michel123456 Posted January 11, 2012 Posted January 11, 2012 AFAIK, if speed increases proportional to distance then there is only one slope following the relation between them. (Look at the blue line in the latest picture you posted, one single slope for all objects.) The last diagram from wiki is a distance/redshift aka distance/speed. My diagram is a distance/time.
Spyman Posted January 11, 2012 Posted January 11, 2012 The last diagram from wiki is a distance/redshift aka distance/speed. My diagram is a distance/time. If space don't expand and speed of light is constant then distance and time is also proportional to each other.
michel123456 Posted January 11, 2012 Posted January 11, 2012 If space don't expand and speed of light is constant then distance and time is also proportional to each other. Yes, "speed of light is constant then distance and time is also proportional to each other". In my diagram represented speed is not SOL, represented speed is the recessing speed of galaxy clusters.
JustinW Posted January 11, 2012 Author Posted January 11, 2012 Isn't it wierd how you can refrence all three: distance, time, and rate of expansion using the SOL. It might nip confusion in the bud.....or on second thought, might further it.
michel123456 Posted January 11, 2012 Posted January 11, 2012 (edited) Isn't it wierd how you can refrence all three: distance, time, and rate of expansion using the SOL. It might nip confusion in the bud.....or on second thought, might further it. Yes. It is weird. Except that rate of expansion has less to do with SOL. It remains that the 3 concepts are tightly linked. An object observed at a distant location is also far away in time and also receding from us. The largest the distance, the more in the past and the more receding. As seen in the diagram of post #36. ------------------------------------------------------------------------ It is even more amazing: if you know the distance to a galaxy cluster and only the distance, you can plot it on the diagram without knowing a priori neither time nor receding speed. If you know only time, again you can plot the new G accurately on the diagram. If you know only the receding speed, again you can plot the new G exactly, without knowing a priori time or distance. Edited January 11, 2012 by michel123456
JustinW Posted January 11, 2012 Author Posted January 11, 2012 Oh what a tangled web we weave in achieving simplicity.
Spyman Posted January 11, 2012 Posted January 11, 2012 Yes, "speed of light is constant then distance and time is also proportional to each other". In my diagram represented speed is not SOL, represented speed is the recessing speed of galaxy clusters. You are going in circles Michel, discussion is back at my post #48.
michel123456 Posted January 12, 2012 Posted January 12, 2012 Those 2 diagrams represent the same thing: speed is increasing with distance. Left is corresponding to wikipedia redshift diagram. Right is corresponding to diagram of post #36. What is that you don't understand? (or what is wrong?)
Spyman Posted January 12, 2012 Posted January 12, 2012 What is that you don't understand? (or what is wrong?) Now we have returned to post #44 where I asked you to explain why light makes a detour... If space don't expand then T and D are proportional to each other and as such all objects in diagram nr.2 should be on a line, just like in nr.1. Without expanding space, then if G1 emitted light at D=1 billion lightyears and G2 emitted light at D=2 billion lightyears, nevermind their receding speed, it would take 1 billion years for light from G1 to reach us and 2 billion years for light from G2 to reach us. In you #36 diagram you have placed G2 on a 1/1 relation and G3 on a 3/2 relation and G4 on a 6/3 relation, but if lightspeed is constant then they all should have the same relation and be placed on a straight line. So what I don't understand and want you to explain thoroughly is why your galaxies don't have equal proportional relation for T and D.
JustinW Posted January 12, 2012 Author Posted January 12, 2012 When you think about it, it should curve when time is conserned. If a galaxy is moving away from you and you are moving away from it, the faster your moving away the longer it takes for that light to reach you. It would show more of a curve the closer you get to the SOL. But time should be the only thing that curves on the graph.
michel123456 Posted January 12, 2012 Posted January 12, 2012 (edited) Now we have returned to post #44 where I asked you to explain why light makes a detour... If space don't expand then T and D are proportional to each other and as such all objects in diagram nr.2 should be on a line, just like in nr.1. Without expanding space, then if G1 emitted light at D=1 billion lightyears and G2 emitted light at D=2 billion lightyears, nevermind their receding speed, it would take 1 billion years for light from G1 to reach us and 2 billion years for light from G2 to reach us. In you #36 diagram you have placed G2 on a 1/1 relation and G3 on a 3/2 relation and G4 on a 6/3 relation, but if lightspeed is constant then they all should have the same relation and be placed on a straight line. So what I don't understand and want you to explain thoroughly is why your galaxies don't have equal proportional relation for T and D. because V is not the speed of light. V is redshift. Edited January 12, 2012 by michel123456
Spyman Posted January 12, 2012 Posted January 12, 2012 because V is not the speed of light. V is redshift. The redshift or receding speed does not alter the speed of the image towards Earth, it doesn't matter what redshift an object has, if it sends out light from an distance of 2 billion lightyears then it will take 2 billion years for the image to reach us.
