Kopeikin and "the speed of gravity"

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Magoo]: So let me see if I have this straight. The acceleration of light is different from the speed of light and the acceleration could be faster than the velocity?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

No, you don't have it straight. Acceleration cannot be "faster" because it isn't speed, nor is it anything for which the adjective "faster" has meaning. Acceleration can be bigger or smaller. But an acceleration <i>cannot</i> be compared to a speed any more than a centimeter can be compared to a gram.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>I'm not really sure how it is that acceleration is not a speed since it is derived from a speed.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

E = 1/2 m v^2 is kinetic energy. It is also derived from a speed. Does that mean that energy must be a speed too? [I really hope you are not going to say "yes" to that question!] For purposes of this discussion, how is acceleration's relation to speed any different than energy's relation to speed? If you say "because it also depends on mass", then I'll just ask about energy per unit mass "e", which does not: e = 1/2 v^2. Besides, acceleration is a combination too, of speed change (not speed itself) and time interval.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>: what are your calculations of:
1. The SPEED of gravity? (I think you say 20 billion c)
2. The ACCELERATION of gravity? (??)<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

1. SOG > 20 billion c
2. g = 10 m/s^2 at Earth's surface, and other values between zero and infinity for other locations near other masses.

There is no way to compare one with the other because the dimensions are incompatible (like centimeters and grams). For question #2, I might have said g = 10 trillion picometers/s^2, which is also the acceleration of gravity at the Earth's surface. Does that make it "faster than the speed of gravity" because I wrote a bigger number in front of the units? Common sense should tell you "no". -|Tom|-

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

For a constant "v", if the interval "t" is very small, the acceleration "a" can be very large -- without limit. -|Tom|-

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

That's not necessarily true as far as I know. acceleration is defined as lim (deltaV/deltat) as deltat goes to zero. Not lim (V/deltat) as deltat goes to zero. in your example, as deltat goes to zero, deltaV is zero and a is 0. if v is not constant but continuous and differentiable, deltaV approaches zero also and the ratio goes always to a finite acceleration.

If you have a pulse acceleration of v/deltat that would mean an impusle function or a Dirac delta function in the limit as deltat goes to zero... but as deltat approaches zero, the Dirac delta function is

delta(t-t1) = V(t)xlim delta(t-t1) as t goes to zero and the property is

+inf
/
|
| v(t) x delta(t-t1)deltat = v(t1) (property of Dirac)
|
/
-inf

acceleration is then defined as dv(t)/dt. In order for the derivative of a function v(t) to be infinite at time t, the function must have a singularity at time t (the slope of V must be infinite...thus, no infinite accelerations are possible...hope so...

in special relativity it can be shown that as v approaches c the apparent acceleration is less than the intrinsic one (a'=c/t)...

Mark

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>acceleration is defined as lim (deltaV/deltat) as deltat goes to zero. Not lim (V/deltat) as deltat goes to zero. in your example, as deltat goes to zero, deltaV is zero and a is 0.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

If you read the original context carefully, my words "constant V" did not mean constant speed. It meant that speed could increase from zero to any given limiting V (< c, of course) in a long time with a small acceleration, or in a short time with a large acceleration. "Constant V" meant holding the limiting velocity constant, not "acceleration = 0" constant.

So in that context, your last sentence does not apply to my example. Suppose we are in an interval where acceleration is constant (however brief that interval might be). We can then drop the limit and simply write a = (deltaV/deltat). Clearly, a can be any constant between 0 and infinity and still keep that relationship.

This has nothing to do with SR or approaching the speed c. Acceleration can become infinite and still produce only a small speed, as in my earlier example of passing through a singularity. Nothing in relativity prevents that -- which gets back to my original point, far above in this topic. Even though speed has limits, acceleration (the derivative of speed) does not. -|Tom|-

Dr. VanFlandern, how did you come up with your answer for the ACCELERATION of the gravity? Since all objects travel towards earth, within the atmosphere, at the same "CONSTANT" rate then how can there even be an acceleration of gravity?

It seems as though you are using a different definition of acceleration then I am. I keep using velocity/time=acceleration, the "UNITS" are [velocity] and [time]. I'm having trouble understanding where the units are different. I understand that there is a diffence between acceleration and velocity in that "change" is present in acceleration but that does not change the type of units being used. The way I understand acceleration is that acceleration is simply the "change" (plus or minus) in the velocity over time. And again, the velocity UNIT being used could not surpass its maximum speed limit. Back to your example: (given- the maximum speed is 60mph) Yes, your car could accelerate(an increase in velocity) quickly but at no time interval could the car be traveling faster then the max of 60mph.

Could you please tell me the definition you are using for acceleration? Also, is my definition wrong? If so, how?

Thanks!

p.s. Please remember that I'm a Dummy so please give me an easy to understand (uncomplicated) answer.

Just learning!
Magoo

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote> Then why some people are trying to prove that c_g=c ?
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Quoting myself<img src=icon_smile.gif border=0 align=middle>
But I would like to understand why those evidences are not accepted in the scientific community.
Someone have an explaination?

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[jacques]: But I would like to understand why those evidences are not accepted in the scientific community. Someone have an explaination?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

That gets into human psychology and special interests. No good reason in physics for rejection of the experiments is still standing following our <i>Foundations of Physics</i> paper last summer. -|Tom|-