The Fundamental Postulate Of Special Relativity Is Self-Contradictory

[StarThrower]

There is no intertial reference frame in which a photon is at rest. The only way to get a contradiction here is if you assume that there is one. But that is not an assumption made by special relativity.

Assume that there is at least one inertial reference frame in which a photon's speed isn't equal to zero, and refer to this reference frame as F1.

According to the fundamental postulate of the theory of special relativity, the speed of the photon in this inertial reference frame is c, where c = 299792458 m/s.

To Prove: The rest frame of the photon is an inertial reference frame.

Let the photon be moving along the positive x-axis of F1, in the direction of increasing x coordinates of F1.

By Newton's first law, the photon's direction will not change, unless the photon is acted upon by an outside force. Presume that over some period of time, the sum of all forces on this photon is zero. Hence, the photon will move through F1 in a straight line, by Newton's first law.

Now, define reference frame F2 as follows:

The origin of F2 is this photon, and the axes of F2 are not rotating with respect to the axes of F1. Furthermore, let the x-axis of F2 be parallel to the x-axis of F1, and the same for the y,z axes of F2.

Consider the position vector of the origin of F1, as defined in F2.

Let the origin of F1 currently have coordinates (x1,y1,z1) in F2.

The position vector of the origin of F1 as defined using F2 is the vector from the origin of F2 to (x1,y1,z1).

Hence, the position vector of the origin of F1 in F2 is:

(x1-0)i^ + (y1-0) j^ + (z1-0)k^ = <x1,y1,z1>

Now, the position vector of the origin of F1 as defined in F2 is changing in time. Thus, the velocity of the origin of F1 in reference frame F2 is:

V = d/dt [ <x1,y1,z1>] = <d(x1)/dt,d(y1)/dt,d(z1)/dt>

Now, the origin isn't moving in the y or z direction in F2, hence

d(y1)/dt=0 and d(z1)/dt = 0

Thus


V = <d(x1)/dt,0,0>

And since speed is relative,

V=c

And c is constant.

Hence, the origin of F1 moves through F2 in a straight line. And there was no force acting to accelerate F1. Hence F2 is an inertial reference frame.

QED

PS (this was a fast proof) For a more rigorous proof, investigate linear transformations from F1 to F2.

Kind regards,

The Star

 

[Hurkyl]

Visually the two lines converge. But due to the nature of Time Dilation the length units on the ct' axis approach infinity, and the length units on x' approach length zero. If you finally reach v=c, then it's reasonable to assume you'll no longer have two axies, but one (ct'). And furthermore you won't ever be able to progress along this axis because every incriment along that axis has a length of infinity:

You've identified the problem quite nicely. I was kinda hoping I could get ST to do the exercise to realize the same thing.

 

[Severian596]

Furthermore, let the x-axis of F2 be parallel to the x-axis of F1, and the same for the y,z axes of F2.

Assuming you can even define an X and Y axis for F2, how can they possibly be PARALLEL to F1's axes??

 

[StarThrower]

Assuming you can even define an X and Y axis for F2, how can they possibly be PARALLEL to F1's axes??

Parallel by stipulation. Furthermore, it is not necessary to make this stipulation, it is merely convenient. Axes are imaginary things which cannot bend.

Kind regards,

The Star

 

[Hurkyl]

Thus, the velocity of the origin of F1 in reference frame F2 is:

V = d/dt

Excellent; now try doing it in relativistic mechanics instead of classical mechanics. (In particular, relativistically, the velocity is the derivative of position with respect to proper time, not coordinate time)

 

[Integral]

You are getting off track. The question is this:

Is a reference frame whose origin is a photon an inertial reference frame?

Kind regards,

The Star

The simple answer is no. Because in order to have a reference frame one must have dimensions. Photons do not understand length or time, therefor it is impossible to differentiate time and distance. A reference frame implies a coordinate system, there is no way to establish a coordinate system without time and space. Therefore a photon cannot be a reference frame.

 

[StarThrower]

Excellent; now try doing it in relativistic mechanics instead of classical mechanics. (In particular, relativistically, the velocity is the derivative of position with respect to proper time, not coordinate time)

Ummm no.

recall:

V = <d(x1)/dt, 0, 0 >

and |V|= c, and the direction of motion is constant by stipulation.

