Martillo talks about the twin paradox

[martillo]

Tach. You're very rude.
And mean.
You're obviously a bad man. I'm fairly certain that kittens tremble in fear around you.
Now, Martillo...:
...
...
...


Well, I will not call you a crank, Sir. I wouldn't stoop so low. I'm not a jerk like Tach. I will call you a douche, instead.

Wouldn't it have been easier to just do the math, see the explanation and admit to not having understood the issue?


Mods: I went overboard and threw a long Ad Hom attack. I don't deny it.

With no doubt you are the one rude, isn't it?
But doesn't hurt you know, your words just show who you really are. May be I could find a proper one to call you but no time at the moment.

I will comment your final sentence the only one relevant to the topic of the thread just to give the answer on why y continue finding the twins' paradox as a contradiction/inconsistency in Relativity.

I have already seen the done math in this thread, also the math in the solution to the classic twins paradox where one stays at Earth and I have even discussed a lot about it in forums in the past. I always find the following error/mistake which seems to be stubbornly ignored and which is not a mathematical problem: The real problem is not to find out which would be the right solution to the age of the twins at the end of the trip, the problem is what each twin observes during the trip. Whatever the math you can do there's something cannot be avoided and is that during the trip each twin (with a "rest" reference attached to him) will see the other(s) twins travelling at some velocity and so aging less. This is the real point on the problem. All twins, from their own frame, have contradictory observations of the same phenomenon involving all the participants. It gets much more clear if in spite of reading clocks we allow them to see recorded images of what they will observe on themselves and on the other(s) twin(s) about their aging at any instant and compare them. The true is that different and contradictory observations are obtained. No hard math is actually needed to verify this.

The other problem I know by experience exist is that most people in forums do trust Relativity...

 

[Tach]

Whatever the math you can do there's something cannot be avoided and is that during the trip each twin (with a "rest" reference attached to him) will see the other(s) twins travelling at some velocity and so aging less.

This is FALSE. I can easily prove to you that the accelerated twin sees the inertial one aging MORE.
Look, Martillo
You have been spewing your antirelatvistic garbage on many forums, for many years now. Why don't you learn something instead of wasting your life away.

 

[martillo]

This is FALSE. I can easily prove to you that the accelerated twin sees the inertial one aging MORE.

That doesn't sound right.
Do the otherones here agree with this?

Why don't you learn something instead of wasting your life away.

Why is that I feel the same about all of you?

 

[origin]

This is FALSE. I can easily prove to you that the accelerated twin sees the inertial one aging MORE.

That doesn't sound right.
Do the otherones here agree with this?.

Of course.

Look, Martillo
Why don't you learn something instead of wasting your life away.

Why is that I feel the same about all of you?

Why do you feel that way? I believe Neverfly rather indelicately pointed that out already.

 

[Tach]

That doesn't sound right.
Do the otherones here agree with this?

I really don't care that "it doesn't sount right". I really don't care how it "sounds" to you.
To calculate the elapsed time $$\Delta t$$ of the inertial observer K as a function of the elapsed time $$\Delta \tau$$ of the non-inertial observer K', the following formula can be used

$$\Delta t^2 = \left[ \int^{\Delta\tau}_0 e^{\int^{\bar{\tau}}_0 a(\tau')d \tau'} \, d \bar\tau\right] \,\left[\int^{\Delta \tau}_0 e^{-\int^{\bar\tau}_0 a(\tau')d \tau'} \, d \bar\tau \right], \ $$
where $$a(\tau)$$ is the proper acceleration of the non-inertial observer K' . The Cauchy–Schwarz inequality shows that:
$$\begin{align}
\Delta t^2 & = \left[ \int^{\Delta\tau}_0 e^{\int^{\bar{\tau}}_0 a(\tau')d\tau'} \, d \bar\tau\right] \,\left[\int^{\Delta \tau}_0 e^{-\int^{\bar\tau}_0 a(\tau')d \tau'} \, d \bar\tau \right] \\
& > \left[ \int^{\Delta\tau}_0 e^{\int^{\bar{\tau}}_0 a(\tau')d\tau'} \, e^{-\int^{\bar\tau}_0 a(\tau') \, d \tau'} \, d \bar\tau \right]^2 = \left[ \int^{\Delta\tau}_0 d \bar\tau \right]^2 = \Delta \tau^2.
\end{align}$$

That is:

$$\Delta t > \Delta \tau$$

 

[przyk]

That doesn't sound right.
Do the otherones here agree with this?

