The Fundamental Postulate Of Special Relativity Is Self-Contradictory

[russ_watters]

This is not a refutation of my argument. This is just you telling everyone here that you believe the theory of special relativity is self-consistent. We already knew you believed that.

So... you assert that our assertion that your assertion is erroneous is in error? Ok, I assert you are wrong. Your move.

StarThrower, you started this thread. Presumably you want to convince us of something. So do it. Otherwise, what is the point of continuing to post these threads?

Lastly, the phrase "bread and butter stuff" is total nonsense. I know of no analog for this phrase in my language.

Essentially, "bread and butter stuff" is the basics.

 

[russ_watters]

Severian:

That concepts of distance, time, and motion are unapplicable from the point of view of a photon. I figure that because of that, the idea of the speed of light is only true with reference frames moving below it. The speed of light is a good reference, but not a good observer.

You're still holding on - yeah, that's correct. The concepts StarThrower is trying to apply to light simply do not apply.

To extend a little, "quantum weirdness" is likely due to the special nature of the photon. For example, a photon "knows" its path before its path even exists (yes, this has been experimentally proven). Why? Well, for a photon, there is no "before" since time doesn't exist for a photon. Its path just is.

 

[Chen]

StarThrower, you started this thread. Presumably you want to convince us of something. So do it. Otherwise, what is the point of continuing to post these threads?

http://info.astrian.net/jargon/terms/t/troll.html

 

[StarThrower]

I'm not sure anyone answered this succinctly. The answer is NO.

Therefore, your example is invalid for an examination of SR. It quite simply doesn't apply.
Yes, JJ.

Prove that the answer is no Russ.

Hey JJ, ask Russ to prove that the answer is no. Since the answer is yes, it is impossible for him to prove the answer is no. A frame in which a photon is at rest is an inertial reference frame.

Kind regards,

The Star

 

[Severian596]

http://info.astrian.net/jargon/terms/t/troll.html

LOL, excellent, Chen. In my position I could feel abused by StarThrower's trolling, but I'm actually pretty thankful I followed this thread because it has helped me look at SR from a perspective I hadn't yet considered; that of the perspective of the photon.

 

[StarThrower]

So... you assert that our assertion that your assertion is erroneous is in error? Ok, I assert you are wrong. Your move.

StarThrower, you started this thread. Presumably you want to convince us of something. So do it.

An inertial reference frame is a reference frame in which Newtons law of inertia is true.

Newton's law of inertia is true in a coordinate system in which a photon is at the origin.

Hence, a frame in which a photon is at rest is an inertial reference frame.

Kind regards,

The Star

 

[StarThrower]

LOL, excellent, Chen. In my position I could feel abused by StarThrower's trolling, but I'm actually pretty thankful I followed this thread because it has helped me look at SR from a perspective I hadn't yet considered; that of the perspective of the photon.

Absolutely no trolling here, I am simply saying that you can prove that the rest frame of a photon is an inertial reference frame.

Kind regards,

The Star

 

[Severian596]

Newton's law of inertia is true in a coordinate system in which a photon is at the origin.

Prove this please.

 

[quantumdude]

An inertial reference frame is a reference frame in which Newtons law of inertia is true.

Newton's law of inertia is true in a coordinate system in which a photon is at the origin.

Hence, a frame in which a photon is at rest is an inertial reference frame.

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.

 

[Severian596]

For a spacetime diagram for your run-of-the-mill reference frame S, you can define up to 4 axes, but let's just use 2 and assume no y or z movement takes place. You then define the x-axis and the y axis. Like this example:

http://vishnu.mth.uct.ac.za/omei/gr/chap1/img35.gif

The red dotted line represents the path of light. The blue line represents a superimposed axis of reference S' that has undergone a transformation. Notice that only ct' is superimposed. If x' were also superimposed it would fall between the x-axis and the red dotted line.

So what happens as v approaches c? Could someone let me know if I've got this part wrong? 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:

\frac{\infty}{x} = \infty

It's ugly, but I believe it starts to illustrate the problem with defining a reference frame that has no eligible (or meaningful) coordinates to define.

UPDATE:
I must say I have the sneaking feeling that this is an incorrect way of representing it, and that this problem is similar to dividing by zero in mathematics...it's simply undefined. My example smacks a bit of Zeno's paradox, because any attempt to move only finds that you have moved no where compared to the scale of a "single step."

 

[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>
[/size]

 

[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>
[/size]

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!