Beaming effect?

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Would the beaming effect suggest that if a light source is moving at close to speed of light (shining toward me) as seen by me (so I am in the lab frame) than I will see the angle of the beam larger than what a person in the moving source frame would see it?
 
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  • #2
pivoxa15 said:
Would the beaming effect suggest that if a light source is moving at close to speed of light and I am stationary wrt to it than I will see the angle of the beam larger than what a person in the moving source frame would see it?
have a look at

Relativistic beaming
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Relativistic beaming is the process by which the relativistic effect modifies the apparent luminosity of a relativistic jet. Beaming is common in many Active Galactic Nuclei (AGN) galaxies. When relativistic beaming occurs in such galaxies, a central supermassive black hole is the ultimate source of energy for twin jets of intensely energetic plasma. Relativistic beaming is important in astronomy to understanding the nature and evolution of galaxies.

Relativistic beaming is of particular importance for a jet which is oriented close to the line of sight from the AGN to Earth. In the simplest case the jet can be considered to be a series of spherical clouds or blobs, each emitting a high luminosity. In the rest frame of the Earth each blob is approaching at speeds which can be in the range of 95% to 98% of the speed of light and, because of Special Relativity and some physical properties of the jet, the luminosity observed on Earth will be higher than the intrinsic luminosity measured in the rest frame of the jet clouds.

At very small angles to the line of sight such beaming effects become quite large and the observed luminosity of a jet may be a few hundred or a thousand times higher than the intrinsic luminosity. The jet on the other side, however, moving away from Earth at relativistic speeds is affected quite differently. The observed luminosity from this “counter-jet” is less than the intrinsic luminosity.

Relativistic beaming effects can greatly alter the appearance of a distant galaxy. Two AGN which are identical except for different angles to the line of sight can seem to us on Earth to be very different, unrelated galaxies. This is, in fact, what seems to be the case with blazars.

Contents [hide]
1 A Simple Jet Model
1.1 Synchrotron Spectrum and the Spectral Index
1.2 Beaming Equation
1.2.1 Relativistic and Non-Relativistic Aberration
1.2.2 Time Dilation
1.2.3 Blue- (Red-) Shifting
1.2.4 Lorentz Invariants
2 Terminology
2.1 Physical Quantities
2.2 Mathematical Expressions
3 External links
 
  • #3
Doc Al
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pivoxa15 said:
Would the beaming effect suggest that if a light source is moving at close to speed of light and I am stationary wrt to it than I will see the angle of the beam larger than what a person in the moving source frame would see it?
I'll restate it this way: Bob is moving at a high speed towards Alice. Bob shines a beam of light--which has some angular spread according to Bob--towards Alice. As seen by Alice, the angle of that beam will be narrower in the direction of motion, and thus the beam appears brighter to Alice than it would to someone stationary wrt Bob. (This is also called the "headlight effect".)
 
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Doc Al said:
I'll restate it this way: Bob is moving at a high speed towards Alice. Bob shines a beam of light--which has some angular spread according to Bob--towards Alice. As seen by Alice, the angle of that beam will be narrower in the direction of motion, and thus the beam appears brighter to Alice than it would to someone stationary wrt Bob. (This is also called the "headlight effect".)
What do you mean by in the direction of motion? Do you mean when Alice view the beam when it is either moving toward or away from her? Or just towards her?

I realised that I had made a mistake with my opening post. I have corrected it now. But it stiill seem to contradict your statement. I know that if I am in the lab frame shining a light in all directions, another person in a frame moving close to c will see the beam in my frame become narrower.
The mathematical relationship is [tex]\cos(a')=\frac{\cos(a)+v/c}{1+(v/c)\cos(a)}[/tex]
a'<a suggesting a narrower beam. where a' is in S' and a in S.

But if we were to reverse the argument
[tex]\cos(a)=\frac{\cos(a')-v/c}{1-(v/c)\cos(a')}[/tex]
then if a beam is shown in a moving frame S' with beam angle of a' than in my lab frame S I measure an angle of a with a>a' which suggest that I see the beam in S' wider than what it actually is (a').
 
