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

[Polymath257]

Yes, But that was what I meant by, they don't come up with the actual/real/true physical length of 5 1/2". All the ships came up with the true length when it was actually on board with them. That is how we know the true length.

OK, so if we see a star that is 4ly away and is moving toward us at 86% of c, what is the 'true distance' between us and the star?

I understand and do agree with the above as far as the math is concerned.

OK, that's at least a good start.

I just don't believe that time and distance actually become zero for everything in any reference frame. That has got to be a make believe world

, because as I have said so many times - a light year is an actual physical length. It's like the pen, if they all could have actually taken a tape and measured the pen on board they would have all had one value they agreed on.

Again, that isn't a valid frame, but we will get there.

I agree that a number can be different in another frame of reference, But one number wouldn't equal every other number you can come up with in the other reference frame, which is what happens with your limiting frame.

To me when you are working in the metric system , that would be like being in another frame of reference compared to our system.
And there are conversion factors you use to go back and forth between the 2 systems. And there is only one value that relates to another value in the other system.

NO. That is NOT what we mean by a different reference frame. That is a different system of units. The lengths are simply a conversion between each other, like you said. But that is NOT what happens between different reference frames. It isn't a simple unit conversion to go between reference frames.

I'll give an example of different reference frames.

Let's have one reference frame be the one where the Earth is at rest. Let the second one be the frame of a car moving down the road at 50mph (a very slow speed). Now, for this new frame, the relativistic effects are minimal, so I'm going to ignore them.

So, the earth sees the car going one way at 50mph. The *car* sees the *earth* moving in the opposite direction at 50mph. But, if the car is moving smoothly, and we drop a ball, it falls directly down *as seen in the car*. From the frame of the earth, that ball moves some distance forward as it falls. The two descriptions are completely equivalent, but there is not a simple conversion of units between them.

In the same way, a ball dropped on the earth (at rest) will fall directly down as seen from the earth, but as seen from the car, it will travel backwards at 50mph as it falls. Again this is a perfectly correct and accurate description of the motion of the ball from the point of view (reference frame) of the car.

Now, suppose that another car is moving along the same road at 60mph. it sees the earth going in the opposite direction at 60mph. Once again, if we are in this second car and drop a ball, the ball drops directly down *from the point of view of those in the car*. But, from the point of view of the earth, the ball moves forward along with the car as it falls.

Those in the second car see a ball dropped at rest on the earth as moving backward at 60mph as it falls. Again, this is the difference in descriptions from two different reference frames.

Now, suppose we look at things from the point of view of the first car. How does *it* see the second car? Well, if they are going the same direction, those in the first car will see those in the second car as going past at 60mph-50mph = 10mph. The ball dropped in the second car is seen as going 10mph forward as well as falling and so keeps up with the second car.

Now, just to emphasize, relativistic effects are being ignored here because they are very, very small. So all distances and times are seen to be exactly the same for all the reference frames we are dealing with here. If the ball on earth takes 1 second to fall for the reference frame of the earth, then it also takes 1 second to fall for either of the cars.

OK, do you agree with these different descriptions so far? Do you understand that differing reference frames are NOT just different points of view, nor are they different systems of units. We can use the same units (feet, meters, light years) in all frames. Similarly, time is a part of all reference frames and they can all use the same unit of time (second day, year).

Is this good so far? Do you have any questions for this situation?

/E: I also want to point out that everyone here agrees with how long it takes light to go to the sun (500 seconds). They also agree with how long it takes for light to go to a star 30 million ly away (30 million years). And they all agree that everyone's pens are 5 1/2" long.

 

[TrueBeliever37]

If that alien is at rest with respect to the earth, they would find the time for the light to get from the sun to the earth is 500 seconds, which is 93 million miles/ 186000 mps = dist/time.

The problems come when the alien is NOT at rest with respect to the earth.

The alien can, in fact, do so.

Again, I am assuming that star is at rest (mot moving towards or away from us) and that all measurements are made in the frame where that star and us are at rest. In that case, distance = 30 million light years and time = 30 million years and dist/time = c.[/QUOTE]

Both of the things above were things you said.

