What happens in the restframe with lightsource?

[ghwellsjr]

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.

Fredrik's comment:

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.

and my similar comment:

I'm glad you mentioned both bodies experiencing acceleration because we can have another variant of the Twin Paradox in which both of them accelerate exactly the same except that one returns home immediately while the other one continues far away from home before matching the acceleration of his twin and returning home a lot later. This clearly shows that it's not the acceleration that causes the differential aging but rather time spent at the relatively higher speed that causes the differential aging.

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.

 

[PAllen]

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]

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.

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]

"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]

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]

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]

Here's your example in post #24 of Dumb twin paradox question:

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):

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]

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]

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.)

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 occurred in one frame and give you how much time has occurred 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 dilation 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]

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:

...I'm going to use the process I described in post #16:

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:

τ_A = 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:

τ_B1 = 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:

τ_B2 = 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:

τ_B = τ_B1 + τ_B2 = 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]

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]

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]

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]

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.

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 occurred in one frame and give you how much time has occurred 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.

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.

Also if you do short substitution for the time dilation 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]

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]

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.

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).

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 dilation 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 dilation 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]

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.

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.

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?

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?

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.

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 dilation 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 dilation 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 into where you are getting all this from. I have no idea.

 

[harrylin]

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]

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 similar 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.

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.

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 separate 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 dilation 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 dilation 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.

 

[ghwellsjr]

Yes, but a lot of it comes from my own simple proof of relativity that is just similar 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.

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.

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.

The wiki article is not presenting a proof. You seem to be taking exception with it because it is an inadequate proof but you have a fundamental misunderstanding of what it is doing. There can be no proof of the speed of a photon or of the propagation of light as is used in the article. So rather than try to prove how the light moves in the light clock under different circumstances, they use Einstein's (unprovable) postulate that light propagates at c in any Frame of Reference. That's why they talk about viewing the light clock from two different Frames of Reference, first from a Frame in which the light clock is stationary and then from a Frame in which the light clock is moving. This enables them, without any proof, to assert that the light is propagating at c in both cases.

There is nothing wrong with the article but, as I said before, you have to understand what their variables stand for. If you think there is something wrong with the article, please don't mix up your complaint with your own version, just point out where you think it is in error and we can deal with that separately from your own version.

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.

Your simple proof starts out confusing me and I have several problems with it.

First off, I know that you cannot observe a photon or measure its progress so you're going to have to show me how you plan on doing this before I can get motivated to try to understand the rest of your proof.

Secondly, it seems like you have it backwards. Why do you say an observer in motion observes the photon traveling straight out but the observer at rest measures it at an angle? I'm trying to associate your scenarios with the diagrams in the wiki article and maybe that's a mistake. Maybe you are presenting something totally different. So I cannot make sense of your equations that follow. You need to provide your own pictures and relate your equations to them so that I can follow your proof. But remember, you cannot observe the motion of a photon so deal with that before continuing on.

(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 separate 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 dilation 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 dilation 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.

I really have a problem taking seriously anyone who thinks after more than a century of thousands of the best minds in science coming to the same conclusions about MMX and SR, that he has discovered a fundamental mistake and wants to publish his own theory of Special Relativity. I can help you understand Special Relativity but I have no interest in trying to understand your alternative theory, only in discovering where you are making your own errors, and the first one is thinking that you know how a photon propagates.

 

[John232]

It is Newton's equation for distance... I don't really know what more there is to say about it. d=vt. I am saying that you can replace v with the speed of light in order to find the distance the photon has traveled. Both observers would then use their own time to measure this distance. So then if the photon traveled for one secound then you would multiply that times the velocity and get about 300,000 km.

