Derivations

[gluon]

E^2-(pc)^2=(Mc^2)^2 where the expression (Mc^2)^2 is by definition, the squared mathematical precision of an ''invariant mass'', hence, Mc^4.

\rightarrow Mv(\frac{E}{M})=Mc^2(v)

allow v=c then this simplifies to Mv^3=Mc^3 (Just to show that these are relativistic equivalances without the need of gamma function. This now leads me to calculate:

Mv(E)(\frac{D}{v})=Mc^2.v

The Second Part

my equation, albiet as simple as it is, will show its importance throughout the metric work:

[1] M(1+M)=2M if

[2] -(\frac{E}{c})^2+mv^2=p^2

then combine by division of [1] and [2] equations, allowing the relativistic proof:

\frac{\eta^{\mu v}p_{\mu}p_{\mu}}{2m}=\frac{p^2}{2m}=\frac{p^2}{2} \frac{E}{c^2}

which then follows

p^2=\eta^{\mu v}p_{mu}p_{v}=-(\frac{E}{c^2})^2+p^2

I need to go the now, i will finish this later, but sooner than later

[AlphaNumeric]

If v=c then M=0. And the second line of your post requires p=0, which is not true for v=c. You don't define D either. M(1+M) = 2M is only true if M=1. And the second equation, labelled [2] is not true at all. Then your equation p^{2} = p^{\mu}p_{\mu} is sloppy because you use the same symbol for 3 and 4 momenta. Also, all you've done is end up with the equation you assumed at the beginning. All in all you've gone round in a circle, along the way using incorrect equations. It's a case of 8 wrongs seem to have made a right.

[gluon]

Strange, because i posted this to a physicist who said it was mathematically correct, but he asked what derivation this was going to lead to.

Nevertheless, the p=o is an assumption that gamma function is being used in a differential format... that's all. Remember, this derivation is not finished, and there is a reason for the circular reasoning, if one gives me a chance to finalize this work. You will see, hopefully where this will end, because i have written the work down already, so a little patience begs.

[AlphaNumeric]

Strange, because i posted this to a physicist who said it was mathematically correct, but he asked what derivation this was going to lead to. Was he real or imaginary? And if he's real is he older than 12? And if he's real and older than 12 is he you?

Given anyone can see M(1+M) = 2M needs M=1 or 0, I find it hard to believe someone well versed in relativity thought your post was worth anything. I've also noticed further mistakes because you mix up the 4-momentum and 3-momentum more. You also fail to realise that the square of the 4 momentum is the mass suqared, so

\frac{p^{a}p_{a}}{2m} = \frac{-m}{2}. The expression \frac{p^{2}}{2m} which appears in things like the Schrodinger equation is in terms of the 3-momentum.
Nevertheless, the p=o is an assumption that gamma function is being used in a differential format... that's all.. No, it's the same as v=0 and so \gamma=1. No momentum means not moving which means no relativistic dilation.
Remember, this derivation is not finished, and there is a reason for the circular reasoning, if one gives me a chance to finalize this work. You will see, hopefully where this will end, because i have written the work down already, so a little patience begs. Doesn't matter if you pumped out 200 pages of complicated algebra, graphs, functional plots and topped it off with a supposed proof of the non-existence of God, if you start with a wrong equation you'll completely invalidate all of the rest of the work.

A well known physicist once told me about a PhD he'd been the external examiner on. Upon reading the introduction he thought "That's weird" and flicked to the conclusions and realised the thesis attempted to prove something he'd proven impossible by a No Go Theorem years ago. He worked backwards until he found the error, in the opening chapter. One of the central equations was wrong and somehow the person hadn't noticed. For 3 years. Despite the huge effort put into the work and all the work which derived from that central equation was right, because the equation itself was wrong, the thesis was worthless. Now while you don't put that much effort into your work, you do squander vast quantities of time on playing about with the wrong equations. Wrong is wrong, no matter how you try to dress it up or how much BS you try to spin.

[gluon]

Actually, it was a mathematicain from ''physics forum- homework help''

[gluon]

Oh, and the post is interesting, if you give me a bloody chance to finish thne work.

[gluon]

No, it's the same as v=0 and so . No momentum means not moving which means no relativistic dilation.


And, what? YUou just stated the same statement i made, essentially.

[AlphaNumeric]

And, what? YUou just stated the same statement i made, essentially.No, mine was coherent.

[gluon]

What, coherently-nonesense? I'll proove it, just stop fucking my mind.

[gluon]

E^2-(pc)^2=(Mc^2)^2 where the expression (Mc^2)^2 is by definition, the squared mathematical precision of an ''invariant mass'', hence, Mc^4.

