The Fundamental Postulate Of Special Relativity Is Self-Contradictory

[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

[Phobos]

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... [;)]

[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%20Kinematics.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?

[StarThrower]

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... [;)]

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.

[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

[StarThrower]

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%20Kinematics.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?

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.

[StarThrower]

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

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

[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.

[StarThrower]

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.

This is just you saying things.


Kind regards,

The Star

[DrChinese]

This is just you saying things.

Kind regards,

The Star

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

[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.

[StarThrower]

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.

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

[DrChinese]

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.

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.

[ahrkron]

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.

[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! :wink:

[Integral]

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 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.

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.

the speed of photon B must equal 299792458 meters per second in any inertial reference frame,

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.

[StarThrower]

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.

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

[StarThrower]

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.

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

[StarThrower]

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! :wink:


Ummm no. :cool:

[StarThrower]

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.

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

[Chen]

Ummm no. :cool:
Oh, the eloquence! :rolleyes:

[Integral]

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


Do the math and weep. Photons do not know time or distance. Therefore do not know velocity. That is proper applcation of SR. If you do not use the tools correctly you do not get meaningful results. SR does not specify a number for c it only says that it is constant. In the frame of reference of a photon the velocity of all photons is zero. That is a constant and satisfies the postulate of SR.

Edit:
Added a lost negation.

[Eyesaw]

This is just another example of the violation of simple logic by the
SR postulates. Inside an arbitrary set of "rest coordinates", if J is moving at c with respect to M and M is moving at v with respect to L then J- whether it represent jumping frogs, waffling ducks, bald eagles, flying beagles, or photons- must be moving at c +/-v with respect to L, else it would lead to reductio absurdum- i.e. you start
out with the assumption that there is velocity between M and L but because
of the logic of SR, you end up with the conclusion that there is no velocity between
M and L.

A photon, after emitted in vacuum, is not in anyone's inertial frame unless we pondered an ether- thus, it is painfully obvious that different observers moving
through the empty vacuum at different velocities must have a different velocity
with respect to the photon because 2+2 = 4, 3+3 = 6, 4+4 = 8, 5+5 =10 and 6+6
= 12. Sure, one can still obtain the speed of the photon through space as a constant
after subtracting the velocity of the observer through space but this was not the
method chosen by Einstein in SR, even though it is the most simple way to achieve
constancy of c that is independent of source. In lieu of using simple math and
commmon sense, the physics community have decided warping space and time is cooler so nevermind that it is wrong.

[Integral]

Eyesaw,
Do you by any chance speak English? You do not make a single coherent sentence. I refuse to waste my time interpreting garbled language.

[Integral]

Star,
Let me but this way, your assertion that you can measure the velocity of a photon from the frame of reference of a photon is incorrect. In SR there are no meaningful measurements that can be made in the frame of reference of a photon. So your example is erronous, a photon cannot see the speed of a second photon.

But then since you have convinced yourself that SR is invalid without the ability to correctly apply it or even understand it. I do not expect you to accept it.

But then since it is incorrect in would not any application of it lead to incorrect results. You applied it therefore your results are incorrect.

[outandbeyond2004]

Let's be realistic. An inertial reference frame is one that an observer like you or me can set up. Not one that can be set up in imagination like Starry's fast moving frame. Just in reality. No observer can move at c (because he has some rest mass, unlike photons). So, no such thing as an inertial rf that moves at c --except in Starry's imagination (just joshing, mind you).

[Hurkyl]

I find it interesting that (to my knowledge) there are exactly zero anti-SR arguments that assign space-time coordinates to events and analytically derive a fallacious conclusion.

[outandbeyond2004]

Hurkyl, by 'analytically' you mean 'with sound logic'?

[Hurkyl]

That's a given. :smile:

But no, I meant that they never attempt to do any sort of geometry or calculus to derive results.

[outandbeyond2004]

Hurkyl, perhaps you meant 'derive a contradiction' for 'derive a fallacious conclusion'?

[Chen]

I really hope StarThrower hasn't spend all ten years studying Special Relativity. :smile:

[Michael F. Dmitriyev]

Certainly, light always travel in c. Here SR right.
But Einstein had not made a suitable explanation, why it so.
Naturally, absence of a convincing explanation results in doubts in validity of this postulate. The constant mess with frames of reference does not promote the understanding of this the postulate also. In my opinion, there is the best way for an explanation of the light’s speed independence from the speed of radiator.
It is necessary to take into account, that the photon can be found out only face to face. It can’t be observed, and therefore can’t be measured from the “sideways”.
ST (StarThrower),
You are right too (though it can sound strange).
The photons radiated by a short laser pulse will achieved the target
SIMULTANEOUSLY. This an experience shows, that they have passed this way with an identical
speed.
The concept "relativity" is applicable to any material objects, except of light.
They exist in RELATIVE TIME.
Speed of Light is ABSOLUTE.
Light exist in ABSOLUTE TIME.

[Integral]

Michael,
You fail to understand the meaning of postulate. Einstein is not obligated to prove anything about his postulate. That is the nature of a postulate. You most certainly cannot learn anything other then the results of a constant c by studying Relativity. It is after all a development that explores the implications of a constant speed of light.

If you wish to find the roots of the constancy of c you need to study and understand the work of Clerk Maxwell. The origins of a velocity that is independent of the source, is Maxwell's equations cast in the form of the wave equation. When this result was published in the late 1860s the world of Physics was changed forever. How to rectify the source independence of the speed of Electromagnetism with the accepted and well understood precepts of Classical Mechanics was the single largest issue in Physics of that era. Due to that Einstein was able to postulate a constant c. Physicist of that era did not blink an eye at the postulate because they had spend a generation attempting to disprove the constancy of c. They failed.

[Integral]

Ok Eyesaw, you win, lets attempt to make some sense of this post.


This is just another example of the violation of simple logic by the SR postulates.



What is an example of the SR postulates. Or are you saying the SR postulates are an example of a violation of logic.

Do you even know what the postulates of SR are?


Inside an arbitrary set of "rest coordinates", if J is moving at c with respect to M and M is moving at v with respect to L then J- whether it represent jumping frogs, waffling ducks, bald eagles, flying beagles, or photons- must be moving at c +/-v with respect to L, else it would lead to reductio absurdum- i.e. you start
out with the assumption that there is velocity between M and L but because
of the logic of SR, you end up with the conclusion that there is no velocity between
M and L.

That is one heck of a sentence, but what does it say? This looks to be some kind stawman construction that really means nothing.

What I say is that you cannot measure time or space from the frame of reference of a photon, therefore you cannot measure the speed of a photon.

A photon, after emitted in vacuum, is not in anyone's inertial frame unless we pondered an ether- thus,

it is painfully obvious that different observers moving
through the empty vacuum at different velocities must have a different velocity
with respect to the photon because 2+2 = 4, 3+3 = 6, 4+4 = 8, 5+5 =10 and 6+6
= 12.
[ Sure, one can still obtain the speed of the photon through space as a constant
after subtracting the velocity of the observer through space but this was not the
method chosen by Einstein in SR, even though it is the most simple way to achieve
constancy of c that is independent of source. In lieu of using simple math and
common sense, the physics community have decided warping space and time is cooler so never mind that it is wrong.

Wow, an even better sentence. Are you saying that the speed of light is not constant to all observers?

Does the piece in bold really say what I think it does? Welcome to the world of Aristotle, your logic is impeccable your science is non existent.

