Light speed using circumference

[Vmedvil]

36 minutes ago, studiot said:

Yes in the linear case there need be no acceleration.

In the rotating case you cant do without it.

 

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

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

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

Edited November 27, 2017 by Vmedvil

[swansont]

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.

[Vmedvil]

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 = MC²  + 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 November 27, 2017 by Vmedvil

[swansont]

3 minutes ago, Vmedvil said:

I'll take that bet calculate for Linear Momentum  E₁ = MC²  + 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 = MC²  + 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?

[Vmedvil]

4 minutes ago, swansont said:

E = MC²  + 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 + p²C²  either way.

Edited November 27, 2017 by Vmedvil

[swansont]

Just now, Vmedvil said:

I just rooted the E²  = M²C⁴ + P²C²

Then you did it incorrectly.  (MC² + PC)² ≠ M²C⁴ + P²C²

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

[Vmedvil]

35 minutes ago, swansont said:

Then you did it incorrectly.  (MC² + 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  (MC² + PC)² ≠ M²C⁴ + P²C²

Edited November 27, 2017 by Vmedvil

[studiot]

 

[Vmedvil]

13 minutes ago, studiot said:

 

It happens when addressing multiple threads at once constantly hearing PING PING PING PING, Medvil does severe multiforuming and tasking, but yes use E² = M²C⁴ + 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 November 27, 2017 by Vmedvil

[swansont]

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 E² = M²C⁴ + 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)

[Vmedvil]

2 minutes ago, swansont said:

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)

Meters obvious but just use the correct form.

[swansont]

2 minutes ago, Vmedvil said:

Meters obvious but just use the correct form.

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.

[Vmedvil]

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. 

[swansont]

2 hours ago, Vmedvil said:

Well, for that E²  = M²C⁴ + p²C²  

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

Prove it.

[Vmedvil]

20 minutes ago, swansont said:

Prove it.

MV = p , p = R x L, L = I ω  , I = MR²  , ω = V/R

Edited November 27, 2017 by Vmedvil

[swansont]

15 hours ago, Vmedvil said:

MV = p , p = R x L, L = I ω  , I = MR²  , ω = V/R

and?