Seeing things in their past? You are full of beans!

[TrueBeliever37]

Once again, there is no reference frame for the photon. Talking about coordinate distances and time only makes sense in a reference frame. The closest you can get for a photon is for proper time, which is always zero for a photon. There is no conversion factor going from proper time to coordinate time.

You are trying to do a conversion (proper time to coordinate time) that makes no sense.

We see differences in time: those are coordinate time. Those are NOT directly related to proper time for the photon path, which is all that makes sense in the case of the photon. NO CONVERSION FACTOR RELATES PROPER TIME AND COORDINATE TIME!

Polymath, I am not asking for a reference frame for the photon. I am just asking if a photon experiences no time or distance, then WHY are we able to detect differences in time for a photon to travel different expanses in space?

What does proper time even mean then? What does it really tell you?

 

[Polymath257]

Then what does it even mean to say t=0, and d=0 for a photon? It makes no sense to say it travels at c, yet time and distance are both 0.

It travels at c in every reference frame. But a photon does not have a reference frame where it is at rest. So, you are right, no time of distance makes sense ina nd of themsevles: there are no coordinates for a photon frame. The *only* way to make sense of the question of what the photon 'experiences' is by asking about the proper time between events at the ends of a photon path. And that proper time is 0. This isn't t=0 (since that would be a coordinate time) and there is no d=0 (that would be coordinate distance).

 

[Polymath257]

Polymath, I am not asking for a reference frame for the photon. I am just asking if a photon experiences no time or distance, then WHY are we able to detect differences in time for a photon to travel different expanses in space?

When you ask what a photon experiences, you are asking about the proper time along a photon path. When you are asking about *we* detect differences in time, you are asking about coordinate time in a reference frame. Those are just two very different things.

What does proper time even mean then? What does it really tell you?

Well, for objects moving slower than light, it tells how much time they experience. For a photons, it is a calculation mechanism to help determine photon paths. Since it it rather silly to ask what a photon 'experiences' and since it has no reference frames for coordinate times or distances, the proper time is useful for answering silly questions.

 

[TrueBeliever37]

When you ask what a photon experiences, you are asking about the proper time along a photon path. When you are asking about *we* detect differences in time, you are asking about coordinate time in a reference frame. Those are just two very different things.



Well, for objects moving slower than light, it tells how much time they experience. For a photons, it is a calculation mechanism to help determine photon paths. Since it it rather silly to ask what a photon 'experiences' and since it has no reference frames for coordinate times or distances, the proper time is useful for answering silly questions.

Polymath, Thanks for all the discussion.

The conclusions I have reached after all of this are that:

Light travels at a constant speed of 186,000 miles/sec

It takes a photon longer to travel a longer distance, than it does a shorter distance, as can be seen when using our frame of reference. Same result using any valid reference frame for that matter, so a time difference is always involved.

It is basically worthless to discuss t=0, and d=0 for a photon, because it has no valid meaning, no reference frame, no conversion to coordinate times or distances, etc. It's only use appears to be to battle questions challenging this theory.

 

[gnostic]

If t=0 and d=0 for a photon to travel, isn't that what is implied? No matter if it is traveling what we would call 30 million light years or 500 light seconds, the photon is said to experience no time or distance.

As far as I know, since I am no expert in Relativity, there is no such thing as “light second”.

As indicated in my past replies, the light year is a unit of length or distance taken for light to travel in one year. It isn’t a measure of time.

A light-year (symbol: ly), sometimes written light year or lightyear is an astronomical unit of length (not time) equal to just under 9.5 trillion kilometres (or about 6 trillion miles).

If you look at the unit for light year. When converting light year into units that we normally used, they are are given in metres, kilometres or miles; the unit conversion are not time.

Light-year
Unit system astronomical units
Unit of length
Symbol ly
Unit conversions
1 ly in... is equal to...

SI units 9.4607×10^15 m​

imperial & US units 5.8786×10^12 mi

other astronomical

6.3241×10^4 AU units
0.3066 pc​

All astronomy textbook will say similar thing.

Your example, “30 million light years” is equal to 283.8x10^12 km.

