8
   

A speed beyond light question ^^

 
 
High Seas
 
  1  
Reply Sat 3 Jan, 2009 12:55 pm
@Brandon9000,
And you, Brandon, are demonstrating a terminal lack of capacity to understand simple mathematics, not to mention basic English.

Never once did I say anything other than I'm a mathematician. You are evidently not one, nor have you made the least effort to work your way through the Lorentz transformations - which have many applications in physics, as even we mathematicians have heard. That you have no clue on Molniya orbits is evident by your mention of "circular orbit" (sic!!) when there's no such thing......

The only reason Newtonian gravity came up in this thread - since you can't read any better that you can count - was your challenge to Gunga to explain how high-school physics might account for his statement on what would happen to planets in our system if our sun collapsed inwards. Gunga was right, and you were wrong; attempts to explain to you why that is have evidently failed so far, and your attempted digression into satellite orbits has been shown up for what it is, a blind man's attempt at a "hail Mary" pass.

Re-read the thread, read the Feynman book on gravity - you haven't done that - and then come back to discuss them, if you wish. Good bye till then.
Brandon9000
 
  1  
Reply Sun 4 Jan, 2009 01:23 am
You asked me to "sober up enought to learn basic Newtonian gravity" and asked me to "start showing at least some grasp of basic concepts."

High Seas wrote:
Read the damn book, Brandon, exact reference was posted above. Sober up enough to learn basic Newtonian gravity (infinite GRAVITY propagation speed MUST apply IF angular momentum is to be conserved, and the conservation of angular momentum wasn't doubted by Newton!) first, though.

I really don't see where you add value to this discussion if you can't even grasp complete BASICS. Sorry I'm not going to waste any more time until you start showing at least SOME grasp of basic concepts, never mind relativistic gravity field equations.


I acceded to your wishes, and asked you to demonstrate the basic familiarity you were asking me to show. I posed the following high school physics problem in Newtonian mechanics and gravity to you.

Brandon9000 wrote:
Although Newtonian mechanics has been superseded by relativistic mechanics, and, therefore, is not relevant here, I am willing to respond to this post of yours. Neither of us can add value unless he understands the basics. With two physics degrees, I think I do. Do you? I'll ask you an elementary question about gravity and circular motion in Newtonian mechanics, which anyone who completed a high school physics class with good grades could work out in 10 or 15 minutes. If you can answer, we can continue the discussion. If not, I'll give the solution in detail, and you can stop talking about physics.

Your post was number 3519687. I'll use that number. A satellite has a circular orbit around the Earth at a height of 3519 miles. What is its speed and kinetic energy? Give your units, and show your work.

and

Brandon9000 wrote:
For the kinetic energy question, you can assume that the satellite weighs 10 pounds, or else keep your equations in symbolic form.

Certainly a great expert on Newtonian mechanics and satellite orbits like yourself should be able to operate at the high school level. According to your own criteria:

High Seas wrote:
I really don't see where you add value to this discussion if you can't even grasp complete BASICS.


We aren't discussing philosophy or sociology. There actually is an absolute correct and incorrect here. You shouldn't claim expertise on a subject you know little to nothing about. Certainly, you cannot say much about a subject so complicated that it's over the heads of most physicists, if you lack even the physics knowledge of a bright high school student.

Here is the solution of the problem I posed above to you. Please forgive any arithmetic errors it might contain.


I'll start by defining my terms:

Me = mass of the Earth
Ms = mass of the satellite
G = the gravitational constant
r = the distance between the satellite and the center of the Earth
v = the velocity of the satellite
F = the force exerted by gravity to hold the satellite in orbit


v and F refer to the magnitude of these vector quantities, since this is a scalar calculation. As stated above, the Force between two masses, in this case, the Earth and the satellite, is:

G Me Ms / r ^ 2

This is equal to the centripetal force on the orbiting satellite:

G Me Ms / r ^ 2 = Ms v ^ 2 / r

Therefore,

v = SQRT ( G Me / r )

Before I go much further, I'm going to convert everything to metric units, so I don't have to convert the gravitational constant into imperial units.

