The Twin Paradox and the Equivalence Principle

[Simon Bridge]

No nononono; it is you who has not understood what you are being told.
You are on your rocket ship and so it the captain and the navigator ... so you all agree about what age she is "right now" just like you all agree on what the ship's time is. YOu just have to decide what you mean by "right now".

If you didn't know how to work out her age yourself, then you maybe ask them to do it for you right? Then they would be justified in asking which coordinate system you wanted the calculation done for: the ship's, her's, whatever? Each are equally "valid" certainly for the purposes of "imagining what she's doing right now".

Usually - to work what you are doing "right now" I just look at you ... if you are a long way-away I may want to factor in the time it takes light to travel to me. That the sort of thing you mean? So you imagine looking at her with a telescope and you see that she is 20yo but you are 3ly away so she must be 23? We can ask: how old are you when you made the observation?

 

[harrylin]

If the rocket driver told me that she didn't have some age right then, I wouldn't believe him. I wouldn't believe anybody who told me that. [..] If I was on a rocket trip, every time I thought about my sister back home, I would know that she was somewhere right then, and I would know that she was doing something right then. [..] She has some age right then, and it's not just some choice or pick that people on that rocket can make. I might not know how to figure out what her age is, but she has some particular age right then, and everyone on that rocket should figure the same age for her, I'm sure of that.

Yes of course, if everyone on that rocket uses the same assumptions and reference (=the same reference frame), then indeed everyone on that rocket should figure the same age for her. And if they use other assumptions, then they will figure a different age for her; they will not claim that she doesn't have some age right then.

Elaborating on what Simon said:
They decide (by convention) what her age is based on their assumption about how long it takes for light from her to reach them. For example they receive a message from her in which she says that now she's 30 years old. Assuming that the speed of light is 3E8 m/s relative to the rocket, they calculate that the radio signal took 1 year to reach them so that "now" she's 31 years old according to their reckoning. But according to her reckoning, if they are moving way from Earth it takes for example 2 years for her call to reach them so that according to her they receive her call when she's 32 years old.

 

[Dale]

But the fact that there are a variety of ways to construct a noninertial coordinate system means to me that none of them is really telling things "from the point of view of the accelerated twin".

The term "point of view" isn't rigorously defined either. Pretty much any coordinate system can be taken as the point of view of some object if you define it that way. It is simply a matter of definition.

 

[Underwood]

Usually - to work what you are doing "right now" I just look at you ... if you are a long way-away I may want to factor in the time it takes light to travel to me.

That makes a lot of sense to me. I think that's the answer I've been looking for. Any time on that rocket that I'm thinking about my sister, I just look at her on a TV screen, and I can see how old she was when that TV picture started toward me. And she can even be telling me on the TV exactly how old she was then. But that's not her age right now. But if I can figure out how much older she got while that TV picture was getting to me, I can just add that amount and then I'll know how old she is right now. That's the answer! Now, I just have to figure out how much older she got while the TV picture was getting to me. Anyone know how to do that?

... I may want to factor in the time it takes light to travel to me.

Why do you say "may want"? Seems to me that you'd always have to do that. Otherwise you could be very wrong about her age now. She does have some particular age, I'm sure of that. You just have to figure out what it is, and the above way gets it right. That's the answer.

 

[harrylin]

That makes a lot of sense to me. [..] if I can figure out how much older she got while that TV picture was getting to me, I can just add that amount and then I'll know how old she is right now. That's the answer! Now, I just have to figure out how much older she got while the TV picture was getting to me. Anyone know how to do that?

I already told you how to do that in my answer.

 

[stevendaryl]

The term "point of view" isn't rigorously defined either.

I know, it's an informal notion. But I think there is a sense in which, for an inertial observer, the description using an inertial coordinate system in which that observer is at rest can be thought of as describing things from his point of view. For an observer with constant proper acceleration, describing things using Rindler coordinates can be thought as describing things from the point of view of the accelerated observer. But for complicated motions of the observer, there really is no sensible notion of describing things from the point of view of that observer, except locally.

