# What motivated Einstein to start thinking about a General Theory of Relativity?

#### Nugatory

Mentor
The "rate of aging" of a twin is a coordinate-dependent quantity. In the inertial coordinate system of frame A, T2 ages slowest afterward. In the inertial coordinate system of frame B, T1 ages slowest.
But note that although the "rate of aging" is a coordinate-dependent quantity, the actual amount of aging that a twin will experience on his path between two points in spacetime is not coordinate-dependent. It will be the same no matter which coordinates we use and what coordinate-dependent rate of aging those coordinates suggest. The key here is that although the "rate of aging" will be different in different coordinate systems, so will the "time" during which this aging is happening, and the two effects always balance out to give the same total amount of aging along the journey.

[Of course stevendaryl knows this already. I'm just trying to stop someone else who doesn't know this from being confused by this aging-faster/aging slower right now thing]

#### Mentz114

Gold Member
I thought it was "the proper interval between two events on a clocks worldline is the time elapsed on the clock ( between those events)"

#### PeterDonis

Mentor
The Usenet Physics FAQ has a good explanation:

http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html

Basically, it's the postulate that a clock's "rate of time flow", as seen by an observer, is not affected by its acceleration; it's only affected by the clock's velocity relative to the observer.

#### nitsuj

The Usenet Physics FAQ has a good explanation:

http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html

Basically, it's the postulate that a clock's "rate of time flow", as seen by an observer, is not affected by its acceleration; it's only affected by the clock's velocity relative to the observer.
Cool, Thanks PeterDonis #### Mentz114

Gold Member
This diagram is the scenario where the twins T1 (blue) and T2 (green) comove, then part company. The measurements are made between times t0 and t2.

The proper times of interest are AB, along T2's worldline, and CD+DE along T1's worldline. Without doing any calculations ( I think these proper times are stevendaryl's τ2 and τ1), it is obvious that whether AB is greater than or less than CD+DE, this relationship will be true from any inertial frame.

The reason being, that proper time is invariant under LT.

Definition of proper time
$$d\tau^2 = dt^2-dx^2$$
Transform the intervals dx->dX, dt->dT with a Lorentz transformation
$$dT=\gamma dt + \gamma\beta dx,\ \ dX=\gamma dx + \gamma\beta dt$$
Calculate new proper time

\begin{align} dT^2-dX^2 &= (\gamma dt + \gamma\beta dx)^2 - (\gamma dx + \gamma\beta dt)^2\\ &= \gamma^2(1-\beta^2)dt^2- \gamma^2(1-\beta^2)dx^2\\ &= dt^2 - dx^2 \end{align}

Any decent, simple book on relativity tells us this.

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#### PeterDonis

Mentor
This diagram is the scenario where the twins T1 (blue) and T2 (green) comove, then part company.
This is what we thought the scenario was, but now we're not sure. You have T1 and T2 separated in the x direction and moving in the x direction; but we think arindamsinha meant to have them separated in the y direction (no initial separation in the x direction) and moving in the x direction.

#### PeterDonis

Mentor
Is that a correct description of the scenario as seen from Frame A?
I'm going to assume that it is and go ahead and post the analysis, since it's pretty simple.

We have the following events (coordinates t, x, y are given relative to Frame A):

#1: $(0, 0, 0)$ T1 starts the experiment, moving in the x direction at velocity v.

#2: $(0, 0, 1)$ T2 starts the experiment, moving in the x direction at velocity v.

#3: $(t_1, v t_1, 0)$ T1 stops moving.

#4: $(t_2, v t_2, 1)$ T2 stops moving and ends the experiment.

#5: $(t_2, v t_1, 0)$ T1 ends the experiment.

We have, by hypothesis, $t_2 > t_1$, and for convenience I will define $\delta t = t_2 - t_1$.

The proper times in Frame A are then:

$$\tau_1 = \frac{t_1}{\gamma} + \left( t_2 - t_1 \right) = \frac{t_1}{\gamma} + \delta t$$

$$\tau_2 = \frac{t_2}{\gamma} = \frac{t_1 + \delta t}{\gamma}$$

This makes it obvious that $\tau_1 > \tau_2$.

Now let's look at things in Frame B. Here are the event coordinates t', x', y' in that frame, obtained by Lorentz transforming the coordinates given above (note that we have assumed the origins of both frames are the same, at event #1):

#1: $(0, 0, 0)$ T1 starts moving in the x direction at velocity v.

#2: $(0, 0, 1)$ T2 starts moving in the x direction at velocity v.

#3: $(t_1 / \gamma, 0, 0)$ T1 stops moving.

#4: $(t_2 / \gamma, 0, 1)$ T2 stops moving.

#5: $(t_1 / \gamma + \gamma \delta t, - \gamma v \delta t, 0)$ T1 ends the experiment.

The proper times in this frame are then:

$$\tau_1 = \frac{t_1}{\gamma} + \frac{t_1 / \gamma + \gamma \delta t - t_1 / \gamma}{\gamma} = \frac{t_1}{\gamma} + \delta t$$

$$\tau_2 = \frac{t_2}{\gamma} = \frac{t_1 + \delta t}{\gamma}$$

In other words, the proper times are the same in both frames, as they should be. The key thing to note, of course, is that in Frame B, event #5 happens *later* than event #4, and the additional coordinate time that this adds to T1's "moving" segment in that frame more than compensates for the fact that T1 is moving while T2 is at rest. This is basically the same resolution as the previous scenario; the y coordinate drops out of the analysis since all the motion is in the x direction, but there is still separation in the x direction at the end of the experiment (even though there isn't at the start), so relativity of simultaneity still comes into play in making event #5 later than event #4 in Frame B.

#### D H

Staff Emeritus

Metz114 et al: Please remember to use the report button to let the mentors know about nonsense such as that which you highlighted.

OK. Cleanup complete. I deleted 50 posts. That's a bit much, perhaps too much. Those posts are still here; I soft-deleted them. Let me know if there's anything that you members strongly feel needs to be restored.

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