Roger Dynamic Motion Posted June 1, 2017 Share Posted June 1, 2017 Here is an example; of a space ship cruising at .25 second the speed of light with a source up front ''beam On.''I'm right to estimate the arrival time of the ship at 3 quarter of a second after the light; at target? Link to comment Share on other sites More sharing options...
Delta1212 Posted June 1, 2017 Share Posted June 1, 2017 Assuming the target is 1 light second away from the ship when the beam is turned on, and all of this is from the frame of the target, then yes. Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 1, 2017 Author Share Posted June 1, 2017 Assuming the target is 1 light second away from the ship when the beam is turned on, and all of this is from the frame of the target, then yes.+ one for you! + one for you!so the light, to a point by ,changing the timing, will arrive at the same time with the ship Link to comment Share on other sites More sharing options...
Delta1212 Posted June 1, 2017 Share Posted June 1, 2017 Light emitted from a ship will never arrive at its destination at the same time as the ship, assuming everything is traveling in straight lines. The light will always arrive first. The question is simply one of by how much. Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 1, 2017 Author Share Posted June 1, 2017 (edited) Light emitted from a ship will never arrive at its destination at the same time as the ship, assuming everything is traveling in straight lines. The light will always arrive first. The question is simply one of by how much.how much ? I like that ! by the way,,, how much before ? Edited June 1, 2017 by Roger Dynamic Motion Link to comment Share on other sites More sharing options...
Janus Posted June 1, 2017 Share Posted June 1, 2017 how much ? I like that ! by the way,,, how much before ? Again, it depends on who's measuring the time. For a someone at rest with respect to the target, with the ship 1 light sec from the target at the moment the light is fired, the light will take 1 sec to reach the target, And it will take 4 sec for the ship to reach the target for a difference of 3 sec. As far as the ship is concerned, the light will reach the target after ~0.7746 sec, and the ship and target will meet after ~3.873 sec, a difference of 3.098 sec. Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 1, 2017 Author Share Posted June 1, 2017 Again, it depends on who's measuring the time. For a someone at rest with respect to the target, with the ship 1 light sec from the target at the moment the light is fired, the light will take 1 sec to reach the target, And it will take 4 sec for the ship to reach the target for a difference of 3 sec. As far as the ship is concerned, the light will reach the target after ~0.7746 sec, and the ship and target will meet after ~3.873 sec, a difference of 3.098 sec. how So! << what is causing the delay in time ? Link to comment Share on other sites More sharing options...
Janus Posted June 2, 2017 Share Posted June 2, 2017 how So! << what is causing the delay in time ? In the target frame: The ship is one light sec away from the target when the light is emitted. 1 light sec is the distance light will travel in 1 sec, so it will take 1 sec for the light to make the trip. The ship is moving at 0.25c, this is 1/4 as fast as the light, so it will take 4 times longer for the ship to travel the same distance, so it arrives 4 secs after the light is first emitted or 3 sec after the light arrives. From the ship frame: Due to length contraction, the same distance that the target frame measures as 1 light sec, the ship measures as being 0.9682 light sec. The light is traveling away from the ship at c and the target is coming towards the ship at 0.25c, so they will meet in 0.9682/(1+.25)=0.7746 sec. The target, approaching at 0.25c, will take 0.9682/0.25=3.945 sec to cross the same distance. Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 3, 2017 Author Share Posted June 3, 2017 (edited) Here is an example; of a space ship cruising at .25 second the speed of light with a source up front the beam at 0 second the light comes on , ''I'm right to estimate the arrival time of the ship at 3 quarter of a second after the light; at target? Quote: Read it again, please Edited June 3, 2017 by Roger Dynamic Motion Link to comment Share on other sites More sharing options...
Janus Posted June 3, 2017 Share Posted June 3, 2017 Here is an example; of a space ship cruising at .25 second the speed of light with a source up front the beam at 0 second the light comes on , ''I'm right to estimate the arrival time of the ship at 3 quarter of a second after the light; at target? Quote: Read it again, please "a space ship cruising at .25 second the speed of light" makes no sense as written, so I assumed you meant ".25 the speed of light". If you meant something else, you are going to have to be a lot clearer as to what that is. Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 3, 2017 Author Share Posted June 3, 2017 (edited) Ok I was referring to time. not speed my error sorry . got to go Edited June 3, 2017 by Roger Dynamic Motion Link to comment Share on other sites More sharing options...