 

[Dale]

I know, it's an informal notion. But I think there is a sense in which, for an inertial observer, the description using an inertial coordinate system in which that observer is at rest can be thought of as describing things from his point of view. For an observer with constant proper acceleration, describing things using Rindler coordinates can be thought as describing things from the point of view of the accelerated observer. But for complicated motions of the observer, there really is no sensible notion of describing things from the point of view of that observer, except locally.

Personally, I think that, if Einstein's simultaneity convention is taken as the "point of view" of an inertial observer, then Dolby and Gull's convention should be taken as the "point of view" of a non-inertial observer. It works for arbitrarily complicated motions, and it is the logical extension of Einstein's convention. It reduces to Rindler coordinates for a constantly accelerating observer and to Minkowski coordinates for an inertial observer.

http://arxiv.org/abs/gr-qc/0104077

 

[Simon Bridge]

That makes a lot of sense to me. I think that's the answer I've been looking for. Any time on that rocket that I'm thinking about my sister, I just look at her on a TV screen, and I can see how old she was when that TV picture started toward me. And she can even be telling me on the TV exactly how old she was then. But that's not her age right now. But if I can figure out how much older she got while that TV picture was getting to me, I can just add that amount and then I'll know how old she is right now. That's the answer! Now, I just have to figure out how much older she got while the TV picture was getting to me. Anyone know how to do that?

I see - you have already been told of course ... your problem is that you need to define your terms.

You can apply time dilation to work out how old she sees herself to be when you are viewing the image of her and then how old she measures you to be when you are viewing the image and so on back and forth and get very confused if you like.

You need to be careful with your terminology and reasoning in relativity - which is what everyone is trying to impress upon you.

Why do you say "may want"? Seems to me that you'd always have to do that. Otherwise you could be very wrong about her age now. She does have some particular age, I'm sure of that. You just have to figure out what it is, and the above way gets it right. That's the answer.

I mean that not everyone thinks that way... I mean you don't normally take account of the time it take light to get to you right? Some people would define "right now" to be whatever they just experienced.

 

[ghwellsjr]

That makes a lot of sense to me. I think that's the answer I've been looking for. Any time on that rocket that I'm thinking about my sister, I just look at her on a TV screen, and I can see how old she was when that TV picture started toward me. And she can even be telling me on the TV exactly how old she was then. But that's not her age right now. But if I can figure out how much older she got while that TV picture was getting to me, I can just add that amount and then I'll know how old she is right now. That's the answer! Now, I just have to figure out how much older she got while the TV picture was getting to me. Anyone know how to do that?

Let's say that your sister has several brothers and they all take a trip in the same direction in separate spaceships but they leave at different times and travel at different speeds. The first one to leave travels at the slowest speed and each one thereafter travels at a faster speed. There are several years between the first one leaving and the last one leaving.

Then your sister sends her video and it just so happens that all the brothers are at the same distant location when they receive the video. If they each use their own reference frame and the time on their own clock since they left your sister, they will each determine that she is a different age. They will all be different ages too.

The only way that they could agree on her age is if they use your sister's reference frame but if they do that, they don't need a video or any signal from your sister, they just have to assume that she stays put and doesn't do any space traveling herself.

 

[Underwood]

I mean that not everyone thinks that way... I mean you don't normally take account of the time it take light to get to you right? Some people would define "right now" to be whatever they just experienced.

We all know that the TV picture doesn't immediately get to the rocket. So it's an old TV picture, and it doesn't tell me what my sister's age really is when I'm looking at that TV picture. She had to get older while the TV picture was getting to me. I don't know yet how to figure out what that extra time was, but it was some particular time. It's not some extra time that I have a choice about. It is what it is.

 

[Underwood]

Let's say that your sister has several brothers and they all take a trip in the same direction in separate spaceships but they leave at different times and travel at different speeds.

I understand that. My brothers won't agree about her age. I'm just saying that everyone on my rocket has to get the same answer for her age, and they don't get to choose anything.

 

[Underwood]

I already told you how to do that in my answer.

I wasn't able to do it after I read what you wrote. Can you show me how to figure out how much older she gets while the TV pictures are getting to me, and then use that to tell me how much older she gets while I'm turning around? That's what I really want to be able to do, and I can't yet.

