Attention Paid To Accelerating Reference Frames Overthrows SR

[StarThrower]

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.

Regards,

Star

 

[Hurkyl]

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.

 

[StarThrower]

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

 

[Hurkyl]

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.

 

[StarThrower]

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

 

[Hurkyl]

Yes, yes I would.

 

[Hurkyl]

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?

 

[StarThrower]

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

 

[StarThrower]

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

 

[Severian596]

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

 

[StarThrower]

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

 

[Hurkyl]

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.

 

[StarThrower]

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

 

[Hurkyl]

You're seriously asking me why D² = 0 is false?

 

[StarThrower]

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

 

[Hurkyl]

There's a whole grab-bag of reasons why we conclude D = 0. The problem is that D doesn't have to be zero in the thought experiment.


Or, I suppose we can conclude that there is no distance between mirrors and photon guns instead of there being no inertial reference frames with relative velocity c.

 

[StarThrower]

There's a whole grab-bag of reasons why we conclude D = 0. You presumed D was not zero, though, in the thought experiment, did you not?

Yes, D is nonzero by stipulation. We have a big experiment set up in our lab, involving a laser gun, and a distant mirror. We are absolutely certain that not (D=0) in reality. But here we are investigating theory. And as we are investigating theory, we have come across something odd. We have come across the result that:

$$(\Delta t = 0) \wedge (D=0)$$

(the wedge symbol ^ means AND)

We now quickly ask ourselves how we arrived at this obviously incorrect result.

Let us briefly tabulate all unclosed assumptions:

1. c=c`
2. v=c

Thus, we are here:

If c=c` and v=c then (D=0 and not (D=0)).

(Keep in mind, that at the beginning of this reasoning event, we stipulated as true, that not (D=0). Then, during this reasoning event, we opened the scope of several assumptions, which allowed us to conclude that D=0. We can now close the scope of all the assumptions, to arrive at a statement which is true, and not true by assumption. Just plain old true.)

Then we use reductio ad absurdum to quickly get here:

not (c=c` and v=c )

Then DeMorgan's law of conjunction gets us here:

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

And here is where I wanted to be, watching the world wake up to history.

Kind regards,

The Star Thrower

 

[Hurkyl]

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

If this is all you're trying to prove, we could have gotten here much sooner, and in a much less sloppy manner. Oddly enough, if you've paid attention, you'd've noticed we've been telling you not(v=c), so you wouldn't have wasted all this time proving the above statement which we already take to be true.

 

[StarThrower]

If this is all you're trying to prove, we could have gotten here much sooner, and in a much less sloppy manner. Oddly enough, if you've paid attention, you'd've noticed we've been telling you not(v=c), so you wouldn't have wasted all this time proving the above statement which we already take to be true.

You have been telling me that not (v=c) is true?

No, that is not what we currently can be said to know epistemologically speaking. We do not know that not (v=c). What we know is this:


Current Knowledge: not (c=c`) OR not (v=c)

P.S. This is not all I am trying to prove, I will prove not (c=c`). Additionally, the sloppy manner you speak of has tremendous pedagogical value.

 

[Hurkyl]

This whole thought experiment is supposed to be done in the context of SR, remember? And since SR => not(v=c), it is thus true that not(v=c).

 

[StarThrower]

This whole thought experiment is supposed to be done in the context of SR, remember? And since SR => not(v=c), it is thus true that not(v=c).

I guess I should repeat my point, it is a logico/deductive one.

A theory of physics is a conjunction of many statements. If anyone of the statements is false, the whole theory is internally inconsistent.

The special theory of relativity is composed of many statements obviously. Some of which a reasoning agent is certain is true, and some of which a reasoning agent is not certain of the truth value of.

Suppose a theory of physics T consists of exactly 8 statements whose truth value we are uncertain of. We can define the portion of the theory which is uncertain to be the conjunction of these 8 statements. The rest of the theory we are certain of, and we cannot be certain of that which is false, hence the rest of the theory is consistent.

$$\Im = S_1 \wedge S_2 \wedge S_3 ... \wedge S_8$$

Now, if the uncertain portion of this theory leads us to a contradiction, then we know that at least one of the eight statements is false, though we will not yet know which one.

Let us define the special theory of relativity to be a theory of physics which consists of exactly one uncertain postulate, namely its fundamental postulate:

The Fundamental Postulate Of The Special Theory Of Relativity

The speed of a photon in any inertial reference frame is 299792458 meters per second.

Now, I introduced a do-able experiment, designed to measure the speed of light in two inertial reference frames moving relative to each other at a constant speed v. (This is exactly the 'thought experiment' to consider, in order to falsify SR)

Now, we reached the following true statement:

not [ c = c` and v=c ]

Where c,c`, and v were defined for this experiment.

The previous statement is logically equivalent to the following statement:

If c=c` then not (v=c)

But the case where v=c is where two photons are fired at right angles to one another, and this experiment is do-able in reality.

You have a dilemma on your hands, I do not.

Kind regards,

StarThrower

 

[Pergatory]

This whole thought experiment is supposed to be done in the context of SR, remember? And since SR => not(v=c), it is thus true that not(v=c).

Thank you!

StarThrower, despite all the math that has been thrown around you have only restated the same thing over and over again which is that you believe inertial reference frames can have v=c. Unless you address this SPECIFICALLY, you are not making progress, all you have done so far is tell us why SR is invalid if v=c.

