The death of "Modern Physics". Prepair it's funeral!
- Thread starter martillo
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UnderWhelmed
Registered Senior Member
Ok I'm back!
Nevermind, this is retarded...I'm out again
martillo said:
Nevermind, this is retarded...I'm out again
The word 'age' can be used as a noun and a verb. You have used it as a noun.
To 'age' means to grow older - used as a verb.
If I say that I am aging, it means I am growing older.
You however said 'younger" - and no one grows younger, not even near the speed of light.
Hence, it is one twin 'aging' slower than the other. One twin is growing older slower than the other.
Kapeesh?
To 'age' means to grow older - used as a verb.
If I say that I am aging, it means I am growing older.
You however said 'younger" - and no one grows younger, not even near the speed of light.
Hence, it is one twin 'aging' slower than the other. One twin is growing older slower than the other.
Kapeesh?
martillo
Registered Senior Member
I think is not so easy.
In Relativity velocities and accelerations are both relative magnitudes and they cannot make difference.
May be UnderWhelmed was right in that I should have presented a more detailed problem...
I have thought an example to illustrate how the contradiction arises.
Give me just a moment to present it.
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martillo
Registered Senior Member
Here is the thought experiment.
To not consider the movement of Hearth we will think in a mother-ship that goes to the most "fixed place" you can imagine. May be some point at a fixed position relative to the known "fixed stars of the Universe.
The mother-ship goes there brakes and stop remaining there. After that, two small space-ships with twins accelerate in opposite directions, travel some time and brake in the same manner making a perfect symmetric travel to stop at some far distance.
After that, they turn their space-ships in the opposite direction and at some time (may be synchronized by the mother-ship that is at equal distance from them) they accelerate and travel in a second symmetrical flight deviating a negligible little (to not collide) just to pass very near of them and the mother-ship at the same instant but they don't brake!.
The intention is to capture the movement as they are flying at some considerable velocity to detect some relativistic effects.
We must consider that the state of both twins can be directly observed by them and by the people in the mother-ship! For example photographs can be taken at the instant of "crossing" and be sent to everybody, even the twins!
Now the situation is:
Both are travelling at some velocity v but in opposite directions just in front of the mother-ship.
Now we will apply Lorentz Transform to the twins to see how they are aging. For simplicity we will consider time zero this instant they are in front of the mother-ship.
First we choose a referential in the mother-ship pointing in the same direction as the velocity of one of the twins. We must replace x=+vt and x=-vt for each twin in the equations of time.
We assume k = (1-v2/c2)exp-1/2
Then for one twin we will have (x=+vt):
t' = k(1-v2/c2)t = t/k
and for the other (x=-vt):
t' = k(1+v2/c2)t
We can see that for each twin time t' is different what means they age differently. The first one is smaller what means the first twin will get younger than the other.
But the direction of the referential was arbitrary choused with the velocity of one of the twin! If we select the other twin the equations are inverted and that twin will get now older than the other!
This means opposite contradictory results.
Now we will consider the problem as seen by the twins themselves. they see each other travelling at a velocity w (classicaly is 2v but with the relativistic addition of velocities is something different) chousing the directions of the referentials as the directions of the relative velocity.
For them we must consider k = (1-w2/c2)exp-1/2
Then for both twins we will have the same:
t' = t/k
This means that for each one the other twin is getting younger than himself.
This means also opposite contradictory results.
We must also note that the rate of aging is different as seen in the mothership than seen by the twins.
We must pay attention that they have made a perfect symetrical travel, they accelerated the same amount, they traveled at the same velocity for the same time so there's no privileged direction in the experiment to decide for one of the cases.
We also note with some surprise that the results for the mother-ship referential is not the same aging as we would expect in this totally symmetric problem.
Do you "kapeesh" the contradictions of the problem now?
I wish I was clear enough.
To not consider the movement of Hearth we will think in a mother-ship that goes to the most "fixed place" you can imagine. May be some point at a fixed position relative to the known "fixed stars of the Universe.