What he's saying is related to a well known result that is nowadays called "gravitational time dilation". Basically, if you accelerate toward an object, it starts ageing very rapidly in your accelerating reference frame, and the degree with which that happens increases both with magnitude of acceleration and distance.

The name "gravitational" time dilation comes from how this effect carries over to gravity via general relativity's equivalence principle. According to the equivalence principle, gravitational fields like the one we feel on Earth are locally indistinguishable from accelerating reference frames, so if we get this extra time dilation effect in accelerating frames we also expect objects higher up in a gravitational field to age more rapidly. This is a famous (and experimentally observed) prediction of general relativity.

 

[brucep]

What he's saying is related to a well known result that is nowadays called "gravitational time dilation". Basically, if you accelerate toward an object, it starts ageing very rapidly in your accelerating reference frame, and the degree with which that happens increases both with magnitude of acceleration and distance.

The name "gravitational" time dilation comes from how this effect carries over to gravity via general relativity's equivalence principle. According to the equivalence principle, gravitational fields like the one we feel on Earth are locally indistinguishable from accelerating reference frames, so if we get this extra time dilation effect in accelerating frames we also expect objects higher up in a gravitational field to age more rapidly. This is a famous (and experimentally observed) prediction of general relativity.

Nice post. As an 'otherone' I agree.

 

[mikelizzi]

What he's saying is related to a well known result that is nowadays called "gravitational time dilation". Basically, if you accelerate toward an object, it starts ageing very rapidly in your accelerating reference frame, and the degree with which that happens increases both with magnitude of acceleration and distance.

The name "gravitational" time dilation comes from how this effect carries over to gravity via general relativity's equivalence principle. According to the equivalence principle, gravitational fields like the one we feel on Earth are locally indistinguishable from accelerating reference frames, so if we get this extra time dilation effect in accelerating frames we also expect objects higher up in a gravitational field to age more rapidly. This is a famous (and experimentally observed) prediction of general relativity.

Nicely stated. What a relief to read a correct statement about time and SR for a change. There is another relativity forum where so-called experts are claiming there is no way that one can conclude clocks run faster in SR.

 

[martillo]

I really don't care that "it doesn't sount right". I really don't care how it "sounds" to you.
To calculate the elapsed time $$\Delta t$$ of the inertial observer K as a function of the elapsed time $$\Delta \tau$$ of the non-inertial observer K', the following formula can be used

$$\Delta t^2 = \left[ \int^{\Delta\tau}_0 e^{\int^{\bar{\tau}}_0 a(\tau')d \tau'} \, d \bar\tau\right] \,\left[\int^{\Delta \tau}_0 e^{-\int^{\bar\tau}_0 a(\tau')d \tau'} \, d \bar\tau \right], \ $$
where $$a(\tau)$$ is the proper acceleration of the non-inertial observer K' . The Cauchy–Schwarz inequality shows that:
$$\begin{align}
\Delta t^2 & = \left[ \int^{\Delta\tau}_0 e^{\int^{\bar{\tau}}_0 a(\tau')d\tau'} \, d \bar\tau\right] \,\left[\int^{\Delta \tau}_0 e^{-\int^{\bar\tau}_0 a(\tau')d \tau'} \, d \bar\tau \right] \\
& > \left[ \int^{\Delta\tau}_0 e^{\int^{\bar{\tau}}_0 a(\tau')d\tau'} \, e^{-\int^{\bar\tau}_0 a(\tau') \, d \tau'} \, d \bar\tau \right]^2 = \left[ \int^{\Delta\tau}_0 d \bar\tau \right]^2 = \Delta \tau^2.
\end{align}$$

That is:

$$\Delta t > \Delta \tau$$

I didn't considered accelerated frames. The symmetric twins' paradox preciselly avoids the problem of the effect of acceleration since the twins experience exactly symmetric travels and so exactly same relativistic effects.

