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Light speed using circumference


3118beach

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36 minutes ago, studiot said:

Yes in the linear case there need be no acceleration.

In the rotating case you cant do without it.

Angular+Momentum,+cont.jpg

 

No, the Radius term is just missing in pC being 1 for E = MC2 + pC

Difference-Between-Linear-Momentum-and-A

AngularMomentumInCM.png

Which would be the radius from the observer frame R or l, which is 1 in linear but in angular is not always 1.

length%20contraction%201.PNG

Which does require an acceleration to do, but is not the reason why.

Edited by Vmedvil
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1 hour ago, Vmedvil said:

Is it really different besides being over a radius?

Yes, it is really different. Were you to actually try and apply the concepts in solving physics problems, it would become obvious.

48 minutes ago, Vmedvil said:

E = MC2 + pC

That's not a valid equation.

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7 minutes ago, swansont said:

Yes, it is really different. Were you to actually try and apply the concepts in solving physics problems, it would become obvious.

I'll take that bet calculate for Linear Momentum  E1 = MC + PC , where R = 1, then do the same for angular momentum and see if it is different when E2 = MC2 + PC  , P = RL , L = Iω , I = MR2  you will find E1 = E2

Edited by Vmedvil
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3 minutes ago, Vmedvil said:

I'll take that bet calculate for Linear Momentum  E1 = MC + PC , where R = 1, then do the same for angular momentum and see if it is different when E2 = MC2 + PC  , P = RL , you will find E1 = E2

E = MC + PC is not a valid equation (i.e. it does not hold true in general), and of what value would an equation be that's only valid for R=1?

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4 minutes ago, swansont said:

E = MC + PC is not a valid equation (i.e. it does not hold true in general), and of what value would an equation be that's only valid for R=1?

I just rooted the entire equation to not have to type all the exponents E2  = M2C4 + p2C either way.

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35 minutes ago, swansont said:

Then you did it incorrectly.  (MC+ PC)2 ≠ M2C4 + P2C2

i.e. there is a missing cross-term of 2MPC2

Damn it, I don't care which way you do it, it will be the same as long as R = 1 on Angular, but yes I typed that quickly but it does not  (MC+ PC)2 ≠ M2C4 + P2C2

Edited by Vmedvil
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13 minutes ago, studiot said:

headinclouds.jpg.e8936dfac125c9ccdea86f5bd2d9906f.jpg

 

:)

It happens when addressing multiple threads at once constantly hearing PING PING PING PING, Medvil does severe multiforuming and tasking, but yes use E2M2C4 + P2C2    or your answer will be wrong, but it would not matter as both sides were by the same amount wrong so it would be right for the E1 = E2 scenario either way.

Edited by Vmedvil
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1 hour ago, Vmedvil said:

It happens when addressing multiple threads at once constantly hearing PING PING PING PING, Medvil does severe multiforuming and tasking, but yes use E2M2C4 + P2C2    or your answer will be wrong, but it would not matter as both sides were by the same amount wrong so it would be right for the E1 = E2 scenario either way.

It will actually matter a great deal. 

Is R 1 meter, or 1 cm, or 1 foot? (which only begins to illuminate why the claim doesn't work)

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6 minutes ago, swansont said:

It's not obvious. If it only works in one unit system then it's a happy accident, and not something that's universally true.

Well, for that E2  = M2C4 + p2C2  

It only works at one point but for other kinematics equations universally true. 

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