So then say Michealson stays with his experiment and then measures the speed of light traveling in it. All he would have to do is find the amount of time it took to travel across the experiment and the distance across the experiment to find the speed of light. So then c=d/t. Now say Morley pass's the experiment in his car and watches it through the window. He notices that the experiment still measured the photons to travel in a straight line in the experiment to reach the end at the same time. Traveling at a relative velocity doesn't effect the outcome of the experiment. He then draws the path of the photon on his car window as it passes by with a magic marker. The line he just drew was moveing at an angle along with the motion of the car. Morley then concludes that the photon traveled a larger distance in his frame of reference so then he solves for the speed of light. So then he takes this longer distance and the time he measured it to take for it to travel and he gets the same answer for the speed of light. So then light has been seen to travel two different distances and comes out to be the same speed. They then compare their watches to find that the amount of time it took to reach across the experiment was different when Morley was in the car. So then they could both set up a right triangle to give their relation to the time that is experienced in the car and just sitting next to the experiment. But, this could get too confusing because I switched the observer at rest and the observer in motion, so I will stop here. But the point is mainly that the observer in motion wouldn't observe the results of this experiement to come out differently. In Einsteins world the experiment would have been seen to shoot the photon at an arc, but that wasn't the case. An observe in motion can't change the outcome of a Michealson-Morley experiment just by accelerating and looking at it from afar or by accelerating the experiment further. That is one of the main differences of my theory.

 

[harrylin]

[..] That is one of the main differences of my theory.

Sorry if I'm a game breaker here, but that sounds as if you are not playing by the Physicsforums rules. Which theory that is presented in the scientific literature do you want to discuss or ask questions about?

 

[ghwellsjr]

All he would have to do is find the amount of time it took to travel across the experiment...

How does he do that?

 

[John232]

You guys are really no fun at all, I wasn't aware that physics forums was not to include anything about the scientific process. You need to go back to high school and relearn the first property of physics, I fear this is just going to be way over your head...

 

[PAllen]

You guys are really no fun at all, I wasn't aware that physics forums was not to include anything about the scientific process. You need to go back to high school and relearn the first property of physics, I fear this is just going to be way over your head...

Redirect this comment at yourself, where it belongs. ghwellsjr simply asked you how something is measured, and you suggest that is a bad question. It is the most important question in physics - you can' measure it, it isn't physics.

 

[John232]

Redirect this comment at yourself, where it belongs. ghwellsjr simply asked you how something is measured, and you suggest that is a bad question. It is the most important question in physics - you can' measure it, it isn't physics.

The speed of light hasn't been measured now? Do I really need to get into all that? It seems like it would be something for another forum. Why don't you actually do the math of what I am saying and not just assume that it is wrong without even checking into it? I see why now how this has come to be over a hundred years this has been overlooked.

Okay, a pitcher throws a ball it takes 2s to reach the catcher. He threw the ball at 10m/s. How far is the catcher from the pitcher? (2s)(10m/s) = 20 meters The value for secounds cancels and then you are left with only the unit meters...

 

[Passionflower]

I fear this is just going to be way over your head...

It is, but not the way you envision it. :)

 

[ghwellsjr]

The speed of light hasn't been measured now? Do I really need to get into all that? It seems like it would be something for another forum. Why don't you actually do the math of what I am saying and not just assume that it is wrong without even checking into it? I see why now how this has come to be over a hundred years this has been overlooked.

Okay, a pitcher throws a ball it takes 2s to reach the catcher. He threw the ball at 10m/s. How far is the catcher from the pitcher? (2s)(10m/s) = 20 meters The value for secounds cancels and then you are left with only the unit meters...

When a pitcher throws a ball, you can watch it, you can see when he throws it and you can see when the pitcher catches it. But you are using light which is traveling millions of times faster than the ball to see those two events and so your error in timing is negligible and you don't have to factor out the time it takes for the light to travel from those two events to your eye.

But when Michelson's experiment emits a flash of light, how does he see it to know when that happened? When the light pulse hits the other end of the experiment, how does he see it to know when it arrived? He can only watch it with light, correct? So I need for you to tell me how he can factor out the light travel time from those two events so that he can get a meaningful measurement.

