Attention Paid To Accelerating Reference Frames Overthrows SR

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The discussion centers on the implications of accelerating reference frames on the synchronization of clocks, challenging the validity of Special Relativity (SR). It posits that when one clock accelerates while the other remains inertial, the time dilation formula indicates that the accelerating clock ticks slower, leading to a divergence in their readings. The argument suggests that once the accelerating clock returns to an inertial frame, it will never synchronize again with the stationary clock, creating a contradiction in SR's predictions. Critics argue that the time dilation formula remains valid and that SR accounts for acceleration, emphasizing that the assumptions made in the argument overlook key principles of relativity. Ultimately, the debate highlights the complexities of applying SR to scenarios involving acceleration and the need for careful consideration of reference frames.
  • #91
Severian596 said:
You need to understand relativity better. It never caught on because actual physicists (who know more on this topic than I do) evaluated it, pondered it, and dismissed it. Why? Because they know more than you.

They cannot make me believe that which is false. They can confuse me about it for awhile, but they can never make me believe it.

Regards,

StarThrower
 
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  • #92
StarThrower said:
He concluded that photons can move relative to each other.

This line should be changed to read "He concluded that photons can move relative to himself." I can only mildly apprecite your candor.

Imagine that you have two trains A and B that leave the station C like this:

A
^
|
|
|
|
C----------->B

And they just so happen to be traveling at the speed of light. They'll never observe each other, so they'll never determine how fast they're traveling relative to each other.
 
  • #93
And...I shall exit stage left, as I should've done from square one.

Goodbye, StarThrower
 
  • #94
Severian596 said:
To suppliment my post about your incorrect use of Special Relativity, peruse http://van.hep.uiuc.edu/van/qa/section/New_and_Exciting_Physics/Relativity/20021019150128.htm to the twin paradox. The paradox is not a paradox at all. The individual cites his source and mentions that General Relativity is needed to correctly answer the paradox because Special Relativity does not supply a means to deal with acceleration and/or a preferred frame.

Ok I've looked at the site. The first thing to say, is that I understood Dolan's question.

If you have ever been to the carnival, there is a ride where you are spun around really fast, and then the floor drops out, but you don't fall to the earth. In this ride, you are pinned to the wall by a force which has arisen because you are in uniform circular motion.

Principle of equivalence: It is impossible to tell if you are at rest on the surface of a body exerting a gravitational force F on you, or accelerating uniformly through space.

Dolan asks about the following experiment.

You have two vials of some radioactive substance, with a precisely known half life. For example, suppose that the half life is one hour. The samples are identical, in the sense that they contain the same initial population. So for example, suppose that P0 = 1,000. Thus, if the vials are left at rest, then in one hour, there will be (approximately) 500 radioactive particles left in both samples.

Now, perform this experiment, but this time put one of the samples in a device like the carnival ride described before.

The particles in the centrifuge are obviously subjected to a force which the particles in the vial on the bench are not subjected to.

Dolan's question then, is whether or not the experiment has ever been done, and whether or not any time dilation was observed.
 
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  • #95
Severian596 said:
This line should be changed to read "He concluded that photons can move relative to himself." I can only mildly apprecite your candor.

Imagine that you have two trains A and B that leave the station C like this:

A
^
|
|
|
|
C----------->B

And they just so happen to be traveling at the speed of light. They'll never observe each other, so they'll never determine how fast they're traveling relative to each other.

Umm but they are moving relative to each other, even if they cannot measure the relative speed.
 
  • #96
StarThrower said:
Umm but they are moving relative to each other, even if they cannot measure the relative speed.

That's it! That's your flaw. Why should train A or B assume anything based on YOUR measurements. They will never detect each other's presence, so therefore they are NOT moving relative to each other. YOU may determine that they're moving at X speed relative to each other, or Y speed, or whatever speed, but YOU have the luxury of experiencing time and space in a way completely removed from what either of these trains must experience. To them, YOUR claim that they are moving relative to each other is completely preposterous! No matter how long they vigilantly wait for information that let's them draw a conclusion about their motion RELATIVE TO EACH OTHER it will never come.

There IS no absolute preferred frame, and that means YOUR FRAME is not the divine frame from which to ordain all conclusions about the relative motion of any two bodies, or the nature of space and time.

Believe whatever you like. I'm not wasting another moment on you. I've wasted enough.
 
  • #97
Severian596 said:
That's it! That's your flaw. Why should train A or B assume anything based on YOUR measurements. They will never detect each other's presence, so therefore they are NOT moving relative to each other. YOU may determine that they're moving at X speed relative to each other, or Y speed, or whatever speed, but YOU have the luxury of experiencing time and space in a way completely removed from what either of these trains must experience. To them, YOUR claim that they are moving relative to each other is completely preposterous! No matter how long they vigilantly wait for information that let's them draw a conclusion about their motion RELATIVE TO EACH OTHER it will never come.