In view of learning anything about the modern state of physics you choose to talk nonsense.

[Hurkyl]

Hurkyl, perhaps you meant 'derive a contradiction' for 'derive a fallacious conclusion'?

I didn't want to say contradiction because, in principle, a counter-argument could simply derive a disagreement with reality instead of a contradiction within SR. Fallacious was probably the wrong word because it's generally inferred to be speaking about the logic and not the result, but I suppose most adjectives used there are the same.

[StarThrower]

Do the math and weep. Photons do not know time or distance. Therefore do not know velocity. That is proper applcation of SR. If you do not use the tools correctly you do not get meaningful results. SR does not specify a number for c it only says that it is constant. In the frame of reference of a photon the velocity of all photons is zero. That is a constant and satisfies the postulate of SR.

Edit:
Added a lost negation.

No sir. Photons, like other things, move relative to each other. It certainly is meaningful to view motion from the reference frame of a photon, regardless of whether or not the reference frame is inertial. So, regardless of whether or not a coordinate system whose origin is a photon is inertial, certainly other photons move in this frame relative to the origin with different velocities.

Kind regards,

The Star

[Integral]

Star.
What is the basis of your words? Are these your personal thoughts? What experimental evidence do you base your ideas on? Please provide some references to your sources.

[StarThrower]

Star,
Let me but this way, your assertion that you can measure the velocity of a photon from the frame of reference of a photon is incorrect. In SR there are no meaningful measurements that can be made in the frame of reference of a photon. So your example is erronous, a photon cannot see the speed of a second photon.

But then since you have convinced yourself that SR is invalid without the ability to correctly apply it or even understand it. I do not expect you to accept it.

But then since it is incorrect in would not any application of it lead to incorrect results. You applied it therefore your results are incorrect.

The issue in this thread isn't about measuring photon speed in the frame of reference of a photon. It is about whether or not the fundamental postulate of the special theory of relativity self-contradicts. At any rate, when a photon moves in any reference frame, it moves through that frame with some velocity, regardless of whether or not that velocity can be measured. And so we can use binary logic to at least check the consistency of the theory, because we are supposed to know the meaning of the term 'velocity'.

Kind regards,

The Star

[Integral]

Then why are you attempting use the incorrect example of a photons frame of reference. It is YOU who have proposed that argument. I have only shown you why your argument is incorrect. You are not applying binary logic you are attempting to tell us how one photon perceives another. I have provided you with what SR has to say about that situation. That is a photon cannot perceive motion, therefore does not see other photons moving at c. Your basic hypothesis is incorrect. Therefore your conclusions are invalid.

Are you

[StarThrower]

Then why are you attempting use the incorrect example of a photons frame of reference. It is YOU who have proposed that argument. I have only shown you why your argument is incorrect. You are not applying binary logic you are attempting to tell us how one photon perceives another. I have provided you with what SR has to say about that situation. That is a photon cannot perceive motion, therefore does not see other photons moving at c. Your basic hypothesis is incorrect. Therefore your conclusions are invalid.

Are you


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

[Eyesaw]

Ok Eyesaw, you win, lets attempt to make some sense of this post.




What is an example of the SR postulates. Or are you saying the SR postulates are an example of a violation of logic.

Do you even know what the postulates of SR are?


That is one heck of a sentence, but what does it say? This looks to be some kind stawman construction that really means nothing.

What I say is that you cannot measure time or space from the frame of reference of a photon, therefore you cannot measure the speed of a photon.


Wow, an even better sentence. Are you saying that the speed of light is not constant to all observers?

Does the piece in bold really say what I think it does? Welcome to the world of Aristotle, your logic is impeccable your science is non existent.

In view of learning anything about the modern state of physics you choose to talk nonsense.


You are the one repeating the nonsense that a "moving observer" will make the same observations of any event, be it photon speed or whatever, observed by the "stationary observer". What a load of crap- there is no difference between a moving observer and a stationary one then, therefore everything is standing still just like Zeno said. You are really stupid, I mean really stupid to think the time dilation and length contraction effects are real changes in physical space and time and not just artificacts of the transformation method. The theory you support is nothing short of mysticism.

[Severian596]

You are getting off track. The question is this:
Is a reference frame whose origin is a photon an inertial reference frame?
According to the University of Dallas, the second postulate of SR (I'm not calling it the fundamental postulate, that sounds very flowery as there are two equally-important postulates) is:

Light is always propagated in empty space with a definite velocity c, which is independent of the state of motion of the emitting body.

A second source on this postulate is Wikipedia's (http://en.wikipedia.org/wiki/Special_relativity#Postulates_of_Special_Relativit y) definition, which states:

The speed of light in vacuum is the same to all inertial observers. This postulate has been verified experimentally.

As far as you're concerned, StarThrower, the part about EXPERIMENTAL VERIFICATION will be totally useless, so you'll ignore that. In any case we have to define what an inertial observer is. I found this rather expansive definition provided by a retired nuclear engineer here (http://www.madsci.org/posts/archives/apr2000/956866429.Ph.r.html):

To be inertial, the observer, or coordinate system must satisfy the
following properties:

1) The distance between any two points in the coordinate system must be
time independent, the distance is not different for different times,

2) The clocks, assumed distributed throughout the coordinate system, are
synchronized (the time reported is not dependent on the location of the
clock) and run at the same rate throughout the coordinate system (the
observer), and

3) The geometry of space at any constant t is Euclidean. The simplest
way to think of a Euclidean space is that it is a space (a coordinate
system) in which the geometry most people are familiar with is
applicable. If you were a mathematician, you would probably take issue
with that last sentence, claiming it to be an oversimplification, but it
should work in this instance.

There are further stipulations placed on inertial frames of reference, namely they must take up VERY small volumes of space (because otherwise parallel lines tend to converge due to the gravitational affect of nearby massive bodies and the curvature of space...this is addressed by #3 above stipulating that space must be "flat" in the inertial frame), and they must be nonrotating with respect to distant cosmic mass.

I can accept that we could concentrate on a very small distance travelled by a photon, so #3 COULD apply to a photon, but how are #1 and #2 applied to a photon? Photons don't even experience time, so there is no distance between any two points as far as a photon is concerned. Therefore it could not define a frame of reference (which is a system of coordinates) because the length of the ct, x, y, z axes would be zero. As far as a clock, would it ever tick once (assuming it's massless and traveling at c)? Could there ever be more than one clock in the reference frame, because there are no points in the reference frame OTHER than the origin?

[Severian596]

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.

This is from your first post, StarThrower. I challenge you to tell me the difference in their velocity vectors if photons A and B are travelling in

a) perpendicular directions
b) directions separated by angle \theta where \theta = 45^\circ
c) opposite directions

EDIT:
Show your work, please.

[Hurkyl]

And let's further expand upon this. Suppose photonic rest frames are valid.

Allow me to specify the worldlines of two photons in the reference frame of an ordinary observer:

In the coordinate system (t, x) of the ordinary observer:

Photon 1's worldline is given by the equation x = ct.
Photon 2's worldline is given by the equation x = ct.

Now, transform into the coordinates (t', x') of photon 1's rest frame. What are the worldlines of photon 1 and photon 2?

(again, please show your work)

[StarThrower]

According to the University of Dallas, the second postulate of SR (I'm not calling it the fundamental postulate, that sounds very flowery as there are two equally-important postulates) is:

Light is always propagated in empty space with a definite velocity c, which is independent of the state of motion of the emitting body.

