thorkil2

2008-Oct-15, 06:51 PM

For the Moderators: I'm not sure where to put this, as it is not, strictly speaking, a new question. But I made a comment in the Black Hole? thread that ended up in an argument that hijacked the thread. It isn't ATM, and I don't see another easy fit, but feel free to place it where you will if this is inappropriate. I've opened this thread to answer objections raised without taking over Mike842's OP any further than we have already.

I've quoted much of the argument below for background.

I don't see it as strange. For the photon or the wave front, the speed of light is infinite (edit) within that frame of reference (end edit). Zero distance from point a to point b (because it takes zero time to get there). Not hard then to see the Universe as a singularity from that frame of reference.

Ahhhhh, So, when 'science'/we/they/mainstream does the Lunar Ranging experiments, the photon/wave front is traveling 'instantly' to and from the moon??? Either? Neither?

So far as the observer is concerned, no. We measure a finite time of travel. But the photon, in it's own frame of reference, takes zero time to get from any point a to any point b anywhere in the Universe. It is only seen as finite speed from a frame of reference that is not moving at c.

So, when 'science' measures photons to be traveling at a Constant "c", 186,282.40 miles per second, that means that when we send a light signal to the moon, in 1 tenth of a second, that 'wave front' has traveled ~18,628 miles......and will reach the 186,282.40 mile mark in 9 more tenths of a second

SO, how do you explain those photons in your definition outracing the 'wave front' to get to the moon "Instantaneously"/in 0 time?

The zero travel time is only from the frame of reference of the photon, not from your position as a measuring entity. You experience and measure the time it takes for a photon to get from the earth to the moon. From the photon's frame of reference, traveling at c, no time elapses and there is no distance. If you could travel at c, no time would elapse for you either, for any distance traveled. It's a consequence of time dilation. The photon you see takes a finite travel time as you measure it. If you could place a hypothetical clock at the wavefront, traveling with the wavefront, it would take your identical rest frame clock an infinite number of seconds to equal the traveling clock's one second (or any fraction thereof, for that matter). Of course, travel at c is impossible for anything but the photon/light wave, but it still is a definable frame of reference, and what happens there is not the same as what you observe from your rest frame.

To get a clearer picture, try this: You have two identical clocks. One travels at 0.5c, the other remains with you in your rest frame. It will take your clock 1.1547 seconds to equal one second on the traveling clock. At 0.9c, it will take your clock 2.2941 seconds to equal one second of the traveling clock, and the interval grows as the speed increases toward c. At c (if you could achieve c) the difference would be infinite. Length contraction does not apply only to the moving object, but (because of time dilation) to the subjective distance traveled. If you're the traveler, the Universe gets shorter for you as you approach c. Again, if you could achieve c, distance (from your moving frame) would disappear. It would take zero time to get from any point a to any point b in the Universe. That's not ATM; I think you'll find the effect described in any physics book.

To put this another way, for c to be constant, time and space must be variable in the ways noted above.

First of all, THIS is absolutely False!

To make a flat out statement like this is absolutely irresponsible!

"c" can be Constant and Time can be Constant as well as 'Space'.

Not since Newton was expanded by Einstein. In fact, much of the point of SR is that space and time are frame-dependent.

How do you suggest we discuss redshift? Perhaps we call it an instant shift in the energy of the photon, which never had time nor place to gain or lose energy in the first place.

Let's teach it starting tomorrow, sounds like fun.

The only way we can say a photon is instantaneous is if it has an instant effect, which has not been found.

For timb and alainprice--come on, this is text book SR. If you have a complaint, take it to Einstein. The red shift is measured from some relative rest frame. Your measurements are valid from that frame and only that frame. Change frames, and you get different results. I'll say it differently: if I could use the word "experiences" loosely in this context, for lack of a better word, I would say the photon "experiences" no elapsed time from source to destination. That has nothing to do with what you measure because you are doing the measuring from a different frame of reference. It isn't instantaneous from any frame of reference you have access to, so no, there won't be instant effect. To the photon, however, the effect is instantaneous. Different frames of reference. Now, I've been dealing with this for 50 years, but I won't make you take my word for it. A quote from Paul Davies (Professor of Natural Philosophy at the University of Adelaide in Australia at the time this was written) in his book "About Time" (1995), p.190:

From the point of view of the [light] pulse, no time at all elapses as, in our frame of reference, it sweeps across the solar system.

Will that do, or do I need to dig out more references?

SO, when we send a light beam to the moons reflector (Let's assume for ease that the moon is exactly 186,282.40 Miles) we wind up with 2 maths for the 'same' beam IE; that beam is both traveling at "c" and takes one second to reach the moon AND that same beam traveled 0 distance (Because 'space was contracted) in 0 time, umm or is it 186,282.40 miles traveled "Instantly"?and BOTH maths are considered "Simultaneous"?

Don't you think that this causes a "Duality" problem?

IF, there was a mirror at 186,282.40 miles, and if those photons were 'Really' getting there 'instantaneously', then would light travel time there and back only be 1/2 of what it should be ie; 1 second rather than the round trip time of 2 seconds?

So, to repeat my former question (and open with a question), how many other sources do I need to dig up? Paul Davies is a well-known and highly respected Physicist, not some crackpot I pulled off the wall. This is textbook Special Relativity. I can dig up more, but that's time-consuming, and I don't see how the conversation can go forward productively until you've read up on SR and can make your points in that context. But to answer your last question: No. The photons get there and back instantaneously--from their frame of reference. From yours, as the observer (a different frame of reference), they take time to get there and back. I can't seem to repeat often enough that the frames are different.

