Actually the derivative still is fixed on the expanding wave. Doing the partial derivative for fixed y and z just means the author isn't doing it for fixed $$\theta$$ and $$\varphi$$: i.e. he's considering a point that's moving around the expanding sphere in such a way that its y and z coordinates always stay the same instead of expanding radially outward. In some cases such a point will actually move faster than c, which could easily lead to the sort of causality reversals the author seems to be confused by. Note that the author also found t' decreasing, for increasing t, under exactly the same circumstances.
Special Relativity Is Refuted
- Thread starter chinglu
- Start date
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.
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.
Pseudoscience?!?
I said several times and I pointed out that has nothing to do with math.Who starts on this path, make a strategic mistake.
The only problem is the speed of light relative to a moving object.
Researchers in the field applied, not even bothering to contradict.
They don't have time to deal with such nonsense. If you want to live in this Alice's Wonderland is your business.
You know any device that the operation is based on length contraction or time dilation?
If you're so sure I propose to finance the realization of the following device.(I warn you will lose the money.)
"Solar panels that provide electricity converted from the light.
Refractive indices of air 1.000277 speed of light in air c/1.000277
Refractive indices of ethyl alcohol (ethanol) 1.36 speed of light in ethanol c/1.361 (http://en.wikipedia.org/wiki/List_of_refractive_indices)
Due to length contraction, the light from ethanol "sees" length 28.8 times higher that the light from the air. (0.6783339/0.0235372)
Surface 28.8X28.8=829,4 times higher.
So if we dive in ethanol the panel, we obtain an energy of 829.4 times higher."
If you want I can give the evidence for the absence of length contraction.
I said several times and I pointed out that has nothing to do with math.Who starts on this path, make a strategic mistake.
The only problem is the speed of light relative to a moving object.
Researchers in the field applied, not even bothering to contradict.
They don't have time to deal with such nonsense. If you want to live in this Alice's Wonderland is your business.
You know any device that the operation is based on length contraction or time dilation?
If you're so sure I propose to finance the realization of the following device.(I warn you will lose the money.)
"Solar panels that provide electricity converted from the light.
Refractive indices of air 1.000277 speed of light in air c/1.000277
Refractive indices of ethyl alcohol (ethanol) 1.36 speed of light in ethanol c/1.361 (http://en.wikipedia.org/wiki/List_of_refractive_indices)
Due to length contraction, the light from ethanol "sees" length 28.8 times higher that the light from the air. (0.6783339/0.0235372)
Surface 28.8X28.8=829,4 times higher.
So if we dive in ethanol the panel, we obtain an energy of 829.4 times higher."
If you want I can give the evidence for the absence of length contraction.
Sure - GPS networks have to account for it.
Please a link.
It is really hard to find this info - you have to google :time, dilation, GPS.
Try this link.
So is this site a conspiracy, are they liars, are they sheeple, or do they just not have the intelligence of you and motor daddy?
About all I can say is that is really, really stupid, for any number of reasons. What does the refractive index of a material have to do with length contraction? Hint - nothing.
Go for it.
Motor Daddy
Valued Senior Member
....or do they just not have the intelligence of you and motor daddy?
It's not about intelligence, it's about what's right. You mention intelligence as if it's measurable as an absolute?
Can a wave be spherical?
Then refute the link.
Sound waves are.
Sound waves.
Yes, I suppose they go in all directions once they have been emitted.
Are light waves the same?
Yes, I suppose they go in all directions once they have been emitted.
Are light waves the same?
Sound waves are not in theirselves spherical, though they do propagate 3-dimensionally.., spherically.
This is probably what the OP had meant or intended, light propagating from a point of origin in a sperical wave front. Though this is possible. We "see" such when we turn on a common light bulb, but what is missing is that that wave front is not the same as the wave front we find in the case of sound. It is comprised of individual photons, each of which propagates in its own line away from the point of origin.
If the OP were correct we should expect that the photon essentially propagates away and toward the point of origin. (that was awkward) This just does not make sense and is not consistent with experience.
The mathematical issue seems to be being addressed.
Motor Daddy
Valued Senior Member
If the question is what is a nanosecond compared to.., that would be one second.
Motor Daddy
Valued Senior Member
So there is a standard second, of which all measures of time are to be taken?
How come a light wave can be seen as being made up of particles, but a sound wave can't?
It's all energy isn't it?
I did not say there was a single standard second, only that a nanosecond is defined as a specific fraction of a second.