Phi for All Posted March 17, 2005 Posted March 17, 2005 Where did you get that idea from? "Accelerate all parts of the body equally' date=' with no significant internal stresses" this seems correct.[/quote']But so far it isn't science, just science fiction. "Inertial dampeners" like Star Trek had would keep your organs from being crushed against your spine under such conditions. No matter how well you secured the body, we're still mostly fluid inside.
J.C.MacSwell Posted March 17, 2005 Posted March 17, 2005 Where did you get that idea from? "Accelerate all parts of the body equally' date=' with no significant internal stresses" this seems correct.[/quote'] Just a "blurt". I see no way of achieving it except in a "science fiction" sense as Phi pointed out. (alignment of all electron spins, Yadda, Yadda...) How did the sub propulsion system work? Or maybe charging the bodies and have an electrostatic field within the ship. (Careful not to electrocute them. )
Sayonara Posted March 17, 2005 Posted March 17, 2005 it's impossible for the speed of any particular object to be the same in all frames, I don't have to bother reading anything to understand that. How about understanding that it appears the same to observers in those frames. Stop taking this thread off-topic.
Johnny5 Posted March 17, 2005 Posted March 17, 2005 Just a "blurt". I see no way of achieving it except in a "science fiction" sense as Phi pointed out. (alignment of all electron spins' date=' Yadda, Yadda...)How did the sub propulsion system work? Or maybe charging the bodies and have an electrostatic field within the ship. (Careful not to electrocute them. )[/quote'] I thought about your "blurt" last night. Answer the following question for me if you can. You said create a "human magnet." I didn't understand that yesterday, nor do I understand it now. But here is my question, we can move it to another thread if necessary: In whatever you were conceiving a "human magnet" to be, suppose what you are thinking of has been done inside a hollow shell. Here is my question: The ship is accelerating at 100g. The human magnet is holding an ordinary pencil. How much does the pencil weigh? Or to put this another way, suppose there is an ordinary wooden table inside the shell. The human magnet walks around inside the shell just fine. The shell is accelerating at 100g. Does the table snap to smithereens? In other words, is it just the human being we have to turn into a magnet, or the way this idea works, must everything inside the shell be turned into a magnet? The reason I ask this is obvious. If all objects inside the shell would have a weight one thousands times that on earth, then everything would be crushed, except the human magnet, which means this approach won't work. PS: Electric field inside an electrostatically charged spherical shell is zero. Use Gauss' law. See Faraday cage.
J.C.MacSwell Posted March 17, 2005 Posted March 17, 2005 How about understanding that it appears the same to observers in those frames. Stop taking this thread off-topic. Are you saying the speed of light is constant in all frames' date=' or do you mean just the ones that scientists agree that the speed of light is constant in? Or are you perhaps [b']defining[/b] a "frame" as that which the speed of light is constant in? Just a suggestion as I think this is a great forum with some very knowledgable and smart people on it (yourself included) but if we had a list of agreed definitions it could save us some grief and keep us on topic. I know I am as guilty as anyone in using the wrong terminology at times but I do not do it intentionally and try to "flag" it in some way if I'm unsure. Outside of inertial frames I cannot think of any others offhand that the speed of light is constant in.
J.C.MacSwell Posted March 17, 2005 Posted March 17, 2005 I You said create a "human magnet." I didn't understand that yesterday' date=' nor do I understand it now. But here is my question, we can move it to another thread if necessary: I.[/quote'] I didn't understand that yesterday, nor do I understand it now either. Open up a thread in pseudo science. We can discuss it there.
Johnny5 Posted March 17, 2005 Posted March 17, 2005 I didn't understand that yesterday' date=' nor do I understand it now either. Open up a thread in pseudo science. We can discuss it there.[/quote'] Ok done. It's called appropriately enough, "What is a human magnet?"
Sayonara Posted March 17, 2005 Posted March 17, 2005 Do you want me to move the relevant posts over to the new thread?
Johnny5 Posted March 17, 2005 Posted March 17, 2005 Do you want me to move the relevant posts over to the new thread? Yeah that's a good idea.
