# What happens in the restframe with lightsource?

• faen
In summary, during the conversation, the concept of time dilation was discussed and how it is observed differently in two frames of reference. The conversation also touched upon an experiment involving mirrors and light traveling between them at different speeds in two frames of reference. The equation for time dilation was mentioned and how it can be used to predict opposite results depending on the location of the experiment. However, this apparent contradiction is due to the fact that time dilation is defined differently in different frames of reference. It only applies to objects/observers/clocks that are moving in a given frame of reference, not to the rest frame itself.
faen
So I was reading the proof of time dilation. It explains how the observer in a rest frame observes a longer path of the light than the observer in the moving frame.

However, I made a drawing of a slightly different experiment, where I can't predict the result of it. I was hoping somebody could help me out. In the picture below, the person in frame of reference A is moving close to the light speed. In the rest frame an observer is sending light between two vertical points as shown in the picture. The person in the rest frame would see that the light goes between the two points in time t determined by the given distance d and velocity c. Now, I am wondering what would the person in frame of reference A observe? If time is really slower for him, he would have to observe a shorter path with the same speed. However as far as I can see, the c and d variables is the same for the people in both frame of references giving equal time? I hope somebody can help me out with my confusion here, thanks a lot :)

In A's rest frame, the mirrors are moving, so the light going from the top mirror to the bottom is going "diagonally". The distance between the mirrors is the same in both frames. Light going from the top mirror to the bottom travels a distance d=ct in B's frame, and ct' in A's frame (where t' is the time measured by A). The pythagoran theorem tells us that ##(ct')^2=(ct)^2+(vt')^2##. Solve this for t, and you get
$$t=\frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma t'.$$.

Fredrik said:
In A's rest frame, the mirrors are moving, so the light going from the top mirror to the bottom is going "diagonally". The distance between the mirrors is the same in both frames. Light going from the top mirror to the bottom travels a distance d=ct in B's frame, and ct' in A's frame (where t' is the time measured by A). The pythagoran theorem tells us that ##(ct')^2=(ct)^2+(vt')^2##. Solve this for t, and you get
$$t=\frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma t'.$$.

Yes I thought about this too. But the result seems strange to me, because it now predicts that time goes slower in the rest frame and not in the moving frame, just because we changed the location of the experiment.

For example if we had done the experiment in frame A instead of B, t and t' would change place in the equation predicting opposite results, as such:

$$t'=\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}.$$.

How is this possible?

It's possible because t and t' mean different things in the two equalities. Consider two inertial coordinate systems S and S' with a common origin. Denote the velocity of S' in S by v. We assume that v>0. Denote the event at the origin by O. Let E be an event on the time axis of S, and denote the time coordinate of E in S by t. Now the time dilation formula tells us that t'=γt. Here t' is the time coordinate of E in S'. Note that E is not on the time axis of S'.

But when you use the time dilation formula to go from S' to S instead, t' is the time coordinate of an event F on the time axis of S', and t is the time coordinate of F in S. So we're not just doing the calculation we did before "in reverse". It's a calculation that involves a different event.

This is what I said about the apparent contradiction in another thread:
Fredrik said:
"B's clock is slow relative to A" appears to contradict "A's clock is slow relative to B". To understand the problem here, it's essential that you understand that these statements are actually defined to mean something different from what they appear to be saying. What they actually mean is this:

"The coordinate system associated with A's motion assigns time coordinates to B's world line that increase faster along B's world line than the numbers displayed by B's clock"

"The coordinate system associated with B's motion assigns time coordinates to A's world line that increase faster along A's world line than the numbers displayed by A's clock"​

faen said:
Yes I thought about this too. But the result seems strange to me, because it now predicts that time goes slower in the rest frame and not in the moving frame, just because we changed the location of the experiment.

For example if we had done the experiment in frame A instead of B, t and t' would change place in the equation predicting opposite results, as such:

$$t'=\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}.$$.

How is this possible?
According to Special Relativity, time does not go slower in any rest frame you choose, it only goes slower for objects/observers/clocks that are moving in any rest frame you choose. It's no different than saying that the rocket is moving in the ground's rest frame and the ground is moving in the rocket's rest frame. How is that possible?

But it does help to consider that in the ground's rest frame, the clock on the rocket is ticking slower and it is moving away from the clock on the ground. So between each tick of the rocket clock, as it observes the ground clock, it has moved farther away and so the image of the ground clock's ticks have farther to travel resulting in a longer time compared to its own ticks. It turns out that the ratio of the ground clock's tick interval compared to the rocket clock's tick interval as determined by the rocket is the same as the ratio of the rocket clock's tick interval compared to the ground clock's tick as determined by the ground. Work it out, you'll see that this is true.

ghwellsjr said:
According to Special Relativity, time does not go slower in any rest frame you choose, it only goes slower for objects/observers/clocks that are moving in any rest frame you choose. It's no different than saying that the rocket is moving in the ground's rest frame and the ground is moving in the rocket's rest frame. How is that possible?

