Light Clock Query: Explaining why Light Does Not Escape

[BlumRat]

So my uncle recently asked me about a phenomenon that I have been unable to explain, but there that there must surely be a good answer for.

A light clock is being observed on an airplane, with the light bouncing back and fourth between the plates of clock. The plates of the clock have the same velocity as the airplane, and the light is moving back and fourth at light speed. Why then, does the light not slowly move horizontally and escape the light clock?

I have a few thoughts on the matter, but I would appreciate input so that I could come to a more concise answer.

 

[schaefera]

The light and the plates in the moving frame have 0 x-component of velocity, so it doesn't escape simply because its velocity doesn't point in a direction in which it CAN escape.

In the Earth's frame of references, the light pulse has a nonzero x-component of velocity, but it is exactly equal to the plate's x-velocity, so again the light simply isn't moving in a way that it is able to escape.

 

[James S Saint]

The trajectory of a photon is determined by the orthogonal velocity of its source. The speed of the photon has nothing to do with the source, but the vector does.

When the light is originally shot toward the other plate, because the source is moving with the plates, the light doesn't travel perpendicular to the "non-moving" frame of reference, but instead angles exactly enough to meet the other plate in the exact right location so as to reflect again at that same angle (all with respect to that "non-moving" frame).

In other words, from the perspective of the non-moving frame, the light was never aimed perpendicular to the plates. If it had appeared to be aimed truly perpendicular, the plate would move out of its path.

From the perspective of the airplane, it is the Earth that is moving under it. The light source is aimed directly at the plate, so the Earths movement has nothing to do with the light or plate.

If you had shot the light originally while the plane wasn't moving, but could somehow accelerate the plane very quickly, the light would actually escape.

 

[BlumRat]

The light and the plates in the moving frame have 0 x-component of velocity, so it doesn't escape simply because its velocity doesn't point in a direction in which it CAN escape.

In the Earth's frame of references, the light pulse has a nonzero x-component of velocity, but it is exactly equal to the plate's x-velocity, so again the light simply isn't moving in a way that it is able to escape.

Granted,but from the Earth's frame of reference, when the plane is accelerating, does the light also appear to accelerate? That seems bizarre, because I thought the speed of light is constant.

I understand there is some fundamental truth I am overlooking here, is it that the "speed limits" and "laws" for light only act when you are in the same reference point as the light?

 

[BlumRat]

In other words, from the perspective of the non-moving frame, the light was never aimed perpendicular to the plates. If it had appeared to be aimed truly perpendicular, the plate would move out of its path.

I like this explanation, but in my head I am picturing a nozzle where the light came from pointing downwards. I feel that if I looked at the same nozzle from outside the airplane passing by that it would still be point downwards, which begs the question: how did the light come out at an angle?

 

[schaefera]

I was picturing it shooting upward, but let's go with your picture!

The nozzle does shoot it straight down- relative to the nozzle. But the nozzle is also moving to the side, and you can't remove that sideways velocity from the light. So we need to think of velocity as a vector, which separate components, and the total length of the vector velocity (which I'm calling speed) is always c... but because of the way velocities transform at speeds near c, the speed remains c even if it is acceleration or has components being added together.

 

[BlumRat]

I was picturing it shooting upward, but let's go with your picture!

The nozzle does shoot it straight down- relative to the nozzle. But the nozzle is also moving to the side, and you can't remove that sideways velocity from the light. So we need to think of velocity as a vector, which separate components, and the total length of the vector velocity (which I'm calling speed) is always c... but because of the way velocities transform at speeds near c, the speed remains c even if it is acceleration or has components being added together.

In essence, you're saying that the light trades some of its downward velocity for horizontal velocity? Also, it's horizontal velocity is such that it is equivalent to the speed of the craft, and that is because the craft was moving at a certain speed when it was fired? (This is all from the outside reference frame as I understand.)

 

[James S Saint]

Well okay, if you need to get more into the detail;

A photon is produced by a specific mechanism and its geometry, for example; when an electron falls from a higher orbit and releases a photon. The production of a photon is not an instantaneous event. Now if that geometry is in motion while producing that photon, each tiny bit of the production of that photon is shifting in time, thereby shifting the "front" of the photon from is "back".

The amount of shifting of the position of the front and back is exactly equal to the amount of shifting of the atom as it was traveling. Thus the photon is actually manufactured already angled with consideration of the velocity of the atom that produced it. If the photon was released in the forward direction, the wavelength of the light would be shorter (blue-shifted). Frankly, I suspect that a "non-moving" frame would detect a polarization of the light coming out from the side.

..probably not the best explanation... pictures would help.

