Special relativity, light direction and wave source.

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SUMMARY

This discussion centers on the implications of special relativity regarding the propagation of light and the perception of time. Participants explore how light travels at a constant speed (c) regardless of the observer's frame of reference, leading to phenomena such as time dilation and length contraction. The concept of a "light clock" is introduced to illustrate how moving observers perceive time differently. The conversation emphasizes that while light travels diagonally from a stationary observer's perspective, it appears vertical to an observer in motion, demonstrating the relativity of simultaneity and the consistency of light speed across all frames.

PREREQUISITES
  • Understanding of Einstein's theory of special relativity
  • Familiarity with the concept of time dilation
  • Knowledge of light propagation and wavefronts
  • Basic grasp of inertial reference frames
NEXT STEPS
  • Research the "light clock" concept to understand time dilation in moving frames
  • Study the implications of length contraction in special relativity
  • Explore the relationship between light speed and observer motion in various reference frames
  • Examine real-world experiments that validate Einstein's predictions about light propagation
USEFUL FOR

Students of physics, educators teaching special relativity, and anyone interested in the fundamental principles of light and time in the context of modern physics.

  • #31
Banana Joe said:
I know you can't see the flash traveling, but let's say you can, and let's say the observer hasn't seen the setup, then from his perspective the light will seem to travel slower than c, because he (like the rocket man) erroneously thinks that the light only traveled up and down the height of the triangle, and not along the oblique sides.
There is no "erroneous" here. Either viewpoint is perfectly valid. When you add the requirement that lightspeed be constant for all observers, you can derive the equations for time dilation and length contraction.

To see the point about both viewoints being valid, imagine two identical rockets with lightclocks, traveling in opposite directions. According to one rocket, its own lightclock is working vertically and the other is working on the diagonal. The other rocket can say the same thing the other way round. Which one is right? There's nothing to choose. Both viewpoints are equally valid.
 
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  • #32
Mentz114 said:
Nothing related to motion can un-happen it.
It would be double plus ungood if it could. Cause could be effect and effect, cause.
 
  • #33
A.T. said:
That is no error. The light does travel only up and down in his frame.

You've misread the conditions that Banana Joe proposed:
Let's assume we have a light source and a sensor at the opposite extremes of the base of an isosceles triangle and a mirror at the third vertex, all stationary. We also have two stationary observers, one looking from the side, the other far away positioned on the extension of the triangle's base, so that he only sees a line which coincides with the height of the triangle. Now, if the light source fires an impulse so that it travels up along the oblique side of the triangle, hits the mirror and travels down along the other oblique side and hits the sensor, I would expect the observer on the side to see the impulse travel on this diagonal path at c, but I would expect the other to see the impulse go straight up and down like it was traveling along the height of the triangle, at a lower speed than c.
Here all of the observers, mirrors, sources, and detectors are at rest relative to one another, and Banana Joe is trying to use the projection of the distance traveled onto a plane not parallel to the direction of travel as the distance.
 
  • #34
Nugatory said:
Banana Joe is trying to use the projection of the distance traveled onto a plane not parallel to the direction of travel as the distance.
I see. He is confusing the visual impression of the observer due to perspective with what happens in the rest frame of the observer. I blame the use of the words "observer/observes/sees" instead of "reference frame/measures" for such misconceptions.
 
  • #35
A.T. said:
I see. He is confusing the visual impression of the observer due to perspective with what happens in the rest frame of the observer. I blame the use of the words "observer/observes/sees" instead of "reference frame/measures" for such misconceptions.

Many people, myself included, share the blame for that confusion :smile:
 
  • #36
bahamagreen said:
I had a feeling it might be like that, but just to be complete and eliminate all loop holes, there are no such attributes whose variation with direction of emission could be measured at different parts of the wavefront? Frequency, amplitude, wave shape, ...?

So it would be impossible to create a spherical waveform that has some "encoding" impressed into one or more combined attributes such that a measurement could identify the direction from which the measuring detector was relative to the source?

Maybe emitting the light from a "magnetic bubble" or some other contrivance so that some attribute(s) are skewed and detection reveals the source direction?

Even if the answer to all these is no... what about an emission that comprised a series of wavefront bursts sent in different directions, each direction series coded by differential attributes... maybe using an array of lasers on the surface of a ball to approximate the wavefront?

Or is it that there might be such a thing in principle, but SR always invokes invariance so there is null measured difference?



Of course we can mark different parts of the wave front. By putting a fancy lamp shade over the light source, for example.

If we do that, we notice that a certain part of the wave hits a certain part of the rocket wall. If we change the velocity of the rocket, the pattern on the wall does not change.

(A lamp attached to rocket wall is shining light to the opposite rocket wall in this scenario)
(Or: A source of a wave front attached to rocket wall is sending a wave to the opposite rocket wall)
 
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  • #37
jartsa said:
Of course we can mark different parts of the wave front. By putting a fancy lamp shade over the light source, for example.