But it does help to consider that in the ground's rest frame, the clock on the rocket is ticking slower and it is moving away from the clock on the ground. So between each tick of the rocket clock, as it observes the ground clock, it has moved farther away and so the image of the ground clock's ticks have farther to travel resulting in a longer time compared to its own ticks. It turns out that the ratio of the ground clock's tick interval compared to the rocket clock's tick interval as determined by the rocket is the same as the ratio of the rocket clock's tick interval compared to the ground clock's tick as determined by the ground. Work it out, you'll see that this is true.

That was a surprising result. I think I more or less understand it now. Thanks a lot for clearing it up for me, to both of you. I have one more question though.

If a person travels close to the light speed, will he become older slower than the observer not traveling or is this a myth then?

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I think the confusion might lie in the use of the word ‘really’. If ‘time is really slower for him’, in an absolute sense, there’d be a problem because it would imply that time in the ‘rest’ frame was running faster than his own. Time is only running more slowly in A’s frame as observed from the ‘rest’ frame on the ground. As far as A is concerned, his time is just fine: to him, it is time in the ground’s frame that is running slower than his own. The slowing of time isn’t an absolute thing in the ordinary sense - it is a conclusion reached by one observer on time in the other frame.

It’s best to avoid saying time ‘really’ is running more slowly on the ship because this might suggest that the situation is one-sided. It seems to imply that the ship is ‘really’ moving, and the ground is ‘really’ stationary. Relativity doesn’t allow us to say this – instead, all we can assert is that the ship and the ground are moving relative to each other.

To observer A, the light clock on the ground is moving past him, so as far as he’s concerned the light which leaves the source and reaches the mirror hasn’t just covered distance d in the vertical direction – it must appear to follow a diagonal path, which, therefore, is longer than d. Since, as you rightly say, all observers measure light to travel at velocity c, A must conclude that the light has had to take longer to go from the source to the mirror than if the clock were stationary in his own frame. So A concludes, from observing B’s light clock, that time must be running slower in B’s frame than his own.

So why does observer B say that time in A’s frame is running slower? Suppose A had an identical light clock on his ship. He would observe the light moving a vertical distance d, in his frame, from the source to the mirror. But B would observe, because of the relative motion of the ship, the light tracing a diagonal path. This path would be longer than d, leading B to conclude that, since the light, traveling at velocity c, takes longer to cover this diagonal distance than it would take to travel the just the vertical distance d, time on the ship must be running slower than time on the ground.

It’s always tempting to think of something ‘moving’, but a more accurate expression (albeit a clumsier one) would be something like ‘moving relative to a certain frame of reference’. In special relativity, this helps keep at bay the completely natural, though wrong, notion that something is ‘really’ moving and that something else is ‘really’ still.

Sorry – didn’t see the last posts – I should pay more attention.

Somebody traveling at close to light speed (or any speed relative to us) would appear to age more slowly than us. But that’s only as measured by us. To the person traveling, it is we who are doing the moving relative to them, so they would see us as ageing more slowly than them.

Again, the situation is perfectly symmetric (so long as nobody changes velocity). Both are right.

How is it possible that the length contraction equation is inverted in relation to the Lorenzt transformation as compared to the time dilation equation? I think I recall haveing to take the inverse of the result in order to actually get a smaller value for the length of time that the object seen to travel at a velocity would experience. For instance, you can't have x/z and (x)(z) and get smaller values for both when z is the same value.

It has made me wonder if the real time dilation equation could look identical to the length contraction equation. I think the distance ct' would have to be assigned to the vertical distance. This would assume that the object in question measures their time by seeing the light travel straight up and down this vertical distance. Their time would then have to slow down in order for them to measure the same speed for light makeing this trip giving a smaller value for t to account for the smaller distance traveled. Then the distance the object traveled would have to be vt, this says that the observer watching the object travel used their own time to determine this as how they measure time. Then the other side of the triangle ct would be how an observer at rest measured the photon traveling at na angle using their own time as they measure it at rest. This would be a longer distance traveled and then would require a longer duration of time for the photon to be seen to travel this distance for an observer at rest. So in effect all you would be doing is comparing the relation of two sides of a triangle from one frame of reference to another side of a triangle that exist in another frame of reference. The amount of time t' would have to "really" go slower in order for them to measure the same speed of light when it is seen to travel a shorter distance.

faen said:
That was a surprising result. I think I more or less understand it now. Thanks a lot for clearing it up for me, to both of you. I have one more question though.

If a person travels close to the light speed, will he become older slower than the observer not traveling or is this a myth then?
As Goodison_Lad said, they are both aging slowly in the other one's rest frame. (In A's frame, B is aging slower. In B's frame, A is aging slower). There's no contradiction here, because of what I said in my previous post. If one of them turns around so that they eventually meet again, the one who turned around will be younger. This result can't be calculated with the time dilation formula alone, because there are at least three inertial frames involved.