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  • #5
Doc Al
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pivoxa15 said:
What do you mean by in the direction of motion? Do you mean when Alice view the beam when it is either moving toward or away from her? Or just towards her?
Assuming Bob shines his beam in the direction he is moving, Alice will see the beam to be concentrated in a smaller angle.

I realised that I had made a mistake with my opening post. I have corrected it now.
It's still ambiguous. Is the beam directed forward or backward wrt the motion of the source?

But it stiill seem to contradict your statement. I know that if I am in the lab frame shining a light in all directions, another person in a frame moving close to c will see the beam in my frame become narrower.
If you shine a light uniformly in all directions, then I (moving wrt you at some high speed) will see the light concentrated towards the direction that you are moving (with respect to me). If you are moving in my +x direction, I will see your light bunching together in the +x direction.
 
  • #6
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Doc Al said:
Assuming Bob shines his beam in the direction he is moving, Alice will see the beam to be concentrated in a smaller angle.


It's still ambiguous. Is the beam directed forward or backward wrt the motion of the source?
So it matters if the light source is shining into a direction that is opposite the motion of the light and a bystander will see the light differently as when if the light was shining in the direction of motion which is approaching him. Correct? In fact it should be the reverse of what will be seen in the direction of motion, i.e light become broader or expand.


Doc Al said:
If you shine a light uniformly in all directions, then I (moving wrt you at some high speed) will see the light concentrated towards the direction that you are moving (with respect to me). If you are moving in my +x direction, I will see your light bunching together in the +x direction.
Your argument seems to make sense suggesting what I wrote in the previous post was wrong but why was it wrong. How did I misuse the formulas?
 
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  • #7
aberration

pivoxa15 said:
Would the beaming effect suggest that if a light source is moving at close to speed of light (shining toward me) as seen by me (so I am in the lab frame) than I will see the angle of the beam larger than what a person in the moving source frame would see it?[/Quote/]
Have please a critical look at

arXiv.org > physics > physics/0503003
physics/0503003
Comments: 13 pages, 9 figures
Subj-class: Physics Education

The paper presents a relativistic space-time diagram, which displays in true values the space (Cartesian and polar) and the time coordinates of the same event detected from two inertial reference frames in relative motion related by the Lorentz-Einstein transformations, the aberration angles and the Doppler shifted periods and wavelengths. We use it in order to illustrate the reflection of light on moving mirrors (horizontal and vertical) and the way in which a single observer could measure the length of a moving rod. It displays in true values the space-time coordinates of the same event generated by a light signal.
the best things a physicist can offer to another one are information and constructive criticism
 
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Could you provide a link, it seems to be hard to find.
 
  • #9
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bernhard.rothenstein said:
pivoxa15 said:
Would the beaming effect suggest that if a light source is moving at close to speed of light (shining toward me) as seen by me (so I am in the lab frame) than I will see the angle of the beam larger than what a person in the moving source frame would see it?[/Quote/]
Have please a critical look at

arXiv.org > physics > physics/0503003
physics/0503003
Comments: 13 pages, 9 figures
Subj-class: Physics Education

The paper presents a relativistic space-time diagram, which displays in true values the space (Cartesian and polar) and the time coordinates of the same event detected from two inertial reference frames in relative motion related by the Lorentz-Einstein transformations, the aberration angles and the Doppler shifted periods and wavelengths. We use it in order to illustrate the reflection of light on moving mirrors (horizontal and vertical) and the way in which a single observer could measure the length of a moving rod. It displays in true values the space-time coordinates of the same event generated by a light signal.
the best things a physicist can offer to another one are information and constructive criticism

Could you provide a link, it seems to be hard to find. Thanks.
 