Now, the alien was able to calculate both scenarios accurately using the same formula C = distance/time?
He calculated the time for light from sun to earth at the same value we say it is, about 8 minutes and 20 seconds (500 seconds) and it relates to the same distance we say it is.
Then he calculates time for light from star to planet 30 million ly away, as t = 30 million years. So both scenarios work using C=distance/time


I had asked the following in an earlier post. - So are you telling me, that if I was somehow able to ride like a jockey on a photon. What is for us about an 8 min. 20 second trip from the Sun to the Earth, would be no different than a trip 8 million ly away? I would see no difference in the time or the distance I traveled?

The alien was able to provide the answer. He seems to say that there would be a difference in time. Proving my initial point that time and distance are involved.

 

[TrueBeliever37]

And nobody has said any other formula is required. But the distance and the time involved have to be measured in the same reference frame.

Because you are wanting to know what the 'light experiences'. That would involve looking at a reference frame moving at the speed of light. There is no real reference frame that does so. The only way to make sense of the question is to take the limit of frames going slower than light as the speed goes to that of light.

Polymath257,

Here is what I think is happening.

First thing is C is a constant.
Let's look at this as Time and Distance approach 0.

Speed of Light = C = Distance/Time
= (1/2) Distance / (1/2)Time [This would represent a point were you have 1/2 the time left to travel because you are 1/2 way there. ]
= (1/4) Distance / (1/4) Time [This would be a point where you have 1/4 the distance left to travel, so you only need 1/4 the time to arrive.]
= (o) Distance / (0) Time [ This is a point where you have reached your destination. So there is no distance left to travel, and hence no more time is needed.]
The equation therefore had become indeterminate, because you can't divide by 0. Yet we know that C is a constant.

Similar to the following scenario involving electricity:

R = V / I Let's choose a constant R = 10 ohms resistance, and a 10 Volt Battery.

10 ohms = 10 Volts /1 Amp
= 5 Volts / 1/2 Amp [ representing an aging battery that no longer can supply the 10 volts]
= 0 volts / 0 amps [ representing the totally discharged battery that can supply no current.]
Hence the equation had become indeterminate because you can't divide by 0 , yet we know that R is a constant of 10 ohms.

 

[Polymath257]

The alien can, in fact, do so.

Again, I am assuming that star is at rest (mot moving towards or away from us) and that all measurements are made in the frame where that star and us are at rest. In that case, distance = 30 million light years and time = 30 million years and dist/time = c.

Both of the things above were things you said.

Now, the alien was able to calculate both scenarios accurately using the same formula C = distance/time?
He calculated the time for light from sun to earth at the same value we say it is, about 8 minutes and 20 seconds (500 seconds) and it relates to the same distance we say it is.
Then he calculates time for light from star to planet 30 million ly away, as t = 30 million years. So both scenarios work using C=distance/time

I want to emphasize that this alien is at rest with respect to the earth, the sun, and that star.

I had asked the following in an earlier post. - So are you telling me, that if I was somehow able to ride like a jockey on a photon. What is for us about an 8 min. 20 second trip from the Sun to the Earth, would be no different than a trip 8 million ly away? I would see no difference in the time or the distance I traveled?

I want to emphasize that if you 'jockey' on a photon, you are NOT at rest with respect to the earth, sun, or star. THAT IS THE CRUCIAL DIFFERENCE.

The alien was able to provide the answer. He seems to say that there would be a difference in time. Proving my initial point that time and distance are involved.

Being an alien is not the important factor. Relative motion *is*.

So, let's go back to our examples. We have a star 4 light years away from earth and at rest with respect to earth. We agreed that it takes 4 years for the light to go from the star to the earth in the reference frame of the earth. We also agreed that a spaceship going at 50% of c will take 4/.5 = 8 years to get to the star.

Now, let's look at *this same situation* from the frame of the spaceship.

In the frame of the spaceship, the spaceship is at rest. It isn't moving. That is what it means to be in the reference frame of the spaceship. That star is approaching the ship at 50% of the speed of light and the earth is moving away at 50% of c. Are we good so far?

As the spaceship passes the earth, it sends out a light signal. The spaceship measures that light signal as moving at the speed of light using dist/time. From the earth's reference frame, that light takes 4ly/c = 4 years to reach the star.

