# The Fundamental Postulate Of Special Relativity Is Self-Contradictory

1. Mar 31, 2004

### StarThrower

Fundamental Postulate Of Special Relativity: The speed of a photon in any inertial reference frame must be measured as c, where c = 299792458 meters per second.

It is provable within the framework of the special theory of relativity, that

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.

Let F1 be an inertial reference frame. Let two photons be moving in the same direction in F1.

By the fundamental postulate of the theory of special relativity, the speed of photon A in inertial reference frame F1 is 299792458 meters per second.

By the fundamental postulate of the theory of special relativity, the speed of photon B in inertial reference frame F1 is 299792458 meters per second.

Since they are moving in the same direction, the difference in their velocity vectors (as defined in F1) is equal to zero. Thus, the two photons are not moving relative to each other.

Define coordinate system F2, to have photon A as origin, and let the unit vector on the positive x axis point to photon B. By the previously mentioned theorem of SR, F2 is an inertial reference frame.

Now, since the photons are not moving relative to each other, the speed of photon B in F2 is equal to zero. And F2 is an inertial reference frame.
By the fundamental postulate of the special theory of relativity, the speed of photon B must equal 299792458 meters per second in any inertial reference frame, hence by the fundamental postulate of SR, the speed of photon B in F2 must equal 299792458 m/s, therefore the speed of photon B in F2 isn't equal to zero.

Hence, we arrive at the following explicit contradiction:

The speed of photon B in F2 is equal to zero, and the speed of photon B in F2 isn't equal to zero.

Therefore, the fundamental postulate of the theory of special relativity is false.

QED

Respectfully,

The Star

Last edited: Mar 31, 2004
2. Mar 31, 2004

### Phobos

Staff Emeritus
In your first example, the two photons are in the same reference frame, so it's a non-issue.

In the second example, where you separate the two photons into two reference frames (no apparent reason why since they're both still moving in the same way relative to each other), there is no information exchange between the two reference frames so A can't measure anything from B. Non-issue.

There seems to be some confusion between measured speed and actual speed. A and B are still moving at c (same ref frame) even though they don't notice any difference between themselves. Not that photons can notice anything...

3. Mar 31, 2004

### Severian596

StarThrower, unlike most of your posts this actually piqued my interest. I typed "photon as origin reference frame" in Google.com as I'd never considered trying to create an inertial frame of reference for a photon. The first link that came up (posted below) dealt with photon kinematics, and it seems that certain measures must be taken when using a photon's reference frame.

Consider the idea that perhaps two photons traveling parallel to the X-axis could never measure each others' speed because they would never observe each other. Any light leaving photon A would have to travel some distance $\Delta y \$ at speed c. If $\Delta y \$ > 0, it will never reach photon B.

Anyway those were my ramblings and I'm at work right now. Here's that link about photon kinematics:

http://www.comcity.com/distance-time/Photon Kinematics.html

Unlike your other posts this one did NOT involve changing frames of reference, acceleration, or other GR topics, claiming that they "debunked" SR. So at least I gave it more attention than I gave the others.

BTW, if you would explain who you are and what formal education you've had on the topics of SR and/or GR I'd be VERY curious. Every post you've created had some "SR is wrong" theme...would you mind elaborating on why we should believe that you have the knowhow to claim such things?

Last edited: Mar 31, 2004
4. Mar 31, 2004

### StarThrower

You must have misunderstood something, since there is only one "example". In fact, it is not even an example. I will re-explain.

We have two photons, with the same velocity in inertial referemce frame F1.

Photon A *____________________________* Photon B
Velocity: 299792458 m/s ---> Velocity: 299792458 m/s --->

The difference in the velocity vectors is zero in this frame, hence it is zero in all reference frames. Thus, these two photons are at rest with respect to each other in all reference frames.

I then define reference frame (really rectangular coordinate system) F2 as follows:

The origin of F2 is photon A.
The direction of the i^ unit vector points from photon A to photon B. Thus, photon B lies on the positive x axis of F2.

I then state (without proof yet) that by a theorem of special relativity, F2 is an inertial reference frame. It then follows by the fundamental postulate of the theory of special relativity that BOTH photons must be moving in F2, at a speed of 299792458 m/s (simply because F2 is an inertial reference frame).

An explicit contradiction follows. It doesn't get better than this.

5. Mar 31, 2004

### jdavel

Severian596: "....I gave it (star thrower's post) more attention than I gave the others."