I indicated a height of 3519 miles

3519 mi = 3519 mi x 1.6093 km/mi = 5663 km = 5.663 x 10 ^ 6 m

However, presumably this is above the surface of the Earth, not its center, so the radius of the Earth must be added in. The radius of the Earth is 6.38 x 10^6 m.

r = 5.663 x 10 ^ 6 m + 6.38 x 10^6 m = 1.2043 x 10 ^ 7 m

The weight of the satellite was given as 10 lbs. Obviously, I meant it's weight on the Earth's surface. The mass in a 1 lb weight is equivalent to 0.45359 kg. Note that pounds are units of force, but a kilogram is a unit of mass.

Ms = 4.5359 kg

The mass of the Earth, Me, is 5.97 x 10^24 kg. The gravitational constant, G, is 6.67 x 10 ^ -11 N-m^2/kg^2 (N is the unit Newtons).

Therefore,

v = SQRT ( G Me / r ) = SQRT ( 6.67 x 10 ^ -11 N-m^2/kg^2 x 5.97 x 10^24 kg / 1.2043 x 10 ^ 7 m ) = SQRT( 3.3065 x 10^ 7 m^2 / s^2 ) = 5.75 x 10 ^ 3 m/s

The kinetic energy is 1/2 m v ^2 = 1/2 ( 4.5359 kg) ( 5.75 x 10 ^ 3 m/s )^2 = 7.4984 x 10 ^ 7 kg - m^2/s^2 = 7.4984 x 10 ^ 7 joules
Brandon9000
 
  0  
Reply Sun 4 Jan, 2009 01:47 am
@High Seas,
High Seas wrote:

And you, Brandon, are demonstrating a terminal lack of capacity to understand simple mathematics, not to mention basic English.

Never once did I say anything other than I'm a mathematician. You are evidently not one, nor have you made the least effort to work your way through the Lorentz transformations - which have many applications in physics, as even we mathematicians have heard...

Oh, you mean these Lorentz transformations?

x' = x + vt / SQRT(1 - v^2 / c^2)
y' = y
z' = z
t' = t + vx/c^2 / SQRT(1 - v^2 / c^2)

for a frame F' moving at speed v in the x direction relative to frame F.

For God's sake, I worked my way through them in high school. I couldn't have gotten even my first degree in physics without being competent in Special Relativity. I repeat, don't talk about what you don't know about.
Brandon9000
 
  -1  
Reply Sun 4 Jan, 2009 02:28 am
@Brandon9000,
Sorry, I need parentheses:

x' = (x + vt) / SQRT(1 - v^2 / c^2)
y' = y
z' = z
t' = (t + vx/c^2) / SQRT(1 - v^2 / c^2)

for a frame F' moving at speed v in the x direction relative to frame F.
High Seas
 
  2  
Reply Sun 4 Jan, 2009 02:41 am
@Brandon9000,
Brandon9000 wrote:

............................
Certainly a great expert on Newtonian mechanics and satellite orbits like yourself ....

Source, please; certainly not a claim I ever made!
0 Replies
 
High Seas
 
  2  
Reply Sun 4 Jan, 2009 04:42 am
@Brandon9000,
Brandon9000 wrote:

Sorry, I need parentheses:

x' = (x + vt) / SQRT(1 - v^2 / c^2)
y' = y
z' = z
t' = (t + vx/c^2) / SQRT(1 - v^2 / c^2)

for a frame F' moving at speed v in the x direction relative to frame F.


Brandon - the problems with your equations exceed lack of parentheses: specifically, you left out an additional term. The extra term (which, if you had remembered, you would have preceded by SQRT, which is conventionally written as {^1/2}, when, as on this website, mathematical notation is unavailable) is:

[1-(v/c)^2]^1/2

Testing the new term presents vast difficulties in our current state of technology. Mathematically, though, we can construct a relativistic spacetime metric with more than one characteristic speed. A single Lorentz transformation would not suffice in that case: Poincare (a mathematician building on the work of another mathematician, Riemann) pointed this out back in 1905, btw, and if he was right, then conventional relativity fails, and the propagation speed of the gravitational field is vastly greater than c.

