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.