JustinW Posted January 12, 2012 Author Posted January 12, 2012 (edited) So what I said earlier about there being a curve when graphing the time duration the closer to the speed of light, is that right or wrong? The time duration of lights travel shown on a graph should get lower the closer the object gets to moving away at the SOL. Edited January 12, 2012 by JustinW
michel123456 Posted January 12, 2012 Posted January 12, 2012 The redshift or receding speed does not alter the speed of the image towards Earth, it doesn't matter what redshift an object has, if it sends out light from an distance of 2 billion lightyears then it will take 2 billion years for the image to reach us. Yes, that is correct. But the speed under discussion is not the speed og light, it is the recessing speed. What I do is simply this: G1 at the origin Take a galaxy cluster G2: what's its distance: say 1 Billion LY. what's its recessing speed: say 1 BLY/10 Billion Years or 1/10 (note 1BLY/1BY is SOL) plot G2 on the distance/time diagram. Take G3 distance say 3 billion years. recessing speed: 3 BLY/20 BY or 3/20 > 1/10 plot G3 Take G4 distance say 6 BLY recessing speed: 6 BLY/30 BY or 6/30 > 3/20 > 1/10 plot G4 And so on. You get a curve.
Spyman Posted January 13, 2012 Posted January 13, 2012 So what I said earlier about there being a curve when graphing the time duration the closer to the speed of light, is that right or wrong? The time duration of lights travel shown on a graph should get lower the closer the object gets to moving away at the SOL. Justin don't get confused by my and Michels discussion, Michel like to explore other possibilities and right now I think he is trying to make a model which would simulate what we see but without space expansion. The current scientific consensus is the graph I posted in my post #41 on top of page 3. What I do is simply this: ... And so on. You get a curve. Michel, I don't understand what your curve is supposed to show, you have not plotted the locations where we see the objects emitt the light from and it does not have a proportional recessing speed to distance. What does the plotted locations represent and why change the 10, 20 & 30 numbers? If it should fit with what we observe and without expanding space, then G2 with a distance of 1 billion lightyears should be plotted at the time of 1 billion years, G2 of 3 billion lightyears at the time of 3 billion years and G3 of 6 billion lightyears at the time of 6 billion years. The you have to choose on one slope for the proportion of distance versus speed, if you for example pick the one for G3 then recessing speed for G3 could be 3/20 but that would also mean that recessing speed for G2 would be 2/20 and for G4 it would be 6/20.
michel123456 Posted January 13, 2012 Posted January 13, 2012 (edited) Justin don't get confused by my and Michels discussion, Michel like to explore other possibilities and right now I think he is trying to make a model which would simulate what we see but without space expansion. The current scientific consensus is the graph I posted in my post #41 on top of page 3. Michel, I don't understand what your curve is supposed to show, you have not plotted the locations where we see the objects emitt the light from and it does not have a proportional recessing speed to distance. What does the plotted locations represent and why change the 10, 20 & 30 numbers? If it should fit with what we observe and without expanding space, then G2 with a distance of 1 billion lightyears should be plotted at the time of 1 billion years, G2 of 3 billion lightyears at the time of 3 billion years and G3 of 6 billion lightyears at the time of 6 billion years. The you have to choose on one slope for the proportion of distance versus speed, if you for example pick the one for G3 then recessing speed for G3 could be 3/20 but that would also mean that recessing speed for G2 would be 2/20 and for G4 it would be 6/20. you are sticking to SOL. This diagram is about observed velocity. Maybe you should do it for yourself. on a simple diagram Distance/Time like this one You have to plot 4 moving objects, G1, G2, G3, G4. the placement upon the diagram indicates velocity: meters/sec. _You know from Hubble's law that velocity of G4> vG3 > vG2 > vG1 _You know from Hubble's law that distance of G4 from origin is > G3 > G2 > G1 _You know from Hubble's law that velocity is proportional to distance. _You know that velocity at origin is null. Edited January 13, 2012 by michel123456
Spyman Posted January 13, 2012 Posted January 13, 2012 Maybe you should do it for yourself. I already did, look at the green dots in my graph at post #41 on page 3.
