Twin Paradox - difference between outbound and inbound

[Enovik]

Lorentz formulas are obtained for one simple case: motion in one direction with constant speed. And they are true for this case only. Travelling twin moves in two direction at least: there and back. And two different formulas are needed at least: for there motin and back motion. Time accelerates, if twin moves away, and time slows down, if twin moves back. Is that so?
Thank you.

 

[PeroK]

Time accelerates, if twin moves away, and time slows down, if twin moves back. Is that so?
Thank you.

That's not right. It's fundamentally about spacetime geometry.

I suggest you start a new thread if you want to ask a question about relativity.

 

[Ibix]

Lorentz formulas are obtained for one simple case: motion in one direction with constant speed.

No. If you have a set of clocks and rulers moving along together at the same constant speed, and another set of clocks and rulers moving along together at another constant speed, the Lorentz transforms describe the relationship between measurements made with one set of instruments and those made by the other set. They are a generalisation of the Galilean transforms that do the same thing (approximately) at low speeds.

You can use the Lorentz transforms to produce a simple formula (the time dilation formula) that someone using one set of clocks and rulers can use to determine the tick rate of a clock in arbitrary motion. However, the converse is not true; there is no simple formula letting a clock in arbitrary motion describe the tick rate of any other clock (there is a formula, just not a simple one). The Twin Paradox is a demonstration of the truth of those statements: if you blindly apply the simple formula when you aren't moving at constant speed then you end up with a wrong answer.

And two different formulas are needed at least: for there motin and back motion.

No. There are several ways to resolve the paradox, but the intended learning point is that simple formulas don't work except in simple cases.

Time accelerates, if twin moves away, and time slows down, if twin moves back.

No. The Doppler shift dominates direct observations, and means that you will see a clock ticking slowly as it moves away from you and more quickly as it moves towards you, but this is simply the same phenomenon as a car engine note sounding different as it approaches you, passes you, and recedes from you.

 

[Enovik]

Dear sirs, thank you for your answers.
As far I understand you think, that moving point time will be slow from the stationary point time at any rest. And it will be so if moving point is going to stationary point also from stationary point. OK!
But let us consider two simple scenarios.
View attachment 346925
View attachment 346926
The first case is the base for Lorentz common formulas. But it is the only one scenario from many.
What is the sense of the symbol "v" in Lorentz formulas?
There is my mistaken?
Thank you

 

[Ibix]

There is my mistaken?

The formula you have in the diagram is $$t' =t\sqrt{\frac{1-v/c}{1+v/c}}$$which is not the Lorentz transform. It is the Doppler effect, which includes both the time dilation effect and the distance change effect. So it does not relate time in two frames, but rather relates emission rate to arrival rate of light pulses. That is a different thing, and it does depend on whether the emitter and receiver are moving apart. In the form you supplied ($t'$ is the period of the emitter and $t$ is the period measured by the receiver), $v>0$ implies the two are moving apart and $v<0$ implies they are moving towards each other.

You can do a Doppler analysis of the Twin Paradox, and it's a good way to convince yourself that the traveller's non-inertial frame isn't equivalent to the stay-at-home's inertial one. But it is not an anslysis using the Lorentz transforms or time dilation, and shouldn't be confused with one.

 

[Enovik]

The formula you have in the diagram is $$t' =t\sqrt{\frac{1-v/c}{1+v/c}}$$which is not the Lorentz transform. It is the Doppler effect, which includes both the time dilation effect and the distance change effect.

Lorentz formula $$t' =\frac{t-vx/c^2}{\sqrt{1-v^2/c^2}} = t\sqrt{\frac{1-v/c}{1+v/c}}$$ if x=ct.
And equalities x=ct and x'=ct' are the base conditions for Lorentz formula derivation.
So, equation $$t' =t\sqrt{\frac{1-v/c}{1+v/c}}$$ is quite right Lorentz formula,
Value t' is the time in the point, approaching to the stationary one.
If the point is moving with the same speed v ( module) away from the stationary one, then $$t' =t\sqrt{\frac{1+v/c}{1-v/c}}$$.
This are simple kinematics results.
And what do you name Doppler effect? Dilation effect and distance change effect are not related to Doppler effect. It are simple school kinematics effects.
I think you have studied Wikipedia physics. Sorry, but I do not know such science.

