Five questions re: SR

[TakenItSeriously]

Based on the widely diverse solutions to the twin paradox, Twin paradox solutions almost seem like an opinion poll and they seem to disagree on certain aspects, so I'm lost on what's the most accepted.

 

Five questions on SR:

1) is acceleration ≡ gravity still the most accepted solution for the twin paradox?

2) Which is more accepted: Clocks moving towards observer run faster or slower than the local clock?

3) Is there any proof that shows that length could not exceed proper length regardless of direction of movement (towards/away)?

4) What are the consequences of the invariant spacetime interval?

5) is there a specific finding or proof that triggered controversy?

Edited August 27, 2016 by TakenItSeriously

[swansont]

While there are a number of ways to look at the problem, acceleration is what causes one twin to move to a new frame of reference, and accounts for the asymmetry in elapsed time.

 

Moving clocks (in inertial frames) run slow.

[Strange]

1) is acceleration ≡ gravity still the most accepted solution for the twin paradox?

 

 

I have never come across gravity used in the explanation. Do you have a reference? I suppose that because SR is a subset of GR, an explanation in GR could be in terms of gravitational effects, but I really don't know.

 

 

 

2) Which is more accepted: Clocks moving towards observer run faster or slower than the local clock?

 

Slower. Universally accepted.

 

But (as noted in another thread about this) the approaching clock may be seen to run faster because of the Doppler effect.

 

 

 

3) Is there any proof that shows that length could not exceed proper length regardless of direction of movement (towards/away)?

 

Is there any reason to think it could?

 

 

 

4) What are the consequences of the invariant spacetime interval?

 

This is the basis of the Lorentz transformation - see the recent thread on "objects and frames" for a great explanation from ajb.

 

 

 

5) is there a specific finding or proof that triggered controversy?

 

I am not aware of any controversy regarding any aspect of SR. (Or even GR.)

[ajb]

4) What are the consequences of the invariant spacetime interval?

Invariant under the Poincare group to be a little more careful - anyway all the strange and wonderful effects we have in special relativity really come down to the space-time interval being invariant when changing between inertial frames. This really is the key thing to keep in mind.

[TakenItSeriously]

Based on the widely diverse solutions to the twin paradox, Twin paradox solutions almost seem like an opinion poll and they seem to disagree on certain aspects, so I'm lost on what's the most accepted.

 

Five questions on SR:

1) is acceleration ≡ gravity still the most accepted solution for the twin paradox?

I have never come across gravity used in the explanation. Do you have a reference? I suppose that because SR is a subset of GR, an explanation in GR could be in terms of gravitational effects, but I really don't know.

I think you got the gist of it. They use the Equivalence Principle to equate acceleration to gravity then apply GR's time dilation to the accelerated turn around event to explain the time deviation. The following youtube video does a good job explaining it short and to the point.

https://youtu.be/0iJZ_QGMLD0

 

2) Which is more accepted: Clocks moving towards observer run faster or slower than the local clock?

Slower. Universally accepted.

But (as noted in another thread about this) the approaching clock may be seen to run faster because of the Doppler effect.

Thanks, I'll look for it.

 

3) Is there any proof that shows that length could not exceed proper length regardless of direction of movement (towards/away)?

Is there any reason to think it could?

I think there is, but I'm still working out the details before I commit to anything, I hope to have an answer soon.

 

4) What are the consequences of the invariant spacetime interval?

This is the basis of the Lorentz transformation - see the recent thread on "objects and frames" for a great explanation from ajb.

Ok, great, thanks.

 

5) is there a specific finding or proof that triggered controversy?

I am not aware of any controversy regarding any aspect of SR. (Or even GR.)

Thats fine, it just seemed like a bunch of different twin paradox solutions started denying acceleration as the cause.

 

Invariant under the Poincare group to be a little more careful - anyway all the strange and wonderful effects we have in special relativity really come down to the space-time interval being invariant when changing between inertial frames. This really is the key thing to keep in mind.

