michel123456's relativity thread (from Time dilation dependence on direction)

[michel123456]

On 4/26/2017 at 11:43 AM, Janus said:

A third clock travels from your position to the second clock at 0.8c

Using the formula above we find that you will see it tick 1/3 as fast as your own. It will arrive at the second clock in 1.25 hrs when the second clock reads 01:15

 

Why 01:15 and not 01:25?

 

 

!

Moderator Note

(this is referencing https://www.scienceforums.net/topic/105185-time-dilation-dependence-on-direction/)

 

This thread is now the only place where michel123456 may discuss topics related to time and relativity

 

 

[michel123456]

On 4/26/2017 at 6:43 PM, Janus said:

So let's work out a couple of examples:

Assume you are situated 1 light hr from a stationary(to you) clock that is synchronized to your own clock (when your clock reads 12:00, it reads 12:00)

You looking at this clock when your clock reads 12:00 will visual see the clock as reading 11:00. (since it took the light carrying that image 1 hr to travel from the other clock, to get to you when your clock reads 12:00, it had to leave the other clock when it read 11:00)

 

A third clock travels from your position to the second clock at 0.8c

Using the formula above we find that you will see it tick 1/3 as fast as your own. It will arrive at the second clock in 1.25 hrs when the second clock reads 01:25. But because of the light travel time lag between you and the second clock, you will not see this until until your clock reads 02:25. This means the you will watch the third clock recede for 2.25 hrs by your clock while seeing it ticking 1/3 as fast and thus if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45. So in the 1:25 it took to make the trip it only ticked off 45 min. (remember, even though you didn't see the third clock arrive until your clock read 02:15, it had actually arrived an hour earlier.)

You are considering as if the travel was 1 hour long, but the travel was 1.25 long.

 

And you have explained to the community how to travel 1 LightHour in 45 minutes.

[J.C.MacSwell]

1 hour ago, michel123456 said:

And you have explained to the community how to travel 1 LightHour in 45 minutes.

A light hour in one frame during a duration of 45 minutes in another frame.

We're all doing it right now.

[Halc]

4 hours ago, michel123456 said:

Quote

 

Why 01:15 and not 01:25?

Because one and a quarter hours is an hour and 15 minutes, not an hour and 25 minutes.

Edited September 16, 2020 by Halc

[swansont]

1 hour ago, michel123456 said:

You are considering as if the travel was 1 hour long, but the travel was 1.25 long.

 

And you have explained to the community how to travel 1 LightHour in 45 minutes.

It's one light-hour only in the rest frame. 

[michel123456]

I am confused.

first of all, hours have 60 min, not 100. So 60/0.8=75 or 1h15min (and not 1h 25 min.). Correct?

 

On 4/26/2017 at 6:43 PM, Janus said:

You looking at this clock when your clock reads 12:00 will visual see the clock as reading 11:00. (since it took the light carrying that image 1 hr to travel from the other clock, to get to you when your clock reads 12:00, it had to leave the other clock when it read 11:00)

 

A third clock travels from your position to the second clock at 0.8c

Using the formula above we find that you will see it tick 1/3 as fast as your own. It will arrive at the second clock in 1.25 hrs when the second clock reads 01:15. But because of the light travel time lag between you and the second clock, you will not see this until until your clock reads 02:15. This means the you will watch the third clock recede for 2.25 hrs by your clock while seeing it ticking 1/3 as fast and thus if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45. So in the 1:15 it took to make the trip it only ticked off 45 min. (remember, even though you didn't see the third clock arrive until your clock read 02:15, it had actually arrived an hour earlier.)

Generally I am confused with the mix of 1.25 & 1.15 in the text above.

[Halc]

Yes, it takes 1:15 in the frame of 1st/2nd clock for the 3rd clock to get one LH away.

In the frame of that 3rd clock, it is stationary, so it isn't traveling anywhere in the 45 minutes it logs, so it isn't moving faster than light.

Edited September 16, 2020 by Halc

[michel123456]

11 minutes ago, Halc said:

Because one and a quarter hours is an hour and 15 minutes, not an hour and 25 minutes.

Oh, I got it: you mean 1.25 hour means 1h and a quarter? That's a bad mix of notation.

2 hours ago, michel123456 said:

On 4/26/2017 at 6:43 PM, Janus said:

if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45. So in the 1:25 it took to make the trip it only ticked off 45 min. (remember, even though you didn't see the third clock arrive until your clock read 02:15, it had actually arrived an hour earlier.)

You are considering as if the travel was 1 hour long, but the travel was 1.25 long

Still not understanding this: if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45.