If we do that, we notice that a certain part of the wave hits a certain part of the rocket wall. If we change the velocity of the rocket, the pattern on the wall does not change.

(A lamp attached to rocket wall is shining light to the opposite rocket wall in this scenario)
(Or: A source of a wave front attached to rocket wall is sending a wave to the opposite rocket wall)
The sphercal wave analogy breaks down here. There is no way to distinguish the light that hits the mirror in the rest frame and the moving frame. It is the same mirror and the same light. If we could detect a difference due to motion this would violate the principles of relativity.

Maybe this is what you are saying ... not sure though.
 
  • #38
Let’s consider this experiment: we have a straight rail with a rocket attached, on the rocket we have two arms (in the same orientation of the track) with a light source in the middle, each arm has at its extreme a light detector, plus we have one stationary detector at each end of the track. The rocket starts at one end of the track, reaches the maximum speed possible, and then the light source fires an impulse exactly in the middle of the track. If the speed of light was affected by the motion of the light source, I would expect the two sensors on the rocket to detect the light at the same time, and the stationary sensor at the end of the track in the rocket’s direction to detect it before the other one. If instead the light traveled at c from the point of emission regardless of the speed of the light source, I would expect the two stationary sensors to receive the light at the same time, and the sensor on the back of the rocket to receive it before the one in the front. According to relativity, instead, not only the moving sensors detect it simultaneously, but also the stationary ones do, correct? Doesn’t that mean that light would be traveling at two speeds at the same time, at c relative to the track and the stationary sensors, and at c relative to the moving sensors (so at (c + the velocity of the rocket) relative to the track)?

If I’m in the rocket and everything is slower and I can’t notice, and I see the light traveling at c, shouldn’t that mean that it’s actually moving slower? If it was actually moving at c relative to me, shouldn't it appear to move faster than c?
 
  • #39
Banana Joe said:
Let’s consider this experiment: we have a straight rail with a rocket attached, on the rocket we have two arms (in the same orientation of the track) with a light source in the middle, each arm has at its extreme a light detector, plus we have one stationary detector at each end of the track. The rocket starts at one end of the track, reaches the maximum speed possible, and then the light source fires an impulse exactly in the middle of the track. If the speed of light was affected by the motion of the light source, I would expect the two sensors on the rocket to detect the light at the same time, and the stationary sensor at the end of the track in the rocket’s direction to detect it before the other one. If instead the light traveled at c from the point of emission regardless of the speed of the light source, I would expect the two stationary sensors to receive the light at the same time, and the sensor on the back of the rocket to receive it before the one in the front. According to relativity, instead, not only the moving sensors detect it simultaneously, but also the stationary ones do, correct? Doesn’t that mean that light would be traveling at two speeds at the same time, at c relative to the track and the stationary sensors, and at c relative to the moving sensors (so at (c + the velocity of the rocket) relative to the track)?

If I’m in the rocket and everything is slower and I can’t notice, and I see the light traveling at c, shouldn’t that mean that it’s actually moving slower? If it was actually moving at c relative to me, shouldn't it appear to move faster than c?

I think it is simpler to analyse this if there is a light emitter in center of the rocket and a sensor at each end. Someone in the rocket will conclude that the light hits both sensors at the same time by his clock. If the rocket is moving ( inertially) along the track, then an observer on the ground ( the track frame) will see the back sensor detect the light before the front sensor. It is easy to calculate those times if we write equations for the motion of back of the rocket, the light moving towards the back, the light moving forwrd and the front of the rocket.
Using x and t in the ground frame, and with the end of the rocket being at x=0 at t=0 ( the time of the light emission) we get

back of rocket x = v*t
light going back x = L/(2*γ) - c*t

front of rocket x = L/γ + v*t
light forward x = L/(2*γ) + c*t

From which I get from t0 = L/(2*γ*(v+c)) and t1 = L/(2*γ*(c-v))

L is the rest length of the rocket, v is the relative velocity of rocket and ground, γ is the usual factor.

If v = 0 then t0 = t1 = L/(2*γ*c)

This does not prove anything but it is correct to keep c in both frames as the speed of light, and this makes the times different in the moving frame.

If you repeat the calculation with additive velocities, the times will be the same t0 = t1 .

So we get non-simultaneity if the speed of light is the same in both frames.
 
