-Demosthenes- Posted January 29, 2004 Posted January 29, 2004 Okay, a man is in a train that is going 25 mph. If he runs at 5 mph in the train relative to the train he's going 5 mph. But relative to the earth he's going 30 mph (if he's running the same direction). So if light always goes the same speed, what is it relative to? If it's going the speed of light relative to the earth, then how fast is it going relative to the sun? or the other planets? How can we even classify speed at all? I'm confued, will someone talk some sence into me?
JaKiri Posted January 30, 2004 Posted January 30, 2004 -Demosthenes- said in post # :So if light always goes the same speed, what is it relative to? ALL observers. You shine a torch at me, you'll see the light going at c (if it's a vacuum). If I'm running away, I'll see the light going past me at c. And you're actually incorrect about the man on the train. He's actually going ever so slightly less than 30MPH. The equation for adding velocities isn't a + b = velocity, it's (a+b)/(1+ab/c^2)
-Demosthenes- Posted January 30, 2004 Author Posted January 30, 2004 Why is that(about the train and the man) And about the light, it's relative to all observers, you mean everyone who can see it? Does it have to be a person? Or is it relative to all things? Thanks!
JaKiri Posted January 30, 2004 Posted January 30, 2004 -Demosthenes- said in post # :Why is that(about the train and the man) And about the light, it's relative to all observers, you mean everyone who can see it? Does it have to be a person? Or is it relative to all things? Thanks! Relative to anything possible ever. Even if you 'were' a light beam, light going the other way would still be going off at the speed of light. And as to, the train and the man it's called a relativistic effect. It's a consquence of the thing about the speed of light.
JaKiri Posted January 30, 2004 Posted January 30, 2004 -Demosthenes- said in post # :Relative to everything at once? Whereever you are, whatever speed you're going at, the speed of light in a vacuum will still be 299792458ms^-1
Cap'n Refsmmat Posted January 30, 2004 Posted January 30, 2004 If you are going the speed of light, and shine a flashlight, the light will not move (relative to you!) because it is going the speed of light, and so are you, so it is going the same speed as you, and relative to you, is not moving.
JaKiri Posted January 30, 2004 Posted January 30, 2004 Cap'n Refsmmat said in post # :If you are going the speed of light, and shine a flashlight, the light will not move (relative to you!) because it is going the speed of light, and so are you, so it is going the same speed as you, and relative to you, is not moving. Nope. It'll go away from you at lightspeed.
Cap'n Refsmmat Posted January 30, 2004 Posted January 30, 2004 But to someone going at 0, it will go at lightspeed too, relative to them? But then... it would be going twice as fast as you, yet to them, as fast as you? Elaborate.
JaKiri Posted January 30, 2004 Posted January 30, 2004 Cap'n Refsmmat said in post # :But to someone going at 0, it will go at lightspeed too, relative to them? But then... it would be going twice as fast as you, yet to them, as fast as you? Elaborate. Lightspeed is the same to all observers. All rest frames are equally valid. Look at it this way; when you're going at 30mph, what are you doing it relative to? The planet. However, it's equally valid to say that the planet is going 30mph with you at rest (assuming no acceleration occurs). Another way of looking at it is... You're in space. You think you're stationary, yes? Then some other guy drifts past you. To HIM, it feels like he is stationary, and you are drifting past HIM.
Cap'n Refsmmat Posted January 30, 2004 Posted January 30, 2004 That still doesn't make sense. Your argument is true, it just doesn't explain it. The light from your flashlight will go the speed of light past you, so from the observers point of view, the light is going TWICE the speed of light. That's just not possible.
NavajoEverclear Posted January 30, 2004 Posted January 30, 2004 Light Squared? I'm also interested to the explanation for this.