 

[harrylin]

I understand that. My brothers won't agree about her age. I'm just saying that everyone on my rocket has to get the same answer for her age, and they don't get to choose anything.

They can choose whatever reference system they like - see post #37 (again). Usually astronauts use the Earth's ECI-frame system.

 

[harrylin]

I wasn't able to do it after I read what you wrote. Can you show me how to figure out how much older she gets while the TV pictures are getting to me, and then use that to tell me how much older she gets while I'm turning around? That's what I really want to be able to do, and I can't yet.

From her Earth perspective/ ECI reference system: at the time that she sends the message, the rocket is at a distance dx flying with a certain speed v (let's assume constant speed exactly away from Earth). Then the radio wave will catch up with the rocket with a relative speed c-v (also called "closing speed"). Thus the time of transit according to her is dx/(c-v).
And according to the rocket's rest frame, the rocket is in rest and so the time of transit is dx'/c.

But I don't understand what you mean with "while I'm turning around"; in the simplest scenarios that turn-around is assumed to be relatively fast, so that it can be neglected. Perhaps you refer to Einstein's "induced gravitational field"? That's not much appreciated nowadays, as that field is just a fake field to make a correction which by definition equals the effect based on special relativity (see post #7 and on).

 

[Nugatory]

I understand that. My brothers won't agree about her age. I'm just saying that everyone on my rocket has to get the same answer for her age, and they don't get to choose anything.

If by "everyone on my rocket has to get the same answer" you mean "if everyone on my rocket will choose to use the same reference frame (presumably our rocket frame because it is right under our noses and is so natural/obvious that we sometimes forget it's not the only possibility) they have to get the same answer" then yes, you are correct.

 

[Underwood]

My brothers won't agree about her age. I'm just saying that everyone on my rocket has to get the same answer for her age, and they don't get to choose anything.

They can choose whatever reference system they like - see post #37 (again). Usually astronauts use the Earth's ECI-frame system.

Anytime I hear anyone talking about the time dilation formula, they say that on the outward part of the trip, the home sister says that her rocket brother is aging slower than she is, and that the rocket brother says that his home sister is aging slower than he is. I never hear anyone say that the rocket brother needs to choose a reference frame before he can tell you if his home sister is aging slower or faster than he is.

Why should the rocket brother need to choose a reference frame before he can figure out how much older his home sister gets while the TV picture is getting to him, but he doesn't need to choose a reference frame before he can tell you about time dilation?

 

[pervect]

Anytime I hear anyone talking about the time dilation formula, they say that on the outward part of the trip, the home sister says that her rocket brother is aging slower than she is, and that the rocket brother says that his home sister is aging slower than he is. I never hear anyone say that the rocket brother needs to choose a reference frame before he can tell you if his home sister is aging slower or faster than he is.

Why should the rocket brother need to choose a reference frame before he can figure out how much older his home sister gets while the TV picture is getting to him, but he doesn't need to choose a reference frame before he can tell you about time dilation?

You basically always need to choose a reference frame. So when the home sister says that her brother is aging more slowly, she's talking about it from her own reference frame.

 

[Underwood]

From her Earth perspective/ ECI reference system: at the time that she sends the message, the rocket is at a distance dx flying with a certain speed v (let's assume constant speed exactly away from Earth). Then the radio wave will catch up with the rocket with a relative speed c-v (also called "closing speed"). Thus the time of transit according to her is dx/(c-v).
And according to the rocket's rest frame, the rocket is in rest and so the time of transit is dx'/c.

Your first part is the home sister's view of how much older she gets while the TV picture is getting to the rocket brother. I want the rocket brother's view of that. I think your saying that the rocket brother's view is dx'/c, and I guess you mean x' is the rocket brother's view of the distance between them when she sent the TV picture, the length contracted distance. If that's what you meant, I don't think your answer is right. It would say that the home sister doesn't get any older when the rocket brother turns around, because x' doesn't change then. But if the rocket brother uses the time dilation formula and says that his sister ages less than he does on both the outward and the inward part of his trip, his sister has to get a lot older when he turns around, because otherwise she couldn't be older than he is when he gets home.

