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What happens in the restframe with lightsource?

by faen
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PAllen
#37
Mar6-12, 10:13 AM
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Quote Quote by ghwellsjr View Post
Fredrik's comment:

and my similar comment:

were both made in the context of both twins experiencing the same acceleration and is perfectly clear English. The whole point of these statements is to show that acceleration is not what causes the difference in aging. I fail to understand what you think you are offering to help people who think that it is the acceleration that is what makes the difference because they have heard that "it is the one who accelerates that ages less" (which is true if only one accelerates). We're trying to help novices who may not yet even know what Proper Time is to take a small step from a point of misunderstanding to a better understanding and I don't know why you think it is helpful to create a debate in the middle of this attempt.
You're the only one making a debate about it. I added an additional item of information. Frederik had no problem with this and noted:

- within the specific example he posed, there was no ambiguity in his wording

- for the general case, you cannot use such wording (agreeing with me about this).

At this point, there was perfect closure and mutual understanding. You chose to try and turn it into a debate. For some reason, you don't like the idea that there is anything wrong with the statement "the twin that spends more time at higher speed ages less"; I believe, most everyone else understands that for the general case, interpreted as words are normally used, it is wrong. For suitably constrained cases, it is accurate.
ghwellsjr
#38
Mar6-12, 11:08 AM
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Quote Quote by PAllen View Post
You're the only one making a debate about it. I added an additional item of information. Frederik had no problem with this and noted:

- within the specific example he posed, there was no ambiguity in his wording

- for the general case, you cannot use such wording (agreeing with me about this).

At this point, there was perfect closure and mutual understanding. You chose to try and turn it into a debate. For some reason, you don't like the idea that there is anything wrong with the statement "the twin that spends more time at higher speed ages less"; I believe, most everyone else understands that for the general case, interpreted as words are normally used, it is wrong. For suitably constrained cases, it is accurate.
At that point, you had provided a link in which my name was prominently displayed and in which you said, "This is really good example of the inherent pitfalls". You dragged me into the previous debate once again and I'm not going to let it go unchallenged. So my response to your link was to provide the link for the entire thread, not just one post by you taken out of context, and I politely asked you to back up your claim that my simple statement wouldn't work or would lead to errors, but instead of doing that, you continued to rehash the previous debate, leading to the conclusion that my simple statement (as I explained in context) was OK.
Quote Quote by ghwellsjr View Post
Personally, saying "accumulate proper time along the paths" is no different than saying "the time on the clocks as they travel" and doesn't off[er] any additional explanation unless you say how you calculate the time on the clocks based on the speed in a frame.
And that's what I'm indicating when I say "more time at higher speed" and if you still think there is a case in SR where this won't work or leads to errors, please present it, I beg you (once again).
PAllen
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Mar6-12, 11:23 AM
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Quote Quote by ghwellsjr View Post
"Personally, saying "accumulate proper time along the paths" is no different than saying "the time on the clocks as they travel" and doesn't off[er] any additional explanation unless you say how you calculate the time on the clocks based on the speed in a frame."