\rightarrow Mv(\frac{E}{M})=Mc^2(v)

allow v=c then this simplifies to Mv^3=Mc^3 (Just to show that these are relativistic equivalances without the need of gamma function. This now leads me to calculate:

Mv(E)(\frac{D}{v})=Mc^2.v

The Second Part

my equation, albiet as simple as it is, will show its importance throughout the metric work:

[1] M(1+M)=2M if

[2] -(\frac{E}{c})^2+mv^2=p^2

then combine by division of [1] and [2] equations, allowing the relativistic proof:

\frac{\eta^{\mu v}p_{\mu}p_{\mu}}{2m}=\frac{p^2}{2m}=\frac{p^2}{2} \frac{E}{c^2}

which then follows

p^2=\eta^{\mu v}p_{mu}p_{v}=-(\frac{E}{c^2})^2+p^2

I need to go the now, i will finish this later, but sooner than later

Now the continuation:

define a 4-momentum vector p^{\mu}=E,P where energy has a 4-momentum and P has a 3-momentum. Remove the metric signature, we then have (-1,1,1,1) WHICH gives us a flat spacetime. The proof is as following:

-M^2=\eta^{\mu v}p_{mu}p_{v} where we replace \eta with g we then have a relativistic curvature. This is defined and proved by:

-p^0p^0+|p|^2, and by rearranging, we can have:

E^2=M^2+|p|^2

if M^2=(\frac{E}{c^2}), then we find E^2=\frac{E^2}{c^2}+|p|^2, which leads to my conclusion that

|p|^2-E^2=(\frac{E^2}{c^2}

then the new equation is found:

|p|^2=\frac{E^2}{c^2}+E^2

We should know this, since the only reason why relativity brings this into mathematical consideration, is because the ''difference'' between the positive energy and the negtaive matter constitutes a relation between the squared values differently.

Thus a usual metric changes a signature if

-M^2=+E^2-(p^1)^2+p^2_{\bot}

[prometheus]

E^2=\frac{E^2}{c^2}+|p|^2

I'm sure it isn't the only error, but this equation is wrong because on the LHS you have Kg^2 and in the right you have Kg

[gluon]

I nevere put it in by mitsake.

Now, you're extremely big-headed young-one. You are sure this isn't the only error? What?

You are the PHD student, surely you can tell if there is?

[prometheus]

Does it matter how many there are? Surely 1 is enough?

[D H]

I'm sure it isn't the only error, but this equation is wrong because on the LHS you have Kg^2 and in the right you have Kg
Even worse, the two terms on the right are dimensionally inconsistent.

[gluon]

Plus, if you used a little logic, there cannot be any errors, minus my small mistake of not adding that dimension.... alphenumeric has stfu over this thread, knowing fine well i can do this sort of stuff.

[prometheus]

Even worse, the two terms on the right are dimensionally inconsistent.

Damn! I quoted the wrong wrong equation!

I meant M^2 = \frac{E}{c^2}

[gluon]

DH, latex is not my speciality. Either cooperate, or continue to be an arsehole. Simple.

[prometheus]

Plus, if you used a little logic, there cannot be any errors, minus my small mistake of not adding that dimension.... alphenumeric has stfu over this thread, knowing fine well i can do this sort of stuff.

being wrong in the maths invalidates any result you get. Your claim is meaningless.

[gluon]

Remedial inccantations from someone who tried to make a point in the beginning...



what was that again... oh yes... trying to mkae my math look bad.

[James R]

gluon/Reiku:

You have previously been banned for spamming the Physics forum with crap.

You are warned that you will be banned again if you continue to pollute the Physics forum with your incoherent nonsense mathematics.

[gluon]

How are these derivations wrong?????????

[gluon]

Also, answer this if you willl.

How come alphanumeric found no errors in the math initially, so you have basically chosen to take his side, over the fact that it was right anyway, the derivations, that is.

[James R]

How are these derivations wrong?????????

They are meaningless nonsense. There's no physics in them - just the appearance of physics borrowed from cut-and-paste website fragments and/or textbook fragments. There's no comprehension on your part about what any of your source equations mean. There's no logical progression from one step to the next. It's like putting a bunch of equations in a blender and jumbling them about at random.

[gluon]

no physics in them.... are you being serious???????????

[James R]

Totally serious.

[gluon]

Then you know not of the expressions, equations or even elements behind them. YUou have made yourself a total fool.

[James R]

Meh.

[gluon]

Then you know not of the expressions, equations or even elements behind them. YUou have made yourself a total fool.

unto which, a ''meh'' was applied. Tell me, sir, you are qujite immature for yoour age, which is almost synonymous with mine.

[AlphaNumeric]

Plus, if you used a little logic, there cannot be any errors, minus my small mistake of not adding that dimension.... alphenumeric has stfu over this thread, knowing fine well i can do this sort of stuff.Actually I have been at work all day. At the time you posted what I've just quoted I was teaching. I got home about 20 minutes ago.

I 'STFU' because I have other things to do and I am confident that people like DH or Guest or Ben can and will mop the floor with you if you put your foot in it. Which you did. Again.

/edit

And I found flaws in the maths immediately. I said so. I pointed them out. I explained why. You retorted none of them.