  • #10
aberration

pivoxa15 said:
Would the beaming effect suggest that if a light source is moving at close to speed of light (shining toward me) as seen by me (so I am in the lab frame) than I will see the angle of the beam larger than what a person in the moving source frame would see it?
go on google please and give it to look for
[physics/0503003] Graphical aids for relativistic optics
 
  • #11
robphy
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bernhard.rothenstein said:
go on google please and give it to look for
[physics/0503003] Graphical aids for relativistic optics
What you call "relativistic space-time diagrams" appear to be only spatial diagrams.

You can see the beaming/headlight effect in the "circular light clock" animations on "[URL [Broken]
physics.syr.edu/courses/modules/LIGHTCONE/LightClock/[/URL]
 
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  • #12
Doc Al
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pivoxa15 said:
So it matters if the light source is shining into a direction that is opposite the motion of the light and a bystander will see the light differently as when if the light was shining in the direction of motion which is approaching him. Correct?
Correct.
In fact it should be the reverse of what will be seen in the direction of motion, i.e light become broader or expand.
Exactly.
Your argument seems to make sense suggesting what I wrote in the previous post was wrong but why was it wrong. How did I misuse the formulas?
I suspect you are getting your frames mixed up.

pivoxa15 said:
The mathematical relationship is [tex]\cos(a')=\frac{\cos(a)+v/c}{1+(v/c)\cos(a)}[/tex]
a'<a suggesting a narrower beam. where a' is in S' and a in S.

But if we were to reverse the argument
[tex]\cos(a)=\frac{\cos(a')-v/c}{1-(v/c)\cos(a')}[/tex]
then if a beam is shown in a moving frame S' with beam angle of a' than in my lab frame S I measure an angle of a with a>a' which suggest that I see the beam in S' wider than what it actually is (a').
Let me try to restate it with the frames clearly defined. Let (as is usually done) the "moving" frame of the light source be the primed frame; let the lab frame be the unprimed frame. The primed frame moves with speed v in the +x direction of the unprimed frame. If a ray of light makes an angle [itex]\theta'[/itex] with the x'-axis in the primed frame (the proper frame of the light source), it will be seen to make an angle [itex]\theta[/itex] with the x-axis in the unprimed (lab) frame, where:
[tex]\cos\theta=\frac{\cos\theta'+v/c}{1+(v/c)\cos\theta'}[/tex]

As you know, for angles [itex]\theta' < 90[/itex] degrees, [itex]\theta < \theta'[/itex]: In the lab frame, forward going light beams appear at a narrower angle.

You can certainly reverse this, viewing things from the primed frame:
[tex]\cos\theta' = \frac{\cos\theta-v/c}{1-(v/c)\cos\theta}[/tex]

When you interpret this as light being shined at some angle in the unprimed frame (but viewed in the prime frame), realize that from the view of the primed frame, the unprimed frame is moving away and is shining light backwards! :wink:
 
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  • #13
robphy said:
What you call "relativistic space-time diagrams" appear to be only spatial diagrams.

You can see the beaming/headlight effect in the "circular light clock" animations on "[URL [Broken]
physics.syr.edu/courses/modules/LIGHTCONE/LightClock/[/URL]
I think that the diagram is not only a spatial diagram but it is a genuine space-time diagram because if r represents a position vector then r/c represents a time coordinate of an event generated by a light signal and characterized by a space coordinate r.
sine ira et studio
 
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  • #14
robphy
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bernhard.rothenstein said:
I think that the diagram is not only a spatial diagram but it is a genuine space-time diagram because if r represents a position vector then r/c represents a time coordinate of an event generated by a light signal and characterized by a space coordinate r.
sine ira et studio
If so, then your "genuine space-time diagram" is not the "Minkowski spacetime diagram". If, from a given point P on your diagram, an arrow relates two lightlike related points, how do you represent events that are spacelike related to P? timelike related to P?
 