BUT, and this is important, the star is NOT 4ly away when the spaceship passes the earth! There is a time dilation effect with a dilation factor of 86% so instead of 4ly away, the ship, using the best measurements it can do, finds the star to be 4*.86=3.54ly away at the point when the spaceship passes the earth. THIS IS LENGTH CONTRACTION AGAIN.

Now, the star is approaching the spaceship at 50% of c and the star was 3.54ly away, so it take dist/speed = 3.54/.5 = 7.08 years for the star to reach the spaceship.

To determine how long it takes for the light to get to the star, we have to take into account that the star is *moving towards the spaceship* at 50% of c. So the combined speed of the light and the star give c+.5c=1.5c as the combined speed, so the time it takes for the light to get to the star is 3.54ly/1.5c = 2.36 years.

Notice that in *all* cases, we use the formula speed=dist/time or its variant time=dist/speed to do our calculations. But we use the distance, time, and speed from that reference frame.

Next, instead of a spaceship going past earth at 50% of c, let's do the same scenarios with a spaceship going past earth at 99.995% of c. Again, a pulse of light is sent to the star just as this spaceship passes the earth.

In the frame of the earth, the light still takes 4 years to go 4ly. The spaceship takes time = dist/speed = 4/.99995 =4.0002 years to travel to the star.

Once again, make sure you agree here. We are *always* using speed = dist/time or its variant time = dist/speed.

Now, what happens from the reference frame of this spaceship?

Well, in the reference frame of this spaceship, the spaceship is at rest (again, the definition of a reference frame) and the star is moving toward it at 99.995% of c. The earth is moving away at that same speed. Check to make sure you agree here.

Now, the dilation factor for a speed of 99.995% of c is .01. So, that star is measured to be 4*.01 = .04ly from the spaceship as it passes the earth. This is about 14.6 days.

Since the star is approaching at 99.995% of c, the time it takes for the star to reach the spaceship is going to be .04/.99995 = .040002 years.

Next, when does that light sent out just as the spaceship passed the earth arrive at the star? Once again, we have to add the speed of light to that of the star, so the speed is not .99995c+c = 1.99995c. The distance to travel is .04ly, so this happens in .0200005 years, so a bit over 7 days.

Once again, we always use the dist=speed*time formula and its variance. But we need to use the distances, speeds, and times as measured in the frame we are working in. The answers will be different, but each individual frame has a completely consistent description.

 

[Polymath257]

Polymath257,

Here is what I think is happening.

First thing is C is a constant.
Let's look at this as Time and Distance approach 0.

Speed of Light = C = Distance/Time
= (1/2) Distance / (1/2)Time [This would represent a point were you have 1/2 the time left to travel because you are 1/2 way there. ]
= (1/4) Distance / (1/4) Time [This would be a point where you have 1/4 the distance left to travel, so you only need 1/4 the time to arrive.]
= (o) Distance / (0) Time [ This is a point where you have reached your destination. So there is no distance left to travel, and hence no more time is needed.]
The equation therefore had become indeterminate, because you can't divide by 0. Yet we know the C is a constant.

Similar to the following scenario involving electricity:

R = V / I Let's choose a constant R = 10 ohms resistance, and a 10 Volt Battery.

10 ohms = 10 Volts /1 Amp
= 5 Volts / 1/2 Amp [ representing an aging battery that no longer can supply the 10 volts]
= 0 volts / 0 amps [ representing the totally discharged battery that can supply no current.]
Hence the equation had become indeterminate because you can't divide by 0 , yet we know the R is a constant of 10 ohms.

Yes, that is what is going on. You have just computed a limit! But we have to remember that different frames measure different distances and times. But in all cases, the speed of light will be c. As the speed approaches c, the measured distances and the measured times both approach 0. The ratio stays constant at c.

 

[TrueBeliever37]

Yes, that is what is going on. You have just computed a limit! But we have to remember that different frames measure different distances and times. But in all cases, the speed of light will be c. As the speed approaches c, the measured distances and the measured times both approach 0. The ratio stays constant at c.