Save your time on this one too; once you cut through star thrower's obfuscation (deliberate or not) his argument is:

1) The postulate says all photons move at c

2) Suppose I have a photon that's not moving at c

3) Therefore the postulate is wrong

6. Mar 31, 2004

### StarThrower

Since the photons must have the same speed in reference frame F1, it necessarily follows that if they are moving in the same direction in F1, that they are not in relative motion (whether an experiment could prove they aren't in relative motion is another issue). But, I have stipulated that they have the same velocity precisely so that they aren't in relative motion.

Now, if there is to be some measurement of the relative speed, that measurement has to be zero, in order to be a perfect measurement. If photon C (next to and moving parallel to photon A) emitted photon D in the direction of photon B, the speed of photon D in reference frame F1 would exceed the speed of light (299792458 m/s), contrary to the fundamental postulate of SR. Hence, if SR is correct then the hypothetical experiment is impossible to carry out.

The entire question then, is whether or not F2 is an inertial reference frame. All attention must then be shifted to the theorem I stated without proof yet.

As for who I am... that doesn't matter right now.

As for my familiarity with SR/GR, that does matter.
I studied at an excellent university in the northeastern US, for approximately ten years. For now, this is all I'll say on this.

Last edited: Mar 31, 2004
7. Mar 31, 2004

### StarThrower

No, that isn't the argument, in fact what you said doesn't make sense.

First of all, the postulate doesn't say that all photons move at speed c=299792458 m/s. The postulate says that in any INERTIAL REFERENCE FRAME the speed of a photon must be 299792458 m/s. Certainly, there are non-inertial reference frames in which the speed of a photon isn't equal to 299792458 m/s.

The issue then becomes whether or not a reference frame moving along with a photon is an inertial reference frame or not.

I then clearly say that if the fundamental postulate of SR is true, then any reference frame moving along with a photon is an inertial reference frame. In any such frame the speed of a photon is zero, and not c=299792458 m/s.

It now follows that SR self contradicts.

Kind regards,

The Star

Last edited: Mar 31, 2004
8. Mar 31, 2004

### DrChinese

There is no inertial reference frame that can move at c=299792458m/s. Even if you try to define one at such a velocity, it does not exist. There are no reference frames at 1000000000 m/s either.

SR is not self contradictory, but it does have limits of applicability. No secret that.

9. Mar 31, 2004

### StarThrower

This is just you saying things.

Kind regards,

The Star

10. Mar 31, 2004

### DrChinese

Me saying things? You are the one violating the speed limit. Write yourself a ticket. Don't worry: it won't affect your insurance.

11. Mar 31, 2004

### JJ

A high schooler butting in here:

CAN a photon be an inertial reference frame? The fact that the photon always travels at c means that it cannot accelerate or otherwise be affected by a force, only distortions; therefore, according to my limited knowledge, it cannot not be an inertial reference system. So the photon would be an absolute speed limit. Saying that photons are moving at the identical speed is incorrect, since their speed is imaginary. You can't force it beyond c, or force it back, so for all intents and purposes, the second photon is always travelling at the speed of light relative to the first.

I admit I'm out of my league, so I'm only asking for a simple yes or no answer. And I do realize that what I've written is incoherent.

12. Mar 31, 2004

### StarThrower

The first thing to say is that it isn't a fact that a photon always moves at speed c.

Yes, a photon can be in an inertial reference frame. An inertial reference frame is a reference frame in which Newton's law of inertia is valid.

In order to discuss the motion of something using mathematics, physicists use coordinate systems. These are extensively studied in mathematics, and the concept of the cartesian (rectangular) coordinate system dates back to Rene Descartes.

A rectangular coordinate system consists of three mutually perpendicular number lines, with a unit of distance in real space chosen. The international unit of distance is the meter.

So, the three lines are number lines, and are called the axes of the coordinate system.

Now, the fundamental postulate of the special theory of relativity is that the speed of a photon in any inertial reference frame must be 299792458 meters per second. Suppose you have a reference frame set up, which you know for a fact is an inertial reference frame, and that currently there is a photon located at the origin of this reference frame (or coordinate system).

ASSUMING that the special theory of relativity is correct, it must be the case that after one second has elapsed, the distance from the origin of this coordinate system to the location of the photon in this coordinate system must be 299792458 meters. And the time measurement (which I am saying is one second) is to be measured by a clock which isn't moving in this coordinate system.

The point is, that certainly a photon can move through some inertial reference frame/coordinate system.

