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Hello everyone,
I have a quick question for you. Do you have any preference as to whether or not I continue this discussion on this forum?
Let me explain. In this post I continue a discussion with Dick on an astronomy question I asked a long time ago about what we are really seeing when we look at distant galaxies. Depending on the response to my question to you, I will either continue the discussion on this forum, or I will, instead, continue it via private e-mail with Dick. You get to decide. Here are the rules:
If there is no response to my question, it goes to e-mail.
If there are only non-negative replies, even a single NT will do, it stays here.
If the number of positive replies exceeds 10% of the number of negative replies, it stays here. Thanks for reading this far.
Warm regards,
Paul
To Dick,
Thank you for your response to my question As usual you go too fast for my slow brain. Let me slow you down a little.
Before we start, though, I want to lay a Cartesian coordinate system over the diagram so I can talk about it using high school math. Let's put the origin at B, the x-axis along BR, and the y-axis along BE. (I am hoping to be able to express those hyperbolas you mention in an equation so that I will be sure I know what you are talking about.)
***At this point [my step 13.], what you should do to your diagram is add the trajectories of some assorted distant galaxies.***
Great suggestion. It is essentially what I was trying to do by the introduction of G, but instead of locating the position of galaxy G "now" with a single point, which as you point out is made ambiguous by the ambiguity of the term 'now', the trajectory is a much better idea. I'm not sure an assortment of more than one galaxy at this point would made things more clear, though. For now, let's stick with an assortment of just one galaxy, G, to make sure I get the diagram right. I'm having trouble drawing the hyperbola you described. I can add other galaxies to the assortment later.
***Those trajectories would be straight lines from B terminating on the line segment EC.***
I'm not so sure about that.
I agree that the trajectory would terminate on EC because that would mean that the galaxy is somewhere within the big bang's light cone "now". On this diagram, "now" is defined as y = 14.43, which contains the segment EC. The big bang's light cone is the segment BC rotated about BE. Thus, if the trajectory did not intersect EC now, it would be outside the big bang's light cone and we would not have chosen that galaxy as our example G. Let us designate the point of intersection with EC as Gn. (I realize that I have been using the symbol G ambiguously: sometimes as a label for the galaxy, and sometimes as a, yet to be determined, point on the diagram. By introducing point Gn, which is the point on the diagram corresponding to the location of galaxy G now, I hope to avoid any problems resulting from the ambiguity of the symbol G.)
Next, I wondered whether or not the trajectory of G would be a straight line on the diagram. I think it would depend on whether the expansion of the universe has been constant (in which case I agree the trajectory would be straight) or if the expansion has been slowing or speeding up (in which cases it would bend one way or the other). For the purposes of this enquiry, it would be reasonable to assume a constant rate of expansion, so, I agree that the line would be straight.
Finally, I wondered whether or not the trajectory would include B. My thinking was that all galaxies visible to us now must have originated somewhere within the triangle BME. Galaxies didn't actually form until some time after the big bang so the trajectory wouldn't actually start at B. But the question is, if you extended the trajectory in a straight line, would it intersect B? Hmmm. Using my own thought experiment of identifying the location of galaxy G as the center of mass of all stable particles which comprise it, and then trace all those particles back to their origin, I guess the trajectory of that center of mass must go back and approach B along a straight line.
So, I conclude that you are right. The trajectory of galaxy G is a straight line going through B and intersecting EC at Gn. (Sorry it took me so long to catch up.)
***They would appear to be receding from the trajectory of the earth (line BE) at some velocity less than light.***
Yes, I agree. It means that the angle GBE < 45 degrees. (If HTML wiped out the less-than symbol between GBE and 45, I will quit using that symbol in my text, but that's what I typed in there.)
At this point, Dick, let me interrupt and give you a brief description of my intuitive view of what is going on thus far. Then I will get back to your response to me.
As I indicated, the big bang's light cone is generated by rotating BC about BE. Similarly, earth's light cone now is generated by rotating ER (and its extension into the future) about BE. The future part of earth's light cone, i.e. the portion above E, confines all information emanating from earth now. The portion below E, contains all information about the past that is available to earth now. Also, a light cone for earth at any past point in time could be generated by rotating a 45 degree line intersecting BE at the point representing that moment in past time.