Thanks, I need to dig into this a bit more. I had always just based relativistic effects on the constancy of c and the linking of spacetime which I found to be pretty self-consistent but it sounds like there's a bit more to it than that.

[Mordred]

umm you have to be careful in thinking acceleration. I will dig up a lesson in one of my textbooks. This particular textbook is one Ajb mentioned to me. It has the best solution. Just as soon as I dig it out and have my coffee

[studiot]

Just remember that acceleration is the second time derivative of distance and both quantities depend upon how you measure them.

 

Oh!

 

crossed with Mordred.

 

Edited August 27, 2016 by studiot

[Mordred]

OK The statement "acceleration isn't the cause" is what I am specifically addressing. However I took a moment and looked at the video posted and although 100% accurate, doesn't involve some key math details to show the frame rotation. As this is important to understanding the twin paradox I will include this detail. (leading to the Sagnac effect).

 

The twin paradox is essentially if a see's b time as slow and b see's a time as slow, which twin ages slower? when both observers see the same time dilation between each reference frame. One might think the twins will be the same age on the return trip. However this isn't correct. The video explains the solution as the accelerating observer is the one who will be younger. Which is correct, however it may lead one to assume its the acceleration in GR that's important. (not saying the OP is under this misconception, however we do have other readers.)

 

So lets start with the elevator... This is important as it distinquishes the principle of equivalence.

 

the equivalence principle states the equality between gravitational mass and inertial mass, the problem with this statement occurs when you consider two particles in the elevator. Keeping in mind there is two types of acceleration. Change in direction but not velocity. Change in velocity (magnitude but not direction.)

 

So in the case of the elevator with the two particles. First take a pen and paper, draw a box, with two particles separated on the same vertical elevation. Then draw a line from each particle to the center of the mass.

 

We can see from this drawing that the falling particles undergo two forms of acceleration, change in magnitude and change in direction. The two particles approach one another as they freefall. This is the tidal effect due to the gravitational field. (the deviation from the parallel paths,coincidentally is also used to determine your geodesic equations) This being an example of parallel transport deviation. (curvature).

 

So the equivalence is an expression that holds true only if you treat it at a single point in the gravitational field. Put another way it is locally equivalent. However it isn't equivalent over long distances due to changes in direction. So lets throw some numbers in and describe the tidal effect.

 

[latex]F_a=F_b=\frac{mMG}{r^2}[/latex]

 

let x be the distance between a and b, let [latex]\alpha[/latex] represent the angle between one test body and the center line (center of gravity vertical axis) so from frame a, b experiences a force directed toward a.

 

[latex]F=2F_asin\alpha=2F_a*\frac{x}{2r}=\frac{mMG}{r^3}x[/latex]

 

a then observes b to be accelerating towards him by

 

[latex]F=-md^2x/dt^2[/latex]

[latex]\frac{d^2x}{dt^2}=\frac{MG}{r^3}x[/latex]

 

the 1/r^3 is characteristic of tidal forces.

 

That clears up the principle of equivalence a bit.... the key importance is to understand when this relation holds true.

 

the next part will take me a bit to type in...

 

the twin paradox which isn't a paradox at all, essentially can be broken down to the following statement.

 

"we should never have considered the age of a and b to be the same, as the frame of b undergoes accelerations that a does not undergo."

 

so lets look at the acceleration

 

first define the four velocity. [latex]u^\mu[/latex]

[latex]u^\mu=\frac{dx^\mu}{dt}=(c\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau})[/latex]

 

this gives in the SR limit [latex]\eta u^\mu u^\nu=u^\mu u_\mu=-c^2[/latex]

the four velocity has constant length.

