Lorentz Transformations Are Wrong

[DarkStar]

Something has me puzzled about the theory of relativity.

At time 0, a photon is emitted from the origin of a rectangular coordinate system. At time t, the photon is at position x, on the positive x axis. Therefore in amount of time t-0=t, the photon has traveled a distance of x. The speed of the photon in this frame is by definition:

x/t

Let c denote the speed of the photon in this frame. Thus

c=x/t

Now, consider things from the photon's point of view. Let time in this system be represented by t`. Let the origin of both frames overlap when t`=0.

When t`=0 let the origin of the other frame begin to move onto this systems positive x` axis. Thus, the speed of the origin of the other system in this frame is:

x`/t`

And speed is relative hence

x/t=x`/t`

Suppose that the Lorentz transformation is correct.
Thus, x`= x (1-v^2/c^2)^1/2.

That leads to the following result:

1/t = (1-v^2/c^2)^1/2 divided by t`

But since v=c, it follows that (1-v^2/c^2)^1/2=0, from which it follows that

1/t = 0

from which it follows that

1=0, which is false.

Thus, the Lorentz transformations are invalid.

Where is my mistake?

 

[Chen]

Now, consider things from the photon's point of view.

Not again, StarThrower.

 

[OneEye]

Classic Albgebraic Proof

This is a classic "division by zero" case. I remember my 10th-grade math teacher pulling something like this on me on a field trip. He used algebra to "prove" that 1=2 by sneaking a division by zero in. WATCH YOUR VALUES!

It is the case the the Lorentz transform breaks down when v=c. Everyone acknowledges this. You have simply shown this to be true.

Finally, this has got to be the SHORTEST "1=0 proof" I have ever seen. Congratulations!

 

[DarkStar]

The logic looks good to me, I can't seem to spot my error. And I never divided by zero.

 

[Chen]

The unit vector of the X axis from the photon's point of view is of zero length. So how exactly can a photon perceive space?

 

[Chen]

Suppose you were traveling through the universe at an infinite speed. You would cover any distance in no time, literally. Could you then measure the speed of other objects in relation to you?

 

[quantumdude]

StarThrower, I have already told you that this belongs in Theory Development.

 

[quantumdude]

Where is my mistake?

Right here:

Now, consider things from the photon's point of view. Let time in this system be represented by t`. Let the origin of both frames overlap when t`=0.

Once again, you regard the photon as the origin of a frame that can be considered stationary, which of course cannot be done.

 

[OneEye]

Divide by zero error

The logic looks good to me, I can't seem to spot my error. And I never divided by zero.

The correct expression of the Lorentz transform for t' is:

t'=(t-(v/c^2)x)/(1-v^2/c-2)^1/2

So, if v=c (as it does in your model), v^2/c^2=1, so 1-v^2/c^2=0, the square root of which is zero which, as a denominator, renders the equation meaningless.

Yes, I am sorry, you did divide by zero.

 

[DarkStar]

The unit vector of the X axis from the photon's point of view is of zero length. So how exactly can a photon perceive space?

Show me your proof that the unit vector of the X axis from the photon's point of view is of zero length.

 

[OneEye]

Show me your proof that the unit vector of the X axis from the photon's point of view is of zero length.

Lorentx-transform the x in the same way that you did t:

1/x'=(1-v^2/c^2)^1/2/(x-vt)
=> 1/x' = (1-c^2/c^2)^1/2/(x-ct)
=> 1/x' = (1-1)^1/2/(x-ct)
=> 1/x' = 0/(x-ct)
=> 1/x' = 0

Again, a meaningless result because v must never exceed c in SR.

 

[Chen]

Are you avoiding my question for a reason?

 

[DarkStar]

Suppose you were traveling through the universe at an infinite speed. You would cover any distance in no time, literally. Could you then measure the speed of other objects in relation to you?

Definition: speed = v = D/t

D denotes distance travelled.
t denotes amount of time to travel distance D

let D be positive; hence D is nonzero.
let t=0

Hence you get here:

v = \frac{D}{0} = \infty

if D was zero, then speed would be indeterminate, but I stipulated that D is nonzero.

