Shape of a light flash

[kleinwolf]

Are those both facts true :

 

a ) Suppose a circular wave is made in water, when seen from a plane, this is a moving circle

 

b) Suppose a light flash was emitted in vacuum, when seen from a fast moving observer, this should be an ellipse (longer in the direction of the speed) ?

[Martin]

Are those both facts true :

 

a ) Suppose a circular wave is made in water, when seen from a plane, this is a moving circle

 

b) Suppose a light flash was emitted in vacuum, when seen from a fast moving observer, this should be an ellipse (longer in the direction of the speed) ?

 

You could try explaining how one would see a lightwave. A water wave can be seen because light is reflected off the wavy surface of the medium but how do you see the peaks and troughs of a light wave?

 

A ring of detectors? Better specify.

 

Setting that aside for the moment, I want to comment on "longer in the direction of the speed".

 

If a circular hoop whizzes past me, I believe it's shape, for me, is oval but shorter in the direction of motion. Squashed in the direction of motion. Am I wrong? It seems inconsistent with what you said about the wave.

 

Hopefully others will respond and help clear this up.

[Bob_for_short]

You could try explaining how one would see a lightwave. A water wave can be seen because light is reflected off the wavy surface of the medium but how do you see the peaks and troughs of a light wave?

 

A lightwave is easily visualizable in a dusty but nearly transparent medium (foggy atmosphere, for example).

 

Bob.

[Sisyphus]

Yeah, I think you need to specify what the set up is. Anyway, the pertinent facts as I see them would be:

 

1) Distances compress in the direction of travel.

2) Light always travels at the same rate, C, in every direction relative to the observer, no matter what the relative velocity of the source to the observer.

3) You can't directly observe an expanding sphere of light like you would an expanding water wave.

[kleinwolf]

I reasoned with point (2) : speed of light is "always" c...consider a circular flash at rest, observed when you travel at speed vx in direction x, and consider the part of the light flying "perpendicularly" in direction y, let say with speed vy, then vx²+vy²=c², hence vy is "smaller" than c, and hence it should grow as an ellipse, because in the direction x, the speed of the light is c.

 

However how to "observe" this, could maybe with photodetectors placed at long distances..but I have no idea about technical feasibility thereof.

[Sisyphus]

I don't really follow your logic, but I was thinking the implication of point 2 was that it was necessarily a perfect sphere, as viewed from any reference frame.

 

So, from your perspective, you are motionless, and the light source is moving towards you at your relative velocity, say 0.5C. At some point, it flashes, and from that point, the light wave expands in a sphere moving at C in every direction, and the light source continues on its way moving at 0.5C, the forward wave moving ahead of it at 0.5C, the backward one moving behind it at 1.5C.

 

In the reference frame of the light source, however, the source is at rest, you're the one moving, and the light is still moving at C in every direction, approaching you at 1.5C if you're still moving towards the source at the time of the flash (adding your velocities), or 0.5C if you're moving away.

 

(Bear in mind, I'm not certain about any of this.)

[NowThatWeKnow]

My question is, are you talking about the actual shape of the light as if you could see it in real time? Or as it would be seen by your eyes using only the light that did reach you and when it reached you?

 

I thought the speed of light is constant, but only relative to the frame it is in. Since the only light you can see is what reaches your eyes, it is in your frame and the speed will be constant. The actual shape would be distorted by gravity and other things.

[Sisyphus]

Alright, how about this:

 

You arrange a series of detectors in a sphere (equidistant from a single point), all at rest to one another. Then you fire a projectile such that it will pass through the center of the sphere moving uniformly at a significant fraction of C, and as it does it flashes a light. What happens?

 

This is, I think, equivalent to the original question. And the answer depends on your frame of reference. In the frame where the detectors are at rest, they will all receive the light signal at the same time. Of that I'm almost positive. Since that is entirely equivalent to passing a light source while moving yourself (deciding who is moving is entirely arbitrary), it is clear that the "ball of light" is neither "elongated" nor "flattened."

 

As for what happens in the reference frame where the projectile is at rest and the detectors are moving, I think that's a bit more complicated. The sphere of detectors would be flattened in the direction of travel, for one thing, and when the light is emitted one side of the "sphere" will be traveling towards the source and one away, and the one traveling towards will receive the signal first. The ball of light itself, obviously, would still just be a uniformly expanding sphere.

[NowThatWeKnow]

...The ball of light itself, obviously, would still just be a uniformly expanding sphere.

Are you saying the light would NOT be influenced by gravity?

[Sisyphus]

Are you saying the light would NOT be influenced by gravity?

 

Oh, sure, I'm pretty much discounting gravity. But it seems like any effect would be symmetrical, no? Maybe not. Anyway, it would still be a "sphere," in that every photon would be traveling at the same speed in "straight lines." It's just that those straight lines might not look straight from every perspective...

[NowThatWeKnow]

...But it seems like any effect would be symmetrical, no? ...

 

If you could isolate a brief segment of the Sun's light as it left the Sun, it would seem almost symmetrical, for awhile. If you followed that brief segment (in real time) until it reached the outer limits of the milky Way on one side and the huge black hole at the center on the other, it would be far from symmetrical.

 

When light enters a high gravity area, an outside observer will see the light slow down. I look at it as space is curved and the light has further to go to cover the same distance. if you are in the same frame with the light, your clock slows down so the light speed is still constant.

 

 

If my layman vision of relativity is wrong, please straighten me out. I am here to learn.

 

Edit - Another way to put it is that if time slows down when you approach a black hole, the speed of light must slow down to maintain 186,000 miles per second.

[Sisyphus]

Hehe, was there a black hole added to this question when I wasn't looking?

[NowThatWeKnow]

Hehe, was there a black hole added to this question when I wasn't looking?