 

[BlumRat]

Well okay, if you need to get more into the detail;

A photon is produced by a specific mechanism and its geometry, for example; when an electron falls from a higher orbit and releases a photon. The production of a photon is not an instantaneous event. Now if that geometry is in motion while producing that photon, each tiny bit of the production of that photon is shifting in time, thereby shifting the "front" of the photon from is "back".

The amount of shifting of the position of the front and back is exactly equal to the amount of shifting of the atom as it was traveling. Thus the photon is actually manufactured already angled with consideration of the velocity of the atom that produced it. Frankly, I suspect that a "non-moving" frame would detect a polarization of the light.

..probably not the best explanation... pictures would help.

That is actually brilliantly explained, thank you!
But then... what happens when the plane accelerates @.@

 

[schaefera]

I like James' explanation, but to simplify a bit (in case it helps) imagine throwing a tennis ball in the air while sitting in a car. It travels with the same sideways velocity as the car! This is how kinematics works, and it's all due to the way vectors add.

The interesting thing with light is that, no matter what the components are, the total vector sum in magnitude of the velocity comes out to be c.

 

[James S Saint]

Acceleration changes the game. The light would escape if the plane accelerates. Merely the travel time between the plates could be enough to allow the plate to move out of the line of fire.

 

[BlumRat]

Thank you both, I'm glad to finally have an answer! haha

 

[mananvpanchal]

The situation is same as I am struggling with. So, I have a question here.

If source's velocity vector is pointing to east. Light vector pointing to north-west locally to the source. Then for stationary observer, at some source speed, light vector pointing to north?

 

[harrylin]

The situation is same as I am struggling with. So, I have a question here.

If source's velocity vector is pointing to east. Light vector pointing to north-west locally to the source. Then for stationary observer, at some source speed, light vector pointing to north?

Exactly!

 

[mananvpanchal]

Exactly!

Ok, now we are with this. Please, look at below image.

If we think about spherical light source, then is above situation true?
Left circle shows source light vectors locally. And right circle shows source light vectors while source is moving right relative to an observer.

 

[harrylin]

Ok, now we are with this. Please, look at below image.

View attachment 48233

If we think about spherical light source, then is above situation true?
Left circle shows source light vectors locally. And right circle shows source light vectors while source is moving right relative to an observer.

Yes, it's called the headlight or relativistic beaming effect. :tongue2:
- https://en.wikipedia.org/wiki/Relativistic_beaming

 

[mananvpanchal]

Yes, it's called the headlight or relativistic beaming effect. :tongue2:
- https://en.wikipedia.org/wiki/Relativistic_beaming

So, intensity will be increased in front of light source and will be decreased in back of light source. Right?

 

[harrylin]

So, intensity will be increased in front of light source and will be decreased in back of light source. Right?

Right.

 

[mananvpanchal]

Right.

Ok, If there are two intensity detector at front of light source and end of light source in light source's frame. And same two detector in observer's frame.

Will the source's detectors measure equal intensity? And observer's detectors measure unequal intensity?

 

[yuiop]

Ok, If there are two intensity detector at front of light source and end of light source in light source's frame. And same two detector in observer's frame.
Will the source's detectors measure equal intensity? And observer's detectors measure unequal intensity?

Yes.

 

[mananvpanchal]

Thanks harrylin. Thanks yuiop.

 

[A.T.]

Yes, it's called the headlight or relativistic beaming effect. :tongue2:
- https://en.wikipedia.org/wiki/Relativistic_beaming

I don't understand the time dilation part:
http://en.wikipedia.org/wiki/Relativistic_beaming#Time_dilation

The picture with the caption "Time Dilation" seems to have nothing to do with time dilation.

http://upload.wikimedia.org/wikipedia/en/4/4d/AGN_Jet_Dilation.png
http://en.wikipedia.org/wiki/File:AGN_Jet_Dilation.png

In fact it is called "Jet Dilation" and described as follows:

A homogeneous source emits photons in all directions equally,but if the source is moving than it will appear to an outside observer in front of the object that photons are being emitted more frequenctly. The result is the object looks brighter.

So the picture explains why a resting detector in front of the moving source will encounter more photons/time than a resting detector behind it. But what does this have to do with time dilation?

Time dilation of the moving source would mean that the source is emitting photons less frequently in the rest frame of the detectors, than in its own frame. But that factor alone would affect both detectors equally.

 

[harrylin]

I don't understand the time dilation part:
http://en.wikipedia.org/wiki/Relativistic_beaming#Time_dilation

The picture with the caption "Time Dilation" seems to have nothing to do with time dilation.
[..] In fact it is called "Jet Dilation" [..] So the picture explains why a resting detector in front of the moving source will encounter more photons/time than a resting detector behind it. But what does this have to do with time dilation?

Time dilation of the moving source would mean that the source is emitting photons less frequently in the rest frame of the detectors, than in its own frame. But that factor alone would affect both detectors equally.

