1. Jul 17, 2016

Electric to be

I understand that the twin paradox isn't truly a paradox, and that the traveling twin ages since the universe can be "aware" of his traveling. However, my question is how does the time dilation formula break down? Why can't you integrate dT' = dT * Gamma for each velocity as the traveler accelerates, and then actually reach the so called paradox? As both integrals for the traveler and earth stayer would be the same. Does the time dilation formula break down during acceleration?

2. Jul 17, 2016

Staff: Mentor

The time dilation formula doesn't break down, and you can certainly integrate it to get the total time, regardless of the acceleration of the object.

The issue is that the time dilation formula only applies in inertial frames. So you can use it in any inertial frame, but not in a non inertial frame. You can integrate along the path of a non inertial object, but only using the coordinates in an inertial frame.

3. Jul 17, 2016

Orodruin

Staff Emeritus
4. Jul 18, 2016

stevendaryl

Staff Emeritus
If you want to know how old a traveling twin will be when he gets to his destination, you can always use the formula:

$\Delta \tau = \int_{t_A}^{t_B} \sqrt{1-\frac{v^2}{c^2}} dt$

where $t_A$ is the time at the start of the journey, $t_B$ is the time at the end of the journey, and $v$ is the speed of the traveler (which might be changing---it doesn't matter). You can pick any inertial coordinate system to measure $t_A, t_B, v$ and $dt$, and you will get the same answer. The critical point is the word "inertial". If the one twin turns around and returns, then there is no inertial reference frame in which he is always at rest.

You can use a noninertial coordinate system, also, but the formula has to be modified:

$\Delta \tau = \int \sqrt{\sum_{\alpha \beta} g_{\alpha \beta} \frac{dx^\alpha}{ds} \frac{dx^\beta}{ds}} ds$

where $g_{\alpha \beta}$ is the components of the metric tensor in that coordinate system, and $x^\alpha$ are the coordinates in that reference frame, and $s$ is any parametrization of the journey (it could be proper time, or it could be anything that increases smoothly along the journey).

5. Jul 18, 2016

Electric to be

That second part is a bit beyond my understanding, and I would have to look into it more. However, as for the first bit about integrating in an inertial frame. When time derivation is derived using Lorentz Transformations, doesn't it assume that you are transforming time space coordinates between two inertial frames of reference, for example S, and another frame moving with a constant velocity S'

So, aren't you then describing transforming coordinates between an inertial and a non inertial frame, which violates the assumptions of the Lorentz transform?

6. Jul 18, 2016

Orodruin

Staff Emeritus
No, there is only one frame in that statement and no Lorentz transform. Along with the definition of the Minkowski metric.

It is also a common misconception that SR cannot handle accelerated frames.

7. Jul 18, 2016

Electric to be

I see that it can handle acceleration, based on his second equation. However, how is there only one frame in that statement? There are two time coordinates being compared. Tau is the time that you observe to pass in the moving frame of reference, and t is the time passed in the stationary frame?

Also time dilation is derived from Lorentz transformation (other equivalent methods can be used but they're all essentially a simplified version of Lorentz transformation) , which can transform the coordinates of some moving frame S' to a stationary frame S', however the usage of this specific transformation requires inertial reference frames.

This is one such derivation : http://resonanceswavesandfields.blogspot.com/2011/07/derivation-of-length-contraction-and.html?m=1

8. Jul 18, 2016

Orodruin

Staff Emeritus
No, there are not. There is a time coordinate and a proper time.

You have to be more careful when you say "moving frame of reference". If it is an inertial frame, fine, but if the object is accelerating, there are several different coordinate charts that can be used. Tau is the proper time of an observer following that particular world line.

No it is not. It is perfectly possible to derive it from first principles without any reference to the Lorentz transformation.

9. Jul 18, 2016

Electric to be

Well I guess I think of it as comparing the difference between two events of two temporary inertial frames. In the small moment dT that the object is currently in a single velocity prior to moving past and accelerating to the next, I think of it as using Lorentz transforms to find the difference in the temporary coordinate system events in the moving frame (and in some current inertial frame at its temporary velocity) for some proper time dT'. Then integrating over all of these velocities and temporary coordinate transformations.