A different question is whether or not a reference frame which is 'attached' to a photon is an inertial reference frame.

If special relativity is correct then the answer is no.
If the answer is yes then special relativity is incorrect.

Also, photons can be accelerated. Keep in mind that if a photon changes direction of travel, it has been accelerated. Thus, when photons strike a mirror, they were accelerated.

Regards,

Star

13. Mar 31, 2004

### DrChinese

Sounds pretty good to me.

P.S. I would not take Star's comments as gospel. There is a good reason why this thread was moved to Theory Development.

14. Mar 31, 2004

### ahrkron

Staff Emeritus
Very good post JJ.

As DrChinese wrote, StarThrower is not the best guide you can find (for physics at least). Relativity is extremely well established (both in terms of internal consistency and of agreement with experiment); physicists are nowadays working on quite different problems. Special relativity is just your basic "bread and butter" stuff.

15. Mar 31, 2004

### jdavel

star thrower: "....the postulate doesn't say that all photons move at speed c=299792458 m/s.

Einstein: "We will....also introduce another postulate....namely that light is always propagated through empty space with a definite velocity c which is independent of the state of motion of the emitting body"

Your argument falls apart as soon as you say "the speed of photon B in F2 is equal to zero" According to the postulate of SR, photons don't EXIST at zero speed; they don't even exist at c/10 or c/2 or .999c. They only exist at speed c. If it's not going at speed c, it's not a photon!

Is that counter intuitive? Of course it is! Does it seem wrong based on our everyday experience? Of course it does! But it's just a postulate. The only way to disprove it, is to find an example in the physical world where it's not true. You don't get to disprove it with a thought experiment. Gallileo didn't disprove that the earth was at the center of the solar system by just saying "Let the sun be at the center of the solar system. Now doesn't that seem more reasonable?" He disproved it by building himself a telescope and looking at the solar system. That's how physcial science is done.

In a more positive vein, you seem interested in this theory. Why not just agree with yourself to pretend you believe the postulate. Believe it conditionally for awhile. Then get yourself an introductory textbook on SR and read it from cover to cover. Do all the problems until you can get the answers the book says are right (even if you think they're wrong). At that point, you'll understand the theory well enough to decide whether or not to abandon it. Keep us posted!

16. Mar 31, 2004

### Integral

Staff Emeritus
In the frame of reference of a photon there is no distance or time. As far as the photon is concerned it is adsorbed the instant it is emitted. Since it has traveled no distance in no time there is no problem.

Any argument made from the frame of reference of a photon must take this into consideration. We live in the world of distance and time, the photon does not. So when you

With this in mind let us look at Stars argument.
Since photons do not know about length there is no such thing as direction. Since photons do not know motion the conclusion that the are not moving with respect to each other is trivial.

Ok lets measure the speed of photon B in the frame of reference of photon A. It moves no distance in no time, remember time and distance do not exist for a photon.

now lets measure the speed of photon A in the frame of reference of photon B. It moves no distance in no time, because, again, time and distance do not exist for a photon.

A key to understanding SR is the ability to use the Lorentz transforms. We can compute what the photon know of our time by

$$t_{photon} = \sqrt { 1 - {\frac v c}^2}t_{us}$$
similar for the length
$$x_{photon} = \sqrt { 1 - {\frac v c}^2}x_{us}$$

insert v=c in these relationships to see that real time and distance does not exist for a photon. If you wish to measure time and distance as known by a photon you need to use these formulas.

SR is self consistent.

17. Mar 31, 2004

### StarThrower

My comments shouldn't be taken as gospel, nor randomly dismissed. My advice to JJ would be to continue thinking for himself, and do it by using binary logic correctly.

And the reason the thread was moved, is because the moderators believe the the theory of special relativity is correct. They want to be in the majority. That is the herd instinct in them winning out over binary logic.

Kind regards,

The Star

18. Mar 31, 2004

### StarThrower

Ahrkron, you don't know what kind of guide I am. For all you know, I am an alien from a world approximately 83 million light years away, that came here on a spaceship which can break the speed of light. Obviously, my civilization knows the time dilation formula is incorrect, and to us you sound foolish.

The whole point of this thread, is to inform your species that the theory of special relativity is not internally consistent. Where is your head?

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

Kind regards,

The Star

19. Mar 31, 2004

### StarThrower

Ummm no.

20. Mar 31, 2004

### StarThrower

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.

Kind regards,

The Star