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Everything posted by Janus

  1. 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.
  2. Fashion?

    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.
  3. Light in a vacuum

    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.
  4. 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.
  5. 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.
  6. No. If you look at the standard time dilation equation: t0 =tf(1-2GM/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(1-Ve2/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: v2/2-3GM/2R = 0 v2/2 = 3GM/2R v2 = 3GM/R Thus: t0 =tf(1-3GM/Rc2)1/2
  7. First Interstellar Asteroid Observed

    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.
  8. First Interstellar Asteroid Observed

    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."
  9. First Interstellar Asteroid Observed

    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".
  10. Best places in space too place computers?

    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.
  11. Best places in space too place computers?

    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.
  12. The Solution to Minkowski Spacetime

    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.
  13. The Solution to Minkowski Spacetime

    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.
  14. Incorrect. g ( gravitational force) is zero at the center, but the Specific(per unit mass) Gravitational potential is -3GM/2R where M is the mass of the planet, and R is its radius. At the surface of the planet (or any point above it) the Specific Gravitational potential is -GM/r where r is the distance from the center of the planet ( on the surface r=R) Specific Gravitational Potential Energy tells us how much energy it would take to move a unit mass from one point in the field to another. It takes energy to lift a mass from the center of the Earth to the surface, just like it takes energy to lift the same unit mass from the surface of the Earth to a point above it. It is the difference in potential gravitational energy is responsible for gravitational time dilation, not the difference in g. An easy way to demonstrate this is to calculate the gravitational time dilation factor for the surface of the Earth vs the surface of a planet with twice the radius and 4 times the mass of the Earth. You will get two different answers even though the value of g will be the same at the surface of both planets. A clock at the center of the Earth is at a lower potential than one on the surface and thus will run slower than one on the surface, even though it feels 0g compared to the surface clock at 1g.
  15. The Problem With Centripetal Force

    Even if this were true, (which it isn't due to the vast difference between Sagittarius A and the galaxy as a whole. Remove Sag A, and the galaxy would hardly notice it), your claim of a difference in weight for objects on the leading and trailing sides of the Earth is wrong. In this statement: The key phrase is "abruptly starts going fast". You feel pulled back due to the fact that the car is changing velocity in that direction. Once you reach speed, you no longer feel that pull. The Earth, is not abruptly changing is velocity in the direction it is traveling around the galaxy, it maintains a pretty constant pace. The only acceleration is has with respect to the galaxy center is towards the center. ( if you want to use the car example, it would like when you feel pushed to the side when you go around a sharp corner at speed.) But even then, the magnitude of that acceleration depends not only on the velocity, but the degree of "bend" in the curve. Even with the high speed that the solar system travels with respect to the center of the galaxy, the curve is so gentle, that the acceleration only works out to 0.000000000018 g (For a 175 lb person, this would result in a difference of 0.000000005 ounces). And even then, the other key word is "orbiting" Objects in orbit are in a free fall path. they may be accelerating but all parts are accelerating equally in response to gravity, so they feel no net difference between their individual parts ( other than that due to the early mentioned tidal force). So even that small acceleration in the above paragraph isn't felt. ( a falling elevator is constantly accelerating downward, but someone in it would not be pined against the ceiling, but would just float around like they were weightless)
  16. Uranium Split

    Or, As I should have probably said, the two neutrons produced by a fission event could potentially induce two more fission events.
  17. Uranium Split

    A nuclear fission reaction doesn't occur by one atom at a time. First fission event produces 2 neutrons (or sometimes 3). Each of these neutrons go on to induce fission in two other atoms. Each of these atoms produce 2 neutrons each, so now you have 4 free neutrons, each capable of inducing fission in a nucleus. So it goes something like this: 1 atom produces 2 neutrons leading to 2 atoms producing 4 neutrons (2 each) leading to 4 atoms producing 8 neutrons, leading to 8 atoms producing 16 neutrons. etc.
  18. Uranium Split

