Kopeikin and "the speed of gravity"

n/a3
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21 years 10 months ago #3104 by n/a3

Replied by n/a3 on topic Reply from

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>

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.

<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Tom Van Flandern...

0/0 and infinity/infinity are indefined ratios... in order then to have an infinite acceleration you must devide a finite change in velocity by a zero change in time...this means that the slope of the velocity function is infinite at that point...if you have a continuously differentiable velocity function with a limit defined at time t, what you are saying is impossible

a cannot take any value you desire... a can take only one value an that is a...this is because any attempt to decrease deltat results in a decrease in deltaV in such a way as to bound the ratio to the limit...

let me give you an example...free fall...

v=16t feet/sec

let v+dv=16(t+dt)=16t+16dt = v+16dt > dv=16dt > dv/dt =16

where do you see any choice other than 16?

the only way one could get infinite accelerations is to have a function of the form:

dv/dt = k/dt and threreafter by choice of k and dt making dv/dt take values from o to infinity

this type of function cannot exist in continuous 3-d euclidean space-time...energy conservation etc...

it can exist however in non-euclidean, non-standard space-time...let's not get into that here...

do you subscribe to a euclidean, isotropic and uniform space time continuoum? if yes, i expect you reformulate your statements accordingly...

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tvanflandern
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21 years 10 months ago #3931 by tvanflandern

Replied by tvanflandern on topic Reply from Tom Van Flandern

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>[Magoo]: 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?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

This might be just a trick of language. In ordinary physics, the "acceleration of gravity" means the acceleration of objects falling in a gravitational field. Near Earth's surface, that acceleration is a constant 10 meters per second per second. For example, if the object falls from a high tower and observes a tape measure as it falls, the tape reads "0" meters as the object is released, "5" after one second, "20" after 2 seconds, "45" after 3 seconds, "80" after 4 seconds, etc. Note that constant acceleration still produces rapidly increasing speed.

This is a whole different thing from what it now appears you meant to ask, which was about the acceleration of the speed of gravity, or the acceleration of the speed vg > 20 billion c. That is zero. Gravity itself has some average speed, and that speed does not change, so there is no acceleration in it.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>It seems as though you are using a different definition of acceleration then I am.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Not a different definition of acceleration. There is only one meaning for that. But we might not have been talking about the same thing that accelerates.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>I keep using velocity/time=acceleration, the "UNITS" are [velocity] and [time]. I'm having trouble understanding where the units are different.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

The correct definition is (change in speed) divided by (time interval for that change to occur). The numerator has the same units as speed. The denominator has units of time. So acceleration has units of length over time squared -- different from the units of speed.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>The way I understand acceleration is that acceleration is simply the "change" (plus or minus) in the velocity over time.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Language might be playing tricks here again, but this appears wrong. Acceleration is not the change in speed that happens over some time. It is the change in speed divided by the time interval, which is the rate of change of speed, not the change itself. Do you appreciate this difference? See my example above. The speed is constantly changing, but the acceleration (the rate of change of speed) is constant. In fact, the speeds at those same moments are:
t = 0 v = 0 m/s
t = 1 v = 10 m/s
t = 2 v = 20 m/s
t = 3 v = 30 m/s
t = 4 v = 40 m/s
The rate of change of speed is constant at (10 meters per second) per second. The speed is variable and could go up indefinitely with no change in the acceleration.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>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.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Agreed. But the acceleration could have been 60 billion mph per second if it lasted only for a nanosecond.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Please remember that I'm a Dummy so please give me an easy to understand (uncomplicated) answer.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

At some point, if you are not getting it from one teacher, try another, or pick up a physics book. (There are hundreds to choose from.)

In any case, might I request that follow up to this matter, and off-topic questions in general, go into a separate topic? I don't want to have so much extraneous material to wade through for people who already know this stuff and are just interested in Kopeikin and the speed of gravity. When anyone wants to discuss a new topic, they should feel free to start one. And thanks for that consideration. -|Tom|-

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rbibb
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21 years 10 months ago #4442 by rbibb

Replied by rbibb on topic Reply from Ron Bibb

Thank you tremendously Dr. VanFlandern.

If you wouldn't mind just a follow-up to this statement:
"Agreed. But the acceleration could have been 60 billion mph per second if it lasted only for a nanosecond."

Isn't 60 billion mph per second for 1 nanosecond(1 billionth of a second) the same as 60mph?
How is that speed any different from the maximum limit?

But if it is different and a snapshot was taken during that billionth of a second then wouldn't the speed show to be at 60 billion mph, it just can't be...

Oh well, I'll get back to the books. Thanks for all your help.

