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Is gamma factor a coefficient of non-simultaneity?


DimaMazin

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  • 3 months later...

It's part of it, but not the whole thing. If two events are separated in space by [math]\Delta x[/math] and happen simultaneously, then in a reference frame moving at [math]v[/math] there is a time [math]\Delta t'= \frac{\gamma v}{c^2} \Delta x[/math] between them.

 

 

Gravitational energy creates simultaneity.

Kinetic energy creates non-simultaneity.

Interaction of gravitational energy and kinetic energy creates different time in simultaneity. :P

 

This is just nonsense.

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  • 2 weeks later...

It's part of it, but not the whole thing. If two events are separated in space by [math]\Delta x[/math] and happen simultaneously, then in a reference frame moving at [math]v[/math] there is a time [math]\Delta t'= \frac{\gamma v}{c^2} \Delta x[/math] between them.

 

Thanks Elfmotat.

If v/c is speed and unit of distance is 299792458 m then gamma is quantity of non-simultaneity(in seconds) per the unit of distance.

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Thanks Elfmotat.

If v/c is speed and unit of distance is 299792458 m then gamma is quantity of non-simultaneity(in seconds) per the unit of distance.

 

No, because [math]\frac{v}{c^2} \Delta x[/math] has units of time (the same as [math]\Delta t'[/math]). So [math]\gamma[/math] is a dimensionless number.

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No, because [math]\frac{v}{c^2} \Delta x[/math] has units of time (the same as [math]\Delta t'[/math]). So [math]\gamma[/math] is a dimensionless number.

 

v - special meter/s

c - special meter/s

gamma - no unit

dx - special meter

 

But dt'=v/c*quantity of special meters * gamma

Then gamma is factor of non-simultaneity and has units which are second / special meter. :P

Edited by DimaMazin
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No, gamma is a dimensionless number:

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

Wrong also.

v/c - have units special meter/s

c/c - have units special meter/s

then gamma has units s/special meter ;)

1special meter=299792458 m

Edited by DimaMazin
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1special meter=299792458 m

 

1 light-second?

 

But even if you use that as the unit of speed, then v/c is still dimensionless because both v and c use the same units. And therefore gamma is dimensionless.

 

speed should have units

 

It does. Usually metres/second.

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1 light-second?

 

But even if you use that as the unit of speed, then v/c is still dimensionless because both v and c use the same units. And therefore gamma is dimensionless.

 

 

It does. Usually metres/second.

Then what is units of 0.5c? v/c=0.5 then v=0.5c

Edited by DimaMazin
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dima, the units should cancel leaving a dimensionless number. v and c must have the same units.

 

v^2/c^2 = (.5c)^2/c^2 = (.5*3e8 m/s)^2 /(3e8 m/s)^2 = 2.25e16 m^2/s^2 / 9e16 m^2/s^2 = ~.25

 

*sorry for lack of latex know how*

Edited by andrewcellini
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The units of 0.5c is still m/s.

 

v/c does not equal 0.5c. It might equal 0.5. Or v might equal 0.5c. But v/c can never equal 0.5c.

Yes,I have confused.Even when c=1 it still has units.

Edited by DimaMazin
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It's part of it, but not the whole thing. If two events are separated in space by [math]\Delta x[/math] and happen simultaneously, then in a reference frame moving at [math]v[/math] there is a time [math]\Delta t'= \frac{\gamma v}{c^2} \Delta x[/math] between them.

 

 

 

And so the formula shows :

non-simultaneity=momentum*dx/gravitational energy

Would you like to tell how do you define non-simultaneity of satellite or half-satellite?

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And so the formula shows :

non-simultaneity=momentum*dx/gravitational energy

Would you like to tell how do you define non-simultaneity of satellite or half-satellite?

 

The formula shows no such thing, and I have no idea what you're asking.

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The formula shows no such thing, and I have no idea what you're asking.

I see no problem with such thing in the formula. dt'=gamma*vm*dx/mc2

Does relative simultaneity exist between you and the Moon?

Does relative simultaneity exist between you and asteroid which is temporarily grasped by gravitation of the Earth?

What is formula do you use for definition of relative simultaneity in gravitational field when path of relative object is curved?

Edited by DimaMazin
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I see no problem with such thing in the formula. dt'=gamma*vm*dx/mc2

Does relative simultaneity exist between you and the Moon?

Does relative simultaneity exist between you and asteroid which is temporarily grasped by gravitation of the Earth?

What is formula do you use for definition of relative simultaneity in gravitational field when path of relative object is curved?

 

I think gravitation reduces relative simultaneity when gravitation M is in reference frame of observer:

dt'=gamma*vm*dx / (mc2+GMm/r)

 

You're making up nonsensical equations. The equation I posted applies exclusively to inertial observers with no gravitation. What you're posting has nothing to do with either special or general relativity.

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