But there is always time involved, because you can't use a reference frame that doesn't really exist. There is a difference in time involved, any time you use a valid reference frame.
Therefore there is a difference in time when you travel a short distance versus a long distance such as 30 million light years. As was proven by the aliens measurements.

 

[Polymath257]

But there is always time involved, because you can't use a reference frame that doesn't really exist. There is a difference in time involved, any time you use a valid reference frame.
Therefore there is a difference in time when you travel a short distance versus a long distance such as 30 million light years. As was proven by the aliens measurements.

And you cannot 'jockey' on a photon of light. That is the limiting case where distances and times go to 0.

Please look at post #1044

 

[TrueBeliever37]

Yes, that is what is going on. You have just computed a limit! But we have to remember that different frames measure different distances and times. But in all cases, the speed of light will be c. As the speed approaches c, the measured distances and the measured times both approach 0. The ratio stays constant at c.

But you are using t=0 and d=0, when there is still time needed to travel , and a huge physical distance left to travel, such as a 30 million ly gap between a star and a planet.
To me that's the problem. It's not that I don't believe in different reference frames, which have different values. Although I don't think you can use a limiting reference frame, that has all times and distances as 0.

There has to be a difference in time for a photon to travel 1ly versus 30 million light years in any valid reference frame. So when you say t=0 and d=0 for every physical distance we are considering, that can't be valid.

 

[TrueBeliever37]

And you cannot 'jockey' on a photon of light. That is the limiting case where distances and times go to 0.

Please look at post #1044

I could be an imaginary jockey on a photon of light from the sun to the earth. And we could figure out the time, and distance I would have traveled. Why would it not be the limiting case in that instance?

 

[Polymath257]

But you are using t=0 and d=0, when there is still time needed to travel , and a huge physical distance left to travel, such as a 30 million ly gap between a star and a planet.
To me that's the problem. It's not that I don't believe in different reference frames, which have different values. Although I don't think you can use a limiting reference frame, that has all times and distances as 0.

There has to be a difference in time for a photon to travel 1ly versus 30 million light years in any valid reference frame. So when you say t=0 and d=0 for every physical distance we are considering, that can't be valid.

And once again, there is not a valid frame for the photon.

Again, the distance you gave are in some frame, that where the earth, sun, and star are at rest. But if you take a frame that is moving at 99.995% of c, the distances and times will be *very* different. But, when you divide distance by time, you will always get c for the speed of light.

So, we look at light going from the earth to the sun (500 light seconds) and a star 30 million light years away (30 million years) from the point of view of 4 different reference frames:

1. one that is at rest with respect to the earth: it takes 500 seconds fro the light to go to the sun and 30 million years for light to go to the star.

2. One that is moving at 50% of c in the appropriate direction. It takes (.86)*(500)/(1c +.5c) = 287 seconds for the light to get to the sun and (.86)*(30)/(1c+.5c) = 17.2 million years to get to the star.

3. One that is moving at 99.995% of c in the appropriate direction. It now takes (.01)*(500)/(1c+.99995c) = 2.5 seconds to get to the sun and (.01)*30)/(1c+.99995c) = 150,000 years to get to the star.

4. On that is moving at 99.9999995% of c in the appropriate direction. It now takes (.0001)*500)/(1.999999995)=.025 seconds to reach the sun and (.0001*(30 million)/1.999999995 = 1,500 years to reach the star.

As the speed of the spaceship increases up to c, the distance measured goes to 0 and the time it takes also goes to 0. And, just as in your example for resistance, in the limit, the speed is still c.

 

[Polymath257]

I could be an imaginary jockey on a photon of light from the sun to the earth. And we could figure out the time, and distance I would have traveled. Why would it not be the limiting case in that instance?

And the limiting distance as measured by that jockey would be 0. The limiting time as measured by that jockey would also be 0. But, the ratio would stay constant at c.

The dilation factor when measuring things at a relative velocity of v is going to be

sqrt(1 -(v/c)^2 )

As v-->c, this factor goes to 0. And this is what gets multiplied by both the distance and the time to get the measured distance and time in the moving frame.

 

[TrueBeliever37]

And the limiting distance as measured by that jockey would be 0. The limiting time as measured by that jockey would also be 0. But, the ratio would stay constant at c.