Here's what Riemann wrote in the mid-19th century:
Quote:
A similar path to the same goal could also be taken in those manifolds in which the line element is expressed in a less simple way, e.g., by a fourth root of a differential expression of the fourth degreeā€¦
Now look what happens as you keep changing parameters for v' of gravity (to be added to the original speed of gravity transmission, where v=c) greater than c:
http://www.mathpages.com/rr/s9-03/9-03_files/image006.gif

Nobody KNOWS if what is mathematically possible is ACTUALLY THE CASE. What good are physicists if they can't design a decent experiment to test mathematical hypotheses I don't know, but as Mechsmith pointed out (quoting Minkowski) on page 7 of this thread, NASA's Lisa may provide some indication on whether v' is higher than zero, ie if gravity fields move faster than c, and by how much. Nobody doubts that the light cone itself expands at c, btw, as the source of the above graphic makes clear:
http://www.mathpages.com/rr/s9-03/9-03.htm

Quote:
Similarly for the super-light cones (or spheres), there would be a single state of motion with respect to which all of those null surfaces would be spherically symmetrical. Only the accumulation shell, i.e., the actual light-cone itself, would be spherically symmetrical with respect to all states of motion.


Note that the above, while addressed to Brandon, is written as a courtesy to the original poster and to the many other posters interested in the subject. Brandon is kindly requested not to waste any more of our time with irrelevancies ever again. If that wasn't clear, I don't know what is.




High Seas
 
  2  
Reply Sun 4 Jan, 2009 05:06 am
@akaMechsmith,
PS to Mechsmith, Crayon, Gunga, others here: the mathpages site from which I took the graphic has a complete (and reasonably accessible to non-mathematicians) collection of articles on the subject. Here is the link:
http://www.mathpages.com/rr/rrtoc.htm

As I'll be out of the country for a while won't post until I'm back, so hope that link will answer some of your queries.
High Seas
 
  2  
Reply Sun 4 Jan, 2009 05:21 am
@High Seas,
PPS hope it's clear that the hastily written phrase in my previous post but one: "..Now look what happens as you keep changing parameters for v' of gravity (to be added to the original speed of gravity transmission, where v=c) greater than c:..." means that IF there is an ADDED term (greater than zero) THEN the total speed of gravity would exceed c. On the graphic "v" is the reference to the added term, so don't let the notation confuse anyone.
Brandon9000
 
  0  
Reply Sun 4 Jan, 2009 07:14 am
@High Seas,
High Seas wrote:

Brandon9000 wrote:

Sorry, I need parentheses:

x' = (x + vt) / SQRT(1 - v^2 / c^2)
y' = y
z' = z
t' = (t + vx/c^2) / SQRT(1 - v^2 / c^2)

for a frame F' moving at speed v in the x direction relative to frame F.


Brandon - the problems with your equations exceed lack of parentheses: specifically, you left out an additional term. The extra term (which, if you had remembered, you would have preceded by SQRT, which is conventionally written as {^1/2}, when, as on this website, mathematical notation is unavailable) is:

[1-(v/c)^2]^1/2

.....

And, what do you suppose I meant by "SQRT?" As a mathematician, one would think you would know that raising something to the power of 1/2 is equivalent to taking its square root. You can't even figure out that:

SQRT(1 - v^2 / c^2)

is the same as [1-(v/c)^2]^1/2 ???

The equations as I gave them are precisely correct. What's next, you do surgery on yourself based on a popular health book?
0 Replies
 
Brandon9000
 
  0  
Reply Sun 4 Jan, 2009 07:33 am
Since you appear to only understand things you can Google on the Web:

Quote:
The transformations apply to a four-dimensional coordinate system, with three spatial coordinates (x, y, & z) and one time coordinate (t). The new coordinates are denoted with an apostrophe, pronounced "prime," such that x' is pronounced x-prime. In the example below, the velocity is in the xx' direction, with velocity u:

x' = ( x - ut ) / sqrt ( 1 - u2 / c2 )
y' = y
z' = z
t' = { t - ( u / c2 ) x } / sqrt ( 1 - u2 / c2 )


From: http://physics.about.com/od/relativisticmechanics/a/relativity_3.htm
0 Replies
 
akaMechsmith
 
  2  
Reply Wed 7 Jan, 2009 08:50 am
@High Seas,
Very nice link, Thanks
gungasnake
 
  2  
Reply Thu 8 Jan, 2009 07:40 am
@High Seas,
Thanks!
0 Replies
 
Mr BOB
 
  1  
Reply Thu 1 Oct, 2009 10:53 pm
@crayon851,
Although Einstein theorized that the speed of light as well as time is constant...and that the speed of light is the fastest known speed within our universe..