And that's what I'm indicating when I say "more time at higher speed" and if you still think there is a case in SR where this won't work or leads to errors, please present it, I beg you (once again).
Accumulate proper time and "time on clocks as they travel" are both perfectly good, along with the formula (1/gamma * coordinate time) to compute them. "More time at higher speed" is not the same, and I have given several examples in the other thread where "more time at higher speed" leads to an incorrect conclusion: the twin that spent more time at higher speed (in the English language meaning of the words) aged more rather than less. Your only response is that you use the words differently than their normal English meaning. Doing so is guaranteed to lead to confusion.
ghwellsjr
#40
Mar6-12, 11:59 AM
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Quote Quote by PAllen View Post
Accumulate proper time and "time on clocks as they travel" are both perfectly good, along with the formula (1/gamma * coordinate time) to compute them. "More time at higher speed" is not the same, and I have given several examples in the other thread where "more time at higher speed" leads to an incorrect conclusion: the twin that spent more time at higher speed (in the English language meaning of the words) aged more rather than less. Your only response is that you use the words differently than their normal English meaning. Doing so is guaranteed to lead to confusion.
OK, now I see what you are saying. I should have said "more coordinate time at higher speed". Yes, the way I said it does sound rather contradictory--like saying the older I get, the younger I am. I will take your advice in the future. Thanks for hanging in there and finally getting through to me.
PAllen
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Mar6-12, 12:21 PM
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Quote Quote by ghwellsjr View Post
OK, now I see what you are saying. I should have said "more coordinate time at higher speed". Yes, the way I said it does sound rather contradictory--like saying the older I get, the younger I am. I will take your advice in the future. Thanks for hanging in there and finally getting through to me.
Well, actually, the issue isn't so much coordinate time as 'higher speed'. My example in #24 (I think) in that other thread has the twin with more coordinate time at higher speed aging more, not less. The issue is that (in the general case) you cannot say anything better than "accumulate coordinate time * 1/gamma", or some equivalent formulation; the one with smaller amount of this ages less. I do not accept that " less accumulated coordinate time * 1/gamma" has the same meaning in English as "more coordinate time at higher speed". The former is true, the latter is false, in general, though true in suitably constrained cases.
ghwellsjr
#42
Mar6-12, 01:09 PM
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Here's your example in post #24 of Dumb twin paradox question:
Quote Quote by PAllen View Post
Consider a twin variant where neither twin is ever moving inertially, but they separate and come back together with different ages.
...
Consider that, while neither twin is ever inertial (due to continuous changes in direction), twin A is always moving at speed .4c in this chosen inertial frame. Suppose twin B is moving .1c for 80% of the coordinate time between separate and meet up, and at .99999c for 20% of the coordinate time.
And here is your preferred method for analyzing the accumulated proper times of the two twins (post #31 of the same thread):
Quote Quote by PAllen View Post
In my mind, "accumulate proper time" is associated with a notion of line element, and this one notion applies with full generality to SR or GR; whether the line element is:

d tau^2 = d t^2 - (dx^2 + dy^2 + d z^2)/ c^2 [which trivially gives (1/gamma) dt by rearrangement]

or the more general GR metrics, you have one method, one concept.
Please show us how you use this one method to determine the accumulated proper times of the two twins in your example.
PAllen
#43
Mar6-12, 01:34 PM
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Quote Quote by ghwellsjr View Post
Here's your example in post #24 of Dumb twin paradox question:

And here is your preferred method for analyzing the accumulated proper times of the two twins (post #31 of the same thread):

Please show us how you use this one method to determine the accumulated proper times of the two twins in your example.
Start with:

d tau^2 = d t^2 - (dx^2 + dy^2 + d z^2)/ c^2

From this:

d tau/ dt = sqrt (1 - ((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)/c^2)

d tau/dt = sqrt (1 - v^2 /c ^2) = 1/gamma

Then, for any segment of constant v, compute sqrt(1-v^/c^2) times coordinate time of that segment, and add them up (no need to actually integrate if v is constant).

So, in the given example we have, we have let's make total coordinate time be 10. Then for twin A we have:

10 * sqrt (1- .4^2) = 9.165 appx.

For twin B we have:

8 * sqrt( 1-.1^1) + 2 * sqrt(1 - .99999^2) = 7.9688 appx

Thus twin A ages more even though they were going 4 times faster 80% of the time, and only about 2.5 times slower 20% of the time.

What does any of this have to do with the point under discussion except to establish the obvious fact that the line element contains all information needed to compute proper times?
John232
#44
Mar6-12, 01:48 PM
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Quote Quote by ghwellsjr View Post
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'=yγt 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.)
I didn't say anything about the wiki before, but I thought it would be easier to discuss if we sticked to the same proof/equation. If you looked at it like any other geometry problem then t would be the amount of time it takes for v or c to travel a certain distance. I think you should be able to solve for c in the proof by only using t and the length of the sides. It turns out that if you do that in the current settup you get the wrong value for c. I don't agree that a larger value for t should imply that time is going slower. I think a larger value for t would mean that there are more ticks on a clock, for more ticks to happen time would have to go faster. The distance the object traveled is ticks of the clock times velocity.

