einsteinium Posted March 3, 2009 Share Posted March 3, 2009 I was was just wondering about what would happen if a signal was being transmitted toward us from a transmitter moving at 0.5C. Could the final velocity of the signal be 1.5C. Einstein did state that no matter how fast you are moving, if you turn a lamp on the light will still MOVE AWAY from you at 3E08 m/s so if the transmitter was moving at 0.5C then the signal being transmitting would move away from it at 1.0C giving the signal a resultant velocity of 1.5C relative to your velocity. If this were to happen then the wavelength will be two thirds its original making its frequency 1.5 times the original, kind of like how the sound waves coming from an emergency vehicle sound different as it passes by. Just have a think and you get an idea about what I'm talking about. Link to comment Share on other sites More sharing options...
swansont Posted March 3, 2009 Share Posted March 3, 2009 The speed of light is independent of the speed of the source or receiver. What will happen, as you imply, is a Doppler shift, but the speed will be the same. Link to comment Share on other sites More sharing options...
einsteinium Posted March 3, 2009 Author Share Posted March 3, 2009 The speed of light is independent of the speed of the source or receiver. What will happen, as you imply, is a Doppler shift, but the speed will be the same. That doesn't make sense...I give you an example of what I'm talking about: Say we had two people with identicle guns and both had to fire from the same spot, one is standing still on the ground and the other was standing on top of a car moving at full speed. Now whos bullet will have the greatest impact velocity with the target? answer: The guy on the car U1 + U2 = V Now lets exchange the guns with radio transmitters now the guy on the moving vehicle will reach the reciever quicker because his initial velocity is added to the initial velocity of the photons moving away from him. Link to comment Share on other sites More sharing options...
npts2020 Posted March 3, 2009 Share Posted March 3, 2009 That doesn't make sense...I give you an example of what I'm talking about: Say we had two people with identicle guns and both had to fire from the same spot, one is standing still on the ground and the other was standing on top of a car moving at full speed. Now whos bullet will have the greatest impact velocity with the target? answer: The guy on the car U1 + U2 = V Now lets exchange the guns with radio transmitters now the guy on the moving vehicle will reach the reciever quicker because his initial velocity is added to the initial velocity of the photons moving away from him. You are equating mass with photons. Until the time when some mechanism converts one to the other, they behave in very different manners. Link to comment Share on other sites More sharing options...
Sisyphus Posted March 3, 2009 Share Posted March 3, 2009 What you're talking about is the reason relativity comes into play. No matter what the source is, light always moves at the same speed relative to anything else. If you're moving towards me at 0.5C and you turn on a flashlight pointed towards me, the beam will move away from you at C from your perspective and move towards me at C from my persective. From your persective, the beam will seem to move towards me at 1.5C, and from my perspective, the beam will seem to move away from you at 0.5C. This absolutely seems contradictory at first glance, but nonetheless it is what happens. The contradiction is resolved when you account for special relativity, in which it turns out that time and distance are not constant but depend on one's frame of reference. Link to comment Share on other sites More sharing options...
swansont Posted March 3, 2009 Share Posted March 3, 2009 That doesn't make sense...I give you an example of what I'm talking about: Say we had two people with identicle guns and both had to fire from the same spot, one is standing still on the ground and the other was standing on top of a car moving at full speed. Now whos bullet will have the greatest impact velocity with the target? answer: The guy on the car U1 + U2 = V Now lets exchange the guns with radio transmitters now the guy on the moving vehicle will reach the reciever quicker because his initial velocity is added to the initial velocity of the photons moving away from him. This is a reason why relativity was not discovered until after the behavior of electromagnetic waves was described by Maxwell's equations. The solution to the wave equation requires that the speed be the same in all reference frames. Your reasoning is not unsound, but assumes the world behaves according to Galilean coordinate transforms. It's not obvious at low speeds, but that turns out to be wrong. Link to comment Share on other sites More sharing options...
cameron marical Posted March 4, 2009 Share Posted March 4, 2009 If you're moving towards me at 0.5C and you turn on a flashlight pointed towards me, the beam will move away from you at C from your perspective and move towards me at C from my persective. From your persective, the beam will seem to move towards me at 1.5C, and from my perspective, the beam will seem to move away from you at 0.5C. i dont understand why it wouldnt be the other way around. wouldnt the light seem to move away from the projecter at 1.5C and the light will seem to go towards the target at just normal C, therefore light should travel faster than usuall. .5 faster. i know this is contradictory to relativity, but im just trying to grip this. Link to comment Share on other sites More sharing options...