  • #15
events generated by light signals

robphy said:
If so, then your "genuine space-time diagram" is not the "Minkowski spacetime diagram". If, from a given point P on your diagram, an arrow relates two lightlike related points, how do you represent events that are spacelike related to P? timelike related to P?
The diagram I propose involves events that in I' are located on a circle whereas in I they are located on an elipse. In the case of events generated by light signals do we make distinction between time like and space like?
sine ira et studio
 
  • #16
robphy
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bernhard.rothenstein said:
The diagram I propose involves events that in I' are located on a circle whereas in I they are located on an elipse. In the case of events generated by light signals do we make distinction between time like and space like?
sine ira et studio
Essentially, you are projecting down the events on the light cone of the event corresponding to the origin of your diagram. Your circle and ellipse are the observer-dependent spatial-projections of the wavefronts seen by each observer.

Indeed, in a 2+1 Minkowski spacetime [which is the arena of your situation], your diagram is a 2-dimensional space with some certainly non-Minkowskian metric on it. So, your diagram is not what most people would call a "spacetime diagram".
 
  • #17
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Doc Al said:
Correct.

Exactly.

I suspect you are getting your frames mixed up.


Let me try to restate it with the frames clearly defined. Let (as is usually done) the "moving" frame of the light source be the primed frame; let the lab frame be the unprimed frame. The primed frame moves with speed v in the +x direction of the unprimed frame. If a ray of light makes an angle [itex]\theta'[/itex] with the x'-axis in the primed frame (the proper frame of the light source), it will be seen to make an angle [itex]\theta[/itex] with the x-axis in the unprimed (lab) frame, where:
[tex]\cos\theta=\frac{\cos\theta'+v/c}{1+(v/c)\cos\theta'}[/tex]

As you know, for angles [itex]\theta' < 90[/itex] degrees, [itex]\theta < \theta'[/itex]: In the lab frame, forward going light beams appear at a narrower angle.

You can certainly reverse this, viewing things from the primed frame:
[tex]\cos\theta' = \frac{\cos\theta-v/c}{1-(v/c)\cos\theta}[/tex]

When you interpret this as light being shined at some angle in the unprimed frame (but viewed in the prime frame), realize that from the view of the primed frame, the unprimed frame is moving away and is shining light backwards! :wink:
My primes and unprimes are defined just like yours but my formula is the opposite of yours. In your example, S' is always behind S on the x axis isn't it. So that is the confusion because in my formulas, it assumed S is always behind S' but I used it as if S' is behind S.
This is all assuming that light is always shown in the direction of motion.
 
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  • #18
Doc Al
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pivoxa15 said:
My primes and unprimes are defined just like yours but my formula is the opposite of yours. In your example, S' is always behind S on the x axis isn't it. So that is the confusion because in my formulas, it assumed S is always behind S' but I used it as if S' is behind S.
This is all assuming that light is always shown in the direction of motion.
If we are using the same definition for the primed and unprimed frames, and the same definition of angle, yet your formula is "opposite" of mine, then your formula is incorrect.

S and S' are frames, not things, so I'm not sure what you mean about one being behind the other. According to S, the S' frame moves in the +x direction; According to S', the S frame moves in the -x' direction.

In my equations the angle is always with respect to the +x' (or +x) axis. So for angles less than 90 degrees, a ray of light in S' points forward as seen by S. But a ray of light in S points backward as seen by S', since S moves in the -x' direction. I believe this is what is messing you up.
 
  • #19
aberration of light

robphy said:
If so, then your "genuine space-time diagram" is not the "Minkowski spacetime diagram". If, from a given point P on your diagram, an arrow relates two lightlike related points, how do you represent events that are spacelike related to P? timelike related to P?
Having a look at

arXiv.org > physics > physics/0510113

Physics, abstract
physics/0510113


Illustrating Einstein's special relativity: A relativistic diagram that displays in true values the components of a four vector
you will see how the diagram works in the case of space and time like physical quantities. critical comments are highly appreciated.
sine ira et studio
 
  • #20
bernhard.rothenstein said:
The diagram I propose involves events that in I' are located on a circle whereas in I they are located on an elipse. In the case of events generated by light signals do we make distinction between time like and space like?
sine ira et studio
having a look at

arXiv.org > physics > physics/0510113
you could see fow the diagram works in the case of time like and space like physical quantities. critical comments are highly appreciated,
sine ira et studion
 
  • #21
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Doc Al said:
If we are using the same definition for the primed and unprimed frames, and the same definition of angle, yet your formula is "opposite" of mine, then your formula is incorrect.

