Special Relativity Is Refuted

[chinglu]

Can a wave be spherical?

Well, probably not.

But, it is close enough.

 

[AlexG]

Pseudoscience is where this belongs.

A load of crap by a couple of cranks.

 

[chinglu]

Hello, back from a long drive and dinner.
There are five glaring mistakes on page 2, which is exactly what we expect when the level of discussion is "the article used calculus"

The Five mistakes I would like to unravel are:
1) Incorrect manipulation of expressions, which ignore the postulates. Thus he stops talking about the physics of light.
2) Ignoring the rest of the Lorentz transform
3) The absurd assertion that radially propagating light also propagates along a vertical line with fixed y.
4) The misuse of multivariable calculus that asserts that just because "$$\frac{\partial x'}{\partial x} > 0$$, if x increases, then x' increased."
5) This misuse of multivariable calculus that asserts that $$\frac{\partial x'}{\partial x} > 0$$ and $$\frac{\partial x}{\partial t} > 0$$ and $$x' < 0$$ are enough to conclude that in the primed frame the light is moving toward the coordinate origin.

Correct manipulation of expressions
Given $$z = 0 \, , \; 0 < x \, , \; 0 < y \, , \; 0 < v < c $$, $$c^2 t^2 = x^2 + y^2$$ describes a two-dimensional surface. So solutions are necessarily parameterized by two independent variables.
$$t(x,y) = \frac{1}{c} \sqrt{x^2 + y^2}$$ is one possible parameterization.
$$x(\theta,t) = ct \, \cos \theta \, , \; y(\theta,t) = ct \, \sin \theta$$ is another.
$$x(y,t) = \sqrt{c^2 t^2 - y^2}$$ is another.

So when we parameterize x', t' and y' in terms of x and t, we have the related choices:
$$x'(x,y) = \left( x - v t(x,y) \right) \gamma = \frac{c x - v \sqrt{x^2 + y^2}}{c \sqrt{1 - \frac{v^2}{c^2}}} = \frac{c x - v \sqrt{x^2 + y^2}}{\sqrt{c^2 - v^2}} \\ y'(x,y) = y \\ t'(x,y) = \left( t(x,y) - \frac{v x}{c^2} \right) \gamma = \frac{\frac{1}{c} \sqrt{x^2 + y^2} - \frac{v x}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{c \sqrt{x^2+y^2}-v x}{c \sqrt{c^2-v^2}} $$

Partial derivatives are only partial derivatives

Now we can take partial derivatives of these expressions, remembering to do the work Mr Banks ignored:
$$ \begin{array}{rclrclrclrcl} \frac{\partial x'(x,y)}{\partial x} & = & \frac{c - \frac{v x}{\sqrt{x^2 + y^2}}}{\sqrt{c^2 - v^2}} & \frac{\partial y'(x,y)}{\partial x} & = & 0 & \frac{\partial t'(x,y)}{\partial x} & = & \frac{\frac{cx}{\sqrt{x^2 + y^2}} - v}{c \sqrt{c^2 - v^2}} & \frac{\partial t(x,y)}{\partial x} & = & \frac{x}{c\sqrt{x^2 + y^2}} \\ \frac{\partial x'(x,y)}{\partial y} & = & \frac{- v y}{\sqrt{x^2 + y^2}\sqrt{c^2 - v^2}} & \frac{\partial y'(x,y)}{\partial y} & = & 1 & \frac{\partial t'(x,y)}{\partial y} & = & \frac{y}{\sqrt{x^2 + y^2}\sqrt{c^2 - v^2}} & \frac{\partial t(x,y)}{\partial y} & = & \frac{y}{c\sqrt{x^2 + y^2}} \end{array}$$

So here we can see that even though $$\frac{\partial x'}{\partial x} > 0$$, $$\frac{\partial x'}{\partial y} < 0$$ and so the sign of $$\partial x'$$ is indeterminate unless we add a constraint on $$\partial x \, , \; \partial y$$. Mr. Banks sees to hold y constant, but then he is no longer describing the propagation of rays of light but where the expanding sphere of light meets the line $$y = y_g$$ which is akin to a searchlight beam hitting a wall and not akin to a particle. Like a spot on the wall lit up by a laser pointer, it is not constrained to move slower than light.
$$\frac{\partial x}{\partial t} = c \sqrt{1 + \frac{y^2_g}{x^2}} > c$$

Directions need more than one coordinate
Even with all the problems, does the imaginary choice move away from the new coordinate origin?
To answer this, we must take the dot product of the velocity with the position in the new coordinate system and check the sign.
$$x'(x,y) \frac{\partial x'}{\partial t'} + y'(x,y) \frac{\partial y'}{\partial t'} = x'(x,y) \frac{\partial x'}{\partial x} \frac{\partial x}{\partial t'} + y'(x,y) \frac{\partial y'}{\partial x} \frac{\partial x}{\partial t'} = c \frac{c \sqrt{x^2+y^2}-v x}{\sqrt{c^2-v^2}} $$ which is always greater than zero, indicating that even in the bad physics, Mr. Banks has also erred when "the article used calculus."

