Alternate derivation of Lorentz Trans.

[ghwellsjr]

Just been looking over some old physics books of my student days.
In Special Relativity, Rindler,W. 1960, section 3 describe how Lorentz and Fitzgerald derived L-F length Contraction formula based on an ether, which is familiar from Einstein's Relativity theory. The same section in Rindler also shows how Lorenz derived the time-dilation (the same one as in Relativity), based on either theory and the constancy of observed light speed. The derivations are easy...much easier than the derivations I have seen in Relativity theory.
Exercise 12 on p24 Rindler says "as far as it goes, the Lorentz theory (with ether) is parallel to the Einstein theory (with no ether, but with the relativity principle).

Actually, both LET and SR are based on the same first postulate, the principle of relativity, but they have different second postulates. LET's second postulate is that light propagates at c only in a single frame, the rest state of the immovable ether, whereas SR's second postulate is that light propagates at c in any inertial frame you want to pick. Note that the first postulate concerns things that can be measured and observed, like measuring the round-trip speed of light, while the second postulate concerns that which we can neither measure nor observe, the propagation of light.

But, on further reading, I find I still do not understand the purported resolutions of the twin paradox.
The twin scenario is: (i) two twins move apart at a constant speed relative to each other.
(ii) By relativity they both observe, by em-signals, that the other seems to be ageing faster than themselves.

You've got this backwards, they each observe the other ageing more slowly than themselves while they are traveling away from each other.

(iii) One of the twins misses the other, turns around, and returns to their twin at the same relative velocity.

And as soon as he turns around and travels back he will observe the twin that remained stationary as ageing faster than himself and this will continue for the entire trip back. When he gets back he will see that the sum of the equal intervals of observed slow ageing and fast ageing adds up to the actual amount that the stationary twin aged during the trip.

Now what does the stationary twin observe of the traveling twin? Well, he's not going to see the traveling twin turn around until long after he actually turns around because he has to wait for the em (light) signal or image of the turn-around to propagate back to him and until it reaches him, he will continue to see the traveling twin ageing at a slower rate than himself. Eventually, he will see his twin turn around and at that point, he will see him age faster than himself. So because he observes the traveling twin aging at a slower rate during most of the trip, he agrees that the traveling twin actually did age less than himself at the end of the trip.

To summarize: the traveling twin sees the stationary twin age slow for 1/2 the time and fast for 1/2 the time while the stationary twin sees the traveling twin age slow for, say, 3/4 of the time and fast for 1/4 of the time and this is why there is an imbalance in their ages when they reunite.

When they meet again the one who turned around has aged less.

Yes, and hopefully it makes sense from the viewpoint of what each twin actually sees and observes but the acceleration at the turn around isn't what made it happen.

The turnaround, which we may assume instantaneous, is the only difference between the twins, since they are both in inertial frames during the rest of their separation.

This isn't quite an accurate statement for two reasons:

1) Everybody and everything is in all inertial frames. Usually when we say that a twin is in his inertial frame, we mean he is at rest in that inertial frame, but his other twin is also in the inertial frame, it's just that he is moving.

2) The stationary twin is at rest in a single inertial frame during the entire scenario but this isn't true for the traveling twin. The traveling twin is at rest in one inertial frame during the outbound portion of the trip and then he is at rest in a different inertial frame during the inbound portion of the trip. So now we have three inertial frames to consider but we could also consider any other frame and they will all give the same answer as to amount that each twin aged during the scenario.

The derivation of the age difference usually considers one twin stationary on earth, and the other moving away. In moving away, the L-transformations shorten lengths and dilate times of the mover.

And this is the easiest frame in which to analyze the scenario: moving twin's clock runs slower, therefore he ages less than the stationary twin when they reunite. Problem solved. No need to consider any other frame.

But if you want, you can transform all the relevant events in this first frame into one of the other frames to see how it looks there. What you will find is that while the traveling twin is at rest during one half of the trip, the "stationary" twin ages less, but during the other half of the trip, the traveling twin has to travel at a higher speed than the "stationary" twin and so he experiences even more time dilation and ends up younger.

The time-space measurement coordinate frame of the mover tighten up on the null/light cone.
But, why is this reasonable, relativistically speaking? In all inertial frames the speed of light is the same, so why is one coordinate measurement frame more lightlike than any other?
Some (including Mach?) justify the asymmetry by invoking the distant universe towards which the mover moves, and the stayer does not. This seems to be invoking a kind of "ether" in terms of the distant stars. But we know they are not fixed, but moving and accelerating away! Not convincing.
Some explanations note the red or blue shift observed in light from the partner, and this does seem to correspond to differing relative rate of aging behaviour. But, why do such considerations overcome the inertial frame equivalence of the two twins on the bulk of their journeys?
Sorry if these are all well warn and ignorant considerations...as I am sure they are. A reference to a really clear and solid resolution of the twin paradox would be appreciated.

I cannot relate to these last comments of yours. Hopefully, I have steered you in the right direction to be able to understand the resolution of the twin paradox. If not, ask for clarification on any points that are still unclear.

 

[keithR]

Thanks ghwellsjr, for your careful explanations, relating to the twin paradox.
Almost all of what you have explained I accept, and in fact your explanations are the same as those I have read elsewhere as I have been trying to find clarification.
However, I still remain with the same unease that I don't really "see" the complete adequacy of the explanations in answering my specific misgivings. Some of these you have not commented on, and others you say you do not relate to, presumably because my words are non-sensical within the SR context. Let me focus on some of these specific misgivings.
Sorry if I just continue to grovel in my own mess.