    While spontaneous fission is always occurring and in turn producing some induced fission reactions, This isn't enough to cause a chain reaction if you don't have enough uranium 235 contained in a small enough region. For pure U 235, this requires a sphere 52 kg in mass. With a smaller amount, too many of the produced neutrons make their way out of the sphere without encountering a nucleus.1 If you break the 52 kg into small pieces and separate them this has the same effect, it allows for more neutrons to escape without inducing a reaction. Bring the pieces suddenly together and you have a critical mass that will produce your nuclear explosion. 2 1 you can get away with using less if you surround the uranium with a neutron reflector which bounces the neutrons back through the mass, giving them another chance to interact with a nucleus. 2 This presents its own problems. As the pieces are brought closer together, the rate of induced fission events increases. This creates an increasing energy output from the Uranium. If they aren't brought together properly, or fast enough, this release of energy can be enough to blow the pieces apart from each other before they can get close enough to form that fully critical mass and produce the explosive fission event. The bomb will fizzle out. This type of bomb has to be designed properly so that the critical mass is formed before the individual pieces are blown apart again. It's not a matter of keeping the bomb from exploding, it's getting it to explode properly.
  19. Uranium Split

    I answered that in the post. Not all instances of U 235 fission needs to be induced by absorbing a neutron. One of the ways Uranium 235 can decay is to spontaneously undergo fission. It not the most likely way, but you have some 2.5e21 Atoms per gram of Uranium. Even with nuclear fuel only enriched to 5% U 235, you still are left with 1.25e20 nuclei; a large enough number that even with only a 2e-7% chance of any given nuclei decaying by spontaneous fission, you are going to have a good number of nuclei undergoing such fission per minute.
  20. Probably just a screw-up using the quote system, but neither of these are quotes of statements made by me.
  21. Uranium Split

    Elements high on the periodic table contain lots of neutrons and protons in their nucleus. This makes them less stable. Uranium 235 is one of those elements. Also, an excess of neutrons to protons can add to instability. Most isotopes undergo simple radioactive decay (alpha or beta) to reach a more stable state. Uranium 235 does this. It decays by alpha decay into Thorium 231. But because of the particular arrangement of neutrons and protons in its nucleus, it can also split into two smaller nuclei and some free neutrons by in an act called spontaneous fission instead. This very rare event can happen all by itself.( All radioactive decay processes are statistical by nature, and some nuclei can take more than one route. You can never say with certainty which route a particular nuclei will take, only the odds of it following one or the other*. Spontaneous fission is very low on the probability list. ) If one of the free neutrons from this fission is absorbed by another U-235 nucleus, it briefly becomes a U-236 nucleus. The U 236 nucleus is even more unstable and much more likely to decay by fission. This of course can lead to further induced fission in other nuclei. So the introduction of the neutron makes an already unstable nucleus even more unstable. And the fission inducing neutron could have been produced by the spontaneous (non-induced) fission of a U 235 Nucleus. * Even with fission, there is a small chance of breaking into three smaller nuclei rather than two.
  22. No one here is trying to rewrite the theory of Hawking radiation here except for you. You are the only one claiming that antiparticles will lower the mass of a black hole, presently accepted physics does not. If you don't want to believe in the accepted physics, that's up to you, but don't try to pass your own ideas off as accepted physics.
  23. No. A virtual particle pair forms outside the event horizon. One of them crosses the horizon, it doesn't matter which one. So for example. if it is a electron positron pair, either the electron or positron can fall past the event horizon. This leaves the over of one the pair without a partner to recombine with and you are now left with a real particle, either a electron or positron. But the energy for the creation of this particle can't just come from nowhere, do it comes from a loss of mass in the black hole. It is like the particle that falls in becomes a negative mass particle (not an antiparticle. Antimatter has positive mass just like regular mass does.)
  24. Photons have zero rest mass. But gravity doesn't rely on on rest mass alone but comes form the stress momentum tensor, of which rest mass is only one component. Light has both energy and momentum and so also produces gravity. When you convert matter to photons, you are removing the rest matter component, but increases in other components increase to compensate. 1 gram of matter could be converted into ~9e13 joules of photons, and those photons will have the same gravitational effect as the 1g of matter that created them. If this is news to you, then it means you haven't studied the subject deeply enough.
  25. Why light speed?

    So according to you, the Earth should collapse into a black hole just because some distant object is traveling at a high fraction of c relative to us. Again, relative motion will not produce a singularity in any frame. If you think it will, you are seriously misunderstanding something.