 

[Hurkyl]

But the case where v=c is where two photons are fired at right angles to one another, and this experiment is do-able in reality.

Sure, we can fire two photons at right angles. However, this experiment which is doable in reality is not the thought experiment we were discussing. Where are the clocks attached by a rod? Where is the inertial frame of reference? Where is the triangular photon path?

 

[StarThrower]

The case where v=c is do-able. The theory of SR says that the experiment is not do-able. Therefore, SR has been falsified.

Kind regards,

StarThrower

Clock B is a photon, and clock C is a photon. Under SR's assumption, they move at the same speed relative to the photon gun. Hence, the distance between them isn't changing, in other words photon B isn't moving relative to photon C. So it is as if photon B and photon C are attached by a rigid rod according to SR. That having been said, the rest frame of the photon gun was stipulated to be an inertial reference frame. So there is at least one inertial reference frame. Now, consider the do-able experiment.

Two photons are simultaneously emitted from the origin, at right angles in inertial reference frame F1.

To be clear, photon A travels in the j^ direction (direction of increasing y coordinates), and photon B travels in the -i^ direction (direction of decreasing x coordinates.

The photon gun is located at the origin of F1 obviously, and there is a clock at rest there, clock A, which ticks in units of seconds. Now, 299792458 meters in the j^ direction, there is a mirror stationed. Every time a photon has been fired at that mirror, it took two seconds to return to the photon gun, as measured by clock A.

Similarly, let there be a 2nd mirror stationed 299792458 meters in the -i^ direction.

Under the operating assumption of SR, when we fire the two photons, they should return to the photon gun simultaneously.

Clearly, the two photons are in relative motion during the experiment. I now ask a very simple question.

What is the speed of photon A relative to photon B?

The result which you must inevitably get, is that:

$$\Delta t = \Delta t^\prime$$

Kind regards,

StarThrower

 

[Hurkyl]

The case where v=c is do-able.

On what grounds do you assert that?

 

[StarThrower]

On what grounds do you assert that?

I've done the experiment.

Regards,

StarThrower

 

[Hurkyl]

Excellent! What was your experimental setup? What were the lengths of the sides of the triangle? What was the the reading on the clocks at each event of interest? What was the experimental error?

And what bearing does the experiment have on the internal consistency of SR?

 

[Severian596]

I'm also curious on how he attained an inertial reference frame during the experiment...

 

[StarThrower]

Excellent! What was your experimental setup? What were the lengths of the sides of the triangle? What was the the reading on the clocks at each event of interest? What was the experimental error?

And what bearing does the experiment have on the internal consistency of SR?

The distance to the mirror was about 20 feet or so, and an oscilloscope was used to make a time calculation. The laser used was helium neon. As for all the juicy details, I don't remember them now. But, you don't need them anyways, to check the internal consistency of SR. All you need to do, is compute the relative speed of two photons fired at right angles to one another simultaneously.

As someone correctly pointed out, SR says the relative speed should be c. So under the single assumption of SR, the triangle in the rest frame of the photon gun is an equilateral triangle. Thus, the angle is 60, when it is 90 in reality. Thus, the theory of relativity ends in contradiction. It is internally inconsistent.


Kind regards,

StarThrower

P.S. Of course you do need to prove that the photon's frame is inertial.

 

[Hurkyl]

Er, so the oscillosope was moving at light speed?

And what definition are you using for "relative velocity of photons" and what bearing does it have on the validity of SR?

 

[Hurkyl]

You seem to have lost all of your attention to detail. Don't you find it telling that you get sloppy every time you try to make your conclusion that SR is inconsistent?

 

[StarThrower]

Er, so the oscillosope was moving at light speed?

And what definition are you using for "relative velocity of photons" and what bearing does it have on the validity of SR?

The experiment was designed to simply measure the speed of light. We couldn't even get 299792458 exactly. There was only one mirror, there didn't need to be two. There was only one clock, and it was at rest in the lab frame.

But of course you are just sidestepping the issue.

Two photons are emitted from the origin of inertial reference frame F1, at right angles. What is their relative speed?

Assume the fundamental postulate of SR is correct.

Thus, each photon is moving at speed 299792458 meters per second relative to the photon gun.

Finish the analysis, tell me what the relative speed of these two photons is, assuming SR is correct.

Kind regards,

The Star

 

[Hurkyl]

But of course you are just sidestepping the issue.

You claimed to have carried out the thought experiment we were describing, with v = c. Whether or not I'm sidestepping, you are being intentionally misleading, at best.

Two photons are emitted from the origin of inertial reference frame F1, at right angles. What is their relative speed?

The only definition of relative speed of which I'm aware does not apply to this situation. Would you care to provide one?

 

[StarThrower]

You seem to have lost all of your attention to detail. Don't you find it telling that you get sloppy every time you try to make your conclusion that SR is inconsistent?

You know what they say... you can lead a horse to water, but you can't make him drink.

And no I'm not getting sloppy. I have led you to where I wanted to. Consider two photons fired at right angles to one another. Assume SR is correct. What is their relative speed?

Regards,

Star

 

[StarThrower]

The only definition of relative speed I'm aware does not apply to this situation. Would you care to provide one?

Speed in an inertial frame = distance traveled in inertial frame divided by time of travel in inertial frame

Regards,

StarThrower