Ok this proves my point, at least to me - The limiting distance is 0, which we know to be inaccurate, because our common sense lets us know there really is a physical distance between the sun and us here on earth. And we also know that it definitely takes time for the light to get here. And if it is inaccurate in this case , it is inaccurate in the case of 30 million light years also.

 

[Polymath257]

Ok this proves my point, at least to me - The limiting distance is 0, which we know to be inaccurate, because our common sense lets us know there really is a physical distance between the sun and us here on earth. And we also know that it definitely takes time for the light to get here. And if it is inaccurate in this case , it is inaccurate in the case of 30 million light years also.

Common sense doesn't apply in special relativity. Our common sense is forged at slow speeds, not at those approaching the speed of light.

And let's look at how long it takes a spaceship to go from the earth to the sun at 99.5% of c.

From the point of view of the earth, it would be 500/.995 = 502.5 seconds. Are you ok with this?

But, from the point of view of the spaceship, the distance to the sun is NOT 500 light seconds, it is .1*500 = 50 light seconds. And the sun is approaching the spaceship at 99.5% of c, so in the frame of the spaceship, it will take 50/.995 = 50.25 seconds to reach the sun.

If the spaceship was moving at 99.995% of c, then the distance from the earth to the sun *in the frame of the spaceship) would be 500*.01 = 5 light seconds. And, it would take 5/.99995 = 5.00025 seconds to complete the journey *from the reference frame of the spaceship*. It would take 500/.99995 = 500.025 seconds *from the reference frame of the earth*.

These are ALL accurate descriptions of the situation.

Do you see what happens as the speed of the spaceship gets closer to that of light? The measured distance *in the frame of the spaceship* goes to 0. And so does the time it takes. This is the *limit*.

 

[TrueBeliever37]

Common sense doesn't apply in special relativity. Our common sense is forged at slow speeds, not at those approaching the speed of light.

And let's look at how long it takes a spaceship to go from the earth to the sun at 99.5% of c.

From the point of view of the earth, it would be 500/.995 = 502.5 seconds. Are you ok with this?

But, from the point of view of the spaceship, the distance to the sun is NOT 500 light seconds, it is .1*500 = 50 light seconds. And the sun is approaching the spaceship at 99.5% of c, so in the frame of the spaceship, it will take 50/.995 = 50.25 seconds to reach the sun.

If the spaceship was moving at 99.995% of c, then the distance from the earth to the sun *in the frame of the spaceship) would be 500*.01 = 5 light seconds. And, it would take 5/.99995 = 5.00025 seconds to complete the journey *from the reference frame of the spaceship*. It would take 500/.99995 = 500.025 seconds *from the reference frame of the earth*.

These are ALL accurate descriptions of the situation.

Do you see what happens as the speed of the spaceship gets closer to that of light? The measured distance *in the frame of the spaceship* goes to 0. And so does the time it takes. This is the *limit*.

Polymath257,

I don't think the problem is with the math. I think the problem is with the application of the math, and the conclusions drawn.

You guys think the limiting reference frame, where t=0, and d=0 applies to the whole trip for the photon, while I think it only applies when the trip has ended/finished/been completed.

Just like in the examples I gave, regarding speed of light, and the battery. (Post 1043)

I think t=0, and d=0 applies only when the photon has arrived. You guys want to take that as the situation even while the journey is occurring or still has to take place.

V=0, I=0 applied when the battery had reached the absolute end of it's life. (the end of it's journey if you will)

So I think you can look at time from the perspective of the photon itself. I don't agree that it experiences no time or distance.

 

[Polymath257]

Polymath257,

I don't think the problem is with the math. I think the problem is with the application of the math, and the conclusions drawn.

You guys think the limiting reference frame, where t=0, and d=0 applies to the whole trip for the photon, while I think it only applies when the trip has ended/finished/been completed.

No, that is definitely NOT what is going on.

Just like in the examples I gave, regarding speed of light, and the battery.

(Post 1043)

I think t=0, and d=0 applies only when the photon has arrived. You guys want to take that as the situation even while the journey is occurring or still has to take place.

V=0, I=0 applied when the battery had reached the absolute end of it's life. (the end of it's journey if you will)

So I think you can look at time from the perspective of the photon itself. I don't agree that it experiences no time or distance.