 

[John232]

He could move an entanglement experiment next to the Michealson-Morley experiment and then put an atomic clock on the end of it. Then have a switch linked to both experiments so that it is the same distance to each. When the switch turns on it measures the entangled particle and turns on the Michealson-Morley experiment. The atomic clock would then have a detector to see when the photon reaches it, it then turns on when the other entangled particle changes it's spin from being measured on the other side. The atomic clock stops and read time when the photon reaches the detector. That would be a bunch of trouble just to find that a particle follows d=vt. I thought it should be a given. Or he could just shoot a photon at a piece of material that alters when hit and then time how long it takes it to do it, other way could be a lot of trouble. I thought they already have done this and should be a given. Like when they found the wave properties of light. Why does it take over a hundred years for everyone to know that it was measured to always travel at the same speed? Why would particles travel with anything other than their velocity? Is there some kind of particle velocity I have not heard about? Maybe you could try using my equation to find the velocity in respect to time and get the same answer as the velocity you originally put into it and find that the other equation doesn't. If v=v', then gamma will always cancel so then it doesn't matter if you used the equation for dialated time or not they will reduce to the same equations. Sounds like that should be something you should see wrong with accepted equation.

 

[John232]

I found one link to particle velocity but I don't think it applies here, because I said it was in a vacuum with no resistance. Laws of motion don't even take into account resistance anyways...

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

 

[John232]

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

Here is another link, do these equations look familiar?

 

[ghwellsjr]

He could move an entanglement experiment next to the Michealson-Morley experiment and then put an atomic clock on the end of it. Then have a switch linked to both experiments so that it is the same distance to each. When the switch turns on it measures the entangled particle and turns on the Michealson-Morley experiment. The atomic clock would then have a detector to see when the photon reaches it, it then turns on when the other entangled particle changes it's spin from being measured on the other side. The atomic clock stops and read time when the photon reaches the detector. That would be a bunch of trouble just to find that a particle follows d=vt. I thought it should be a given.

We are not trying to measure how long it takes a massive particle to travel some distance, that would be like a baseball where you use light, which travels faster than the massive particle to identify when the particle left one end of the experiment and you use light at the other end to identify when the particle arrives there and so that both light signals can be used to start and stop a timer with only a minimal error caused by the ratio of the speed of light to the speed of the particle being less than infinity.

Or he could just shoot a photon at a piece of material that alters when hit and then time how long it takes it to do it, other way could be a lot of trouble.

How do you time how long it takes for the photon to leave the photon gun and arrive at the target? Let's assume that you have at the gun a very fast electronic circuit that produces a pulse precisely when the photon is fired and you have at the target another very fast electronic circuit that produces a pulse precisely when the photon hits the target. The problem is how do you use these two pulses separated in space to start and stop a timer to make your measurement? That's what I need for you to describe for me.

I thought they already have done this and should be a given. Like when they found the wave properties of light. Why does it take over a hundred years for everyone to know that it was measured to always travel at the same speed? Why would particles travel with anything other than their velocity? Is there some kind of particle velocity I have not heard about?

All measurements of the speed of light involve a round-trip for the light so that the two fast electronic circuits that I described earlier are located at the same place, at the photon gun. So the gun shoots a photon and starts the timer. The photon hits a reflector some measured distance away and a photon returns. This photon hits the second detector and stops the timer. Now we can calculate the "average" speed of light. It turns out that this value is a constant equal to c as long as the experiment is inertial. But we cannot know whether it took the same amount of time for the photon to travel from the gun to the reflector as it took for the photon to travel from the reflector back to the detector colocated with the gun.

Maybe you could try using my equation to find the velocity in respect to time and get the same answer as the velocity you originally put into it and find that the other equation doesn't. If v=v', then gamma will always cancel so then it doesn't matter if you used the equation for dialated time or not they will reduce to the same equations. Sounds like that should be something you should see wrong with accepted equation.