 

[Ibix]

And equality x=ct is the base condition for Lorentz formula derivation.

I have no idea where you get that from. It's wrong.

Value t' is time in the point, approaching to the stationary one.

Except that you added the constraint that $x=ct$, which means that your formula does nothing except relate the time coordinates in the two frames along that arbitrarily chosen line. Note that it's a null line, so no clocks can travel along it, so this formula does not relate the times on any one clock to anything else.

And what do you name Doppler effect?

The formula you gave is correct for the relativistic Doppler effect. Apparently that was just by chance.

I think you have studied Wikipedia physics.

Sure. You go on believing that.

 

[PeroK]

Sorry, but I do not know such science.

You do not know relativity, I think we can all agree on that.

 

[Enovik]

You do not know relativity, I think we can all agree on that.

Oh yes. It is not possible to understand, it is possible to believe only.

 

[Ibix]

It is not possible to understand,

Please don't confuse you not understanding with everybody else not understanding. If you want to learn, there are a lot of knowledgeable people here who can help you. If you want to believe "relativity is a cult" then you're wrong, and I'd suggest re-reading the site rules (specifically the part about "non-mainstream theories") before posting further.

 

[Enovik]

I have no idea where you get that from. It's wrong.

Except that you added the constraint that $x=ct$, which means that your formula does nothing except relate ...

The conditions x=ct and x'=ct' are used in any Lorentz formulas derivation. This conditions are consequences from the second STR postulate.

 

[Ibix]

The conditions x=ct and x'=ct' are used in any Lorentz formulas derivation.

Wrong. See, e.g. Palash Pal, "Nothing but relativity".

This conditions are consequences from the second STR postulate.

Wrong. Requiring $x=ct$ implies everything travels at lightspeed. Perhaps even "nothing exists except at the single point $x=ct$".

I suspect you've come across a derivation of the Lorentz transforms that starts with the null interval (like, ironically, the one at Wikipedia) and misunderstood it.

 

[Enovik]

Please don't confuse you not understanding with everybody else not understanding. If you want to learn, there are a lot of knowledgeable people here who can help you. If you want to believe "relativity is a cult" then you're wrong, and I'd suggest re-reading the site rules (specifically the part about "non-mainstream theories") before posting further.

I do not deny relativity. I want to know what is the sense of the symbol v in Lorentz formulas.
Is it possible to choice true answer from the next?
1. v is the module of vector speed, which can be directed in any direction.
2. v is the velocity of the moving away along axis x.
3. v is any velocity along axis x, it may be moving away and moving back, v>0 in both cases.
4. v is any velocity along axis x, it may be moving away and moving back. v>0 in the away case, v<0 in the back case.
5. It is not a tactful question.
Thank you.

 

[Ibix]

I want to know what is the sense of the symbol v in Lorentz formulas.

$v$ is the velocity of one frame with respect to the other. By convention both frames have their $x$ axes in the same direction and parallel to the relative velocity of the primed frame with respect to the unprimed frame, so it is in the $+x$ direction. You can make other choices and construct Lorentz transforms for arbitrary frame velocities, but it just makes the maths messier; better to just rotate your coordinate system so that the above is true.

You will not get any sense out of this if you arbitrarily impose a requirement that $x=ct$ on top, though. And none of this imposes any restrictions on the direction of motion of any object you wish to describe with these frames.

 

[Dale]

I want to know what is the sense of the symbol v in Lorentz formulas.

The $v$ in the Lorentz transform is the relative velocity between two specified inertial frames.

The usual formula that you are probably familiar with has the $x$ and $x’$ axes aligned and the $y,z$ axes parallel to the $y’,z’$ axes. This is called the standard configuration. However, there are more general formulas available which have the axes in arbitrary orientations.