1. First, I would prefer to set aside accounts of the twin aging process in terms of what how they see each other aging. I accept the correctness of the account, but would prefer to deal with how a person actually is aging, rather than how they are seen to be aging.
(i) possibly, you will be unhappy with this preference, since simultaneity in SR is defined in terms of such light/em signals.
(ii) but I do feel the "observer-based" account is misleading.
Reasons:
(a) It does not matter if the velocity is away or towards the "fixed" observer since the effects depend on v2. The time units, or the ticks of the clock in the moving inertial frame are longer than they are in the fixed frame by a factor of β(1+v2)1/2where β=(1-v^2)-1/2 and c=1.
(b) I get this from the usual twins diagram, with the "y" axis being t and the "x" axis being x for the stationary frame of reference. Hence (0,1) refers to the stationary x-origin, after unit time. (0',1') refers to the moving x'-origin, when t'=1. Calculate (x,t) corresponding to (0',1') using the inverse LT (i.e. t=β(t'+vx') , x=β(x'+vt') to get β(v,1) which defines the direction fo the time axis in terms of the stationary frame of ref. It is alond the direction of movemement.
Taking a numerical example which I have seen used in the twin discussion elsewhere, let v=3/5, so that the rocket travels out 3 light years during 5 years in the stationary time frame.
Then β=5/4, which is the time-dilation factor, yielding a moving flight time of 4 years.

β(1+v^2)1/2=√34/4, and since the "length" of the space-time trajectory, in the stationary frame is √34. So again, it is confirmed that the time experienced on the outgoing rocket is 4 years.
The same applies on the return journey.
What is seem through mutual light based tv's, or the spectrum shifts involved, is interesting, but to my mind, confuses the issue rather.

Similarly to above, we find that the space-like (1',0') in the rocket, is β(1,v) in the stationary frame. This, and the β(v,1) for the rocket time-like (0',1') are what I meant when in my last paragraph previously, when I said the time and space basis vectors close up (symetrically) on the null cone as v→c.
One of my previous comments on this issue, which you did not relate to, was as to how this squeezing up of moving (space,time) frames to the light cone, with increasing speed, consistent with the relativity principle: in any frame, the speed of light is still c, so why is squeezing up with speed possible? This is anther form of the "twin-paradox" I think.
Penrose (fabric of Reality p409, in his discussion of the "paradox") mentions this squeezing up of the time-like and space-like (orthoganal) axis to the null cone as v→c.
But I don't think penrose's use of the triangle inequality really addresses the paradox, since he just says with the longest side of the space-time triangle (in proper time) is in the frame that is stationary. But how do we say what frame is stationary!?

(c) All the above, right or wrong, just says I do accept the correctness of the age reduction in the twin example, from the LT equations, and form space time diagrams.

Explaining the life shortening in another way does not relieve me of the feeling of paradox.

The paradox only arises when we realize that the results should be symmetric between the two frames, as inertial frames, during the both the steady motion periods between the departure and turn-round points.
If you do not see this is paradoxical, you do not see the twin paradox.
Rindler, page 30 says: "If the two clocks were the whole content of the universe this would indeed be a paradox", so, he sees the paradox. Do you?
Rindler then says, as you do, one of the clocks has jumped inertial frames, so there is a lack of symmetry between the two twins. But, what has this got to do with the issue, if we can say that we assume the turn-around can be considered immediate. Rindler mentions "preferred frames"!. This sounds like "Ether" in another form.
As I mentioned in my first post, Mach's answer to the paradox was the "distant fixed stars". Another illusionary preferred frame.

Sorry, I don't know to quote your message points, ghwellsjr.
Thanks for your stimulating discussion.

 

[ghwellsjr]

I'm not sure which paradox you are referring to.

1) Are you talking about the "paradox" of two observers traveling with respect to each other, neither one changing their speed or direction and they each see the other ones clock running slower than their own?

2) Or are you talking about the "Twin Paradox" where both twins always see the other ones clock running slower than their own and by the same amount, yet when they return, one of them has aged less than the other one?

Virtually all paradoxes in Special Relativity come about as a result of conflating the parameters determined by two different inertial Frames of Reference or by not correctly applying the Lorentz Transformation to get from one inertial Frame of Reference to another one. You said:

The paradox only arises when we realize that the results should be symmetric between the two frames, as inertial frames, during the both the steady motion periods between the departure and turn-round points.

But I can't tell whether the two frames you are talking about here are the rest frame of the stationary twin and the rest frame of the traveling twin during the outbound portion of the trip OR if you are talking only about the rest frame of the traveling twin during the outbound portion of the trip and the rest frame of the traveling twin during the inbound portion of the trip OR if you are talking about the traveling twin being in a single inertial rest frame during the entire trip (except for the negligible instant of turn-around) and the rest frame of the stationary twin.

Please tell me which of the these three options you are talking about in the above quote with regard to the two frames (or maybe some other option).

 

[harrylin]

[..] why do such considerations overcome the inertial frame equivalence of the two twins on the bulk of their journeys?
Sorry if these are all well warn and ignorant considerations...as I am sure they are. A reference to a really clear and solid resolution of the twin paradox would be appreciated.

Hi keith,

The first clear and solid resolution of the twin paradox was perhaps in the first discussion of that scenario (without any "paradox"), here:
http://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time

It's very long-winding, but in this context mainly the sections of p.47 + p.50-52 are of interest.