Yes I agree, it looks as if there the Doppler frequency effect is confounded with time dilation. Moreover, I would say that the headlight or beaming effect and the forward frequency increase effect are two different aspects of the Doppler effect. While with "Doppler" we usuallly mean something to do with frequencies, the headlight effect is not even frequency dependent.

I gave a link to it as it shows that the OP refers to what is known as the headlight effect, and the intro gives a clear explanation - but as you noticed, that article needs important correction further on.

 

[A.T.]

Yes I agree, it looks as if there the Doppler frequency effect is confounded with time dilation. Moreover, I would say that the headlight or beaming effect and the forward frequency increase effect are two different aspects of the Doppler effect. While with "Doppler" we usuallly mean something to do with frequencies, the headlight effect is not even frequency dependent.

Okay, but what about those points:

- Aside from the increased frequency, is the number of received photons/time also increased for an approaching source? This wouldn't make sense to me. The "higher frequency, with same amplitude" is the wave-model. While "same number of more energetic photons" is the particle-model.

- What effect does the time dilation of the moving source have on the light detected by stationary detectors? If the light source emits 1kW in its own frame, does it emit less than that in the frame where it moves?

 

[harrylin]

Okay, but what about those points:

- Aside from the increased frequency, is the number of received photons/time also increased for an approaching source? This wouldn't make sense to me. The "higher frequency, with same amplitude" is the wave-model. While "same number of more energetic photons" is the particle-model.

This is somewhat new to me as well, but my guess is that the same number of more energetic photons corresponds to what I earlier distinguished as the Doppler frequency effect; I would add to that the headlight effect, which, I think, distributes more of the photons forward, so that also slightly more photons/time should arrive at a certain surface in front of it. (But see next about time dilation).
ADDENDUM: also the number of photons/time increases, see post 29.

- What effect does the time dilation of the moving source have on the light detected by stationary detectors? If the light source emits 1kW in its own frame, does it emit less than that in the frame where it moves?

Yes I think so. The simplest example is radioactive particles that give off heat: at high speed their half life is much increased, so in the same time they should give off less heat.

PS: it looks to me that time dilation implies less emitted photons/time as well as lower frequency. I haven't worked that out. The Wikipedia article seems to suggests that the forward beaming effect is stronger than the time dilation effect, so that still more photons/time are beamed forward. Perhaps someone else knows this stuff?

 

[A.T.]

Aside from the increased frequency, is the number of received photons/time also increased for an approaching source? This wouldn't make sense to me. The "higher frequency, with same amplitude" is the wave-model. While "same number of more energetic photons" is the particle-model.

This is somewhat new to me as well, but my guess is that the same number of more energetic photons corresponds to what I earlier distinguished as the Doppler frequency effect; I would add to that the headlight effect, which, I think, distributes more of the photons forward, so that also slightly more photons should arrive at a certain surface in front of it.

I don't want to conflate it with aberration, so let's say it is a laser beam pointing exactly forward on a moving source. Will a co-moving detector detect a different number of photons/time than a stationary one?

 

[harrylin]

I don't want to conflate it with aberration, so let's say it is a laser beam pointing exactly forward on a moving source. Will a co-moving detector detect a different number of photons/time than a stationary one?

Supposing that you mean from a moving source, then I can repeat my basic answer: Yes I think so.
The simplest example is radioactive particles: at high speed their half life is much increased, so in the same time less particles decay and for example less beta photons are emitted.

 

[A.T.]

Supposing that you mean from a moving source, then I can repeat my basic answer: Yes I think so.
The simplest example is radioactive particles: at high speed their half life is much increased, so in the same time less particles decay and for example less beta photons are emitted.

Okay, that is the effect of time dilation, which would reduce the number of photons emitted in both directions by the same factor. So this should not contribute to the headlight effect.

But given two identical (in the source frame) laser beams from a moving source, pointing back and forward. Would stationary detectors at the back and front record different numbers of photons/time?

 

[harrylin]

Okay, that is the effect of time dilation, which would reduce the number of photons emitted in both directions by the same factor. So this should not contribute to the headlight effect.

But given two identical (in the source frame) laser beams from a moving source, pointing back and forward. Would stationary detectors at the back and front record different numbers of photons/time?

In fact yes, and this can easily be seen. This is classical Doppler. Take two consecutive 10Hz wave trains of 1 s with 1 s interval:

S -_-_-_-_-_-_-_-_-_-_____________________-_-_-_-_-_-_-_-_-_-_____________________ D

As each following wave crest is formed closer to the detector in front, that detector receives each wave crest less than 1/10 of a second after the preceding one. But that is also true for the time interval to the next wave train: the first wave crest of the next wave train is formed much closer to the detector so that it will be received much less than 1 s after that of the preceding one. And the opposite is true for a detector at the back.