Take the previous equation seriously. Thus we have:

\frac{D}{0} = \infty

There is your equation for infinite speed chen.

Focus on the LHS.

As you can see, we have a nonzero numerator, and a zero denominator.

That is the division by zero error of algebra.

Thus, the concept of infinity is meaningless.

Infinite speed has thus been rendered logically impossible. And that which is logically impossible is impossible. Since you cannot travel through space at an infinite speed, there is no need to formulate a "what if" statement.

What if pigs could fly, and the sky was green instead of blue, and when apples fell off trees, they went into orbit about earth, instead of falling down.

 

[DarkStar]

Right here:



Once again, you regard the photon as the origin of a frame that can be considered stationary, which of course cannot be done.

What do you mean of course that cannot be done. I just did it. Consider motion from the photon's point of view. Consider things on the photon's worldline or whatever.

 

[Chen]

Take the previous equation seriously. Thus we have:

\frac{D}{0} = \infty

There is your equation for infinite speed chen.

Focus on the LHS.

As you can see, we have a nonzero numerator, and a zero denominator.

That is the division by zero error of algebra.

Thus, the concept of infinity is meaningless.

Perhaps our algebra is wrong? How do you jump to the conclusion that it is not possible to travel through space at an infinite speed, only because you cannot grasp the idea of doing so?

Just so you know, I was born in a galaxy thousands of light years away, and I traveled here to Earth at an infinite speed, and my people have been doing so for as long as we can remember.

 

[quantumdude]

What do you mean of course that cannot be done. I just did it. Consider motion from the photon's point of view. Consider things on the photon's worldline or whatever.

You know exactly why it cannot be done. A postulate of SR is that there is no frame in which the speed of light is anything other than 'c'. When you assume the negative of that postulate, you are no longer doing relativity.

 

[OneEye]

Tempest in a teapot

Put another way, precisely because the Lorentz equations break down wqhen v=c, c is a fundamental limiting velocity.

Darkstar, you have demonstrated this fact. Perhaps you have elaborated on it somewhat. But that is all that you have done.

You have not proven the Lorentz transform wrong. You have only publicized a conclusion of it: That c is unattainable within SR (by anything except light).

This is essential to the math of the Lorentz transform. Dividing by (1-v^2/c^2)^1/2 results in a logarithmic decrease in the denominator of the transform as v approaches c - hence, a logarithmic increase in t and x dilation WR K'.

That's all.

Don't mean to dis you here - but all you are doing is demonstrating what everyone already knew - that v must always be less than c.

Hope this is useful to you.

 

[DarkStar]

Perhaps our algebra is wrong? How do you jump to the conclusion that it is not possible to travel through space at an infinite speed, only because you cannot grasp the idea of doing so?

Just so you know, I was born in a galaxy thousands of light years away, and I traveled here to Earth at an infinite speed, and my people have been doing so for as long as we can remember.

It is impossible for our algebra to be wrong.

Let A,B,C denote arbitrary numbers.

Axiom: not (0=1)
Axiom(transitive property of equality): if A=B and B=C then A=C
Axiom(reflexive property of equality): A=A
Axiom(symmetric property of equality): if A=B then B=A

Axiom(closure of addition): If A+B=D then D is a number.
Axiom(closure of multiplication): If A*B=D then D is a number.

Axiom(commutativity of addition): A+B=B+A
Axiom(associativity of addition): (A+B)+C=A+(B+C)
Axiom(additive identity): There is at least one number 0, such that for any number A, 0+A=A.
Axiom(negative numbers): For any number A, there is at least one number (-A), such that A+(-A)=0

Axiom(commutativity of multiplication): A*B=B*A
Axiom(associativity of multiplication): (A*B)*C=A*(B*C)
Axiom(multiplicative identity): There is at least one number 1, such that for any number A, 1*A=A
Axiom(reciprocal numbers) For any number A, if not (A=0) then there is at least one number 1/A such that A*(1/A) = 1
Axiom(distributivity): A*(B+C)=A*B+A*C

You won't find a contradiction.