A second source on this postulate is Wikipedia's (http://en.wikipedia.org/wiki/Special_relativity#Postulates_of_Special_Relativit y) definition, which states:

The speed of light in vacuum is the same to all inertial observers. This postulate has been verified experimentally.

As far as you're concerned, StarThrower, the part about EXPERIMENTAL VERIFICATION will be totally useless, so you'll ignore that. In any case we have to define what an inertial observer is. I found this rather expansive definition provided by a retired nuclear engineer here (http://www.madsci.org/posts/archives/apr2000/956866429.Ph.r.html):

There are further stipulations placed on inertial frames of reference, namely they must take up VERY small volumes of space (because otherwise parallel lines tend to converge due to the gravitational affect of nearby massive bodies and the curvature of space...this is addressed by #3 above stipulating that space must be "flat" in the inertial frame), and they must be nonrotating with respect to distant cosmic mass.

I can accept that we could concentrate on a very small distance travelled by a photon, so #3 COULD apply to a photon, but how are #1 and #2 applied to a photon? Photons don't even experience time, so there is no distance between any two points as far as a photon is concerned. Therefore it could not define a frame of reference (which is a system of coordinates) because the length of the ct, x, y, z axes would be zero. As far as a clock, would it ever tick once (assuming it's massless and traveling at c)? Could there ever be more than one clock in the reference frame, because there are no points in the reference frame OTHER than the origin?

The source which you cited is not reputable. The definition given is not the internationally accepted definition. I did a search on google (inertial reference frame definition) here you go:

Web Definition (http://www.google.com/search?q=define:inertial+reference+frame)

Definition 1: a coordinate system in which Newton's first law of motion is valid

Source 2 (http://www.thefreedictionary.com/reference%20frame)

Definition 2: inertial frame, inertial reference frame - a coordinate system in which Newton's first law of motion is valid

Source 3 (http://appletree.mta.ca/courses/physics/4701_97/EText/Inertial.html)

Definition 3: An inertial reference frame is one in uniform motion (all accelerometers read zero). In special relativity we think of inertial reference frames as those which are moving at a constant velocity. An observer, without being able to make reference to the rest of the universe (e.g. in a windowless room) cannot determine his or her velocity. Within special relativity we think of inertial referenceframes as reference frames in constant velocity motion.

Source 4 (http://id.mind.net/~zona/mstm/physics/mechanics/framesOfReference/inertialFrame.html)

Definition 4: An inertial frame of reference has a constant velocity. That is, it is moving at a constant speed in a straight line, or it is standing still. Understand that when something is standing still, it has a constant velocity. Its velocity is constantly zero meters per second.

The following source (Stanford university) is reputable:

Source 5 (http://plato.stanford.edu/entries/spacetime-iframes/)
Definition 5:

Space and Time: Inertial Frames
A “frame of reference” is a standard relative to which motion and rest may be measured; any set of points or objects that are at rest relative to one another enables us, in principle, to describe the relative motions of bodies. A frame of reference is therefore a purely kinematical device, for the geometrical description of motion without regard to the masses or forces involved. A dynamical account of motion leads to the idea of an “inertial frame,” or a reference frame relative to which motions have distinguished dynamical properties. For that reason an inertial frame has to be understood as a spatial reference frame together with some means of measuring time, so that uniform motions can be distinguished from accelerated motions. The laws of Newtonian dynamics provide a simple definition: an inertial frame is a reference-frame with a time-scale, relative to which the motion of a body not subject to forces is always rectilinear and uniform, accelerations are always proportional to and in the direction of applied forces, and applied forces are always met with equal and opposite reactions. It follows that, in an inertial frame, the center of mass of a system of bodies is always at rest or in uniform motion. It also follows that any other frame of reference moving uniformly relative to an inertial frame is also an inertial frame. For example, in Newtonian celestial mechanics, taking the “fixed stars” as a frame of reference, we can determine an (approximately) inertial frame whose center is the center of mass of the solar system; relative to this frame, every acceleration of every planet can be accounted for (approximately) as a gravitational interaction with some other planet in accord with Newton's laws of motion.

This appears to be a simple and straightforward concept. By inquiring more narrowly into its origins and meaning, however, we begin to understand why it has been an ongoing subject of philosophical concern. It originated in a profound philosophical consideration of the principles of relativity and invariance in the context of Newtonian mechanics. Further reflections on it, in different theoretical contexts, had extraordinary consequences for 20th-century theories of space and time.

End of source quotes

As you can see, some sources say that an inertial reference frame is a reference frame in which Newton's first law is true, other more reputable sources say that an inertial reference frame is a reference frame in which all three of Newton's laws are true.

What matters here, is the definition which Einstein used in his formulation of the fundamental postulate of the special theory of relativity, which is this:

An inertial reference frame is a reference frame in which a body which is not subjected to an outside force will either

A. Remain at rest (if at rest).
B. Continue to move in a straight line (if already moving in a straight line).

And a body which is subjected to an outside force F, will obey the following equation:

F = dP/dt = d(mV)/dt

Where m is the object's mass, and V is the objects velocity vector.

And there will be an action/reaction pair for this object. The other object will experience the same force F, but in the opposite direction.


Regards,

The Star

[Hurkyl]

The source which you cited is not reputable. The definition given is not the internationally accepted definition.

You'll notice, however, that the three facts listed are true in both Newtonian and SR inertial reference frames. You'll recognize that they (partially) specify a spatial reference frame together with some means of measuring time.

[Hurkyl]

And a body which is subjected to an outside force F, will obey the following equation:

F = dP/dt = d(mV)/dt

Assuming, of course, P = mV.

[outandbeyond2004]

Maxwell's equations for electromagnetism assumed the very same inertial frames that Newtonian mechanics assumed. Don't let all the different definitions of inertial frames fool you. (Though, to be sure, one definition may be better than another.) Yet, Maxwell's equations predicted just one value, c, for the speed of light. Logically, StarThrower has to assert that Maxwell's equations are invalid, contrary to 200 years of experiment.

[StarThrower]

Assuming, of course, P = mV.

That is the equation to focus on Hurkyl. When Einstein originally formulated the fundamental postulate of the special theory of relativity, he had not yet developed the concept of mass which depends upon speed.

This early oversight on Einstein's part does not alter the definition of inertial reference frame which he used, and this brings me back to the original question.

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

Kind regards,

The Star

[StarThrower]

Maxwell's equations for electromagnetism assumed the very same inertial frames that Newtonian mechanics assumed. Don't let all the different definitions of inertial frames fool you. (Though, to be sure, one definition may be better than another.) Yet, Maxwell's equations predicted just one value, c, for the speed of light. Logically, StarThrower has to assert that Maxwell's equations are invalid, contrary to 200 years of experiment.

I would argue as follows:

Maxwell's equations predict one universal value for the speed of light relative to the source. They do not say that the speed of light is independent of the motion of the source.

Kind regards,

The Star

[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:

discussion (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) that happened on this very forum[/url]

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...

:wink:

[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, thats 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?
troll v.,n. (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]

troll v.,n. (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.

[Tom Mattson]

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. :smile:

[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. :cool:

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 :smile:

[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 :smile:

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 travelling 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 travelling 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?

[Severian596]

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?

And if the velocity vectors of A and B are not parallel? Not equal in magnitude?

[StarThrower]

Your definition for a coordinate system is correct as long as our common sense idea of space applies.