In the standard configuration $v$ can be any real number, positive or negative. In the general formulas $v$ can be any real 3D vector.

The concepts of “towards” and “away” don’t make any sense in this context. Inertial frames cover all of space and all of time. So if you pick any $x$ there are an infinite number of $x’$ going towards it and an infinite number of $x’$ going away from it, regardless of $v$.

$x=ct$ is a pulse of light emitted from the origin in the positive $x$ direction. It is not a general condition for the Lorentz transform.

 

[phinds]

It is not possible to understand, it is possible to believe only.

As @Ibix has already pointed out, this is nonsense, it's just that YOU don't understand it. The attitude that no one CAN understand it will lead you astray.

I can sympathize with your finding it a bit hard to figure out, but keep trying.

 

[Ibix]

It's perhaps worth explaining what the Lorentz transforms really mean. Simply draw a displacement-time graph (by convention in relativity, time goes up the page, not across). Since we interpret space and time as a single 4d-manifold, we now interpret this as a map of a "slice" through spacetime. As with any map, we can draw a coordinate grid:

So far, so boring. But if you look at the Lorentz transforms (without adding any $x=ct$ nonsense) then you find that the $t'=0$ and $x'=0$ lines are straight lines through the origin with slope $1/v$ and $v$ respectively. That means we can draw the primed coordinate grid on this diagram (picking $v=0.6c$ in this case):

I've removed the black coordinate grid for clarity. Note that the red lines are orthogonal to each other in the Minkowski sense. To see this, consider a vector along the $t'$ axis expressed in terms of the unprimed coordinate system, $(v\Delta t,\Delta t)^T$, and a vector along the $x'$ axis, $(\Delta x,v\Delta x)^T$ (note that I've picked units where $c=1$ and suppressed the $c$s - you can put them back in by dimensional analysis if you want). The Minkowski sense of orthogonal is that the product of the x components minus the product of the t components is zero, and sure enough: $(v\Delta t)(\Delta x)−(\Delta t)(v\Delta x)=0$. The axes only appear non-orthogonal on the diagram because the page obeys Euclidean geometry and we cannot represent a Minkowski geometry plane exactly accurately.

Since the primed axes are orthogonal we can draw the same diagram using that frame, making the black unprimed frame appear skewed (and suppressing the red grid this time):

All that the Lorentz transforms do is this: pick a point on the chart (representing an event in spacetime) and read off its coordinates using the black grid and then the red grid. Apply the Lorentz transforms (with $v=0.6c$ in the case I've drawn) to the $(x,t)$ values you read from the black grid and you will get the values you read from the red grid.

And now you can see what is: it's the inverse gradient of the time axis of the primed coordinate system in the unprimed coordinates system, and is also the speed the unprimed coordinate system assigns to objects at rest in the primed frame.

The whole of the content of special relativity is working out the implications of the "natural" coordinates transforming like the above instead of the Galilean transforms, which skew the coordinate grids in different way. The equivalent to the second and third images, if Newtonian physics were anywhere near correct at $0.6c$ would be these two:

Note that in these last two diagrams the coordinate grids share a definition of "space at one time" (the spatial axes of both frames are parallel) and a clock tick rate (the time coordinate labels are at the same heights). Neither of these apply in relativity.

 

[Enovik]

$v$ is the velocity of one frame with respect to the other. By convention both frames have their $x$ axes in the same direction and parallel to the relative velocity of the primed frame with respect to the unprimed frame, so it is in the $+x$ direction. You can make other choices and construct Lorentz transforms for arbitrary frame velocities, but it just makes the maths messier; better to just rotate your coordinate system so that the above is true.

Thank you. It is the expecting answer.
Some more questions, please. How to use Lorentz transforms if frames are approaching along the axis x? Is it needed to use v<0 in Lorentz transforms, or v>0 at any rate?
And did you saw Lorentz formulas derivation for the approaching frames?

 

[Ibix]

It is the expecting answer.