Janus Posted June 3, 2017 Share Posted June 3, 2017 Ok I was referring to time. not speed my error sorry . got to go "refering to time" doesn't help. You are going to be a lot more specific. You are talking about a ship that fires a light at a target that it, itself is also heading towards, correct? And you want to know how long after the light reaches the target the ship reaches the target, right? To answer this you need to know how fast the ship is moving relative to the target, and either how far it is from the target when it emits the light or how long it takes for the light to reach the target, and the reference frame from which these measurements are made. Link to comment Share on other sites More sharing options...
studiot Posted June 3, 2017 Share Posted June 3, 2017 (edited) Speed of light independent of the source ? This is true of all waves. That is the speed of the wave is independent of the speed of the source. Light is no different in that respect. I would add that the big breakthrough by Einstein with regard to light was the proporition that the speed is also independent of the speed of the observer. The observed speed of other waves does depend upon the speed of the observer. However this forum is entitled classical physics. Did you really mean to put the question there? Edited June 3, 2017 by studiot Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 3, 2017 Author Share Posted June 3, 2017 I rephrased it. A space ship cruising at .25 the speed of light with a source of light up front,the light is turned on at 1 second away to target (1860001 Miles away, the arrival time of the ship will be 3 quarter of a second after the light at target. I'm I right? Link to comment Share on other sites More sharing options...
studiot Posted June 3, 2017 Share Posted June 3, 2017 (edited) I rephrased it. A space ship cruising at .25 the speed of light with a source of light up front,the light is turned on at 1 second away to target (1860001 Miles away, the arrival time of the ship will be 3 quarter of a second after the light at target. I'm I right? Well I could think of circumstances in which that is true. But then I could think of circumstances in which it is untrue. But then you did not do what Janus asked. I would suggest you use several short sentences. Each sentence should contain only one statement. That would help you a great deal. Edited June 3, 2017 by studiot Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 3, 2017 Author Share Posted June 3, 2017 Studiot: thanks for the expose, but You are not telling if my quote is right Well I could think of circumstances in which that is true. But then I could think of circumstances in which it is untrue. But then you did not do what Janus asked. I would suggest you use several short sentences. Each sentence should contain only one statement. That would help you a great deal. Well I could think of circumstances in which that is true. But then I could think of circumstances in which it is untrue. But then you did not do what Janus asked. I would suggest you use several short sentences. Each sentence should contain only one statement. That would help you a great deal. Studiot: thanks for the expose, but You are not telling if my quote is right -1 Link to comment Share on other sites More sharing options...
Janus Posted June 3, 2017 Share Posted June 3, 2017 (edited) I rephrased it. A space ship cruising at .25 the speed of light with a source of light up front,the light is turned on at 1 second away to target (1860001 Miles away, the arrival time of the ship will be 3 quarter of a second after the light at target. I'm I right? Your question still doesn't add up. You say the ship is 1 sec away from the target, but as traveled by the ship, or by the light?( if by the light, then it is one light-sec from the target when the light is turned on, and my previous answer holds if this is the distance measured by the target frame). Then you go on to give a distance of 1860001 miles, which would take light ~10 sec to travel and the ship ~40 sec to travel and in no way matches up with your "one second away statement"( the only way to make these two statements consistent with each other is for the distance to be measured in the target frame, the time measured in the Ship frame and the relative speed of ship to target being much greater than 0.25c). In addition, you have still failed to specify whether this distance is being measured by someone at rest with respect to the ship or by someone at rest with respect to the target(an important piece of information when dealing with relativistic speeds.) If you meant the the ship would take 1 sec after turning on the light to reach the target as measured by someone at rest with respect to the target, then the ship would be 1/4 light sec or ~46500 miles from the target as measured by that same person. Then, as measured by that person, the light would take 1/4 from the moment it is turned on to reach the target and you would get your 3/4 of a second difference. For someone measuring this same scenario and riding in the ship, the light will take 0.1936 sec to reach the target, and the ship and target will meet 0.9682 sec after the light is turned on, meaning the ship and target meet 0.7746 sec after the light reaches the target. Again the problem is that you are not being careful enough in your description of the problem, meaning that people have to guess at what you mean. This means we can't give you a straight yes or no answer in case we've guessed wrong. Edited June 3, 2017 by Janus Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 3, 2017 Author Share Posted June 3, 2017 (edited) Your question still doesn't add up. You say the ship is 1 sec away from the target, but as traveled by the ship, or by the light?( if by the light, then it is one light-sec from the target when the light is turned on, and my previous answer holds if this is the distance measured by the target frame). Then you go on to give a distance of 1860001 miles, which would take light ~10 sec to travel and the ship ~40 sec to travel and in no way matches up with your "one second away statement"( the only way to make these two statements consistent with each other is for the distance to be measured in the target frame, the time measured in the Ship frame and the relative speed of ship to target being much greater than 0.25c). In addition, you have still failed to specify whether this distance is being measured by someone at rest with respect to the ship or by someone at rest with respect to the target(an important piece of information when dealing with relativistic speeds.). Janus ,,sorry I just realized I meet 186000 miles not 1860001 Edited June 3, 2017 by Roger Dynamic Motion Link to comment Share on other sites More sharing options...