Hence we have the following scalar equation:

c = d(x1)/dt

which implies

c dt = d(x1)

Now, integrate both sides of this equation:

$$\int{c dt} = \int{ d(x1) }$$

Since c must be constant:

$$c \int{dt} = \int{ d(x1)$$

Now, the integral is to be performed in the rest frame of the photon.

Let the final position of the origin of F1 in F2 be (x2,y2,z2). And there cannot be motion in the j^ or k^ directions. Hence, the final position of the origin of F1 in F2 is:

Final position: (x2,0,0)

also,

Initial position: (x1,0,0)

Thus, the change in position is:

$$\Delta x = x2-x1$$

Hence, we must have the following equation be satisfied:

$$c \int{dt} = \Delta x$$

From which it follows that:

$$c \Delta t = \Delta x$$

From which it follows that

$$ct2 - ct1 = c\Delta t = \Delta x$$

The difference in the time coordinates are made using a clock at rest in the photon's reference frame.

From the previous equation, it follows that the trajectory of the origin of F1 in reference frame F2 is a straight line, and the speed of F1 through F2 is a constant. Hence, the rest frame of the photon is inertial.

QED

Kind regards,

The Star

 

[Hurkyl]

recall:

V = <d(x1)/dt, 0, 0 >

Recall that this is the classical definition of 3-velocity.


The relativistic definition of 4-velocity is

u = dx / d&tau;

where &tau; is proper time (not coordinate time)

Or, in component form:

u = <dt/d&tau;, dx₁/d&tau;, 0, 0>

 

[StarThrower]

Now, try doing it again using relativistic mechanics instead of classical mechanics.

The time coordinates are observed by a clock that is moving along with the photon, and so this is the time of the event in the photon's reference frame. The velocity of the origin of F1 in F2 must be defined using this time, and not any other.


Kind regards,

The Star

 

[StarThrower]

Recall that this is the classical definition of 3-velocity.


The relativistic definition of 4-velocity is

u = dx / dτ

where τ is proper time (not coordinate time)

Or, in component form:

u = <dt/dτ, dx₁/dτ, 0, 0>

The proper time in whose frame Hurkyl??

Technically speaking, from the photonic frame, the proper time of the event would be the time of the event in frame F1, but the velocity of the origin of F1 through F2 should not be defined using the time of the event in F1, it must be the time of the event in F2. Or more simplistically, the time measurement must be made by a clock at rest in reference frame F2.

The whole point is that the motion of the origin of F1 in F2 is linear.

$$c = \frac{\Delta x}{\Delta t}$$

Clearly c is a constant, and so the trajectory is linear, at a constant speed, exactly as Newton's law of inertia says. Hence, the rest frame of the photon is an inertial reference frame. The origin of F1 isn't moving up or down, its moving in a straight line at a constant speed.

Kind regards,

The Star

 

[Hurkyl]

The proper time in whose frame Hurkyl??

Er, all frames compute the same proper time interval over any trajectory.


And, BTW, to what do you think &Delta;t is equal?

 

[StarThrower]

Er, all frames compute the same proper time interval over any trajectory.


And, BTW, to what do you think Δt is equal?

The proper time of this event Hurkyl, is Δt since there is no time dilation of any kind, since SR is wrong. The notion of proper time is a misguided relativistic remnant.

Kind regards,

The star

P.S. You are missing the whole point anyways, which is that the motion of F1 through F2 is a straight line at a constant speed, regardless of what clock is making what measurement.

Actually, no I don't think you are missing the whole point, I think you just want to see what I will say. Cheers

 

[Hurkyl]

So you are starting with the assumption that SR is wrong? Silly me I thought you were trying to prove SR is wrong.

P.S. You are missing the whole point anyways, which is that the motion of F1 through F2 is a straight line at a constant speed, regardless of what clock is making what measurement.