Odd that it was none of the five options you suggested, then.

How to use Lorentz transforms if frames are approaching along the axis x?

As Dale said, "approaching" makes no sense in this context. This is a relationship between frames that overlap everywhere, not between bodies that have specific locations in space.

Assuming the standard configuration (others are possible), positive $v$ means that any point that is fixed in the primed frame is moving in the $+x$ direction according to the unprimed frame. Negative $v$ means that any point that is fixed in the primed frame is moving in the $-x$ direction according to the unprimed frame. Neither of these imply "approaching" in any meaningful sense.

 

[PeroK]

The conditions x=ct and x'=ct' are used in any Lorentz formulas derivation. This conditions are consequences from the second STR postulate.

$x = ct$ mathematically is a single line through the origin. Physically it represents a specific light signal.

The Lorentz Transformation is a coordinate Transformation for all of spacetime. Mathematically this is an example of a Linear Transformation (applied to any $x$ and $t$).

 

[Enovik]

Odd that it was none of the five options you suggested, then.

You answer coincides with my second variant:

2. v is the velocity of the moving away along axis x.
The moving away in -x direction is not a new scenario.

 

[Ibix]

2. v is the velocity of the moving away along axis x.

But, as we have repeatedly told you "away" and "apart" and "approaching" are meaningless here. So that sentence is nonsense, and not the same as what I wrote.

 

[Enovik]

... Neither of these imply "approaching" in any meaningful sense.

Lorentz formulas are one way ticket? What will you say travelling twin on the back way? What will show his watch?

 

[Ibix]

Lorentz formulas are one way ticket? What will you say travelling twin on the back way? What will show his watch?

People are not frames. People can be approaching one another or moving apart. Frames cannot do either. And the $v$ relates to frames, not people.

 

[Enovik]

People are not frames. People can be approaching one another or moving apart. Frames cannot do either. And the $v$ relates to frames, not people.

Relativity acts on frames only? OK! Let's navigate and calculate in one stationary frame and our watches and sizes still invariable at any speed. Is that so?

 

[Ibix]

Sorry, this reference doesn't contain any formulas

There are 29 numbered formulas in the paper.

My computer doesn't open it (for safety).

Download the TeX source if you don't trust PDFs.

The rest of your post is simply nonsense. You pull random formulae out of nowhere with no justification and claim that they somehow mean something. Show what (you think is) an actual derivation, and then we can explain whatever you are misunderstanding that leads to the nonsense you keep getting.

If you know another result write, please, formulas, not a row of references

In an inertial frame, an object moving inertially has equation of motion $x=ut+x_0$. In another inertial frame it has $x'=u't'+x'_0$. This is a definition of an inertial frame. Therefore, if there exist global inertial frames any transform between them must map straight lines to straight lines. The most general transform that does this is an affine transform. The transforms can therefore be written in matrix form$$\begin{eqnarray*}
\vec x'&=&\Lambda\vec x\\
\left(\begin{array}{c}x'\\t'\end{array}\right)
&=&
\left(\begin{array}{cc}A&B\\C&D\end{array}\right)
\left(\begin{array}{c}x\\t\end{array}\right)
\end{eqnarray*}$$where the capital letters are unknowns that may be functions of the velocity between the frames but cannot be functions of the coordinates (or the transform would not have the required properties).

Now note the following things.

1: if the velocity of the primed frame is $v$, an object moving at velocity $v$ in the unprimed frame must have speed zero under transform. That is, $(x,t)^T=(vT_1,T_1)^T$, where $T_1$ is an arbitrary time, must map to $(x',t')^T=(0,T'_1)^T$.

2: Similarly, the inverse transform must map $(x',t')^T=(-vT'_2,T'_2)^T$, where $T'_2$ is an arbitrary time, to $(x,t)^T=(0,T_2)^T$.

3: If you set $T_1=T'_2$ then $T'_1$ must equal $T_2$.

