Rigidity of a light beam

[vanholten]

Hello, just a question.

I searched the internet for an answer to how exactly a beam of light behaves when it’s source is moved perpendicular to the direction of its own propagation, in  vacuum space. I didn’t find a clear explanation, only a few remarks. Perhaps the Ligo interferometers give details on this?

Thanks.

[swansont]

Light beams are not rigid.

If we're talking about motion at a constant velocity, then the behavior of the light is exactly the same is if it were stationary, and everything else was moving.

[vanholten]

 

17 minutes ago, swansont said:

Light beams are not rigid.

If we're talking about motion at a constant velocity, then the behavior of the light is exactly the same is if it were stationary, and everything else was moving.

Do you mean that the light source being part of everything else moves away from the light that is already on its way?

[swansont]

27 minutes ago, vanholten said:

Do you mean that the light source being part of everything else moves away from the light that is already on its way?

A moving source and stationary target will give exactly the same results as a moving target and stationary source.  Everyone will measure the light to be moving at c, though they may disagree on what the frequency of the light is.

 

[vanholten]

I am still not sure if what you say is in line with my own thoughts on this.

When a light source in constant motion emits a first photon, it will travel away from the source at c. Assuming the light beam is not rigid, then the source and the photon loose connection. Because the light source has no means to transfer its motion retrospectively to the photon, I would say the photon perseveres in its perpendicular path independent of the motion of the light source.

If the light source would be able to transfer its motion to the photon, then the photon would seem to act as if it had inertia, which it can’t have since it has no mass. Another explanation for inertia could be that it's not mass related but energy related, and has to do with energy conservation of the light waves, resulting in some kind of rigidity. However you explained light has no rigidity to it, so that is no option.

Next a second photon is shot away and like the first, perpendicular to the motion of the light source. I would say both photons wouldn’t follow the exact same perpendicular track. And that a stream of photons traveling astronomical distances would show the light beam bending away from the light source. Indeed due to the fact the beam has neither inertia, nor rigidity.  I assume that such bending not necessarily affects the propagation velocity of light, but that it might cause some frequency shifts.

The lasers of LIGO interferometers are among to the most accurate on earth. I wonder if the laser beams while maintaining c of course, show any signs of deflection caused by the motion of earth.

[Janus]

light behaves as follows:

 We start with sources A and B which have a relative motion relative of 0.5c to each other, have just met and each has emitted a pulse of light upwards relative to themselves.

The top image show the moment of emission according to both A and B.

The second image shows 1 sec later according to A.  B has moved 0.5 light sec to the left. The pulse emitted by A (red sphere) is 1 light sec directly above A, and the Pulse emitted by B(blue sphere) is directly above B and one light sec from where it was emitted. (which puts its 0.866 light sec above B)

The bottom image show 1 sec after emission according to B.  A has moved 0.5 light sec to the right. The pulse emitted by B is 1 light sec directly above B, and the pulse emitted by A is directly above A and 1 light sec from where it was emitted (putting it 0.866 light sec above A)

 

[vanholten]

Thanks for the relativistic explanation. But I was looking a bit closer to home. If I ride my bicycle and throw an apple straight in the air, it falls back in my hand, when not too much wind or wobbling. When I point a laser pen straight up, why should a photon at a light second away keep track of me?

[Ghideon]

36 minutes ago, vanholten said:

When I point a laser pen straight up, why should a photon at a light second away keep track of me?

The photon does not keep track. You have to keep moving at a constant speed to "stay under" the photon you did shoot straight up. Which is just another way of saying what @swansont said above. If you turn or brake the photons does not track you and obviously the apple does not track you either. 

An attempt to use an analogy with your equipment from above. Do not attempt this IRL, you may crash
-Ride the bicycle.
-With one hand, throw an apple straight up. and catch it again. (no wind or wobbling)
-With the other hand, point a laser pen straight up at the apple.  (no wobbling)

1: Why should photons miss the apple?

The above analogy does have limits; for instance that gravity affect light.

 

Edited October 30, 2018 by Ghideon
spelling

[Strange]

1 hour ago, vanholten said:

Thanks for the relativistic explanation. But I was looking a bit closer to home. If I ride my bicycle and throw an apple straight in the air, it falls back in my hand, when not too much wind or wobbling. When I point a laser pen straight up, why should a photon at a light second away keep track of me?

