1m/s^2 gravity? For example can I say an electron with KE proportional to 1m/s to a hovering observer loses all of its upward kinetic energy in one second (to the hovering observer), but the kinetic energy lost in one second by the 1m/s electron is much less than the kinetic energy lost in one second by an electron with V=299792457.9999608591483m/s to the hovering observer?

In other words can I say in a gravity field which is to a good approximation uniform 1m/s^2 gravity, the V=299792457.9999608591483m/s electron loses 993.75GeV kinetic energy in one second, and the V=1m/s electron loses 0.0000000000028428ev in one second, to the hovering observer?
 
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I can’t strip off decimals without drastically changing the energy. The specified velocity = 1000GeV to a hovering observer.
 

kimbyd

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...cannot be analyzed; you need General Relativity if gravity is present (i.e., if spacetime is curved).
Sure you can, using a pseudoclassical approximation. The acceleration will be off by a factor of two at high velocity (if the simplest approximation is used), but the overall concepts aren't really broken by looking at it in this way.
 

kimbyd

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Assuming what you say is all correct, can I then say that the more kinetic energy an electron initially has to the hovering observer, the more it will lose in the first second in the region which is to a good approximation uniform 1m/s^2 gravity? For example can I say an electron with KE proportional to 1m/s to a hovering observer loses all of its upward kinetic energy in one second (to the hovering observer), but the kinetic energy lost in one second by the 1m/s electron is much less than the kinetic energy lost in one second by an electron with V=299792457.9999608591483m/s to the hovering observer?

In other words can I say in a gravity field which is to a good approximation uniform 1m/s^2 gravity, the V=299792457.9999608591483m/s electron loses 993.75GeV kinetic energy in one second, and the V=1m/s electron loses 0.0000000000028428ev in one second, to the hovering observer?
No. Because as I pointed out above, the acceleration doesn't change ##\vec{v}##. It changes ##\gamma\vec{v}##.

If the particle is moving at 0.995c, then ##\gamma = 10##. So the value ##9.995c## gets reduced by 1m/s, meaning that the speed only drops by about 0.1m/s instead.

Now, as I mentioned earlier in response to PeterDonis, this isn't quite right. This approximation doesn't really work properly for relativistic objects (gravity is twice as strong as you'd expect from the Newtonian approximation for relativistic objects). But the basic concept remains reasonable: the force of gravity subtracts from the momentum, not the velocity. Close to the speed of light, large changes in momentum only translate to small changes in velocity.
 

kimbyd

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What is a "pseudoclassical approximation"?
There are a few different ones. In this case I mean combining Newtonian gravity with special relativity. Which means using ##F = dp/dt## for Newton's second law, Newtonian gravity for the force, and the relativistic momentum.

As I mentioned, it isn't correct: as speeds get near the speed of light, the deflection in a gravitational field you would expect from this approximation is a factor of two off (the actual deflection is twice that predicted here).

You can fix this up to get the right answer for photons by simply multiplying the force by a factor of two. But I don't think you can fix it in general because it requires a velocity-dependent force term, which is a radical departure from Newtonian behavior, and would likely have a number of unexpected side effects if you tried to do anything. I think at that point it's better to just drop the approximation entirely.

Another semiclassical interpretation that's available involves taking Newtonian gravity and adding extra terms to the force equation (the first being proportional to ##1/r^3##). This works for non-relativistic particles in fairly strong gravitational fields when plain Newtonian gravity starts to break down, and was successfully used to predict the correct orbit of Mercury for the first time.
 
Actually, it isn't; @kimbyd's point is well taken, that "deceleration" actually affects γvγv\gamma v, not vvv.
In other words can I say in a gravity field which is to a good approximation uniform 1m/s^2 gravity, the V=299792457.9999608591483m/s electron loses 993.75GeV kinetic energy in one second, and the V=1m/s electron loses 0.0000000000028428ev in one second, to the hovering observer?
Can I say the upward V=299792457.9999608591483m/s electron (1000gev to hovering observer):
A) loses approximately 0.0000000000028428ev in one second to the hovering observer in a gravitational field which closely approximates uniform 1m/s^2
B) loses a significantly different amount of energy than 0.0000000000028428ev in one second to the hovering observer in a gravitational field which closely approximates uniform 1m/s^2
 
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In this case I mean combining Newtonian gravity with special relativity.
AFAIK you can't do that consistently, since Newtonian gravity has instantaneous action at a distance and is incompatible with Lorentz invariance.

Do you have a reference for this? I have not seen anything like this in a relativity textbook.
 
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a gravitational field which closely approximates uniform 1m/s^2
It has to be uniform over a length of one light-second, since that's how far the electron will travel in one second (to a very high accuracy). You can't just wave your hands and say this happens; you have to actually do the math and see what kind of gravitational source would make it possible, and within what range of radial coordinates from the source.
 

kimbyd

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AFAIK you can't do that consistently, since Newtonian gravity has instantaneous action at a distance and is incompatible with Lorentz invariance.

Do you have a reference for this? I have not seen anything like this in a relativity textbook.
I've only seen it used in tests for gravitational lensing which demonstrate that the Newtonian-like setup predicts the wrong answer.

I used it here because it's a simple enough explanation that at least captures the essence of what's going on, even though it's wrong in detail.
 

kimbyd

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Can I say the upward V=299792457.9999608591483m/s electron (1000gev to hovering observer):
A) loses approximately 0.0000000000028428ev in one second to the hovering observer in a gravitational field which closely approximates uniform 1m/s^2
B) loses a significantly different amount of energy than 0.0000000000028428ev in one second to the hovering observer in a gravitational field which closely approximates uniform 1m/s^2
The right way to talk about this phenomenon is to ignore the concept of uniform acceleration entirely, and instead examine gravitational redshift. If your particles are highly-relativistic, you an treat them as if they were photons with the same energy and your answer won't be far off. The calculations for gravitational redshift are rather different from anything discussed in this thread.
 

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