Factor Equation: (1-v^2/c^2)^3/sqr((1-v^2/c^2)^2)

[dextercioby]

So u can simplify the sqrt with the denominator and be left with a product.

Daniel.

 

[eNathan]

Ok, with all this said, is ther any way to simplify

\sqrt{|x|} * \frac{x} {|x|}

Or is that the simplest it can get?

 

[dextercioby]

|x|=\left(\sqrt{|x|}\right)^{2}

So,unless that "x" is 0,u could simplify.

Danie.

 

[eNathan]

|x|=\left(\sqrt{|x|}\right)^{2}

So,unless that "x" is 0,u could simplify.

Danie.

From what it looks like, you just said that abs(x) = sqr( abs(x) )^2

How is that a simplfication of the equation? I am not being mean, I am just woundering ...

 

[matt grime]

he didn't say that was a simplification he said you cuold use it to simplify.

 

[eNathan]

And how would I use it to simplify?

.../

 

[dextercioby]

On a second thought,there's no point in simplifying,as the square root would end up in the denominator and that wouldn't be pretty.Trying to fix it,would lead you back where u started...

Daniel.

 

[eNathan]

hmn..OK then, it has been simplified enough and this thread has been dragging on forever lol.

Thanks for the help everyone!

 

[baby_garfield]

isn't it just 1?!

 

[Moo Of Doom]

isn't it just 1?!

\frac{x}{|x|}?

No. That's the signum function, sgn(x), which is -1 for negative values and 1 for positive ones.

On that note, you could say \frac{x}{|x|}=\frac{|x|}{x} also.

 

[eNathan]

hmn, I will look into that sgn function. If anyone has not noticed my equation is very similar to the Lorenz Transformation. And I was the one who made the equation to do something that the traditional Lorenz Transformation could not. I don't wish to argue the validity of my equation, because I did start a thread on it once and nobody could understand what my equation was. But if you want me to show you why the mathematics of my equation does something that the tradition one cannot I will.

And I guess I could use the sgn(x) function as compared to dividing x by abs(x)

Thx!

 

[gregmead]

what does your equation do that the traditional lorenz transofrmation can't ?

 

[master_coda]

what does your equation do that the traditional lorenz transofrmation can't ?

It gives negative values when v > c instead of complex ones.

 

[eNathan]

It gives negative values when v > c instead of complex ones.

THANK YOU! It just seemed that people over here https://www.physicsforums.com/showthread.php?t=65446 could not get that (althoug half of it was my terrible latex formate). The reason why I like the equation is for theoretical v > c travel. My equation states that if you travel faster than light, "negative time" will have passed, hence you traveled into the past. I know relativity forbids it.

 

[gregmead]

ah ok. any reason why you made this ?

 

[master_coda]

THANK YOU! It just seemed that people over here https://www.physicsforums.com/showthread.php?t=65446 could not get that (althoug half of it was my terrible latex formate). The reason why I like the equation is for theoretical v > c travel. My equation states that if you travel faster than light, "negative time" will have passed, hence you traveled into the past. I know relativity forbids it.

You probably would have gotten a better response if you actually explained why you think traveling faster than c should cause you to travel backwards in time.

 

[whozum]

eNathan, the reason people don't get your writing is because you don't get their explanations. Realize that most of the people really DO know what theyre talking about, as most of them are graduate students or above in physics. They have degrees in this kind of thing. When they tell you something, take it as correct.

What are your qualifications? Do you think something as trivial as an absolute value sign could answer our questions about time travel? If it was that easy, do you think Einstein wouldn't have come up with it? Any equation can be manipulated to do what you want it to, but that's not the point. The point is to find an equation that describes what happens.

I can say:

v = -d/t [/tex], so then -d = v*t [/tex]<br /> <br /> and claim "hey look, you can have negative distances!", but that equation no longer describes the physical phenomenon its intended to.

 

[eNathan]

eNathan, the reason people don't get your writing is because you don't get their explanations. Realize that most of the people really DO know what theyre talking about, as most of them are graduate students or above in physics. They have degrees in this kind of thing. When they tell you something, take it as correct.

What are your qualifications? Do you think something as trivial as an absolute value sign could answer our questions about time travel? If it was that easy, do you think Einstein wouldn't have come up with it? Any equation can be manipulated to do what you want it to, but that's not the point. The point is to find an equation that describes what happens.

I can say:

v = -d/t [/tex], so then -d = v*t [/tex] <br /> <br /> and claim "hey look, you can have negative distances!", but that equation no longer describes the physical phenomenon its intended to.

<br /> <br /> I see your point. My equation probably does not describe anything physical, but I said "if I were to assume v > c = -t (backwards in time) then how in the world do I calculate it? Hence, I made the equation. I guess the traditional \sqrt {1-(v/c)^2} would be "better" in the respect that it limits the equation to only v < c and v = c travel. I am fully aware that my equation does not abide by relativity, but it seemed that the "mathematical" aspect of my equation was in question, when in fact the equation works fine. Some tried to simply the v > c equation but they could not. For instance [\sqrt{1-(\frac{c}{v})^{2}}]^{-1} gives a positive result for v > c and a negative (imaginary) result for v < c. <br /> <br /> I assume all this is cleard up now. My equation works if you were to assume <br /> v > c means negative time (over infinite mass and negative length). But it would be best to keep the original LT because the theory of relativity forbids it.<br /> <br /> <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />

 

[whozum]

Your equation works but it has no meaning. You can't assume v>c means negative time when you are talking about your equation, it represents nothing in physical reality:

It is equivalent to

f(x) = \sqrt{1-\frac{x^2}{k}}

Why would someone consider that function to have anything to do with reality, until he observes reality to follow that pattern? If they would, why not

f(x) = \frac{e^xsin(x)}{(x^3)(x^2-2x+e^xln(.47sin(x)))}? I could say when ln(.47sin(x)) is larger than e^x then such and such happens, but there's no reason to say that the pattern of increase of those two functions will describe how something works.