But now, the question is, what is the light cone for galaxy G at some point G along its trajectory BGn? I think it is generated by the rotation of a line intersecting BGn at G at 45 degrees about the axis BGn. In my paragraph 13. where I said "BG forms the reference frame for galaxy G", this is what I meant. When I read popular books (e.g. Roger Penrose's "Shadows of the Mind", pp. 219-223) describing the tilting of light cones, I think that is the same tilt, and for the same reason, as the tilt in my diagram caused by generating G's light cone about BGn rather than about BE. Now I realize that this means that relative to earth, the light that leaves G in a direction exactly opposite earth is going faster than light from earth's frame. I.e. on my diagram, the right hand intersection of G's light cone with the diagram is at an angle greater than 45 degrees to BE. But, it cannot intersect our light cone so I guess it doesn't matter.
But, in pondering the implications of this tilted light cone, the question immediately comes up as to whose time are we using anyway to talk about galaxy G? It is not the time defined by the y axis. So is it uniform markings along the line BGn? If so, how are those marks spaced with respect to the units along the y axis?
So, with these puzzles in my mind, Dick, I will rejoin my analysis of your response.
***The second thing you should do (just to clarify the picture in your mind) is to add in the lines which would indicate the time as measured locally at those distant galaxies.***
As you can see from the preceding paragraphs, we need to do more than clarify the picture in my mind; we need to establish a picture in the first place. I'm not sure how lines would "indicate the time as measured locally at those distant galaxies". In thinking about it, it seems to me that you could indicate local time on the diagram by putting tick marks along the trajectory of the local system at, say, billion year intervals. Thus, for earth's local time, there would be 14 tick marks on BE and E would be 0.43 tick-mark-lengths of the way to the next one. Similarly, there would be evenly spaced tick marks along G's trajectory, BGn. So, I wonder how I would plot them. It would be an obvious error to plot them with the same y values as those for earth (i.e. the intersection with horizontal lines through earth's tick marks.)
I might guess that the tick-mark-interval length for G would be the same as that for earth but laid out on BGn. That would give a y value for the interval as earth's interval times the cosine of the angle between earth's trajectory and G's trajectory.
*** If you divide the BE line segment up into equal time steps, the corresponding time steps along the trajectories of those distant galaxies are longer (the local clocks seem to take longer to mark off those equivalent time periods).***
I think what I just described is consistent with what you say here. I just now used some of the remnants of my memory of High School math to confirm that my definition of tick-mark-interval length would give the y component of each interval on my diagram of exactly the factor sqrt(1-v*v) times the interval length for earth along BE.
***Those lines are not too difficult to estimate as the clocks in the distant galaxies run slow by exactly the factor sqrt(1-v*v) [since, in the units of measure in our diagram, c is unity].***
So I think I'm on the right track, at least about plotting the local time interval points on the trajectory. Now let me proceed to try to understand your suggestion. Here, let's suppose I took your advice and plotted the trajectories of several assorted galaxies, and plotted tick marks on them again representing billion-year intervals in their respective local times. Now, if you drew curves connecting all the corresponding tick marks for each of these galaxies, you would get curves that might be called isochronic lines, or isochrons. (I just made that term up because I don't know any better.)
Are these isochrons the lines you are talking about "which would indicate the time as measured locally at those distant galaxies"?
***The resultant curves are simple hyperbolas with the line BC as the limiting tangent at infinity. Note that the amount of time which appears to pass for an object moving very close to BC is very close to zero!***
Oh, oh. Here I have a problem. My isochrons are arcs of concentric circles centered at B. They are not hyperbolas. I need help. Please describe these hyperbolas a little more specifically. An equation in x and y would be most useful. I don't understand exactly what you mean by the phrase "the amount of time which appears to pass". I would like you to explain it in terms of my diagram if you would. Do you mean "appears" from the perspective of an observer on earth? If so, is "the amount of time" the y component of the time interval along the trajectory of the other galaxy?
I am running out of time and I am sort of stuck here anyway. I will send this off and wait for your answers before I continue. Thank you for your continued patience with me, Dick.
Warm regards,
Paul |