 

[DarkStar]

The correct expression of the Lorentz transform for t' is:

t'=(t-(v/c^2)x) /(1-v^2/c-2)^1/2

So, if v=c (as it does in your model), v^2/c^2=1, so 1-v^2/c^2=0, the square root of which is zero which, as a denominator, renders the equation meaningless.

Yes, I am sorry, you did divide by zero.

First off, I didn't use the Lorentz transformation from t to t`.
At any rate, I will have a look at your work. You don't necessarily have the division by zero error of algebra if the numerator is also zero. That being the case would give:


0=t-(v/c^2)x

From which it would follow that:

t=xv/c^2

From which it would follow that:

tc^2=xv

from which it would follow that

c^2=(x/t)v

and x/t=v so it follows that

c^2 = v^2

From which it would follow that c=v.

So if c=v, which it does in this case, you get

t` = 0/0 which is indeterminate.

That is not the division by zero error of algebra.

I still don't see any error.

 

[matt grime]

So you admit you have an indeterminate quantity in your calculations?

Not big, not clever, but if it weren't so depressing it would be funny.

 

[DarkStar]

So you admit you have an indeterminate quantity in your calculations?

Not big, not clever, but if it weren't so depressing it would be funny.

No, I do not admit that because I didn't use the time transformation.

 

[matt grime]

But if you wished to demonstrate the inconsitency from within the theory then you must have used something equivalent to this transform, otherwise you aren't "playing by the rules"

 

[DarkStar]

But if you wished to demonstrate the inconsitency from within the theory then you must have used something equivalent to this transform, otherwise you aren't "playing by the rules"

Matt, I must be making some dumb kind of mistake. Just help me isolate it, that's all i ask.

Hopefully, you followed the problem, and can get here:

\frac{x}{t} = \frac{x^\prime}{t^\prime}

The equation above isn't wrong. Let me pick up the argument from here:

x^\prime = x (\sqrt{1-v^2/c^2}
Open scope of first assumption
\frac{x}{t} = \frac{x \sqrt{1-v^2/c^2} }{t^\prime}
Substitute equivalent expressions for one another.
\frac{1}{t} = \frac{ \sqrt{1-v^2/c^2} }{t^\prime}
divide both sides by x (x is positive, since it is a distance traveled)

Take the limit of both sides of the above equation as v approaches c, and you get:
\frac{1}{t} = \frac{ \sqrt{1-c^2/c^2} }{t^\prime} ^2

\frac{1}{t} = \frac{ \sqrt{0}{t^\prime} ^2

\frac{1}{t} = \frac{0}{t^\prime} ^2

\frac{1}{t} = 0 ^2

1=0

And one of the axioms of algebra is not (1=0).

Thus, using the Lorentz transformation, we can reach the following explicit contradiction:

1=0 and not (1=0)

Now, close the scope of your only assumption, to reach the following absolute truth:

If x` = x (1-v^2/c^2)^1/2 then (1=0 & not(1=0).

The following absolute truth now follows using reductio ad absurdum:

not[x` = x (1-v^2/c^2)^1/2 ]

Thus, the Lorentz transformations are wrong.

 

[Hurkyl]

Matt, I must be making some dumb kind of mistake. Just help me isolate it, that's all i ask.

I would say it's ignoring the fact that (given a fixed t) the limit of t' is 0 as v approches c.

 

[matt grime]

You go wrong at the first statement where you write

x^\prime = x (\sqrt{1-v^2/c^2}

since you know v^2=c^2 in your thought experiment, that is you are using equations in situations that are explicitly not allowed by the postulates.

Lorentz transforms apply to situations where explicitly v is not equal to c. As people keep telling you, under whichever personality you are posting.

Why does it bother you that the theory goes wrong when you don't obey the axioms. Surely it would be more troubling if ther opposite were true.

 

[OneEye]

Return of the Tempest

Matt, I must be making some dumb kind of mistake. Just help me isolate it, that's all i ask.