By the way, this board is AWESOME! (I almost want to add Dude, but I won't....oops <img src=icon_smile_blush.gif border=0 align=middle>)

Just learning!
Magoo

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tvanflandern
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21 years 10 months ago #3935 by tvanflandern

Replied by tvanflandern on topic Reply from Tom Van Flandern

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>[mark]: 0/0 and infinity/infinity are undefined ratios...<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Nonetheless, they can take on a specific value in the limit. For example, what is (5x/x) in the limit as x --> 0? The limit is undefined, but the approach to the limit defines a unique solution, 5.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>in order then to have an infinite acceleration you must divide a finite change in velocity by a zero change in time...<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Not necessarily. In dynamics, the most common place infinite accelerations arise is at singularities such as point masses. Then Newton's acceleration law, a = GM/r^2, gives (a = infinity) at (r = 0). Mathematically, the motion of a point particle in the gravitational field of the point mass is perfectly well-defined even if the singularity is a point along the particle's orbital path.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>this means that the slope of the velocity function is infinite at that point...if you have a continuously differentiable velocity function with a limit defined at time t, what you are saying is impossible<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Look at v = a*t. If we apply an infinite impulse at an instant, we have the product of zero and infinity on the right, which as you surely know well is indeterminate. (You seem to be familiar with the math of infinities and indeterminates.) So once again, we must look at what happens in the limit to break the indeterminacy. (I haven't supplied enough information in this example to do that.)

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>"a" cannot take any value you desire... "a" can take only one value and that is "a"...this is because any attempt to decrease deltat results in a decrease in deltaV in such a way as to bound the ratio to the limit...<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

This is just not true. Acceleration can in fact be anything, which BTW is not my opinion, but standard physics and dynamics. (The field of studying orbits that can encounter or approach singularities is called "regularization".)

Given that acceleration is unconstrained, why do you say that deltaV must decrease if deltat does? That's the part that is false. It could be anything.

I hope we are not arguing over whether accelerations could become infinite in physical reality. Of course they cannot. Nothing finite can ever become infinite. But in math there is no problem handling real infinities. And in physics, infinity is a convenient shorthand for "indefinitely large" or "so large that no bound can be set". The speed of gravity is a good example. I would never maintain that it was infinite. That's impossible. But it is too large to measure yet.

However, in many areas of physics, including dynamics, the concept of "impulses" is handy and useful. An impulse is an instantaneous velocity change. It therefore requires an infinite acceleration by definition. If gravity is Le Sagian in nature, then large bodies appear to accelerate smoothly. But at the graviton level, their motion is being changed suddenly in tiny increments by each graviton impact. We model these as impulses because we know nothing about the rigidity of gravitons.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>let me give you an example...free fall...
v=16t feet/sec
let v+dv=16(t+dt)=16t+16dt = v+16dt > dv=16dt > dv/dt =16
where do you see any choice other than 16?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Your initial assumption, v = 16t ft/s, has a velocity on the left side and the product of a velocity and a time on the right side. But velocity times time is distance. So you have equated a velocity to a distance. <img src=icon_smile_dissapprove.gif border=0 align=middle> I don't think you meant that. You probably meant the 16 to be an acceleration, such as 16 ft/sec/sec. Is that a good guess? At least, it would then be consistent because acceleration times time is velocity.

If my assumption is right, you have a velocity increasing linearly with time. That means, by definition, you assume a constant acceleration of 16 ft/s^2. Then in your last line, everything is a velocity until you get to the last inequality, where you switch units back to acceleration. Things in inconsistent units cannot be compared in an inequality. (You can't say 10 centimeters is greater than 5 grams, and you can't say 10 m/s > 5 m/s^2.)

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>do you subscribe to a euclidean, isotropic and uniform space time continuum? if yes, i expect you reformulate your statements accordingly...<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

I've reformulated as far as I dare go, because it's your example, and I don't know where you are headed with it. With the corrections just given, perhaps you can find your own answers. Singularities and impulses cannot exist in physical reality; but they are useful approximations of reality. In that sense, it's worth noting that the math is all consistent. If you are merely saying that acceleration cannot be infinite in reality, well, then we've been talking past one another. -|Tom|-

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n/a3
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21 years 10 months ago #4029 by n/a3

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good day to all of you! it's a cold Sunday morning..a perfect opportunity to saty inside and do physics 101 and calculus 101...

Tom Van Flandern...

obviously, I meant 32 in place of 16, just a typo...

let me give you an example...free fall...

v=32t feet/sec
let v+dv=32(t+dt)=32t+32dt = v+32dt > dv=32dt > dv/dt =32 feet/s^2
where do you see any choice other than 32 (feet/s^2)?

i always put the units next to the function...in order to avoid any confusion arising from such basic matters...

your original claim was that acceleration can get to infinity while speed is finite...that's a wrong claim to make...