[Janus]

3 hours ago, michel123456 said:

Oh, I got it: you mean 1.25 hour means 1h and a quarter? That's a bad mix of notation.

Still not understanding this: if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45.

How is using 1.25 hrs any different than saying 1 and a quarter hrs?  or 1 1/4 hrs.  The fact that the number is followed by "hrs" indicates that the numerical value is applies to the one unit, hours, while with 1:15  the ":" denotes a marker between units, or 1 hr, 15 min. 

I'm sorry if this confuses you, but when working with such problems it is easier to the work the math out in decimal hrs and then convert back to Hr and min afterwards if you need to.

If the clock left you moving at 0.8c to a point 1 light hr away ( as measured by you), while both it and your clock both read 12:00, then it will take, by your determination, 1 hr 15 min to reach that point. You, however, will not see him arrive when your clock reads 1:15, as It will take an another hr for the light from his arrival to reach you, so you will see the image of his arrival 2 hr and 15 min after he left, when your clock reads 2:15. In other words, you will "see" his entire trip as being stretched out over 2 hrs 15 min.

As you watched him recede from you, you will see him via relativistic Doppler shift. This does not only effects the measured frequency of the light you get from him, but also the rate at which you would "see" his clock tick.  The Relativistic Doppler shift rate for 0.8c is 1/3.   Since you see this happening for the entire 2 hr and 15 min, you will see that clock ticking off only 1/3 of that, or 45 min.  Thus when you see the clock arrives at that point 1 light hr away, you will see it as reading 12:45.

 

[swansont]

4 hours ago, michel123456 said:

I am confused.

first of all, hours have 60 min, not 100. So 60/0.8=75 or 1h15min (and not 1h 25 min.). Correct?

 

Generally I am confused with the mix of 1.25 & 1.15 in the text above.

They are not mixed.

The text:
Using the formula above we find that you will see it tick 1/3 as fast as your own. It will arrive at the second clock in 1.25 hrs when the second clock reads 01:15. But because of the light travel time lag between you and the second clock, you will not see this until until your clock reads 02:15. This means the you will watch the third clock recede for 2.25 hrs by your clock while seeing it ticking 1/3 as fast and thus if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45. So in the 1:15 it took to make the trip it only ticked off 45 min. (remember, even though you didn't see the third clock arrive until your clock read 02:15, it had actually arrived an hour earlier.)

 

1.15, for example, is not used. All of the time-of-day readings (as one might see on a digital clock) or elapsed times denoting hours and minutes use a colon, per the ISO 8601 standard (further, all clock readings are denoted as such)

The decimals are used for elapsed time calculations, as one might expect when the values used don't result in a whole number

The two notations are used consistently. Decimal point used for decimal numbers, colon used for hours:minutes

 

 

 

 

 

[michel123456]

45 minutes ago, swansont said:

They are not mixed.

The text:
Using the formula above we find that you will see it tick 1/3 as fast as your own. It will arrive at the second clock in 1.25 hrs when the second clock reads 01:15. But because of the light travel time lag between you and the second clock, you will not see this until until your clock reads 02:15. This means the you will watch the third clock recede for 2.25 hrs by your clock while seeing it ticking 1/3 as fast and thus if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45. So in the 1:15 it took to make the trip it only ticked off 45 min. (remember, even though you didn't see the third clock arrive until your clock read 02:15, it had actually arrived an hour earlier.)

 

1.15, for example, is not used. All of the time-of-day readings (as one might see on a digital clock) or elapsed times denoting hours and minutes use a colon, per the ISO 8601 standard (further, all clock readings are denoted as such)

The decimals are used for elapsed time calculations, as one might expect when the values used don't result in a whole number

The two notations are used consistently. Decimal point used for decimal numbers, colon used for hours:minutes

 

 

 

 

 

Well understood now. Thx.

 

1 hour ago, Janus said:

How is using 1.25 hrs any different than saying 1 and a quarter hrs?  or 1 1/4 hrs.  The fact that the number is followed by "hrs" indicates that the numerical value is applies to the one unit, hours, while with 1:15  the ":" denotes a marker between units, or 1 hr, 15 min. 

I'm sorry if this confuses you, but when working with such problems it is easier to the work the math out in decimal hrs and then convert back to Hr and min afterwards if you need to.

If the clock left you moving at 0.8c to a point 1 light hr away ( as measured by you), while both it and your clock both read 12:00, then it will take, by your determination, 1 hr 15 min to reach that point. You, however, will not see him arrive when your clock reads 1:15, as It will take an another hr for the light from his arrival to reach you, so you will see the image of his arrival 2 hr and 15 min after he left, when your clock reads 2:15. In other words, you will "see" his entire trip as being stretched out over 2 hrs 15 min.