Let N denote the maximum number of infinitely long straight lines that can meet at a point, such that the lines are mutually perpendicular.

The answer is independent of common sense.
The answer is called the dimensionality of space.
The answer is three... i.e. N=3.

There is a unique distance between any two points in space.

Let A denote one point in space, let B denote another point in space. The basic idea of a straight line segment comes from the following distance axiom:

Let |AB| denote the distance from point A to point B.

Let C denote any point on the straight line segment AB.

|AC| is the distance from A to C
|CB| is the distance from C to B.

Notice that |AC|+|CB| = |AB|

Now, suppose that C is off of straight line segment AB. The following is now the case:

|AC| + |CB| > |AB|

Recall: Triangle Inequality

Kind regards,

The Star

[Michael F. Dmitriyev]

Michael,
You fail to understand the meaning of postulate. Einstein is not obligated to prove anything about his postulate. That is the nature of a postulate. You most certainly cannot learn anything other then the results of a constant c by studying Relativity. It is after all a development that explores the implications of a constant speed of light.

If you wish to find the roots of the constancy of c you need to study and understand the work of Clerk Maxwell. The origins of a velocity that is independent of the source, is Maxwell's equations cast in the form of the wave equation. When this result was published in the late 1860s the world of Physics was changed forever. How to rectify the source independence of the speed of Electromagnetism with the accepted and well understood precepts of Classical Mechanics was the single largest issue in Physics of that era. Due to that Einstein was able to postulate a constant c. Physicist of that era did not blink an eye at the postulate because they had spend a generation attempting to disprove the constancy of c. They failed.
Excuse me, but I can’t consider your answer differently than pseudoscientific demagogy.
You do not give the concrete answer, except for the general reasonings.
Do you have the answer to a question – why?

[Severian596]

Let |AB| denote the distance from point A to point B.


I will not ask how we define distance, but rather I'll ask you directly if the spacial distance between points A and B is absolute to all reference frames.

EDIT:
In addition you have not accounted for time in your coordinate system. Is the distance \Delta t between two events in one frame absolute--that is, equal in all frames in motion with respect to events' frame?

If this isn't so, and if time is not absolute, how can you define absolute space?

[Severian596]

The answer is called the dimensionality of space.

This is not fully understood. The nature of space and time (and how they are related) is more complex than length, width, height, and how long it takes me to go to the bathroom. Furthermore to fully understand their relationship I believe it takes general relativity...but I don't know anything about GR.

[StarThrower]

I will not ask how we define distance, but rather I'll ask you directly if the spacial distance between points A and B is absolute to all reference frames.



Yes, the distance between two points is absolute.

Regards,

The Star

[Severian596]

Yes, the distance between two points is absolute.

Well, we can stop talking now. I'll keep an eye out for your "StarThrowerian Physics" publication, if it ever hits the shelves.

In relativistic mechanics, there is no such thing as an absolute length or an absolute time since length and time depend on the frame of reference of the observer.

[StarThrower]

Well, we can stop talking now. I'll keep an eye out for your "StarThrowerian Physics" publication, if it ever hits the shelves.

In relativistic mechanics, there is no such thing as an absolute length or an absolute time since length and time depend on the frame of reference of the observer.

Relativistic mechanics is wrong, hence I fail to see your point.

Regards,

The Star

[StarThrower]

And if the velocity vectors of A and B are not parallel? Not equal in magnitude?

A,B are points in F2. Therefore, the distance from A to B is constant.

If the velocity of A in F2 is different from the velocity of B in F1, then the distance from A to B would vary. In other words, the vector from A to B is rigid.

Regards,

The Star

[Severian596]

Hence, I'll leave you to your theories and wait for your work to be published by the experts. I'm tired of comparing apples to oranges.

You're entitled to believe what you like, and so am I.

The conversation was stimulating...

[Severian596]

If the velocity of A in F2 is different from the velocity of B in F1, then the distance from A to B would vary. In other words, the vector from A to B is rigid

You didn't stipulate this, which didn't account for any possible rotation or stretching of F2 in comparisson to F1.

[StarThrower]

In order to understand the concept three dimensional Euclidean space, one has to understand the definition of a straight line segment. In order to understand this, one needs the axioms of the Euclidean distance function.

Here they are for convenience.

Euclidean Distance Function

Let x,y,z denote arbitrary points in a coordinate system.

Definition: For any points A,B in a coordinate system, |AB| denotes the distance from A to B.

Let x,y,z denote arbitrary points in some coordinate system.

Axiom I: |xy| \geq 0
Axiom II: |xy|=0 \leftrightarrow x=y
Axiom III: |xy| = |yx|
Axiom IV: |xy|+|yz| \geq |xz|

The straight line segment from xy consists of all points c, such that:

|xc|+|cy| = |xy|

Keep in mind that in principle, the distance from one point to another can be measured by a device called a ruler. Thus, distance doesn't need to be defined logically, instead it can be defined operationally.

[Severian596]

In order to understand the concept three dimensional Euclidean space...

It's too bad that because we don't live in Euclidean space your efforts are wasted.

I'm curious how you, StarThrower, explain the fact that every other galaxy is redshifted if you also believe in absolute space. Oh and you never answered my question about absolute time, do you also hang on to that?

PS I'll wait for a new-age newtonian purist physics system from you, if that's the case. I assume you're hip with the ether, too then.

[StarThrower]

It's too bad that because we don't live in Euclidean space your efforts are wasted.

I'm curious how you, StarThrower, explain the fact that every other galaxy is redshifted if you also believe in absolute space. Oh and you never answered my question about absolute time, do you also hang on to that?

PS I'll wait for a new-age newtonian purist physics system from you, if that's the case. I assume you're hip with the ether, too then.

Absolute space, absolute time, yes that's me. Redshift and Blueshift can be explained by Newtonian mechanics.

Kind regards,

The Star

[StarThrower]

Once you are armed with the concept of a straight line segment, it is easy to generalize that concept to an infinite straight line.

The straight line segment from A to B consists of all points C, such that:

|AC|+|CB|=|AB|

Now, consider THE infinite straight line through A,B.
It will consist of the straight line segment from A to B, and also, it will consist of all points D such that:

|AB|+|BD| = |AD| or |DA|+|AB|=|DB|

So in principle, the concept of the distance from one point to another has been made as intuitive as possible.

Next comes the notion of two infinite straight lines that contain one point in common.

Any two such lines determine a unique plane.

In Euclid it is proved that vertical angles are congruent. And when all four angles are equivalent, each is a right angle, and the lines are said to be perpendicular.

Now, suppose that we have two infinite straight lines that meet at right angles.

Now, send a third infinite straight line through the intersection point of the first two, such that all vertical angles are congruent. There is one and only one infinite straight line which can meet the other two, such that all three lines are mutually perpendicular.

Thus, real space is three dimensional Euclidean.

Anyone who wants a fancier proof of this is welcome to try.
The fact that space is three dimensional was known by the ancient greeks.

I find it surprising that this fact was ever doubted.

Kind regards,

The Star


P.S. At this point, it is a hop, skip, and a jump to a rectangular coordinate system.

[Severian596]

"The fact that space is three dimensional was known by the ancient greeks."

... welcome to the 21st century?

"I find it surprising that this fact was ever doubted."

The fact is simply incomplete without considering time, and how time and space interact.

[Hurkyl]

What is the distance between A and B where:
A is the origin at time 0
B is the origin at time 1
?