A better way
Because of the conservation of angular momentum, we should use a better parameterization to see the fate of actual rays of light. $$x(\theta,t) = ct \, \cos \theta \, , \; y(\theta,t) = ct \, \sin \theta$$ is natural.
So when we parameterize x', t' and y' in terms of theta and t, we have the related choices:
$$x'(\theta,t) = \left( x(\theta,t) - v t \right) \gamma = \frac{c \, \cos \theta - v}{\sqrt{1 - \frac{v^2}{c^2}}} t \\ y'(\theta,t) = y(\theta,t) = ct \, \sin \theta \\ t'(\theta,t) = \left( t - \frac{v x(\theta,t)}{c^2} \right) \gamma = \frac{1 - \frac{v}{c} \, \cos \theta }{\sqrt{1 - \frac{v^2}{c^2}}} t $$

$$ \begin{array}{rclrclrclrclrcl} \frac{\partial x'(\theta,t)}{\partial \theta} & = & \frac{-c \, \sin \theta }{\sqrt{1 - \frac{v^2}{c^2}}} t & \frac{\partial y'(\theta,t)}{\partial \theta} & = & ct \, \cos \theta & \frac{\partial t'(\theta,t)}{\partial \theta} & = & \frac{\frac{v}{c} \, \sin \theta }{\sqrt{1 - \frac{v^2}{c^2}}} t & \frac{\partial x(\theta,t)}{\partial \theta} & = & - c t \, \sin \theta & \frac{\partial y(\theta,t)}{\partial \theta} & = & c t \, \cos \theta \\ \frac{\partial x'(\theta,t)}{\partial t} & = & \frac{c \, \cos \theta - v}{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{\partial y'(\theta,t)}{\partial t} & = & c \sin \theta & \frac{\partial t'(\theta,t)}{\partial t} & = & \frac{1 - \frac{v}{c} \, \cos \theta }{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{\partial x(\theta,t)}{\partial t} & = & c \, \cos \theta & \frac{\partial y(\theta,t)}{\partial t} & = & c \, \sin \theta \end{array}$$

So how fast is light moving in the unprimed frame (holding theta constant)?
$$\sqrt{\left( \frac{\partial x}{\partial t} \right)^2 + \left( \frac{\partial y}{\partial t} \right)^2} = c$$
And in the unprimed frame?
$$\sqrt{\left( \frac{\partial x'}{\partial t'} \right)^2 + \left( \frac{\partial y'}{\partial t'} \right)^2} = \sqrt{\left( \frac{\partial x'}{\partial t} \frac{\partial t}{\partial t'} \right)^2 + \left( \frac{\partial y'}{\partial t} \frac{\partial t}{\partial t'} \right)^2} = \sqrt{\left( \frac{c \, \cos \theta - v}{1 - \frac{v}{c} \, \cos \theta } \right)^2 + \left( c \sin \theta \frac{\sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{v}{c} \, \cos \theta } \right)^2} = c$$

So is the light moving towards the origin in the primed frame?
$$\left( \frac{c \, \cos \theta - v}{\sqrt{1 - \frac{v^2}{c^2}}} t \right) \left( \frac{c \, \cos \theta - v}{1 - \frac{v}{c} \, \cos \theta } \right) + \left( ct \, \sin \theta \right) \left( c \sin \theta \frac{\sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{v}{c} \, \cos \theta } \right) = \frac{c^2 t (v-c \cos \theta)^2}{\sqrt{c^2-v^2} (c-v \cos \theta)} + \frac{c^2 t \sqrt{c^2 - v^2} \sin^2 \theta}{c - v \cos \theta} = \frac{c^2 t (c-v \cos \theta)}{\sqrt{c^2-v^2}} > 0$$
Of course not. Only someone without the ability to understand special relativity would think so. Only someone who doesn't understand calculus would calculate so.

I will address two more glaring errors in your presentation.

3) The absurd assertion that radially propagating light also propagates along a vertical line with fixed y.

Directions need more than one coordinate

1) You are claiming that a spherical light pulse does not propagate along a fixed y line because it magically disappears along that line.

2) You confuse yourself with directions. The article clearly states the light like space time interval is invariant. Therefore, the direction of motion is measured from the origin of the frame to the intersection of the SLW with the fixed yg line.

