The Fundamental Postulate Of Special Relativity Is Self-Contradictory

[Severian596]

Are you planning on answering mine and Hurkyl's questions, StarThrower?

 

[Hurkyl]

Maxwell's equations predict one universal value for the speed of light relative to the source.

In order to argue this, you really should point out something source related in the wave equation.

 

[Severian596]

I found this through google.com searching:

http://www.google.com/search?q=cache:QDMd3MVoHcoJ:www.physicsforums.com/archive/topic/11184-1.html+photon%27s+reference+frame&hl=en&ie=UTF-8

This topic (and the one we're in) is fun for me because it only reinforces my understanding of SR. Unfortunately for StarThrower the discussion isn't novel and apparently he isn't the genius (or alien from 83 million light years away) that he thinks he is.

But anyway I'm still wondering if he's going to oblige us with some answers to our questions...and I won't tell you my theory on whether StarThrower = Tempest.

 

[Severian596]

Theorem Of Special Relativity: If all coordinates of reference frame F2 are moving in a straight line at a constant speed in some inertial reference frame F1, then F2 is also an inertial reference frame.

Ah-hA! Very interesting that StarThrower started off this entire thread with this "theorem of SR." You know why it's interesting? I was reading through the post I mentioned in my last message and came across StarThrower's contribution to the conversation. He admitted that, "This post has really got me thinking," and then he summarized his ideas with the following paragraph:

That being said, if you can now show that [the photon's rest frame] MUST be an inertial reference frame, you would likely be onto something. And so this is why you have gotten me thinking. Here is what would have to be done. You would have to prove that if X is an inertial reference frame, and Y is a reference frame whose origin is moving at a constant speed relative to the origin of X, and whose axes aren't rotating in frame X, then Y is an inertial reference frame.

Then, because the photon is moving at a constant speed in the atomic frame, it would follow by a theorem not yet proven, that a frame in which the photon is at rest, is an inertial reference frame. And since one of the consequences of the special theory is that in any inertial reference frame the speed of any photon is c, you would have accomplished something.

Have you proved this then, StarThrower? You made a powerful statement in your opening message by saying that it's a "theorem of SR." Whose theorem is it? Can you explain it? I'm still reading the other thread (click on my link in my last post) to see what's said, but I thought the content very appropriate to post here...

 

[outandbeyond2004]

You found a contradiction after assuming that the photon's rest frame is an inertial frame of reference. Congragtulations! Very well, then: Let's agree that every photon's rest frame is NOT an inertial reference frame. Let's please not push SR out of the boundaries of its validity. It was never meant to be a theory for ALL situations. There is after all a reason why Einstein went on to develop GR. And, have you never heard of efforts to develop GUTs? Have you got no gut feeling for such things whatsoever (sorry couldn't resist).

 

[Severian596]

I just want to make sure...you're addressing more than one person in your post, right? You use 'you' a few times...

 

[JJ]

Jesus Christ, this thread is confusing the hell out of me. Can I safely assume what Integral is saying is right?

 

[Severian596]

What did he say, JJ? I mean, which thing that he said are you referring to?

 

[russ_watters]

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

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.

Jesus Christ, this thread is confusing the hell out of me. Can I safely assume what Integral is saying is right?

Yes, JJ.

As for trying to follow this discussion, its probably best to let it go: in your first post, you demonstrated an understanding of SR that far exceeds StarThrower's. But trying to make sense of this confusing mess is like trying to untangle spaghetti and not worth your effort.

Keep asking the good questions and welcome to the forum.

 

[JJ]

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.

 

[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!

 

[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.