Einstein's clock

[Aqueous Id]

The only data that can be observed is that which has already happened.

Within a common reference frame, yes. In opposing reference frames, "when" loses its link. So now you're stuck with the problematics. You have have to apply the Lorentz transform or you're never going to get the right answer.

 

[Motor Daddy]

My understanding is consistent with science: my observation of light within its frame of reference is consistent with its actual spectrum. If I am moving away, I will observe redshift. (It's related to velocity, not distance.)

Redshift? There is no frequency to observe. The light is emitted and when it hits you the timer is stopped. There is no frequency for redshift or blueshift to be observed. The light travels a distance to reach you and when it does the timer stops. Game over.

Do you understand that velocity affects the pitch change of a passing siren, not distance? Evidently not.

Again, you are speaking of changing frequency when you are speaking of pitch changing. There is no frequency!!! The sound travels a distance to reach you, PERIOD! We are not throwing many baseballs at a specific interval, we are throwing ONE BASEBALL, and measuring the time of flight until it hits you and the timer is stopped! I say again, you don't know the difference between frequency and distance!

 

[Motor Daddy]

Within a common reference frame, yes. In opposing reference frames, "when" loses its link. So now you're stuck with the problematics. You have have to apply the Lorentz transform or you're never going to get the right answer.

I get the right answers and I don't use Lorentz transform. Maybe a pic will let you see the error of your ways?

 

[Aqueous Id]

Redshift? There is no frequency to observe. The light is emitted and when it hits you the timer is stopped. There is no frequency for redshift or blueshift to be observed. The light travels a distance to reach you and when it does the timer stops. Game over.

There is no light without frequency, and there is no frequency stability with relative velocity. Game on.

Again, you are speaking of changing frequency when you are speaking of pitch changing. There is no frequency!!! The sound travels a distance to reach you, PERIOD!

Delete PERIOD, insert COMMA: however, you can hear the pitch change when the firetruck passes, whereas the firemen hear a continuous frequency, and this is the problem that begs to be solved. How does the discrepancy arise between the two observers? Note, if the truck creeps by slowly, you hear no pitch change, therefore distance is irrelevant to your perception of pitch.

We are not throwing many baseballs at a specific interval, we are throwing ONE BASEBALL, and measuring the time of flight until it hits you and the timer is stopped!

This analogy will work for pulsed radar, in which I measure range to target by counting the delay from sending a pulse until the return of the echo, dividing that into 2c to get distance. However, it tells me nothing about speed of the approaching aircraft. In this case (referencing the earliest, "continuous wave" [CW ] radars) I can simply emit a continuous frequency, and scan for the return signal. When I find it offset by a frequency Δf I can calculate the relative velocity Δv. It's relative velocity, so it's how fast he is closing on me or moving away. Δv<0 means he's pulling away. Δv=0 means his range is holding steady. And Δv>0 means he is closing. And these correspond to detected frequency shifts at Δf<0, Δf=0, and Δf>0, respectively.

Until you concede that relative velocity forms the basis for observing frequency shift, you will never be able to explain why the siren wails or how the doppler effect works. Beyond that, you only need to accept that the wave moves at a constant speed regardless of the velocity of the emitter, and you're on your way to batting on an actual sand lot.

I say again, you don't know the difference between frequency and distance!

I say again, you don't know the difference between range and speed!

 

[Motor Daddy]

There is no light without frequency, and there is no frequency stability with relative velocity. Game on.

Really? What is the frequency of one photon departing the sun and arriving at earth? The photon travels a specific distance in a duration of time, and there is no frequency of that one photon. We are not talking about the duration of time between multiple photons impacting the receiver on earth, we are talking about the time of travel for one photon to travel from the sun to the earth. Not two or more, but one. Comprehend?

Delete PERIOD, insert COMMA: however, you can hear the pitch change when the firetruck passes, whereas the firemen hear a continuous frequency, and this is the problem that begs to be solved. How does the discrepancy arise between the two observers? Note, if the truck creeps by slowly, you hear no pitch change, therefore distance is irrelevant to your perception of pitch.

You still don't comprehend the difference between one baseball and many baseballs, so let me try again to enlighten you on the differences. We are not talking about you hearing a sound for a duration of time, we are talking about a sound being emitted from the truck a distance away from you, you not hearing it yet, the sound traveling towards you, you still not hearing it yet, and finally, the sound hits you and the clock stops. You did NOT hear a sound for a duration of time. The clock STOPPED when and exactly when the sound reached you. In order for you to hear a change in pitch, multiple sound waves must impact you over a duration of time.

