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

[james blunt]

No desire to learn deepities

Ciao

- viole

So you don't like deepities within you ?

 

[viole]

So you don't like deepities within you ?

No. I do not like deepities in general. Do you?

Ciao

- viole

 

[TrueBeliever37]

Nope. If I am moving at 99.5% of the speed of light with respect to you, we will *both* see the light as moving at 100% of c away from us.

The original experiment along these lines was done in the 1800's by Michelson and Morley. They wanted to measure the speed of the earth by looking at how much the speed of light changed in one direction vs another. They expected that in the direction of the Earth's motion, it would look like it was going away from us slightly slower and in the opposite direction, that it would look like it was going away from us slightly faster (the differences in each case being due to the motion of the Earth). The experiment they did was fully capable of doing these measurements, but they got a null answer: the measured speed of light in every direction was the same.

THAT is what it means for the speed of light to be constant in all frames! Even someone 'moving' will see light going away from them at the speed of light. Now, at least, you are beginning to see the strangeness of special relativity.

Polymath, I saw the strangeness right away, that was why I didn't believe it.

Would a ship moving at 0.5C take twice as long to go 30 million ly, as it would to go 15 million ly?

Would a ship moving at 0.75C take twice as long to go 30 million ly as it would to go 15 million ly?

Would a ship moving at 0.9C take twice as long to go 30 million ly as it would to go 15 million ly?

Would a ship moving at 0.99C take twice as long to go 30 million ly as it would to go 15 million ly?

 

[Polymath257]

Polymath, I saw the strangeness right away, that was why I didn't believe it.

Would a ship moving at 0.5C take twice as long to go 30 million ly, than it did to go 15 million ly?

Would a ship moving at 0.75C take twice as long to go 30 million ly than it did to go 15 million ly?

Would a ship moving at 0.9C take twice as long to go 30 million ly than it did to go 15 million ly?

Would a ship moving at 0.99C take twice as long to go 30 million ly than it did to go 15 million ly?

Yes, in all cases. ALL of these questions happen in a reference frame from which those distances and speeds are measured. But your basic question (concerning 0 distance and time) happens in a different frame than the one where the distance is 30 million ly. You have to be *very* careful when switching reference frames. Here's a point to ponder: what's twice 0?

Each of those ships would see light going past them at 100% of c *in their reference frame*. So, the ship going at 99% of C past the earth turns on a light and measures the light going away from it at 100% of c. The earthbound observer *also* measures that light as going away from earth at 100% of c.

 

[Thermos aquaticus]

Would a ship moving at 0.5C take twice as long to go 30 million ly, as it would to go 15 million ly?

Assuming they don't change velocity, yes. However, what you measure as 30 million ly from Earth would be 26 light years for the ship. 15 light years from Earth would be 13 ly for the ship. I found a handy online calculator that does the math for you, if you are interested:

Length Contraction Calculator - Omni

It has the speed of light in km/s which through me off at first, so make sure that the speed of light is a percentage of 298,000 km/s.

Would a ship moving at 0.75C take twice as long to go 30 million ly as it would to go 15 million ly?

Yes. However, the distances as measured on the ship would be 20 million ly and 10 million ly.

Would a ship moving at 0.9C take twice as long to go 30 million ly as it would to go 15 million ly?

Yes. However, the distances as measured on the ship would be 6.7 and 3.4 million ly.

Would a ship moving at 0.99C take twice as long to go 30 million ly as it would to go 15 million ly?

Yes. However, the distances as measured on the ship would be 5.3 and 2.7 million ly.

 

[Polymath257]

Assuming they don't change velocity, yes. However, what you measure as 30 million ly from Earth would be 26 light years for the ship. 15 light years from Earth would be 13 ly for the ship. I found a handy online calculator that does the math for you, if you are interested:

Length Contraction Calculator - Omni

It has the speed of light in km/s which through me off at first, so make sure that the speed of light is a percentage of 298,000 km/s.



Yes. However, the distances as measured on the ship would be 20 ly and 10 ly.



Yes. However, the distances as measured on the ship would be 6.7 and 3.4 ly.



Yes. However, the distances as measured on the ship would be 5.3 and 2.7 ly.

Each of those should be in millions of light years.

 

[Thermos aquaticus]

Each of those should be in millions of light years.

Details . . . details.

Thanks for the correction!

 

[Polymath257]

Assuming they don't change velocity, yes. However, what you measure as 30 million ly from Earth would be 26 light years for the ship. 15 light years from Earth would be 13 ly for the ship. I found a handy online calculator that does the math for you, if you are interested:

Length Contraction Calculator - Omni

It has the speed of light in km/s which through me off at first, so make sure that the speed of light is a percentage of 298,000 km/s.

Your numbers below for the higher speeds are off. I suspect the calculator works with 298,792 km/s. That makes a difference in the percentage and calculation for speeds close to 298,000.

/E: Nope....seems to be using 300,000 km/s....bad.

Yes. However, the distances as measured on the ship would be 20 million ly and 10 million ly.



Yes. However, the distances as measured on the ship would be 6.7 and 3.4 million ly.



Yes. However, the distances as measured on the ship would be 5.3 and 2.7 million ly.

I got 4.23 million ly and 2.12 million ly, resp.

 

[TrueBeliever37]

Yes, in all cases. ALL of these questions happen in a reference frame from which those distances and speeds are measured. But your basic question (concerning 0 distance and time) happens in a different frame than the one where the distance is 30 million ly. You have to be *very* careful when switching reference frames. Here's a point to ponder: what's twice 0?

