Solving the Time Dilation Paradox - Special Relativity Explained

[hprog]

Consider A and B are moving in a linear motion, A claims B to move, and B claims that A is in motion, now when they look at each other they would clearly see which clock is running slower.
Let us denote the resting object by x and the moving object by y.
The special relativity explains, that since y is also moving away from the light and the light will take some time to catch up with y, so y will also see the x's clock slowing down.

1)Unfortunately I can't get the math correctly, let the velocity be 0.99 of the speed of light then y will see x's clock slowing down to 0.01 seconds for each second of y, which is a factor of 100.
So even after taking in account that y's clock has already been slowed down by a factor of about 7, and also taking in account that y claims x to move at 0.99c which means another factor of about 7, so we still have a factor of about 50 and not 100.
But y sees x's clock slowing down with a factor of 100.

2) But even if will forgive on that, there is still a much stronger question.
Just as y sees x's clock slowing down, for the same reason he also sees x's velocity slowing down with a factor of 100.
So even after we take y's time dilation into account we remain that y sees x's clock slowing down with a factor of 14 for x's 0.01c velocity.

So I am clearly missing something.

And in plain English, just as the moving object sees the resting objects clock slower he also sees the resting objects speed much slower, and the more the speed is slowing down the more is the clock slowing down, something that doesn't fit with what special relativity explains.

What do I am missing?

 

[Janus]

Consider A and B are moving in a linear motion, A claims B to move, and B claims that A is in motion, now when they look at each other they would clearly see which clock is running slower.
Let us denote the resting object by x and the moving object by y.
The special relativity explains, that since y is also moving away from the light and the light will take some time to catch up with y, so y will also see the x's clock slowing down.

You're describing the Doppler effect, which is entirely different from SR.

1)Unfortunately I can't get the math correctly, let the velocity be 0.99 of the speed of light then y will see x's clock slowing down to 0.01 seconds for each second of y, which is a factor of 100.

This is completely wrong. If you use the non-relativistic Doppler shift formula you get a apparent slow down factor of 1.99, not 100.

So even after taking in account that y's clock has already been slowed down by a factor of about 7, and also taking in account that y claims x to move at 0.99c which means another factor of about 7, so we still have a factor of about 50 and not 100.

You don't double apply time dilation. According to y, x is moving at 0.99c, thus according to y, x's clock runs slow by a factor of 7. If you add in the additional Doppler shift, and use the Relativistic formula, y sees x's clock run slow by a factor of 14.1[/quote]

But y sees x's clock slowing down with a factor of 100.

[/quote]No he doesn't

2) But even if will forgive on that, there is still a much stronger question.
Just as y sees x's clock slowing down, for the same reason he also sees x's velocity slowing down with a factor of 100.

The fact that x's clock runs slow according to y has no effect on the velocity x has according to y.

So even after we take y's time dilation into account we remain that y sees x's clock slowing down with a factor of 14 for x's 0.01c velocity.

So I am clearly missing something.

And in plain English, just as the moving object sees the resting objects clock slower he also sees the resting objects speed much slower, and the more the speed is slowing down the more is the clock slowing down, something that doesn't fit with what special relativity explains.

Again, no.

What do I am missing?

A heck of a lot. You appear to be totally confused about what SR says and what it doesn't.

 

[hprog]

You're describing the Doppler effect, which is entirely different from SR.This is completely wrong. If you use the non-relativistic Doppler shift formula you get a apparent slow down factor of 1.99, not 100. You don't double apply time dilation. According to y, x is moving at 0.99c, thus according to y, x's clock runs slow by a factor of 7. If you add in the additional Doppler shift, and use the Relativistic formula, y sees x's clock run slow by a factor of 14.1

No he doesn't The fact that x's clock runs slow according to y has no effect on the velocity x has according to y. Again, no.

A heck of a lot. You appear to be totally confused about what SR says and what it doesn't.

Thanks for your information, but I don't actually understand since I am actually missing this information.
And I also don't understand why the velocity is not affected, actually we see this happenning with stars moving away from us, and actually this is how scientist first realized that light has a finite speed by seeing jupiter's moon moving from us away slower then when it comes back.
Also what happenes when A and B move "toward" each other, in this case the moving object should see the resting object's clock getting faster.
Can you provide me a source that explains this subject (how each object views the others clock and velocity as well as the details of the duppler affect and how it relates to SR)?
Thanks.