 

[latex]d/d\tau(u^\mu u_\mu)=0=2\dot{u}^\mu u_\mu[/latex]

 

the acceleration four vector [latex]a^\mu=\dot{u}^\mu[/latex]

[latex]\eta_{\mu\nu}a^\mu u^\nu=a^\mu u_\mu=0[/latex]

 

so the acceleration and velocity four vectors are

 

[latex]c \frac{dt}{d\tau}=u^0[/latex]

 

[latex]\frac{dx^1}{d\tau}=u^1[/latex]

 

[latex]\frac{du^0}{d\tau}=a^0[/latex]

 

[latex]\frac{du^1}{d\tau}=a^1[/latex]

 

both vectors has vanishing 2 and 3 components.

 

using the equations above

[latex]-(u^0)^2+(u^1)2=-c^2 ,,-u^0a^0+u^1a^1=0[/latex]

 

in addition [latex]a^\mu a_\mu=-(a^0)^2=(a^1)^2=g^2[/latex] Recognize the pythagoras theory element here?

 

the last equation defines constant acceleration g. with solutions

 

[latex]a^0=\frac{g}{c}u^1,a^1=\frac{g}{c}u^0[/latex]

 

from which [latex]\frac{da^0}{d\tau}=\frac{g}{c}\frac{du^1}{d\tau}=\frac{g}{c}a^1=\frac{g^2}{c^2}u^0[/latex]

hence

[latex]\frac{d^2 u^0}{d\tau^2}=\frac{g^2}{c^2}u^0[/latex]

similarly

[latex]\frac{d^2 u^1}{d\tau^2}=\frac{g^2}{c^2}u^1[/latex]

 

so the solution to the last equation is

[latex]u^1=Ae^{(gr/c)}=Be^{(gr/c)}[/latex]

hence

[latex]\frac{du^1}{d\tau}=\frac{g^2}{c^2}(Ae^{(gr/c)}-Be^{gr/c)})[/latex]

 

with boundary conditions [latex]t=0,\tau=0,u^1=0,\frac{du^1}{d\tau}=a^1=g [/latex] we find A=_B=c/2 and [latex]u^1=c sinh(g\tau/c)[/latex]

 

so

[latex]a^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex]

hence

[latex]u^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex]

and finally

[latex]x=\frac{c^2}{g}cosh(g\tau /c)[/latex],

[latex] ct=\frac{c^2}{g}sinh(g\tau /c)[/latex]

 

the space and time coordinates then fall onto the Hyperbola during rotation

[latex]x^2-c^2\tau^2=\frac{c^4}{g^2}[/latex]

 

 

So regardless of how fast or slow the acceleration is, the rotation itself causes changes to reference frames

 

this detail is often overlooked, by the statement :"it doesn't matter how fast observer B accelerates or decelerates" most articles don't mention the rotation as to the reasoning behind that statement....

 

A lot of attempted solutions try to use the instant accelerated, decelerated argument but ignore frame rotation.

(as mentioned by Studiot)

 

this is the problem with placing two rotated frames (outgoing,incoming) onto the same timeline, leads to errors...not to mention ignoring the spatial separation components of doppler vs relativistic doppler.

 

edit::oops almost forgot to reference the above...

 

:Introduction to General Relativity" by Lewis Ryder.

 

Its an excellent book, as it explains details in an easy math level (well easy to me lol) step by step. The author assumes the reader has less previous knowledge, than in many other GR textbooks. (thanks Ajb for bringing it to my attention, its a great addition to my huge collection lol)

Edited August 27, 2016 by Mordred

[Mordred]

1) is acceleration ≡ gravity still the most accepted solution for the twin paradox?

 

side note, in a free falling particle scenario (Einstein elevator) where is the force? Subject for another thread lol but boils down to "is gravity due to the stress tensor/curvature " vs acceleration argument. Prior answer being the correct one.

 

In the Einstein elevator scenario, just for those interested. Please start a new thread if you have questions on this portion. I worked this up for another thread but its handy to show free fall and how parallel transport relates to geodesics.