I've quoted much of the argument below for background.

I don't see it as strange. For the photon or the wave front, the speed of light is infinite (edit) within that frame of reference (end edit). Zero distance from point a to point b (because it takes zero time to get there). Not hard then to see the Universe as a singularity from that frame of reference.

Ahhhhh, So, when 'science'/we/they/mainstream does the Lunar Ranging experiments, the photon/wave front is traveling 'instantly' to and from the moon??? Either? Neither?

So far as the observer is concerned, no. We measure a finite time of travel. But the photon, in it's own frame of reference, takes zero time to get from any point a to any point b anywhere in the Universe. It is only seen as finite speed from a frame of reference that is not moving at c.

So, when 'science' measures photons to be traveling at a Constant "c", 186,282.40 miles per second, that means that when we send a light signal to the moon, in 1 tenth of a second, that 'wave front' has traveled ~18,628 miles......and will reach the 186,282.40 mile mark in 9 more tenths of a second

SO, how do you explain those photons in your definition outracing the 'wave front' to get to the moon "Instantaneously"/in 0 time?

The zero travel time is only from the frame of reference of the photon, not from your position as a measuring entity. You experience and measure the time it takes for a photon to get from the earth to the moon. From the photon's frame of reference, traveling at c, no time elapses and there is no distance. If you could travel at c, no time would elapse for you either, for any distance traveled. It's a consequence of time dilation. The photon you see takes a finite travel time as you measure it. If you could place a hypothetical clock at the wavefront, traveling with the wavefront, it would take your identical rest frame clock an infinite number of seconds to equal the traveling clock's one second (or any fraction thereof, for that matter). Of course, travel at c is impossible for anything but the photon/light wave, but it still is a definable frame of reference, and what happens there is not the same as what you observe from your rest frame.

To get a clearer picture, try this: You have two identical clocks. One travels at 0.5c, the other remains with you in your rest frame. It will take your clock 1.1547 seconds to equal one second on the traveling clock. At 0.9c, it will take your clock 2.2941 seconds to equal one second of the traveling clock, and the interval grows as the speed increases toward c. At c (if you could achieve c) the difference would be infinite. Length contraction does not apply only to the moving object, but (because of time dilation) to the subjective distance traveled. If you're the traveler, the Universe gets shorter for you as you approach c. Again, if you could achieve c, distance (from your moving frame) would disappear. It would take zero time to get from any point a to any point b in the Universe. That's not ATM; I think you'll find the effect described in any physics book.

To put this another way, for c to be constant, time and space must be variable in the ways noted above.

First of all, THIS is absolutely False!

To make a flat out statement like this is absolutely irresponsible!

"c" can be Constant and Time can be Constant as well as 'Space'.

Not since Newton was expanded by Einstein. In fact, much of the point of SR is that space and time are frame-dependent.

How do you suggest we discuss redshift? Perhaps we call it an instant shift in the energy of the photon, which never had time nor place to gain or lose energy in the first place.

Let's teach it starting tomorrow, sounds like fun.

The only way we can say a photon is instantaneous is if it has an instant effect, which has not been found.

For timb and alainprice--come on, this is text book SR. If you have a complaint, take it to Einstein. The red shift is measured from some relative rest frame. Your measurements are valid from that frame and only that frame. Change frames, and you get different results. I'll say it differently: if I could use the word "experiences" loosely in this context, for lack of a better word, I would say the photon "experiences" no elapsed time from source to destination. That has nothing to do with what you measure because you are doing the measuring from a different frame of reference. It isn't instantaneous from any frame of reference you have access to, so no, there won't be instant effect. To the photon, however, the effect is instantaneous. Different frames of reference. Now, I've been dealing with this for 50 years, but I won't make you take my word for it. A quote from Paul Davies (Professor of Natural Philosophy at the University of Adelaide in Australia at the time this was written) in his book "About Time" (1995), p.190:

From the point of view of the [light] pulse, no time at all elapses as, in our frame of reference, it sweeps across the solar system.

Will that do, or do I need to dig out more references?

SO, when we send a light beam to the moons reflector (Let's assume for ease that the moon is exactly 186,282.40 Miles) we wind up with 2 maths for the 'same' beam IE; that beam is both traveling at "c" and takes one second to reach the moon AND that same beam traveled 0 distance (Because 'space was contracted) in 0 time, umm or is it 186,282.40 miles traveled "Instantly"?and BOTH maths are considered "Simultaneous"?

Don't you think that this causes a "Duality" problem?

IF, there was a mirror at 186,282.40 miles, and if those photons were 'Really' getting there 'instantaneously', then would light travel time there and back only be 1/2 of what it should be ie; 1 second rather than the round trip time of 2 seconds?

So, to repeat my former question (and open with a question), how many other sources do I need to dig up? Paul Davies is a well-known and highly respected Physicist, not some crackpot I pulled off the wall. This is textbook Special Relativity. I can dig up more, but that's time-consuming, and I don't see how the conversation can go forward productively until you've read up on SR and can make your points in that context. But to answer your last question: No. The photons get there and back instantaneously--from their frame of reference. From yours, as the observer (a different frame of reference), they take time to get there and back. I can't seem to repeat often enough that the frames are different.