So, to recap, I am talking about the wavefront hitting you and the clock stopping, and you are talking about multiple waves hitting you after the initial wave hits you. One wave (me) and many waves (you). Do you see the light yet, or do you need further schooling?

I say again, you don't know the difference between range and speed!

I say again, you don't know the difference between frequency and distance. You don't know the difference between a wavefront and many waves with a distance between them (with possibly changing distances between them) over a duration of time.

 

[BWE1]

Read what I wrote down, the equation derived from the effective potential of the Schwarzschild equation of motion, accounts for BOTH gravitational and relative motion time dilation. It gives the same result as if you did them individually and summed the results together. If you don't understand how to transform from a coordinate frame to the invariant proper frame then you should be learning that first.

this pretty much sums up this whole thread.

Well put.

 

[arfa brane]

What is the frequency of one photon departing the sun and arriving at earth?

It depends on the energy of the photon.

You aren't confusing frequency with how many times something happens, are you? You realise that radio and TV stations broadcast at different frequencies, that a microwave oven generates a fairly narrow band of frequencies, and that a prism or a rainbow separates different frequencies of visible light, because light has ... wait for it ... different frequencies? So does sound, incidentally.

 

[Mazulu]

It depends on the energy of the photon.

You aren't confusing frequency with how many times something happens, are you? You realise that radio and TV stations broadcast at different frequencies, that a microwave oven generates a fairly narrow band of frequencies, and that a prism or a rainbow separates different frequencies of visible light, because light has ... wait for it ... different frequencies? So does sound, incidentally.

Energy of the photon is E = hf, where h is the Planck constant. This is one of the easiest equations in the world. I don't even have to use the math editor.

 

[Aqueous Id]

Really? What is the frequency of one photon departing the sun and arriving at earth?

The frequency at departure times the Lorentz factor.

The photon travels a specific distance in a duration of time, and there is no frequency of that one photon.

See above.

We are not talking about the duration of time between multiple photons impacting the receiver on earth, we are talking about the time of travel for one photon to travel from the sun to the earth.

See above.

Not two or more, but one. Comprehend?

See above.

You still don't comprehend the difference between one baseball and many baseballs, so let me try again to enlighten you on the differences.

See above.

We are not talking about you hearing a sound for a duration of time, we are talking about a sound being emitted from the truck a distance away from you, you not hearing it yet, the sound traveling towards you, you still not hearing it yet, and finally, the sound hits you and the clock stops.
You did NOT hear a sound for a duration of time. The clock STOPPED when and exactly when the sound reached you. In order for you to hear a change in pitch, multiple sound waves must impact you over a duration of time.

Explain doppler in a siren.

So, to recap, I am talking about the wavefront hitting you and the clock stopping, and you are talking about multiple waves hitting you after the initial wave hits you.

Same-same.

One wave (me) and many waves (you).

Same-same.

Do you see the light yet, or do you need further schooling?

Funny you should mention school. And light.

I say again, you don't know the difference between frequency and distance.

I say again, you don't understand the correspondence between frequency and velocity.

You don't know the difference between a wavefront and many waves.

You don't know the difference between superposition and...and... superposition.

with a distance between them

Same-same. See above.

(with possibly changing distances between them) over a duration of time.

Careful. Propagation velocity is constant.



Meanwhile, think relativity.

 

[Motor Daddy]

Why bother. Your responses are disingenuous and out of context with the discussion, and lack any kind of real substance. I could get a parrot to repeat "See above" and I'd have the same quality of conversation with him as I do you. Have a nice day. I'm not gonna waste my time with you.

 

[Aqueous Id]

I gave you a wavelet with the frequency (wavelength) of a photon to help you out. The other thing you needed was an easy way to understand relativity. I offered you doppler as a function of velocity, to simplify it further. I gave you a simple graphical depiction. All of these are pretty basic and I was willing to help you out. I understand that's not what you want. Sorry it's not panning out for you. I put you on "ignore". Feel free to do the same.

 

[chinglu]

You've just defined an observer-dependent frame of reference. In order for the observer to be "stationary" they must be moving with the system of coordinates. If by "barycenter" you mean the sun, then the observer is stationary wrt to sun, which is moving around the center of the galaxy.

And no, different observers in relative motion will not see the same location or time for light emitted from the earth. This does not mean the light is at different locations, that's just asinine. It means there are two coordinate systems, one for each observer.

Your sentence "one light sphere with the same center is located at two different positions" just does not follow, since the center is also observer dependent.

Yes, there is an observer stationary with the sun coords.