 

[AdamMc]

Hello HProg,

Im currently reading the book called About Time, found https://www.amazon.com/dp/0684818221/?tag=pfamazon01-20, And i urge you to find this book in a book store and quickly browse to Chapter 2. If you can: skim to the part about Betty and Anne. The Author goes onto explain several points, one being specifically the issue you're having!

If i had the book with me at work I'd explain it to you in his words, but because I am no physicist (just really intrigued by relativity) i couldn't possibly explain it from scratch with 100% accuracy.

Im not telling you to go BUY the book - just head to a local bookstore that has it and browse real quick. Chapter 2. Betty/Anne - Twins example.

 

[Passionflower]

Thanks for your information, but I don't actually understand since I am actually missing this information.

You really only need two formulas, the time dilation formula:

$${\frac {1}{\sqrt {1-{\frac {{v}^{2}}{{c}^{2}}}}}}$$

This formula gives the other observer's clock rate after taking into account the time it takes light to arrive to the observer.

And the relativistic Doppler formula:

$$\sqrt { \left( 1+{\frac {{v}^{2}}{{c}^{2}}} \right) \left( 1-{\frac { {v}^{2}}{{c}^{2}}} \right) ^{-1}}$$

This formula is the 'raw' information, it is the other observer's clock rate that a person sees or measures.

In short when two observers are in relative motion each observer concludes the other observer's clock will run slow conform the first formula. However this result is calculated not actually measured. If you want to know what is actually measured you use the second formula. If they approach each will measure the other observer's clock to go faster and if they retreat they will measure the other observer's clock to go slower.

The relativistic Doppler formula also has a generic form where two observers move at an angle. It is interesting to note that if the angle is pi/2, e.g. when the movement is lateral, this formula becomes formula 1 regardless whether they approach or retreat from each other.

$$\sqrt {1-{\frac {{v}^{2}}{{c}^{2}}}} \left( 1+{\frac {v\cos \left( \theta \right) }{c}} \right) ^{-1}$$

 

[hprog]

I am still very confused.
Can some one give me a good online source for time dilation as it being viewed from moving objects and for the duppler effect?
Actually I need to understand first what is the duppler effect and what it is doing to the light and the view.
what I know about the duppler effect is, that for a moving object will the the light in front of him be differnet then the light in back, but here the question is how the moving object will see the resting object, so from where is the duppler effect coming from?

 

[Janus]

Thanks for your information, but I don't actually understand since I am actually missing this information.
And I also don't understand why the velocity is not affected, actually we see this happenning with stars moving away from us, and actually this is how scientist first realized that light has a finite speed by seeing jupiter's moon moving from us away slower then when it comes back.

No. What was measured was the period between eclipses of the Moon. While Jupiter was moving either away or towards the Earth. When Jupiter was moving away from us the Eclipses appeared farther apart, and when it was approaching, they appeared closer together. This was due to the increasing or decreasing time lag caused by the changing distance between Jupiter and Earth. It was an application of the Doppler effect.

Also what happenes when A and B move "toward" each other, in this case the moving object should see the resting object's clock getting faster.

It is important to distinguish between what A and B "see" vs. what they determine is happening to each other's clock. For example, in the above example, we may "see" the eclipses as being further apart or closer together, but we know that they actual occur at evenly spaced intervals.
Since the distance between A and B is decreasing, they would "see" each others clock as running fast, but they would determine that each other's clocks are running slow.

 

[AdamMc]

Doppler Effect is the change in frequency of a wave viewed by an observer with relation to a source.

If the Observer and Source are stationary - The observer sees the wave in the same form it was when it left the source.
If the observer and Source are moving away from one another - The Observer will see the wave LOWER in frequency.
If the observer and Source are moving Toward each other - The Observer will see the wave HIGHER in frequency.

Real world example: Stand on the side of a highway. Listen to the sound waves emitted by vehicles as they drive up to, and then pass you while you're stationary on the side of the road.

The car is emitting the same sound regardless of it's position. but the observer will hear the sound increase in frequency as the car approaches him. Once the car has passed him, the observer will hear the sound Decrease in frequency.