Once again, *from the perspective of someone making the journey*, the distance and the amount of time it takes to go that distance get smaller and smaller as the speed increases.

This is analogous to taking smaller and smaller potentials across your resistor and getting smaller and smaller currents. In each case, V/I = R, but as V-->0, I-->0 also with the ratio staying at R.

But, if you took another resistor with a much higher resistance, the current would always be lower than that of the first resistor and V/I would be the higher resistance. But, as V-->0, we still have I-->0, but this time with V/I the larger resistance.

So, if I compare the times *according to the spaceship making the journey* for a trip to a star 30 million light years away, and look at ships going various speeds approaching that of light, the measured distances *according to the ships* gets closer and closer to 0 as the speed of the ship gets closer and closer to c.

If, instead, we looked at spaceships going those same speeds and going to a star only 4ly away, the distances and times would be correspondingly smaller, always by a factor of 30 million/4 = 7.5 million. But, as the speed approaches that of light, both dist-->0 and time-->0.

 

[Subduction Zone]

Polymath257,

I don't think the problem is with the math. I think the problem is with the application of the math, and the conclusions drawn.

You guys think the limiting reference frame, where t=0, and d=0 applies to the whole trip for the photon, while I think it only applies when the trip has ended/finished/been completed.

Just like in the examples I gave, regarding speed of light, and the battery. (Post 1043)

I think t=0, and d=0 applies only when the photon has arrived. You guys want to take that as the situation even while the journey is occurring or still has to take place.

V=0, I=0 applied when the battery had reached the absolute end of it's life. (the end of it's journey if you will)

So I think you can look at time from the perspective of the photon itself. I don't agree that it experiences no time or distance.

Let me give this a shot. You do not seem to understand the concept of a limit being applied. So let's look at some different situations.

First the set up. We measure the distance from the Earth to a star that is stationary relative to our system at ten light years. And it takes ten years for light to get from their to here.

Now we have an astronaut in a spaceship that is going towards the Earth from that star at 0.86c (86% of the speed of light). It will take him 11.6 years to cross that distance. But he will measure it as only 5.8 years. And he will measure the distance as 5 light years.

Do you understand this? This sort of time compression and length compression has been observed and verified.

 

[TrueBeliever37]

No, that is definitely NOT what is going on.



Once again, *from the perspective of someone making the journey*, the distance and the amount of time it takes to go that distance get smaller and smaller as the speed increases.

This is analogous to taking smaller and smaller potentials across your resistor and getting smaller and smaller currents. In each case, V/I = R, but as V-->0, I-->0 also with the ratio staying at R.

But, if you took another resistor with a much higher resistance, the current would always be lower than that of the first resistor and V/I would be the higher resistance. But, as V-->0, we still have I-->0, but this time with V/I the larger resistance.

So, if I compare the times *according to the spaceship making the journey* for a trip to a star 30 million light years away, and look at ships going various speeds approaching that of light, the measured distances *according to the ships* gets closer and closer to 0 as the speed of the ship gets closer and closer to c.

If, instead, we looked at spaceships going those same speeds and going to a star only 4ly away, the distances and times would be correspondingly smaller, always by a factor of 30 million/4 = 7.5 million. But, as the speed approaches that of light, both dist-->0 and time-->0.

Let's say the distance to be traveled is 186,000 miles.

C = approximately 186,000 miles/second = distance/time [means a photon will experience 1 second to travel the 186,000 miles]

= 93,000 miles/(1/2)second [ at the halfway point the photon has experienced 1/2 second and traveled 93,000 miles, and still has 1/2 second left to complete the journey]

=0 miles / 0 seconds [The equation is indeterminate just like in the examples I gave. But C is still a constant. The photon has arrived, and therefore d=0, and t=0, since there is no distance left to travel, nor time needed.]

This let's us see why you always had t=0, and d=0, no matter what the distance was. Because it's at the end of the journeys that the distance is always 0, and no more time is needed.

I don't claim that it is impossible that I could be wrong. But that's what I believe to be the way it is.

 

[Polymath257]

Let's say the distance to be traveled is 186,000 miles.

