The Fundamental Postulate Of Special Relativity Is Self-Contradictory

[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:

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

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

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!

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!

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

 

[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?

 

[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.