Sayonara Posted March 17, 2005 Posted March 17, 2005 OK - I'm about to leave work so I'll do it when I get home. They'll go in by order of post time, so they'll be in the right order.
5614 Posted March 17, 2005 Posted March 17, 2005 That's a bit better, what is an inertial observer? Inertial Observer: An observer occupying an inertial frame of reference. Inertial Reference Frame: A reference frame in which a free particle experiences no force. Frame of Reference: The coordinate system to which a particular observer refers his or her measurements. ========= About the whole c=constant thing I think Johnny5's point is that if c = 299,792,458 metres per second and I am not moving that is fine. If I accelerate to 100mph then c is still 299,792,458 metres per second relative to me. The fact that I accelerated and light remained the same speed relative to me means that light must have changed it's speed too. That is what I think Johnny5 is trying to say.
Johnny5 Posted March 17, 2005 Posted March 17, 2005 Inertial Reference Frame: A reference frame in which a free particle experiences no force. The definition above is practically meaningless. A free particle is a particle which isn't experiencing any force, so its as if you said, a free particle is a free particle. Although quibbling about this is dumb. Additionally, what if there are no free particles in nature... but that's not even my main objection. The definition of inertial reference frame is so important, that something more technical is required. Why didn't you say the following: An inertial reference frame is a reference frame in which a free particle isn't accelerating. Just curious. PS: I know that's not your definition, its found all over the web. Kind regards
5614 Posted March 17, 2005 Posted March 17, 2005 The definition above is practically meaningless. A free particle is a particle which isn't experiencing any force, so its as if you said, a free particle is a free particle. Which is precisely why when I said: The second postulate of special relativity is that c is the same for all inertial observers or[/b'] the speed of light is measured as constant in all frames of reference. It had effectively an identical meaning.
Johnny5 Posted March 17, 2005 Posted March 17, 2005 Which is precisely why when I said: Originally Posted by 5614 (me!!!!) The second postulate of special relativity is that c is the same for all inertial observers or the speed of light is measured as constant in all frames of reference. It had effectively an identical meaning. You know something, why don't you spell out for me what you mean when you say frame of reference. I saw the definition you spewed out. From memory, that means I am not going back to look at what you wrote, it said, "A frame of reference is a system of coordinates to which an observer refers to his/her measurements." Something like that. Is that what you mean by reference frame, when you say "the speed of light is c in all reference frames?" I get the first part of your OR correctly. The speed of light is c for all inertial observers. Why don't you just come out and say that an inertial reference frame is a reference frame in which a free particle is either at rest or moving in a straight line at a constant speed.
swansont Posted March 17, 2005 Posted March 17, 2005 I already did define it. Any three non-collinear points in S are moving in a straight line at a constant speed in reference frame S. So if there was a solid body in frame S`' date=' and the three points were on the surface of the body, then the body isn't rotating in reference frame S`. And if that body was being viewed in reference frame S, then those three points on the surface of that body would be moving in straight lines at constant speeds in S, and the three points would be at rest in S`. Was that clear enough?[/quote'] No, I need a mathematical transform, and it needs to conform to the laws of physics.
Johnny5 Posted March 17, 2005 Posted March 17, 2005 No, I need a mathematical transform, and it needs to conform to the laws of physics. Do you know why I chose three non-collinear points?
