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868 Glorious LeaderAbout Janus

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Two Theoretical Bodies in Space, Time to Collide.
Janus replied to Zemy's topic in Astronomy and Cosmology
You have to use integral calculus to account for the fact that the force, and thus the acceleration, changes as the distance decreases. 
Two Theoretical Bodies in Space, Time to Collide.
Janus replied to Zemy's topic in Astronomy and Cosmology
There's an equation that will give you the exact answer. it is [latex] T = \frac{\cos ^{1} \sqrt{\frac{x}{r}}+\sqrt {\frac{x}{r} \left ( 1 \frac{x}{r} \right )} }{ \sqrt {2G (m_1+m_2)}} r^{\frac{3}{2}}[/latex] Here r is the distance between the center of the balls at the start and x the distance between the centers when they collide (2 times the radius of the balls themselves) Using this method, I get an answer of ~2.9 days. 
Why is the speed of gravity (curvature of spacetime) equal to electromagnetic disturbances
Janus replied to Romeo22's topic in Modern and Theoretical Physics
Both electromagnetic waves and gravitational waves (the aspect of gravity that travels at c) are massless entities, and as such are required to travel at c, the invariant speed of the universe. When neutrinos were first proposed and and later discovered, they also were assumed to be massless and thus were expected to travel at c. This is a consequence of Relativity. 
What exactly does "fashion sense" even mean? If it were a real thing then the same type of dress would always be "in fashion". But it isn't; What was in fashion 20 years ago would be considered going against "fashion sense" today. So I guess fashion sense is just the ability to see what everyone else is wearing and then copy it.

I Think that it would be well served if you gave your thoughts on the matter and what your reasoning is for it. You went to the trouble of giving the distance between the walls and the activation period for the laser, so I 'm assuming that you think this is important to the answer. Assuming that the whole set up isn't under some extreme acceleration which either severely Doppler shifts the laser or causes its path to bend enough that it hits one of the other surfaces in the room rather than the one opposite it, There is no reason for the laser not to strike the sensitive wall.

What if Einsteins Definition of Simultaneity is incorrect?
Janus replied to vanholten's topic in Classical Physics
Right, he is saying that the speed of light in both directions is the same with respect to any inertial frame as measured from that frame. So in the following example we have two observers. One standing along the tracks and the other traveling along the tracks in a railway car. Two flashes are emitted from two points along the tracks that are equal distance from the track observer. the light from these flashes arrive at the midpoint observer at the same moment as the railway observer is passing him. Thus both observers detect the light from the flashes at the same time. Like this: For the midpoint observer ( or anyone at rest with respect to the tracks) these flashes were emitted simultaneously, as shown by the expanding circles: However, for the railway car observer, events have to occur differently. He still detects the light from both flashes simultaneously, and they arrive when he is adjacent to the track observer. But unlike the track observer he has not remained halfway between the emission points the entire time. He is not an equal distance from the emission points when either of the flashes was emitted. But he must also measure the speed of light for each of the flashes as being the same relative to himself. But since the distances each of these flashes travel relative to him are not the same, in order for the light of the flashes to reach him simultaneously, they must have left at different times. And the sequence of the events for him occur like this: For the track observer, the flashes are emitted simultaneously, but for the railway observer they are not. This is the relativity of simultaneity: Events that are simultaneous in one inertial frame are not so according to another which in relative motion with respect to the first frame. 
What if Einsteins Definition of Simultaneity is incorrect?
Janus replied to vanholten's topic in Classical Physics
No. You have three clocks, A, B and C all at rest with respect to each other. For this example we will put them in a straight line, with B between A and and twice as far from C as it is A. When Clock B reads 12:00 it sends a light flash towards both A and C and then waits for the flash to reflect back to, and when each reflection returns, it records the reading it sees on each of the other clocks when its flash returns. So if the flash from clock A returns when clock B reads 2:00, B knows that clock A is 1 light hr away. And if the accompanying image of Clock A shows 1:00, Clock B knows that that image left 1 hr ago, clock A has advanced 1 hr in that time and at at the moment that B sees the returning flash reads 2:00. Thus clock A is in sync with clock B. Further, the flash from clock C returns when clock B reads 4:00, meaning clock C is 2 light hrs away. If the accompanying image of Clock C reads 2:00, Clock B knows that it left 2 hrs ago, Clock C has advanced 2 hrs since then, reads 4:00 at the moment B sees the flash and is in sync with clock B. It also knows that Clock A has also advanced to read 4:00, so all three clocks are in sync. While it is convenient to use equally spaced clocks to determine simultaneity, as it removes the need to determine the distance between clocks, it is not a requirement for determining simultaneity. 
What is gravitational slowing of time in center of mass?
Janus replied to DimaMazin's topic in Relativity
No. If you look at the standard time dilation equation: t0 =tf(12GM/Rc2)1/2 One thing stands out. It is that 2GM/R also appears in the escape velocity equation: Ve= (2GM/R)1/2 and so, 2GM/R = Ve2 Ergo, we could rewrite the gravitational time dilation equation as: t0 =tf(1Ve2/c2)1/2 So what we need to do is substitute the escape velocity from the center of the Earth for Ve Escape velocity is reached when the sum of the kinetic energy and gravitational potential energy equal zero. Using specific energies: v2/23GM/2R = 0 v2/2 = 3GM/2R v2 = 3GM/R Thus: t0 =tf(13GM/Rc2)1/2 
Even if we use the fastest estimated rotation rate and the longest estimated length, you only get the equivalent of ~4/1,000,000 of a g at the ends.

