# TIME DILATION. WHY do clocks that are

by abbott287
Tags: clocks, dilation, time
P: 25
 Quote by ghwellsjr I can see you like round numbers. Unfortunately, your example is flawed. You would have to go to a speed of 0.866c to get each twin to age two times slower than his twin, so with your permission, I'd like to illustrate what happens at a different speed, 0.6c because it makes the arithmetic come out in nice easy numbers.
As i stated in my first post i am not a physicist and the last time i used higher math was 12 years ago, so my use of numbers was just abstract to illustrate my point.

 Quote by ghwellsjr Let's say that after many days, the traveler turns around. He will immediately now see the flashes from his earth bound twin come in at double his own rate. But what will the earth bound twin see? Well, he won't see anything different until several days later because the sight of his twin turning around is subject to the delay in the light travel time. Eventually though, he will see the light flashes coming from his twin suddenly go from one half his rate to double his rate. But this will happen near the end of the trip. It is this imbalance in the ratio of the observed rates by each twin of the other twin's flashes that accounts for the difference in the total count that each one makes of his twin's flashes and thus the amount that each one has aged during the trip. Look up Relative Doppler for more details.
So from what i understand it is the turnaround that gives the difference. What i don't understand is how come only the Earth bound twin takes into account this turnaround? I mean, the moving brother might not see the Earth bound one turn around, but from his point of view he would have stopped moving away from him, so for that amount of time the flashes would come at the same rate as his own flashes going out. But the same should be said for the Earth bound brother. He should see a transition from the 1/2 rate to 1 rate to 2 (double rate). It was even said in a previous post that time delay would change during a half circle turnaround. So basically both brothers should still experience the same changes.

Or it would be even simpler if we take out the whole turnaround event. Since this is purely theoretical example we can put an event of instant turnaround where the moving brother goes through one portal and comes out another at exactly the same distance but moving in reverse direction (towards Earth). That way the Earth bound brother won't have the turnaround event to mess things up for him. What happens in that situation then?

DaleSpam i opened that page you linked, and i still find it difficult to agree with the explanations. I take into consideration the fact that i don't understand most of the math in there, but still... the example that you prefer with the graph still takes into account only the POV of the earth bound and does its math only from there. I opened another explanation which thankfully didn't have any math in it and the way they used to explain it there was with the whole inertial frame missing from the one brother. But even there they write that a possibility of the moving brother not experiencing any acceleration and gravitational forces with an instant turnaround could be considered and the explanation of that was left to the "reader". I take that as a failure to explain the problem fully and explaining it only when certain conditions are met and others aren't.
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P: 4,536
 Quote by Denius1704 As i stated in my first post i am not a physicist and the last time i used higher math was 12 years ago, so my use of numbers was just abstract to illustrate my point.
The math in Special Relativity is very simple. If you have a calculator with a square root function, that's all you need. General Relativity, which involves gravity, requires very complex math. So please be content to leave gravity out of the discussions until you have mastered SR.
 Quote by Denius1704 So from what i understand it is the turnaround that gives the difference. What i don't understand is how come only the Earth bound twin takes into account this turnaround? I mean, the moving brother might not see the Earth bound one turn around, but from his point of view he would have stopped moving away from him, so for that amount of time the flashes would come at the same rate as his own flashes going out. But the same should be said for the Earth bound brother. He should see a transition from the 1/2 rate to 1 rate to 2 (double rate). It was even said in a previous post that time delay would change during a half circle turnaround. So basically both brothers should still experience the same changes.
If you want the traveler to stop first and then turn around and start his journey home, then yes, there would be the transitions from 1/2 rate to the 1 rate to the double rate, and you are correct that the traveler sees this as soon as he stops and turns around. And the earth twin will see exactly the same thing but not at the moment it happens because he has to wait for the flashes in travel over a very great distance to reach him which will take a long time.
 Quote by Denius1704 Or it would be even simpler if we take out the whole turnaround event. Since this is purely theoretical example we can put an event of instant turnaround where the moving brother goes through one portal and comes out another at exactly the same distance but moving in reverse direction (towards Earth). That way the Earth bound brother won't have the turnaround event to mess things up for him. What happens in that situation then?
If I'm understanding you correctly, this is essentially what I was describing originally, that is, I didn't worry about how the traveling twin actually got turned around, I just assumed that he instantly changed from going away from the earth (seeing 1/2 flash rate) to going toward the earth (double flash rate).

