Why is the speed of light the same for all observers?

[phyti]

ghwellsjr;4611058

Many people think that you measure the same speed for light because both the distance and the time are reduced by the same factor and "cancel" each other out but the exact opposite is what is true as you can see in the diagram. And it's important to use the Coordinate Time because that is the frame in which the light travels at c. Because the mirror is moving away from the location where you started the stopwatch, it takes longer for the light to reach the mirror (although you have no awareness of this). And because you are moving towards the location of the reflection, it takes less time for the reflection to get back to you (again, you have no awareness of this). So, in fact it is a distance expansion divided by the Time Dilation factor that cancels each other out and results in the speed of light continuing to be the same. The distance expansion is shown in the diagram as the sum of 12 feet for the light to get from you to your mirror and 3 feet for the reflection to get from the mirror back to you for a total distance the light has to travel of 15 feet and it takes 15 nanoseconds resulting in a speed of 1 foot per nanosecond. It is important to realize that the distance to your mirror must contract in order for the measurement to come out the same. If it didn't, the light would have to go farther in both directions taking longer and you would get a smaller measurement for the speed of light.

The question is: “why do all observers measure light speed as constant?”.
All your drawings are showing the different cases from the U perception.
That’s already settled since U is not moving and exempt from lc and td.
It would be more convincing showing perception of the moving observer.

In the following the A perception is overlaid onto the U perception using the same time axis,
and providing a visual comparison.
A has a stick 4 units long (L) with a mirror at the front end.
In each case, a light signal is sent from the origin to reflect from the mirror and return to A
In each case U measures stick length as L/gamma, and A-time equals U-time/gamma.

In fig.1 A is at rest in the U frame.
Light is emitted at A(0, 0), and detected at A(0,8.0).
A calculates* coordinates as A(4.00, 4.00).

In fig.2, A moves at .6c.
Light is emitted at A(0, 0), and detected at A(0, 8.0).
A calculates* coordinates as A(4.0, 4.0).

In fig.3, A moves at .8c.
Light is emitted at A(0, 0), and detected at A(0, 8.0).
A calculates* coordinates as A(4.0, 4.0).

* using SR simultaneity convention

https://www.physicsforums.com/attachments/65069

 

[ghwellsjr]

The question is: “why do all observers measure light speed as constant?”.
All your drawings are showing the different cases from the U perception.

Yes, if by "the U perception", you mean the rest frame of the light source. That's my understanding of what the OP asked about so it shouldn't be surprising that I limited my answer to his question.

That’s already settled since U is not moving and exempt from lc and td.

Yes, if by "U", you mean the light source, since it is at rest in the U coordinate system.

It would be more convincing showing perception of the moving observer.

If by "perception" you mean the rest frame of an observer, then I'm not sure what more there is to show than either your first diagram or my first diagram in post #26 which are essentially identical, don't you agree?

In the following the A perception is overlaid onto the U perception using the same time axis, and providing a visual comparison.

Your overlaid diagrams make it look like the light flash is taking two paths. Is that intentional?

A has a stick 4 units long (L) with a mirror at the front end.
In each case, a light signal is sent from the origin to reflect from the mirror and return to A
In each case U measures stick length as L/gamma, and A-time equals U-time/gamma.

Although that is true, it obfuscates the real relationship of LC and TD. U is the coordinate system. L is the Proper Length and A-time is the Proper Time. So your statement is:

The Coordinate Length is the Proper Length divided by gamma and the Proper Time is the Coordinate Time divided by gamma.

But if we compare apples to apples, we would say:

The Coordinate Length is the Proper Length divided by gamma and the Coordinate Time is the Proper Time multiplied by gamma. That makes the reciprocal (or inverse) relationship between LC and TD more obvious.

In fig.1 A is at rest in the U frame.
Light is emitted at A(0, 0), and detected at A(0,8.0).
A calculates* coordinates as A(4.00, 4.00).

In fig.2, A moves at .6c.
Light is emitted at A(0, 0), and detected at A(0, 8.0).
A calculates* coordinates as A(4.0, 4.0).