As you watched him recede from you, you will see him via relativistic Doppler shift. This does not only effects the measured frequency of the light you get from him, but also the rate at which you would "see" his clock tick.  The Relativistic Doppler shift rate for 0.8c is 1/3.   Since you see this happening for the entire 2 hr and 15 min, you will see that clock ticking off only 1/3 of that, or 45 min.  Thus when you see the clock arrives at that point 1 light hr away, you will see it as reading 12:45.

 

Thx, that is now clear as crystal water.

now the return trip:

On 4/26/2017 at 6:43 PM, Janus said:

Now let's do a reverse trip. you see the second clock read 01:15 and the third clock reads 12:45 when your clock reads 02:15, however this image of those two clocks left 1 hr ago, so the third clock left the second clock when your clock read 01:15 (in other words, if the third clock turns around and heads back the moment it reaches the second clock, by the time you see this, the third clock is already well along its trip back to you.). The return trip takes the same 1.25 hrs, which means it arrive back at you when your clock reads 02:30. This means that you will see its entire return trip occur during the 15 min between 02:15 and 02:30. At a accelerated tick rate of 3 you will see the third clock tick off 45 min during that period, and read 01:30 upon arrival, when your clock reads 02:30. So again, it ticked off 45 min during the 1.25 hr trip, the same as for the outbound trip and accumulated at total of 1.5 hrs for your 2.5 hrs.

Please explain the bold part, does it come from the dilation formula you posted above?

[Janus]

26 minutes ago, michel123456 said:

Well understood now. Thx.

 

Thx, that is now clear as crystal water.

now the return trip:

Please explain the bold part, does it come from the dilation formula you posted above?

Again. Relativistic Doppler effect. (which is what you get by combining both light propagation and time dilation)   If the clock is receding at 0.8 c, you see it ticking 1/3 as fast as your own.   If it is approaching at 0.8 c, you see it running 3 times faster.

One way to look at it is that by the time you visually see that clock begin its return trip, the clock has already been on its return leg for an hr, and only 15 min remain until it arrives. All the light emitted from it during the return leg is squeezed into that 15 min by your clock.  If you see the returning clock tick 3 times faster, 3*15 min = 45 min is the time you see pass on that clock during its return leg.

If you factor out the light propagation delay you will conclude that the clock left reading 12:45, when your clock read 1:15.  It took 1 hr 15 min to cross the 1 light hr at 0.8c,  and arrives when your clock reads 2:30.   The clock read 12:45 when it left, and reads 1:30 when it arrives, thus ticked off 45 min in that 1 hr 15 min by your clock, or  6/10 of  1 hr 15 min(75 min)

The time dilation factor for 0.8c is 0.6.

[michel123456]

7 minutes ago, Janus said:

Again. Relativistic Doppler effect. (which is what you get by combining both light propagation and time dilation)   If the clock is receding at 0.8 c, you see it ticking 1/3 as fast as your own.  If it is approaching at 0.8 c, you see it running 3 times faster.

So it depends on direction after all.

[Janus]

35 minutes ago, michel123456 said:

So it depends on direction after all.

The Doppler effect does, but the time dilation does not.

The relativistic Doppler shift is due to two compounding effects:

Changing light propagation delay due to changing distance, the effect of which is determined whether the propagation delay is increasing or decreasing.

Time dilation, which only depends on relative speed and is independent of direction.

So with the above example, Doppler shift gives a value of 1/3 on the for the receding leg and 3 for the return leg, but time dilation has a factor of 0.6 for both legs.

[michel123456]

22 minutes ago, Janus said:

The Doppler effect does, but the time dilation does not.

The relativistic Doppler shift is due to two compounding effects:

Changing light propagation delay due to changing distance, the effect of which is determined whether the propagation delay is increasing or decreasing.

Time dilation, which only depends on relative speed and is independent of direction.

So with the above example, Doppler shift gives a value of 1/3 on the for the receding leg and 3 for the return leg, but time dilation has a factor of 0.6 for both legs.

So, from your example I can conclude that:

1. the 3rd clock traveled 1 LightHour in 45 minutes, but that was not faster than SOL

2. as seen from A, the same clock popped in 15 minutes from 1 LH away, but that is not considered faster than SOL.

[Halc]

21 minutes ago, michel123456 said:

So, from your example I can conclude that:

1. the 3rd clock traveled 1 LightHour in 45 minutes, but that was not faster than SOL

2. as seen from A, the same clock popped in 15 minutes from 1 LH away, but that is not considered faster than SOL.