Anyway, you calculated the elapsed time from one frame's perspective of observation. To see the key feature of the paradox now you must interchange roles and consider the other frame as "at rest" and the first one moving. You will find exactly the same relation but now for the symmetric elapsed times what means the opposite result. There is the paradox.
As I mentioned in the above post always the error/mistake is to not consider all observations possible in all considered frames and compare observations. You made one observation, the symmetric other one is missing.

 

[Tach]

Tach, I know that you are not stupid,

But I know that you are. Not only stupid but dishonest as well, you keep trying to cover your basic mistakes.

My question was explicitly about how time dilation would be different for the two travelers,

There is no time dilation, get with the program. Learn about elapsed proper time.

 

[Tach]

Anyway, you calculated the elapsed time from one frame's perspective of observation. To see the key feature of the paradox now you must interchange roles and consider the other frame as "at rest" and the first one moving. You will find exactly the same relation but now for the symmetric elapsed times what means the opposite result. There is the paradox.

Nope, the other way around, contrary to your crank claims, you get exactly: $$\Delta \tau < \Delta t$$

 

[przyk]

There is another relativity forum where so-called experts are claiming there is no way that one can conclude clocks run faster in SR.

Well in fairness gravitational time dilation is really an artifact of how coordinate systems (and in particular simultaneity) are defined in relativity. It's a bit like standing on a clear night with the moon right in front of you, then turning 90 degrees on the spot and finding the moon on your left, and saying it has "moved" half a million kilometres in the (rotating) reference frame attached to you. Gravitational time dilation isn't actually directly seen or measured by the accelerating twin. They could only deduce and describe it in their accelerating coordinate system.

martillo seems really interested in what both twins actually see, which is actually the Doppler shift factors rather than the relativistic time dilation factors. So let's work that out.

In SR, if either of the twins looks at the other through a telescope, say, then what they actually see is also affected by the fact that light takes more and more or less and less time to travel between them. The result is that each twin sees the other apparently age at the rate $$\sqrt{\frac{1 - v/c}{1 + v/c}$$, and not by $$1/\gamma = \sqrt{1 - v^{2}/c^{2}}$$, where $$v$$ is the velocity they're moving apart.

Suppose we look at the scenario where, according to the twin on Earth, the plan is for the travelling twin to move away at velocity $$v$$ for a time $$T$$, then come back at velocity $$-v$$. So when they meet up, the earthbound twin has aged $$2T$$ and the travelling twin by $$2\tau = 2T/\gamma$$. This is the result we expect.

What does the travelling twin see? They see their Earthbound twin apparently age at the rate $$\sqrt{\frac{1 - v/c}{1 + v/c}$$ while they've moving away from Earth. Then, when they've aged by $$\tau = T/\gamma$$, they turn back. During the trip back, during which they age $$\tau$$ more, they see their Earthbound twin apparently age at the rate $$\sqrt{\frac{1 + v/c}{1 - v/c}$$. So when they return they expect their Earthbound twin to have aged by

$$\sqrt{1 - v^{2}/c^{2}} T \biggl( \sqrt{\frac{1 - v/c}{1 + v/c}} \,+\, \sqrt{\frac{1 + v/c}{1 - v/c}} \biggr) \,,$$​

which using a bit of algebra (and the fact that $$\sqrt{1 - v^{2}/c^{2}} = \sqrt{1 + v/c} \sqrt{1 - v/c}$$), one can easily check simplifies to $$2T$$ - exactly as expected.