In the presence of matter or when matter is not too distant physical distances between two points change. For example an approximately static distribution of matter in region D. Can be replaced by tve equivalent mass

[latex]M=\int_Dd^3x\rho(\overrightarrow{x})[/latex] concentrated at a point [latex]\overrightarrow{x}_0=M^{-1}\int_Dd^3x\overrightarrow{x}\rho(\overrightarrow{x})[/latex]

Which we can choose to be at the origin

[latex]\overrightarrow{x}=\overrightarrow{0}[/latex]

Sources outside region D the following Newton potential at [latex]\overrightarrow{x}[/latex]

[latex]\phi_N(\overrightarrow{x})=-G_N\frac{M}{r}[/latex]

Where [latex] G_n=6.673*10^{-11}m^3/KG s^2[/latex] and [latex]r\equiv||\overrightarrow{x}||[/latex]

According to Einsteins theory the physical distance of objects in the gravitational field of this mass distribution is described by the line element.

[latex]ds^2=c^2(1+\frac{2\phi_N}{c^2})-\frac{dr^2}{1+2\phi_N/c^2}-r^2d\Omega^2[/latex]

Where [latex]d\Omega^2=d\theta^2+sin^2(\theta)d\varphi^2[/latex] denotes the volume element of a 2d sphere

[latex]\theta\in(0,\pi)[/latex] and [latex]\varphi\in(0,\pi)[/latex] are the two angles fully covering the sphere.

The general relativistic form is.

[latex]ds^2=g_{\mu\nu}(x)dx^\mu x^\nu[/latex]

By comparing the last two equations we can find the static mass distribution in spherical coordinates.

[latex](r,\theta\varphi)[/latex]

[latex]G_{\mu\nu}=\begin{pmatrix}1+2\phi_N/c^2&0&0&0\\0&-(1+2\phi_N/c^2)^{-1}&0&0\\0&0&-r^2&0\\0&0&0&-r^2sin^2(\theta)\end{pmatrix}[/latex]

Now that we have defined our static multi particle field.

Our next step is to define the geodesic to include the principle of equivalence. Followed by General Covariance.

Ok so now the Principle of Equivalence.

You can google that term for more detail

but in the same format as above

[latex]m_i=m_g...m_i\frac{d^2\overrightarrow{x}}{dt^2}=m_g\overrightarrow{g}[/latex]

[latex]\overrightarrow{g}-\bigtriangledown\phi_N[/latex]

Denotes the gravitational field above.

Now General Covariance. Which use the ds^2 line elements above and the Einstein tensor it follows that the line element above is invariant under general coordinate transformation(diffeomorphism)

[latex]x\mu\rightarrow\tilde{x}^\mu(x)[/latex]

Provided ds^2 is invariant

[latex]ds^2=d\tilde{s}^2[/latex] an infinitesimal coordinate transformation

[latex]d\tilde{x}^\mu=\frac{\partial\tilde{x}^\mu}{\partial x^\alpha}dx^\alpha[/latex]

With the line element invariance

[latex]\tilde{g}_{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g_{\alpha\beta}x[/latex]

The inverse of the metric tensor transforms as

[latex]\tilde{g}^{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g^{\alpha\beta}x[/latex]

In GR one introduces the notion of covariant vectors [latex]A_\mu[/latex] and contravariant [latex]A^\mu[/latex] which is related as [latex]A_\mu=G_{\mu\nu} A^\nu[/latex] conversely the inverse is [latex]A^\mu=G^{\mu\nu} A_\nu[/latex] the metric tensor can be defined as

[latex]g^{\mu\rho}g_{\rho\nu}=\delta^\mu_\mu[/latex] where [latex]\delta^\mu_nu[/latex]=diag(1,1,1,1) which denotes the Kronecker delta.

Finally we can start to look at geodesics.

Let us consider a free falling observer. O who erects a special coordinate system such that particles move along trajectories [latex]\xi^\mu=\xi^\mu (t)=(\xi^0,x^i)[/latex]

Specified by a non accelerated motion. Described as

[latex]\frac{d^2\xi^\mu}{ds^2}[/latex]

Where the line element ds=cdt such that

[latex]ds^2=c^2dt^2=\eta_{\mu\nu}d\xi^\mu d\xi^\nu[/latex]

Now assunme that the motion of O changes in such a way that it can be described by a coordinate transformation.