As to how long for photon to travel 30 million light-year?

That would be

t = d / c
t = 283.8x10^15 (m) / 299.792x10^6 (m/s)
t = 299x10^9 seconds = 57,0381 years​

So light will take 57,0381 years to reach 30 million ly.

 

[Subduction Zone]

As far as I know, since I am no expert in Relativity, there is no such thing as “light second”.

As indicated in my past replies, the light year is a unit of length or distance taken for light to travel in one year. It isn’t a measure of time.



If you look at the unit for light year. When converting light year into units that we normally used, they are are given in metres, kilometres or miles; the unit conversion are not time.



All astronomy textbook will say similar thing.

Your example, “30 million light years” is equal to 283.8x10^12 km.

As to how long for photon to travel 30 million light-year?

That would be

t = d / c
t = 283.8x10^15 (m) / 299.792x10^6 (m/s)
t = 299x10^9 seconds = 57,0381 years​

So light will take 57,0381 years to reach 30 million ly.

A light second is how far a photon travels in a second. Time does not exist for the photon itself. A difficult concept to grasp. Nor does distance. When one is talking about a light second or a light year one is talking as an observer of a photon, not as the photon itself:

"It is defined as the distance that light travels in free space in one second, and is equal to exactly 299792458 metres by the definition of metre (just over 186282 miles)."

https://en.wikipedia.org/wiki/Light-second

 

[Polymath257]

As far as I know, since I am no expert in Relativity, there is no such thing as “light second”.

As indicated in my past replies, the light year is a unit of length or distance taken for light to travel in one year. It isn’t a measure of time.

The phrase 'light second' is standard. It means the distance light goes in one second, so 186,282 miles.

 

[TrueBeliever37]

As far as I know, since I am no expert in Relativity, there is no such thing as “light second”.

As indicated in my past replies, the light year is a unit of length or distance taken for light to travel in one year. It isn’t a measure of time.



If you look at the unit for light year. When converting light year into units that we normally used, they are are given in metres, kilometres or miles; the unit conversion are not time.



All astronomy textbook will say similar thing.

Your example, “30 million light years” is equal to 283.8x10^12 km.

As to how long for photon to travel 30 million light-year?

That would be

t = d / c
t = 283.8x10^15 (m) / 299.792x10^6 (m/s)
t = 299x10^9 seconds = 57,0381 years​

So light will take 57,0381 years to reach 30 million ly.

Some of the guys used light seconds in their discussions with me, so I assumed that was also a term used.

I realize a light year is a distance. But 30 million light years is supposed to be the distance that light will travel in 30 million years, from what they have said in these discussions.

Here is the problem, they use an equation where observer time = proper time/ square root of (1-(v/c)^2 )
In this equation, if I understand correctly,

observer time is the time an observer A on earth would record, for say a ship B to get from earth to the Sun at a certain velocity.

proper time is the time an observer B on ship B would record for the ship at that velocity, to get from earth to the Sun. (So at this point in the discussion, proper time represents an actual time that the ship B would see itself as taking to get from earth to Sun.)

But they keep looking at this equation as the velocity of the ship approaches c. When the velocity reaches c, the proper time has a limit of 0. Observer time on earth would be recorded as 500 seconds. (Which if you will notice, is the time it took the ship to get from earth to Sun, from observer A on earth's reference.)

If they kept the same meaning for proper time, as representing the time it took to get from the earth to the Sun, from observer B's reference, then at least it would make sense. But suddenly when proper time = 0, it no longer represents the time it took for the ship (or photon) to get from earth to Sun. Because if it did take 0 time to get from earth to Sun, it would also take 0 time to get from any star to earth, because proper time always ends up being 0 when you use this limit.

But I have asked them about a photon race between photon A going from Sun to earth, versus photon B going from sun to a planet 30 million ly away. I wanted to know if the race would be a tie, if both photons were emitted at the same instant, from the same source. Because if proper time = 0 does represent the time it would take to get from sun to earth, or sun to planet, it should be a tie. I believe I was told it would not be a tie.