iglak Posted January 30, 2004 Posted January 30, 2004 actually, what would the speed of light be relative to you if you were going the speed of light relative to a stationery observer? the speed of light is only able too look like the speed of light to all observers because all observers are going slower than the speed of light, so the light can look that speed if certain visual spacial and temporal (i think that's the word for time-ial) restrictions are put in place. but if you are going the speed of light, the exact speed of light relative to a stationery observer, the light could not look like it was going the speed of light, because it isn't going faster than you. i think you wouldn't be able to move. i think that as one gets infinitely close to the speed of light that length contraction has such a big effect that it would look to you as if you crossed the galaxy in an instant (it would just look like a very short galaxy). so, if you were going the exact speed of light, then it would appear to you as if the universe was nothing. you wouldn't be able to percieve it since it would be going by you all at once (i thought of a mathematical term for this before: "everything"). thus, you would never be able to stop. since you can't even move... in graphical terms, easier to understand: think of a line that stretches to infinity, and put markers on it at specific intervals. the faster you go, the closer the markers get together, but since the line is infinitely long, you just keep seeing it forever. as you approach the speed of light, the markers get infinitely closer, so that the line looks like 2d versions of the markers stacked on top of eachother with absolutely no space inbetween, but the line is still infinitely long. at the point you reach the speed of light, the infinite line becomes nothing. this is because you are passing by all points on the infinite line at the same time. you are passing by "everything" ("everything" can also be used to identify things divided by zero ). it is not infinitely close to zero distance, for it is infinite distance, and can never look finite. thus, it is zero, or "everything". thus, you can never stop going that speed once you start since time no longer applies to you, you pass everything all at once in zero time. <edit> P.S. i just thought of an addition to this. also, relative to a stationery observer, it would appear that you are infinitly thin as you approach the speed of light, and at the speed of light you would appear to be gone, because others would see your "everything" pass in zero time too, because of relativity. so then the question is now... do you simply cease to exist? or does each atom/sub atomic particle become a photon? P.P.S. in case you haven't noticed, "everything" only gives meaning to "undefined". without my "everything", "undefined" is misenterpereted as "impossible", and only remains undefined.
JaKiri Posted January 30, 2004 Posted January 30, 2004 iglak said in post # :the speed of light is only able too look like the speed of light to all observers because all observers are going slower than the speed of light No.
JaKiri Posted January 30, 2004 Posted January 30, 2004 Cap'n Refsmmat said in post # :The light from your flashlight will go the speed of light past you, so from the observers point of view, the light is going TWICE the speed of light. That's just not possible. You only observe velocities from your own rest frame.
YT2095 Posted January 30, 2004 Posted January 30, 2004 ok, 2 flash lights back to back, switch them on and both beams travel in opposite directions, how fast is the front of the 1`st beam moving away from the front of the other beam?
JaKiri Posted January 30, 2004 Posted January 30, 2004 YT2095 said in post # :ok, 2 flash lights back to back, switch them on and both beams travel in opposite directions, how fast is the front of the 1`st beam moving away from the front of the other beam? Assuming it's a vacuum, c.
YT2095 Posted January 30, 2004 Posted January 30, 2004 so beam A is moving away from the torch at half C?
JaKiri Posted January 30, 2004 Posted January 30, 2004 YT2095 said in post # :so beam A is moving away from the torch at half C? Nope. c. [edit] If it WAS c/2, the combined speed would only be 3/4 c.
YT2095 Posted January 30, 2004 Posted January 30, 2004 how do you work that out? or arrive at 3/4 (.75). I see it as, there`s 2 cars back to back and strating point X car A and car B take off in opposite directions from point X at 50 mph. car A is moving from car B at 100 mph, rellative to each other, is that right so far?
JaKiri Posted January 30, 2004 Posted January 30, 2004 YT2095 said in post # :how do you work that out? or arrive at 3/4 (.75). I see it as, there`s 2 cars back to back and strating point X car A and car B take off in opposite directions from point X at 50 mph. car A is moving from car B at 100 mph, rellative to each other, is that right so far? No. They're actually seperating at 99.999999999998887mph. [edit] And read my FIRST POST IN THE THREAD, I went over this.
YT2095 Posted January 30, 2004 Posted January 30, 2004 I did and it didn`t make any sense why 99.99988 blah blah, it`s 100 mph surely, it`s just basic maths, 50+50 is 100.
JaKiri Posted January 30, 2004 Posted January 30, 2004 YT2095 said in post # :I did and it didn`t make any sense why 99.99988 blah blah, it`s 100 mph surely, it`s just basic maths, 50+50 is 100. If you want to add speeds a and b (well, velocities strictly speaking) it IS NOT a + b. It's (a+b)/(1+ab/c^2)
YT2095 Posted January 30, 2004 Posted January 30, 2004 so when they callibrate the speedo in a car, that`s how they do it?
JaKiri Posted January 30, 2004 Posted January 30, 2004 YT2095 said in post # :so when they callibrate the speedo in a car, that`s how they do it? No. a+b is a fine working approximate until you reach higher velocities.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now