What does the Earthbound twin see? Their travelling twin turns around after a time $$T$$, but by that point the twin is a distance $$D = vT$$ away, and light from that far away takes an additional time $$D/c = (v/c) T$$ to reach Earth, so the Earthbound twin only sees their travelling twin turn around and start heading back for Earth after a total time of $$(1 + v/c)T$$, and they see their twin finally return to Earth an additional time $$(1 - v/c)T$$ later. So by the time the travelling twin returns to Earth, their age should be

$$(1 + v/c) T \sqrt{\frac{1 - v/c}{1 + v/c}} \,+\, (1 - v/c) T \sqrt{\frac{1 + v/c}{1 - v/c}} \,,$$​

which simplifies to $$2 \sqrt{1 - v^{2}/c^{2}} T$$ - again exactly as expected.

Both twins always see reciprocal apparent time dilation factors when they look at each other: $$r_{1} = \sqrt{\frac{1 - v/c}{1 + v/c}$$ when the other is receding and $$r_{2} = \sqrt{\frac{1 + v/c}{1 - v/c}$$ when the other is approaching. The thing that breaks the symmetry is that when the travelling twin turns their rocket around, they see their Earthbound twin's apparent ageing rate change from $$r_{1}$$ to $$r_{2}$$ immediately, while the Earthbound twin only sees the change from $$r_{1}$$ to $$r_{2}$$ in their travelling twin after a delay because of the time it takes light to reach Earth from the twin's rocket in deep space.

 

[Emil]

martillo, these guys do not understand basic physics. How can you ask that they understand SRT?

Let's recap SRT.
The condition you can apply SRT is the two postulates to be true.
Between two inertial reference frames, which have a speed relative to each other, there is a time dilation and length contraction.
Physical quantities involved are time, distance (length) and speed, their derived.

But this leads to the paradox of time.
What were they thinking? Let's modify what Einstein said and there is no need for two inertial reference frames, is sufficient only one and they introduces the concept of acceleration.
Now they say that this is the "mainstream".

They do not even realize how they destroy SRT.
I quote my post no 70

Yes, and I think in terms strictly SR, you're right.

That we can analyze, I give four scenarios and please say your opinion on them.

scenario 1:
a and b in the same place at rest. (a and b have equal masses)
A force acting upon b during one month, which finally lead to a speed V.
After a month, the force acting on b reverses the direction ,
during two months.
Then the force changes direction again for a month and now b is found at original place together with a.
(b always had an accelerated motion)

scenario 2:
Same as scenario 1 with the difference:
after a month of acceleration, for one year does not act upon b any force, so there is a movement with a constant speed relative to b, during one year.
Same difference when b returns.
(So b had a motion relative to a with constant speed V, during two years.)

scenario 3:
a and b in the same place at rest. (a and b have equal masses)
Two forces F / 2 but with opposite directions, one acting on a, and the other acting on b,during one month.
After a month, the forces disappears and a and b are moving with constant speed V relative to each other during one year.
Now the two forces (F/2) start reverse acting upon a and b, during two months.
Same on the way back, they move with constant speed during a year, but now is approaching, then reduce speed during one month and they meet each other.

scenario 4:
Like scenario 3), but the two forces are not equal, but their sum is still F.
For instance 2F / 3 and F / 3.

Every scenario has a single question. When a and b are meeting again who is "the older" and who is "the younger".

If you notice in the scenarios 2,3 and 4 have identical parts when a and b have identical relative speed between them, namely V.
But they say the time dilation is not identical in these scenarios due to different accelerations.

So if there are two reference frames, with a relative speed between them, I can not apply SRT because I also need to know the previous conditions(accelerations).
And now the question arises of these previous conditions (accelerations), up to go? Up to Big Bang?

 

[martillo]

Nope, the other way around, contrary to your crank claims, you get exactly: $$\Delta \tau < \Delta t$$

May be you didn't considered the symmetry properly. I will try to make it clear but I will use the more common case where the moving twin age less to mantain the equivalence with the problem as proposed by the author of the thread.
Remember as both twins make a totally symmetric travel exactly the same relativistic effects on them are considered.

Situation 1:
Twin A "at rest" and so with attached frame " K " and twin B moving and so attached frame " K' ". Elapsed time for twin A: " dt ". For twin B: " dt' ".
You get dt'<dt and so you conclude that "dt of twin B" smaller than "dt of twin A" (twin B aging less).

