Lorentz Transformations Are Wrong

[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.