Hi ICANT,
ICANT writes:
NoNukes writes:
In fact they can both be true as long as "taking on the motion" does not mean increasing or decreasing the speed of a photon in a vacuum in any inertial reference frame.
If the pulse of light is traveling at 299,792,458 meters per second, how does it add any speed without breaking the speed of light.
I explained this in detail in the part of the post for physics lovers, and I worked a couple of examples. As is explained in Einstein's paper, when C is combined with other velocities, the result is speed C. However, the direction of the vector may be different.
What I find extremely amusing, is that my initial statement, which you quoted above ought to have been enough to put you on notice that my explanation does not involve "breaking the speed of light" What did you think I meant when I said "does not mean increasing or decreasing the speed of a photon in a vacuum in any reference frame"?
Do you have an equation that shows that the speed of the beam causes the cycle to stop in mid vaccume?
No, because I never said that such a thing happens. By the way, vacuum is spelled with only one "c", a doubled "u", and no "e". We've been spelling it at least three different ways here.
As measured in any inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body.
That says "light is always propagated in empty space with a definite velocity c."
Yes, a definite velocity c,
as measured in any and every reference frame. Don't skip over that vital point.
That particular point has some consequences that are simply non-intuitive. Because you insist on using v+w to combine velocities rather than the equation from Einstein's paper, you'll never understand how the same light beam appears to have the same velocity "c" despite being measured by observers in relative motion to each other.
Maybe I don't understand what empty space is. As I understande what Einstein was talking about he was talking about a vaccum as he mentions several times.
Your understanding of what constitutes empty space is correct. But that does not stop us from understanding what happens in when light travels through other mediums.
In my cycle experiment with the open clock.
What open clock?
The path of the photon in the light clock has been enclosed in a vacuum chamber almost since I introduced the thing. Not that it makes any difference, but it is my light clock. The interplanetary space between earth and planetX is a great vacuum, but the light clock is not exposed to it, because I didn't want the added headache of explaining that to you.
But even in the case where the light beam is traveling in air, Einstein's equations will show that the speed of a light beam can be increased only by tiny amount due to the motion of an observer relative to the source.
For example, the speed of light in air according to at least one source is 298,925,574 meters per second. I'm not stuck on that number, but let's use it.
http://cadlab6.mit.edu/2.009.wiki/anchor/index.php?title=...
So, the speed of light in air is about 0.99711c. So if we were looking at the headlight of the rail car moving at 0.25c, we would find that the speed of light in the frame of reference of a ground observer is (sorry for the rough appearance, but latex tags seem to be on the blink)
(0.99711c + 0.25c)
----------------------------- = 0.9983c
1 + (0.99711c *0.25c)/c^2
Of course the speed of light as measured in the train conductors frame is 0.99711. So postulate #2 is approximately approximately correct if even if we substitute air for vacuum. If the train were moving at a mere 200 mph, the same equation tells us that the speed of travel of a light beam traveling in air would be indistinguishable from 0.99711c in both inertial frames.
Why can't the pulse travel at 299,792,458 meters per second while the cycle is traveling at 149,896,229 meters per second?
It does do exactly that. But the speed of light must be also be 299,792,458 meters per second in vacuum as measured in every inertial frame including, to your chagrin, the frame of the space cycle.
I'm going to attempt to address your questions on the thought experiment. Up until now, I haven't paid all that much attention to it. One caveat though...
ICANT writes:
In Message 689 to Son which you replied to I ask that the math be shown where it was wrong.
In Message 710 You replied with a lot of smoke and mirrors but you did not show where the math was wrong.
I'm not going to play the silly game of showing where your math is wrong. I have no qualms about your ability to add or subtract But your arithmetic does not jibe with postulate #2, so it is wrong or irrelevant. If the velocities are all along a single direction, then try the equation above.
Instead, I'll do the problem the proper way. You can look at my numbers and figure out how much or little you got right.
When the velocities are not in a line, as is the case in your thought experiment then things are more difficult. But we should still be able to get through them by applying postulates #1 and #2. Of course, that appears to be the case in your thought experiment.
In my cycle experiment with the open clock or the train flatcar with the open clock the light pulse is traveling at a 90 angle to the two mirrors which are 1 meter apart and 18 inches long.
In what reference frame is the 90 degree angle measured?