swansont Posted March 4, 2009 Share Posted March 4, 2009 i dont understand why it wouldnt be the other way around. wouldnt the light seem to move away from the projecter at 1.5C and the light will seem to go towards the target at just normal C, therefore light should travel faster than usuall. .5 faster. i know this is contradictory to relativity, but im just trying to grip this. Any observer in an inertial reference frame can consider him/herself to be at rest. You always measure the speed of light to be c, relative to you and your reference frame. Link to comment Share on other sites More sharing options...
alan2here Posted March 13, 2009 Share Posted March 13, 2009 Am I right in thinking that the man moving verry fast on the train will experance space-time compression and this will strech out what is in front of him to make things consistant such that the light originating neer him apears to be behaving corectly? Link to comment Share on other sites More sharing options...
swansont Posted March 14, 2009 Share Posted March 14, 2009 Am I right in thinking that the man moving verry fast on the train will experance space-time compression and this will strech out what is in front of him to make things consistant such that the light originating neer him apears to be behaving corectly? It will compress what is in front of him, in the direction of movement. He will think the trip was shorter. This will reconcile the observation from a stationary observer that the train's clock was running slow. Link to comment Share on other sites More sharing options...
Janus Posted March 14, 2009 Share Posted March 14, 2009 That doesn't make sense...I give you an example of what I'm talking about: Say we had two people with identicle guns and both had to fire from the same spot, one is standing still on the ground and the other was standing on top of a car moving at full speed. Now whos bullet will have the greatest impact velocity with the target? answer: The guy on the car U1 + U2 = V Now lets exchange the guns with radio transmitters now the guy on the moving vehicle will reach the reciever quicker because his initial velocity is added to the initial velocity of the photons moving away from him. Actually, the correct expression is [math]\frac{U_1+ U_2}{1+\frac{U_1U_2}{c^2}} =V[/math] Where c is the speed of light in a vacuum. When U1 and U2 are small when compared to c then the answer does come out to be close to being equal to U1+U2. However, if either equals c, you get: [math]\frac{U_1+ c}{1+\frac{U_1c}{c^2}} =V[/math] [math]\frac{U_1+ c}{1+\frac{U_1}{c}} =V[/math] Which gives an answer of c for any value of U1 Link to comment Share on other sites More sharing options...
simplejack Posted March 21, 2009 Share Posted March 21, 2009 Suppose Bob and John are in a box that is moving through space at the speed of light. Bob is at the front of the box, and John is at the back of the box. To them it feels as though they are not moving at all, but would Bob be able to see John who is at the back of the box since they are moving at the speed of light? Link to comment Share on other sites More sharing options...
iNow Posted March 21, 2009 Share Posted March 21, 2009 Suppose Bob and John are in a box that is moving through space at the speed of light. Bob is at the front of the box, and John is at the back of the box. To them it feels as though they are not moving at all, but would Bob be able to see John who is at the back of the box since they are moving at the speed of light? It's a meaningless question. Bob and John have mass, ergo your first sentence is complete fantasy, as it's impossible for any object with non-zero mass to travel at the speed of light. Once you suspend the laws of physics, you can do damn well anything you want. The point being, in order for question to warrant an answer, you must suspend the laws of physics, and when you do that, you can give any answer you want and still be correct. You may as well be asking if purple unicorns cause flatulence in leprechauns. It's a meaningless question. Link to comment Share on other sites More sharing options...