I'm aware of that. I was addressing the fact that coffesippin seemed to think that we had actually determined that the object was only 1mm thick and from that hypothesized that it could be an alien craft. Whereas the reasoning goes " If we assume that the accelerations are due to it being some type of light sail, then the object would need to be very thin for that assumption to hold."

There is no measurement or observation that indicates that it is that thin. While that would be a requirement for the solar sail hypothesis to valid, this does not mean that it actually is that that thin. There nothing about the measurements we have made of it that indicates that it is made from so thin a material. You got the argument backwards. It is "IF the probe were a light sail then it would need to be very thin", not "It is very thin, so it could be a light sail".

Best places in space too place computers?
Janus replied to Question about supercomput's topic in Computer Science
Yes, you are confused. If I have a computer that does 1 trillion calculations per sec. and put it somewhere where time runs 1/2 as fast as where I am, then, for me that the computer only performs at 500 billion calculations per sec. I will have doubled my wait time, not halved it. 
Best places in space too place computers?
Janus replied to Question about supercomput's topic in Computer Science
Even if you were to remove your computer to a point as far from any mass as possible, it would only gain ~ 2/100 of a sec of computing time per year as compared to being on the surface of the Earth. You would waste much, much much more time communicating with it due to signal delay. It just isn't a viable idea. While there are places in the universe where time runs much slower than it does on the Earth ( which is the opposite of what you would want), there is no place where time runs significantly faster. 
No. The measurements I was talking about are after we have accounted for the delay due to light propagation. It what is left over after you have factored this effect out. Thus in this example, including the light propagation delay, observer 1 will visually see, clock 2 running at a rate ~1.2% as fast as his own, but after taking into account the time delay due to the increasing distance between them will conclude that clock 2 is actually running ~21% as fast as his own. You are again confusing light delay effects with Relativistic ones. If the receiver and sender are not moving relative to each other, each of them will agree as the simultaneity of events. So for example. If you you the source sends a signal to two receivers , each 1 light sec from him in opposite directions he will conclude that the light will arrive at both receivers simultaneously. Someone at either of the receivers, will also agree that the light reached both at the same time even though each of them will not actually see the signal arrive at the other receiver until 2 sec after his own signal arrived. This is not what Relativity of simultaneity means. Relativity of Simultaneity is this: As clocks 1and 2 pass the midpoint between A and B, a light is emitted from the midpoint. According to the observer that remains at the midpoint, this light travels outward at c and reaches points A and B simultaneously. However according to the observer with clock 1, this light expands outward at c from him. The distance between point A and himself is decreasing at 0.8c and the distance between point B and himself is increasing at 0.8c. Point A runs into the light before the light in the other direction catches up to point B. The light does not reach both points simultaneously. According to the observer with clock 2, the light expands outward at c relative to him, The distance to point B is decreasing at 0.8c and the distance to point A increasing at 0.8c. the light runs into point B before it catches up to point A. Again the light does not reach both points simultaneously, but the order is reversed to that which happened according to clock 1's observer.

By " agreed fixed distance" I will assume that you mean that as clock 1 and clock 2 pass each othe,r then both clocks will agree that point A( toward which clock 1 is moving) is an equal distance away as point B ( the point clock 2 is moving). The problem with your assertion, is that while both the observers will agree that both Clock 1 and 2 will have ticked off the same number of reflector hits upon reaching their respective point, they will not agree that both clocks reached those points simultaneously. Each observer will determine that his clock reached its point before the other clock did. This is result of the relativity of simultaneity. It can also be demonstrated by considering the relativistic addition of velocities. Assume that each clock is moving at 0.8 c relative to the point half way between points A and B ( the point where they pass it other). Each observer will measure the other clocks velocity relative to themselves as being (0.8c+0.8c)?(1+0.8c(0.8c)/c2) = 0.9756098c And thus the difference in velocity between the midpoint and the other clock as being 0.1756098c compared to their own velocity relative to the midpoint of 0.8c. Since we also know that each observer measures the distance between midpoint and either point A or B as being the same, he also has to conclude that he will reach his point first. Thus when he reaches his point and registers X reflector hits he also knows that the other clock will have to registered fewer than X reflector hits at that moment ( since the other clock has yet to reach its point). Thus the other clock has to have ticked slower than his clock, and he measures time dilation in the other clock.