The first of these is the definition of "moving at velocity $v$" for an inertial frame. The second and third are applications of the principle of relativity (Einstein's first postulate). If you work through the maths, you will find that you can express $B$, $C$ and $D$ in terms of $A$ and $v$.

Now add a new requirement that a composition of these transforms must be a transform too. Add a subscript $v$ to get $\Lambda_v$ whose elements are all functions of $A_v$ and $v$, and define a transform $\Lambda_u$ whose elements are obtained from $\Lambda_v$ by changing all $v$s to $u$s. This is the same type of transform but to a different velocity, $u$, Write the composition, $\Lambda_w=\Lambda_u\Lambda_v$, which is a boost to a frame with some velocity $w$ relative to the unprimed frame. Multiply that out. You will have found that the leading diagonal components of $\Lambda_v$ are equal, so insist on this for $\Lambda_w$. You will get a messy equality that you can tidy up so that all the $u$ and $A_u$ are on one side and all the $v$ and $A_v$ are on the other. It must be true for arbitrary $u$ and $v$, so pick $A$ to eliminate the dependency.

You will find three ways to do this: $A=0$, $A=1$, and $A=(1-Ev^2)^{-1/2}$, where $E$ is an unknown constant. The first is a spurious solution that makes the matrix singular. The second and third produce the Galilean and Lorentz transforms (albeit with an unknown constant). The sole application of the second postulate is here: the Galilean transforms do not respect it so we reject them, and the Lorentz transforms only do so if $E=1/c^2$, so we require this.

That concludes the derivation of the Lorentz transforms, reasoning from first principles. Note that none of your formulas 1-4 will appear (because they have nothing to do with the Lorentz transforms), although 5 and 6 are one option for deducing the value of $E$.

 

[Ibix]

Relativity acts on frames only?

The Lorentz transforms are simply transforms between frames. As I already wrote, relativity is the implications of those being the correct transforms.

Let's navigate and calculate in one stationary frame

You certainly can do all physics in SR working in one frame.

our watches and sizes still invariable at any speed.

That would be Galilean relativity.

If you want to work out the implications of the Lorentz transforms for bodies in motion you can do so. You cannot, however, do it by ramming in an irrelevant $x=ct$ and then trying to repurpose the $v$ as the speed of the body, which is what you have been doing.

 

[Dale]

Thank you. It is the expecting answer.
Some more questions, please. How to use Lorentz transforms if frames are approaching along the axis x? Is it needed to use v<0 in Lorentz transforms, or v>0 at any rate?
And did you saw Lorentz formulas derivation for the approaching frames?

As you have already been told, “approaching” makes no sense in this context. An inertial frame covers all of space and all of time. Objects can approach or recede, but not frames.

The sign of $v$ is straightforward to determine, as follows: If an object at rest in the primed frame has a negative velocity in the unprimed frame then $v$ is negative.

 

[Dale]

Sorry, this reference doesn't contain any formulas. It contains a row of another references only. My computer doesn't open it (for safety).
On the other hand, Lorentz formulas may to be obtained by simplest calculations in some lines. And it is published in many reviewed manuals of physics.
The main results are:
1. Time t' on the moving point, approaching with speed v to the stationary one, is slowed down and calculated by formula:
(1) t' = t *((1-v/c)/(1+v/c))^1/2
2. All Lengths l' in the moving direction is reduced and calculated by formula:
(2) l' = l * ((1-v/c)/(1+v/c))^1/2
Equalities (1), (2) are obtained by use the second STR postulate, which is written as:
(3) l = ct
(4) l' = ct'
Equalities (1)- (4) are the base for Lorentz formulas.
Build stationary frame with axis x along motion, and moving frame with axis x' along motion.
Then we have:
(5) x = ct
(6) x' = ct'
We will obtain Lorentz formulas if will set (5), (6) to (1), (2). It means that Lorentz formulas are right only by conditions (5), (6).
Lorentz formulas can't to transfer any (x,t) to (x',t').
If you know another result write, please, formulas, not a row of references.