In fig.3, A moves at .8c.
Light is emitted at A(0, 0), and detected at A(0, 8.0).
A calculates* coordinates as A(4.0, 4.0).

* using SR simultaneity convention

I don't disagree with any of this but since the OP asked about measuring the value of the speed of light (a round-trip) I didn't bother to mention that the propagation of light is defined to be c (one-way) in any Inertial Reference Frame.

 

[bhobba]

I never understand why the speed of light is the same for all observers irrespective of their motion relative to the source of light.

Well suppose there is a maximum speed, C, information can be sent - it may be infinite which seems the obvious choice - but for the heck of it we will allow for the situation where nature is such that, for some reason, only allows information to be sent at some finite maximum speed.

Now from the Principle Of Relativity that speed must be the same in all inertial reference frames. Since inertial frames are all traveling at constant velocity wrt each other it means if we have a transmitter of information at that maximum speed stationary in a frame, in another frame it will be moving but the speed of that information will be exactly the same ie its speed it independent of the speed of the source.

In fact you can derive the Lorentz transformations using this C - which I have cunningly chosen to suggest the speed of light. It shows there can only be one speed that is the same regardless of the speed of the source.

Now have a look at Maxwell's Equations. The speed of EM radiation does not depend on its source, and it is well known that speed is the speed of light. So, C must be the speed of light.

The bottom line here is C, in the way I have derived the Lorentz transformations above, is simply a parameter whose value needs to be fixed by experiment - it just turns out to be the speed of light.

You will find this way of looking at it in Rindler - Introduction To Special Relativity:
https://www.amazon.com/dp/0198539525/?tag=pfamazon01-20

It also shows how Bell can be accommodated in SR - even though that suggests non local effect can happen instantaneously it can't be used to send information

Thanks
Bill

 

[phyti]

Proper Length divided by gamma and the Coordinate Time is the Proper Time multiplied by gamma.

Yes in the U frame, not in the A frame, which represents all other frames. If light speed is measured in the U frame once as c, it won't be any different the next time. Thus I show from the perspective of A that it is c also regardless of his speed. Overlaying the A view eliminates a 2nd drawing and gives a direct visual comparison.

I don't disagree with any of this but since the OP asked about measuring the value of the speed of light (a round-trip) I didn't bother to mention that the propagation of light is defined to be c (one-way) in any Inertial Reference Frame.

I see the convention as defining equal path lengths since light speed is already declared constant. This should be obvious in Minkowski plots and the MMX, where the question is path length, not light speed. It's also the reason why LENGTH contraction solves the problem.

 

[phyti]

ghwellsjr;
You stated you didn't understand one of my recent drawings, so here is one of yours to show how easy it is to add the moving observers description of events in the same drawing. The sequence of details (red) is: the arc centered on the origin, the vertical from the detection event to the arc, the horizontal to the time axis, a light path to the initial light path, and a vertical to the x axis. With coordinate axes, calculations are not usually needed, and um is not relevant as long as they match, year:light yr, sec:light sec, and in your method, nanosecond:light ns.
Once you have the coordinates, the construction lines can be erased to eliminate clutter.


https://www.physicsforums.com/attachments/65112

 

[ghwellsjr]

The Coordinate Length is the Proper Length divided by gamma and the Coordinate Time is the Proper Time multiplied by gamma.

Yes in the U frame, not in the A frame, which represents all other frames.

If we are talking about inertial observers that pass through the origins of both frames where their clocks are synchronized to zero, then my statement is true in all frames, even in the rest frame of a particular observer. In that case, the speed is zero and gamma is 1 so the Coordinate Length is equal to the Proper Length and the Coordinate Time is equal to the Proper Time. If you think my statement is not true in the A frame (the rest frame of the observer) then what is true?

Also, I have no idea what you mean by the A frame represents all other frames. It sounds like you are giving preference to the rest frame of an observer.

If light speed is measured in the U frame once as c, it won't be any different the next time.