[latex]d\xi^\mu=\frac{\partial\xi^\mu}{\partial x^\alpha}dx^\alpha, x^\mu=(ct,x^0)[/latex]

This and the previous non accelerated equation imply that the observer O, will percieve an accelerated motion of particles governed by the Geodesic equation.

[latex]\frac{d^2x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}(x)\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=0[/latex]

Where the new line element is given by

[latex]ds^2=g_{\mu\nu}(x)dx^\mu dx^\nu[/latex] and [latex] g_{\mu\nu}=\frac{\partial\xi^\alpha}{\partial\xi x^\mu}\frac{\partial\xi^\beta}{\partial x^\nu}\eta_{\alpha\beta}[/latex]

and [latex]\Gamma^\mu_{\alpha\beta}=\frac{\partial x^\mu}{\partial\eta^\nu}\frac{\partial^2\xi^\nu}{\partial x^\alpha\partial x^\beta}[/latex]

Denote the metric tensor and the affine Levi-Civita connection respectively.

pS note the equation for "non accelerated motion"

Edited August 27, 2016 by Mordred

[TakenItSeriously]

Thanks Mordred, Impressive post, I'll take your word on the math.

OK The statement "acceleration isn't the cause" is what I am specifically addressing. However I took a moment and looked at the video posted and although 100% accurate, doesn't involve some key math details to show the frame rotation. As this is important to understanding the twin paradox I will include this detail. (leading to the Sagnac effect).

 

The twin paradox is essentially if a see's b time as slow and b see's a time as slow, which twin ages slower? when both observers see the same time dilation between each reference frame. One might think the twins will be the same age on the return trip. However this isn't correct. The video explains the solution as the accelerating observer is the one who will be younger. Which is correct, however it may lead one to assume its the acceleration in GR that's important. (not saying the OP is under this misconception, however we do have other readers.)

Your right, I was only asking to see which way the wind was blowing. I didn't see acceleration as being a logical solution. Though my reasoning wasn't as impressive, but then logic is never impressive.

 

So lets start with the elevator... This is important as it distinquishes the principle of equivalence.

 

the equivalence principle states the equality between gravitational mass and inertial mass, the problem with this statement occurs when you consider two particles in the elevator. Keeping in mind there is two types of acceleration. Change in direction but not velocity. Change in velocity (magnitude but not direction.)

 

So in the case of the elevator with the two particles. First take a pen and paper, draw a box, with two particles separated on the same vertical elevation. Then draw a line from each particle to the center of the mass.

 

We can see from this drawing that the falling particles undergo two forms of acceleration, change in magnitude and change in direction. The two particles approach one another as they freefall. This is the tidal effect due to the gravitational field. (the deviation from the parallel paths,coincidentally is also used to determine your geodesic equations) This being an example of parallel transport deviation. (curvature).

 

So the equivalence is an expression that holds true only if you treat it at a single point in the gravitational field. Put another way it is locally equivalent. However it isn't equivalent over long distances due to changes in direction.

This seems like a big deal. Was this seen as a big blow against Equivalence? I mean, wouldn't it put it's Principle status in question?

 

So lets throw some numbers in and describe the tidal effect.

 

[latex]F_a=F_b=\frac{mMG}{r^2}[/latex]

 

let x be the distance between a and b, let [latex]\alpha[/latex] represent the angle between one test body and the center line (center of gravity vertical axis) so from frame a, b experiences a force directed toward a.

 

[latex]F=2F_asin\alpha=2F_a*\frac{x}{2r}=\frac{mMG}{r^3}x[/latex]

 

a then observes b to be accelerating towards him by

 

[latex]F=-md^2x/dt^2[/latex]

[latex]\frac{d^2x}{dt^2}=\frac{MG}{r^3}x[/latex]

 

the 1/r^3 is characteristic of tidal forces.