(a) You're computing classical velocity, not relativistic velocity.
(b) You're making a crucial assumption that &Delta;x is defined and &Delta;t is nonzero

 

[StarThrower]

So you are starting with the assumption that SR is wrong? Silly me I thought you were trying to prove SR is wrong.


You seem to be relying on the assumption that Δt and Δx are both nonzero. You can prove this is classical mechanics, but in SR...

No, obviously I don't want to assume SR is wrong. Don't miss the point. Regardless of the length of the time interval, the motion of the origin of F1 through F2 is in a straight line at a constant speed, and F1 isn't subjected to any forces, and so Newton's first law is satisfied, and so the rest frame of the photon is an inertial reference frame, contrary to the fundamental postulate of SR, hence SR self-contradicts, hence "proper time is a un-needed relativistic remnant..." etc etc

Kind regards,

The Star

 

[Severian596]

The proper time of this event Hurkyl, is Δt since there is no time dilation of any kind, since SR is wrong. The notion of proper time is a misguided relativistic remnant.

Kind regards,

The star

P.S. You are missing the whole point anyways, which is that the motion of F1 through F2 is a straight line at a constant speed, regardless of what clock is making what measurement.

Actually, no I don't think you are missing the whole point, I think you just want to see what I will say. Cheers

What little interest I had left for this thread is now gone. Your last 10 or so posts have not made sense and I'm convinced you're just mentally masturbating all over this thread, and marvelling at your own ability to throw terms around.

If your ability to convince someone who IS interested in hearing what you have to say lacks this much, imagine what everyone else thinks.

Good luck with trying to reconstruct physics from scratch since you're assuming SR is wrong and you refuse to use any of its equations. You just lost half of your audience.

EDIT:
Oh, hopefully reality won't unravel itself when you prove to someone, somewhere that photons have inertial reference frames.

 

[StarThrower]

So you are starting with the assumption that SR is wrong? Silly me I thought you were trying to prove SR is wrong.




(a) You're computing classical velocity, not relativistic velocity.
(b) You're making a crucial assumption that Δx is defined and Δt is nonzero

Response To (a):

I am doing nothing more than using the definition of velocity.

Response to (b):

Δx is defined and Δt is nonzero

 

[StarThrower]

What little interest I had left for this thread is now gone. Your last 10 or so posts have not made sense and I'm convinced you're just mentally masturbating all over this thread, and marvelling at your own ability to throw terms around.

If your ability to convince someone who IS interested in hearing what you have to say lacks this much, imagine what everyone else thinks.

Good luck with trying to reconstruct physics from scratch since you're assuming SR is wrong and you refuse to use any of its equations. You just lost half of your audience.

EDIT:
Oh, hopefully reality won't unravel itself when you prove to someone, somewhere that photons have inertial reference frames.

Severian... I am not screwing around. Watch and see who wins, Hurkyl or me. If he wins then relativity is correct.

Kind regards,

The Star

 

[Hurkyl]

Let's make this more explicit. You've defined the spatial axes for your photonic frame. I have no problem with those.

You seem to imply that a photon-centered clock is used to define the temporal axis.

The problem is that you give no reason to believe the photon-centered clock ever changes its reading. Thus your temporal axis is ill-defined. The origin of F1 does not move along a line; it occupies an entire line simultaneously.

 

[StarThrower]

Let's make this more explicit. You've defined the spatial axes for your photonic frame. I have no problem with those.

You seem to imply that a photon-centered clock is used to define the temporal axis.

The problem is that you give no reason to believe the photon-centered clock ever changes its reading. Thus your temporal axis is ill-defined. The origin of F1 does not move along a line; it occupies an entire line simultaneously.

Hurkyl, you seem to be convinced that a clock at rest with respect to the photon doesn't tick in the photon rest frame. That would mean that the position of the origin of F1 is changing in F2, even though the clock isn't ticking. Then, in the definition of the speed of F1 through F2, you have a non-zero numerator, and a denominator of zero, which is the division by zero error of algebra.

 

[Hurkyl]

Then, in the definition of the speed of F1 through F2, you have a non-zero numerator, and a denominator of zero, which is the division by zero error of algebra

Bingo! Thus, you can't say the speed of F1 is constant, because the speed isn't even defined!