Hence, your statement above and extensive failed math after that is worthless.

 

[chinglu]

Pseudoscience is where this belongs.

A load of crap by a couple of cranks.

The calculus has been presented.

You can either refute it or submit to it.

 

[AlexG]

rpenner refuted it quite nicely.

 

[chinglu]

rpenner refuted it quite nicely.

I can see you do not understand math.

Why don't you summarize his argument and show you understand it.

 

[chinglu]

Wow, I have yet to see a poster here that understands the proper use of the partial derivative.

Any attacker to this thread would need to understand that since the partial derivative is used extensively in SR.

I will expect any attacker that defends SR to show they understand multivariable calculus since that is SR.

 

[AlexG]

Simply ignore experiment and observation in favor of your misapplication of calculus.

There is no point in continuing to argue with you, because you ignore reality.

 

[arfa brane]

I would expect anyone who mentions blackbody radiation as part of an astronomical measurement to understand it too.

Oops, I got one crank confused with another.

 

[chinglu]

Simply ignore experiment and observation in favor of your misapplication of calculus.

There is no point in continuing to argue with you, because you ignore reality.

The calculus is applied correctly. Otherwise prove it is applied incorrectly.
To defend SR, you will need to prove calculus is false.

Can you do that?

 

[AlexG]

To defend SR, you will need to prove calculus is false.

To defend SR, we simply look at experiment, observation and application.

Reality trumps you.

 

[chinglu]

Here I will show everyone how it works.
Picture 1 is the SLW when it hits the point (0,yg,0) in the unprimed frame.
X
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UP...P

Picture 2 the SLW has expanded and intersected the line y = yg further from the point (0,yg,0) in the unprimed frame but is now closer to the primed frame origin. Thus the expansion of the SLW in the unprimed frame translates to contraction in the primed frame which contradicts the light postulate in the primed frame making SR a contradiction.
| X
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|
|
|
|.........|
UP.....P

 

[arfa brane]

The calculus is applied correctly. Otherwise prove it is applied incorrectly.

The calculus is not applied correctly. If it was it would agree with actual measurements, and it doesn't.
Prove that calculus can show that actual measurements are all wrong, go on.

It's like saying you can prove that petrol isn't a fuel that powers internal combustion in an engine. Or that birds don't fly due to aerodynamic principles, but some other (mathematical) reason.
Or say, that light doesn't propagate through space due to an electromagnetic principle, which has been established, and accepted as an explanation that works, for well over a century.

It's just dumb, end of story.

 

[chinglu]

To defend SR, we simply look at experiment, observation and application.

Reality trumps you.

No it does not.

Math and calculus wins.

You sound like a religious bigot that hates math and calculus results.

 

[chinglu]

The calculus is not applied correctly. If it was it would agree with actual measurements, and it doesn't.
Prove that calculus can show that actual measurements are all wrong, go on.

It's like saying you can prove that petrol isn't a fuel that powers internal combustion in an engine. Or that birds don't fly due to aerodynamic principles, but some other (mathematical) reason.
Or say, that light doesn't propagate through space due to an electromagnetic principle, which has been established, and accepted as an explanation that works, for well over a century.

It's just dumb, end of story.

That is the thing about math.

Any idiot can prove a misapplication or error.

I see you cannot do that. LOL

 

[AlexG]

No it does not.

Math and calculus wins.

You sound like a religious bigot that hates math and calculus results.

Only an idiot thinks that reality doesn't matter.

Cranks like to ignore reality.

They live in a universe of their own.

 

[Trippy]

I agree

Not what I said - you're being dishonest.

I disagree.
The corrections which are made ​​are described here: Global Positioning System, Correcting a GPS receiver clock

We're talking about corrections applied to the satelite, not the receiver. Try again, try harder (actually, I'll settle for you just trying at this point).

There is a relativistic correction which is applied before launch, which is the correction I was referring to, and there are corrections applied to the whole network due to changes in the way the earth rotates about its own axis, and then there are corrections applied to the recievers.

 

[arfa brane]

Any idiot can prove a misapplication or error.

I see you cannot do that. LOL

Any idiot can claim they can use calculus to prove that measurements are wrong, I see you have no idea how to do that.

What's that stuck on your chin, btw?

 

[chinglu]

Only an idiot thinks that reality doesn't matter.

Cranks like to ignore reality.

They live in a universe of their own.

So, since any idiot can prove an application of calculus is false and you cannot, then .....

 

[chinglu]

Any idiot can claim they can use calculus to prove that measurements are wrong, I see you have no idea how to do that.

What's that stuck on your chin, btw?

I proved that the calculus is correct. Otherwise, prove it is false.

So, the calculus stands.