One disproof of EP?

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21 years 11 months ago #3775 by makis
Reply from was created by makis
Cindy

Are you trying to disprove a principle? You cannot do that within the context of its own conclusions, which are the very same laws you are using.

Second, and more important

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>

Let's consider m_i
By definition m_i = F/a =dP/adt
Where F is a force; a is acceleration of m.

<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

You have a very wrong definition of force.

F = d(mv)/dt and not F = mdv/dt

More importantly:

Did your analysis (although wrong) prove any variation of Mi/Mg from unity? Absolutely none.

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21 years 11 months ago #3522 by Cindy
Replied by Cindy on topic Reply from
Hi Makis,
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote> You cannot do that within the context of its own conclusions, which are the very same laws you are using.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
I am using SR to disprove EP. They are not the same laws.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
You have a very wrong definition of force.

F = d(mv)/dt and not F = mdv/dt

<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
In my analysis, I define F= dP/dt. Where P is relativitis momentum, P = gamma.m.v
Thus F = d(m.v.gamma)/dt
If the force is always perpendicular to the velocity v, you could show that F= gamma.m.a
If the force is always parallel to the velocity v, you have F = (gamma^3)m.a
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>More importantly:

Did your analysis (although wrong) prove any variation of Mi/Mg from unity?
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Yes, I think so.
For somebody who believe that gravitational force = GMm/R^2,
they have m_g = m
For somebody who beleive in SR or LR, gravitational force = GMm.gamma/R^2,
they have m_g = gamma.m
Anyway, <b>they have only one value of m_g </b>
While m_i is different, <b>m_i may have any values,</b> which are between gamma.m and (gamma^3).m, depent on angle between force and velocity.

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21 years 11 months ago #3523 by makis
Replied by makis on topic Reply from
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>

For somebody who believe that gravitational force = GMm/R^2,
they have m_g = m
For somebody who beleive in SR or LR, gravitational force = GMm.gamma/R^2,
they have m_g = gamma.m

<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

You are not using proper coordinates,time and definition for rest mass. You must use a Lorentz transformation. When you do that properly (usually o sophomore level problem in Physics) you will see that there is not any issue of EP being violated. -

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21 years 11 months ago #3524 by MarkVitrone
Replied by MarkVitrone on topic Reply from Mark Vitrone
Are you going to attempt Einstein GR/SR or Lorentz relativity? Don't mix apples and oranges.

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21 years 11 months ago #3525 by Cindy
Replied by Cindy on topic Reply from
Hi Makis,
I am trying to show that:

1/ For m_g
By definition, m_g = gravitational force / g
where g = GM/R^2,
Therefore, no matter you are using SR, or LR, or not,(but not both), you always have <b>only one value</b> of m_g for each speed of m. Direction of v doesn't matter.

2/ For m_i
By definition, m_i = F/a
If you apply SR, you will get <b>more than one value</b> of m_i for each speed of m. Values of m_i vary form gamma.m to (gamma^3)m, depent on direction of v.

If you agree to my both issues, I can claim that m_g differ from m_i in value.
If you don't agree to either issue, please show me in detail your calculation. Thank you so much,

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21 years 11 months ago #3967 by makis
Replied by makis on topic Reply from
There is a third alternative. I give you a hint and you work on it:

F = d(mv)/dt

m = m0 x gamma

m0= rest mass

Then:

F = m0 x v x d(gamma)/dt + m0 x gamma x dv/dt



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