So the bottom line is that the traveler counts the half-rate flashes from the earth twin for the same amount of time that he counts the double-rate flashes, but the earth twin counts the low-rate flashes from the traveler for most of the trip and doesn't start counting the high-rate flashes until near the end so he ends up with a much smaller count than the traveler does. And remember, they are each counting the one-second flashes from the other twin.
P: 117
 Quote by ghwellsjr .................................. And the earth twin will see exactly the same thing but not at the moment it happens because he has to wait for the flashes in travel over a very great distance to reach him which will take a long time. ...................................... So the bottom line is that the traveler counts the half-rate flashes from the earth twin for the same amount of time that he counts the double-rate flashes, but the earth twin counts the low-rate flashes from the traveler for most of the trip and doesn't start counting the high-rate flashes until near the end so he ends up with a much smaller count than the traveler does. And remember, they are each counting the one-second flashes from the other twin.
ghwellsjr, both brothers will see absolutely the same in a mirror turned image and will count flashes in absolutely the same ratio, since they are at same distance with same velocity relative to each other.
You'll eventually need to include the acceleration of the flying brother, but I don't see how that would resolve the time difference, because the brother on Earth (or in a motionless space craft) also sees himself as accelerating away. If the G-force during the acceleration helps somehow, then you will be right :)
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P: 16,485
 Quote by Denius1704 DaleSpam i opened that page you linked, and i still find it difficult to agree with the explanations.
Well, the question was how relativity would explain it. All of those explanations are valid under relativity. It sounds like you do not understand relativity. Unfortunately for you, the experimental evidence is overwhelming. There is no avoiding relativity.

 Quote by Denius1704 the explanation of that was left to the "reader". I take that as a failure to explain the problem fully and explaining it only when certain conditions are met and others aren't.
I think this is a little unfair to call a "failure". It is not possible for any document to cover everything. So any document must make a choice about what to cover and what not to cover. Then everything else is left to the reader or to other documents.
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P: 4,536
 Quote by sisoev ghwellsjr, both brothers will see absolutely the same in a mirror turned image and will count flashes in absolutely the same ratio, since they are at same distance with same velocity relative to each other. You'll eventually need to include the acceleration of the flying brother, but I don't see how that would resolve the time difference, because the brother on Earth (or in a motionless space craft) also sees himself as accelerating away. If the G-force during the acceleration helps somehow, then you will be right :)
The ratios that I'm talking about are not the 1/2 and double rates but rather the length of time that each twin sees those rates coming from the other twin. I will be 50-50 for the traveler and 80-20 for the earth twin.

Earlier in this thread you have been asking about the light travel time but now you seem to believe that it is non-existent. Are you saying that when the traveler is far away from the earth twin, the earth twin will still see him change direction half way through the trip and he won't have to wait for the light from that action to reach him? If you believe this, then you are believing in action-at-a-distance or an infinite speed for light. Are you sure you want to maintain this position?
P: 117
 Quote by ghwellsjr The ratios that I'm talking about are not the 1/2 and double rates but rather the length of time that each twin sees those rates coming from the other twin. I will be 50-50 for the traveler and 80-20 for the earth twin. Earlier in this thread you have been asking about the light travel time but now you seem to believe that it is non-existent. Are you saying that when the traveler is far away from the earth twin, the earth twin will still see him change direction half way through the trip and he won't have to wait for the light from that action to reach him? If you believe this, then you are believing in action-at-a-distance or an infinite speed for light. Are you sure you want to maintain this position?
ghwellsjr with the twin brothers example you better think of the speed as of velocity, because in absence of a third object the speed is irrelevant.

Now try to imagine what happens to both brothers and you'll see no difference for both of them. One of them will do the real traveling, but the other will see himself as traveling the same path with the same acceleration, speed, turning around and arriving back.

When the "brother in motion" turns back, he will keep seeing the flashes sent to him before he turned back, and some time latter he will start seeing the flashes sent after he turned around.
Same for the other brother; he will see himself as turning around some time later after the "real turn" was made from his brother.

I do not imply limitless light speed in this example, for there is no space for such suggestion.
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P: 16,485
 Quote by sisoev I cannot find any reason of why the delay will be proportional to the distance. It makes more sense to say that it is proportional to the light shift, because the light shift represents the distance.
Your comment here does not make sense. Do you understand that a light pulse travels a distance of $\Delta d = c \Delta t$ therefore the distance is proportional to the time and vice versa.

 Quote by sisoev I cannot also imagine that the light from a moving in circle object will be red shifted.
A failure of imagination on your part does not change the facts. The light from an object moving in a circle, or tangentially to you, is red shifted. Did you not read the link I posted?