The equation finds the relation between the times of each side, it doesn't consider how many times the photon goes to the top and bottom of a clock. So then the answer should give you a direct translation of how much time has occured in one frame and give you how much time has occured in another frame, since c is the same on two sides.

I am sorry you have not concinced me yet, because I have gotten really good at algebra. They made me retake it twice from switching colleges and then I took it in high school and junior high 3 times. I got an A in every course, if they hadn't have done that we may not have had this problem... The theory here just doesn't seem to flow like older well done proofs. I apoligize for not being that forum savy.

Also if you do short substitution for the time dialation equation you don't get the length contraction equation. If you substitute t' in the equation L'=vt', then say L=ct you get the wrong relation between the two equations to gamma, and v≠v'. If a different value for v is found then put back into the equation it doesn't give the same value's for L and t. But, both observers should agree on the relative velocity.
ghwellsjr
#45
Mar6-12, 03:40 PM
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Quote Quote by PAllen View Post
Thus twin A ages more even though they were going 4 times faster 80% of the time, and only about 2.5 times slower 20% of the time.

What does any of this have to do with the point under discussion except to establish the obvious fact that the line element contains all information needed to compute proper times?
So I guess when I say "time spent at a higher speed" and when you include three different speeds, that entitles you to pick which two speeds to use, ignoring the effect from the third speed.

Here's how I analyzed your scenario:
Quote Quote by ghwellsjr View Post
...I'm going to use the process I described in post #16:
Quote Quote by ghwellsjr View Post
But this thread and all the discussion up to this point has been about the Twin Paradox where they start out together, separate, and come back together and I'm saying that if you agree to ignore gravity, then 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 and no time disappeared or needs to be accounted for.
I know that's a mouthful but it's really very simple to analyze using Einstein's formula to get the proper time interval, τ, (tau, the time interval on a clock) as a function of its speed, β, (beta, the speed as a fraction of the speed of light), and the coordinate time interval, t, as specified in the Frame of Reference:

τ = t√(1-β2)

First we analyze Twin A who travels at 0.4c for 100% of the time:

τA = 100%√(1-0.42) = 100%√(1-0.16) = 100%√(0.84) = 100%(0.9165) = 91.65%

Now we analyze the first part of Twin B's trip at 0.1c for 80% of the time:

τB1 = 80%√(1-0.12) = 80%√(1-0.01) = 80%√(0.99) = 80%(0.995) = 79.6%

And the last part of Twin B's trip at 0.99999c for 20% of the time:

τB2 = 20%√(1-0.999992) = 20%√(1-0.99998) = 20%√(0.00002) = 20%(0.00447) = 0.09%

Finally we add the two parts of Twin B's trip to get the total time:

τB = τB1 + τB2 = 79.6% + 0.09% = 79.69%

So we see that Twin B with 79.69% of the coordinate time of the scenario ages less than Twin A with 91.65% of the coordinate time.
Notice how I included all three speeds and got the correct answer. I didn't ever say or imply that you could ignore the effect of one of those speeds or make some erroneous attempt to "average" two of those speeds like you did. If you follow the precise procedure I outlined earlier in post #16, there won't be any errors.

Remember, the whole point of this discussion is to counter the idea that acceleration alone is what results in an aging difference and my only point is although accleration causes a change in speed, the twin has to accumulate time at that speed in order to effect his aging, the more (coordinate) time and the more (coordinate) speed, the lower his aging during that interval.