 

[Nugatory]

Why should the rocket brother need to choose a reference frame before he can figure out how much older his home sister gets while the TV picture is getting to him, but he doesn't need to choose a reference frame before he can tell you about time dilation?

We need to choose a reference frame, always. It's just that people are sometimes careless in their statement of the conditions of the problem, and do not explicitly specify a reference frame when they're assuming the reference frame that they're at rest in. There's not a single place, anywhere in this thread or in the twin paradox, where a reference frame hasn't been selected either explicitly or by assumption... any statement that uses the words just now or at the same time must state or assume some reference frame to make sense.

 

[Simon Bridge]

We need to choose a reference frame, always. It's just that people are sometimes careless in their statement of the conditions of the problem, and do not explicitly specify a reference frame when they're assuming the reference frame that they're at rest in.

I just want to second this. It is very important to understand this when you have anything to do with any kind of relativity. As soon are we make the reference frame explicit a lot of the confusion clears up.

What's been happening all through this thread is that someone asks a question without making the reference frame explicit and people reply using examples of different reference frames. State the reference frame in the question, and always do this, and watch what happens.

 

[harrylin]

[..] I think your saying that the rocket brother's view is dx'/c, and I guess you mean x' is the rocket brother's view of the distance between them when she sent the TV picture, the length contracted distance.

His view of the distance is dx' if he created an "inertial rocket frame" on which he based his view. I added d to suggest "delta", in order to distinguish from the x' coordinate of the Lorentz transformation - but as the length is very large, I should have written delta to distinguish from little dx. A determined distance is (x2'-x1') at a certain time t'.

If that's what you meant, I don't think your answer is right. It would say that the home sister doesn't get any older when the rocket brother turns around, because x' doesn't change then. But if the rocket brother uses the time dilation formula and says that his sister ages less than he does on both the outward and the inward part of his trip, his sister has to get a lot older when he turns around, because otherwise she couldn't be older than he is when he gets home.

The standard modern variant of the Twin paradox has the standard SR answer: when the rocket turns around it starts to move very fast in the inertial frame that the brother chose earlier. The brother can continue to calculate in accordance with his earlier reference system, but as he supposedly has no instruments at rest in it, he'll have to continuously "transform" between his new readings and the ones that correspond to the earlier one. Or he can create a new reference system that corresponds to switching to a new rest frame, in which distant time is measured differently. If he had made a system with a clock in the front and another one in the back, he'll have to change their synchronization. Consequently the sister is perceived to be much older in that frame. She doesn't "get" any older due to a change of perspective or clock manipulation by the brother. He could one day after the turn-around realize that it's easier to switch frames. The moment that he switches, she'll appear to him much older "now" as the radio waves that reach the rocket were emitted much longer ago according to the newly chosen reference frame.

 

[Underwood]

I found this link on a physics news group that gave me the answer I've been looking for, how to figure out how much older my home sister gets while I'm turning around. I haven't looked at very much of that web sight yet, its long, but I did find a formula on there that gives me the answer. I was surprised that its easy to do.

https://sites.google.com/site/cadoequation/cado-reference-frame

 

[Simon Bridge]

You've also been provided with that answer several times - though not as baldly as that. Notice how the opening statement is that the accelerating twin gets to pick any reference frame they like? That's what everyone has been saying from the start.

There's also [urlhttp://www.physicsguy.com/ftl/html/FTL_part1.html]a more detailed overview[/url]... treating the matter graphically and more how you may be used to.

The main problem all through has been communication - you need to be very carefl about your language when you are talking about relativity. Notice how this works in the CADO reference frame formulation - eg. the L in the formula is the distance between the twins explicitly in the non-accelerated twin's reference frame.

When you make the reference frames explicit, you make more progress.

 

[Underwood]

You've also been provided with that answer several times - though not as baldly as that.

Its the first time that anyone has shown me how to actually calculate how much older my home twin gets while I'm turning around. Thats what I wanted to see how to do. And they give a simple example, with all the numbers given.