C = approximately 186,000 miles/second = distance/time [means a photon will experience 1 second to travel the 186,000 miles]

= 93,000 miles/(1/2)second [ at the halfway point the photon has experienced 1/2 second and traveled 93,000 miles, and still has 1/2 second left to complete the journey]

=0 miles / 0 seconds [The equation is indeterminate just like in the examples I gave. But C is still a constant. The photon has arrived, and therefore d=0, and t=0, since there is no distance left to travel, nor time needed.]

This let's us see why you always had t=0, and d=0, no matter what the distance was. Because it's at the end of the journeys that the distance is always 0, and no more time is needed.

I don't claim that it is impossible that I could be wrong. But that's what I believe to be the way it is.

No, I am NOT talking about half-way points. I am talking about the distance the spaceships measure to be the distance from the earth to the sun/star/whatever.

If a star is 10ly away from the earth as measured by the earth and if a ship goes past the earth on the way to that star at 99.5% of c, then *at the instant the ship passes the earth*, the ship measures the distance to the star as being 1 ly (not 10ly). The time it takes for that ship to get to the star as measured by the earth is 10.5 years. That is when the earth measures the ship and the star to be in the same location. In contrast, the ship measures the time of travel to b 1.05 years.

The *whole journey* from the point of view of the ship is 1ly.

There is not a unit conversion. The clock on the spaceship and an exact replica on the earth can be used to measure the times. The exact same methods can be used to figure out the relevant distances.

But, from that moving frame, the distances and times are NOT the same as the distances and times as measured from the earth.

Furthermore, this is an actual, measured effect. this is NOT just 'theory': it has been verified in many different ways over the last century.

 

[TrueBeliever37]

No, I am NOT talking about half-way points. I am talking about the distance the spaceships measure to be the distance from the earth to the sun/star/whatever.

If a star is 10ly away from the earth as measured by the earth and if a ship goes past the earth on the way to that star at 99.5% of c, then *at the instant the ship passes the earth*, the ship measures the distance to the star as being 1 ly (not 10ly). The time it takes for that ship to get to the star as measured by the earth is 10.5 years. That is when the earth measures the ship and the star to be in the same location. In contrast, the ship measures the time of travel to b 1.05 years.

The *whole journey* from the point of view of the ship is 1ly.

There is not a unit conversion. The clock on the spaceship and an exact replica on the earth can be used to measure the times. The exact same methods can be used to figure out the relevant distances.

But, from that moving frame, the distances and times are NOT the same as the distances and times as measured from the earth.

Furthermore, this is an actual, measured effect. this is NOT just 'theory': it has been verified in many different ways over the last century.

I'm not in disagreement with your math you keep showing as far as differences because of relationships to each other. I can tell you are extremely intelligent and very capable of doing the math.

I am in disagreement that a photon experiences no time or distance. If the photon experiences no time or distance, then why does it only take 8m and 20s to get here from the Sun, but take 30 million years to get here from another star? After all it was traveling at C in both cases.

 

[Polymath257]

I'm not in disagreement with your math you keep showing as far as differences because of relationships to each other. I can tell you are extremely intelligent and very capable of doing the math.

I am in disagreement that a photon experiences no time or distance. If the photon experiences no time or distance, then why does it only take 8m and 20s to get here from the Sun, but take 30 million years to get here from another star? After all it was traveling at C in both cases.

Both the 30 million year and the 8 minutes 20 seconds are *as measured from the earth*. If a spaceship was going past at 99% of the speed of light, the times (and distances) would be completely different. The distances and times *depend on the reference frame*. In the (limiting) frame of the photon, both distances are 0 and both times are 0. But, for any speed slightly less than that of light (a real reference frame), the distance to the star is 3 billion times as much as to the sun. And the times are the same way. It is only in the *limit* that the distances and times go to 0.

Think of it like this. What happens when you multiply 30 million by 0? Now, what happens when you multiply 500 by 0? How can the answers be the same?

In a sense, you are asking how that resistor could have a voltage of 30 V and a current of 1 A and also have a voltage of 0 and a current of 0. The different voltages correspond to the distances for different reference frames and the different currents correspond to times in different reference frames.