Hopefully, you followed the problem, and can get here:

\frac{x}{t} = \frac{x^\prime}{t^\prime}

The equation above isn't wrong. Let me pick up the argument from here:

x^\prime = x (\sqrt{1-v^2/c^2}
Open scope of first assumption
\frac{x}{t} = \frac{x \sqrt{1-v^2/c^2} }{t^\prime}
Substitute equivalent expressions for one another.
\frac{1}{t} = \frac{ \sqrt{1-v^2/c^2} }{t^\prime}
divide both sides by x (x is positive, since it is a distance traveled)

Take the limit of both sides of the above equation as v approaches c, and you get:
\frac{1}{t} = \frac{ \sqrt{1-c^2/c^2} }{t^\prime} ^2

\frac{1}{t} = \frac{ \sqrt{0}{t^\prime} ^2

\frac{1}{t} = \frac{0}{t^\prime} ^2

\frac{1}{t} = 0 ^2

1=0

And one of the axioms of algebra is not (1=0).

Thus, using the Lorentz transformation, we can reach the following explicit contradiction:

1=0 and not (1=0)

Now, close the scope of your only assumption, to reach the following absolute truth:

If x` = x (1-v^2/c^2)^1/2 then (1=0 & not(1=0).

The following absolute truth now follows using reductio ad absurdum:

not[x` = x (1-v^2/c^2)^1/2 ]

Thus, the Lorentz transformations are wrong.

Two issues come to mind.

First off, I cannot see why you would say,
x^\prime = x (\sqrt{1-v^2/c^2})

According to the Lorentz transform, the correct expression is:

x^\prime = \frac { x-vt } { \sqrt{ 1-v^2/c^2 } } , not

x^\prime = x (\sqrt{1-v^2/c^2})

So, this adds some consternation. Further, when you say:

\frac{1}{t} = \frac{ \sqrt{1-v^2/c^2} }{t^\prime}

Let's just do the obvious and invert both equations:

t = \frac { t^\prime } { \sqrt { 1 - { v^2 \over c^2 } } }

Which, through simple substitution of v = c, gives:

t = \frac { t^\prime } { \sqrt { 1 - { c^2 \over c^2 } } }

\Rightarrow t = \frac { t^\prime } { \sqrt { 1 - 1 } }

\Rightarrow t = \frac { t^\prime } { 0 }

And hence, we see the infamous division by zero.

What should be giving you qualms here is that even the equations that you provide work fine at sub-light speeds, just as the Lorentz transform does.

The problem is not that the Lorentz transform is wrong. It is simply that the algebra here becomes an invalid operation when v \ge c.

I don't claim to be a mathematician here. Probably, you are more of a mathematician that I. It just seems farily obvious to me that, with an equation which has a denominator of \sqrt { 1-v^2/c^2 }, you're not going to get any meaningful results when v=c. And, when v>c, you had better get ready for some imaginary results.

That's all.

In case you haven't already read it, may I commend Appendix I of Einstein's Relativity (Crown, New York)? The appendix provides A "Simple Derivation of the Lorentz Transformation". Feel free to check Albert's math.

(BTW, thanks for introducing me to TEX inserts; I was hoping to find a way to write equations on this board. Great help!)

t^\prime = { { t - { v \over c^2 } x } \over \sqrt { 1 - { v^2 \over c^2 } } }

x^\prime = { { x - vt } \over \sqrt { 1 - { v^2 \over c^2 } } }

Heeheehee!

 

[Chen]

You won't find a contradiction.

You will not be able to prove the consistency of that system, either. I refer you to Gödel's Incompleteness Theorem.

(Over at my planet, by the way, we use a completely different type of "algebra".)

 

[RoguePhysicist]

A derivation of the Lorentz transformations can be found here Darkstar:

http://casa.colorado.edu/~ajsh/sr/construction.html

It seems that you presumed that the distance traveled by the unprimed frame is "Lorentz contracted," rather than using the Lorentz coordinate transformation? Why did you do that?

It is obvious that there is division by zero error in the Lorentz transformations if v=c. You are telling us that means the Lorentz transformations are wrong. We are telling you that means that c is a limiting velocity.

I am going to humor you, and work on the problem for awhile, and let you know what I think.

 

[Severian596]

Oh man...every time I see a new account with fewer than 3 posts show up in one of these threads I'm paranoid it's just StarThrower. His aliases include:

StarThrower (220 posts)
MindWarrior (1 post)
DarkStar (9 posts and counting)

Hopefully RoguePhysicist won't be added to this list.

