In twin frame gamma=dt/dt' and in twin' frame gamma=dt'/dt

It all says that gamma isn't a coefficient of relative slowing of time , gamma is here a coefficient of non-simultaneity.

Gravitational energy creates simultaneity.

Kinetic energy creates non-simultaneity.

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

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.

 

This is just nonsense.

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.

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.

 

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.

Edited October 18, 201411 yr by DimaMazin

No, gamma is a dimensionless number:

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

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 October 18, 201411 yr by DimaMazin

v/c and c/c are both dimensionless. Obviously.

 

What do you mean by "special meter/s"?

v/c and c/c are both dimensionless. Obviously.

 

What do you mean by "special meter/s"?

1special meter=299792458 m

speed should have units

Edited October 18, 201411 yr by DimaMazin

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.

 

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 October 18, 201411 yr by DimaMazin

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

 

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.

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 October 18, 201411 yr by andrewcellini

 

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 October 18, 201411 yr by DimaMazin

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?

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.

 

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 October 24, 201411 yr by DimaMazin

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

dt'=gamma_*vm_*dx / (mc²+GMm/r)

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.

When fast particle travels in black hole gravitation then you can't have the same relative simultaneity with the particle.But the particle travels therefore still it has relative simultaneity.