Whenever any inertial observer measures the speed of light from any source (with a single clock, a single ruler, and a single mirror) they don't have to consider any reference frame. They don't have to be aware of any theory such as Special Relativity. So although I agree with your statement, it concerns me that you put stipulations on it. I would just say "if light speed is measured once as c, it won't be any different the next time" or even more simply, "light speed will always be measured as c". Again, since we are talking about a measurement, it has to be a round-trip.

Thus I show from the perspective of A that it is c also regardless of his speed.

I thought that when you say "the perspective of A" you mean the "rest frame of A" in which case his speed is zero so I don't know what you mean by "regardless of his speed". Help me with my confusion.

Overlaying the A view eliminates a 2nd drawing and gives a direct visual comparison.

If by "A view" you mean the "rest frame of A" or the "A frame" then I see no purpose in overlaying it on another frame. Doesn't that discredit all the other frames? Doesn't that promote the preference of a rest frame?

I don't disagree with any of this but since the OP asked about measuring the value of the speed of light (a round-trip) I didn't bother to mention that the propagation of light is defined to be c (one-way) in any Inertial Reference Frame.

I see the convention as defining equal path lengths since light speed is already declared constant. This should be obvious in Minkowski plots and the MMX, where the question is path length, not light speed. It's also the reason why LENGTH contraction solves the problem.

I am confused. You just said that you believe the path lengths are equal and then you said LENGTH contraction solves the problem.

As I have stated over and over again, in a frame where a moving observer is measuring the speed of light the path length of the light is longer than in his rest frame, not equal and not contracted. I explained it all in great detail in post #26.

 

[ghwellsjr]

ghwellsjr;
You stated you didn't understand one of my recent drawings, so here is one of yours to show how easy it is to add the moving observers description of events in the same drawing. The sequence of details (red) is: the arc centered on the origin, the vertical from the detection event to the arc, the horizontal to the time axis, a light path to the initial light path, and a vertical to the x axis. With coordinate axes, calculations are not usually needed, and um is not relevant as long as they match, year:light yr, sec:light sec, and in your method, nanosecond:light ns.
Once you have the coordinates, the construction lines can be erased to eliminate clutter.

Wow, I'm impressed. Did you come up with this construction technique on your own or did you learn about it somewhere?

I can see how a graphical technique could be useful a hundred years ago when there were no computers to draw diagrams or even calculators to help and I will admit that the application I wrote took a very long time to develop but now that I have it, I let the computer do all those pesky calculations and it works for any in-line scenario, not just the simple ones where the observers are inertial and pass through the origins. Can you use your technique for every in-line scenario? Can I give you a more complicated scenario and see how you do it?

However, my diagram that you started with shows everything that the moving observer measures. There's no reason or advantage to add any other lines or to overlay his rest frame. If you count the dots between when he started his stopwatch (at the origin) up to where he detected the reflection, you will see that he measured 12 nanoseconds, just like in his rest frame. Nothing else matters. As I stated in post #26, he is unaware that it took 15 nanoseconds in this frame. He is also unaware that the distance to the mirror is contracted or that the light takes longer to make the round trip.

The OP was asking how Relativity supports the observer's measurement when he is moving and that's what I showed. It seems that you don't want to show it in a frame in which the observer is moving but rather you keep reverting back to the observer's rest frame to assert that the measurement is the same. What is significant in all frames is that all observations and measurements are the same, it's just that the coordinates are different.

 

[ghwellsjr]

In the following the A perception is overlaid onto the U perception using the same time axis, and providing a visual comparison.

Overlaying A's rest frame on all the frames in which A is moving implies that those frames do not contain enough information when in fact, every frame that is created through the Lorentz Transformation process contains exactly the same information as any other such frame and it detracts from the importance and significance of the Lorentz Transformation and I think for that reason should be discouraged. Even a traditional Minkowski plot affirms the Lorentz Transformation process but what you are doing seems to bypass the Lorentz Transformation and diminish its importance and significance.

If anyone wants a visual comparison, they can print each one of my diagrams and look at them side by side.

 

[phyti]

Wow, I'm impressed. Did you come up with this construction technique on your own or did you learn about it somewhere?