Each of those ships would see light going past them at 100% of c *in their reference frame*. So, the ship going at 99% of C past the earth turns on a light and measures the light going away from it at 100% of c. The earthbound observer *also* measures that light as going away from earth at 100% of c.

That was why I avoided that reference frame, as it doesn't really exist anyway. All I was really trying to show was that time was involved in all those cases. Every valid reference frame took twice as long to go twice the distance.

So when a photon has traveled 1/2 the distance to the earth from the sun. You really don't think it has experienced 1/2 the 8m and 20s time it takes to get from the sun to earth?

 

[TrueBeliever37]

Assuming they don't change velocity, yes. However, what you measure as 30 million ly from Earth would be 26 light years for the ship. 15 light years from Earth would be 13 ly for the ship. I found a handy online calculator that does the math for you, if you are interested:

Length Contraction Calculator - Omni

It has the speed of light in km/s which through me off at first, so make sure that the speed of light is a percentage of 298,000 km/s.



Yes. However, the distances as measured on the ship would be 20 million ly and 10 million ly.



Yes. However, the distances as measured on the ship would be 6.7 and 3.4 million ly.



Yes. However, the distances as measured on the ship would be 5.3 and 2.7 million ly.

Hi Thermos,

So does that mean when you get to C, that suddenly you have instant arrival no matter what the distance(since it is always 0)?

 

[Polymath257]

That was why I avoided that reference frame, as it doesn't really exist anyway. All I was really trying to show was that time was involved in all those cases. Every valid reference frame took twice as long to go twice the distance.

So when a photon has traveled 1/2 the distance to the earth from the sun. You really don't think it has experienced 1/2 the 8m and 20s time it takes to get from the sun to earth?

In a word, no. A ship going past at 99% of c watching that same photon sees that same journey of the photon to the sun (or vice versa) as taking 35 seconds. So half is 17.5 seconds.

Does the photon 'experience' 17.5 seconds? Or 250 seconds? Both are valid measurements of time from different real frames.

The 'proper time' for the photon's path is zero.

 

[Thermos aquaticus]

Your numbers below for the higher speeds are off. I suspect the calculator works with 298,792 km/s. That makes a difference in the percentage and calculation for speeds close to 298,000.

/E: Nope....seems to be using 300,000 km/s....bad.

Hopefully it is close enough so that @TrueBeliever37 can stick some numbers in and experiment a bit.

 

[TrueBeliever37]

Assuming they don't change velocity, yes. However, what you measure as 30 million ly from Earth would be 26 light years for the ship. 15 light years from Earth would be 13 ly for the ship. I found a handy online calculator that does the math for you, if you are interested:

Length Contraction Calculator - Omni

It has the speed of light in km/s which through me off at first, so make sure that the speed of light is a percentage of 298,000 km/s.



Yes. However, the distances as measured on the ship would be 20 million ly and 10 million ly.



Yes. However, the distances as measured on the ship would be 6.7 and 3.4 million ly.



Yes. However, the distances as measured on the ship would be 5.3 and 2.7 million ly.

Hi Thermos,

Doesn't it look to you like no matter what valid reference frame is used, the photons involved, experienced twice the time for twice the distance?

 

[Thermos aquaticus]

So does that mean when you get to C, that suddenly you have instant arrival no matter what the distance(since it is always 0)?

I would say that time to arrival is a meaningless question for the photon since time and distance don't exist like we experience it.

 

[Polymath257]

Hi Thermos,

Doesn't it look to you like no matter what valid reference frame is used, the photons involved, experienced twice the time for twice the distance?

When you ask what the photons 'experienced', you get into trouble. The relativistic 'proper time' for the path the light takes is 0. The time in various reference frames is different, but those are not the proper time for the path.

 

[TrueBeliever37]

In a word, no. A ship going past at 99% of c watching that same photon sees that same journey of the photon to the sun (or vice versa) as taking 35 seconds. So half is 17.5 seconds.

Does the photon 'experience' 17.5 seconds? Or 250 seconds? Both are valid measurements of time from different real frames.

The 'proper time' for the photon's path is zero.

But my point is the photon would always experience/take twice the time for twice the distance, no matter what valid reference frame you are using.

 

[Thermos aquaticus]

Hi Thermos,

Doesn't it look to you like no matter what valid reference frame is used, the photons involved, experienced twice the time for twice the distance?

In all of those frames of reference we can describe what an observer will observe in those frames of reference. We are not describing what the photon experiences.

 

[Polymath257]

I would say that time to arrival is a meaningless question for the photon since time and distance don't exist like we experience it.

It *can* be meaningful. The spacetime 'separation' is an analog for arc length for relativistic paths that gives the time as experienced by the observer going over that path. For photons, that separation is always 0.

The proper time between events at (x,t) and (x',t') is given by sqrt( (t-t')^2 - (x-x')^2 /c^2 ). This is 0 for photons.

/E: try seeing what you get when you put x=vt and x'=t'=0. You might be amused.

 

[Polymath257]

But my point is the photon would always experience/take twice the time for twice the distance, no matter what valid reference frame you are using.

And twice zero is zero.

 

[Polymath257]

But my point is the photon would always experience/take twice the time for twice the distance, no matter what valid reference frame you are using.

That is NOT what the photon 'experiences'. It is what the earth, or ship, or whatever *watching* the photon experiences.