How to define momentum of the Earth with the Moon?

[DimaMazin]

 

Same here.

 

 

Gamma in all formulas relating to relativity is always the same:

 

[latex]\gamma = \frac{1}{1-\frac{v^2}{c^2}}[/latex]

No. The same is equation for gamma.Speed defines gamma. Speed of moving of the system relative to foreign observer can be not the same as speed of motion of the Moon around the Earth. For definition of relativistic mass of the Moon you shouldn't use gamma which you should use for definition of momentum of all the system.

[pavelcherepan]

Hence in post #3 I said - what is your frame of reference! This is exactly what you're talking about!

 

 

For definition of relativistic mass of the Moon you shouldn't use gamma which you should use for definition of momentum of all the system.

 

And no. This is totally wrong. You can use Lorentz factor at any velocity and in any inertial FoR.

Edited May 31, 2015 by pavelcherepan

[DimaMazin]

Hence in post #3 I said - what is your frame of reference! This is exactly what you're talking about!

 

 

And no. This is totally wrong. You can use Lorentz factor at any velocity and in any inertial FoR.

For me it is simpler to define speed of system and mass of system than to define every speed of every object of the system relative to me and then sum of their momentums. I am not sure your method is correct .

[Strange]

I don't understand what you try to say. Here only answer of Janus is completely exact. He use relativistic mass, but his relativistic mass doesn't confirm relativistic mass of Strange in that post.

 

We were talking about different things. Janus was, quite rightly, pointing out that you could also take into account orbital and rotational energy. I was pointing out that you use of linear velocity and kinetic energy was unnecessary (but it is quite possible I misunderstood what you were trying to say).

 

 

Please stop doing that in serious discussions.

For me it is simpler to define speed of system and mass of system than to define every speed of every object of the system relative to me and then sum of their momentums.

 

Which is what I suggested in post #5; your original post did appear to try to include the speed of each object.

[DimaMazin]

 

. I was pointing out that you use of linear velocity and kinetic energy was unnecessary

 

Which is what I suggested in post #5; your original post did appear to try to include the speed of each object.

My words and equations in my question didn't make that. That is just your disinformation.