S and S' are frames, not things, so I'm not sure what you mean about one being behind the other. According to S, the S' frame moves in the +x direction; According to S', the S frame moves in the -x' direction.

In my equations the angle is always with respect to the +x' (or +x) axis. So for angles less than 90 degrees, a ray of light in S' points forward as seen by S. But a ray of light in S points backward as seen by S', since S moves in the -x' direction. I believe this is what is messing you up.
Being in front or behind is important in this context? You say "according to S, S' frame moves in the +x direction" Lets assume the rest source is in S'. If the person in S is viewing S' in front of it than he will view the source in S' as having a larger angle than normal since S' will be coming towards him. However, if S is behind S' than S' will be moving away from S so S will view the source as having a larger angle than normal. Normal being angle in S'.

I realise that frames are not things but I am thinking about the person in the frame. So I am first thinking about the person than thinking about their perspectives which is the frames S and S'. One can move in the +x direction but is the person moving towards or away from me? This question has big consequences in this context.


I have read your post again. Are you assuming the observer in S and S' are at the origin of these frames? If so than when you say "According to S, the S' frame moves in the +x direction and
According to S', the S frame moves in the -x' direction." You still need to mention the portion of the coordinate system they are moving in because this will determine if they are moving towards or away from the origin which is very important.
 
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  • #22
Doc Al
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pivoxa15 said:
Being in front or behind is important in this context? You say "according to S, S' frame moves in the +x direction" Lets assume the rest source is in S'. If the person in S is viewing S' in front of it than he will view the source in S' as having a larger angle than normal since S' will be coming towards him. However, if S is behind S' than S' will be moving away from S so S will view the source as having a larger angle than normal. Normal being angle in S'.
Sure, but if you want to understand the formula relating the angle of a ray seen from different frames, think of a particular light ray being aimed forward at a angle theta' wrt the +x' direction. Obviously, if a particular observer in S were behind the light source in S' they wouldn't see anything! :smile: (All they could see are light rays that the source emitted backwards, if any.)

I realise that frames are not things but I am thinking about the person in the frame. So I am first thinking about the person than thinking about their perspectives which is the frames S and S'. One can move in the +x direction but is the person moving towards or away from me? This question has big consequences in this context.
Consider a frame to have observers everywhere. If the source in S' is emitting a forward cone of light, then all observers (who are in the path of that light) in S will see a tighter forward cone of light. Again, if you are behind the person who is shining light forward you see nothing.

In the formula, theta' (measured wrt the +x' axis) refers to any particular ray of light in the source frame S'. Theta is the angle that that same ray of light makes in the S frame, measured from the +x axis. Theta is always less than (or equal to) theta'. Rays with theta' < 90 (they go forward) will appear in S to be closer to the x-axis; Rays with theta' > 90 (they go backward) will appear in S to be further from the x-axis. Whether you, a particular observer in S, can see those rays depends on where you are.


I have read your post again. Are you assuming the observer in S and S' are at the origin of these frames? If so than when you say "According to S, the S' frame moves in the +x direction and
According to S', the S frame moves in the -x' direction." You still need to mention the portion of the coordinate system they are moving in because this will determine if they are moving towards or away from the origin which is very important.
Again, we are talking about observations made in a frame--with many observers. Yes, where you stand makes all the difference in the world as to what you will see--not for any esoteric reason, but just because the light may not reach you. See if my comments above don't clear things up. (That's why in my original post I spoke of Alice and Bob, and what they were doing with respect to each other. Bob moved toward Alice, etc.)
 
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