# Janus

Resident Experts

2049

27

2. ## Red/blue shift question.

But you don't need to run into faster light waves to see a doppler effect any more that you need to run into faster sound waves to hear a doppler shift. If I'm standing on the train platform on a windless day and someone on the platform yells at me, the sound travels at the speed of sound in air towards me. If a train approaches and blows its whistle the instant it passes that person, the sound from it will travel at the same speed as the yell (they will both reach me at the same time). The train whistle will be doppler shifted however.

4. ## Time Travel

To say that Ion engines "go faster" is a bit of a misnomer. They are just more efficient due to the fact that they have higher exhaust velocities than chemical rockets, which means that they can reach higher velocities with the same reaction mass to payload ratio. Oh, and by the way, Ion engines are still rockets as they use the action/reaction principle for propulsion. There is a better system on the drawing board, called the VASIMR (VArible Specific Impulse Magnetohydrodynamic Rocket) It could produce even greater exhaust velocities than the Ion engine and more thrust (producing better acceleration).
5. ## Time Travel

Um no what? Have you even done the math?

9. ## Relativity Refutation

To any observer traveling with the ship, this is what happens: No matter what direction he is facing.
10. ## Pictures of Stars

The stars are there, it is just the they are too dim to be seen with the exposure used. The main objects in the picture (planet etc) are so bright compared to the stars, that the camera setting is such that the stars don't show up. You see the same effect if you go out and look at the night sky during a moonless night compared to a night with a full moon. During a full moon you'll see a lot fewer stars than on a moonless night. If the moon were a little cloeser to the Earth, it would be bright enough that when you looked at it, you wouldn't see any stars at all.
11. ## Relativity Refutation

Here is an animation that shows what happens according to an observer in S for which the ship is moving. The white dots are the light pulses, the yellow lines are the path that the pulses follow with respect to S' the green lines show the paths the pulse's take with respect to S'(the ship}
12. ## Relativity Refutation

It does not matter whether you consider the light as being shot from the S or S', you get the same results either way. That is the whole point of Relativity. As they would be. We don't live in a Newtonian Universe, we live in a Relativistic Universe. And in a Relativistic Universe the speed of light is invariant; Meaning that all observers will measure light in a vacuum as moving at c with respect to themselves regardless of any relative motion thye may have to the source. This is the second postulate of Relativity. there is no problem here, it is just a different aspect of the fact that time is relative. In this case it is the idea of simultaneity that is relative. In S', the three pulses strike the walls at the same time
13. ## Relativity Refutation

You said: ct1 and ct2 are distances the light travels in S before it hits the sphere. because that is not what relativity predicts. If you wish to discuss Relativity you must first learn it, something you've obviously have neglected to do. I stand by my images, as they are accurate. If you cannot see that, then that is your problem. I start out with the assumption that the light is emitted at angle Alpha as seen by the inside observer. I then can calculate the path the light will take with respect to the outside obserever, based on that assumption and the velocity v and knowing the speed of light. one does not have to see something to know that it is true. I didn't have to be around 2,000,000 yrs ago to see the light leaving the Andromeda galaxy to know that the light from it reaching the earth now left it 2,000,000 yrs ago. Thus just because the Outside observer doesn't see the light traveling at angle ALPHA does not mean he cannot know that the light travels at angle alpha as seen by the inside observer. OF course the outside observer in S can see ct1 and ct2 As these are the distances the light travels as seen by him by definition!Sure, I could have determined angle BETA and worked from that, but to determine angle BETA, I have to start from angle ALPHA and determine it from that. Or conversely, I could start with Angle BETA and derive angle ALPHA from it. Either way you go,the two angle are co-dependent. $Y \sin \alpha$ is measured along a line perpendicular to the line of relative motion, Length contraction only applies to distances that are parallel to the line of relative motion. Again, if you had actually tried to learn Relativity, you would know this. $\arccos \left( \frac{v}{c} \right)$ Sure, just imagine that the spaceship is open and the light is passing through a series of glass plates. The light will slightly illuminate each plate as it passes though. If the outside observer draws a line through these spot of illumination they will follow this angle. If you can't follow the example just say so, but your inability to understand it speaks more about you than it does the example.
14. ## Relativity Refutation

Why? What prevents him from knowing it? Unless you are claiming that the sphere is opaque and that the outside observer cannot see what happens in the spaceship, in which case this whole exercise is pointless. What speed $\gamma Y \cos \alpha$? The only speeds in this example are v which can have any value <c and c, the speed of light in a vacuum.