John232 said:
How is it possible that the length contraction equation is inverted in relation to the Lorenzt transformation as compared to the time dilation equation?
The time dilation formula tells you the difference between the t' coordinates that another observer would assign to two events on the t axis of the inertial coordinate system in which you are at rest. The reason why the Lorentz contraction formula is different is that it doesn't tell us the difference between the x' coordinates that another observer would assign to two events on our x axis. Instead, it tells us the difference between the x' coordinates he would assign to two simultaneous (in his rest frame) events on the world lines of the endpoints of an object that's at rest relative to you. So the two events he's comparing are on one of his simultaneity lines, not yours. If he had compared two events on one of your simultaneity lines, the formula would have had the same factor as the time dilation formula.

Fredrik said:
In A's rest frame, the mirrors are moving, so the light going from the top mirror to the bottom is going "diagonally". The distance between the mirrors is the same in both frames. Light going from the top mirror to the bottom travels a distance d=ct in B's frame, and ct' in A's frame (where t' is the time measured by A). The pythagoran theorem tells us that ##(ct')^2=(ct)^2+(vt')^2##. Solve this for t, and you get
$$t=\frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma t'.$$.

In this equation you got the wrong relation to t and the Lorentz factor. t'=yt not t=yt'. Then x'=x/y, where y is the lorentz factor. It would seem that the assigning of t' can be counter intuitive and gives different relations for t and x to the lorentz factor, just to be clear about what I was saying before. I think I meant to say lorentz factor instead, before.

But, then say you where doing the proof on a black board and explaining how these variables are supposed to be assigned, how do you explain that the time variables are set up to be the opposite of what they "should" be? It looks like you have the original equation set up correctly, but then it was worked wrong or was a typo. I think the way you had it set up should give the correct relation to the lorentz factor. But then why would you measure the distance an object traveled as vt' when you are at rest using the moveing object dialated time and then measure ct' as the hypotenus as an object see's light travel at a greater distance at an angle again using the objects that is traveling in motions time, and then saying that it is compared in relation to the pythagrean to your own time as a light beam travels straight up and down to you at rest?

I think their should be a better answer than if you convert the change in time to actual time you end up haveing to take the inverse of the change in time that in turns gives you a smaller value for time for in object in motion, and then how does the act of accouting for the change in time affect how variables are even assigned? Or is accouting for the change in time even necessary in this algebra problem? I think it would be funny if Einstein goofed on setting up this equation and misassigned the time variables. I think if you exchanged the time variables and then considered them as the actual times, then the answer you would get wouldn't be all that much different than if you did a conversion that takes the inverse of the answer that would in turn give lower values for the objects time that was in motion. The only difference would be the difference you would get in an answer by switching the vairiables of an equation and then takeing the inverse of it, but I don't think doing that gives you the same answer.

I would hate to be right about this because if there was something wrong with this equation it would mean that we would all have to start modern physics over. Then again they don't really use this in particle physics anyways...

John232 said:
In this equation you got the wrong relation to t and the Lorentz factor. t'=yt not t=yt'.
Good catch. I seem to have exchanged the meaning of t and t' somewhere. When t and t' are defined as I described, the result should be t'=γt.

Fredrik said:
As Goodison_Lad said, they are both aging slowly in the other one's rest frame. (In A's frame, B is aging slower. In B's frame, A is aging slower). There's no contradiction here, because of what I said in my previous post. If one of them turns around so that they eventually meet again, the one who turned around will be younger. This result can't be calculated with the time dilation formula alone, because there are at least three inertial frames involved.

And what if both decide to return and then meet halfway?

I think my aging process will not change if someone runs away from me.

Fredrik said:
Good catch. I seem to have exchanged the meaning of t and t' somewhere. When t and t' are defined as I described, the result should be t'=γt.

No problem, I have wrestled with this idea for some time. I still haven't been able to get an answer that resovles the confusion of this issue. I have thought about trying to publish my own theory of special relativity that only switchs the time variables and doesn't consider the time as being a change in time, and I don't know how far into the mathematics I would have to go into other theories that depend on this equation. I am sure it would find a lot of opposition since it would mean that everything we know about modern physics could be slightly wrong, but their is a growing opionion that relativity may still need some work done on it inoder to obtain grand unification and that may conclude that relativity as we know is not the final answer. Or, that a more accurate theory could get rid of dark matter or something of the like, I think it could be that something that physics has been looking for...

Tantalos said:
And what if both decide to return and then meet halfway?
Then they'll be the same age when they meet.

I tried to do some research about the time travel thing. As far as I've understood while the person is in the spaceship he turns around, accellerates, and he will feel that he is in a gravitational field. Being in a gravitational field, causes time to go slower. Thus for him time will go slower for him and faster on earth, while observers from Earth predict the same, that time goes slower for him and faster for them. However this accelleration lasts only a very short amount of time compared to the whole trip. Therefor he will only feel as if he is in a gravitational field slowing his time only for as long as he accellerates. After he is finished accellerating both observers will experience that the other persons clock is slower, right?