Now imagine that sound waves are light pulses or clock ticks. If neither the observer or the source are moving with relation to each other, the clock will tick normally. But once they start moving away from each other, the observer will notice a slowing of the clock ticks as it takes longer for each consecutive clock tick to reach him. (Which makes sense: If you're moving away from me and I'm throwing foot balls at you, Each football has a greater distance to travel. If I'm throwing them at a constant pace, You'll actually receive the footballs more slowly the faster away from me you run).

I hope i didn't botch the non-technical explanation too badly

 

[hprog]

Doppler Effect is the change in frequency of a wave viewed by an observer with relation to a source.

If the Observer and Source are stationary - The observer sees the wave in the same form it was when it left the source.
If the observer and Source are moving away from one another - The Observer will see the wave LOWER in frequency.
If the observer and Source are moving Toward each other - The Observer will see the wave HIGHER in frequency.

Real world example: Stand on the side of a highway. Listen to the sound waves emitted by vehicles as they drive up to, and then pass you while you're stationary on the side of the road.

The car is emitting the same sound regardless of it's position. but the observer will hear the sound increase in frequency as the car approaches him. Once the car has passed him, the observer will hear the sound Decrease in frequency.

Now imagine that sound waves are light pulses or clock ticks. If neither the observer or the source are moving with relation to each other, the clock will tick normally. But once they start moving away from each other, the observer will notice a slowing of the clock ticks as it takes longer for each consecutive clock tick to reach him. (Which makes sense: If you're moving away from me and I'm throwing foot balls at you, Each football has a greater distance to travel. If I'm throwing them at a constant pace, You'll actually receive the footballs more slowly the faster away from me you run).

I hope i didn't botch the non-technical explanation too badly

I am asking what will happen when the source is stationary, then will the moving object see any difference?
If so then my question has not been answered since I am asking here how the moving object will see the resting object.

 

[hprog]

No. What was measured was the period between eclipses of the Moon. While Jupiter was moving either away or towards the Earth. When Jupiter was moving away from us the Eclipses appeared farther apart, and when it was approaching, they appeared closer together. This was due to the increasing or decreasing time lag caused by the changing distance between Jupiter and Earth. It was an application of the Doppler effect.

It is important to distinguish between what A and B "see" vs. what they determine is happening to each other's clock. For example, in the above example, we may "see" the eclipses as being further apart or closer together, but we know that they actual occur at evenly spaced intervals.
Since the distance between A and B is decreasing, they would "see" each others clock as running fast, but they would determine that each other's clocks are running slow.

I still not understand why we shouldn't see a difference in speed just because of the light without any Doppler effect.
Why is just the clock slowing down without the entire velocity?
Also I do not understand, since the speed of light is constant what is the difference if the light is blue shifted or red shifted?
Maybe you have an online source for that, then please let me know, thanks

 

[AdamMc]

I am asking what will happen when the source is stationary, then will the moving object see any difference?
If so then my question has not been answered since I am asking here how the moving object will see the resting object.

The moving observer will still see a difference because he is moving RELATIVE to the source. It doesn't matter if the source is moving or the observer is moving.

If the Observer is moving towards a 'stationary' source that is emitting light pulses, he will notice the pulses increase in frequency. Alternatively, if he's moving away - looking backwards - he would notice the pulses slow in frequency. The same effect happens if the source is moving either away or toward the observer.

 

[AdamMc]

About Time: Pg 62-63

"... First let me compute the total duration of the journey as predicted by Einstein for each twin. At 80% the speed of light, it takes 10 years to travel 8 light-years. So Ann, on Earth, Will find that Betty returns in Earth year 2020 (if she left in Earth Year 2000). betty, on return, agrees that it is Earth year 2020 but insists that only twelve years have elapsed for her and her rocket clock - a standard atomic clock carefully synchronized before take off with Ann's identical clock on Earth - confirms this assertion: it reads 2012.

Now suppose we equip our twins with powerful telescopes so that they can watch each other's clocks throughout the journey and see for themselves what is going on. Ann's Earth clock ticks steadily on and betty looks back at it through her telescope as she speeds away into space (at 240,000km/s). According to Einstein, betty should see Ann's clock running at 60% of the rate of her own clock. in other words, during one hour of rocket time, betty is supposed to see the Earth clock advance only 36 minutes. in fact, she sees it going even slower than this. The reason concerns an extra effect, not directly connected with relativity, that is usually left out of the discussions of the twins paradox. it is vital to include the extra effect if you want to make sense of what the twins actually see.

