The death of "Modern Physics". Prepair it's funeral!
- Thread starter martillo
- Start date
What arguments have you presented?martillo said:
I've seen you use relativistic mass to disprove relativity. That is horseshit - there is no such thing as relativistic mass, it is a concept used by pebble brains, you just don't understand (like a lot of people) that relativistic mass is not true per relativity.
I've seen you argue that special relativity is not consistent. In fact special relativity is consistent, but you just don't understand that special relativity introduces the relativity of simultaneity to keep consistency.
What other arguements have you presented that "I haven't given a chance for them to be true?". All of your arguments just show your ignorance.
martillo
Registered Senior Member
I will repeat here:
Even Einstein derived the necessity of a relativistic mass variation in Relativity Theory but you don't. If you don't admit that you are inventing a new theory! But this means nothing to you...
I haven't touched the problem of the relativity of the simultaneity in Relativity Theory. I don't understand how it "turns" Relativity consistent.
I have found inconsistency in the problem I presented in a post above. Tell me how the relativity of simultaneity turns it consistent.
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I understood what you said the first time you said it. You just didn't understand the case against what you are saying.martillo said:
Therefore, a quote from Wikipedia on relativistic mass is in order:martillo said:
<b<a
In the earlier years of relativity, it was the relativistic mass that was taken to be the "correct" notion of mass, and the invariant mass was referred to as the rest mass. Gradually, as special relativity gave way to general relativity and found application in quantum field theory, it was realized that the invariant mass was the more useful quantity and scientists stopped referring to the relativistic mass altogether. [...] the concept of energy has replaced the relativistic mass.
</a></b>
martillo said:
You need to read some modern texts on relativity. It is clear now that you have no conceptual understanding of modern relativity.
martillo
Registered Senior Member
Now I have to agree with you.
Then a new theory called "Moder Relativity" have raised?
I really haven't heard that there's a new theory that replaces Einstein's Relativity.
Can you tell us about it?
Is it successfull?
For example how the famous equation E=mc2 can be derived without assuming m=km0?
You know, the old Relativity derives: p=mv, F= dp/dt, E=integration(Fdx)...E=mc2!
How the new "Modern Relativity" get it?
Then a new theory called "Moder Relativity" have raised?
I really haven't heard that there's a new theory that replaces Einstein's Relativity.
Can you tell us about it?
Is it successfull?
For example how the famous equation E=mc2 can be derived without assuming m=km0?
You know, the old Relativity derives: p=mv, F= dp/dt, E=integration(Fdx)...E=mc2!
How the new "Modern Relativity" get it?
martillo
Registered Senior Member
Well here is the problem I have already presented with some little changes to explain it very well:
Here is a "perfect" experiment thought to show the inconsistence of Relativity Theory.
Is a new version of the well known twins paradox.
Just 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, even to us, to analyze the phenomenon!
Now the situation is:
Both are travelling at some velocity v but in opposite directions just in front of the mother-ship.
For simplicity we will consider time zero this instant they are in front of the mother-ship.
Now we will apply Lorentz Transform to the twins to see how they are aging. Note that age is an intrinsic property of living individuals.
We are going to consider the results in different frames and compare it. Any phenomenon of Nature is independent of the referential we chouse to observe and describe it so the results should be consistent (no contradictions should exist).
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.
Then three inconsistences were found because of contradictory results.
NOTE:
We can 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.
Here is a "perfect" experiment thought to show the inconsistence of Relativity Theory.
Is a new version of the well known twins paradox.
Just 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, even to us, to analyze the phenomenon!
Now the situation is:
Both are travelling at some velocity v but in opposite directions just in front of the mother-ship.
For simplicity we will consider time zero this instant they are in front of the mother-ship.
Now we will apply Lorentz Transform to the twins to see how they are aging. Note that age is an intrinsic property of living individuals.
We are going to consider the results in different frames and compare it. Any phenomenon of Nature is independent of the referential we chouse to observe and describe it so the results should be consistent (no contradictions should exist).
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.
Then three inconsistences were found because of contradictory results.
NOTE:
We can 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.
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The twins paradox I am familar with seemed to be shorter by 20 fold.martillo said:
OK - I am going to ignore all of the incorrect notions in this paragraph and just assume that the mothership exists at some point in spacetime.martillo said:
Whatever - the mothership exists at some point in spacetime, no need to mention breaking and stopping.martillo said:
You are assuming the travel is seen as symmetrical by the mothership. This is where the relativity of simultaneity (which you have admitted to not understanding) comes into play. Lets call them twin A and twin B because those are my favorite names. From A's perspective/frame, when A turns around, B is still travelling. This is NOT an effect of delayed perception, from A's frame, that is what A will conclude. The simultaneous event from the frame of the mothership of A and B switching directions and travelling back to the mothership will not be simultaneous from A's frame. Neither will it be simultaneous in B's frame. The only frame in which it will be simultaneous is in the frame of the mothership - this is the relativity of simultaneity.martillo said:
Now, since both twins A and B accelerated, comparing their clocks when they return will not show the time dilation you probably think each should show with respect to each other. They will have ever show time dilation compared to the clock on the mothership.
martillo
Registered Senior Member
Aer,
You could omitt all the first paragraph since it doesn't matter for the problem if the twins sees the events simultaneously or not.
What does matter is what is happening with them particularly at the cross point.
The correctly application of the Lorentz Transform for each twin gives opposite results but this doesn't matter to you. You simply say "they ...will not show time dilation" and it must be acepted true...
Correctly selecting a frame of observation and applying Lorentz Transform give different results for each twin and as presented simply changing the direction of the frame the obtained results are opposite. But this is irrelevant to you isn't it, what math says...
Relativity proposes the math but you simply ignore it at your convenience.
Sorry but this is not a proper refutation.
You could omitt all the first paragraph since it doesn't matter for the problem if the twins sees the events simultaneously or not.
What does matter is what is happening with them particularly at the cross point.
The correctly application of the Lorentz Transform for each twin gives opposite results but this doesn't matter to you. You simply say "they ...will not show time dilation" and it must be acepted true...
Correctly selecting a frame of observation and applying Lorentz Transform give different results for each twin and as presented simply changing the direction of the frame the obtained results are opposite. But this is irrelevant to you isn't it, what math says...
Relativity proposes the math but you simply ignore it at your convenience.
Sorry but this is not a proper refutation.
No, this is where the relativity of simultaneity comes into play. A will say that B started the return trip before A. B will say that A started the return trip before B. A will see B's clock as dilated. B will see A's clock as dilated. If we assume the clocks started ticking immediately after each respective turnaround, then A and B will each read the same time on their clocks when they meet at the mothership.martillo said:
You keep forgetting the relativity of simultaneity. It is fundamentally important in the setup you posed.
martillo
Registered Senior Member
Aer,
You propose to start the clocks at the turn instances of the space-ships.
But I can propose now to start the clocks simultaneously when they take off the mother-ship and compare the results at the back cross point. It is clear that for each twin the other twin have passed all the time travelling at some speed and this makes its clock run slower and at the cross point it will give a smaller value. This will happen for both twins and the results are opposite.
You propose to start the clocks at the turn instances of the space-ships.
But I can propose now to start the clocks simultaneously when they take off the mother-ship and compare the results at the back cross point. It is clear that for each twin the other twin have passed all the time travelling at some speed and this makes its clock run slower and at the cross point it will give a smaller value. This will happen for both twins and the results are opposite.