And, when the earth is at that observer, a light is flashed.

Then, exactly one year later let's say the earth again meets that observer.

Then, measuring with an axis perpendicular to the line of motion of the earth, the question to be resolved is where is the light sphere in the say "y" direction.

Since the light emission point of both frames is again common, if one clock is not synched with the other clock, either they measure a different speed of light if the light sphere is located in the y direction at the same place in space, or they claim the one light sphere is at a different location along the common non-contracted y axis..

 

[chinglu]

this pretty much sums up this whole thread.

Well put.

If you really know what you are doing, feel free to answer my previous post.

 

[brucep]

I get the right answers and I don't use Lorentz transform. Maybe a pic will let you see the error of your ways?

So you're the comic relief.

 

[arfa brane]

Since the light emission point of both frames is again common, if one clock is not synched with the other clock, either they measure a different speed of light if the light sphere is located in the y direction at the same place in space, or they claim the one light sphere is at a different location along the common non-contracted y axis..

If one observer is stationary wrt the sun, the other observer is on the earth, then their clocks will not be synchronised.
They each record different times and locations when the light signal is emitted, because they are each in their own frame of reference. "Location" is observer dependent, remember? The observers don't agree on "location" and they don't agree on "when" because they are in relative motion, remember?

There is one light signal, but there are two observers who each see different locations for your light sphere. So saying the same light sphere is in two different places does not follow, unless you mean in two observer dependent places.

When will you get the idea? Why can't you see that you aren't using the right logic?

 

[arfa brane]

So you're the comic relief.

In the Motor Daddy universe, doing an experiment means you draw a diagram and add in some numbers.

No need to measure anything, and especially not the speed of light. It's defined, you see. Although that leaves the question how is it defined if it can't be measured. No problemo, the numbers don't lie, it's hard to argue with a number, and if you give it 20 or 30 decimal places, even harder.

 

[BWE1]

Yes, there is an observer stationary with the sun coords.

And, when the earth is at that observer, a light is flashed.

Then, exactly one year later let's say the earth again meets that observer.

Then, measuring with an axis perpendicular to the line of motion of the earth, the question to be resolved is where is the light sphere in the say "y" direction.

Since the light emission point of both frames is again common, if one clock is not synched with the other clock, either they measure a different speed of light if the light sphere is located in the y direction at the same place in space, or they claim the one light sphere is at a different location along the common non-contracted y axis..

If you really know what you are doing, feel free to answer my previous post.

If I understand this is the problem, then I also understand why responding here is futile. Do you know why General relativity is called general relativity?

You cant use two reference frames to calculate one location or event t.

 

[chinglu]

If I understand this is the problem, then I also understand why responding here is futile. Do you know why General relativity is called general relativity?

You cant use two reference frames to calculate one location or event t.

That is certainly correct, EXCEPT, I put the origins at the same place again and used a y axis that is not in the direction of travel.

I certainly hope you understand basic math, when the origins of 2 coordinate systems are common and nothing else determines an axis other than the axis itself, then you can measure events in the same context.

So, I have the 2 origins at the same place again and an un-contracted y axis.

This light sphere was emitted when the 2 origins where common.

Therefore, I may now ask if the light sphere emitted the previous year is at the same place along the y axis.



GR says no and hence, it is crack-pottery because in order for GR to be true, one light sphere must be located at two different y axis positions since the two frames disagree on time.

 

[arfa brane]

one light sphere must be located at two different y axis positions since the two frames disagree on time.

There's your problem. You don't understand how light propagates, you don't understand the difference between two frames of reference, and you don't understand Einstein.

Your logic is faulty, your conclusion does not, therefore, follow.
Your inability to see this suggests you don't have much intelligence. Have you done any postgrad math or logic? Perhaps a course in relativity? Any undergrad stuff?
If not, I suggest you at least consider the possibility that you don't understand relativity, at all.

 

[chinglu]

There's your problem. You don't understand how light propagates, you don't understand the difference between two frames of reference, and you don't understand Einstein.

Your logic is faulty, your conclusion does not, therefore, follow.
Your inability to see this suggests you don't have much intelligence. Have you done any postgrad math or logic? Perhaps a course in relativity? Any undergrad stuff?
If not, I suggest you at least consider the possibility that you don't understand relativity, at all.

Use your brain to calculate the location of the light sphere when the earth is again at the same location as the observer at rest with the sun after 1 year.

Where is the light sphere located along some y axis when the 2 origins are again common?

Your brain must first answer this question in the context of GR.

If your brain does not have an answer to this, then your brain's comments are useless.