So is it only the time during the accelleration that the person in the rockets time is really going relatively slower, or his time is really going relatively slower the whole trip?

Also, if gravity is curvature in space-time, how does such curvature provide matter with energy just by being curves?

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Fredrik said:
As Goodison_Lad said, they are both aging slowly in the other one's rest frame. (In A's frame, B is aging slower. In B's frame, A is aging slower). There's no contradiction here, because of what I said in my previous post. If one of them turns around so that they eventually meet again, the one who turned around will be younger. This result can't be calculated with the time dilation formula alone, because there are at least three inertial frames involved.

Fredrik said:
Then they'll be the same age when they meet.

This is against logic. According to you if B travels first and then A goes that same distance with the same speed as B to meet him then A will be younger.

And what if they decide to travel both in one spaceship the same distance and speed? Logically they will be of the same age, but with the previous result their age depends on whether they travel alone or with a company.

I read somewhere that special relativity and time dilation has been proven to really occur. Isn't this a contradiction according to what has been said so far?

Consider the following experiment:

There is a rest frame of reference, with available clocks anywhere. There is another person with a clock traveling very close to light speed such that the traveling clock should almost stop its ticks relative to the restframe. After traveling in a straight line for some seconds, the person stops traveling to compare the differences of the clocks. According to experiments the traveling clock has been ticking slower. I am wondering why? Because according to the traveling clock, it was the other clock which slowed down?

Is it because the moving frame had to stop and deaccellerate? Doesn't this mean that time dilation does not really occur because of traveling faster, but that accelleration is what causes time dilation? There is some contradiction here about that special relativity slows time, since it is simply a contradiction that both clocks are slow relative to each other.

I would appreciate if somebody could help me out again to understand this better.

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Tantalos said:
This is against logic. According to you if B travels first and then A goes that same distance with the same speed as B to meet him then A will be younger.
I don't think I said anything that can be interpreted that way. If A and B both leave Earth, going at the same speed in opposite directions, and after a while turn around and go back, they'll be the same age when they meet. If instead A stays here while B leaves and later comes back, B will be younger when they meet.

In the scenario that you describe now, where B goes from Earth to some distant location and stays there, and A makes the same journey some time later, then they will certainly be the same age when they meet.

Tantalos said:
And what if they decide to travel both in one spaceship the same distance and speed? Logically they will be of the same age, but with the previous result their age depends on whether they travel alone or with a company.
Huh?

Why is nobody replying to my questions anymore? :/ If there's anything I need to explain better please let me know and I'll try..

faen said:
I read somewhere that special relativity and time dilation has been proven to really occur. Isn't this a contradiction according to what has been said so far?

No. If it weren't observable, then we wouldn't be talking about it and it wouldn't be science.

Consider the following experiment:

There is a rest frame of reference, with available clocks anywhere. There is another person with a clock traveling very close to light speed such that the traveling clock should almost stop its ticks relative to the restframe. After traveling in a straight line for some seconds, the person stops traveling to compare the differences of the clocks. According to experiments the traveling clock has been ticking slower. I am wondering why? Because according to the traveling clock, it was the other clock which slowed down?

The reason depends on which observer you are. The unaccelerated observer attributes the disagreement in the clocks as being due to time dilation while the spacecraft was in motion.

The accelerated observer attributes the difference to a combination of factors, but mainly failure of simultaneity. Suppose your rest observer, with clocks stationed everywhere, in particular has a clock next to himself where the race begins, and a clock at the "finish line" where the traveling observer slows down and stops, and that these clocks are synchronized in his frame.

Then they will not be synchronized in the traveling frame. The traveling observer observes both clocks running at the same slow rate, but the finish line clock will read a value that is vastly ahead of the starting line clock. So it is no surprise to him that when he gets there, it is reading many years ahead of his own clock, even though it runs more slowly.

it is simply a contradiction that both clocks are slow relative to each other.

No it isn't.

ZikZak said:
The reason depends on which observer you are. The unaccelerated observer attributes the disagreement in the clocks as being due to time dilation while the spacecraft was in motion.

The accelerated observer attributes the difference to a combination of factors, but mainly failure of simultaneity. Suppose your rest observer, with clocks stationed everywhere, in particular has a clock next to himself where the race begins, and a clock at the "finish line" where the traveling observer slows down and stops, and that these clocks are synchronized in his frame.

Then they will not be synchronized in the traveling frame. The traveling observer observes both clocks running at the same slow rate, but the finish line clock will read a value that is vastly ahead of the starting line clock. So it is no surprise to him that when he gets there, it is reading many years ahead of his own clock, even though it runs more slowly.

Why will he read that the clock at the finish line is years ahead of his own clock, even if it is running the same speed as the one at the starting line? If there was a person in the rest frame, he would predict that time went slower for the person in the space ship. If this was true as well, why isn't he the one who is younger than the person who traveled?