studiot Posted June 3, 2017 Share Posted June 3, 2017 (edited) studiot But then you did not do what Janus asked. Janus ,,The post is real clear.read again Is that clear enough? Edited June 3, 2017 by studiot Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 3, 2017 Author Share Posted June 3, 2017 (edited) Your question still doesn't add up. You say the ship is 1 sec away from the target, but as traveled by the ship, or by the light?( if by the light, then it is one light-sec from the target when the light is turned on, and my previous answer holds if this is the distance measured by the target frame). Then you go on to give a distance of 1860001 miles, which would take light ~10 sec to travel and the ship ~40 sec to travel and in no way matches up with your "one second away statement"( the only way to make these two statements consistent with each other is for the distance to be measured in the target frame, the time measured in the Ship frame and the relative speed of ship to target being much greater than 0.25c). In addition, you have still failed to specify whether this distance is being measured by someone at rest with respect to the ship or by someone at rest with respect to the target(an important piece of information when dealing with relativistic speeds.). There is only one person in the ship who knows the speed of the ship .25 the speed of light and the light travel with the ship , at one second away from target. the light is turned on,,,so what i'm saying the ship will arrived 3 quarters of a second later than light. obviously. Edited June 3, 2017 by Roger Dynamic Motion Link to comment Share on other sites More sharing options...
swansont Posted June 3, 2017 Share Posted June 3, 2017 There is only one person in the ship who knows the speed of the ship .25 the speed of light and the light travel with the ship , at one second away from target. the light is turned on,,,so what i'm saying the ship will arrived 3 quarters of a second later than light. obviously. That was not clear before, and no, it wasn't obvious. The time it takes depends on who is doing the measuring. Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 3, 2017 Author Share Posted June 3, 2017 (edited) That was not clear before, and no, it wasn't obvious. The time it takes depends on who is doing the measuring.Swansont: The Astronauts Could it be also who is doing the timing ? Edited June 3, 2017 by Roger Dynamic Motion Link to comment Share on other sites More sharing options...
Janus Posted June 3, 2017 Share Posted June 3, 2017 (edited) Janus ,,sorry I just realized I meet 186000 miles not 1860001Then there is no way you can have a 3/4 sec time difference between the arrival of the light and the ship. There is only one person in the ship who knows the speed of the ship .25 the speed of light and the light travel with the ship , at one second away from target. the light is turned on,,,so what i'm saying the ship will arrived 3 quarters of a second later than light. obviously.Obviously not. If the measurements are being made by someone in the ship, and they turn on the light when they are 186,000 miles from the target, then by their reckoning it will take 186,000/(186,000+46500)*= 0.8 sec for the light to hit the target. They will meet up with the target in 186,000/46500 = 4 sec, or 3.2 sec after the light hits it. I don't know what you are doing to get 3/4 of a sec with the numbers you have given, but whatever it is, it's wrong. *The speed of light is 186,000 miles per second and 0.25 c is 46500 miles per sec. For someone in the ship the light is traveling away at 186,000 miles per sec, and the distance between themselves and the target is decreasing at a rate of 46500 miles per sec, and thus the light and target will meet when the target is 148,800 miles from the ship. The light will have traveled this distance at 186,000 miles per sec, which takes 0.8 sec. The distance between ship and target will have decreased by 37,200 miles, which at a relative velocity of 46500 miles per sec takes 0.8 sec to traverse. Edited June 3, 2017 by Janus Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 3, 2017 Author Share Posted June 3, 2017 Then there is no way you can have a 3/4 sec time difference between the arrival of the light and the ship. Obviously not. If the measurements are being made by someone in the ship, and they turn on the light when they are 186,000 miles from the target, then by their reckoning it will take 186,000/(186,000+46500)*= 0.8 sec for the light to hit the target. They will meet up with the target in 186,000/46500 = 4 sec, or 3.2 sec after the light hits it. I don't know what you are doing to get 3/4 of a sec with the numbers you have given, but whatever it is, it's wrong. *The speed of light is 182,000 miles per second and 0.25 c is 46500 miles per sec. For someone in the ship the light is traveling away at 186,000 miles per sec, and the distance between themselves and the target is decreasing at a rate of 46500 miles per sec, and thus the light and target will meet when the target is 148,800 miles from the ship. The light will have traveled this distance at 186,000 miles per sec, which takes 0.8 sec. The distance between ship and target will have decreased by 37,200 miles, which at a relative velocity of 46500 miles per sec takes 0.8 sec to traverse. you mention here that ,,*The speed of light is 182,000 miles per second . Where did you get that ? Link to comment Share on other sites More sharing options...
KipIngram Posted June 3, 2017 Share Posted June 3, 2017 I've always heard 183,000, but neither is exact. Link to comment Share on other sites More sharing options...
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