I will attempt to be more precise in my future descriptions so as to not require you to correct me, OK? I will either stipulate that my statement applies only for those Frames of Reference where two speeds exist, one for each twin, but when analyzed from a different Frame, there will be more than two speeds and you have to calculate the partial aging for each speed segment separately, or for twins that accelerate more than the minimal number of times, you also have to calculate the partial aging for each speed segment separately, and for twins that accelerate over a long period of time instead of instantaneously, you have to actually integrate the acceleration to get a speed profile and do really complicated computations. OK?
John232
#46
Mar6-12, 04:38 PM
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Why don't they just start out teaching relativity using (τ) if they are not going to derive the equation anyways? But, tau doesn't give the same answer as 1/t'. How would you then convert t' into the proper time?
PAllen
#47
Mar6-12, 05:33 PM
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Quote Quote by ghwellsjr View Post
Remember, the whole point of this discussion is to counter the idea that acceleration alone is what results in an aging difference and my only point is although acceleration causes a change in speed, the twin has to accumulate time at that speed in order to effect his aging, the more (coordinate) time and the more (coordinate) speed, the lower his aging during that interval.
For me, this was never the point, because I never advocated anything resembling it. I have advocated the following point of view, I believe very consistently (sticking wholly to SR with normal topology for this discussion):

- Acceleration by at least one twin is needed so the twins can get back together. Thus, acceleration somewhere is a necessary (but not sufficient) condition.

- The acceleration isn't the cause of differential aging, nor can the age difference be localized to the acceleration or any other part of the path, in any objective way.

- This formulation of yours: "my only point is although acceleration causes a change in speed, the twin has to accumulate time at that speed in order to effect his aging, the more (coordinate) time and the more (coordinate) speed, the lower his aging during that interval." is fine, I would never dispute it. It is precise and accurate in noting you have to consider segment by segment in some (any) frame.

- I would add that which segments of a journey are associated with 'slower aging' is frame dependent, and not objectively meaningful. But any frame will come up with the same total for a journey.
ghwellsjr
#48
Mar6-12, 06:31 PM
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Quote Quote by PAllen View Post
For me, this was never the point, because I never advocated anything resembling it. I have advocated the following point of view, I believe very consistently (sticking wholly to SR with normal topology for this discussion):

- Acceleration by at least one twin is needed so the twins can get back together. Thus, acceleration somewhere is a necessary (but not sufficient) condition.

- The acceleration isn't the cause of differential aging, nor can the age difference be localized to the acceleration or any other part of the path, in any objective way.

- This formulation of yours: "my only point is although acceleration causes a change in speed, the twin has to accumulate time at that speed in order to effect his aging, the more (coordinate) time and the more (coordinate) speed, the lower his aging during that interval." is fine, I would never dispute it. It is precise and accurate in noting you have to consider segment by segment in some (any) frame.

- I would add that which segments of a journey are associated with 'slower aging' is frame dependent, and not objectively meaningful. But any frame will come up with the same total for a journey.
Excellent--I'm in total agreement. (I better be, it's correct.)
ghwellsjr
#49
Mar6-12, 10:40 PM
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Quote Quote by John232 View Post
I didn't say anything about the wiki before, but I thought it would be easier to discuss if we sticked to the same proof/equation. If you looked at it like any other geometry problem then t would be the amount of time it takes for v or c to travel a certain distance. I think you should be able to solve for c in the proof by only using t and the length of the sides. It turns out that if you do that in the current settup you get the wrong value for c. I don't agree that a larger value for t should imply that time is going slower. I think a larger value for t would mean that there are more ticks on a clock, for more ticks to happen time would have to go faster. The distance the object traveled is ticks of the clock times velocity.
Just to make sure we're on the same page here, you are referring to the wikipedia article on Time Dilation in the section called "Simple inference of time dilation due to relative velocity", correct?

OK, well in the picture of the stationary light clock, they say that the period of the clock is Δt and they calculate it based on the speed of light and the distance between the mirrors. So I agree that t is the amount of time it takes for light, traveling at c, to travel a certain distance. But I don't see why you bring up solving for c--isn't c a known constant and you're solving for Δt, the time between two ticks? So I don't know why you think this leads to a wrong value for c. I have no idea what your concern is.