You model equations to fit the world, not model the world to fit equations. I understnad your intent wasnt to model anything physical, but when you take a relatively (no pun intended) infamous equation and start manipulating it to do something you want, without changing the variables, you are implying that you are attempting to change that model (for that specific problem). If you had instead used

f(x) = \sqrt{1-\frac{x^2}{k}}

instead, you wouldn't have any arguments as to the validity of the equation for modeling purposes.

 

[eNathan]

Your equation works but it has no meaning. You can't assume v>c means negative time when you are talking about your equation, it represents nothing in physical reality:

It is equivalent to

f(x) = \sqrt{1-\frac{x^2}{k}}

Why would someone consider that function to have anything to do with reality, until he observes reality to follow that pattern? If they would, why not

f(x) = \frac{e^xsin(x)}{(x^3)(x^2-2x+e^xln(.47sin(x)))}? I could say when ln(.47sin(x)) is larger than e^x then such and such happens, but there's no reason to say that the pattern of increase of those two functions will describe how something works.



You model equations to fit the world, not model the world to fit equations. I understnad your intent wasnt to model anything physical, but when you take a relatively (no pun intended) infamous equation and start manipulating it to do something you want, without changing the variables, you are implying that you are attempting to change that model (for that specific problem). If you had instead used

f(x) = \sqrt{1-\frac{x^2}{k}}

instead, you wouldn't have any arguments as to the validity of the equation for modeling purposes.

I think we are both on the same page here. Yes, my equation does not describe anything physical. But, if you look at a graph of time dilation, you can almost predict or assume (logical) that if v > c then -t. But I have not taken any courses on relativity (although I want to), and all of my studies are straight from the NET. And the NET has some very innacurate and misleading statements therein.

 

[whozum]

I think we are both on the same page here. Yes, my equation does not describe anything physical. But, if you look at a graph of time dilation, you can almost predict or assume (logical) that if v > c then -t. But I have not taken any courses on relativity (although I want to), and all of my studies are straight from the NET. And the NET has some very innacurate and misleading statements therein.

You keep doing exactly what we tell you not to do. You are graphing a function that has a physical meaning, and then changing the function and trying to interpret the physical meaning of the new function. there's a reason the absolute value sign isn't in there.

IF
f(x) = \sqrt{|1-\frac{x^2}{k}|} AND x^2 > k

THEN f(x) would be negative. This is 100% correct. Thsi is as far as you can take your problem without starting to **** with relativity theories.

 

[eNathan]

You keep doing exactly what we tell you not to do. You are graphing a function that has a physical meaning, and then changing the function and trying to interpret the physical meaning of the new function. there's a reason the absolute value sign isn't in there.

IF
f(x) = \sqrt{|1-\frac{x^2}{k}|} AND x^2 > k

THEN f(x) would be negative. This is 100% correct. Thsi is as far as you can take your problem without starting to **** with relativity theories.

I see what you are saying. I am fully awear that I have changed the original meaning of the Lorenze Transformation, and therefore we cannot accept the new result as a physical meaning. I am just saying "it would be kewl if v > c was posible, and if it was it would resutl in -t. And if this was so perhaps this equation could predict it". I am in not way accepting my equation as something that actually exists. I just like playing with mathematics.

What is the meaning of f(x) = \sqrt{|1-\frac{x^2}{k}|} AND x^2 > k by the way? It looks like a 'take off' of the LT factor.

 

[whozum]

"it would be kewl if v > c was posible, and if it was it would resutl in -t. And if this was so perhaps this equation could predict it"

I agree.

What is the meaning of f(x) = \sqrt{|1-\frac{x^2}{k}|}
AND x^2 > k
by the way? It looks like a 'take off' of the LT factor.

It is the LT factor just as a regular function f(x).k = c^2, f(x) = \gamma, x = v

 

[dextercioby]

1.It's not Lorenze,it's HENDRIK ANTOON LORENTZ.
2

.\gamma=:\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}​

Daniel.
​

 

[whozum]

He was spelling things wrong left and right, I'm not a dictionary, I don't care.

 

[dextercioby]

U had a problem with the "gamma factor" yourself...

Daniel.

 

[whozum]

I copied the one he pasted thinking it was the original one, didnt see the absolute values. I am only human dex. Sure its inverted too.

We can't all fuse into a physics gurus.

 

[abia ubong]

hey zurtex i need more xplanations

 

[Zurtex]

hey zurtex i need more xplanations

I've given up on eNathan threads, if you look back I've already explained it all and where the confusion came from. What exactly if your issue? Perhaps you could start another thread with a problem you have.