-a truly leery

sev

 

[RoguePhysicist]

Lorentz coordinate transformations

x^\prime = { { x - vt } \over \sqrt { 1 - { v^2 \over c^2 } } }


t^\prime = { { t - { v \over c^2 } x } \over \sqrt { 1 - { v^2 \over c^2 } } }

We can undo the division by zero error that occurs in the Lorentz transformations when v=c, if we multiply both sides of the transformations by

\sqrt{1-v^2/c^2}

This brings us to here:

x^\prime \sqrt { 1 - { v^2 \over c^2 } }= x - vt


t^\prime \sqrt { 1 - { v^2 \over c^2 } } = { { t - { vx \over c^2 }

When v=c, the x x` transformation equation now gives:

x^\prime \sqrt { 1 - { c^2 \over c^2 } }= x - ct

x^\prime \sqrt { 1 - 1 }= x - ct

(x^\prime) 0= x - ct

0= x - ct

x= ct

x/t= c

The previous equation tells us that the photon is moving through the unprimed system at speed c in that system. Unfortunately, x` fell out of the analysis.

When v=c, the t t` transformation equation now gives:

t^\prime \sqrt { 1 - { c^2 \over c^2 } } = { { t - { cx \over c^2 }

t^\prime \sqrt { 0 } = t - { cx \over c^2 }

(t^\prime) 0 = t - { x \over c }

0 = t - { x \over c }

t = {x \over c}

ct = x

c = x/t

Which is the same relationship as before. Thus, in the case where v=c, the Lorentz transformations cannot give you the coordinate of something moving at speed v in a photon's rest frame.

Since we are only analyzing the case where v=c, let us write the previous equation as follows:

v = x/t

It also follows that:

v = {x^\prime \over t^\prime}

Using the Lorentz transformations at the top of this post gives:

v = \frac{x^\prime} {t^\prime} = {{ { x - vt } \over \sqrt { 1 - { v^2 \over c^2 } } } \over { { t - { v \over c^2 } x } \over \sqrt { 1 - { v^2 \over c^2 } } }

which simplifies to

v = \frac{x^\prime}{t^\prime} = {x - vt \over {t - { vx \over c^2 }}

And we are only analyzing the case where v=c, so

v = \frac{x^\prime}{t^\prime} = {x - ct \over {t - { cx \over c^2 }}

So

v = \frac{x^\prime}{t^\prime} = {x - ct \over {t - { x \over c }}

And so we reach the following line of work:

v = \frac{x^\prime}{t^\prime} = \frac{x-ct}{t-x/c}

Multiplying both sides of this equation by (t-x/c) we get to

v (t-x/c) = \frac{x^\prime}{t^\prime} = x-ct

From which it follows that

vt-xv/c = \frac{x^\prime}{t^\prime} = x-ct

From which it follows that

vt-xv/c = x-ct

And since v=c, it follows that

ct-xc/c = x-ct

ct-x = x-ct

From which it follows that:

2ct = 2x

Again, we have:

c= x/t

Again, t`, and x` have fallen out of the analysis, unless we work backwards from v=c. We get here:

v = x/t

And

v = \frac{x^\prime}{t^\prime}

From which the original relationship follows:

\frac{x}{t} = \frac{x^\prime}{t^\prime}

It seems that the case v=c just leads you in circles.

The clearest conclusion seems to be that in the case where v=c, the Lorentz transformations cannot be used.

 

[Chen]

We can undo the division by zero error that occurs in the Lorentz transformations when v=c, if we multiply both sides of the transformations by

\sqrt{1-v^2/c^2}

No, you can't. If you do you must qualify and say that v != c, otherwise you are multiplying two sides of an equaiton by zero.

 

[Chen]

Oh man...every time I see a new account with fewer than 3 posts show up in one of these threads I'm paranoid it's just StarThrower.

I hear you. A moderator could probably tell us if it's him or not, though.

