# Beaming effect?

1. Nov 2, 2006

### pivoxa15

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?

Last edited: Nov 2, 2006
2. Nov 2, 2006

### bernhard.rothenstein

have a look at

Relativistic beaming
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. Nov 2, 2006

### Staff: Mentor

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".)

4. Nov 2, 2006

### pivoxa15

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 $$\cos(a')=\frac{\cos(a)+v/c}{1+(v/c)\cos(a)}$$
a'<a suggesting a narrower beam. where a' is in S' and a in S.

But if we were to reverse the argument
$$\cos(a)=\frac{\cos(a')-v/c}{1-(v/c)\cos(a')}$$
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').

Last edited: Nov 2, 2006
5. Nov 2, 2006

### Staff: Mentor

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?

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. Nov 2, 2006

### pivoxa15

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.

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?

Last edited: Nov 2, 2006
7. Nov 2, 2006

aberration

8. Nov 3, 2006

### pivoxa15

Could you provide a link, it seems to be hard to find.

9. Nov 3, 2006

### pivoxa15

10. Nov 3, 2006

### bernhard.rothenstein

aberration

[physics/0503003] Graphical aids for relativistic optics

11. Nov 3, 2006

### robphy

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]

Last edited by a moderator: May 2, 2017
12. Nov 3, 2006

### Staff: Mentor

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 $\theta'$ with the x'-axis in the primed frame (the proper frame of the light source), it will be seen to make an angle $\theta$ with the x-axis in the unprimed (lab) frame, where:
$$\cos\theta=\frac{\cos\theta'+v/c}{1+(v/c)\cos\theta'}$$

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

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

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!

Last edited: Nov 3, 2006
13. Nov 3, 2006

### bernhard.rothenstein

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

Last edited by a moderator: May 2, 2017
14. Nov 3, 2006

### robphy

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. Nov 3, 2006

### bernhard.rothenstein

events generated by light signals

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. Nov 3, 2006

### robphy

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. Nov 3, 2006

### pivoxa15

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.

Last edited: Nov 3, 2006
18. Nov 3, 2006

### Staff: Mentor

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. Nov 3, 2006

### bernhard.rothenstein

aberration of light

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. Nov 3, 2006

### bernhard.rothenstein

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