In any case, the only way you would get a larger value for Δt (why are you dropping the delta?) is if the distance between the mirrors is greater. And if that is the case, the light clock is taking longer between ticks and therefore ticking at a slower rate.

And then you talk about the distance the object traveled, but there is no object traveling, just the light is traveling between the two stationary mirrors.
Quote Quote by John232 View Post
The equation finds the relation between the times of each side, it doesn't consider how many times the photon goes to the top and bottom of a clock. So then the answer should give you a direct translation of how much time has occured in one frame and give you how much time has occured in another frame, since c is the same on two sides.
Now for the next figure, they are considering what happens with the stationary clock in the frame of a moving observer and here they switch to the nomenclature of Δt' to distinguish it from the previous Δt in the stationary frame. In this case, the mirrors are moving and the light has to travel a longer distance and so the period of the clock will take a longer time. So you are right, it doesn't consider how many times the photon goes to the top and the bottom because they are calculating it for just one time up and down.

But you shouldn't read into this explanation any more than just a simple declaration of the fact that a moving light clock will take longer between ticks than a stationary one. They are not addressing any issue with regard to Lorentz Transformation and you shouldn't associate the Δt and Δt' nomenclature with similar nomenclature in the Lorentz Transform equations or even in any time dilation equation.
Quote Quote by John232 View Post
I am sorry you have not concinced me yet, because I have gotten really good at algebra. They made me retake it twice from switching colleges and then I took it in high school and junior high 3 times. I got an A in every course, if they hadn't have done that we may not have had this problem... The theory here just doesn't seem to flow like older well done proofs. I apoligize for not being that forum savy.
Certainly since you have mastered algebra, you know that every problem can reuse the same nomenclature over and over again with different meanings so this shouldn't be too hard for you to grasp.
Quote Quote by John232 View Post
Also if you do short substitution for the time dialation equation you don't get the length contraction equation. If you substitute t' in the equation L'=vt', then say L=ct you get the wrong relation between the two equations to gamma, and v≠v'. If a different value for v is found then put back into the equation it doesn't give the same value's for L and t. But, both observers should agree on the relative velocity.
You've lost me here, there's nothing in the wiki article about length contraction and I don't know where you got these equations from or what they are supposed to mean. If you want me to comment on them, you're going to have to tell me where you got them from.
ghwellsjr
#50
Mar7-12, 12:30 AM
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Quote Quote by John232 View Post
Why don't they just start out teaching relativity using (τ) if they are not going to derive the equation anyways? But, tau doesn't give the same answer as 1/t'. How would you then convert t' into the proper time?
Well, Einstein started out teaching relativity using τ but he also derived the equation. The wikipedia article on Proper Time does say that Δτ=ΔT√(1-v2/c2) so I would say they also are teaching relativity using τ.

But τ is equal 1/t' if you use the formula for Δt' in the wiki article on Time Dilation and you use the formula for Δτ in the wiki article on Time Dilation (and you equate Δt to ΔT).

However, the main point is that the two articles are talking about two different times that are the reciprocal of each other.

But your last question about converting t' into Proper Time is nebulous because as I pointed out in post #25, the term t' is defined in different ways in different contexts. I tried to clear this up for you in that long post but maybe I misunderstood what your concern was. You just have to recognize that each context can have a different meaning for the same nomenclature.
John232
#51
Mar7-12, 02:39 AM
P: 249
Quote Quote by ghwellsjr View Post
Just to make sure we're on the same page here, you are referring to the wikipedia article on Time Dilation in the section called "Simple inference of time dilation due to relative velocity", correct?.
Correct.