Since you won't say, it appears that I'll have to work two separate problems in two different inertial frames. Well I'm up to it. I'm going to do the math using the speed of light in a vacuum rather than air. I think I've demonstrated that things work only slightly differently in air than they do in a vacuum.
I also assume that by "cycle" experiment, you really mean train experiment. As I recall the train moved at 0.25c rather than 0.5c.
Assume that the 90 angle is measured in the ground frame.
If the 90 degree angle is measured in a reference in which the space cycle is moving at 0.5c, then the light beam CAN miss the mirror. Let's arbitrarily call that frame, frame K. An observer on the cycle as well as an observer at rest in frame K must agree on whether the light hits the mirror or postulate #1 would be violated. The observers must also agree that the light beam travels at c
regardless of whether they can see the light beam. In other words, the observers cannot make measurements that conflict with the speed of light being "c".
In the K frame, the math works out pretty much as you've suggested. Assuming that in the K frame, the beam was directed upwards, the beam will be 1 meter high at time 3.3356 nanosecond, but the train will have moved 0.25c * 3.3356 nanosecs, or a little less that 10 inches (9.8 inches = 1/4th meter).
Now let's consider things from the point of view of the person on the train using an appropriate coordinate system. On the train car, the ceiling, floor and framework have fixed coordinates. Further, in this coordinate system if point A is vertical from point B, then points A and B must be the same difference away from any vertical member on the train.
Since the light beam must miss the mirror in the train coordinate system, then the light beam must hit the top frame at a point that is not directly below the point at which the beam first reached the bottom frame. Accordingly, the light beam cannot travel vertically in the train frame of reference. It must travel on a diagonal in the train coordinate system. Further, the light beam must travel at "c" along the diagonal.
Will a person on the train agree that the beam strikes the ceiling 9.8 inches (0.25meters ) to the right of the point it reached the floor? Well, somewhat surprisingly, the answer is no.
The diagonal from the floor to the ceiling makes a right triangle with the frame height of 1 meter, and the amount of displacement of the beam at the ceiling. The length of that diagonal is sqrt (1m*1m + 0.25m*0.25m) = 1.03077 meters. Light simply cannot travel that distance in 3.3356 nsecs because that would violate postulate #2. This means that an observer on the train will not agree that the light beam reached the ceiling in only 3.3356 nsecs.
We could use the ICANT recursive approach to figure out when the light hits the ceiling, but that isn't necessary. We'll know that we have the right distance and time when d/t = the velocity of light.
In fact, from the point of view of the observer on the train, the light beam strikes the top frame at 0.2582 meters (10.17 inches) to the right of the point at which it reaches the bottom frame. The time to travel that amount is (0.2582 meters)/(0.25c *299,792,458 m/second) = 3.445 nanosecond.
As a final check, let's see if the speed of light as measured in the rail car is "c". The length of the diagonal is sqrt (1m *1m + 0.2582m *0.2582m) = 1.03280 meters. So the speed of light as observed in the light car is 1.03280 meters/3.445 nanosecond = 2.99796 * 10^8 meters per second. That is equal to c to within the accuracy that we carried out the calculations.
So we're half done.
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Let's assume that the light travels vertically in the train coordinate system. We can debate about how such a beam might be produced, but I won't do so here. It is certainly possible to direct a light beam to any point we wish from any other point, absent an obstruction. Such a beam must strike the ceiling directly above the point in the railcar that it leaves from the floor, or it would not be vertical in the rail car frame of reference. The railcar rider would agree say that such a beam traveled upwards exactly one meter and did so in 3.3356 nanosecond. Good ole postulate 2.
So what does the ground observer think occurred here? Well if the light struck the mirror in the railcar frame, it must strike the mirror in the ground frame as well. We can carry out a similar analysis to the above to determine exactly what the ground observer sees, but it should be clear that he cannot see the beam traveling vertically, and that the velocity of light must be c for the ground observer. In fact, the ground observer would see a diagonal pattern. The analysis would be similar to that I carried out in the space cycle example with the difference being that the 0.25c speed of the train relative to frame K produces a 14.47 degree angle instead of the 30 degree angle in the space cycle. The light beam would still move at speed c.
Edited by NoNukes, : No reason given.
if I'm sat in the passanger seat as we're driving along, holding a bag of peanuts up to roof, vertically above by lap, and then let it fall... does it:
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