 Quote by sisoev The red shift claim for circular moving object is very much like the explanation of the ladder paradox; it says that although both doors open simultaneously the ladder will see that the closer door opens first. That's wrong, because the ladder sees the doors as light reflection and since the back door opens simultaneously with the front door, there will not be door at the back, to reflect a ligh which will travel to the lader and show a presence of a door.
Again, you are under the misapprehension that SR is about visual appearances. The light delay is accounted for. SR is about what happens after accounting for the delay due to the finite speed of light.

 Quote by sisoev we will end up with two different wave lengths for same light, measured on same distance in two frames of reference.
What do you mean by this? The wave lengths are different in different frames of reference, but I am not sure what you mean by the qualifier "measured on the same distance".
PF Gold
P: 4,536
 Quote by sisoev ghwellsjr with the twin brothers example you better think of the speed as of velocity, because in absence of a third object the speed is irrelevant.
OK, then let the earth be that third object and change every occurrence of "speed" to "velocity" in my previous explanations.
 Quote by sisoev Now try to imagine what happens to both brothers and you'll see no difference for both of them. One of them will do the real traveling, but the other will see himself as traveling the same path with the same acceleration, speed, turning around and arriving back. When the "brother in motion" turns back, he will keep seeing the flashes sent to him before he turned back, and some time latter he will start seeing the flashes sent after he turned around. Same for the other brother; he will see himself as turning around some time later after the "real turn" was made from his brother.
So much of what you say is true and I don't see why you aren't grasping this simple concept: There is a big difference between what the two brothers see. A brother that actually turns around will immediately see a difference in the other brother's flash rate whereas the other brother won't see it til the pattern of flashes gets transmitted from his distant brother to him.

If you don't agree with this can you please explain where you think I'm wrong?
 Quote by sisoev I do not imply limitless light speed in this example, for there is no space for such suggestion.
Good.
 P: 50 I have a question about time too. Sorry if this is a dumb question: if theres a moving train with a mirror in it, and an observer(guy1) inside the train shines a light into it and measures the time it takes for it to get to the mirror and come back. He will measure a shorter time than an observer(guy2) outside of the train that is standing still. So the observer(guy2) says the time goes slower inside the train because its moving, but can the observer(guy1) in the train not say that time is going slower for the guy2 because in guy1's restframe guy 2 is moving? if so can someone explain this to me? sorry if this is a dumb question, I am just curious
P: 117
 Quote by DaleSpam Your comment here does not make sense. Do you understand that a light pulse travels a distance of $\Delta d = c \Delta t$ therefore the distance is proportional to the time and vice versa.
The same distance can be traveled by light sent from approaching source, from departing source or from still source. Then same distance will be covered respectively by blue, red shifted, or unchanged light.
You'll probably ask me how this change anything.

Few posts earlier I asked you the question "what do we consider as light and how do we measure its speed". One photon is not light and its speed should not represent the speed of light.
The speed of light should be represented by the frequency with which we meet every successive wave. The faster we move through the waves the faster we meet every next one thus changing the frequency, respectively the "speed of light". This way we will see that the speed of light depends on the speed of the observer and on the speed of the source.
If you accept this, the theory of relativity will start to make no sense.

 Quote by DaleSpam Again, you are under the misapprehension that SR is about visual appearances. The light delay is accounted for. SR is about what happens after accounting for the delay due to the finite speed of light.
No, I'm not under such misapprehension, but SR should take in to account that some things work only one way and no relativity can be applied to them.
The example with the two garage doors treats overlapping simultaneous events to which you cannot apply simultaneity for the simple reason that you cannot observe them as two events; you see only one of them and you should treat them as one.
The fact that you KNOW about the second event does not make it present to work with it and to apply values to it for later use.
Ignoring this will create paradoxes like the example I gave you with the two light beams from the back garage door.

 Quote by DaleSpam What do you mean by this? The wave lengths are different in different frames of reference, but I am not sure what you mean by the qualifier "measured on the same distance".
The wave length depends on the speed of the source and it cannot be different if the distance to the observer is not changed, but please ignore this for now.
P: 117
 Quote by ghwellsjr OK, then let the earth be that third object and change every occurrence of "speed" to "velocity" in my previous explanations.
Ha-ha :)
Sometime I also need to adjust the environment and the events to my way of thinking, and then I know that I'm somewhere wrong.