Notice how the opening statement is that the accelerating twin gets to pick any reference frame they like? That's what everyone has been saying from the start.

That first sentence just says that the traveler twin can accelerate any way he wants, and the equation still works. But you always use the same equation. So he's not picking a reference frame. They call his reference frame the CADO frame.

the L in the formula is the distance between the twins explicitly in the non-accelerated twin's reference frame.

Yeah, the numbers you need to put into the CADO equation are from the home twin's viewpoint, because those are the easy ones to get. But the answer the equation is giving you is the traveler twin's viewpoint of the home twin's age. That's what I wanted to get.

 

[DrGreg]

Note that the CADO paper linked to above includes this paragraph:

"More than one reference frame for an accelerating observer have been defined, and there is no consensus about which one is most appropriate. This article describes one such reference frame: the CADO frame."​

 

[Dale]

Note that the CADO paper linked to above includes this paragraph:

"More than one reference frame for an accelerating observer have been defined, and there is no consensus about which one is most appropriate. This article describes one such reference frame: the CADO frame."​

That is fantastic news. It looks like Mike actually did learn the point from all of those arguments about CADO! I am very pleasantly surprised.

 

[Underwood]

Note that the CADO paper linked to above includes this paragraph:

"More than one reference frame for an accelerating observer have been defined, and there is no consensus about which one is most appropriate. This article describes one such reference frame: the CADO frame."​

Yeah, your right. Its further down the page. Sorry, I didn't see it.

The table of contents says they do compare the CADO frame to a couple of other frames in a section toward the bottom, but I haven't read that far yet.

 

[harrylin]

I found this link on a physics news group that gave me the answer I've been looking for, how to figure out how much older my home sister gets while I'm turning around. I haven't looked at very much of that web sight yet, its long, but I did find a formula on there that gives me the answer. I was surprised that its easy to do.

https://sites.google.com/site/cadoequation/cado-reference-frame

What you really asked, apparently, is how to calculate the difference in perceived distant age between different inertial frames - good for you that you found a detailed calculation example.

And I see that -happily- that calculation is consistent with our explanations here:
"It is possible for the traveler, by using only elementary observations and elementary calculations, to determine how much she has aged while that image was in transit, and thus to determine what her actual current age was at the instant that he received that image. If he does that correctly, he will get exactly the same result that the Lorentz equations would have given him (and the same result that the CADO equation would have given him)."*

But I wonder if you realize that you touched on a truly problematic issue of the equivalence principle solution with your phrasing "how much older my home sister gets while I'm turning around", and which surely played a role in it being downgraded. Did you consider such things as cause and effect, as well as when exactly this supposed far-away aging due to your turnaround must have happened? *ADDENDUM: If/when you switch between inertial frames then that corresponds to using the Lorentz transformations, and at first sight the web page that you found gives a shortcut to that. Here is the same directly with the Lorentz transformation for time, which looks to me just as simple:

t'=γ(t-vx/c²)

Using that website's example, in the outbound frame we can define that she is moving fast to the right;
x=vt and γ=2 so that v=+sqr(0.75)c=+0.866c and:
t'=2(t-0.75t)

For his age t=20 year at turn-around:
t'=2(20-15)=10 year = her age at turn-around according to the outbound frame.

If we switch to the return frame at turn-around, it's all the same except that now he reckons that she's moving to the left with v=-0.866c:

t'=2(t+0.75t)=2(20+15)=70 = her age at turn-around according to the inbound frame.
The difference is 70-10= 60 years.

And it's just as simple with the direct method: time delay between sending and receiving t2-t1 = (x2-x1)/(c-v)
Thus for this case:

For v=-0.866c: t2-t1 = sqr(0.75)* 20/(1+sqr(0.75)) = 129.28 yr. (inbound or return rocket frame)
For v=+0.866c: t2-t1 = sqr(0.75)* 20/(1-sqr(0.75)) = .. 9.28 yr. (outbound rocket frame)
Difference is.......... 120 yr.

So he now reckons that the signal left Earth 120 yr. earlier on his clock than originally estimated, which corresponds to her now being 60 yrs. older according to him than in his earlier estimation.