So in my view they are applying the math in a faulty manner. Proper time should always represent the same thing no matter what v is.

 

[Polymath257]

Some of the guys used light seconds in their discussions with me, so I assumed that was also a term used.

I realize a light year is a distance. But 30 million light years is supposed to be the distance that light will travel in 30 million years, from what they have said in these discussions.

Here is the problem, they use an equation where observer time = proper time/ square root of (1-(v/c)^2 )
In this equation, if I understand correctly,

observer time is the time an observer A on earth would record, for say a ship B to get from earth to the Sun at a certain velocity.

proper time is the time an observer B on ship B would record for the ship at that velocity, to get from earth to the Sun. (So at this point in the discussion, proper time represents an actual time that the ship B would see itself as taking to get from earth to Sun.)

But they keep looking at this equation as the velocity of the ship approaches c. When the velocity reaches c, the proper time has a limit of 0. Observer time on earth would be recorded as 500 seconds. (Which if you will notice, is the time it took the ship to get from earth to Sun, from observer A on earth's reference.)

If they kept the same meaning for proper time, as representing the time it took to get from the earth to the Sun, from observer B's reference, then at least it would make sense. But suddenly when proper time = 0, it no longer represents the time it took for the ship (or photon) to get from earth to Sun. Because if it did take 0 time to get from earth to Sun, it would also take 0 time to get from any star to earth, because proper time always ends up being 0 when you use this limit.

But I have asked them about a photon race between photon A going from Sun to earth, versus photon B going from sun to a planet 30 million ly away. I wanted to know if the race would be a tie, if both photons were emitted at the same instant, from the same source. Because if proper time = 0 does represent the time it would take to get from sun to earth, or sun to planet, it should be a tie. I believe I was told it would not be a tie.

So in my view they are applying the math in a faulty manner. Proper time should always represent the same thing no matter what v is.

And, once again, there is no valid reference frame for the photon in which to measure the time from B's perspective if B is going the speed of light. The only alternative is to use the proper time, which is 0 according to the equation.

 

[TrueBeliever37]

And, once again, there is no valid reference frame for the photon in which to measure the time from B's perspective if B is going the speed of light. The only alternative is to use the proper time, which is 0 according to the equation.

But Polymath, once again proper time is meaningless the way you apply it. Why does it mean an actual time when v is less than c, but the answer doesn't represent the same thing once v =c?

Or are you saying there is time involved for the photon, it just can't be measured from it's reference?

 

[TrueBeliever37]

And, once again, there is no valid reference frame for the photon in which to measure the time from B's perspective if B is going the speed of light. The only alternative is to use the proper time, which is 0 according to the equation.

So the only way to determine if it does take a photon more time to travel longer distances is to look at it from the other frames of reference. And in all the valid frames of reference, it takes the photon a longer time to travel a longer distance.

They may be different times recorded in different frames, but they all show a difference in time for different distances.

 

[Polymath257]

But Polymath, once again proper time is meaningless the way you apply it. Why does it mean an actual time when v is less than c, but the answer doesn't represent the same thing once v =c?

Because the phrase 'actual time' only makes sense in a reference frame and only v<c actually have reference frames.

Or are you saying there is time involved for the photon, it just can't be measured from it's reference?

It does not *have* a reference! That is the whole point!

 

[Polymath257]

So the only way to determine if it does take a photon more time to travel longer distances is to look at it from the other frames of reference. And in all the valid frames of reference, it takes the photon a longer time to travel a longer distance.

They may be different times recorded in different frames, but they all show a difference in time for different distances.

Yes, the concept of time you are using is coordinate time. Such only exists in a reference frame.

But, there is *proper* time, which makes sense for a photon and agrees with coordinate time when there is a rest frame. That proper time is 0 for a photon. It isn't useful for computing speeds, but it does make sense and agrees with your intuition for speeds slower than that of light. It is useful for many computational reasons.

 

[TrueBeliever37]

As far as I know, since I am no expert in Relativity, there is no such thing as “light second”.

As indicated in my past replies, the light year is a unit of length or distance taken for light to travel in one year. It isn’t a measure of time.