 

[Severian596]

I wonder what the protocol for created "clone" accounts is? If his ISP uses DHCP his IP may change periodically, but I'm assuming all three of his aliases have the same IP. If it were different people on the same machine that would be one thing, but this is obviously the same person attempting to use multiple identities to gain a clean slate in the eyes of other forum members.

This is purposly misleading and a bit unethical.

 

[Chen]

Multiplying both sides of this equation by (t-x/c) we get to

Having just shown that x = ct, or in other words that (t-x/c) = 0, you cannot perform this operation either.

Your work is very messy.

 

[OneEye]

Not that I necessarily know what I'm talking about, but it seems to me that RoguePhysicist is arguing against DarkStar. If this is a ploy, it's an elaborate one, and seems to be working against DarkStar's thesis.

 

[Severian596]

Not that I necessarily know what I'm talking about, but it seems to me that RoguePhysicist is arguing against DarkStar. If this is a ploy, it's an elaborate one, and seems to be working against DarkStar's thesis.

Oh you know what you're talking about, OneEye, and you stated it quite succinctly. If DarkStar/StarThrower/MindWarrior wants so badly for someone to agree with him/her, what better way than to start an imaginary doubting-thomas who is eventually convinced convinced with the error of his ways.

Rotten? You bet, but he's committed identity fraud and should be carefully watched (or avoided) IMHO

 

[RoguePhysicist]

Let us just start with the following line of work, which starts off by assuming that the Lorentz transformations are correct:

Equation 1: v = \frac{x^\prime}{t^\prime} = \frac{x-vt}{t-vx/c^2}

Lorentz Transformations

x^\prime = { { x - vt } \over \sqrt { 1 - { v^2 \over c^2 } } }


t^\prime = { { t - { v \over c^2 } x } \over \sqrt { 1 - { v^2 \over c^2 } } }

In this problem, the relative speed v happens to equal c. Look at the denominator of the RHS of equation 1.

t - vx/c^2

In the case where v=c, the denominator of equation 1 reduces to:

t - x/c

And we know that x/t=c, from which it follows that x/c=t, from which it follows that t-x/c=0. Thus, the denominator of the RHS of equation one contains a zero. This does not necesarily mean that we have the division by zero error of algebra, we need to consider the numerator. Let us now look at the numerator. The numerator of equation 1 is:

x-vt

And in this case, v=c, hence the numerator of equation 1 is x-ct. Since x/t=c, it follows that x=ct, from which it follows that x-ct=0. Thus, in the case were v=c, it follows that x`/t`=0/0, which is an indeterminate form.

Now, we can write the relative speed v (which happens to equal c) as follows:

v = c = \frac{x-ct}{t-x/c}

Multiplying both sides by (t-x/c) gives:

c(t-x/c) = x-ct

From which it follows that

ct - x = x-ct

From which it follows that:

x-ct+x-ct = 0

From which it follows that

2x-2ct=0

From which it follows again that

x/t=c

Again, from this we can only conclude that

x/t=x`/t`

And we start off knowing this.

The customary way to evaluate an indeterminate form, is to use L'Hopital's rule. However, let us pick up the argument here:

x`/t` = 0/0

We start off with the following equation:

v = x/t=x`/t`

So let us take the limit of both sides of the above equation, as v approaches c. We reach the following conclusion:

lim v--> c [x`/t`] = c

This saves us the effort of using L'Hopital's rule. Does anyone know what this means?


For those who wish to use L'Hopital's rule... in the middle of the previous work we arrived here:

v = \frac{x-ct}{t-x/c}

Since c>0, we can divide both sides of the above equation by c, and obtain:

v/c = \frac{x-ct}{ct-x}

Now, if we take the limit as v approaches c of the LHS, we get 1.

The only thing left to do, is take the limit as v approaches c of the RHS.

Since v=x/t, therefore x = vt. Thus, we can write the following:

v/c = \frac{vt-ct}{ct-vt}

Thus, we can write:

v/c = \frac{t(v-c)}{t(c-v)}

The t's cancel out, and we are left with:

v/c = \frac{v-c}{c-v}

The limit as v approaches c of the LHS is equal to 1. The limit as v approaches c of the RHS needs to be broken up into two cases:

Case 1: v>c

Case 2: v<c

Is there anyone who wants to field these cases?