 

That clears up the principle of equivalence a bit.... the key importance is to understand when this relation holds true.

 

the next part will take me a bit to type in...

 

the twin paradox which isn't a paradox at all, essentially can be broken down to the following statement.

 

"we should never have considered the age of a and b to be the same, as the frame of b undergoes accelerations that a does not undergo."

 

so lets look at the acceleration

 

first define the four velocity. [latex]u^\mu[/latex]

[latex]u^\mu=\frac{dx^\mu}{dt}=(c\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau})[/latex]

 

this gives in the SR limit [latex]\eta u^\mu u^\nu=u^\mu u_\mu=-c^2[/latex]

the four velocity has constant length.

 

[latex]d/d\tau(u^\mu u_\mu)=0=2\dot{u}^\mu u_\mu[/latex]

 

the acceleration four vector [latex]a^\mu=\dot{u}^\mu[/latex]

[latex]\eta_{\mu\nu}a^\mu u^\nu=a^\mu u_\mu=0[/latex]

 

so the acceleration and velocity four vectors are

 

[latex]c \frac{dt}{d\tau}=u^0[/latex]

 

[latex]\frac{dx^1}{d\tau}=u^1[/latex]

 

[latex]\frac{du^0}{d\tau}=a^0[/latex]

 

[latex]\frac{du^1}{d\tau}=a^1[/latex]

 

both vectors has vanishing 2 and 3 components.

 

using the equations above

[latex]-(u^0)^2+(u^1)2=-c^2 ,,-u^0a^0+u^1a^1=0[/latex]

 

in addition [latex]a^\mu a_\mu=-(a^0)^2=(a^1)^2=g^2[/latex] Recognize the pythagoras theory element here?

I'm not sure I get it though, how is it that there are two different vectors for acceleration at the turn around. wouldn't it be one long acceleration pointed back towards Earth?

 

Unless... are you describing the rotation of spacetime past a vertex for the Earth Frame? That would be exciting. I've incorporated rotation into a hypothesis Ive been working on and it required 90⁰ rotation in either direction, which I didn't see before now.

 

How does this align with time dilation at an EH?

the last equation defines constant acceleration g. with solutions

 

[latex]a^0=\frac{g}{c}u^1,a^1=\frac{g}{c}u^0[/latex]

 

from which [latex]\frac{da^0}{d\tau}=\frac{g}{c}\frac{du^1}{d\tau}=\frac{g}{c}a^1=\frac{g^2}{c^2}u^0[/latex]

hence

[latex]\frac{d^2 u^0}{d\tau^2}=\frac{g^2}{c^2}u^0[/latex]

similarly

[latex]\frac{d^2 u^1}{d\tau^2}=\frac{g^2}{c^2}u^1[/latex]

 

so the solution to the last equation is

[latex]u^1=Ae^{(gr/c)}=Be^{(gr/c)}[/latex]

hence

[latex]\frac{du^1}{d\tau}=\frac{g^2}{c^2}(Ae^{(gr/c)}-Be^{gr/c)})[/latex]

 

with boundary conditions [latex]t=0,\tau=0,u^1=0,\frac{du^1}{d\tau}=a^1=g [/latex] we find A=_B=c/2 and [latex]u^1=c sinh(g\tau/c)[/latex]

 

so

[latex]a^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex]

hence

[latex]u^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex]

and finally

[latex]x=\frac{c^2}{g}cosh(g\tau /c)[/latex],

[latex] ct=\frac{c^2}{g}sinh(g\tau /c)[/latex]

 

the space and time coordinates then fall onto the Hyperbola during rotation

[latex]x^2-c^2\tau^2=\frac{c^4}{g^2}[/latex]

So regardless of how fast or slow the acceleration is, the rotation itself causes changes to reference frames

this detail is often overlooked, by the statement :"it doesn't matter how fast observer B accelerates or decelerates" most articles don't mention the rotation as to the reasoning behind that statement....