ZikZak said:
No it isn't.

The theory says that time goes slower for both the person in the spaceship and in the rest frame. It says that time goes slower in rest frame relative to spaceship frame, and time goes slower in spaceship frame relative to rest frame. This can't be true if there is only one truth. Also when the two frames arrives at equal velocity, they predict only one truth. However it seems like some kind of acceleration is necessary for them to predict and arrive at the same truth. Thus I am curious about the role of acceleration as well.

faen said:
Why will he read that the clock at the finish line is years ahead of his own clock, even if it is running the same speed as the one at the starting line?

For the same reason each observer observes the other's clock(s) to run slow: that both of them must observe the speeds of every light ray to be c. That requirement leads logically to the necessity for moving clocks to run slow, for moving meter sticks to be shorter than their rest counterparts, and for clocks synchronized in one frame to be unsynchronized in another. That is to say, the Lorentz Transform.
The theory says that time goes slower for both the person in the spaceship and in the rest frame. It says that time goes slower in rest frame relative to spaceship frame, and time goes slower in spaceship frame relative to rest frame. This can't be true if there is only one truth.

I live in the U.S. If I point in the direction of "up," I point in the opposite direction from someone who lives in Australia. We completely and utterly disagree on which direction is up, and yet we are both right, AND there is one truth to the matter.

John232 said:
No problem, I have wrestled with this idea for some time. I still haven't been able to get an answer that resovles the confusion of this issue. I have thought about trying to publish my own theory of special relativity that only switchs the time variables and doesn't consider the time as being a change in time, and I don't know how far into the mathematics I would have to go into other theories that depend on this equation. I am sure it would find a lot of opposition since it would mean that everything we know about modern physics could be slightly wrong, but their is a growing opionion that relativity may still need some work done on it inoder to obtain grand unification and that may conclude that relativity as we know is not the final answer. Or, that a more accurate theory could get rid of dark matter or something of the like, I think it could be that something that physics has been looking for...
Let me see if I can resolve your confusion so you won't be tempted to publish your own theory of special relativity.

Part of the confusion comes from a misunderstanding of the nomenclature. In the usual presentation of the Lorentz Transform the equations are stated with t' and x' on the left side of the equation and t and x on the right side of the equations. You shouldn't think of the primed symbols as being associated exclusively with just a primed frame like S' and the unprimed symbols associated exclusively with an unprimed frame like S. Rather, you should think of the LT as providing a way to convert the coordinates of any frame into the coordinates of any other frame and back again if you want. I think it is better to refer to each frame with unprimed symbols to avoid this confusion.

So, for example, let's say that you have an event in frame A at t=18 and x=10 and you want to find the coordinates in frame B moving at 0.6c in the x direction with respect to B. (I'm going to use compatible units where c=1.) First we need to calculate gamma, γ and note this is not the English letter y, it is the Greek letter γ. You can get it by going to the advanced editing mode and clicking on the fourth Quick Symbol in the second row. Unfortunately, it looks very much like the letter y but you can tell the difference with the tail going straight down instead of angled.

OK, since β=0.6, we can calculate γ:

γ = 1/√(1-β2) = 1/√(1-0.62) = 1/√(1-0.36) = 1/√(0.64) = 1/0.8 = 1.25

Now we are ready to apply the LT formulas:

t' = γ(t-xβ) = 1.25(18-10*0.6) = 1.25(18-6) = 1.25(12) = 15
x' = γ(x-tβ) = 1.25(10-18*0.6) = 1.25(10-10.8) = 1.25(-0.8) = -1

So if we represent the coordinates as [t,x], then [18,10] in frame A is [15,-1] in frame B.

Now if we want to go back the other way, we do the same thing except that now frame A is moving at -0.6c along the x-axis with respect to frame B. Gamma is of course the same value.

t' = γ(t-xβ) = 1.25(15-(-1)*(-0.6)) = 1.25(15-0.6) = 1.25(14.4) = 18
x' = γ(x-tβ) = 1.25(-1-15*(-0.6)) = 1.25(-1+9) = 1.25(8) = 10

We have confirmed that [15,-1] in frame B transforms to [18,10] in frame A, the original coordinates that we started out with.

So you see, the primed symbols can apply temporarily to either frame, they are just intended to mean the new values in whatever frame you are converting the coordinates into. It's also important to recognize that the values that t and x refer to are coordinate values, that is, coordinate time and coordinate distance.

Now I want to get to the confusion over the Time Dilation factor. Since we use t' and t in the Lorentz Transform for two different frames, we don't want to use both of these symbols in the Time Dilation formula because they are both coordinate times (t' being used just temporarily). The Time Dilation formula applies in just a single frame and allows us to determine the time on a moving clock in that frame so we use Einstein's symbol for Proper Time, the Greek letter tau, τ. Einstein presented this formula as τ=t√(1-v2/c2) but it is more common to see it τ=t/γ. When presented these ways, there is no confusion because τ is always the Proper Time for the moving clock and t is always the Coordinate Time for the stationary clocks.