Duration Posted March 21, 2009 Share Posted March 21, 2009 (edited) Any observer in an inertial reference frame can consider him/herself to be at rest. You always measure the speed of light to be c, relative to you and your reference frame. I am traveling at .5c towards you. I turn on a light aimed at you. In one second, how far did the light travel? 186,000 miles? OK How far did I travel in that same second, 93,000 miles? Good. The light is ahead of me 93,000 miles in one second. The light is 186,000 miles closer to you in one second. I am 93,000 miles closer to you. The light appears to be traveling at .5c away from me from my perspective, as in the one second duration, the light goes from me to 93,000 miles in front of me. That means my perception (illusion) is that the light is traveling at .5c. The light is coming towards you at c, BUT, depending on how far away I am from you when I turn the light on, you will not even see the light until it arrives at you, which is a delay from when I turn on the light and when you perceive the light was turned on. During that delay (which is proportional to the distance I am away from you), I traveled a distance, and the light traveled a distance. Your perception and my perception are wrong, they are both illusions that don't represent the reality that the light traveled 186,000 miles in one second, and I traveled 93,000 miles in that same second. You don't even see the light until it hits you, so your observation doesn't begin until the light hits you, which at that point the light is not a distance away from you. If the light continues on its path beyond you, you can then observe the velocity of the light traveling away from you while I continue to travel towards you. Edited March 21, 2009 by Duration Link to comment Share on other sites More sharing options...
cameron marical Posted March 21, 2009 Share Posted March 21, 2009 Suppose Bob and John are in a box that is moving through space at the speed of light. Bob is at the front of the box, and John is at the back of the box. To them it feels as though they are not moving at all, but would Bob be able to see John who is at the back of the box since they are moving at the speed of light? i think that the box is moving, and of course their in it, but their seperate referance frames in the box themselves also, so the light can move at normal predicted speeds because their referance frames are not directly traveling at the speed of light, the box their in is. Link to comment Share on other sites More sharing options...
NowThatWeKnow Posted March 21, 2009 Share Posted March 21, 2009 (edited) I am traveling at .5c towards you. I turn on a light aimed at you. In one second, how far did the light travel? 186,000 miles? OK How far did I travel in that same second, 93,000 miles? Good. The light is ahead of me 93,000 miles in one second. The light is 186,000 miles closer to you in one second. I am 93,000 miles closer to you. The light appears to be traveling at .5c away from me from my perspective, as in the one second duration, the light goes from me to 93,000 miles in front of me. That means my perception (illusion) is that the light is traveling at .5c. ... You will see the light moving away from you at 186,000 miles per second. I think that what you are not considering is time dilatation. Relative to your starting point it will seem to you that the light is moving at 1.5c but your clock is running slower then the person it is pointed at. 1 second for you will be more then one second for the other person so relative to their time light is moving at c. Edited March 21, 2009 by NowThatWeKnow Link to comment Share on other sites More sharing options...
alan2here Posted March 21, 2009 Share Posted March 21, 2009 Suppose Bob and John are in a box that is moving through space at the speed of light. Bob is at the front of the box, and John is at the back of the box. To them it feels as though they are not moving at all, but would Bob be able to see John who is at the back of the box since they are moving at the speed of light? Sorry about whoever that last person to try and answer the question was. If they are moving verry close to the speed of light inside there box light from there point of view inside the box will behave normally. There will be an equal delay for each person seeing the other which will be the usual verry short one. -1 Link to comment Share on other sites More sharing options...
Klaynos Posted March 21, 2009 Share Posted March 21, 2009 I am traveling at .5c towards you. I turn on a light aimed at you. In one second, how far did the light travel? 186,000 miles? OK How far did I travel in that same second, 93,000 miles? Good. The light is ahead of me 93,000 miles in one second. The light is 186,000 miles closer to you in one second. I am 93,000 miles closer to you. The light appears to be traveling at .5c away from me from my perspective, as in the one second duration, the light goes from me to 93,000 miles in front of me. That means my perception (illusion) is that the light is traveling at .5c. The light is coming towards you at c, BUT, depending on how far away I am from you when I turn the light on, you will not even see the light until it arrives at you, which is a delay from when I turn on the light and when you perceive the light was turned on. During that delay (which is proportional to the distance I am away from you), I traveled a distance, and the light traveled a distance. Your perception and my perception are wrong, they are both illusions that don't represent the reality that the light traveled 186,000 miles in one second, and I traveled 93,000 miles in that same second. You don't even see the light until it hits you, so your observation doesn't begin until the light hits you, which at that point the light is not a distance away from you. If the light continues on its path beyond you, you can then observe the velocity of the light traveling away from you while I continue to travel towards you. WARNING FRAME MIXING!!! You cannot just jump between rest frames like you have done here using the same distances and times, you need to apply Lorentz transformations. What can be stated though, the light will be measured to be moving at c in both frames. If it perfectly legitimate to state that in your description above the light source is stationary and the detector is moving, if you then work through your logic again you will get different conclusions, the logic does not match reality. Link to comment Share on other sites More sharing options...