Let me explain what causes this extra slowing. When betty looks back at earth, she does not see it as it is at that instant, but as it was when the light left Earth some time before. The time taken for light to travel from Earth to the rocket will steadily increase as the rocket gets farther out in space. Thus betty will see events on Earth progressively more delayed, because of the need for the light to traverse an ever-widening gap between Earth and the rocket. for example, after one hour's flight as measured from earth, betty is 0.8 light-hours (48 light-minutes) away, so she sees what was happening on Earth 48mins earlier, that being the time (as measured in Earth's frame of reference) required for the light, which conveys the images of Earth to betty, to reach her at that point in the journey. in particular, Ann's clock would appear to betty - I am referring to it's actual visual appearance - to be slow anyway, irrespective of the theory of relativity. After 2 hours' flight, the Earth clock would appear to betty to lag even more behind. this "ordinary" slowing down of clocks and events generally, as seen by a moving observer, is called the Doppler effect, named for an Australian physicist who first used it to describe a property of sound waves. by ADDING the Doppler effect to the time-dilation effect, you get the combined slowdown factor.

Ann will also see betty's rocket clock slowed by the Doppler effect because light from the rocket takes longer and longer to get back to earth. she will in addition see bettys clock slowed by the time-dilation effect. by symmetry, the combined slowdown factor of the other clock should be the same for both of them.

let me now compute the combined slowdown factor, first from Ann's point of view, then bettys. to do so,i shall focus on the great event of bettys arrival at the star. the outward journey takes 10 years as measured on earth. however, ann will not actually SEE the rocket reach the star in the year 2010, because by this stage, betty is eight light-years away. Since it will take light a further 8 years to get back to earth, it will not be until the year 2018 that ann gets to witness visually bettys arrival at the star.

what is the time of the arrival event as registered on bettys clock? Einsteins formula tells us that bettys clock runs at 0.6 the rate of the clock on earth, so 10 years of Earth time implies 6 years in the rocket. the rocket clock therefore stands at 6 years on bettys arrival at the star. so when ann gets to witness this arrival in 2018, the rocket clock says 2006. thus, as far as the visual appearance of the rocket clock is concerned, ann sees only 6 years having elapsed in her eighteen years... ie: bettys rocket clock has been running at 1/3 the rate of ann's Earth clock. now ann is perfectly capable of untangling the time-dilation and doppler effects, and computing the "actual" rate of bettys clock, having factored out the effect of the light delay. she will find the answer to be 0.6, in accordance with einstein's formula. Thus ann deduces (but does not actually see) that throughout bettys outward journey bettys clock was running at 36 mins to ann's hour.

from bettys perspective, things are the other way about. she agrees of course, that her rocket clock stands at 2006, when she arrives at the star, but what does she see the Earth clock registering at that moment? we know that in the Earth's frame of reference, the arrival event occurs at 2010, but, because the star is 8 light years away, the light that actually reaches the rocket at that moment will be from 8 years previously -- ie.. 2002. so betty will look back at earth, on the arrival at the star and see the Earth clock registering 2002. HER clock says 2006, therefore as far as the actual appearance of the Earth clock is concerned, it records 2 years having elapsed for bettys 6 years. thus betty concludes that the Earth clock has been running at 1/3 the rate of her own rocket clock for the outward part of the journey! this is the same factor that ann perceived betty's clock to be slowed by. so the situation is indeed perfectly symmetric. ..."

Prior info for Calculations:
Betty leaves Earth in year 2000 and travel by rocket to a star 8 Light-years away (as measured by Earth's reference) at a speed of 240,000km/s. to keep things simple, periods where the ship was accelerating have been left out. Assume instant velocity.

The 0.6 comes from:

240,000km/s Divided by C (300,000km/s) = 0.8
0.8 Squared = 0.64
1-0.64=0.36
Sqrt[0.36 = 0.6

1 Earth hour = Earth Hour * 0.6 [Rocket time]
60 Earth mins = 36 Rocket Mins

PS: ON the return trip home, instead of time being at 1/3rd of the observers, it is 3x's the observers (ie: accelerated) because the twins are now headed towards each other.