Well this is NOT so..

If we are to travel beyond the speed of light (you asked what would happen)..
The answer likely depends on how you have obtained a high then light speed ,speed ( known simply as "C" )

If one used a worm hole..then it would appear one is going faster then "C" but in reality it is a short cut that enabled you to go past "C" ..

Using a worm hole to Obtain higher then "C" would take you either within our own universe to some far off point..
Or it can take us possibly to an alternate universe and or a different place of time..

There are however issues with "C" and time being constant however...
Traveling "within" our universe via typical means of movement..then one would not likely be able to exceed "C" ..

However the equations spoken of by Einstein Fail to mention that the Universe itself can and does expand faster then "C" and that this expansions also likely is beyond typical constants of "Time" ..

Regardless of this however...
There are processes beyond what Einstein explained ( or was willing to explain perhaps)..

This is where String Theory and M Theory ( membrane theory)...

It used to be thought that String Theory was the final connection between the theory of the large ( Einstein) and the theory of the small..

However String Theory by and of itself fails to explain and fails to hold up as a true theory of everything..

M Theory however...as we understand it so far...is where everything comes from ( starts from)..
Sort of a floating membrane that is of endless length..

From this membrane sprouts string theory and it is string theory that creates most of the universes as well as alternate dimensions ... ( within these dimensions and universes created from the string theory...then within its strings creations...in or within those creations comes what we realize as "C" and time constants..

However on the infinitely small ( M Theory ) many crazy things can happen and happens regularly...
Including going well past constants of "C" and time..

In answer to your asking..
what happens if we travel beyond speed of "C" ..

"Maybe we'll go back in time? Like super man did when he flew around the earth in the opposite direction of its rotation ?"
That is one possibility...
While also entering different dimensions within our own universe (most people realize 3 dimensions...some recognize 4 dimensions...However in reality there is between 15 to 17 dimensions ( perhaps still counting)..

Now a question for you..
You may know what Dark matter is (or know it exists)..
You may know what Dark Energy is(or know it Exists)..

However..
Do you know That there is another Dark (something)..
So...Do you know what it is and or what it is doing ?
0 Replies
 
High Seas
 
  1  
Reply Tue 6 Oct, 2009 02:00 pm
@akaMechsmith,
Welcome, guys. Looking up the general link I got scared to read this passage >
Quote:
This system of specifying positions is quite stable, but not perfect. Around 150 BC the Greek astronomer Hipparchus carefully compared his own observations of certain stars with observations of the same stars recorded by Timocharis 169 years earlier (and with some even earlier measurements from the Babylonians), and noted a slight but systematic difference in the longitudes. Of course, these were all referenced to the supposedly fixed direction of the line of intersection between the Earth's rotational and orbital planes, but Hipparchus was led to the conclusion that this direction is not perfectly stationary, i.e., that the direction of the Sun at the equinoxes is not constant with respect to the fixed stars, but precesses by about 0.0127 degrees each year. This is a remarkably good estimate, considering the limited quality of the observations that were available to Hipparchus. The accepted modern value for the precession of the equinoxes is 0.01396 degrees per year, which implies that the line of the equinoxes actually rotates completely around 360 degrees over a period of about 26,000 years.

http://www.mathpages.com/rr/s6-02/6-02.htm
> and consider that for 22 of the 25 centuries since those guys came up with their calculations (amazingly precise, considering the technical means of the time) we managed to lose all they had discovered and then had to rediscover it all over again.

I worry that most of our population believes in "dark ages" myths - superstitions, scams, patently absurd constructs - and students skip math and science courses. How can we have a fifth of the population at quasi-literacy and innumeracy levels that would shame much poorer Asiatic nations? They spend less than a 20th of what we spend per pupil but their children learn.
0 Replies
 
 

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