That error has already been pointed out.

The example shows clock 3 traveling the light hour in 5/4 hours. That's what it means to be moving at 0.8c. Due to time dilation, clock 3 just happens to have logged only 3/4 hours during that trip.

Relative to the clock's own frame, only 45 undilated minutes pass, but it is stationary (by definition), which is hardly traveling faster than SoL. In that frame, clock 2 comes to it at 0.8c, and moves 36 light-minutes during those 45 minutes, which is also less than SoL.

2: As seen some observer A, the incoming clock is seen to go from a light hour away to 'here' in 15 minutes, yes. That's the difference between its actual speed in that frame and the speed of the light the observer is observing.

Edited September 16, 2020 by Halc

[Janus]

1 hour ago, michel123456 said:

So, from your example I can conclude that:

1. the 3rd clock traveled 1 LightHour in 45 minutes, but that was not faster than SOL

2. as seen from A, the same clock popped in 15 minutes from 1 LH away, but that is not considered faster than SOL.

1. No, the 3rd clock traveled 1 light hr in 1hr 15 min as measured by you, it just also ran slow by a factor of 0.6 and thus only accumulated 45 min , as measured by you.   Visually, you saw it take 2 hr and 15 min to reach that 1 light hr distance.

2. Again, it took the clock 1hr 15 min to travel the distance.  The fact that it arrived only 15 min after you saw it start its trip is just because it was following closely behind it's own light.

Two cars leave a point 100 miles from you, driving towards you.   One car, traveling at 100 mph carries a message that says the other car is also just leaving. The second car travels at 80 mph.

The faster car arrives at 1:00, while the second car arrives 15 min later at 1:15.   Now just because the message saying that the second car was leaving arrived just 15 min before the second car arrives, this does not mean that you will conclude that the second car only took 15 min to cross the distance and had to be traveling at 400 mph.  The second car took 1 hr 15 min to make the trip, it was just following not too far behind the faster car.( By the time the first car arrives, the second car is only 20 miles away)

Again, you are conflating "what someone visually sees", with what they say is actually happening.

If I am 1 light hour from a clock stationary with respect to me, and see it reading 11:00, I don't think that the clock reads 11:00 at that moment, but that, assuming the clock continued to run, it ticked off an additional hr since the image I am now seeing left it, thus I will say that the clock now reads 12:00.  Conversely, if your clock reads 12:00, you know that the light left when your clock read 11:00 and thus both clocks read 11:00 at that moment.

This works if even if the other clock is moving relative to you.   Light that left it when it was 1 light hr away will carry the image of what that clock read 1 hr ago by your clock.

So if your clock reads 2:15 when you see the image of the 3rd clock carried by light that left it when it was 1hr away,  You know that image left when your clock read 1:15. and if that image is of the 3rd clock reading 12:45, you know that the 3rd clock read 12:45 at the same moment your clock read 1:15.

 

[michel123456]

9 hours ago, Janus said:

1. No, the 3rd clock traveled 1 light hr in 1hr 15 min as measured by you, it just also ran slow by a factor of 0.6 and thus only accumulated 45 min , as measured by you.   Visually, you saw it take 2 hr and 15 min to reach that 1 light hr distance.

But the traveler "lived" the travel for 45 minutes.

Or, as Halc explains, since he considers himself stationary, he sees clock 2 rushing to him from 1 LH away in 36 minutes (I don't understand why it is less than SOL), and sees clock 1 go away from him & reach 1 LH in ...minutes.

If you transfer it grossly instead of minutes to years, it means that at 0.8 c you don't need 10 years to reach  a star 10LY away. The astronaut will live the experience in ... years (6 years?).

12 hours ago, Halc said:

2: As seen some observer A, the incoming clock is seen to go from a light hour away to 'here' in 15 minutes, yes. That's the difference between its actual speed in that frame and the speed of the light the observer is observing.

The concept of the incoming clock "chasing its own image" is quite disturbing.

Edited September 17, 2020 by michel123456

[Halc]

2 hours ago, michel123456 said:

Or, as Halc explains, since he considers himself stationary, he sees clock 2 rushing to him from 1 LH away in 36 minutes (I don't understand why it is less than SOL),

 

I said no such thing. Either your reading comprehension skills are completely absent or your are deliberately twisting everybody's replies. I suspect the latter.

I said that since the 'traveler' considers himself stationary in his own frame (by definition no less), clock 2 rushes towards him from 36 light minutes away, and this takes 45 minute, which is less than SoL. That makes no statement about what he sees while this is going on.