Quote Quote by ghwellsjr View Post
OK, well in the picture of the stationary light clock, they say that the period of the clock is Δt and they calculate it based on the speed of light and the distance between the mirrors. So I agree that t is the amount of time it takes for light, traveling at c, to travel a certain distance. But I don't see why you bring up solving for c--isn't c a known constant and you're solving for Δt, the time between two ticks? So I don't know why you think this leads to a wrong value for c. I have no idea what your concern is.
Okay, it says the observer in motion observers the photon to travel a distance cΔt'. I am saying that it doesn't. The observer at rest measures the hypotenus to be cΔt. The length of they hypotenus divided by the change in time would give the speed of light for the observer at rest and not the observer in motion. The observer in motion would measure the vertical distance divided by time prime to be the speed of light, not the other way around. I think the error comes in when you say that the observer at rest see's the photon to travel a vertical distance, the observer in motion also see's this as well, but is being compared to what an observer at rest see's(the hypotenus).

Quote Quote by ghwellsjr View Post
In any case, the only way you would get a larger value for Δt (why are you dropping the delta?) is if the distance between the mirrors is greater. And if that is the case, the light clock is taking longer between ticks and therefore ticking at a slower rate.
I am saying that the light clock doesn't measure ticks, but in fact that time is actually warped by the same amount as the relation between the sides of two light triangles. Where each side represents how the speed of light is measured from each frame of reference. So by putting cΔt' as the vertical distance it explains how time has to be adjusted to allow for the observer in motion to measure c even though he see's the photon travel a shorter distance. So then he has to experience less time to fill in this shorter side of the triangle. The observer at rest has to measure more time in order to account for the greater distance the photon is seen to travel. That is how it is possible to get the proper time out of the light clock example simply by assigning the time variables differently.

Also, Δt√(1-v^2/c^2) ≠ √(1-v^2/c^2)/Δt

v ≠ ΔL√(1-v^2/c^2)/(Δt/√(1-v^2/c^2)

v ≠ ΔL√(1-v^2/c^2)/(√(1-v^2/c^2)/Δt)

v = ΔL/τ the gamma cancels,

(v ≠ ΔL/Δt) So since this does not equal the velocity put into the equation, if you found a velocity with dialated spacetime using ΔL and Δt, the theory breaks down. You could then take that new velocity and then find new values for ΔL and Δt, and then find another new velocity and so on etc. This is why I think the time dialation and the length contraction equation should both be directly porportional to gamma.

Δt'=Δt√(1-v^2/c^2) solving for the time variables reversed

ΔL'=cΔt' I assume the distance traveled by the photon is a different value
since the speed of light is constant

ΔL'=cΔt√(1-v^2/c^2) substitute Δt' from the first equation

ΔL'=ΔL√(1-v^2/c^2) ΔL=cΔt, this also works for the direction of motion
ΔL'=vΔt' if you assume v=v'

Solving for length in the vertical dimension doesn't assume space contraction in that direction since both observers would agree that the photon reached the same position after being measured. Neither one says the photon traveled a greater vertical distance and is a requirement of this calculation to form the right triangle.

I then found an equation that describes time dialation for an object under constant acceleration. It assumes that because of the Michelson-Morely experiment that an object under acceleration still measures the photon to travel in a straight line since the Earth was accelerating and the effects of gravity where balanced out.