A and B cannot be the same, because they're at different times...

[Integral]

We will raise this conjecture (the purport of which will hereafter be called the ``Principle of Relativity'') to the status of a postulate, and also introduce another postulate, which is only apparently irreconcilable with the former, namely, that light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body.

This quote is taken from On the Electrodynamics of a moving body (http://www.fourmilab.ch/etexts/einstein/specrel/www/) By Albert Einstein. I see no reference to inertial frames. I do see the bolded phrase. Some times when we run into apparent contradictions we need to step back and look at the basic physics.

Here in lies the explanation for your apparent contradiction. Photons simply do not emit photons, so it is nonphysical to refer to a photon as an emitting body. Your argument which assumes that a photon has velocity c wrt to a second photon is doing just that. This is nonphysical assumption, therefore nonphysical conclusions are the only possible result, if correct logic is used.

[StarThrower]

"The fact that space is three dimensional was known by the ancient greeks."

... welcome to the 21st century?

"I find it surprising that this fact was ever doubted."

The fact is simply incomplete without considering time, and how time and space interact.

Time doesn't interact with space, that is a whole lot of nonsense. Intuitive understanding of space comes from the study of the affine geometry that the greeks developed, and mathematical understanding of geometry comes from the analytic geometry of Descarte's (mapping the real number system to points on an infinite straight line). Analytic and Synthetic geometry both aid in the understanding of space, and hence in the understanding of motion.

Understanding of time comes from understanding of the natural number system: N = {1,2,3,4,5,6,...}

First moment in time, second moment in time, third instant, four state, etc.

If time is modeled using the real number system, then between any two moments in time, there are an infinite number of moments in time, therefore the concept of a second moment in time is meaningless. That is incorrect.

Kind regards,

The Star

[StarThrower]

What is the distance between A and B where:
A is the origin at time 0
B is the origin at time 1
?

A and B cannot be the same, because they're at different times...

I didn't understand this post... sorry.

[StarThrower]

This quote is taken from On the Electrodynamics of a moving body (http://www.fourmilab.ch/etexts/einstein/specrel/www/) By Albert Einstein. I see no reference to inertial frames. I do see the bolded phrase. Some times when we run into apparent contradictions we need to step back and look at the basic physics.

Here in lies the explanation for your apparent contradiction. Photons simply do not emit photons, so it is nonphysical to refer to a photon as an emitting body. Your argument which assumes that a photon has velocity c wrt to a second photon is doing just that. This is nonphysical assumption, therefore nonphysical conclusions are the only possible result, if correct logic is used.

You left out part of Einstein's quote:

They suggest rather that, as has already been shown to the first order of small quantities, the same laws of electrodynamics and optics will be valid for all frames of reference for which the equations of mechanics hold good.1 We will raise this conjecture (the purport of which will hereafter be called the ``Principle of Relativity'') to the status of a postulate, and also introduce another postulate, which is only apparently irreconcilable with the former, namely, that light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body. A. Einstein

You left out the first sentence which is that the laws of electrodynamics (and optics) will be valid for all FRAMES OF REFERENCE for which the equations of mechanics hold good.

Einstein was convinced that electrodynamics implied that the speed of light was constant.

(A constant with respect to what he thought?)

His answer: In the reference frames for which the laws of classical mechanics are valid, i.e. the inertial reference frames.

This was summarized in his postulate where he says that the speed of light is independent of the speed the emitter has when moving in an inertial reference frame.

In other words Einstein said this.

Let F1 be an inertial reference frame. (Therefore, by definition of an inertial reference frame, Newton's laws of motion are true. That means that an object subject to no outside force will either remain at rest in this frame, or in straight line motion at a constant speed... the law of inertia is true in this frame).

Set up a coordinate system so that distance measurements have meaning in this frame using rulers. Let time measurements be made using clocks at rest in this frame, so that speed of an object in this frame can be defined.

Now, suppose that the speed of photon emitter 1 is 0 in this frame.
And, suppose that the speed of photon emitter 2 is v in this frame.

Let both emitters emit light. (later the photon model was accepted, but at this time Maxwell and others were propounding a wave theory of light).

Using the concept of photons, Einstein was saying this:

The speed of each photon in F1 is C.

Newtonian mechanics says this:

The speed of the photon emitted by emitter one should be 0+c=c in this frame, and the speed of the photon emitted by emitter two should be v+c in this frame.

Disagreement starts here.

Kind regards,

The Star

[Severian596]

I didn't understand this post... sorry.

First off, Hurkyl's post was very succinctly put: distance between two points in time can be examined just like the distance between two points in space. Even a stationary point within 3-space (and therefore within a coordinate system) is not completely stationary...it moves forward through time. The definition of an event is a point in space with a 4th coordinate, time. This was the root of Hurkyl's question.

Secondly, it's obvious that you're an intelligent person, StarThrower. You depend on logic and deductive reasoning, though in my opinion you depend a bit too much on thought experiment over actual experimental data. So why do you choose to cling to ancient precepts for your beliefs? You must realize that the people who developed the classical systems of physics (i.e. Newton) to which you cling used the knowledge of their predicesors to advance theories of their own age? And corrispondingly physicists and philosophers used their ideas to further progress the state of understanding.

Basics ideas developed by Euclid were developed and refined, used by Newton, and then Newton's ideas were refined and devloped by Einstein. Why do you choose to ignore that?

If we did not depend on the work of others, we would all be stuck reinventing the wheel. The reason for studying classical theory is to better understand modern theory. Not so you can look at modern theory and accuse it of being arcane or mythical.

If that weren't the case you may as well believe that bleeding a person is a way to cure illness. And that these tiny things called "micro organisms" don't exist! You can't SEE them after all, not without using some instruments of science.

[Hurkyl]

Using the concept of photons, Einstein was saying this:

The speed of each photon in F1 is C.

Newtonian mechanics says this:

The speed of the photon emitted by emitter one should be 0+c=c in this frame, and the speed of the photon emitted by emitter two should be v+c in this frame.

Disagreement starts here.

And don't forget that Maxwell's theory of electrodynamics says that (in both frames):

\begin{equation*}\begin{split}\nabla \cdot E &= 0 \\
\nabla \cdot B &= 0 \\
\nabla \times E &= \mu_0 \epsilon_0 \frac{\partial B}{\partial t} \\
\nabla \times B &= - \frac{\partial E}{\partial t} \\
\mbox{from which we can conclude} \\
\nabla \times \frac{\partial B}{\partial t} &= - \frac{\partial^2 E}{\partial t^2} \\
\frac{1}{\mu_0 \epsilon_0} \nabla \times ( \nabla \times E ) &= - \frac{\partial^2 E}{\partial t^2} \\
\frac{1}{\mu_0 \epsilon_0} ( \nabla(\nabla \cdot E) - \nabla^2 E ) &= - \frac{\partial^2 E}{\partial t^2} \\
\frac{\partial^2 E}{\partial t^2} &= \frac{1}{\mu_0 \epsilon_0} \nabla^2 E
\end{split}\end{equation*}


Which, if you'll recall from classical mechanics, describes a wave whose velocity is 1/\sqrt{\mu_0 \epsilon_0}. And note that this derivation is valid in any reference frame in which Maxwell's equations are valid.

[StarThrower]

And don't forget that Maxwell's theory of electrodynamics says that (in both frames):

\begin{equation*}\begin{split}\nabla \cdot E &= 0 \\

\end{split}\end{equation*}




Prove the divergence of the electric field is 0.