 

[quantumdude]

StarThrower: The IP Address is: 199.233.80.161. The host name is: pix161.shore.co.monmouth.nj.us.

MindWarrior: The IP Address is: 199.233.80.161. The host name is: pix161.shore.co.monmouth.nj.us.

DarkStar: The IP Address is: 199.233.80.161. The host name is: pix161.shore.co.monmouth.nj.us.

RoguePhysicist: The IP Address is: 199.233.80.161. The host name is: pix161.shore.co.monmouth.nj.us.

 

[Chen]

Figures.

 

[Severian596]

Yes! I have catlike instincts and reflexes!

 

[Hurkyl]

So let us take the limit of both sides of the above equation, as v approaches c.

Inconsistent. You already specified that, in this problem, v = c. Thus, it's nonsensical to take the limit as v approaches anything.

 

[OneEye]

StarThrower: The IP Address is: 199.233.80.161. The host name is: pix161.shore.co.monmouth.nj.us.

MindWarrior: The IP Address is: 199.233.80.161. The host name is: pix161.shore.co.monmouth.nj.us.

DarkStar: The IP Address is: 199.233.80.161. The host name is: pix161.shore.co.monmouth.nj.us.

RoguePhysicist: The IP Address is: 199.233.80.161. The host name is: pix161.shore.co.monmouth.nj.us.

Live and learn, I guess. I don't feel bad about being trusting, but I am shocked at the level of duplicity.

Does anyone have a clue what this person is on about? No-one here is taken in; TS/MW/DS/RP has been effectively answered any number of ways by any number of people. What's the story here? You would think all the luster had worn off this form of entertainment by now.

 

[Severian596]

Live and learn, I guess. I don't feel bad about being trusting, but I am shocked at the level of duplicity.

Does anyone have a clue what this person is on about? No-one here is taken in; TS/MW/DS/RP has been effectively answered any number of ways by any number of people. What's the story here? You would think all the luster had worn off this form of entertainment by now.

It seems like ST's behavior approaches the fervor of a religious fanatic...he's out to convert people because he is convinced they are damned. Damned because they do not know the truth like he does. I guess that's the only way I can see it, unless he's a knowledgeable physicist/mathematician who is deliberately trying to mislead innocent minds for fun. That's pretty sick, really.

 

[OneEye]

It seems like ST's behavior approaches the fervor of a religious fanatic...he's out to convert people because he is convinced they are damned. Damned because they do not know the truth like he does. I guess that's the only way I can see it, unless he's a knowledgeable physicist/mathematician who is deliberately trying to mislead innocent minds for fun. That's pretty sick, really.

Well, speaking as a religious fanatic, I've got to say that even religious fanatics have their limits.

I'd like to put this directly to STMWDSRP:

What are you up to, and why?

You'd have much more effect if you just came right out and laid out your point. Do you think that relativity is broken? Are you peddling some other theory? At this point, no-one wants to listen to you because you are behaving in a shady way. Your apparent disrespect for honest discourse is disqualifying you. Further, your bad math has got everyone on edge. You're having zero impact here with your attempt at subtlety. You might as well try just being candid.

So, StarThrower/MindWarp/DarkStar,RoguePhysicist:

What are you getting at?[/size]

Your co-operation would be more than appreciated.

 

[OneEye]

Sorry to humor this duck, but I've been examining this a little, and perhaps a comment or two is not out of place.

lim v--> c [x`/t`] = c

This saves us the effort of using L'Hopital's rule. Does anyone know what this means?

This means that you are confused. You have already established that v={x^\prime\over t^\prime} = c. So, you are asking us to evaluate the case of {lim \atop ^{v\to c}} v={x^\prime \over t^\prime}=c - or put shortly, {lim \atop ^{v\to c}} v=c - and worse yet, {lim \atop ^{v\to c}} c=c.

Can you differentiate a tautology? Not good math.

Further, since v=c is not in the differentiable domain of, e.g., { v - c } \over { c - v }, you have no grounds to use this function either.

Again, I don't claim to be a mathematician, but it seems pretty probable to me that your whole argument rests on bad math.

I welcome your correction.