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

ghwellsjr
Gold Member
I never understand why the speed of light is the same for all observers irrespective of their motion relative to the source of light. Now suppose I am sitting at the back of a vehicle which is travelling at the speed of 0.999999999999c and light approaches me from behind the vehicle. i.e. I am going away from the source of light while I can still see the light. Now I attach an instrument to my car for measuring the speed of light. Won't it measure 0.000000000001c.

Please help me understand why will the instrument still record 0.999999999999c and not 0.000000000001c according to the theory of relativity.
I'd like to draw you some spacetime diagrams that illustrate how Special Relativity handles moving observers that all measure the speed of the same flash of light even though they are moving at different speeds relative to the light source but the speed you presented will require a huge diagram so I'll do it with smaller speeds.

As I said in post #9, the way you measure the speed of a flash of light that is coming from behind you while you are sitting in the back of a vehicle is to start a stopwatch when you first see the flash of light coming from behind and then let it reflect off of a mirror some measured distance, I'll use six feet, in front of you and stop the timer when you see the reflection.

Here is a spacetime diagram depicting this scenario starting with you stationary with respect to the light source. You are shown as the blue line and your mirror is shown as the green line. The dots mark off one-nanosecond increments of your time. The flash of light is shown as the black line:

When the flash reaches you, you reset your stopwatch to zero. It takes six nsecs for the flash to continue on to the mirror and another six nsecs for the reflection to get back to you. You stop the stopwatch at 12 nsec. Since the light had to travel double the distance to the mirror, you calculate the speed of light to be 12 feet divided by 12 nsecs or 1 foot per nanosecond.

Now we'll repeat the measurement but this time assuming that you are traveling at 60%c with respect to the same light source and according to the same reference frame as before. Gamma, γ, at this speed for you is 1.25 meaning that the dots marking off 1-nsec intervals of time will be stretched out to 1.25 times the Coordinate Time markings. In other words, the Coordinate Time is dilated from 12 nsecs in the first diagram to 15 nsecs in this diagram. Special Relativity says that the distance to your mirror will be contracted by the reciprocal of gamma or 0.8 times what it was at rest. Since it was 6 feet at rest, the distance to your mirror now will be 4.8 feet of Coordinate Distance as you can see in the following diagram:

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.

But, as I said before, you have no awareness of this Time Dilation from 12 nsecs to 15 nsecs. To you, it is still 12 nsecs because that's what your stopwatch measures. And you have no awareness that in this frame, the distance to your mirror is closer to you than it was while you were at rest because any ruler that you use to measure the distance is contracted by the same amount.

Now let's repeat the measurement for you traveling a little faster, at 80%c. At this speed, gamma is 1.667 (one and two-thirds). Here is a new spacetime diagram:

As you can see, the Time Dilation has grown to 20 nsecs and your mirror is now 3.6 feet away and the distances the light has to travel have increased to 18 feet going and 2 feet returning for a total of 20 feet so the speed of light is 20 feet per 20 nsecs or 1 foot per nsec. Again, you have no awareness of these numbers. To you, it is still a total of 12 feet in 12 nsecs.

Have you noticed the trend in how the diagrams show that the mirror gets closer to you and the dots marking your 1-nsec tick marks get farther apart? Now, in an attempt to get as close to your desired speed as possible and still draw a decent diagram, I have made one more at a speed of 98%c where gamma is 5.025:

Notice that this is at a different scale from the previous diagrams. Notice that your tick marks are spaced slightly more than 5 nsecs apart and that the total time is expanded to 60.3 feet and distance to your mirror is contracted to less than 1.2 feet. The distances the light has to travel have also expanded to just under 60 feet going and just over 0.3 feet returning for a total of 60.3 feet resulting in a speed of 60.3 feet divided by 60.3 nsecs or 1 foot per nanosecond.

If we tried to make a diagram for your target speed of 0.999999999999c, gamma would be 707106.78 which would make the diagram incredibly large if you want to actually see the details. But you could do all the calculations and show that the Time Dilation, Length Contraction and distance expansion for the light paths would still result in the light speed being 1 foot per nanosecond. But again, you would have no awareness of any of this. It's no different for you than it was when you were at rest in the frame.

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ghwellsjr, btw what software do you use to draw these diagram??

ghwellsjr
Gold Member
ghwellsjr, btw what software do you use to draw these diagram??
I wrote my own application using a programming language called LabVIEW.

Your comment was similar to rushikesh's in that the calculation of the speed of light (length divided by time) would come out the same because both parameters were equally "retarded" and I just wanted you to think about the situation where the length for a calculation applied at 90 degrees to the direction of motion would not be retarded and yet the speed of light still comes out to be c. The light source never has any bearing on the speed of light. It is defined to be c in all inertial reference frames.
It's about the comparative geometry between the two, including the difference in claiming what is simultaneous.

I was talking about comparing the spacing of the Proper Time tick marks on a spacetime diagram to the coordinate time markings.

No, I don't see a problem. If you are measuring the round-trip speed of light to a mirror that is 10 feet away and your ruler is actually only six inches long but claims to be 12 inches long, then you will measure the distance to be 20 feet in one direction or 40 feet for the round trip.
Huh? Who's measure of twelve feet? Who's measure of 20 feet. To your point everyone measures c to be the same. If time is retarded, so is length. If you wish to consider angles of approach to direction of motion, then consider simultaneity. Such as an analog clock in motion, the dials don't move at a constant speed around the face.