  • #40
Banana Joe said:
If instead the light traveled at c from the point of emission regardless of the speed of the light source, I would expect the two stationary sensors to receive the light at the same time, and the sensor on the back of the rocket to receive it before the one in the front.
Correct, for an observer stood on the track. For one stood on the rocket, the rocket is stationary and the track is moving backwards, so the logic is reversed, and you would expect the pulses to arrive simultaneously at the rocket sensors and not at the track sensors.
Banana Joe said:
According to relativity, instead, not only the moving sensors detect it simultaneously, but also the stationary ones do, correct?
No. There can exist observers who see one pair of detections as simultaneous, and observers who see the other pair as simultaneous. There are no observers who see both as simultaneous. That there is no global notion of "simultaneous" is a key concept in relativity. You need to say who sees something as simultaneous, just like you need to say who sees something as "here".
Banana Joe said:
Doesn’t that mean that light would be traveling at two speeds at the same time, at c relative to the track and the stationary sensors, and at c relative to the moving sensors (so at (c + the velocity of the rocket) relative to the track)?
No. It means that two observers in motion with respect to each other do not agree on the distance between two events or the time between them. They do not always agree on whether or not they are simultaneous, or even which one happens first. They do agree on the speed of light.

This is basically what Mentz proves in his post above. It's been extensively checked by experiment - check out the sticky thread at the top of this forum.

You seem to me to be stuck on the notion that one of the observers is stationary and the others are moving. This is not correct. Any time you are not accelerating, you can say you are stationary. Everybody else is moving.
 
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  • #41
Banana Joe said:
Let’s consider this experiment: we have a straight rail with a rocket attached, on the rocket we have two arms (in the same orientation of the track) with a light source in the middle, each arm has at its extreme a light detector, plus we have one stationary detector at each end of the track. The rocket starts at one end of the track, reaches the maximum speed possible, and then the light source fires an impulse exactly in the middle of the track.

Here's a spacetime diagram depicting your scenario. The rocket starts out stationary at the left end of the track. The ends of the rocket are shown in blue and red. The ends of the track are shown in black and green. The middle of the rocket and the track are shown as separate grey lines. The dots mark off hundred nanosecond intervals of Proper Time for each end of each object and their middles. The speed of light is one foot per nanosecond.

The Proper Length of the rocket is 600 feet when at rest and it is accelerated to 60%c such that it's Proper Length is maintained at 600 feet. The Proper Length of the track is 1800 feet:

attachment.php?attachmentid=66232&stc=1&d=1391336975.png

When the middle of the rocket reaches the middle of the track (the two grey lines intersect), the light source at the middle of the rocket emits a short burst of light which travels at c in both directions in the Inertial Reference Frame depicted in the diagram. As you can see, the light reaches the two detectors at each end of the track 900 nsecs later.

Banana Joe said:
If the speed of light was affected by the motion of the light source, I would expect the two sensors on the rocket to detect the light at the same time, and the stationary sensor at the end of the track in the rocket’s direction to detect it before the other one.
Or if the speed of light was affected by the Inertial Reference Frame (IRF) you choose to depict your scenario in, specifically, if it is defined to propagate at c in all directions, then you would expect the two sensors on the rocket to detect the light at the same time, just like when we transform to the IRF in which the rocket is at rest:

attachment.php?attachmentid=66233&stc=1&d=1391336975.png

As you can see, the light reaches both rocket detectors in 300 nanoseconds. And note that the Proper Length of the rocket is still 600 feet when at rest in this IRF.

Banana Joe said:
If instead the light traveled at c from the point of emission regardless of the speed of the light source, I would expect the two stationary sensors to receive the light at the same time, and the sensor on the back of the rocket to receive it before the one in the front.
Or if the light travels at c no matter what IRF you depict the scenario in, then in the rest IRF of the track, the detectors at the ends of the track receive the light at the same time, as shown in the first diagram above.

Banana Joe said:
According to relativity, instead, not only the moving sensors detect it simultaneously, but also the stationary ones do, correct?
That is not the best way to say it. A better way to say it is in the rest IRF of the rocket, the rocket sensors detect it simultaneously and in the rest IRF of the track, the track sensors detect it simultaneously. So whichever sensors are at rest (stationary) in the IRF are the ones that detect the light simultaneously.

Banana Joe said:
Doesn’t that mean that light would be traveling at two speeds at the same time, at c relative to the track and the stationary sensors, and at c relative to the moving sensors (so at (c + the velocity of the rocket) relative to the track)?
No, you need to pick one IRF at a time, not two IRF's at the same time. Observers cannot measure the propagation of light. They cannot tell which IRF you are using to describe the scenario.

Banana Joe said:
If I’m in the rocket and everything is slower and I can’t notice, and I see the light traveling at c, shouldn’t that mean that it’s actually moving slower? If it was actually moving at c relative to me, shouldn't it appear to move faster than c?
You cannot see the light traveling at all. It's not a bullet that you can shine a light on to see where it is at any given moment and then plot its position as a function of time. How would you light up the light to see where it is? Instead, the best you can do is reflect the light off an object and then measure its "average" round trip time. When we apply Einstein's convention, we assume that the light takes the same amount of time to get to the target as its reflection takes to get back in any IRF and that allows us define the time it takes for the light to reach both stationary detectors in each IRF to be the same.

Pretty simple, isn't it? Does it make sense? Any questions?
 

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