There IS no absolute preferred frame, and that means YOUR FRAME is not the divine frame from which to ordain all conclusions about the relative motion of any two bodies, or the nature of space and time.

Believe whatever you like. I'm not wasting another moment on you. I've wasted enough.

I thought you exited stage left.

Self-contradiction? Oh now you are totally coming unglued huh. I find a flaw in the theory of SR, and you come apart.
 
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  • #98
Severian596 said:
That's it! That's your flaw. Why should train A or B assume anything based on YOUR measurements. They will never detect each other's presence, so therefore they are NOT moving relative to each other. YOU may determine that they're moving at X speed relative to each other, or Y speed, or whatever speed, but YOU have the luxury of experiencing time and space in a way completely removed from what either of these trains must experience. To them, YOUR claim that they are moving relative to each other is completely preposterous! No matter how long they vigilantly wait for information that let's them draw a conclusion about their motion RELATIVE TO EACH OTHER it will never come.

There IS no absolute preferred frame, and that means YOUR FRAME is not the divine frame from which to ordain all conclusions about the relative motion of any two bodies, or the nature of space and time.

Believe whatever you like. I'm not wasting another moment on you. I've wasted enough.

Let them be tied together with a very long string, with a lot of slack in it. Eventually, that string is going to break, and they will 'know' they are/were in relative motion. It will break at the moment in Universal time at which the distance between them exceeds the string length.

Regards,

The Star
 
  • #99
StarThrower said:
moment in Universal time
If you're going to assume SR (or GR), you must drop the notion of Universal time. Otherwise, you will run into a contradiction. :wink:
 
  • #100
StarThrower said:
Let them be tied together with a very long string, with a lot of slack in it. Eventually, that string is going to break, and they will 'know' they are/were in relative motion. It will break at the moment in Universal time at which the distance between them exceeds the string length.

Regards,

The Star

Ignoring the impossibility of a string (which has mass) "keeping up" with the photon traveling at c, the compression wave triggered by the snapping string would travel at roughly the speed of sound, just as any compression wave. For example if you and your friend stand with thighs against a table and he bumps it, how much time passes before you feel the table move? The speed is not infinite, and in fact travels at the speed of sound through the medium of the solid table. The photon would never learn of the "universal time" at which the string snapped.

I will discuss the theory of SR with you no further, but am happy to educate you on some basic physical phenomenon.
 
  • #101
This equation is valid for v=c is it not?

The equation is well-defined, if that's what you mean. In the spirit of the discussion thus far, I'd have to say we currently have no reason to think that the equation would be correct in SR; v=c violates the hypotheses of our derivation.
 
  • #102
Hurkyl said:
The equation is well-defined, if that's what you mean. In the spirit of the discussion thus far, I'd have to say we currently have no reason to think that the equation would be correct in SR; v=c violates the hypotheses of our derivation.

Well, we have the derivation before us, and there was no reason to forbid v=c in the derivation. v = \infty certainly must be forbidden, but not v=c. The reason v cannot be infinite, is because as the relative speed v increases, the angle between the sides of the isosceles triangle approaches 180 degrees, at which point the distance D would have to equal zero. But D is nonzero by stipulation. In fact, the only stipulation made about the relative speed, was that it be nonzero (so that we have a triangle in F2).

As we perform algebraic steps, we certainly must be careful not to divide by zero, or WE make an error.

The next thing to say, is that you certainly were correct about my assumption about D, namely that there is no length contraction in a direction perpendicular to the velocity. You also correctly saw that the relative velocity vector was perpendicular to the photon's velocity vector in the lab frame. As for D not contracting, that is unobjectionable as far as I can see, and as for the relative velocity vector, the direction is just one of the stipulations of the experiment.

But, I am sorry to say, v=c is not forbidden by nature, nor was it forbidden in this wonderfully simple derivation of the time dilation formula. Special relativity has an insurmountable problem to contend with, which is namely that photons cannot move relative to one another.

The logic finishes off this way:

Premise 1: If c=c` then (photons can't move relative to one another).
Premise 2: Photons can move relative to one another.
Conclusion: Not (c=c`)

The truth of premise 1 is known deductively, you in fact derived the result for us.

The truth of premise two is known inductively, through our common sensory perception.

The conclusion is sequitur, via the natural deduction known as modus tollens.


Kind regards,

The Star Thrower
 
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  • #103
But, I am sorry to say, there is no reason to forbid v=c in this wonderfully simple derivation of the time dilation formula.

Sure there is.