Quote

If you transfer it grossly instead of minutes to years, it means that at 0.8 c you don't need 10 years to reach  a star 10LY away.

The astronaut will live the experience in ... years (6 years?).

10 years away as measured in a different frame than your own, yes. If I have a fast enough ship, I can get to the far edge of the galaxy in my lifetime, despite it being 70,000 LY away in the frame of the galaxy. That speed is called proper velocity (dx/dt') and there is no limit of c to it. Normal velocity (dx/dt, or distance and time measured in the same frame) is bounded locally by light speed.

Edited September 17, 2020 by Halc

[michel123456]

3 hours ago, Halc said:

I said no such thing. Either your reading comprehension skills are completely absent or your are deliberately twisting everybody's replies. I suspect the latter.

I said that since the 'traveler' considers himself stationary in his own frame (by definition no less), clock 2 rushes towards him from 36 light minutes away, and this takes 45 minute, which is less than SoL. That makes no statement about what he sees while this is going on.

I assure you I am not deliberately twisting everybody's replies, so I am afraid the other option may be valid.

I don't understand the bold part: clock 2 rushes towards him from 36 light minutes away

Because at departure time, when clock 3 & 1 are together, clock 2 lies 1LH away. So I suppose that clock 3 sees clock 2 rushing at him at tremendous velocity.

 

[swansont]

15 minutes ago, michel123456 said:

I assure you I am not deliberately twisting everybody's replies, so I am afraid the other option may be valid.

I don't understand the bold part: clock 2 rushes towards him from 36 light minutes away

Because at departure time, when clock 3 & 1 are together, clock 2 lies 1LH away. So I suppose that clock 3 sees clock 2 rushing at him at tremendous velocity.

 

You can’t state these values without acknowledging the frame in which they are measured. You have to keep that straight. Mixing frames won’t give a correct or consistent answer.

”Clock 2 lies 1 LH away” is true in the rest frame, not the moving frame. IOW, the moving clock doesn’t see a trip of 1 LH, since there is length contraction.

[michel123456]

28 minutes ago, swansont said:

You can’t state these values without acknowledging the frame in which they are measured. You have to keep that straight. Mixing frames won’t give a correct or consistent answer.

”Clock 2 lies 1 LH away” is true in the rest frame, not the moving frame. IOW, the moving clock doesn’t see a trip of 1 LH, since there is length contraction.

But all clocks (including clock 3) are in a rest frame. If i understand correctly.

[Janus]

17 minutes ago, michel123456 said:

But all clocks (including clock 3) are in a rest frame. If i understand correctly.

And, in the rest frame of clock 3, it is clocks 1 and 2 that are moving at 0.8 c, and thus are length contracted, and this includes the distance between them.

Thus by the 3rd clock's measure, the distance between clocks 1 and 2 is only 0.6 lh.  So, as it passes clock 1, clock 2 is 0.6 lh away, is coming towards him at 0.8c, and takes 0.6lh/0.8c = 0.75 hr = 45 min.

Edited September 17, 2020 by Janus

[michel123456]

So if I understand correctly, in order for every observer to agree, you have changed everything in such a way that velocity (0.8 c) remains the same. You have changed the distance (if it was a rod it is measured a different length). You have changed the time (the ticking rate, the phase, the delay). You have kept velocity the same. why haven't you changed velocity and kept time & space as they are? Why is it so important to keep velocity the same for all observers?

[md65536]

41 minutes ago, michel123456 said:

You have kept velocity the same. why haven't you changed velocity and kept time & space as they are? Why is it so important to keep velocity the same for all observers?

It's only important if you want to be consistent with what we actually measure of reality.

It's a combination of the definition of speed being relative, and that measured speeds are consistent with that. If you have A moving at 0.8c relative to B, does it make sense that B is moving at a different speed relative to A? If you wanted that, you could define speed differently (eg. define speed to be absolute, and please call it something else), but you would end up with a system of measurements that is either inconsistent with measurement, or more cumbersome than what we have.

2 hours ago, michel123456 said:

I assure you I am not deliberately twisting everybody's replies, so I am afraid the other option may be valid.

I think it's a 3rd option: I think you're determined not to accept relativity and so you're determined not to understand it. I think we could find out with a quiz! Do you think that a) you will accept relativity and understand it together, or b) you will eventually understand it, and then accept it after, or c) if you accept that it's correct first, that will make it easier to understand, or d) you will never accept it and it's more likely that you'll find a flaw in it before anyone convinces you that it's true. Or e) other: ______ ?

Edited September 17, 2020 by md65536