So if we want to find out how much time progresses on a clock moving at 0.6c in any frame compared to the coordinate clocks, it is simply 0.8 times the progress of time on the coordinate clocks (since 1/1.25=0.8).

Now let's consider a clock that starts at the origin in our Frame A from above and moves in a straight line at a constant speed to the event [18,10]. What speed will it move at? That's pretty easy to calculate because it has traveled a distance of 10 in a time of 18 so the speed is 0.555556c. If the time on the clock was 0 at the start of the trip, what time will be on the clock when it reaches 10? First we have to calculate gamma, like we did before:

γ = 1/√(1-β2) = 1/√(1-0.5555562) = 1/√(1-0.308642) = 1/√(0.691358) = 1/0.831479 = 1.202676

Then we divide the coordinate time of 18 by 1.20267 and we get 14.96663.

We could have used the Lorentz Transform calculate the coordinates in Frame C moving at 0.555556c to arrive at this same value as follows:

t' = γ(t-xβ) = 1.202676(18-10*0.555556) = 1.202676(18-5.55556) = 1.202676(12.44444) = 14.96663
x' = γ(x-tβ) = 1.202676(10-18*0.555556) = 1.202676(10-10) = 1.202676(0) = 0

Well, look at this, the distance coordinate came out zero and the reason why is because we have gone from Frame A in which the clock was moving at 0.555556c to Frame C in which it is stationary so the x-coordinate better come out zero because that is where it started and it isn't moving in Frame C.

Now the final point of confusion has to do with length contraction and why the distance coordinate isn't also divided by gamma in the Lorentz Transform. The answer to that question has to do with that zero coordinate for distance. In order to properly arrive at the correct time dilation factor, we had to use an event in Frame A that would result in a coordinate of zero (because the clock started at location zero in Frame A) for distance in Frame C. This happens automatically when we calculate where the clock is, based on its speed in Frame A.

But in order to see the correct length contraction, we have to use two events in Frame A that result in the time coordinates being the same at both ends of the desired length in Frame C. How do we do that? There are several ways but the easiest is to simply multiply the distance coordinate by the speed and use the same x-coordinate as before.

Remember, what we are trying to do is calculate the distance that the clock thinks it has traveled in Frame C. So starting from Frame A the clock moved from [0,0] to [18,10]. In Frame A, the distance is 10 and we expect it to be divided by gamma resulting in a value of 8.314794. All we have to do is multiply the distance coordinate by the speed so that is 10 times 0.555556 which is 5.55556 and we create a new event of [5.55556,10] in Frame A which we transform to Frame C:

t' = γ(t-xβ) = 1.202676(5.55556-10*0.555556) = 1.202676(5.55556-5.55556) = 1.202676(0) = 0
x' = γ(x-tβ) = 1.202676(10-5.55556*0.555556) = 1.202676(10-3.08642) = 1.202676(6.91358) = 8.314794

Here we see the requirement that the time coordinate is the same as it was when the clock started moving and this gives us the same length contraction for the distance the clock traveled as when we simply divide the original length by gamma.

Keep in mind that this is an easy calculation because one end was at the origin. If you are calculating the length of an object or a distance in which neither end is at the origin, you need to multiply the speed by the delta between the two ends and then add that number to the time coordinate of the first event.

I hope if you study this, all your confusions will be cleared up and you won't have to publish your own theory.

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faen said:
As far as I've understood while the person is in the spaceship he turns around, accellerates, and he will feel that he is in a gravitational field.
This is just a much more difficult way of looking at these things. I would recommend that you avoid it. Actually, the whole concept of "gravitational field" is problematic. That concept isn't really used in GR (or SR), and there's no need to involve anything but SR here. The counterintuitive aspects of the problem show up in the SR treatement of the problem as well.

faen said:
However this accelleration lasts only a very short amount of time compared to the whole trip. Therefor he will only feel as if he is in a gravitational field slowing his time only for as long as he accellerates. After he is finished accellerating both observers will experience that the other persons clock is slower, right?
This is not a right or wrong issue. It's a matter of what coordinate system we choose to think of as describing his "experience". If we choose to use the comoving inertial coordinate systems, then he "experiences" that the guy on Earth is aging slowly both before and after the turnaround. But the fact that the simultaneity lines of the two comoving inertial coordinate systems before and after the turnaround are very different means that the moment after the turnaround, he would describe the guy on Earth as much older than before the turnaround.

faen said:
So is it only the time during the accelleration that the person in the rockets time is really going relatively slower, or his time is really going relatively slower the whole trip?
It's the other parts of the trip. The turnaround just makes us associate a different coordinate system with his motion.

faen said:
I read somewhere that special relativity and time dilation has been proven to really occur. Isn't this a contradiction according to what has been said so far?
It has, and no, it's not.

faen said:
Consider the following experiment:

There is a rest frame of reference, with available clocks anywhere. There is another person with a clock traveling very close to light speed such that the traveling clock should almost stop its ticks relative to the restframe.
You shouldn't say "the" rest frame, because the other frame is also a rest frame...of another object.

faen said:
After traveling in a straight line for some seconds, the person stops traveling to compare the differences of the clocks. According to experiments the traveling clock has been ticking slower. I am wondering why? Because according to the traveling clock, it was the other clock which slowed down?
I don't like to think of moving clocks as "slowing down". Instead I would say that every clock always does what it's supposed to, which is to display the proper time of the curve that describes the motion it has done so far. Some curves just have shorter proper times than others. The "straight lines" that represent non-accelerating motion have the longest proper times.

faen said:
Is it because the moving frame had to stop and deaccellerate? Doesn't this mean that time dilation does not really occur because of traveling faster, but that accelleration is what causes time dilation?
No, it's the speed, not the acceleration. The following spacetime diagram, made by DrGreg, following a description by yuiop, illustrates this point.

Here the twins are doing exactly the same acceleration, but spend different amounts of time at the higher speed relative to Earth. The one who spent the most time at the higher speed ends up being younger when they're both back on Earth.

faen said:
There is some contradiction here about that special relativity slows time, since it is simply a contradiction that both clocks are slow relative to each other.
It's not a contradiction. I explained this in post #4.

ghwellsjr said:
Now I want to get to the confusion over the Time Dilation factor. Since we use t' and t in the Lorentz Transform for two different frames, we don't want to use both of these symbols in the Time Dilation formula because they are both coordinate times (t' being used just temporarily). The Time Dilation formula applies in just a single frame and allows us to determine the time on a moving clock in that frame so we use Einstein's symbol for Proper Time, the Greek letter tau, τ. Einstein presented this formula as τ=t√(1-v2/c2) but it is more common to see it τ=t/γ. When presented these ways, there is no confusion because τ is always the Proper Time for the moving clock and t is always the Coordinate Time for the stationary clocks.

I was saying that I think the wiki is wrong.

http://en.wikipedia.org/wiki/Time_dilation

I think this calculation should give the proper time, but it doesn't because it misassigned the time variables. It says the time dilation forumla is t'=yt, but I think it should be t'=t/y. If the wikki is correct there then it would in effect be a different theory, unless you are saying that it is also wrong...

The observer in motion doesn't see the light ray to travel a longer distance, it is the opposite. The observer in motion sees the light ray travel straight up and down. They then use their own time to measure this, that would have to be smaller.

If you used the the time dilation equation given there, you would get a larger value for time not a smaller one. That is because it assumes that the observer in motion see's the light to travel a longer distance, so then they would have to experience more time to allow that to happen.

John232 said:
ghwellsjr said:
Now I want to get to the confusion over the Time Dilation factor. Since we use t' and t in the Lorentz Transform for two different frames, we don't want to use both of these symbols in the Time Dilation formula because they are both coordinate times (t' being used just temporarily). The Time Dilation formula applies in just a single frame and allows us to determine the time on a moving clock in that frame so we use Einstein's symbol for Proper Time, the Greek letter tau, τ. Einstein presented this formula as τ=t√(1-v2/c2) but it is more common to see it τ=t/γ. When presented these ways, there is no confusion because τ is always the Proper Time for the moving clock and t is always the Coordinate Time for the stationary clocks.
I was saying that I think the wiki is wrong.
Really? Where did you say anything about the wiki?
John232 said:
http://en.wikipedia.org/wiki/Time_dilation

I think this calculation should give the proper time, but it doesn't because it misassigned the time variables. It says the time dilation forumla is t'=yt, but I think it should be t'=t/y. If the wikki is correct there then it would in effect be a different theory, unless you are saying that it is also wrong...
It is confusing but if you read the whole article, it's very clear that it is the moving clock that is determined by each observer to have its time dilated, that is, seconds are bigger, which means the clock is running slower. So their formula t'=[STRIKE]y[/STRIKE]γt is meant to show that the time it takes for a tick on the moving clock takes longer than a tick on the stationary clock. But we usually use t to refer to the time showing on a ticking clock rather than its inverse which is the time it takes for a tick to occur. (I wish you would use the correct symbol for gamma.)
John232 said:
The observer in motion doesn't see the light ray to travel a longer distance, it is the opposite. The observer in motion sees the light ray travel straight up and down. They then use their own time to measure this, that would have to be smaller.

If you used the the time dilation equation given there, you would get a larger value for time not a smaller one. That is because it assumes that the observer in motion see's the light to travel a longer distance, so then they would have to experience more time to allow that to happen.
Even in these comments of yours, it's confusing. The way I would say it is, ignoring gravity, every observer cannot determine that his own clock is running abnormally but any other clock that is moving with respect to him is running slower than his own. In Special Relativity, no one ever determines that any other clock runs faster than his own, always slower.

But regarding the rest of my post, does it help you to understand the other points of confusion so that you don't feel the need to publish your own theory of Special Relativity?