 

[StarThrower]

And another nifty thing about degenerate coordinate systems; even if you could convince me that a photon-centered inertial reference frame exists, there is no contradiction between the assertion "The speed of the photon is c" and "The photon is at rest" because the worldline of the photon is a single point.

That isn't a way out.

Postulate: The speed of a photon is 299792458 m/s in ANY inertial reference frame.
The postulate is either true or false.

The postulate is false provided there is at least one inertial reference frame in which the speed of a photon isn't 299792458 m/s. In a photonic reference frame (reference frame in which a photon is at rest), the speed of a photon is 0. Thus, if a reference frame in which a photon is at rest is an inertial reference frame, then the postulate is false.

A reference frame in which a photon is at rest will be an inertial reference frame if Newton's first law of motion is satisfied. Newton's first law of motion is that an object at rest will remain at rest unless acted upon by an outside force, and an object in motion will remain in motion in a straight line at a constant speed, unless acted upon by an outside force. This is precisely the case with reference frame F1 in this thread.

We are standing on top of the whole issue. Will a clock traveling along with a photon tick. The answer is either yes or no. You say no. That is what I have a huge problem with, and won't ever believe.

Kind regards,

The Star

 

[Hurkyl]

I deleted that post because upon reflection, I didn't think it made sense (because, as is brought up, you can't really define speed)

 

[StarThrower]

I deleted that post because upon reflection, I didn't think it made sense (because, as is brought up, you can't really define speed)

To me, this is all a matter for binary logic, and its not really even so hard.

There is one statement we are all investigating, and that statement is world-famous. The statement I am challenging is the fundamental assumption of the special theory of relativity.

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

Using nothing more than binary logic, the statement is false provided there is at least one reference frame which is an inertial reference frame, in which the speed of a photon isn't c.

Clearly, a reference frame with a photon at the origin is a reference frame in which the speed of a photon isn't c. The question now is whether or not such a frame is an inertial frame.

We have a definition for an inertial reference frame. Hence regardless of your intelligence and mine, the answer is decidable. I am trying to show you that Newton's first law is satisfied in a photonic frame.

Kind regards,

The Star

 

[Hurkyl]

Will a clock traveling along with a photon tick. The answer is either yes or no. You say no. That is what I have a huge problem with, and won't ever believe.

The problem is that science doesn't conform to what people want to believe.

Incidentally, I can't imagine how one would go about making a clock that can travel at light speed anyways. If light speed clocks don't exist, then it is vacuously true that all light speed clocks don't tick, and also true that all light speed clocks do tick.

 

[Hurkyl]

We have a definition for an inertial reference frame. Hence regardless of your intelligence and mine, the answer is decidable. I am trying to show you that Newton's first law is satisfied in a photonic frame.

We do have a definition. We haven't discussed if that definition is valid in SR, and we don't even seem to agree on what the definition says.

 

[StarThrower]

The problem is that science doesn't conform to what people want to believe.

Incidentally, I can't imagine how one would go about making a clock that can travel at light speed anyways. If light speed clocks don't exist, then it is vacuously true that all light speed clocks don't tick, and also true that all light speed clocks do tick.

If the number of light speed clocks that exist is zero, then the number of light speed clocks that tick is equal to zero.

SR argument for why there are no light speed clocks:

$$\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}$$

suppose v=c, then we have division by zero error unless $$\Delta t^\prime = 0$$

In a case where v=c what we get is this:

$$\Delta t = \frac{0}{0}$$

The RHS is in an indeterminate form, but not so for the LHS.

delta t is the amount of time passing on a clock in a frame in which the photon's speed is c.

delta t` is the amount of time passing on a clock at rest with respect to the photon.

The conclusion is that there are no light speed clocks.

But this conclusion is not, epistemologically speaking, absolute knowledge; rather it is contingent knowledge. Logically, all we know for certain is this:

If $$\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}$$ then there are no light speed clocks.

But the time dilation formula is derived using the main postulate of SR, which is that the speed of a photon is c in any inertial reference frame. So this centers the issue on whether or not the rest frame of a photon is an inertial reference frame.