And I understand that the time dilation CAN be derived using first principles, but it can also be derived from the Lorentz transformation. So the above is my justification for being able to use the time dilation formula as an integral over changing velocity . (By continuing to adhere to the requirements of the Lorentz transform, which is basically that the transformation must be carried out between two inertial frame)

Last edited: Jul 18, 2016
10. Jul 18, 2016

Staff: Mentor

In that equation $\tau$ is the proper time on the clock, and t is the coordinate time in any inertial frame. The label "stationary" or "moving" is just an arbitrary label and the same formula applies equally well to any inertial frame.

11. Jul 18, 2016

Electric to be

Yes I understand now. Does that explanation make sense though? For some dT in coordinate time, we can find a d$\tau$ for the moving object with the Lorentz transformation by assuming it is in some instantaneous rest frame and speed v(T)? Then with this knowledge we can integrate simply using the mentioned formula?

12. Jul 18, 2016

Orodruin

Staff Emeritus
Yes, you can argue in this way for smooth world lines.

However, you should note that going the way you are going is essentially only arguing why, based on expectations from coordinates in inertial frames, some of the properties of SR hold. In a more modern approach, you would simply define the proper time in that manner and find that for an observer at rest, it coincides with the coordinate time if using Minkowski coordinates based on that observer. The physically measurable quantities are all invariants, such as proper time.

Also, time dilation is ultimately based upon an arbitrary simultaneity convention that is "natural" given a particular inertial frame.

13. Jul 18, 2016

Electric to be

I'm not exactly sure what that last bit about the convention means, but I can look into it. But yes I got it. In fact I can really convince myself of that definition by just using the mirror light clock derivation of time dilation.

Thanks for the help. I'm an EE student and I'm just trying to personally research and have a knowledge of some SR on my own. I've noticed that a lot of SR is taught by using certain things that are derived already for you, like the interval invariance and Lorentz transforms, and lacks derivation from the principle that the speed of light is always constant. Just seems to cause confusion. Hopefully it doesn't bite me if I do end up pursuing a physics minor and a mechanics course that covers relativity!

14. Jul 18, 2016

Staff: Mentor

That frame is called the momentarily comoving inertial frame. It only exists for smooth curves.

Certainly you can calculate this value, but what does it mean physically? What is the meaning of a sum of coordinate times in different reference frames? It isn't proper time, it doesn't correspond to the time on any one clock, it doesn't describe the rate of any physical process.

15. Jul 18, 2016

Electric to be

Well the value would just be the proper time. By the coordinate times all I am doing is finding the time difference in events in the momentarily comoving frame and this quantity is proper time. (This is how time dilation is derived using the Lorentz transforms and in this case the Lorentz transform is applied at each instant for the rest frame and the momentarily comoving frame)

16. Jul 18, 2016

Orodruin

Staff Emeritus
This is not my experience and certainly not the modern way of teaching relativity at university level.

Also, you should not put too much emphasis on the postulate of the invariance of the speed of light. While it may have been in Einstein's original formulation, the field has developed significantly in the last 110 years.

17. Jul 18, 2016

Staff: Mentor

No, it wouldn't (unless I am misunderstanding the quantity you mean). That is kind of the point of the twin paradox. From your verbal description, the value you are describing is the naïve calculation that gives the wrong answer in the twin paradox.

Or if you really do just mean the proper time then just call it the proper time, there is no need nor benefit to refer to the momentarily comoving inertial frames.

18. Jul 18, 2016

Electric to be

I think the misunderstanding is by the mcif I am referring to is that of the moving object. The frame you are sitting in and observing the object in is perfectly inertial, and so as you said earlier, you can apply this integral.

I was basically just justifying to myself why I can use the formula if the object (not you) is accelerating.

19. Jul 19, 2016

Orodruin

Staff Emeritus
I think what he is describing would be the same. It is basically approximating the world line with a polygon train and taking the limit.

20. Jul 19, 2016

Staff: Mentor

Ok, then this is proper time and the use of the MCIF is unnecessary and just over complicates things.

Although historically most derivations go from the two postulates to the Lorentz transform and then to everything else, many modern treatments start with the metric $ds^2=-dt^2+dx^2+dy^2+dz^2$. The Lorentz transform is then the class of transforms preserving the metric, and the invariance of the metric for $ds=0$ gives the second postulate and writing the laws of physics as tensors gives the first postulate.

The time dilation formula falls out easily as $d\tau^2=-ds^2$ and then compute $d\tau/dt$

Last edited: Jul 19, 2016