Exactly what kind of frame is this statement true in?

[Hurkyl]

This is Maxwell's theory of electrodynamics. \nabla \cdot E = 0 wherever there is no net electric charge.

[Eyesaw]

The game the SRists are playing goes like this: Let's change the definitions of the numbers 4 and 3 for example (but be discreet and hope no one will notice) and voila, we dazzle the public with the result that 4 + 3 = 2. A great sleight of hand trick, all SR is.


Basically, units are changed between inertial frames in SR but without using a different terminology as required. An inch becomes a centimeter in a different inertial frame but is mistakenly called an inch by the SR hoaxers. Ditto for the units of time. Everyone except SR elitists are required to give a conversion ratio between scales (e.g. 1 centimeters on a road map = 1 mile on the road) so that a set of values can be accurately compared- deception is the method by which SRians circumvent the logical and physical contradictions arising from SR's two postulates.

If the proper units and terminology were used, SR's postulate of a constant speed of light independent of source must result in contradiction with its second postulate, that physics is the same in all inertial frames, since the speed of light then would be c+v and c-v round trip measured inside an inertial frame moving at v relative to the vacuum.

[Hurkyl]

Eyesaw: How do you reconsize varying speed of light with the wave equation from Maxwell's electrodynamics?

[Eyesaw]

Eyesaw: How do you reconsize varying speed of light with the wave equation from Maxwell's electrodynamics?

Simple- don't transform coordinates. If physics is the same in all inertial frames, why does Maxwel''s equations need to take on different forms?
Kind of redundant don't you think? If our starting postulate is that c is c inside all inertial frames, why do we have to transform its value from one to another?

[Hurkyl]

Basically, units are changed between inertial frames in SR but without using a different terminology as required.

What leads you to believe this? Is it that you are running out of ways to rationalize your disbelief in SR and are reaching for the more and more patently absurd?

[Hurkyl]

If physics is the same in all inertial frames, why does Maxwel''s equations need to take on different forms?

It doesn't take on different forms in all inertial reference frames.

That's the point.

In every inertial frame, the wave equation says (in a vacuum):


\frac{\partial^2 E}{\partial t^2} = \frac{1}{c^2} \nabla^2 E


But it never takes the form


\frac{\partial^2 E}{\partial t^2} = \frac{1}{(c+v)^2} \nabla^2 E


or


\frac{\partial^2 E}{\partial t^2} = \frac{1}{(c-v)^2} \nabla^2 E


which would be required for light travelling at speeds of c+v or c-v respectively.

[Eyesaw]

It doesn't take on different forms in all inertial reference frames.

That's the point.

In every inertial frame, the wave equation says (in a vacuum):


\frac{\partial^2 E}{\partial t^2} = \frac{1}{c^2} \nabla^2 E


But it never takes the form


\frac{\partial^2 E}{\partial t^2} = \frac{1}{(c+v)^2} \nabla^2 E


or


\frac{\partial^2 E}{\partial t^2} = \frac{1}{(c-v)^2} \nabla^2 E


which would be required for light travelling at speeds of c+v or c-v respectively.

Not if light is source dependent.

[Severian596]

I'm curious if any doubters here have sources of experimental evidence that the speed of light is DEpendent of the source's speed.

[StarThrower]

Let some electrically charged object be at rest in reference frame F1.
Let us have set up a rectangular coordinate system.
We desire to compute the electric field E at one of the field points, say (x,y,z), due to this electrically charged object.

Let the charge density of the object be denoted by \rho

The position vector of the field point is:

\vec{r} = xi + yj + zk

Let the position of an arbitrary electrically charged particle inside the body be (x`,y`,z`).

The position vector of this particle in F1 is:

\vec{r^\prime} = x^\prime i + y^\prime j + z^\prime k

Thus, we have a vector triangle. Let us define \vec{R} using the following equation:

\vec{r^\prime} + \vec{R} = \vec{r}

Thus, R is a vector that goes from a charged particle in the object, to an arbitrary field point whose coordinates in F1 are (x,y,z).

The total electric field at (x,y,z) due to the charged object, can be computed by integrating over the charged body. Thus, in the integral, the point that is varying is (x`,y`,z`).

Thus, we have:


\vec{E} = \frac{1}{4 \pi \epsilon_0} \int \rho d\tau \frac{\vec{R}}{|\vec{R}|^3}

Rho is a function of the primed coordinates (x`,y`,z`).

Now, take the divergence of both sides if the equation over the field points (x,y,z).

Theorem:

\frac{\vec{R}}{|\vec{R}|^3} = \nabla (\frac{-1}{|\vec{R}|})

From which it follows that:

\vec{E} = - \nabla \frac{1}{4 \pi \epsilon_0} \int \frac{\rho d\tau}{|\vec{R}|}

Now, define the scalar function V as follows:

V = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho d\tau}{|\vec{R}|}

Thus, we have:

\vec{E} = - \nabla V

Kind Regards,

The Star

[Tom Mattson]

Hurkyl: This is Maxwell's theory of electrodynamics. \nabla \cdot E = 0 wherever there is no net electric charge.

Star Thrower: The total electric field at (x,y,z) due to the charged object, can be computed by integrating over the charged body. Thus, in the integral, the point that is varying is (x`,y`,z`).

Thus, we have:


\vec{E} = \frac{1}{4 \pi \epsilon_0} \int \rho d\tau \frac{\vec{R}}{|\vec{R}|^3}

Rho is a function of the primed coordinates (x`,y`,z`).

Now, take the divergence of both sides if the equation over the field points (x,y,z).


That's not too difficult, since r is identically zero.

[StarThrower]

That's not too difficult, since r is identically zero.

In free space, r is zero. I simply wanted to see him derive the more general result:

\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}

I want to see if he uses the delta function to derive the result, or not.


Kind regards,

StarThrower

P.S. Oh and Hurkyl, is the equation above true in any reference frame, or just inertial reference frames?

Case 1: The electrically charged object is not being subjected to a force.
Case 2: The electrically charged object is being subjected to a force.

[Tom Mattson]

As I highlighted in red, Hurkyl stipulated that there is no net charge. Furthermore, he seems to be discussing the propagation of light, whose EM fields do indeed satisfy the in vacuo Maxwell equations.

[StarThrower]

As I highlighted in red, Hurkyl stipulated that there is no net charge. Furthermore, he seems to be discussing the propagation of light, whose EM fields do indeed satisfy the in vacuo Maxwell equations.

I wasn't challenging Hurkyl. I understood he meant Maxwell's equations in free space, even though he didn't state that right away.

Kind regards,

The Star

[Hurkyl]

Not if light is source dependent.

Too bad we don't have any source dependent theories of electromagnetism.


I want to see if he uses the delta function to derive the result, or not.

I think the most straightforward way is to consider that for any solid:


\begin{equation*}\begin{split}
\oint \vec{E} \cdot d\vec{A} &= \int \frac{\rho}{\epsilon_0} \, dV \\
\int \nabla \cdot \vec{E} \, dV &= \int \frac{\rho}{\epsilon_0} \, dV
\end{split}\end{equation*}


And because the integrals are equal for any region, the integrands must be equal (at least where they're continuous).


I don't see any reason to hypothesize that Maxwell's equations remain unchanged in a non-inertial reference frame.

[StarThrower]

Too bad we don't have any source dependent theories of electromagnetism.