Theorem to prove: Suppose F1 is an inertial reference frame. Let V denote a vector in another reference frame F2. If all the points of V move in a straight line at a constant speed in F1, then F2 is also an inertial reference frame.

This theorem is mathematically provable. Hence the main postulate of SR is false.

Kind regards,

The Star

 

[StarThrower]

We do have a definition. We haven't discussed if that definition is valid in SR, and we don't even seem to agree on what the definition says.

Not agreeing on what the definition says might be a problem.

An inertial reference frame is a reference frame in which all three of Newton's classical laws are satisfied.

In particular, the classical form of his second law is:

F = dP/dt


Kind regards,

The Star

 

[Hurkyl]

Actually, there's more to the proof than that (it's easiest to use the differential formula for proper time), but the details are irrelevant since it's still a consequence of the assumption of a constant speed of light.


Anyways, I'm now convinced that the definition you are using is a bad one, because it doesn't even work for classical mechanics; boosts preserve straight lines, even in classical mechanics, but no inertial frame of classical mechanics is a boosted frame.

 

[Severian596]

StarThrower is beating around the bush without offering answers to his own questions. The title of this thread is "The Fundamental Postulate Of Special Relativity Is Self-Contradictory." If you want to convince anyone of this other than yourself StarThrower, I suggest you at least attempt one of the following:

1) Show the mathematical proof that "F1 is an inertial reference frame. Let V denote a vector in another reference frame F2. If all the points of V move in a straight line at a constant speed in F1, then F2 is also an inertial reference frame." I suspect that 1 will require you to
1.1) Prove that the reference frame of a photon has points.
1.2) Prove that the reference frame can travel at the speed of light and still have meaningful coordinates.
1.3) Do not assume it's given (as you've been doing)

2) Describe the reference frame of a photon in more detail. If an inertial reference frame exists at v=c in your thought experiment, what are its properties?

3) Perphaps approach it from a different angle...assume you're correct. What are the implicaitons?

 

[Severian596]

StarThrower, this is my plea that you provide some concrete information. Some concrete conclusions. Yes I understand your logic:

(1) SR states that photons travel at c through a vacuum in all inertial reference frames.
(2) Photons have inertial reference frames.
(3) Photons do not travel through their own inertial reference frames at c.
(c) Therefore SR is incorrect/inconsistent.

I can accept the first premise because it reflects our current understanding of the universe, and there has been substantial evidence (and the support of uncounted professionals and scholars) to back up. So you don’t have to convince me—or anyone else—of statement one.

So that leaves you with statements (2) and (3). Proving one or both of these is not trivial, and thus far the most I’ve seen from you on proving their truth are comments like these:

"A reference frame in which a photon is at rest happens to be an inertial reference frame, so the whole theory is self-contradictory."

"Clearly, a reference frame with a photon at the origin is a reference frame in which the speed of a photon isn't c. The question now is whether or not such a frame is an inertial frame."

"A reference frame in which a photon is at rest will be an inertial reference frame if Newton's first law of motion is satisfied. Newton's first law of motion is that an object at rest will remain at rest unless acted upon by an outside force, and an object in motion will remain in motion in a straight line at a constant speed, unless acted upon by an outside force. This is precisely the case with reference frame F1 in this thread."

You make assumptions. You assume that photons are blessed with frames of reference, and in fact that they are inertial in nature! If Newtonian physics break down at near-light speeds why should they all of a sudden apply to a reference frame that is traveling at c? And how can you conclude that Newton’s first law applies in such a frame? You’re going to have to convince us of that, because we’re not biting yet.

If you manage to prove (2)—the more difficult in my opinion—then it seems to make sense that concluding (3) shouldn’t be as difficult. But then again who knows. You haven’t proven (2) yet so that’s down the road.

So…regardless of ticking clocks at light speed, or proper time, you have the responsibility of proving that statements (2) and (3) are true. You’ll also have to do this without the benefit of using any equations of SR, since those equations are based on your conclusion being FALSE. You’re trying to prove otherwise.

If you have anything substantial or compelling I suggest you present it now. And two or three catch phrases won’t cut it.