Johnny5 Posted March 18, 2005 Posted March 18, 2005 An inertial observer can never be in a frame where the photon is at rest This is untrue. Suppose that S is an inertial reference frame. Define the frame S` to be a coordinate system with a photon at the origin' date=' and let the axes not be rotating in S. S` is also an inertial frame, and the photon is at rest in it. [/quote'] Go ahead and define it. What is the transformation that gets you from S to S'? Originally Posted by Johnny5 I already did define it. Any three non-collinear points in S are moving in a straight line at a constant speed in reference frame S. So if there was a solid body in frame S`, and the three points were on the surface of the body, then the body isn't rotating in reference frame S`. And if that body was being viewed in reference frame S, then those three points on the surface of that body would be moving in straight lines at constant speeds in S, and the three points would be at rest in S`. Was that clear enough? No, I need a mathematical transform, and it needs to conform to the laws of physics. Ok let S denote an inertial reference frame. Now, let a photon be traveling along the x axis of S, in the direction of increasing x coordinates. Now, let S` denote a reference frame in which the photon is at rest. Let the origin of S` be the location of the photon. Let S and S` be in standard configuration. That means the positive x rays of both systems are parallel and point in the same direction, and the positive y rays are parallel and point in the same direction, and the positive z rays are parallel and point in the same direction. Let the origins of both frames coincide when t=t`=0. Here are the Galilean transforms: x`= x-ct y`=y z`=z t`=t So the axes of S` aren't spinning in S. From which it follows that S` is also an inertial reference frame. Kind regards PS: If this is insufficient, tell me what your objection is.
J.C.MacSwell Posted March 18, 2005 Posted March 18, 2005 Ok let S denote an inertial reference frame.Now' date=' let a photon be traveling along the x axis of S, in the direction of increasing x coordinates. Now, let S` denote a reference frame in which the photon is at rest. Let the origin of S` be the location of the photon. Let S and S` be in standard configuration. That means the positive x rays of both systems are parallel and point in the same direction, and the positive y rays are parallel and point in the same direction, and the positive z rays are parallel and point in the same direction. Let the origins of both frames coincide when t=t`=0. [b']Here are the Galilean transforms:[/b]x`= x-ct y`=y z`=z t`=t So the axes of S` aren't spinning in S. From which it follows that S` is also an inertial reference frame. Kind regards PS: If this is insufficient, tell me what your objection is. Galileo was a genius. He tried to measure the speed of light by having an assistant on a distant hilltop "signal him back" to measure the speed of light. If you try this today they lock you up.
swansont Posted March 18, 2005 Posted March 18, 2005 Here are the Galilean transforms: x`= x-ct y`=y z`=z t`=t So the axes of S` aren't spinning in S. From which it follows that S` is also an inertial reference frame. Kind regards PS: If this is insufficient' date=' tell me what your objection is.[/quote'] My ojection is that the transform doesn't follow the observed laws of physics.
5614 Posted March 18, 2005 Posted March 18, 2005 You know something, why don't you spell out for me what you mean when you say frame of reference Frame of reference: is a set of coordinates which allow an observer to measure aspect, position and motion of another frame relative to the reference frame. Inertial frame of reference: A nonaccelerating frame of reference in which Newton's first law applies. Why don't you just come out and say that an inertial reference frame is a reference frame in which a free particle is either at rest or moving in a straight line at a constant speed. OK, I just have! (If it's nonaccelerating frame where newton's 1st law applies it must be travelling in a straight line at a constant speed.)
Johnny5 Posted March 18, 2005 Posted March 18, 2005 My ojection is that the transform doesn't follow the observed laws of physics. That's a good objection, if it's correct. What is wrong with the transformations then?
Johnny5 Posted March 18, 2005 Posted March 18, 2005 Inertial frame of reference: A nonaccelerating frame of reference in which Newton's first law applies. [/i] What exactly is a "non-accelerating frame"? I guess here is what I want to know. Is an inertial reference frame one in which Newton's first law is true, or one in which all three of his laws are true. Which do you say it is?
5614 Posted March 18, 2005 Posted March 18, 2005 A non accelerating frame would be a frame which is moving at a constant speed (be it 0mph or 10mph, constant). Well if Newton's 1st law applies, well f=ma doesn't include vectors, so I suppose you could argue that f=ma didnt always apply (if you were talking about a vector), I suppose the 3rd law always applies.
swansont Posted March 18, 2005 Posted March 18, 2005 That's a good objection, if it's correct. What is wrong with the transformations then? Galilean transformations don't work. This becomes apparent at high speeds.
Johnny5 Posted March 18, 2005 Posted March 18, 2005 Galilean transformations don't work. This becomes apparent at high speeds. How did you draw the conclusion that the Galilean transformations do not work?
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