For exactly the same reason as the apple does (as Ghideon cleverly points out).

Janus's explanation involves relativistic velocities just so you can easily see where the "moving" photon goes. The diagram would be the same in the case of your bicycle (except you wouldn't be able to see that the path of your photon was not vertical because the angle would be too small).

Edited October 30, 2018 by Strange

[vanholten]

1 hour ago, Ghideon said:

If you turn or brake the photons does not track you and obviously the apple does not track you either. 

You mean if you turn or brake, the photons don't keep track of the pen?

46 minutes ago, Ghideon said:

1: Why should photons miss the apple?

The photons wouldn't miss the apple because they are traveling way too fast to miss it.  If the photons were traveling ultra slow (or you would be able to throw the apple high enough) why should they hit apple? ( Not just because it 's such a very big apple you can hardly miss anyway.) Contrary to the apple they have neither inertia, nor rigidity and they don't keep track of the light source either.

In other words: Why do photons adapt the constant velocity of the light source, that is perpendicular to their own propagation, when the reasons I suggested are all invalid.  You all say they do, but I still miss the physical explanation.

[mistermack]

I don't pretend to know the answer to your point, but I believe it's not simple stuff. I think you would need to study relativity and learn the maths to get a picture of what's going on. 

I think these wiki pages address what you are saying, as a sort of overview, but I don't claim to follow it myself.

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

https://en.wikipedia.org/wiki/Aberration_of_light  

 

[vanholten]

Straight light beams appear not to be so obvious anymore.

The airybeam gives some better understanding of what light is capable of.  Researchers say the are able to create self healing lightbeams that actually curve. The curvation of a light beam due to the absence of inertia might not be such a wierd idea.  It’s difficult stuff though. I don’t pretend to fully understand it. Still trying to get a grip. 

https://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.108.163901

[quiet]

10 hours ago, vanholten said:

how exactly a beam of light behaves when it’s source is moved perpendicular to the direction of its own propagation, in  vacuum space.

There is an essential detail. A photon that has been emitted in a vacuum becomes independent from the source immediately.

Other detail. At sufficient distance from the source, the photons that make up the light beam are mutually independent.

Anything else. In pure vacuum, with no gravity or other fields present, the Euclidean definition of straight line is identical to the trajectory of the photon.

Your question is about a source that moves in a vacuum. The movement of the source does not affect the photon that has become independent of it.

Edited October 31, 2018 by quiet

[swansont]

14 hours ago, vanholten said:

I am still not sure if what you say is in line with my own thoughts on this.

When a light source in constant motion emits a first photon, it will travel away from the source at c. Assuming the light beam is not rigid, then the source and the photon loose connection. Because the light source has no means to transfer its motion retrospectively to the photon, I would say the photon perseveres in its perpendicular path independent of the motion of the light source.

The motion of the light source has no effect after the photon is emitted, so if the the light source changes its motion, it will not affect a photon that has already been emitted.

If you have a light source shining on a target (y direction), and you and that system are in relative motion (x direction), the light will still hit the target. From your reference frame, the laser + target system is moving, and the photon is emitted at an angle that is not perpendicular to its motion.

(light has momentum, BTW)

[Strange]

14 hours ago, vanholten said:

When a light source in constant motion emits a first photon, it will travel away from the source at c. Assuming the light beam is not rigid, then the source and the photon loose connection. Because the light source has no means to transfer its motion retrospectively to the photon, I would say the photon perseveres in its perpendicular path independent of the motion of the light source.

Would it help to consider that you have a mirror above the light source that reflects the light straight back down again. This is then like the example of throwing a ball up and catching it.

If you are stationary, the ball goes straight up and come straight back down to your hand. The photon does the same.

If you are moving (at a steady rate; ie inertially) then the ball will still go up and come back to your hand. And the photon does the same. 

Someone watching you cycle past will see the ball/photon go up at an angle, and come back down at an angel to reach your hand (which as moved while they were up in the air).

14 hours ago, vanholten said:

If the light source would be able to transfer its motion to the photon, then the photon would seem to act as if it had inertia, which it can’t have since it has no mass. Another explanation for inertia could be that it's not mass related but energy related, and has to do with energy conservation of the light waves, resulting in some kind of rigidity. However you explained light has no rigidity to it, so that is no option.

