The Fundamental Postulate Of Special Relativity Is Self-Contradictory

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

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 &amp;= 0 \\<br /> <br /> \end{split}\end{equation*}<br />

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):

<br /> \frac{\partial^2 E}{\partial t^2} = \frac{1}{c^2} \nabla^2 E<br />

But it never takes the form

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

or

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

which would be required for light traveling 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):

<br /> \frac{\partial^2 E}{\partial t^2} = \frac{1}{c^2} \nabla^2 E<br />

But it never takes the form

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

or

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

which would be required for light traveling 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

 

[quantumdude]

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

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.

 

[quantumdude]

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:

<br /> \begin{equation*}\begin{split}<br /> \oint \vec{E} \cdot d\vec{A} &amp;= \int \frac{\rho}{\epsilon_0} \, dV \\<br /> \int \nabla \cdot \vec{E} \, dV &amp;= \int \frac{\rho}{\epsilon_0} \, dV<br /> \end{split}\end{equation*}<br />

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:

<br /> \begin{equation*}\begin{split}<br /> \oint \vec{E} \cdot d\vec{A} &amp;= \int \frac{\rho}{\epsilon_0} \, dV \\<br /> \int \nabla \cdot \vec{E} \, dV &amp;= \int \frac{\rho}{\epsilon_0} \, dV<br /> \end{split}\end{equation*}<br />

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

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*