Δt'=Δt√(1-(vi+vo)^2/4c^2) I still need to look into how this differs from
constant motion
ghwellsjr
#52
Mar7-12, 05:23 AM
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Are you looking at the current English version of the wikipedia.org article on Time Dilation? I cannot see much of what you say you are seeing.
Quote Quote by John232 View Post
Okay, it says the observer in motion observers the photon to travel a distance cΔt'.
I don't see any mention of the word "photon" in the article and I don't see any mention of the expression cΔt'. Tell me where you are getting this from.
Quote Quote by John232 View Post
I am saying that it doesn't. The observer at rest measures the hypotenus to be cΔt.
The observer at rest does not see a hypotenuse. Where are you getting this from?
Quote Quote by John232 View Post
The length of they hypotenus divided by the change in time would give the speed of light for the observer at rest and not the observer in motion.
The section starts off by saying that the speed of light is constant in all reference frames. It doesn't say anything about calculating the speed of light. Where are you getting this from?
Quote Quote by John232 View Post
The observer in motion would measure the vertical distance divided by time prime to be the speed of light, not the other way around. I think the error comes in when you say that the observer at rest see's the photon to travel a vertical distance, the observer in motion also see's this as well, but is being compared to what an observer at rest see's(the hypotenus).
You have to be looking at something different than what I'm seeing. None of what you are talking about is in the article that I'm looking at.
Quote Quote by John232 View Post
I am saying that the light clock doesn't measure ticks, but in fact that time is actually warped by the same amount as the relation between the sides of two light triangles. Where each side represents how the speed of light is measured from each frame of reference. So by putting cΔt' as the vertical distance it explains how time has to be adjusted to allow for the observer in motion to measure c even though he see's the photon travel a shorter distance. So then he has to experience less time to fill in this shorter side of the triangle. The observer at rest has to measure more time in order to account for the greater distance the photon is seen to travel. That is how it is possible to get the proper time out of the light clock example simply by assigning the time variables differently.

Also, Δt√(1-v^2/c^2) ≠ √(1-v^2/c^2)/Δt

v ≠ ΔL√(1-v^2/c^2)/(Δt/√(1-v^2/c^2)

v ≠ ΔL√(1-v^2/c^2)/(√(1-v^2/c^2)/Δt)

v = ΔL/τ the gamma cancels,

(v ≠ ΔL/Δt) So since this does not equal the velocity put into the equation, if you found a velocity with dialated spacetime using ΔL and Δt, the theory breaks down. You could then take that new velocity and then find new values for ΔL and Δt, and then find another new velocity and so on etc. This is why I think the time dialation and the length contraction equation should both be directly porportional to gamma.

Δt'=Δt√(1-v^2/c^2) solving for the time variables reversed

ΔL'=cΔt' I assume the distance traveled by the photon is a different value
since the speed of light is constant

ΔL'=cΔt√(1-v^2/c^2) substitute Δt' from the first equation

ΔL'=ΔL√(1-v^2/c^2) ΔL=cΔt, this also works for the direction of motion
ΔL'=vΔt' if you assume v=v'

Solving for length in the vertical dimension doesn't assume space contraction in that direction since both observers would agree that the photon reached the same position after being measured. Neither one says the photon traveled a greater vertical distance and is a requirement of this calculation to form the right triangle.

I then found an equation that describes time dialation for an object under constant acceleration. It assumes that because of the Michelson-Morely experiment that an object under acceleration still measures the photon to travel in a straight line since the Earth was accelerating and the effects of gravity where balanced out.

Δt'=Δt√(1-(vi+vo)^2/4c^2) I still need to look into how this differs from
constant motion
You're going to have to clue me in to where you are getting all this from. I have no idea.
harrylin
#53
Mar7-12, 05:53 AM
P: 3,184
Quote Quote by ghwellsjr View Post
So I guess when I say "time spent at a higher speed" and when you include three different speeds, that entitles you to pick which two speeds to use, ignoring the effect from the third speed.
[..]
Notice how I included all three speeds and got the correct answer. I didn't ever say or imply that you could ignore the effect of one of those speeds or make some erroneous attempt to "average" two of those speeds like you did. If you follow the precise procedure I outlined earlier in post #16, there won't be any errors.[..]
Just a little support from me: I find it unfair to hang someone up on a sound bite that refers to a certain problem and is meant to clarify the essence of a mathematical analysis. For example "the speed of light is constant" is of course wrong as general statement but I bet that we all use that sound bite now and then correctly with a certain meaning in a certain context.
John232
#54
Mar7-12, 05:37 PM
P: 249
Quote Quote by ghwellsjr View Post
Are you looking at the current English version of the wikipedia.org article on Time Dilation? I cannot see much of what you say you are seeing.