Light has momentum because it has energy.

[vanholten]

5 hours ago, swansont said:

The motion of the light source has no effect after the photon is emitted, so if the the light source changes its motion, it will not affect a photon that has already been emitted.

I agree and with you at this point and with  @quiet.  To picture what I thinks is realistic, I created the image below.

-The SR path corresponds with the idea of photons keeping up with the light source. The image has the same setup as the Lorentz Clock in relative constant motion 0,866c

The SR experiment suggests the light lengthens its path due to relative observation. Since it can only travel at c the path is too long and therefore time has to be adjusted. To account for the lengthening of the path, the formula for time dilation is used.

∆t'= ∆t /√(1-(v/c)²)  = ∆t /√(1-(0,866)²)  = ∆t /√0,25 = ∆t /0,5  = 2 ∆t      

-Classic Physics. The dashed lines are the path of photons moving perpendicular to the light-source and persevering in a straight line.

At the point where photon 10 leaves the light-source, the light source traveled the distance BE. Meanwhile photon 0 already traveled the distance equal to L. For photon 10 to travel L and to reach A, the light source has to travel the distance EC. For photon 10 to bounce back and hit the light-source (the reflector) has to travel the distance CF.

This results in the same lengthening of the “light beams”.  However the path of the photons themselves is not lengthened. The propagation velocity of the photons remains c as usual. Because nothing changed, there is no need to adjust time.

If this corresponds with reality time dilatation is not happening.

(You can still apply the time dilation formula as an analogy if you like, and will get the exact same outcome. I am untidy with numbers so please check this out)

https://en.wikipedia.org/wiki/Time_dilation

Edited October 31, 2018 by vanholten

[swansont]

1 hour ago, vanholten said:

I agree and with you at this point and with  @quiet.  To picture what I thinks is realistic, I created the image below.

-The SR path corresponds with the idea of photons keeping up with the light source. The image has the same setup as the Lorentz Clock in relative constant motion 0,866c

The SR experiment suggests the light lengthens its path due to relative observation. Since it can only travel at c the path is too long and therefore time has to be adjusted. To account for the lengthening of the path, the formula for time dilation is used.

∆t'= ∆t /√(1-(v/c)²)  = ∆t /√(1-(0,866)²)  = ∆t /√0,25 = ∆t /0,5  = 2 ∆t      

-Classic Physics. The dashed lines are the path of photons moving perpendicular to the light-source and persevering in a straight line.

Relativity is classical physics. You have to be more clear here.

1 hour ago, vanholten said:

At the point where photon 10 leaves the light-source, the light source traveled the distance BE. Meanwhile photon 0 already traveled the distance equal to L. For photon 10 to travel L and to reach A, the light source has to travel the distance EC. For photon 10 to bounce back and hit the light-source (the reflector) has to travel the distance CF.

This results in the same lengthening of the “light beams”.  However the path of the photons themselves is not lengthened. The propagation velocity of the photons remains c as usual. Because nothing changed, there is no need to adjust time.

If the source and target are both moving, all of the photons hit the target. There are none traveling vertically. You need to pick one frame and describe it. Otherwise you mix up two frames, and get a nonsensical result.

 

 

[vanholten]

The second case shows that if photons behave like that, the clock doesn’t work at all. It would only work when stationary, and all photons choose the same path upwards.

[vanholten]

I think it’s very confusing that both options have the same requirement for the clock to work.

- SR needs the photons to propagate vertically aligned with the light source, to make the clock work.

- The second option also needs the photons to propagate vertically aligned with the light source, to make the clock work.

Perhaps the clock should be put aside for the moment, so it no longer matters if it works or not. Then the remaining issue is the increased length of the light beam and its direction. The length increase is the same in both options. However the directions compared to the motion are opposite.

Only SR has additional requirements. It needs time-dilatation to explain the increased length and the direction of the beam. The other option needs no further explanation since time is running as usual, and the direction of the beam is natural consequence of the motion of the light source.

Edited October 31, 2018 by vanholten

[vanholten]

On 31-10-2018 at 5:24 PM, swansont said:

If the source and target are both moving, all of the photons hit the target. There are none traveling vertically. You need to pick one frame and describe it. Otherwise you mix up two frames, and get a nonsensical result.