If you look at the unit for light year. When converting light year into units that we normally used, they are are given in metres, kilometres or miles; the unit conversion are not time.



All astronomy textbook will say similar thing.

Your example, “30 million light years” is equal to 283.8x10^12 km.

As to how long for photon to travel 30 million light-year?

That would be

t = d / c
t = 283.8x10^15 (m) / 299.792x10^6 (m/s)
t = 299x10^9 seconds = 57,0381 years​

So light will take 57,0381 years to reach 30 million ly.

Polymath,
Is the math he shows here correct for 30 million light years? Surely the comma is at least off in the years. Does it not actually work out to be 30 million years?

 

[gnostic]

Polymath,
Is the math he shows here correct for 30 million light years? Surely the comma is at least off in the years. Does it not actually work out to be 30 million years?

“30 million years” is a measure of time, not distance.

“30 million light years” is measure of distance, not time.

My calculations is how long it would take to travel the distance of 30 million light years away.

Your problem is that you are confusing “light year” with actual “year”.

 

[Polymath257]

As far as I know, since I am no expert in Relativity, there is no such thing as “light second”.

As indicated in my past replies, the light year is a unit of length or distance taken for light to travel in one year. It isn’t a measure of time.



If you look at the unit for light year. When converting light year into units that we normally used, they are are given in metres, kilometres or miles; the unit conversion are not time.



All astronomy textbook will say similar thing.

Your example, “30 million light years” is equal to 283.8x10^12 km.

As to how long for photon to travel 30 million light-year?

That would be

t = d / c
t = 283.8x10^15 (m) / 299.792x10^6 (m/s)
t = 299x10^9 seconds = 57,0381 years​

So light will take 57,0381 years to reach 30 million ly.

This is badly, badly wrong.

One light year is about 9.5 trillion = 9.5 * 10^12 km. Multiply that by 30 million = 30*10^6 to find that 30 million light years is about 280*10^18 = 2.8 * 10^20 km. So, your computation for the distance of 30 million light years is way off.

Now, divide that by the speed of light, which is 299,792 km/s = 2.99792 * 10^5 km/s to see that the trip takes 9.5 * 10^14 seconds. There are 3.155*10^7 seconds in a year. Dividing, we find
3*10^7 = 30 million years for the trip, as it should be.

 

[Polymath257]

“30 million years” is a measure of time, not distance.

“30 million light years” is measure of distance, not time.

My calculations is how long it would take to travel the distance of 30 million light years away.

Your problem is that you are confusing “light year” with actual “year”.

Your calculation at almost every stage was badly wrong. The distance of 30 million light years wasn't converted to km correctly. The division of the distance by the speed of light wasn't done correctly.

And, if you understood the concept of a light year, the answer would be obvious. It takes light 30 million years to go 30 million light years.

 

[gnostic]

Apologies, TrueBeliever37.

I went over my calculations again. My conversions were off, which also muck up all my other calculations.

My background in science is civil engineering, so my physics most dealing with Newtonian mechanics, hence forces and acceleration. Numbers are lot smaller, and I don’t do much conversions, because being an Australian, all textbooks are metric, so very little conversions are involved.

So astrophysics and Relativity are not my strong points, so dealing with numbers of astronomical distance and speed of light are not something I deal with day-to-day.

And thank you, polymath257, for the corrections. And I hoped that I’ve learned from my mistakes.

 

[TrueBeliever37]

Apologies, TrueBeliever37.

I went over my calculations again. My conversions were off, which also muck up all my other calculations.

My background in science is civil engineering, so my physics most dealing with Newtonian mechanics, hence forces and acceleration. Numbers are lot smaller, and I don’t do much conversions, because being an Australian, all textbooks are metric, so very little conversions are involved.

So astrophysics and Relativity are not my strong points, so dealing with numbers of astronomical distance and speed of light are not something I deal with day-to-day.

And thank you, polymath257, for the corrections. And I hoped that I’ve learned from my mistakes.

No problem, I knew it was wrong because of all my discussions with these guys, but wanted polymath257 to answer because I didn't think you would accept what I said.