I don't see any mention of the word "photon" in the article and I don't see any mention of the expression cΔt'. Tell me where you are getting this from.
Yes, but a lot of it comes from my own simple proof of relativity that is just similiar to the page described. In my own proof I say that the distance traveled by the photon is cΔt. Then the length in a direction that I just used as L not considering direction, so that L=cΔt. This equation suggest that you can take L/Δt and then calculate the speed of light.

Quote Quote by ghwellsjr View Post
The observer at rest does not see a hypotenuse. Where are you getting this from?.
If a photon is sent from a moving object it does in the light clock example, since the object in motion observers the photon to travel straight out perpendicular to its direction of motion. So then a observer at rest relative to this object sending the photon see's it to travel straight out along with the object in motion.
Quote Quote by ghwellsjr View Post
The section starts off by saying that the speed of light is constant in all reference frames. It doesn't say anything about calculating the speed of light. Where are you getting this from?
That is a problem I saw with the proof on the page. In making my own proof I found that the distances of the sides of the triangle on the page where not the correct size. Then I realized that it doesn't account for light being a constant speed in each frame of reference, because the variables where not correctly assigined to do so. So what I did was instead of counting the number of times the light clock ticks, I found the relation between two distances that actually measure the value c to come out correctly.


My simple proof.

An observer in motion observers a photon to travel straight out in a line a distance cΔt'. An observer at rest measures the photon to travel out at an angle along with the objects direction of motion cΔt. The observer at rest also measures the object to travel a distance vΔt. This gives three sides of a right triangle and then the relation to how they measure Δt can be found.

(ct')^2+(vΔt)^2=(cΔt)^2 a^2+b^2=c^2

c^2Δt'^2=c^2Δt^2-v^2Δt^2 distribute the square and subtract both sides by (v^2Δt^2)

Δt'^2=(c^2Δt^2-v^2Δt^2)/c^2 divide both sides by c^2

Δt'^2=c^2Δt^2(1-v^2/c^2)/c^2 factor out a (c^2Δt^2) from (c^2Δt^2-v^2t^2)

Δt'=Δt√(1-v^2/c^2) the c^2 cancels and take the square root of both sides


It then follows that the Michealson-Morley experiment also measured the photon to travel out in a straight line even though the experiment was under constant acceleration due to the Earths movement. So then, the distance traveled by the object in constant acceleration can be replaced with Δt(vi+vo)/2.

s=Δtvo+(aΔt^2)/2

s=Δtvo+(Δt^2(vi-vo)/t)/2 substitute from a=(vi-vo)/Δt

s=Δtvo+(Δtvi-Δtvo)/2 cancel a Δt from the substitution and distribute the Δt

s=2Δtvo/2+Δtvi/2-Δtvo/2 multiply Δtvo by 2/2 and seperate the factor

s=Δt(vi+vo)/2 add like terms and factor out Δt

(cΔt')^2+(Δt(vi+vo)/2)^2 = (cΔt)^2 Pythagorean Theorem

c^2Δt'^2 = c^2Δt^2 - Δt^2(vi+vo)^2/4 distribute the square

c^2Δt'^2 = c^2Δt^2(1-(vi+vo)^2/4c^2) factor out c^2Δt^2 from the right side

Δt'^2 = Δt^2(1-(vi+vo)^2/4c^2) divide both sides by c^2

Δt'=Δt√(1-(vi+vo)^2/4c^2) take the square root


Then you have an equation for the relation between time dialation and constant acceleration. I noticed that by adding the initial velocity and the final velocity that you can get double that amount of velocity than just the time dialation equation alone. Then when it is squared it comes to be counteracted by the 4 in the denominator. This is because 2 in the acceleration equation is squared to give 4. So you have 4c^2 instead of just c^2, that seems to work out. But my proof seems to imply that as a velocity becomes c then the amount of time that is dialated comes to zero instead of being undefined.


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