The mother-ship goes there brakes and stop remaining there. After that, two small space-ships with twins accelerate in opposite directions, travel some time and brake in the same manner making a perfect symmetric travel to stop at some far distance.
After that, they turn their space-ships in the opposite direction and at some time (may be synchronized by the mother-ship that is at equal distance from them) they accelerate and travel in a second symmetrical flight deviating a negligible little (to not collide) just to pass very near of them and the mother-ship at the same instant but they don't brake!.
The intention is to capture the movement as they are flying at some considerable velocity to detect some relativistic effects.
We must consider that the state of both twins can be directly observed by them and by the people in the mother-ship! For example photographs can be taken at the instant of "crossing" and be sent to everybody, even the twins!
Now the situation is:
Both are travelling at some velocity v but in opposite directions just in front of the mother-ship.
Now we will apply Lorentz Transform to the twins to see how they are aging. For simplicity we will consider time zero this instant they are in front of the mother-ship.
First we choose a referential in the mother-ship pointing in the same direction as the velocity of one of the twins. We must replace x=+vt and x=-vt for each twin in the equations of time.
We assume k = (1-v2/c2)exp-1/2
Then for one twin we will have (x=+vt):
t' = k(1-v2/c2)t = t/k
and for the other (x=-vt):
t' = k(1+v2/c2)t
We can see that for each twin time t' is different what means they age differently. The first one is smaller what means the first twin will get younger than the other.
But the direction of the referential was arbitrary choused with the velocity of one of the twin! If we select the other twin the equations are inverted and that twin will get now older than the other!
This means opposite contradictory results.
Now we will consider the problem as seen by the twins themselves. they see each other travelling at a velocity w (classicaly is 2v but with the relativistic addition of velocities is something different) chousing the directions of the referentials as the directions of the relative velocity.
For them we must consider k = (1-w2/c2)exp-1/2
Then for both twins we will have the same:
t' = t/k
This means that for each one the other twin is getting younger than himself.
This means also opposite contradictory results.
We must also note that the rate of aging is different as seen in the mothership than seen by the twins.
We must pay attention that they have made a perfect symetrical travel, they accelerated the same amount, they traveled at the same velocity for the same time so there's no privileged direction in the experiment to decide for one of the cases.
We also note with some surprise that the results for the mother-ship referential is not the same aging as we would expect in this totally symmetric problem.
Do you "kapeesh" the contradictions of the problem now?
I wish I was clear enough.
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To which frame? Remember that there are three reference frames.martillo said:
That's the Lorentz transform equations for shifting coordinates to one of the twins' frames, specifically the twin moving to the 'right' along the x-axis (i.e. with velocity v wrt. the mothershpi). Of course, he doesn't see the other as being in synch with himself. The one moving to the 'left' along the x-axis as seen from the mothership has the following transform for the time coordinate
t' = γ(t + vx/c<sup>2</sup>)
Plug in x=-vt and you get t/γ. So observers at the mothership see them time dilate equally.
Clear?
martillo
Registered Senior Member
funkstar,
I disagree.
We cannot chouse a particular referential for each twin.
We must choose ONE referential on the mother-ship to study the complete problem and here the problem is what happens with both twins at the same time.
Nevertheless still a difference with what the twins observe exists.
I disagree.
We cannot chouse a particular referential for each twin.
We must choose ONE referential on the mother-ship to study the complete problem and here the problem is what happens with both twins at the same time.
Nevertheless still a difference with what the twins observe exists.
Excuse me, seeing as there seems to be a bit of a bun fight going on I thought I'd add my 2 cents worth........now where did I put my wallet..?...hmmmmmm............ahhh here it is.
Martillo,
Can I ask how you theory accomodates the invariance of light to all observers with out a loss of simultaneity?