For instance, you assert:

Now, consider this event as viewed from a reference frame F2 which is moving at speed v relative to F1. In this frame, the path of the photon is not a straight line, but rather the path is triangular

Tell me, how long are the three sides to the triangle? For the sake of argument, let's say D is 1 meter.


There is a second clock (clock B) at rest in F2, and let the time it takes the photon to move from the photon gun to the mirror and back to the photon gun in F2 be measured by clock B to be \Delta t'

If, say, \Delta t happened to be 1 nanosecond, what is \Delta t'?


In order to synchronize clocks B,C, we can imagine that a rigid ruler connects them, and that we have located the midpoint, and then a spherical light pulse goes off there, and travels to both clocks, setting them to zero simultaneously in reference frame F2. They therefore remain in sync forever in reference frame F2.

How long is the ruler? How long does it take to synchronize the clocks?
 
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  • #104
Hurkyl said:
Tell me, how long are the three sides to the triangle? For the sake of argument, let's say D is 1 meter.

The answer comes from Euclidean geometry of course.

We have a right triangle, the hypotenuse of which is S. The base of which is 1/2 v delta t`, and the height of which is D. You want to let D=1 meter.

Additionally, S = 1/2 c` delta t`

Thus

(\frac{c^\prime \Delta t^\prime}{2} )^2 = 1 + (\frac{v \Delta t^\prime}{2} )^2

Give me a relative speed v in meters per second, and I can finish.

Regards,

Star
 
  • #105
Give me a relative speed v in meters per second, and I can finish.

You wanted to set v = c; use that.
 
  • #106
Hurkyl said:
You wanted to set v = c; use that.

Give me a numerical value please Mr Hurkyl, since the speed of light depends upon inertial reference frame. :smile:

Regards,

Star
 
  • #107
I thought you were trying to derive a contradiction within SR, you know, where the speed of light doesn't depend upon inertial reference frame.
 
  • #108
Hurkyl said:
I thought you were trying to derive a contradiction within SR, you know, where the speed of light doesn't depend upon inertial reference frame.

I thought we just figured out that SR is wrong. Therefore, the correct answer to your current question comes right out of Euclid.

Don't you realize that photons can move relative to one another?


I thought we were clear on at least that much.

Kind regards,

StarThrower
 
  • #109
I'm still critiquing your proof; we most certainly have not agreed SR is wrong.

You still require the assumption that inertial reference frames can travel at c. The derivation of the time dilation formula does not work if v >= c; if you would carry out the exercises I described, the reason would be clear.

In the end, this will amount to a proof that inertial reference frames cannot have relative velocity c in SR. Until you can prove inertial reference frames can have relative velocity c in SR, you have not proven SR internally inconsistent.
 
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  • #110
Don't go so fast.

We have a right triangle, the hypotenuse of which is S. The base of which is 1/2 v delta t`, and the height of which is D. You want to let D=1 meter.

Additionally, S = 1/2 c` delta t`

Thus

(\frac{c^\prime \Delta t^\prime}{2} )^2 = 1 + (\frac{v \Delta t^\prime}{2} )^2

In the lab frame, there is an experiment going on, and the experimenters there came up with a value of c = 299792458 meters per second for the speed of light in that frame. The following equation holds:

c = \frac{2D}{\Delta t}

Additionally:

c^\prime = \frac{2S}{\Delta t^\prime}

If my memory serves me.

So do you want me to use v = 299792458 meters per second?


P.S. Got to go to work now, be back tomorrow. I will do the things you ask, you will see that SR contradicts.

Kind regards,

Star
 
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  • #111
Yes, yes I would.
 
  • #112
Shall we make a clarification on the definition of "inertial reference frame?"

In particular, shall we declare that if an event (or state, if you prefer) occurs in one inertial reference frame, then it must occur in all inertial reference frames? Or shall we permit inertial reference frames that violate this condition?
 
  • #113
Hurkyl said:
Shall we make a clarification on the definition of "inertial reference frame?"

In particular, shall we declare that if an event (or state, if you prefer) occurs in one inertial reference frame, then it must occur in all inertial reference frames? Or shall we permit inertial reference frames that violate this condition?

What do you mean?

Regards,

Star
 
  • #114
I am not sure where Hurkyl is going with the numerical examples, but here is the logic:

In the problem that we are analyzing, we have a right triangle, the hypotenuse of which is S. The base of which is 1/2 v delta t`, and the height of which is D.