Let us resolve this basic issue here and now, otherwise...we do not qualify for getting into more advanced concepts like the "speed of gravity"...

proposition 1: If a particle moves at a constant acceleration a, then it velocity is finite for all t and the acceleration is a, at every infinitesimal interval ds

proof: infinite instantenuous velocity at infinitesimal interval ds would imply that ds/dt = inf = v ---> dv/dt = inf in the same interval (inf-inf = inf), then a is infinite which contradicts the premises...

if acceleration can have any other value than a, say a' in an interval ds' in dt' then dV'/dt' = a' ----> v' = a't + v0 (1)

but it is also true that v = at + v0 (2)

from (1) and (2) ---> v-v' = (a-a')t (')

let v-v' = dv" and dt-dt' = dt" then it should be also true that:

v-v" = (a-a")t ('')continue as long as you like you will find that

v-v'''''' = (a-a'''''')t ('''''')

then add all the ('''''') equations to get:

NxV - (v'+v"+....+....) = Nxat - (a'+a''+...+...)

divide by N

V - (v'+v"+....+.....)/N = at - (a'+a"+...+....)/N

now take the limit of both sides as N ----> inf. the result is

v=at then dv/dt = a QED.

proposition 2: if a particle has a finite continuous velocity v(t) for all t, then it's acceleration must also be finite for all t

proof: if a = inf then dv/dt = inf then v = inf x t then v(t)is not continuous at t which violates premises. QED.

now, after having settled with the basics...

proposition 3: infinite accelerations are impossible if speeds are finite.

proof: if instantenuous acceleration could get infinite while instantenuous speed is finite then dv/dt = inf ----> dt = 0 which violates the definition of speed being the limit as dt goes to zero and NOT as dt = O... as dt goes to zero dv also goes to zero and the ratio dv/dt approaches asymptotically a. if the limit is not applied, then the only way to get infinite acceleration is have a finite velocity dv at inteval time dt=0. But if dt = 0 then dv = 0 and dv/dt = 0/0. QED.

in special relativity, Einstein did not explicitely deal with acceleration...many believe he did not do so because he thought it is obvious that when a particle reaches a constant speed of c, its apparent acceleration must drop to zero. if it wouldn't, then it would continue to move. relativistic momentum approaches infinity as the particle gets close to the c and that's because of the gamma factor...but apparent acceleration keeps decreasing asymptotically to zero...

concusion: infinite accelerations are only possible at infinite speeds... but infinite speeds do not require infinite accelerations... of course, accelerations can get very large for finite speeds but NEVER infinite...

good day

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tvanflandern
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21 years 10 months ago #3854 by tvanflandern

Replied by tvanflandern on topic Reply from Tom Van Flandern

Mark, I don't see that we are making any progress here. You seem in denial of what I was trying to teach you of standard dynamics (not MM things), which is your right. There's no "final exam" in this course. <img src=icon_smile_wink.gif border=0 align=middle>

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>v=32t feet/sec<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

This still has inconsistent units. Assuming t in seconds (which you did not specify), acceleration = 32 feet/sec^2. Because t carries its own units (it is not a pure number, but a number factored by a unit), everything multiplying t is the acceleration. But your 32 feet/sec is in velocity units, not acceleration units. I gave you the obvious correction before.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>let v+dv=32(t+dt)=32t+32dt = v+32dt > dv=32dt > dv/dt =32 feet/s^2<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

I objected to this, but you repeated it. I took ">" to have its normal meaning in an euation, "greater than". I'm now guessing you meant it simply to be an arrow. Is that right?

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>where do you see any choice other than 32 (feet/s^2)?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

You started with the assumption of constant acceleration = 32 ft/sec^2 in your first line. Why would you expect some inconsistency?

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>proposition 1: If a particle moves at a constant acceleration a, then it velocity is finite for all t and the acceleration is a, at every infinitesimal interval ds<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

That is obvious and needs no proof, provided t is bounded.

I explained to you how zero*infinity can be finite in the limit. You ignored that. Your argument assumes acceleration is a smooth, differentiable function, which is an opposite assumption to the one I made. Impulses are only instantaneously infinite, then continuously zero.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>proposition 2: if a particle has a finite continuous velocity v(t) for all t, then it's acceleration must also be finite for all t<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

That violates the premise of an impulse. It is also false for singularities, whereat velocity can be infinite at a single instant without creating a discontinuity in the path of the object.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>proposition 3: infinite accelerations are impossible if speeds are finite.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Directly violates the mathematical meaning of "impulse". Look it up.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>in special relativity...<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

We weren't dealing with SR here.

I'm not sure it is productive to continue because we are differing about premises, not steps in the argument. But if you do, I have a REQUEST:

Please start a new topic. This is a side issue having little to do with Kopeikin and the "speed of gravity", but only with whether or not the field of standard dynamics is on sound footing. It should be resolved in a separate topic, and any conclusions drawn can be brought back here when appropriate. People interested in your topic and this thread may often not be the same.

Let's all avoid topic drift. -|Tom|-

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