 

I added one to make my point more obvious. I see the entire universe as a variable gravity frame that does have an impact on light. In most cases the effect the would be small, especially over short distances.

[Sisyphus]

Right, but that's not really relevant to the question at hand, is it? The question is how does being in a different reference frame from a light source affect how the light propagates. The OP suggested that the lighted region would be elongated in the direction of travel. I'm suggesting it would be neither elongated nor compressed. (Although the observers in the two frames would not agree on the borders of the sphere, even if both agreed it was the same shape, because space itself would be compressed).

[NowThatWeKnow]

Right, but that's not really relevant to the question at hand, is it? The question is how does being in a different reference frame from a light source affect how the light propagates. The OP suggested that the lighted region would be elongated in the direction of travel. I'm suggesting it would be neither elongated nor compressed. (Although the observers in the two frames would not agree on the borders of the sphere, even if both agreed it was the same shape, because space itself would be compressed).

 

Your first post in this thread said

"I think you need to specify what the set up is." and "2) Light always travels at the same rate, C, in every direction relative to the observer, no matter what the relative velocity of the source to the observer."

so I crawled into that can of worms when you opened it.

 

The original post said:

 

Shape of a light flash

 

Are those both facts true :

 

a ) Suppose a circular wave is made in water, when seen from a plane, this is a moving circle

 

b) Suppose a light flash was emitted in vacuum, when seen from a fast moving observer, this should be an ellipse (longer in the direction of the speed) ?

 

Not sure what the "circular wave" in "water" has to do with the "shape of a light flash" or part b).

 

Other then that, I agree with you.

[Klaynos]

I think the problem here is where are you observing the light pulse from, if you have a ring of detectors as described above you are effectively doing an on axis measurement, it's cylindrically symetric so the only solution is all the detectors detecting at the same time... in the case of the water wave the detector is not on axis the wave is moving outwards in the plane of the velocity...

[kleinwolf]

Well, there are a lot of posts.

 

For example

 

NowThatWeKnow gravitation : suppose a light beam horizontally emitted, the aequivalence principle says there is a vertical component and hence is bent (but the speed norm remains c),

 

 

Now For a light flash, emitted from an observer.

 

The speed for a moving observer O' with speed vx should be obtained by relativistic speed addition 1D : (&): va & vb =(va+vb)/(1+va*vb/c^2)

 

For example : along x as seen in O' : -vx & c=c and -vx & (-c)=-c

 

along y ... we cannot because the formula is only 1-dimensional.

 

hence : (vx,vy)&(va,vb)<>(vx&va,vy&vb)

 

(this is obvious for (vx,0)&(0,c) were (vx,c) whose norm > c

 

We could

 

a) use a full 2d Lorentz Boost to deduce the speed addition law (with crossed terms)

 

or maybe

 

b) reason like this : the resulting speed seen at angle (a) for the moving obs, is

 

(vx & c*cos(a),wy) and wy is the speed of the photon along y-direction, still unkown, but from the condition

 

(vx & c*cos(a))^2+wy^2=c^2 (speed of light is conserved) ??

[Sisyphus]

Again, I'm not following your math, but if you must be doing something wrong - maybe mixing up reference frames? Remember, the observer is always motionless in his own reference frame.

[NowThatWeKnow]

...NowThatWeKnow gravitation : suppose a light beam horizontally emitted, the aequivalence principle says there is a vertical component and hence is bent (but the speed norm remains c),...

I understand the speed of light is constant in a vacuum and flat space.

[GDG]

Are those both facts true :

 

a ) Suppose a circular wave is made in water, when seen from a plane, this is a moving circle

 

b) Suppose a light flash was emitted in vacuum, when seen from a fast moving observer, this should be an ellipse (longer in the direction of the speed) ?

 

(a) is true; (b) is not. You may detect a Doppler shift in the light frequency, but as the speed of light is constant, the wave front will always be circular/spherical (in the absence of distortion or lensing by massive objects).

[kleinwolf]

The speed of light in vacuum is constant. "Speed" is a vector, or as a scalar ?

 

I found 2 formulas on Wikipedia

 

http://en.wikipedia.org/wiki/Velocity-addition_formula#Special_case:_orthogonal_velocities

 

Speed addition of orthogonal speeds U and V :

 

S=V+Sqrt(1-V²/c²)*U

 

(the case of c in y direction and -v in x ):

 

a) V=c U=-v => S=c(ey) (we hide v speed of the observer in the Lorentz singularity)

b) V=-v U=c => S=-v(ex)+sqrt(1-v²/c²)*c(ey) (check : S²=c²)

 

So it's not a way to decide (?)

Edited May 2, 2009 by kleinwolf

[Sisyphus]

Why are you adding velocities? The velocity of light is C, period.

[kleinwolf]

This is the difference between "velocity" and "speed"

 

speed is c

velocity could be written [math]\vec{c}[/math]

 

Maybe the key point is "when" we consider the velocity distribution

 

*) T=0 : the velocities are distributed on the origin of the flash, a point at O

the norm is the speed c, hence a circle, all direction considered

 

*) T>0 : the velocities are distributed on the previous circle

their norm is c

 

Erratum : the title is wrong chosen for the contain, it should be : shape of a wave-front seen from different observer

[Sisyphus]

So it's a circle at both times, centered where the source was at T=0. What velocities are there to add?

[kleinwolf]

The velocities distribution is centered at O only at T=0

 

If T>0, it should be a "crown" : velocities have their base on the previous "circle", isn't it ?

 

we could add relativistically [math]\vec{v}\oplus_r\vec{c}[/math].

 

with [math]\left\|\vec{v}\oplus_r\vec{c}\right\|^2=c^2[/math].