Fredrik said:
Here the twins are doing exactly the same acceleration, but spend different amounts of time at the higher speed relative to Earth. The one who spent the most time at the higher speed ends up being younger when they're both back on Earth.

The intent of this phrasing is admirable, but can lead to errors if taken too seriously. In fact, I conclude no simple, intuitive statement about which twin ages less can cover all cases, even in SR:

https://www.physicsforums.com/showpost.php?p=3753159&postcount=33

PAllen said:
The intent of this phrasing is admirable, but can lead to errors if taken too seriously. In fact, I conclude no simple, intuitive statement about which twin ages less can cover all cases, even in SR:

https://www.physicsforums.com/showpost.php?p=3753159&postcount=33
I agree that there's no statement similar to the one I made that works for all situations, but it certainly works in this case, where we can say once and for all that "time spent at the higher speed" refers to the description in terms of the inertial coordinate system that's comoving with Earth, there are only two speeds involved (0 and 0.8), and the segments that represent the acceleration phases have identical shapes (in that same coordinate system).

In general, there's no substitute for the statement that the number displayed by a clock is the proper time of the curve that represents the motion the clock has done so far.

PAllen said:
Fredrik said:
Here the twins are doing exactly the same acceleration, but spend different amounts of time at the higher speed relative to Earth. The one who spent the most time at the higher speed ends up being younger when they're both back on Earth.
The intent of this phrasing is admirable, but can lead to errors if taken too seriously. In fact, I conclude no simple, intuitive statement about which twin ages less can cover all cases, even in SR:

https://www.physicsforums.com/showpost.php?p=3753159&postcount=33

ghwellsjr said:

You answers are just restatements of my option #5 - you have to define 'more time at higher speed' in a way that is completely at odds with what the English words mean (as effectively equal to: add up 1/gamma times coordinate time for each segment - possibly via integration - which is my option #5, and does not correspond to the English words). I didn't respond after a certain point because I felt your additional posts added no information or insight to what had already been said.

ghwellsjr said:
[..] But it does help to consider that in the ground's rest frame, the clock on the rocket is ticking slower and it is moving away from the clock on the ground. So between each tick of the rocket clock, as it observes the ground clock, it has moved farther away and so the image of the ground clock's ticks have farther to travel resulting in a longer time compared to its own ticks. It turns out that the ratio of the ground clock's tick interval compared to the rocket clock's tick interval as determined by the rocket is the same as the ratio of the rocket clock's tick interval compared to the ground clock's tick as determined by the ground. Work it out, you'll see that this is true.
Very good George - I've rarely seen such a clear explanation. Thanks I'll try to remember that one!

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PAllen said:
ghwellsjr said:
You answers are just restatements of my option #5 - you have to define 'more time at higher speed' in a way that is completely at odds with what the English words mean (as effectively equal to: add up 1/gamma times coordinate time for each segment - possibly via integration - which is my option #5, and does not correspond to the English words). I didn't respond after a certain point because I felt your additional posts added no information or insight to what had already been said.
You didn't state any options, let alone 5. You did state in post #42:
PAllen said:
This is what I've been saying: pick a frame, accumulate dilated time (=proper time) along different paths. Compare. Result same no matter what frame you use for the analysis. Where I have a problem is attempts to state some other simple rule to predict or explain this - all such that I've ever seen I consider simply wrong.
which is a more succinct version of what I stated in post #16:
ghwellsjr said:
...you can analyze the scenario in any single inertial Frame of Reference and the "time spent at a higher speed" is defined uniquely in that FoR and the "higher speed" is defined uniquely in that FoR and we're talking about the coordinate time of each body in that FoR and we apply Einstein's time dilation formula to convert coordinate time into proper time for each body and then we see how much proper time has accumulated for each body as it travels at different speeds according to the FoR for whatever coordinate time intervals from the time they separated until the time they reunite and we get the amount that each one aged and subtract them and we have the differential aging...
So why is your restatement of my statement acceptable and mine not?

ghwellsjr said:
You didn't state any options, let alone 5. You did state in post #42:

which is a more succinct version of what I stated in post #16:
Originally Posted by ghwellsjr View Post

...you can analyze the scenario in any single inertial Frame of Reference and the "time spent at a higher speed" is defined uniquely in that FoR and the "higher speed" is defined uniquely in that FoR and we're talking about the coordinate time of each body in that FoR and we apply Einstein's time dilation formula to convert coordinate time into proper time for each body and then we see how much proper time has accumulated for each body as it travels at different speeds according to the FoR for whatever coordinate time intervals from the time they separated until the time they reunite and we get the amount that each one aged and subtract them and we have the differential aging...

So why is your restatement of my statement acceptable and mine not?

The restatement is fine, but it bears no resemblance to the ordinary meaning of "more time spent at higher speed". There are any number of ways you can talk about accumulating proper time in a given frame - these are correct - but "time spent at a higher speed" is not a reasonable summary of this unless you remove all normal meaning from the words.

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