Gravitational acceleration

Hi Uncle Tom,

You gave two velocity equations:

1. v = [(1-2GM/rc^2)/(1+2GM/rc^2)^3] sqrt(2GM/r)
2. v = (1-2GM/rc^2) sqrt(2GM/r)

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
Naturally, that is merely the appearance when coordinate time is used. In proper time, things remain Newtonian-|Tom|-
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Do you mean that:
In case I stand at the start point of the neutron ( which is infinity from the neutron star ), I should use one of the formulas above to figure out velocity of the neutron ?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Cindy</i>
<br />Do you mean that: In case I stand at the start point of the neutron ( which is infinity from the neutron star ), I should use one of the formulas above to figure out velocity of the neutron?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Right. -|Tom|-

Hi Uncle Tom,

We just consider two cases:

Case 1: Standing on a surface of a neutron star:

v = sqrt(2GM/r)
Standing on the neutron star, I "see":
At the start point, the neutron move slow, then get faster and faster. The maximum velocity of the neutron is at the end of the trip and potential energy is converted completely into kinetic energy.

I feel good in this case, because it get along with the law of conservation of energy.

Case 2: Standing at infinity:

v = [(1-2GM/rc^2)/(1+2GM/rc^2)^3] sqrt(2GM/r)
v = (1-2GM/rc^2) sqrt(2GM/r)

Standing in infinity, I "see" that:
At first, the neutron move slow, then it get faster and faster until a point on the trip. After it pass this point, it move slow down. The velocity of the neutron at the end of the trip is not the maximum velocity.

The phenomenon seem imply that the potential energy of the neutron is not converted completely into kinetic ennergy ? How can I explain the phenomenon with the law of conservation of energy ?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Cindy</i>
<br />The phenomenon seem imply that the potential energy of the neutron is not converted completely into kinetic ennergy? How can I explain the phenomenon with the law of conservation of energy?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Energy must be measured in terms of proper time, where it is conserved. The view you describe uses coordinate time, which is the proper time for some other frame. In relativity, that does not conserve energy.

If you describe these things in terms of the Meta Model, they get much simpler because there is only one time for the universe and energy is conserved for all observers. Infalling bodies don't really slow down, but some of their potential energy gets dispersed into the elysium (light-carrying) medium, and it can take a long time for the slowed light from the impact to reach an outside observer. That makes the infalling body appear to move slower than it really does. -|Tom|-

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
Energy must be measured in terms of proper time, where it is conserved. The view you describe uses coordinate time, which is the proper time for some other frame. In relativity, that does not conserve energy.

|Tom|-
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Hi Uncle Tom,

I don't think that I can accept the concept of coordinate time. To me, it does not obey the law of conservation then simply it does not exist.

I prefer the MM theory, which have "one time for the universe and energy is conserved for all observers".