I think the most straightforward way is to consider that for any solid:


\begin{equation*}\begin{split}
\oint \vec{E} \cdot d\vec{A} &= \int \frac{\rho}{\epsilon_0} \, dV \\
\int \nabla \cdot \vec{E} \, dV &= \int \frac{\rho}{\epsilon_0} \, dV
\end{split}\end{equation*}


And because the integrals are equal for any region, the integrands must be equal (at least where they're continuous).


I don't see any reason to hypothesize that Maxwell's equations remain unchanged in a non-inertial reference frame.

It is unclear how the charge enclosed by the Gaussian surface got divided by the permittivity of free space. The result seems pulled out of thin air.

[StarThrower]

Let some electrically charged object be at rest in reference frame F1.
Let us have set up a rectangular coordinate system.
We desire to compute the electric field E at one of the field points, say (x,y,z), due to this electrically charged object.

Let the charge density of the object be denoted by \rho

The position vector of the field point is:

\vec{r} = xi + yj + zk

Let the position of an arbitrary electrically charged particle inside the body be (x`,y`,z`).

The position vector of this particle in F1 is:

\vec{r^\prime} = x^\prime i + y^\prime j + z^\prime k

Thus, we have a vector triangle. Let us define \vec{R} using the following equation:

\vec{r^\prime} + \vec{R} = \vec{r}

Thus, R is a vector that goes from a charged particle in the object, to an arbitrary field point whose coordinates in F1 are (x,y,z).

The total electric field at (x,y,z) due to the charged object, can be computed by integrating over the charged body. Thus, in the integral, the point that is varying is (x`,y`,z`).

Thus, we have:


\vec{E} = \frac{1}{4 \pi \epsilon_0} \int \rho d\tau \frac{\vec{R}}{|\vec{R}|^3}

Rho is a function of the primed coordinates (x`,y`,z`).

Now, take the divergence of both sides if the equation over the field points (x,y,z).

Theorem:

\frac{\vec{R}}{|\vec{R}|^3} = \nabla (\frac{-1}{|\vec{R}|})

From which it follows that:

\vec{E} = - \nabla \frac{1}{4 \pi \epsilon_0} \int \frac{\rho d\tau}{|\vec{R}|}

Now, define the scalar function V as follows:

V = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho d\tau}{|\vec{R}|}

Thus, we have:

\vec{E} = - \nabla V

Kind Regards,

The Star

In post 138 I said all you have to do is take the divergence of both sides of the definition of the electric field.

\nabla \bullet \vec{E} = \nabla \bullet ( \frac{1}{4 \pi \epsilon_0} \int \rho d\tau \frac{\vec{R}}{|\vec{R}|^3} )

The derivative is with respect to the unprimed field points, and rho isn't a function of the field points, neither is d tau. Hence, the del operator can pass through the integral sign (this can be proven more rigourously) and pass through rho and d tau, and simply operate on R/|R|^3. In other words, rho and d tau are like constants to a del operation with respect to the field points, instead of the source points. Thus, we have the simplest approach to the derivation of the first of Maxwell's laws, simply take the divergence of both sides of the definition of the electric field. This approach is the straightfoward approach. Understanding that rho and d tau are constants with respect to partial derivatives on field points, we can get to the following line of work rather quickly (after taking the divergence of both sides of the equation which is the definition of the electric field):

\nabla \bullet \vec{E} = \frac{1}{4 \pi \epsilon_0} \int \rho d\tau \nabla \bullet ( \frac{\vec{R}}{|\vec{R}|^3} )

Now all we need to do is prove the following theorem, and the first of Maxwell's laws is proven.

Theorem:

\nabla \bullet ( \frac{\vec{R}}{|\vec{R}|^3} ) = 4 \pi \delta (\vec{R})

Thus, we can write the divergence of the electric field as follows:


\nabla \bullet \vec{E} = \frac{1}{4 \pi \epsilon_0} \int \rho d\tau 4 \pi \delta (\vec{R})

Therefore:

\nabla \bullet \vec{E} = \frac{1}{\epsilon_0} \int \rho d\tau \delta (\vec{R})

Now, the delta(R) above, is a 3 dimensional delta function. It has a value of zero everywhere where R isn't zero, and a value of 1 when R is zero. And we had by definition

r` + R = r

Therefore: R = r - r`

Therefore:

\nabla \bullet \vec{E} = \frac{1}{\epsilon_0} \int \rho d\tau \delta (\vec{r} - \vec{r^\prime})

In the integral, the primed coordinates are varying, not the field point, hence we finally obtain the first of Maxwell's equations:

\nabla \bullet \vec{E} = \frac{\rho}{\epsilon_0}

Now we can use the divergence theorem to obtain Hurkyl's result.

Divergence Theorem (http://mathworld.wolfram.com/DivergenceTheorem.html)

Theorem:

Let \vec{F} denote an arbitrary vector function. The following can be proven analytically:

\int \nabla \bullet \vec{F} d\tau = \oint \vec{F} \bullet d\vec{a}

Multiply both sides of the divergence of E by d tau, and then integrate, and you get (using the divergence theorem, and the definition of charge enclosed):

\oint \vec{E} \bullet d\vec{a} = \int \nabla \bullet \vec{E} d\tau = \int \frac{\rho d\tau}{\epsilon_0} = \frac{Qenc}{\epsilon_0}

Kind regards,

StarThrower

[StarThrower]

I don't see any reason to hypothesize that Maxwell's equations remain unchanged in a non-inertial reference frame.

Let us take the stance that acceleration is relative, and force isn't. Thus, an object is in an inertial reference frame precisely if that object isn't being subjected to any outside forces, electric, or otherwise, and an object is in a non-inertial reference frame precisely when that object is being subjected to a force.

Now, consider an electron in a hydrogen atom. While under the influence of the coulomb force due to the proton, it orbits in a roughly circular shape (most probable trajectory in QM, expectation values and all that).

Question: Why doesn't the electron continuously radiate photons as Maxwell's equations predict?

Answer: Because Maxwell's equations are wrong.

So where is the error Hurkyl?

Hint:

Assume that Maxwell's equations are true.
Consider a hydrogen atom, an electron is in the coulomb field of a proton, the separation distance will be denoted by |R|. The force is attractive, and along the direction of the 'separation' vector. The magnitude of this force is:

|\vec{F}| = \frac{1}{4 \pi \epsilon_0} \frac{Qq}{|\vec{R}|^2}

Where Q denotes the charge of the proton, and q denotes the charge of the electron.

For analytical purposes, presume the mass of the proton is infinite compared to the mass of the electron, so that the center of mass of the system is located wherever the proton is. Then, consider the motion of the electron in a reference frame whose origin is located at the proton. Thus, we are in an inertial reference frame in which the proton is at rest. (we can correct for this later if desired, but its not necessary).

Now, consider the motion. The separation vector R is now the position vector of the electron in this coordinate system, so that we can now write:

|\vec{F}| = \frac{1}{4 \pi \epsilon_0} \frac{Qq}{|\vec{r}|^2} = q |\vec{E}|

Since Maxwell's equation are true in this inertial reference frame, it follows that since the electron is accelerating in this reference frame, it should be radiating electromagnetic waves.

This does not in reality happen, or else hydrogen atoms would not be stable.

Thus, if we stipulate that maxwell's equations are true in this inertial reference frame, then:

The electron isn't 'experiencing' a force. That implies that the acceleration of the electron in this frame is 'artificial'. This is a big big hint.