16. ## Relativity Refutation

Of course it is the same space ship! The oval to the left is the spaceship at the instant when the light hits the mirror. The oval to the right is the Same spaceship at the instant when the light returns to the center. This is as seen by an observer outside of the ship, watching it go by at v. (to keep the image uncluttered, I omitted the spaceship at the instant the light is emitted from the center from the image) To someone in the ship, the light simply travels out at angle alpha (wrt the relative motion) for a distance Y, hits the mirror and returns along the same path to the center. Since the light returns to the center of the ship in S' (the frame of the ship), it must also return to the center in frame S (the frame in which the ship is moving at v) Thus in S' the light follows the path shown in my images. Alpha is the angle the light is launched as measured by someone traveing with the ship. Y is the radius as measured by someone traveling with the ship The length of the ship is only contracted along the direction of motion when measured by someone to which the ship has a relative motion. The ship has zero relative motion wrt to someone traveling with the ship, thus it has no length contraction for that person. To say otherwise shows a complete and utter lack of understanding of what Special Relativity is about. To refute Relativity you must first understand it. The simple fact is that Relativity is a completely self-consistant theory. It contains no mathematical or logical errors. It makes numerous predictions which have held up under every real experimental test put to it. No thought experiment you can ever dream up will ever refute it. If you think you've come up with one, all it means is that you have made an assumption that is contradictory to one of the postulates of Relativity. All you've done in this case is show that Relativity is not compatible with your assumption. This in of itself provides nothing in the way of proving which is correct, your assumption or Relatiivity. The only way to refute Relativity is to produce results from a Actual Physical Experiment that gives results that differ from that which Relativity predicts.
17. ## Relativity Refutation