 

A lot of attempted solutions try to use the instant accelerated, decelerated argument but ignore frame rotation.

(as mentioned by Studiot)

 

this is the problem with placing two rotated frames (outgoing,incoming) onto the same timeline, leads to errors...not to mention ignoring the spatial separation components of doppler vs relativistic doppler.

 

edit::oops almost forgot to reference the above...

 

:Introduction to General Relativity" by Lewis Ryder.

 

Its an excellent book, as it explains details in an easy math level (well easy to me lol) step by step. The author assumes the reader has less previous knowledge, than in many other GR textbooks. (thanks Ajb for bringing it to my attention, its a great addition to my huge collection lol)

I'm sorry, I wasn't able to follow the math adequately. Do the spacetime effects all invert on the return trip? I'd think they would have to in order to work over any distance for a single max velocity.

Edited August 28, 2016 by TakenItSeriously

[Mordred]

The elevator portion follows what is predicted by the equivalence principle.

 

If you drew out the path of the two particles in the elevator you can see the deviation from parallel transport.

This deviation is a simple consequence of path taken.

 

For the twin paradox portion. The later math shows the frame rotation on the turn around. You recall the video you posted.

 

Recall the region at 5 seconds that did not have lines? that triangle region that was blank?

 

That is the hyperboloid region the math above derives. Yes the math format includes the GR tensors, which is why it looks complex.

 

The key relation is the rotation between v plus to v minus. Velocity as a vector. Outgoing/incoming. The two different directions count as two different frames of reference.

 

Most ppl just getting into SR would look at the twin paradox and count two reference frames A and B. However B has more than one reference frame. Technically if you get precise. B has an infinite number of reference frames due to various acceleration changes. We can reduce that number by reducing precision.

 

These two scenarios help show why we use velocity not acceleration in time dilation. Acceleration induces reference frame changes which can be infinite in number. The best you can do is reduce that infinite number of reference frames by time slices. (instantaneous velocity) to get an approximation of changes.

 

This is true in both scenarios. Always remember velocity is a vector. Put succinctly a change in direction is a change in velocity. A change in velocity is a change in reference frames.

Edited August 28, 2016 by Mordred

[Tim88]

Based on the widely diverse solutions to the twin paradox, Twin paradox solutions almost seem like an opinion poll and they seem to disagree on certain aspects, so I'm lost on what's the most accepted.

 

Five questions on SR:

[..]

5) is there a specific finding or proof that triggered controversy?

 

Yes of course there were. A recent one (what is it already 5 years ago?!) you can find here and I find it a bit amusing (but surely it was very frustrating for the people involved!):

http://www.telegraph.co.uk/news/science/8782895/CERN-scientists-break-the-speed-of-light.html

[swansont]

Yes of course there were. A recent one (what is it already 5 years ago?!) you can find here and I find it a bit amusing (but surely it was very frustrating for the people involved!):

http://www.telegraph.co.uk/news/science/8782895/CERN-scientists-break-the-speed-of-light.html

This was hardly a controversy. CERN never claimed the result was legitimate.

[Tim88]

This was hardly a controversy. CERN never claimed the result was legitimate.

 

Agreed, controversy does suggest something bigger.

 

Possibly a few in the following list resulted in some controversy: http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Experiments_not_consistent_with_SR

[imatfaal]

 

Agreed, controversy does suggest something bigger.

 

Possibly a few in the following list resulted in some controversy: http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Experiments_not_consistent_with_SR

 

 

"All that being said, I repeat: as of this writing there are no reproducible and generally accepted experiments that are inconsistent with SR, within its domain of applicability."

[Tim88]

 

 

"All that being said, I repeat: as of this writing there are no reproducible and generally accepted experiments that are inconsistent with SR, within its domain of applicability."

 

Yes, exactly: controversial implies disputed, not generally accepted.