What is your equation that works and what is the other equation that doesn't work?

Gamma was proposed by scientists prior to Einstein as a way to explain how the measurement of light would always yield the same answer even though they thought light was only traveling at c in a fixed ether medium. So if the photon travels at a fixed speed with respect to the ether, that could cause the time for the photon to travel from the gun to the reflector to be shorter than the return trip or vice versa.

 

[ghwellsjr]

I found one link to particle velocity but I don't think it applies here, because I said it was in a vacuum with no resistance. Laws of motion don't even take into account resistance anyways...

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

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

Here is another link, do these equations look familiar?

These two wiki articles are about massive paritlcles and don't apply to the one-way speed of light. You should instead look up and study the wiki article called "One way speed of light".

 

[John232]

Einstein found the two way speed of light just by adding the times it reached both clocks and dividing it by two, in no way does this imply that the speed of light in two directions is different. The Michealson-Morley experiment proves that the two way speed of light is the same as the one way even in different directions. In no way does the Einstein synhronisation imply that velocity is different. I still don't get why you have a problem with this. Pretend your Isaac Newton, you measure a ball to be shot across the room. You find that it always travels a distance vt. Now you check every other distance across the room that it could travel and you find that the distance equals vt in every case. So then you know that d=vt. Now say the photon is the same ball Newton was measuring. You measure the photon to travel the same distance according to d=vt, and then do this at different distances. You find it always travels at a constant speed no matter what distance it traveled, so then you know that d=vt is true. How could you have some other equation where light comes to the same speed and the distance it travles is not inversely related to that velocity? If it traveled at different speeds in different directions then the speed of light wouldn't be constant and Michealson-Morley would have had a whole different story. I think you should look into the experiment into further detail and not mind any mention of aether...

It also says on the two way speed of light that an equation that describes this is the lorentz transformations, well what I did was in effect solving for my own lorentz transformation. But the main difference being that my velocity is different so that the two way speed of light is equal to the one way speed of light. That was actually found by the experiment it comes from. Like I said before v doesn't equal the length contraction equation divided by the time dilation equation you can find this by putting a velcoity in both equations and then solving for length and time then useing that to find the velocity, it doesn't come out to be the same, but in my theory it does. So then my theory it says the two way speed of light is the same.

 

[ghwellsjr]

Einstein found the two way speed of light just by adding the times it reached both clocks and dividing it by two, in no way does this imply that the speed of light in two directions is different.

Can you show me a reputable reference for this?

The Michealson-Morley experiment proves that the two way speed of light is the same as the one way even in different directions.

Can you show me a reputable reference for this?

In no way does the Einstein synhronisation imply that velocity is different. I still don't get why you have a problem with this. Pretend your Isaac Newton, you measure a ball to be shot across the room. You find that it always travels a distance v/t. Now you check every other distance across the room that it could travel and you find that the distance equals v/t in every case. So then you know that d=v/t. Now say the photon is the same ball Newton was measuring. You measure the photon to travel the same distance according to d=vt, and then do this at different distances. You find it always travels at a constant speed no matter what distance it traveled, so then you know that d=vt is true. How could you have some other equation where light comes to the same speed and the distance it travles is not inversely related to that velocity? If it traveled at different speeds in different directions then the speed of light wouldn't be constant and Michealson-Morley would have had a whole different story. I think you should look into the experiment into further detail and not mind any mention of aether...

It also says on the two way speed of light that an equation that describes this is the lorentz transformations, well what I did was in effect solving for my own lorentz transformation. But the main difference being that my velocity is different so that the two way speed of light is equal to the one way speed of light. That was actually found by the experiment it comes from. Like I said before v doesn't equal the length contraction equation divided by the time dilation equation you can find this by putting a velcoity in both equations and then solving for length and time then useing that to find the velocity, it doesn't come out to be the same, but in my theory it does. So then my theory it says the two way speed of light is the same.

Until you show me the two references I asked for, nothing else in your post matters.