Situation 2 (applying symmetry):
Twin B "at rest" and so with attached frame " K " and twin A moving and so attached frame " K' ". Elapsed time for twin B: " dt ". For twin A: " dt' ".
You get the same dt'<dt and so you conclude "dt of twin A" smaller than "dt of twin B" (twin A aging less).

Final conclusion: opposite results and so contradiction/inconsistency.

 

[przyk]

Anyway, you calculated the elapsed time from one frame's perspective of observation. To see the key feature of the paradox now you must interchange roles and consider the other frame as "at rest" and the first one moving. You will find exactly the same relation but now for the symmetric elapsed times what means the opposite result. There is the paradox.

Like I explained before, there is only a paradox if you neglect relativity of simultaneity and/or treat accelerating frames as if they were equivalent to inertial frames. Both are errors in the context of relativity.

Suppose Alice leaves her twin Bob behind on Earth at birth and travels away at constant velocity $$v$$. She does this until she has aged by 10 years. Then Alice can say that in her reference frame, Bob has aged by $$10/\gamma = 10\sqrt{1 - v^{2}/c^{2}}$$ years. That's fine.

But, then, Alice almost instantaneously stops and starts heading back to Earth at velocity $$-v$$. Immediately after Alice has changed velocity, she is in a new inertial frame which, in particular, has a different definition of simultaneity. So the standard by which Alice compared Bob's age to her own has changed. Bob's age in Alice's frame, immediately after the acceleration, is not $$10/\gamma$$ years any more. It's something very different. The twin paradox comes from assuming that Bob's age in Alice's frame is nearly the same just before and just after Alice turns her rocket around. This is an error in the context of relativity which ignores the relativity of simultaneity effect. That is why nobody who understands relativity would ever describe Bob's age in Alice's frame that way - it's just plain wrong.


This is discussing the relativistic time dilation factors. Like I point out above in post #97, the relativistic time dilation factors are not what the twins would actually see (which is instead given by the relativistic Doppler effect). For a detailed account of what both twins actually see if, e.g., they look at each other through telescopes the whole time, see post #97 above.

 

[przyk]

Situation 1:
Twin A "at rest" and so with attached frame " K " and twin B moving and so attached frame " K' ". Elapsed time for twin A: " dt ". For twin B: " dt' ".
You get dt'<dt and so you conclude that "dt of twin B" smaller than "dt of twin A" (twin B aging less).

Situation 2 (applying symmetry):
Twin B "at rest" and so with attached frame " K " and twin A moving and so attached frame " K' ". Elapsed time for twin B: " dt ". For twin A: " dt' ".
You get the same dt'<dt and so you conclude "dt of twin A" smaller than "dt of twin B" (twin A aging less).

Final conclusion: opposite results and so contradiction/inconsistency.

No, this is an elementary misunderstanding of both relativity and how derivatives work in mathematics. The full relation between A and B's space and time coordinates, if both are always inertial and moving at fixed velocity relative to one another, is given by the Lorentz transformation, which in differential form is

$$
\begin{eqnarray}
\mathrm{d}t' &=& \gamma ( \mathrm{d}t \,-\, \frac{v}{c^{2}} \mathrm{d}x )
\mathrm{d}x' &=& \gamma ( \mathrm{d}x \,-\, v \mathrm{d} t ) \,.
\end{eqnarray}
$$​

The relation between $$\mathrm{d}t$$ and $$\mathrm{d}t'$$ depends on the path along which you are making the comparison:

For a clock moving with velocity $$v$$, so that $$\mathrm{d}x = v\mathrm{d}t$$, you get $$\mathrm{d}t' = \gamma(1 - v^{2}/c^{2} ) \mathrm{d}t = \mathrm{d}t / \gamma$$. So the rate of a moving clock in the unprimed frame is

$$
\frac{\mathrm{d}t'}{\mathrm{d}t} \,=\, \frac{1}{\gamma} \,.
$$​

For a clock at rest in the unprimed frame, so $$\mathrm{d}x = 0$$, the Lorentz transformation above just simplifies to $$\mathrm{d}t' = \gamma \mathrm{d}t$$. So you find that the rate at which a clock at rest in the unprimed frame advances in the primed frame is

$$
\frac{\mathrm{d}t}{\mathrm{d}t'} \,=\, \frac{1}{\gamma} \,.
$$​

There is no contradiction because both (equal) time dilation factors are derived from the same equation: the Lorentz transformation. The derivatives are different because the first is evaluated for $$x' = \text{constant}$$ while the second is evaluated for $$x = \text{constant}$$, - i.e. along different paths or sections in spacetime.


Like at least one essay I've read liked to point out, the error you are making is a particular instance of the error in assuming partial derivatives invert in the "expected" way. In general, if $$y_{j}$$ are variables that are cast as (locally invertible) functions of other variables $$x_{i}$$, then in mathematics the partial derivative $$\frac{\partial x_{i}}{\partial y_{j}}$$ is in general not equal to $$\biggl( \frac{\partial y_{j}}{\partial x_{i}} \biggr)^{-1}$$. Understanding this will get you a long way toward understanding the symmetrical time dilation factors in relativity.

 

[martillo]

martillo seems really interested in what both twins actually see, which is actually the Doppler shift factors rather than the relativistic time dilation factors. So let's work that out.

In SR, if either of the twins looks at the other through a telescope, say, then what they actually see is also affected by the fact that light takes more and more or less and less time to travel between them... ... ...

No I don't think would be good to include Doppler shifts factors on the problem. This just complicates it more. When it is said to consider what one twin would see about the other twin it means we want to take the relativistic prediction of the observation for the other twin. That's what Lorentz Transform does, it makes the necessary transformations to observe what is happening on the other frame as it would be seen directly and instantaneously by the twin "at rest".
So I don't think appropiate to consider what would be seen by telescopes. Better is to consider as each twin can record images of what is happening to themselves to interchange data sometime after (for example at a crossing point where fast communications would be possible) and compare results.

Like I explained before, there is only a paradox if you neglect relativity of simultaneity and/or treat accelerating frames as if they were equivalent to inertial frames. Both are errors in the context of relativity.

Suppose Alice leaves her twin Bob behind on Earth at birth and travels away at constant velocity . She does this until she has aged by 10 years. Then Alice can say that in her reference frame, Bob has aged by years. That's fine.

But, then, Alice almost instantaneously stops and starts heading back to Earth at velocity . Immediately after Alice has changed velocity, she is in a new inertial frame which, in particular, has a different definition of simultaneity. So the standard by which Alice compared Bob's age to her own has changed. Bob's age in Alice's frame, immediately after the acceleration, is not years any more. It's something very different. The twin paradox comes from assuming that Bob's age in Alice's frame is nearly the same just before and just after Alice turns her rocket around. This is an error in the context of relativity which ignores the relativity of simultaneity effect. That is why nobody who understands relativity would ever describe Bob's age in Alice's frame that way - it's just plain wrong.

I think the symmetric linear travel avoids the problem you mention. Initially the three twins are together with clocks syncronized and two of them travels symmetrically in opposite direction far enough and stop (in relation to the twin "at rest") at the same time (the symmetry allows this). I consider this point as the better starting of the travels to be considered. As they stopped in relation to the twin "at rest" all twins can have their clocks resinchronized again and start a symmetrical travel back for all them meet again.
What matters now is what each twin would observe at any point of the travel, this means to each one consider and record the relativistic predictions of what would be happening on the other twins and their own observations in their own frame to finally compare observations at a final crossing point (I think is simpler to consider they don't reunite there but just interchange the information while travelling at the constant velocity of the travel.
I think if we make such considerations many problems are avoided and the paradox turns simpler.

