The Twin Paradox and the Equivalence Principle

[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.