Here is another image.

A. If the velocity of C decreases, then the angle of a light beam departing from C increases and holds at 90º. Like in B horizontal velocity of vertically emitted photons = velocity of the light source. Thus the red light beams can only result from a rotation of light source A with respect to the horizontal line of motion. Due to the rotation the path of photons narrows down to an angle of 30º and time dilation no longer compensates. Variable velocity means that A has to emit in two mutual variable angles which depend on A ’s relative motion. Outside observation can’t physically alter A. In contrast to B and C that both emit in opposite directions independent from their relative motion. Conclusion: Time dilation is based on a fictional path of light.

B. Identical to a stationary light source. No lengthening of the light beam. Afterglow is only to illustrate the direction of motion.

C. This allows a realistic number of photons to produce a lengthened light beam (x 1.323) without obstructing the propagation velocity of light. No time dilation required. The light beams build up at C to shoot away in opposite direction relative to the motion of C.

Edited November 2, 2018 by vanholten

[swansont]

18 minutes ago, vanholten said:

A. If the velocity of C decreases, then the angle of a light beam departing from C increases and holds at 90º. Like in B horizontal velocity of vertically emitted photons = velocity of the light source.Thus the red light beams can only result from a rotation of light source A with respect to the horizontal line of motion. Due to the rotation the path of photons narrows down to an angle of 30º and time dilation no longer compensates. Variable velocity means that A has to emit in two mutual variable angles which depend on A ’s relative motion. Outside observation can’t physically alter A. In contrast to B and C that both emit in opposite directions independent from their relative motion. Conclusion: Time dilation is based on a fictional path of light.

B. Identical to a stationary light source. No lengthening of the light beam. Afterglow is only to illustrate the direction of motion.

C. This allows a realistic number of photons to produce a lengthened light beam (x 1.323) without obstructing the propagation velocity of light.No time dilation required.The light beams build up at C to shoot away in opposite direction relative to the motion of C.

Acceleration is a new twist you've added. It's usually best to get the basics down before adding complications.

If the source and target are moving, and the light were emitted perpendicular to the source, the light would miss the target. We do not observe this happening. Thus, any conclusion drawn from this idea is wrong.

[vanholten]

58 minutes ago, swansont said:

Acceleration is a new twist you've added. It's usually best to get the basics down before adding complications.

If the source and target are moving, and the light were emitted perpendicular to the source, the light would miss the target. We do not observe this happening. Thus, any conclusion drawn from this idea is wrong.

I did not add acceleration. Please take another look. Indeed the light would miss it’s target. You don’t observe this happening, because it doesn’t happen. 

There is no time dilation. Your observation corresponds with B that is exactly according to your previous post, in which you stated the photons move along with the light source. 

[swansont]

33 minutes ago, vanholten said:

I did not add acceleration. Please take another look. Indeed the light would miss it’s target. You don’t observe this happening, because it doesn’t happen. 

"If the velocity of C decreases"

Changing velocity is an acceleration

Quote

There is no time dilation. Your observation corresponds with B that is exactly according to your previous post, in which you stated the photons move along with the light source. 

Which means they travel a diagonal path for the other observer, but still move at c. And that requires length contraction and time dilation.

[vanholten]

32 minutes ago, swansont said:

"If the velocity of C decreases"

Changing velocity is an acceleration

Which means they travel a diagonal path for the other observer, but still move at c. And that requires length contraction and time dilation.

I showed you in B the line consisting of  1’s  measuring 1.323c is not the path of light. The light beam is build up from the source.

Exactly like you move the top of a vertical ruler along a horizontal line and push it upwards to increase the length above the horizontal. You can’t say that the path of the ruler corresponds with a line connecting the top positions, nor that the length of the ruler is stretched. However, that is what time dilation’s formula does.

Edited November 2, 2018 by vanholten
To picture it better

[vanholten]

1 hour ago, swansont said:

"If the velocity of C decreases"

This is about uniform motions and has nothing to do with acceleration. I didn't introduce a parameter for acceleration.  It must be my failing translation. If I would have written: " At lower velocities", or "v< 0,866c" which does't change anything to the geometry, then will you accept that I did not introduce acceleration?