 

[Emil]

....So the rate of a moving clock in the unprimed frame is

$$
\frac{\mathrm{d}t'}{\mathrm{d}t} \,=\, \frac{1}{\gamma} \,.
$$​

...So you find that the rate at which a clock at rest in the unprimed frame advances in the primed frame is

$$
\frac{\mathrm{d}t}{\mathrm{d}t'} \,=\, \frac{1}{\gamma} \,.
$$​

Yes, that's what I'm saying.
dt=dt[sup]'[/sup]

 

[martillo]

Let's recap SRT.
The condition you can apply SRT is the two postulates to be true.
Between two inertial reference frames, which have a speed relative to each other, there is a time dilation and length contraction.
Physical quantities involved are time, distance (length) and speed, their derived.

But this leads to the paradox of time.
What were they thinking? Let's modify what Einstein said and there is no need for two inertial reference frames, is sufficient only one and they introduces the concept of acceleration.
Now they say that this is the "mainstream".

They do not even realize how they destroy SRT.

Seems you are someway right on this...

 

[martillo]

No, this is an elementary misunderstanding of both relativity and how derivatives work in mathematics. The full relation between A and B's space and time coordinates, if both are always inertial and moving at fixed velocity relative to one another, is given by the Lorentz transformation, which in differential form is

$$
\begin{eqnarray}
\mathrm{d}t' &=& \gamma ( \mathrm{d}t \,-\, \frac{v}{c^{2}} \mathrm{d}x )
\mathrm{d}x' &=& \gamma ( \mathrm{d}x \,-\, v \mathrm{d} t ) \,.
\end{eqnarray}
$$​

The relation between $$\mathrm{d}t$$ and $$\mathrm{d}t'$$ depends on the path along which you are making the comparison:

For a clock moving with velocity $$v$$, so that $$\mathrm{d}x = v\mathrm{d}t$$, you get $$\mathrm{d}t' = \gamma(1 - v^{2}/c^{2} ) \mathrm{d}t = \mathrm{d}t / \gamma$$. So the rate of a moving clock in the unprimed frame is

$$
\frac{\mathrm{d}t'}{\mathrm{d}t} \,=\, \frac{1}{\gamma} \,.
$$​

For a clock at rest in the unprimed frame, so $$\mathrm{d}x = 0$$, the Lorentz transformation above just simplifies to $$\mathrm{d}t' = \gamma \mathrm{d}t$$. So you find that the rate at which a clock at rest in the unprimed frame advances in the primed frame is

$$
\frac{\mathrm{d}t}{\mathrm{d}t'} \,=\, \frac{1}{\gamma} \,.
$$​

There is no contradiction because both (equal) time dilation factors are derived from the same equation: the Lorentz transformation. The derivatives are different because the first is evaluated for $$x' = \text{constant}$$ while the second is evaluated for $$x = \text{constant}$$, - i.e. along different paths or sections in spacetime.


Like at least one essay I've read liked to point out, the error you are making is a particular instance of the error in assuming partial derivatives invert in the "expected" way. In general, if $$y_{j}$$ are variables that are cast as (locally invertible) functions of other variables $$x_{i}$$, then in mathematics the partial derivative $$\frac{\partial x_{i}}{\partial y_{j}}$$ is in general not equal to $$\biggl( \frac{\partial y_{j}}{\partial x_{i}} \biggr)^{-1}$$. Understanding this will get you a long way toward understanding the symmetrical time dilation factors in relativity.

Seems you didn't get the point or I don't understand you on this. You seem to make the same mistake. You just take one reference frame and make the calculations assuming may be that in the other frames just an inverse Lorentz transform would give the needed results. That's wrong. That is not what must be made.
In relativity any frame of observation is equivalent, right? And we can choose any frame of observation possible, right? Then, first consider one twin "at rest" (attach to him frame K) and make the observation (relativistic prediction) on frame k' attached to the other twin. After consider the other twin "at rest" and attach to him frame K and make the observation (relativistic prediction) on frame K' attached to the other twin. Finally make the proper considerations of what means for instance t and t' in each case and compare results.
The mistake I said I see is always made, written in other words, is to not consider a new pair of frames of observation in each situation. Relativity (and I think any other theory) allows free selection of frames of observation. The paradox preciselly comes out when different frames are attached to the different twins and results compared.
Can you get this?