# Lorentz Transformations Are Wrong

1. Apr 10, 2004

### 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?

2. Apr 10, 2004

### Chen

Not again, StarThrower.

3. Apr 10, 2004

### 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!

4. Apr 10, 2004

### DarkStar

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

5. Apr 10, 2004

### 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?

6. Apr 10, 2004

### Chen

Suppose you were travelling 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?

7. Apr 10, 2004

### Tom Mattson

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

8. Apr 10, 2004

### Tom Mattson

Staff Emeritus
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.

9. Apr 10, 2004

### OneEye

Divide by zero error

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.

Last edited by a moderator: Apr 10, 2004
10. Apr 10, 2004

### DarkStar

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

11. Apr 10, 2004

### OneEye

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.

12. Apr 10, 2004

### Chen

Are you avoiding my question for a reason?

13. Apr 10, 2004

### DarkStar

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.

14. Apr 10, 2004

### DarkStar

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.

15. Apr 10, 2004

### Chen

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 travelled here to earth at an infinite speed, and my people have been doing so for as long as we can remember.

16. Apr 10, 2004

### Tom Mattson

Staff Emeritus
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.

17. Apr 10, 2004

### 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.

18. Apr 10, 2004

### DarkStar

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.

19. Apr 10, 2004

### DarkStar

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.

20. Apr 10, 2004

### 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.