All I'm trying to point out is that if we recognize that time gets "dilated" (or expanded, because that's what the word means) and length gets contracted, then it's less likely that someone will jump to the conclusion that the calculation of the speed of light (being length divided by time) remains at c simply because the effects "cancel out".
We must conceptualize this completely different. It was one of the other pf'ers that mentioned a term "Rate Contraction" imo it's a great term. The effects on time/length do change equally and in turn motion has no impact on mechanics.

I'm poor at wording these things, so here is a copy n paste from South Whales U website

"If they agree on the speed of light, but disagree on measurements of time, they must inevitably disagree on length as well. If you observe someone's clocks run slowly by a factor γ, you will also observe her rulers to be short by a factor of γ: that's the only way that she can measure the speed of light to have the same value you get."

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ghwellsjr
Gold Member
Your comment was similar to rushikesh's in that the calculation of the speed of light (length divided by time) would come out the same because both parameters were equally "retarded" and I just wanted you to think about the situation where the length for a calculation applied at 90 degrees to the direction of motion would not be retarded and yet the speed of light still comes out to be c. The light source never has any bearing on the speed of light. It is defined to be c in all inertial reference frames.
It's about the comparative geometry between the two, including the difference in claiming what is simultaneous.
Of course it's about the geometry, but it's not about simultaneity in this geometry. And in this geometry that I asked you to think about, there is no Length Contraction, only Time Dilation. So why would you claim that the way Special Relativity "explains" how a moving observer can measure the speed of light to be c is that Length Contraction and Time Dilation are both retarded to the same degree? That's my question for you and you didn't address it.

I was talking about comparing the spacing of the Proper Time tick marks on a spacetime diagram to the coordinate time markings.

No, I don't see a problem. If you are measuring the round-trip speed of light to a mirror that is 10 feet away and your ruler is actually only six inches long but claims to be 12 inches long, then you will measure the distance to be 20 feet in one direction or 40 feet for the round trip.
Huh? Who's measure of twelve feet? Who's measure of 20 feet.
Huh? I never said anything about twelve feet. I said, if you have a ruler that is only six inches long (one-half foot) but is marked as being 12 inches long (one foot) and you measure a distance that is 10 feet, you will think it is 20 feet.

To your point everyone measures c to be the same. If time is retarded, so is length. If you wish to consider angles of approach to direction of motion, then consider simultaneity. Such as an analog clock in motion, the dials don't move at a constant speed around the face.
Time isn't retarded, it's expanded. That's what dilated means.

I don't see why you want to confuse this issue by bringing up an analog clock. An analog clock in motion (provided the motion is not parallel to the axis of the shafts of the dials) is not even round, it is oval shaped. Not only do the dials not move at a constant speed, they don't even remain at a constant length. Their lengths change as they go around. So what? We're not talking about that. We're talking about how a moving observer has a Length Contracted ruler to measure the distance to a mirror and a Time Dilated stopwatch (or timer or clock) that is colocated with him. The answer that we supply has to work for all orientations of the ruler and mirror with respect to the direction of motion. Can you please address that issue and not bring up side issues?

All I'm trying to point out is that if we recognize that time gets "dilated" (or expanded, because that's what the word means) and length gets contracted, then it's less likely that someone will jump to the conclusion that the calculation of the speed of light (being length divided by time) remains at c simply because the effects "cancel out".
We must conceptualize this completely different. It was one of the other pf'ers that mentioned a term "Rate Contraction" imo it's a great term. The effects on time/length do change equally and in turn motion has no impact on mechanics.
As I so carefully pointed out in post #26, motion has everything to do with why (according to Special Relativity) a moving observer measures the speed of light to be c even when light is traveling at c in the frame in which he is in motion. It's a combination of his Time Dilated stopwatch, his Length Contracted ruler, and his motion, for the case where he is measuring along the direction of motion. If he's measuring perpendicular to his direction of motion, then it is a combination of just his Time Dilated stopwatch and his motion since his ruler is not Length Contracted in this case.

Did post #26 make sense to you?

I'm poor at wording these things, so here is a copy n paste from South Whales U website

"If they agree on the speed of light, but disagree on measurements of time, they must inevitably disagree on length as well. If you observe someone's clocks run slowly by a factor γ, you will also observe her rulers to be short by a factor of γ: that's the only way that she can measure the speed of light to have the same value you get."
That's a rather surprising comment to make since at the top of their webpage, they gave an example of Zoe's light clock oriented perpendicular to her direction of motion and although she and her car are Length Contracted along her direction of motion, the width of her car and the distance between her mirrors are not Length Contracted.

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

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ghwellsjr
Gold Member
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.

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bhobba
Mentor
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 travelling 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&tag=pfamazon01-20

It also shows how Bell can be accommodated in SR - even though that suggests non local effect can happen instantaneously it cant be used to send information

Thanks
Bill

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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.

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.

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ghwellsjr
Gold Member
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
Gold Member
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
Gold Member
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.

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.

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ghwellsjr
Gold Member
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?

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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.

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

ghwellsjr