 Quote by ghwellsjr So much of what you say is true and I don't see why you aren't grasping this simple concept: There is a big difference between what the two brothers see. A brother that actually turns around will immediately see a difference in the other brother's flash rate whereas the other brother won't see it til the pattern of flashes gets transmitted from his distant brother to him. If you don't agree with this can you please explain where you think I'm wrong?
The trick for the right comprehension is to think for both brothers as symmetrically placed for the events in time.
If the departing brother measures x distance, the other will measure it as x as well, and if one of them measure n velocity it is n for the other as well.
Both of them observe the other with the same delay since they are symmetrically placed from the point in time where the events started.
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P: 16,485
 Quote by sisoev The speed of light should be represented by the frequency with which we meet every successive wave. The faster we move through the waves the faster we meet every next one thus changing the frequency, respectively the "speed of light". This way we will see that the speed of light depends on the speed of the observer and on the speed of the source.
Pure nonsense. The units are not even correct. Frequency is in units of 1/time and speed is in units of distance/time. The frequency cannot possibly be the speed of light.

 Quote by sisoev If you accept this, the theory of relativity will start to make no sense.
Good thing I don't accept it then.

 Quote by sisoev No, I'm not under such misapprehension, but SR should take in to account that some things work only one way and no relativity can be applied to them.
There are indeed many things that are invariant under the Lorentz transform. SR does take those into account. In fact, in some sense SR is primarily the study of these invariant quantities.

 Quote by sisoev The wave length depends on the speed of the source and it cannot be different if the distance to the observer is not changed, but please ignore this for now.
OK, but it is wrong.

PF Gold
P: 4,536
 Quote by sisoev The trick for the right comprehension is to think for both brothers as symmetrically placed for the events in time. If the departing brother measures x distance, the other will measure it as x as well, and if one of them measure n velocity it is n for the other as well. Both of them observe the other with the same delay since they are symmetrically placed from the point in time where the events started.
How can you think that both brothers are symmetrically placed from the point in time where the events started when one of them is causing the event to happen (he's turning around) and the other one is observing it from afar?
P: 117
 Quote by DaleSpam Pure nonsense. The units are not even correct. Frequency is in units of 1/time and speed is in units of distance/time. The frequency cannot possibly be the speed of light.
I said "represented" not "measured", DaleSpam

Pardon me, but the rest of your answer seams very much like a dodging.
The fact that SR takes in to account "many things that are invariant under the Lorentz transform" does not answer how do you see the "overlapping simultaneous events" (the two garage doors)
Not mentioning that my note puts in doubt an important part of SR.
I guess I have to wait until someone else comes with explanation.

I can imagine how annoying a guy like me can be in the midst of a company like you guys.
Later these days I'll post a thought experiment with graphics and after your answer I'll stop bothering you :)

I wish you a great weekend :)
P: 117
 Quote by ghwellsjr How can you think that both brothers are symmetrically placed from the point in time where the events started when one of them is causing the event to happen (he's turning around and the other one is observing it from afar?
The "turning around" event has no special meaning in the experiment, ghwellsjr
It is "experienced" from the other brother the same way as the brother who executes the event. The only difference is that if the real turn is to right, the other brother sees it as to left.
Once they start approaching each other in straight line, the tings are the same.
Both of them will observe the events with the same rate ratio.
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P: 16,485
 Quote by sisoev I said "represented" not "measured", DaleSpam
You will have to explain the difference and the relevance then. How do you transform from the frequency "representation" to the actual speed?

 Quote by sisoev Not mentioning that my note puts in doubt an important part of SR.
The only thing in doubt is your understanding of physics.

 Quote by sisoev Later these days I'll post a thought experiment with graphics
I will look forward to it. Your description of the garage door paradox is unclear, so a picture would be useful.
P: 117
 Quote by DaleSpam You will have to explain the difference and the relevance then. How do you transform from the frequency "representation" to the actual speed?
This will come with the experiment I intend to present with graphics.

 Quote by DaleSpam I will look forward to it. Your description of the garage door paradox is unclear, so a picture would be useful.
DaleSpam, talking to you is like talking to my teacher and I appreciate your patience. I really do.

I cannot add more to the way I see the ladder experiment.
My point is that we cannot treat an overlapping simultaneous events like a simultaneity since the light from them brings information only for one of them.
To elaborate; if the simultaneous events are aligned with the observer, there will not be light information for the back event, because it is simultaneous with the front.
We know about the back event, but we cannot deal with its light information because it is absent for us.
If we take this in to account, we will see that the explanation for the ladder paradox fails.
Therefore the rod contraction is false too.
Hence, the theory of relativity stands incorrect.
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P: 16,485
 Quote by sisoev The "turning around" event has no special meaning in the experiment, ghwellsjr It is "experienced" from the other brother the same way as the brother who executes the event.
This is factually incorrect. An accelerometer carried by one brother detects the g-forces of the turnaround, and an accelerometer carried by the other does not. The frequency of the pulses from the other brother changes immediately for one brother (the same one that detects the g-forces) and not for the other. The experiences of the two brothers are not the same.

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