michel123456 Posted January 13, 2012 Posted January 13, 2012 (edited) I already did, look at the green dots in my graph at post #41 on page 3. green dots represent position. Not relative velocity to the observer. Increasing velocity with distance gives a diagram similar to an accelerating body: a curve. Edited January 13, 2012 by michel123456
Spyman Posted January 13, 2012 Posted January 13, 2012 (edited) green dots represent position. Not relative velocity to the observer. Each green dot represent a speed with redshift 0.25, 0.50, 0.75, 1.00, 1.25, 1.50, 1.75, 2.00, 3.00, 4.00, 5.00, 6.00, 7.00, 8.00 relative us. Increasing velocity with distance gives a diagram similar to an accelerating body: a curve. Increasing velocity with distance gives a straight line if the relation is proportional. "In mathematics, two variable quantities are proportional if one of them is always the product of the other and a constant quantity, called the coefficient of proportionality or proportionality constant. In other words, x and y are proportional if the ratio is constant." http://en.wikipedia.org/wiki/Proportionality_(mathematics) Edited January 13, 2012 by Spyman
michel123456 Posted January 14, 2012 Posted January 14, 2012 (edited) Increasing velocity with distance gives a straight line if the relation is proportional. "In mathematics, two variable quantities are proportional if one of them is always the product of the other and a constant quantity, called the coefficient of proportionality or proportionality constant. In other words, x and y are proportional if the ratio is constant." http://en.wikipedia.org/wiki/Proportionality_(mathematics) YES. VELOCITY is proportional= the rate of change of VELOCITY increases along a straight line = for the same time DISTANCE increases following a curve. Velocity increase: At T0, D0, V0 T1, D1, V1 T2, D2, V2 T3, D3, V3 If V3>V2>V1 this is accelerated motion: there exist a rate of change of VELOCITY Wiki In physics, acceleration is the rate of change of velocity with time Here below a Distance/Time diagram representing increasing velocity. After 1 sec, object G has a velocity of 1m/s (displacement 1m on the horizontal, 1 sec on the vertical) After 2 sec, G has velocity 2m/s (displacement 2m on the horizontal, 1 sec on the vertical) After 3 sec, G has velocity 3m/s (displacement 3m on the horizontal, 1 sec on the vertical) Hubble's law states that VELOCITY increases with distance. Not that velocity remains the same. Edited January 14, 2012 by michel123456
Spyman Posted January 14, 2012 Posted January 14, 2012 (edited) ...there exist a rate of change of VELOCITY The observed rate of change of velocity is CONSTANT. EDIT: Feel I need to add: constant over distance and not over time. Here below a Distance/Time diagram representing increasing velocity. After 1 sec, object G has a velocity of 1m/s (displacement 1m on the horizontal, 1 sec on the vertical) After 2 sec, G has velocity 2m/s (displacement 2m on the horizontal, 1 sec on the vertical) After 3 sec, G has velocity 3m/s (displacement 3m on the horizontal, 1 sec on the vertical) You need to check your math, if acceleration is uniform and initial speed is zero then displacement=a*t2/2. http://en.wikipedia.org/wiki/Acceleration If a=1m/s2 then G would be located at L=0.5m when T=1s, at L=2.0m when T=2s and at L=4.5m when T=3s. Edited January 14, 2012 by Spyman
michel123456 Posted January 14, 2012 Posted January 14, 2012 (edited) The observed rate of change of velocity is CONSTANT. EDIT: Feel I need to add: constant over distance and not over time. You need to check your math, if acceleration is uniform and initial speed is zero then displacement=a*t2/2. http://en.wikipedia.org/wiki/Acceleration If a=1m/s2 then G would be located at L=0.5m when T=1s, at L=2.0m when T=2s and at L=4.5m when T=3s. So you accept finally that there is a rate of change of velocity. You need to check your math, if acceleration is uniform and initial speed is zero then displacement=a*t2/2. http://en.wikipedia.org/wiki/Acceleration If a=1m/s2 then G would be located at L=0.5m when T=1s, at L=2.0m when T=2s and at L=4.5m when T=3s. You are right Here you are Is it a straight line, or a curve? Edited January 14, 2012 by michel123456
Spyman Posted January 14, 2012 Posted January 14, 2012 So you accept finally that there is a rate of change of velocity. We must have misunderstood each other, I never ment to say that there was not a rate of change of velocity. Is it a straight line, or a curve? Yes, of course it is a curve, but the important question is: Is velocity proportional to distance in it?
michel123456 Posted January 14, 2012 Posted January 14, 2012 (edited) We must have misunderstood each other, I never ment to say that there was not a rate of change of velocity. Yes, of course it is a curve, but the important question is: Is velocity proportional to distance in it? good question. Let's check out with Hubble's law. V=HoD where Ho is Hubble's constant with Ho=72 and units of distances as unity Vo=0 V1=72 V2=72.2= 144 V3=72.3= 216 V4=72.4= 288 Correct so far? ------------------ Damn. Got it. ------------------- Actually I got nothing. Can you put that into a graph? Here what I get Note: time=1/72=0.0138888 approx Edited January 14, 2012 by michel123456
Spyman Posted January 14, 2012 Posted January 14, 2012 (edited) Can you put that into a graph? You already made it: Since you have declared that: Here below a Distance/Time diagram representing increasing velocity. After 1 sec, object G has a velocity of 1m/s (displacement 1m on the horizontal, 1 sec on the vertical) After 2 sec, G has velocity 2m/s (displacement 2m on the horizontal, 1 sec on the vertical) After 3 sec, G has velocity 3m/s (displacement 3m on the horizontal, 1 sec on the vertical) We know the speed at T1, T2 and T3, simply replace 1s with 1m/s, 2s with 2m/s and 3s with 3m/s. Is velocity proportional to distance? Edited January 14, 2012 by Spyman
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