Martillo,
Can I ask how you theory accomodates the invariance of light to all observers with out a loss of simultaneity?
martillo,
The Lorentz transforms relate frames. If you want to relate time in the mothership frame to a particular twin's frame, of course you need the Lorentz transforms that relate those two particular frames.
It's not a matter of opinion. If twin A moves to the right along the x-axis wrt. the mother ship S with relative speed v, i.e. velocity v, and their coordinates coincide on (0,0), the transformation equations relating spacetime coordinates (x<sub>S</sub>,t<sub>S</sub>) from the S frame to spacetime coordinates (x<sub>A</sub>,t<sub>A</sub>) in the frame A are
x<sub>A</sub>=γ(x<sub>S</sub>-vt<sub>S</sub>)
t<sub>A</sub>=γ(t<sub>S</sub>-vx<sub>S</sub>/c<sup>2</sup>)
If twin B moves to the left along the x-axis wrt. the mother ship S with relative speed v, i.e. velocity -v, and their coordinates coincide on (0,0), the transformation equations relating spacetime coordinates (x<sub>S</sub>,t<sub>S</sub>) from the S frame to spacetime coordinates (x<sub>B</sub>,t<sub>B</sub>) in the frame B are
x<sub>B</sub>=γ(x<sub>S</sub>+vt<sub>S</sub>)
t<sub>B</sub>=γ(t<sub>S</sub>+vx<sub>S</sub>/c<sup>2</sup>)
That's the way it is.
What you did amounts to nothing more than observing that two separated events that are simultaneous in the S frame, are not so in the A frame. No surprises there, and certainly no contradictions. This becomes extremely clear when you use frame names as subscripts instead of the ambiguous prime ('). You were transforming from S to A. That doesn't let you conclude anything about B.
So, you were mixing frames. It's a classic mistake, but a mistake nonetheless. Accept it, and move on.
The Lorentz transforms relate frames. If you want to relate time in the mothership frame to a particular twin's frame, of course you need the Lorentz transforms that relate those two particular frames.
It's not a matter of opinion. If twin A moves to the right along the x-axis wrt. the mother ship S with relative speed v, i.e. velocity v, and their coordinates coincide on (0,0), the transformation equations relating spacetime coordinates (x<sub>S</sub>,t<sub>S</sub>) from the S frame to spacetime coordinates (x<sub>A</sub>,t<sub>A</sub>) in the frame A are
x<sub>A</sub>=γ(x<sub>S</sub>-vt<sub>S</sub>)
t<sub>A</sub>=γ(t<sub>S</sub>-vx<sub>S</sub>/c<sup>2</sup>)
If twin B moves to the left along the x-axis wrt. the mother ship S with relative speed v, i.e. velocity -v, and their coordinates coincide on (0,0), the transformation equations relating spacetime coordinates (x<sub>S</sub>,t<sub>S</sub>) from the S frame to spacetime coordinates (x<sub>B</sub>,t<sub>B</sub>) in the frame B are
x<sub>B</sub>=γ(x<sub>S</sub>+vt<sub>S</sub>)
t<sub>B</sub>=γ(t<sub>S</sub>+vx<sub>S</sub>/c<sup>2</sup>)
That's the way it is.
What you did amounts to nothing more than observing that two separated events that are simultaneous in the S frame, are not so in the A frame. No surprises there, and certainly no contradictions. This becomes extremely clear when you use frame names as subscripts instead of the ambiguous prime ('). You were transforming from S to A. That doesn't let you conclude anything about B.
So, you were mixing frames. It's a classic mistake, but a mistake nonetheless. Accept it, and move on.
martillo
Registered Senior Member
funkstar,
In this case all what I can say is that you are wrong putting a different transform equation for each twin. Once you choused a referential the equations are determined and must be the same for every object observed!
You are defining two different transforms, one for each twin.
This is wrong.
In this case all what I can say is that you are wrong putting a different transform equation for each twin. Once you choused a referential the equations are determined and must be the same for every object observed!
You are defining two different transforms, one for each twin.
This is wrong.