[Hurkyl]

It is unclear how the charge enclosed by the Gaussian surface got divided by the permittivity of free space. The result seems pulled out of thin air.

I was quoting (the integral form of) Gauss's law. *shrug*


simply take the divergence of both sides of the definition of the electric field.

Allow me to point out that what the equation you are using is not the definition of the electric field; it is merely a particular formula for computing electric fiend strength given a static, continuous charge distribution.


Answer: Because Maxwell's equations are wrong.

So where is the error Hurkyl?

I'm not sure what you mean.

[StarThrower]

I was quoting (the integral form of) Gauss's law. *shrug*




Allow me to point out that what the equation you are using is not the definition of the electric field; it is merely a particular formula for computing electric fiend strength given a static, continuous charge distribution.




I'm not sure what you mean.

Since real bodies are three dimensional, we have a volume charge density rho. Of course there is the built in assumption that rho is continuous, which is why we can write the forumlas using integrals. Otherwise we would just sum the contribution of each charge in the body. It's not really a point of contention anyhow.

The more serious question still needs to be addressed, and that is, what is the error in electrodynamics? To understand what I mean, re-read the post. An electron in an inertial reference frame, is orbiting a proton. The proton is at rest in the IRF, and the electron is accelerating in the IRF. Maxwell's equations say that the electron should radiate EM waves, and thus spiral into the nucleus within a fraction of a second. Therefore, according to EM, hydrogen atoms don't exist. And hydrogen atoms are the most numerous atom in the universe. Therefore, EM contains an error. I was challenging you to locate it.

Regards,

StarThrower

[Integral]

The error is not due to Maxwell, but rather your model of the atom. As usual when you make non physical assumptions you get non physical results. That is the main lesson of all your arguments.

[Hurkyl]

Therefore, EM contains an error. I was challenging you to locate it.

The error is that this model does not agree with experimental evidence. You seemed to want something more. *shrug*

[StarThrower]

The error is not due to Maxwell, but rather your model of the atom. As usual when you make non physical assumptions you get non physical results. That is the main lesson of all your arguments.

The model is basically correct. Statically charged objects either attract or repel. We have macroscopic proof of this. Then a theory was developed, which explained such macroscopic behavior, and more.

Something about the model is right, since Bohr was able to derive the Balmer formula from his assumptions. He added a quantum physical assumption, in order to counteract the Maxwellian Catastrophe (namely unstable orbit).

We have an electron that would move in a straight line at a constant speed, or remain at rest. This is true because the electron is in an IRF, and the law of inertia is true in any IRF.

However, the electron isn't doing that, therefore it is experiencing a force.

The Coulomb law is the experimental magnitude of that force, the constant comes from experiment.

This being said, if we now further apply Maxwell's results, this electron should radiate EM waves and spiral into the nucleus. That doesn't happen.
Regards,

StarThrower

[Integral]

In the same vein of your arguments.

Non physical assumptions yield non physcial results.

[jdavel]

Maxwell's equations say that the electron should radiate EM waves, and thus spiral into the nucleus within a fraction of a second. Therefore, according to EM, hydrogen atoms don't exist. And hydrogen atoms are the most numerous atom in the universe. Therefore, EM contains an error.
StarThrower

I hope you enjoy being wrong; you're so good at it! :wink:

Classical electromagnetic theory can be applied with tremendous accuracy within the atom. The fields predicted by Maxwell's equations work just fine. The inability of classical physics (Newton + Maxwell) to explain how electrons behave in atoms is the result of errors in the Newton part, not the Maxwell part.

[StarThrower]

I hope you enjoy being wrong; you're so good at it! :wink:

Classical electromagnetic theory can be applied with tremendous accuracy within the atom. The fields predicted by Maxwell's equations work just fine. The inability of classical physics (Newton + Maxwell) to explain how electrons behave in atoms is the result of errors in the Newton part, not the Maxwell part.


Precisely what part of Newton is wrong?

Aristotle noticed that most things just kind of sit where they are. Most of us notice this fact. Things just sort of stay where they are. Hence, an object at rest will remain at rest. This is violated when some kind of force comes into play, such as us lifting the stone against the gravitational pull of the earth.

As for the part which says, and an object in motion will continue to move in a straight line at a constant speed forever, unless acted upon by an outside force, well this part is not really too obvious. But consider the experiments of Galileo. This part of Newtonian mechanics comes right out of Galileo's experiments.

Galileo's Dialogue Concerning Two New Sciences (http://galileoandeinstein.phys.virginia.edu/tns_draft/index.html)

[russ_watters]

Aristotle noticed that most things just kind of sit where they are. Most of us notice this fact. Things just sort of stay where they are. Hence, an object at rest will remain at rest. This is violated when some kind of force comes into play, such as us lifting the stone against the gravitational pull of the earth. Aristotle also believed that an object would stop moving if the force on it was removed. Physics has come a long way since then. You seem to have come as far as Newton and stopped there. Physics has come a long way since Newton as well: in fact, Newton was essentially the beginning of physics (mathematically), not the end.Precisely what part of Newton is wrong? Quite a bit (if not wrong, at least limited in domain), starting with f=ma, but you don't seem to buy any of that. Tough to continue if you don't accept much of anything of modern physics.

[Severian596]

Tough to continue if you don't accept much of anything of modern physics.

He is a genius in his own mind. He neglects that Newtonian physics yield approximate results for any situation where Einsteinian physics will yield a more accurate reflection of experimental data.

Better yet he ignores any post that spells this sort of thing out...like a horse with blinders.

Ignorance is bliss? :biggrin:

[StarThrower]

Aristotle also believed that an object would stop moving if the force on it was removed. Physics has come a long way since then. You seem to have come as far as Newton and stopped there. Physics has come a long way since Newton as well: in fact, Newton was essentially the beginning of physics (mathematically), not the end. Quite a bit (if not wrong, at least limited in domain), starting with f=ma, but you don't seem to buy any of that. Tough to continue if you don't accept much of anything of modern physics.

Russ, experiments on billiard balls suffice to verify Newton's laws. In fact, Newton formulated his statements from concentrating on bodies in relative motion.

Kind regards,

StarThrower


P.S. Unless you have actually read Aristotle, you shouldn't quote him second hand. Lots of words have been put into the mouth of Aristotle over the years. Not to mention, all we have of his works are translations, not one of us speaks ancient Greek fluently.

As for my quote of Aristotle's, I took it right out of "Physics" which was translated by translated by R. P. Hardie and R. K. Gaye. It's available on the web, at MIT.

I found the quote I was referring to. Here is the translation:


Further, in point of fact things that are thrown move though that which gave them their impulse is not touching them, either by reason of mutual replacement, as some maintain, or because the air that has been pushed pushes them with a movement quicker than the natural locomotion of the projectile wherewith it moves to its proper place. But in a void none of these things can take place, nor can anything be moved save as that which is carried is moved.

Further, no one could say why a thing once set in motion should stop anywhere; for why should it stop here rather than here? So that a thing will either be at rest or must be moved ad infinitum, unless something more powerful get in its way. Aristotle, Physics, Book 4

The bolded part is a formulation of the law of inertia, which predates Galileo by some 1800 years or so. As you can see, Aristotle was referring to motion in the vacuum.

Additionally, it appears that Aristotle did not believe what you said he did Russ. It looks to me like Aristotle was talking about what others believed, rather than himself, right up until he stated his own position, which is Aristotle's Law Of Inertia.