A study of geometry and algebra enables you to understand the relations between different forms, such as the hyperbola and its inverse the circle.

Can you use your technique for every in-line scenario? Can I give you a more complicated scenario and see how you do it?

I use a CAD system and also let the software do the calculations. Give me a problem to try.

The OP was asking how Relativity supports the observer's measurement when he is moving and that's what I showed. It seems that you don't want to show it in a frame in which the observer is moving but rather you keep reverting back to the observer's rest frame to assert that the measurement is the same. What is significant in all frames is that all observations and measurements are the same, it's just that the coordinates are different.

I still feel you have to show from the moving observers frame/perspective that regardless of his speed and his different coordinates, he still measures light speed as c, the same as the rest frame.

I'll respond to your other posts later, but wanted to finish this with the math answer to the original question with the attached doc.

https://www.physicsforums.com/attachments/65136

 

[ghwellsjr]

I use a CAD system and also let the software do the calculations. Give me a problem to try.

I still feel you have to show from the moving observers frame/perspective that regardless of his speed and his different coordinates, he still measures light speed as c, the same as the rest frame.

I'll respond to your other posts later, but wanted to finish this with the math answer to the original question with the attached doc.

I'll have to wait til after the New Year when I get back to work to read your document. The readers I have produce garbage. After that I'll start a new thread to give you a challenge problem because it is off topic for this thread.

However, what is not off topic is the implication that your graphical approach does something different than what I have done. All your technique does is provide a graphical means to determine the Proper Time interval of the observer's measurement between the two light signals, the one leaving the observer on the way to the mirror and the one reflected back from the mirror. I am also doing that using the Lorentz Transformation. We're just taking the coordinates of the end point (since the starting point is at the origin) and transforming them to the rest frame or we could do it using the Spacetime Interval:

t=15, x=9

Proper Time interval = √(t²-x²) = √(15²-9²) = √(225-81) = √144 = 12

There's lots of ways to do it. But that only shows how the observer makes the same measurement with his clock no matter what his speed is in any particular frame.

Since he previously measured the distance to the mirror with his ruler as 6 feet, he assumes that the round trip distance the light had to take was 12 feet and since it did it in 12 nanoseconds, he measures the speed of light to be 1 foot per nanosecond, just like in he does in any frame.

When you draw the light path down at the 45-degree angle and intersect with the light path coming up at the 45-degree angle from the origin, you are also assuming that the light path is 12 feet.

So we agree that the transformation process (or any other equivalent process) works in both directions from the rest frame to a frame in which the observer is moving and vice versa, but what I still cannot get you to focus on is that the combined round-trip light path in a frame in which the observer is moving takes longer than it does in the rest frame (although you did draw some pictures in which you show this). I'm still focused on your comment:

The moving frame is a scaled version of the 'rest' frame, therefore
the expressions involving x and t are equivalent (including light speed). This requires x and t to change by the same proportion (γ).

Can't you acknowledge, especially after you made two drawings that show it, that the light has to travel a longer round-trip distance in the frame in which the observer is moving than in his rest frame?

 

[phyti]

Can't you acknowledge, especially after you made two drawings that show it, that the light has to travel a longer round-trip distance in the frame in which the observer is moving than in his rest frame?

I agree with you but that is according to U (the rest frame). A (the moving observer) does not measure this or is aware of it, and the task is to show why, mathematically or geometrically. That's why the overlay and the extra drawing (made from the overlay) as seen from the A perspective. The alternate method just avoids using the x & t transforms.

I also agree, either method shows the same info.

I also encourage you to keep emphasizing the difference between doppler effect and time dilation.

 

[adjacent]

This thread has turned into a discussion but where's the OP?

 

[ghwellsjr]

I agree with you but that is according to U (the rest frame). A (the moving observer) does not measure this or is aware of it, and the task is to show why, mathematically or geometrically. That's why the overlay and the extra drawing (made from the overlay) as seen from the A perspective. The alternate method just avoids using the x & t transforms.

I also agree, either method shows the same info.

I also encourage you to keep emphasizing the difference between doppler effect and time dilation.

Thanks, we are in complete agreement.