In order to examine this claim that light emmited at different angles somehow changes the results let's try the the folowing exercise. Let's call the frame of the sphere frame S' and the frame in which the sphere moves at v, S In S the light travels a distance of ct1 while our sphere travels a distance of vt1. Now considering image 1: Assuming the light is emitted at at an angle of alpha from the center of the sphere as measured in S' , then the point at which the light hits the sphere is as measured in S is: $( Y \sin \alpha , vt_1+ \gamma Y \cos \alpha )$ where Y is the radius of the sphere as measured in S'. and that the center of the sphere at the instant of impact is at (0,0) $\gamma = \sqrt{1- \frac{v^2}{c^2}}$ The gamma factor is needed because the sphere is length contracted in S Thus: $c^2t_1^2 = Y^2 \sin^2 \alpha + ( vt_1 + \gamma Y \cos \alpha)^2$ $c^2t_1^2 = Y^2 \sin^2 \alpha +v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha + \gamma^2 Y^2 \cos^2 \alpha$ $c^2t_1^2 = Y^2 \sin^2 \alpha +v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha + \left( 1- \frac{v^2}{c^2} \right)Y^2 \cos^2 \alpha$ $c^2t_1^2 = Y^2 \sin^2 \alpha +v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha + \left( \cos^2 - \frac{ \cos^2 v^2}{c^2} \right)Y^2\alpha$ re-arranging: $c^2t_1^2 = v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( \sin^2 \alpha + \cos^2 - \frac{ \cos^2\alpha v^2}{c^2} \right)$ $c^2t_1^2 = v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right)$ Gathering the t1^2 terms together: $c^2t_1^2-v^2t_1^2 = 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right)$ $t_1^2 (c^2-v^2) = 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right)$ Moving all the terms to the left side: $t_1^2 (c^2-v^2) - 2 \gamma Y v t_1 \cos \alpha -Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) =0$ We get an equation that can be solved for t1 using the quadriatic equation thusly: $t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm \sqrt{ 4 \gamma^2 Y^2 v^2 \cos^2 \alpha + 4 (c^2-v^2) Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }$ $t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm \sqrt{ 4\left( 1- \frac{ v^2}{c^2} \right)Y^2 v^2 \cos^2 \alpha -+4 (c^2-v^2) Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }$ $t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm 2 Y \sqrt{ \left( 1- \frac{ v^2}{c^2} \right) v^2 \cos^2 \alpha + (c^2-v^2) \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }$ $t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm 2 Y \sqrt{ \left( 1- \frac{ v^2}{c^2} \right) v^2 \cos^2 \alpha + \frac{(c^2-v^2)}{c^2} c^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }$ $t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm 2 Y \sqrt{ \left( 1- \frac{ v^2}{c^2} \right) v^2 \cos^2 \alpha + \left( 1- \frac{ v^2}{c^2} \right) \left( c^2- \cos^2\alpha v^2 \right) } }{2(c^2-v^2) }$ $t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ v^2 \cos^2 \alpha + \left( c^2- v^2 \cos^2\alpha \right) } }{(c^2-v^2) }$ $t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ v^2 \cos^2 \alpha + c^2- v^2 \cos^2\alpha } }{(c^2-v^2) }$ $t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ c^2 }} {(c^2-v^2) }$ $t_1 = \frac{ \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {(c^2-v^2) }$ Now consider image 2 which shows the path the light takes on its return to the center of the sphere: Here we can see the light travels a distance of ct2 in frame S, while the sphere moves a distance of vt2. As a result we can see that: $c^2 t_2^2 = Y^2 \sin^2 \alpha + ( \gamma Y \cos \alpha - vt_2 )^2$ and following the same steps as above we find that solving for t2 we get: $t_2 = \frac{- \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 }$ The total time T it takes for the light to leave the emitter and return is $T = t_1 + t_2$ Thus $T=\frac{ \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {(c^2-v^2) } + \frac{- \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 }$ $T=\frac{ \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}} - \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 }$ $T=\frac{ \pm cY \sqrt {1- \frac{ v^2}{c^2}} \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 }$ atr this point we have a choice to make. There is only one possible answer for T, but there are four ways to solve this equation due to the "plus or minus" signs. Luckily we can automatically elliminate 3. If both are minus we end up with a negative time for an answer, so that doesn't work. And if one is minus and the other plus we end up with a answer of 0 for T and that makes no sense. this leaves us with both being plus giving us: $T=\frac{ cY \sqrt {1- \frac{ v^2}{c^2}} + cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 }$ $T=\frac{ 2cY \sqrt {1- \frac{ v^2}{c^2}} } {c^2-v^2 }$ Now let's relate T to T' the total round trip time in S' In S' the light travels a distance of Y to the sphere and then travels a distance of Y back again. this gives us: $T' =\frac{ 2Y}{c}$ and $\frac{T'}{T}= \frac{\frac{ 2Y}{c}}{\frac{ 2cY \sqrt {1- \frac{ v^2}{c^2}} } {c^2-v^2 } }$ $\frac{T'}{T}= \frac{ 2Y}{c}} \frac {c^2-v^2 }{ 2cY \sqrt {1- \frac{ v^2}{c^2}} }$ $\frac{T'}{T}= \frac {c^2-v^2 }{ c^2 \sqrt {1- \frac{ v^2}{c^2}} }$ divide top and bottom by c²: $\frac{T'}{T}= \frac {1- \frac{v^2}{c^2} }{ \sqrt {1- \frac{ v^2}{c^2}} }$ simplify and re-arrange: $T = \frac{T'}{\sqrt{1- \frac{v^2}{c^2}}}$ And we see that no matter what angle alpha the light is aimed, the stardard time dilation holds, thus falsifying the claim.
18. ## Different approach to E=MC2

$c^2 t_2^2 = Y^2 \sin^2 \alpha + ( \gamma Y \cos \alpha - vt_2 )^2$ $t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ c^2 }}}{(c^2-v^2)}$ $t_1 = \frac{\gamma v Y \cos \alpha \pm cY\sqrt{1-\frac{v^2}{c^2}}}{(c^2-v^2)}$
19. ## War of the Worlds!!

Definitely read the book, if for no other reason than a sense of history. Written in 1898 by H.G. Wells, a man ahead of his time who more or less invented many of the "themes" of modern SF. (Alien Invasion, time travel, antigravity, invisibility)

21. ## Twin Paradox (yet again!)

Yes it can. According to brother B, it is only the relative velocity that causes A to age less.

23. ## A distinction to be made in Einstein's Relativity

But that reference point is can only be consider "stationary" for convenience's sake we can know nothing about our absolute motion through space A speed through that distance is absolute. Since there is no absolute reference to judge that speed by, speed can never be absolute.
24. ## A distinction to be made in Einstein's Relativity

Sorry, but you are mistaken. This is easily demonstrated every day with particle accelerators. The Earth itself accelerates as it orbits the Sun, thus the Earth's orbital velocity of 30km/sec would be added and substracted to the particle's final velocity as it leaves the accelerator during different points of its orbit. Today's accelerators can reach velocities close enough to c that this 60 km/sec variation would cause a noticeable difference in relativistic effects seen in those particles over the course of a year. No such variations are ever noticed.

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