 

[StarThrower]

StarThrower is beating around the bush without offering answers to his own questions. The title of this thread is "The Fundamental Postulate Of Special Relativity Is Self-Contradictory." If you want to convince anyone of this other than yourself StarThrower, I suggest you at least attempt one of the following:

1) Show the mathematical proof that "F1 is an inertial reference frame. Let V denote a vector in another reference frame F2. If all the points of V move in a straight line at a constant speed in F1, then F2 is also an inertial reference frame." I suspect that 1 will require you to
1.1) Prove that the reference frame of a photon has points.
1.2) Prove that the reference frame can travel at the speed of light and still have meaningful coordinates.
1.3) Do not assume it's given (as you've been doing)

2) Describe the reference frame of a photon in more detail. If an inertial reference frame exists at v=c in your thought experiment, what are its properties?

3) Perphaps approach it from a different angle...assume you're correct. What are the implicaitons?

It is impossible to answer all your questions simultaneously. This particular post of yours practically begs the question, "how do you know space is three dimensional?"

You ask: 1.1) Prove that the reference frame of a photon has points.

Set up a rectangular coordinate system with a photon fixed at the origin. This coordinate system assigns 3 numbers to each point in real space. Photons exist somewhere in real space. In principle, this reference frame has as many coordinate assignments as there are point in real space. Real space consists of an infinite number of places. Your question has been answered. Since real space consists of more than one point, a coordinate system with a photon at the origin will consist of a plurality of points.

You ask: 1.2) Prove that the reference frame can travel at the speed of light and still have meaningful coordinates.

I am not sure what you mean by "meaningful coordinates."


You ask: 1.3) Do not assume it's given (as you've been doing)

Do not assume (what) is given?

You ask: 2) Describe the reference frame of a photon in more detail. If an inertial reference frame exists at v=c in your thought experiment, what are its properties?

A rectangular coordinate system consists of three infinitely long straight lines that meet at a single point, such that the three straight lines are mutually perpendicular. The lines are number lines, with units that correspond to some chosen standard of distance. Points in space are assigned three numerical coordinates which have meaning in this system and no other. No two points in a coordinate system can move relative to each other. Thus, the distance between any two points in a system is constant.

Kind regards,

The Star

 

[Severian596]

The lines are number lines, with units that correspond to some chosen standard of distance.

Your definition for a coordinate system is correct as long as our common sense idea of space applies. In this case you are asserting that our common sense idea of space must apply to a frame traveling at c.

Why?

Our common sense idea of space does not apply at relativistic speeds.

 

[Severian596]

And let me ask this. If considering two frames of reference S and S', observerations in one frame must exactly coincide with observations made from the other. This means the frames are interchangeable (that is we could call S the frame "in motion" relative to S', or we could call S' the frame "in motion"), and that neither frame is preferred over the other.

So how can we assert that a photon in "rest frame" S' be at rest and that S be moving at the speed of light?

 

[Severian596]

Finally, I found this concise quote that reflects my thoughts:

"Things traveling at the speed of light don't have rest frames the way that other things do. The Lorentz transformation equations that relate coordinates in different frames are singular if the relative velocity of the two frames is c."

This summarizes my point earlier, about the nature of space and transforming coordinates as your velocity increases.

 

[StarThrower]

StarThrower is beating around the bush without offering answers to his own questions. The title of this thread is "The Fundamental Postulate Of Special Relativity Is Self-Contradictory." If you want to convince anyone of this other than yourself StarThrower, I suggest you at least attempt one of the following:

1) Show the mathematical proof that "F1 is an inertial reference frame. Let V denote a vector in another reference frame F2. If all the points of V move in a straight line at a constant speed in F1, then F2 is also an inertial reference frame."

Let F1 be a rectangular coordinate system that is also an inertial reference frame.
Let F2 be a rectangular coordinate system.

Let A, B be points in F2.
Let the vector from A to B be symbolized as: $$\vec{AB}$$

Let A be moving in a straight line at a constant speed through F1.
Let B be moving in a straight line at a constant speed through F1.

Theorem: F2 is an inertial reference frame.

Will this theorem be sufficient?