Additionally, S = 1/2 c` delta t`

Thus

(\frac{c^\prime \Delta t^\prime}{2} )^2 = D^2 + (\frac{v \Delta t^\prime}{2} )^2

In the lab frame, there is an experiment going on, and the experimenters there came up with a value of c = 299792458 meters per second for the speed of light in that frame. The following equation holds:

c = \frac{2D}{\Delta t}

Additionally:

c^\prime = \frac{2S}{\Delta t^\prime}

Using this information, Hurkyl was able to derive for us the following result:

\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t\'\;

No one has a problem with the formula above, and c is substitutable for v in the formula. When v=c, the formula predicts that delta t is equal to zero, which means that no time passes in the lab frame, and that leads to an explicit contradiction. So, here is the logic up to this point in the argument:

If c=c` and v = c then (x and not x), for some statement x.

At that point, we have the following absolute fact (independent of relativity), which I am sure Hurkyl will agree to:

Not [ c=c` and v = c]

where c,c`, and v are defined as stipulated in this problem.

Hence, we can all use demorgan's theorem of logic to arrive at:

not (c=c`) OR not (v=c)

In order to let v=c in this problem, let two photons have been emitted from the photon gun at right angles to one another. Thus, clock B is moving away from the photon gun at speed 299792458 meters per second.

There are THREE frames you can analyze the relative motion of the photons in. The frame where the gun is at rest, the frame where photon 1 is at rest, and the frame where photon 2 is at rest.

I will save everyone a lot of time.

In order to see where the error in the special theory of relativity is, consider the relative speed of the two photons. In a reference frame in which photon 1 is at rest, photon two has speed c2. In a reference frame in which photon 2 is at rest, photon one has speed c1.

Since speed is relative, we have c2=c1. It now follows that:

\Delta t = \Delta t^\prime

The problem in the theory of special relativity should now be visible to all.

Kind regards,

The Star
 
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  • #115
StarThrower said:
What do you mean?

Regards,

Star

An event is something that happens at a position in the inertial frame at a given time in that frame. Thus an event is located by three position coordinates and one time in each inertial frame. Of course, the coordinates of the events will be different in different inertial frames moving relative to one another at constant velocity, but the event itself is absolute. Examples of events are birth, death, explosions, detectors going off, etc. In other words, if you are born in one inertial frame, you are also born in any other inertial frame - the time and location of your birth may be different in different frames, but the fact of your birth is absolute.

And there you go
 
  • #116
Hurkyl said:
Shall we make a clarification on the definition of "inertial reference frame?"

In particular, shall we declare that if an event (or state, if you prefer) occurs in one inertial reference frame, then it must occur in all inertial reference frames? Or shall we permit inertial reference frames that violate this condition?

No reference frame can violate that condition (assuming I understand you correctly).

Regards,

StarThrower
 
  • #117
The point of my numerical examples was to demonstrate that the conditions of the thought experiment fail in the case where v = c.

IF the photon does indeed traverse a triangular path (or even a degenerate one) from the gun to the mirror and back to the gun, then we can form the equation:

(\frac{c \Delta t^\prime}{2} )^2 = D^2 + (\frac{c \Delta t^\prime}{2} )^2

which simplifies to

0 = D^2

Which is clearly false.

Our assumption that, within SR, there exists a reference frame with velocity c relative to F1 in which the photon under analysis traverses this triangular path is fase.

So only does the derivation of the time dilation formula fail, we also have that F2 fails to be an inertial reference frame, because it is impossible for it to observe the three events (or states, if you prefer) where the photon leaves the gun, strikes the mirror, and returns to the gun.
 
  • #118
Hurkyl said:
The point of my numerical examples was to demonstrate that the conditions of the thought experiment fail in the case where v = c.

IF the photon does indeed traverse a triangular path from the gun to the mirror and back to the gun, then we can form the equation:

(\frac{c \Delta t^\prime}{2} )^2 = D^2 + (\frac{c \Delta t^\prime}{2} )^2

which simplifies to

0 = D^2

Which is clearly false.

How is it false?

I believe you arrived at the following formula, which I will just refer to as the Hurkelian result:

\Delta t &= \sqrt{1 - \left( \frac{v}{c} \right)^2 } \Delta t\'\;

So when v=c the previous equation leads us to the following equation:

\Delta t = 0 {\Delta t^\prime}

From which we can see that the left hand side must be zero. On the other hand, Delta t` doesn't have to be zero, and the equation is still true.

Did you mean something else?

Regards,

Star
 
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  • #119
You're seriously asking me why D2 = 0 is false?
 
  • #120
Hurkyl said:
You're seriously asking me why D2 = 0 is false?

Yeah, I want to see your reasoning.

Here is one line of reasoning:

c = \frac{2D}{\Delta t}

Hence:

c \Delta t= 2D

So if delta t is equal to zero (and it is if v=c) then D=0.

But if you have a different line of reasoning, I want to see it. The reason I assumed you did, is because you wrote a formula which involved delta t` and D, and then implied the result that D=0 followed from that formula.

Kind regards,

StarThrower
 
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