Essentially none of this is correct, as you have already been told. Please, before posting further, review the rules regarding personal speculation and unpublished theories.

References are preferred over formulas because

1) the reference shows that the formula is not personal speculation but is part of the professional scientific literature

2) the reference provides details and background that are impractical to repeat in a post

3) the reference helps people become familiar with reading the professional scientific literature

When posting formulas instead of references, please use LaTeX to make the formulas more readable.

 

[Ibix]

The another "solution" is:
A= ((1-v/c)/(1+v/c))^1/2,
D= ((1-v/c)/(1+v/c))^1/2
B=0
C=0

That transformation is a simple scaling. It represents a choice to measure distance in units that are $\sqrt{(c+v)/(c-v)}$ times the length of a meter (or whatever distance unit you used in your unprimed coordinates) and time in units the same factor larger than a second (or whatever unit). You are, of course, at liberty to do this, but it's not a frame change and it's pointless complication.

At this point I think I'm going to bow out of this thread. I've sketched a correct derivation for you, drawn explanatory diagrams, and linked to experimental evidence supporting relativity. You seem determined to do things wrong and uninterested in learning how to do them right.

 

[Dale]

The non correct task is "solved" in the article. One "solution" is shown there.
The another "solution" is:
A= ((1-v/c)/(1+v/c))1/2,
D= ((1-v/c)/(1+v/c))1/2
B=0
C=0

That is indeed a linear transform, but it is not a boost of any sort and it is specifically not a Lorentz transform. As @Ibix mentioned, it is a rescaling transform. I cannot think of what this transformation would be useful for, since there doesn’t appear to be any natural meaning for $v$ in that context.

Please remember that all posts should be consistent with the professional scientific literature. If you believe this transform deserves further consideration, please post the peer reviewed scientific publication that derives and explains it.

 

[PeterDonis]

Lorentz formulas are one way ticket?

No, every Lorentz transform has an inverse.

What will you say travelling twin on the back way? What will show his watch?

In the stay-at-home twin's frame, the traveling twin is moving at speed $- v$ on the return leg. So just do the Lorentz transform with $-v$ instead of $v$.

 

[Enovik]

No, every Lorentz transform has an inverse.


In the stay-at-home twin's frame, the traveling twin is moving at speed $- v$ on the return leg. So just do the Lorentz transform with $-v$ instead of $v$.

OK. And Lorentz transform will give the speeding-up of the time $t'$. Is that so? And there is the derivation for the appropriate formulas?

 

[PeterDonis]

And Lorentz transform will give the speeding-up of the time $t'$. Is that so?

No.

And there is the derivation for the appropriate formulas?

Look in any relativity textbook. Or in any number of places online. This information is easy to get.

 

[Sagittarius A-Star]

OK. And Lorentz transform will give the speeding-up of the time $t'$. Is that so? And there is the derivation for the appropriate formulas?

Note down the Lorentz transformation for differences between the coordinates of 2 events:
(1) $\ \ \ \ \Delta x' = x_2' - x_1' = \gamma (x_2-vt_2) - \gamma (x_1-vt_1) = \gamma (\Delta x-v \Delta t)$
(2) $\ \ \ \ \Delta t' = t_2' - t_1' =\gamma(t_2 - \frac{v}{c^2}x_2) - \gamma(t_1 - \frac{v}{c^2}x_1) = \gamma(\Delta t - \frac{v}{c^2}\Delta x)$
with $\gamma=1/\sqrt{1-v^2/c^2}$.

Source:
http://www.scholarpedia.org/article...nematics#Galilean_and_Lorentz_transformations

If the 2 events are ticks of a clock at rest in the unprimed frame, then $\Delta x = 0$.
Time dilation with respect to the primed frame of the clock with $\Delta x = 0$ in equation (2):
$\Delta \tau/\Delta t' = \Delta t_{\text{clock}}/\Delta t' = 1/ \gamma$.

The time dilation-factor is independent of the direction, in which the clock moves. The $\gamma$ factor does not depend on the sign of $v$.