Duration Posted March 21, 2009 Share Posted March 21, 2009 (edited) WARNING FRAME MIXING!!! You cannot just jump between rest frames like you have done here using the same distances and times, you need to apply Lorentz transformations. What can be stated though, the light will be measured to be moving at c in both frames. If it perfectly legitimate to state that in your description above the light source is stationary and the detector is moving, if you then work through your logic again you will get different conclusions, the logic does not match reality. Could you explain to me how you obtained the measurements of the distance I traveled, and the distance the light traveled in that one second, to arrive at my velocity and the lights velocity? Merged post follows: Consecutive posts mergedYou will see the light moving away from you at 186,000 miles per second. How so? I know the light traveled at a velocity of 186,000 miles per second, but what I observe is that in the duration of 1 second, the light is 93,000 miles in front of me, because I also traveled 93,000 miles in that same second. Being alert to the problems of illusions, I quickly pull out my pocket calculator and calculate where I first turned the light on (93,000 miles ago), and I calculate the distance the light is in front of me one second later, which is 93,000 miles, and get a total distance the light traveled in that one second duration, which means the light must have traveled 186,000 miles in that one second, BUT, the fact remains, the light is 93,000 miles in front of me at the end of that one second duration. Illusion for me, not a reality. All I observe is that the light is 93,000 miles in front of me one second after I turned it on. If I disregard the 93,000 miles I traveled in the calculations, that is my mistake. I should have been smart enough to realize the light actually traveled 186,000 miles in that one second, because it started its journey 93,000 miles behind me and is 93,000 miles in front of me one second later. Edited March 21, 2009 by Duration Consecutive posts merged. Link to comment Share on other sites More sharing options...
Klaynos Posted March 21, 2009 Share Posted March 21, 2009 First of all you need to pick a rest frame and stick with it. You then have to apply the relativistic velocity sum equation: To save reinventing the wheel, it'd be best if you read: http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html In the equation if you put in the summation of 0.5c and c you will end at the velocity c, so photons are always measured to be travelling at c in any rest frame. Link to comment Share on other sites More sharing options...
swansont Posted March 21, 2009 Share Posted March 21, 2009 Could you explain to me how you obtained the measurements of the distance I traveled, and the distance the light traveled in that one second? No, nobody has specified the details about how the actual measurement was done. The answer was what you would get if you did the measurement properly. What Klaynos was pointing out was that by mixing the frames, you cannot get the correct answer, regardless of what the details are. How so? I know the light traveled at 186,000 miles per second, but what I observe is that in the duration of 1 second, the light is 93,000 miles in front of me. No, it's not. It's 186,000 miles in front of you. But you can only measure that against a marker that's in your frame of reference. You are describing what an outside observer would measure — light traveling at c, and you moving at c/2. IOW, if your ship was 186,000 miles long, it would take 1 second for the light to get from the front to the back (or vice versa), as long as you were not accelerating. Link to comment Share on other sites More sharing options...
Duration Posted March 21, 2009 Share Posted March 21, 2009 First of all you need to pick a rest frame and stick with it. You then have to apply the relativistic velocity sum equation: To save reinventing the wheel, it'd be best if you read: http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html In the equation if you put in the summation of 0.5c and c you will end at the velocity c, so photons are always measured to be travelling at c in any rest frame. Could you stick to my scenario so I can see where I go wrong? Thanks. Link to comment Share on other sites More sharing options...
Klaynos Posted March 21, 2009 Share Posted March 21, 2009 How so? I know the light traveled at a velocity of 186,000 miles per second, but what I observe is that in the duration of 1 second, the light is 93,000 miles in front of me, No, you don't in your rest frame you are not moving, so you observe the light to have travelled 186,000 miles. Only an external observer in a rest frame in which you are travelling at 0.5c would observe you to have moved 93,000miles. This is the misunderstanding that is resulting in the problem. because I also traveled 93,000 miles in that same second. Being alert to the problems of illusions, I quickly pull out my pocket calculator and calculate where I first turned the light on (93,000 miles ago), and I calculate the distance the light is in front of me one second later, which is 93,000 miles, and get a total distance the light traveled in that one second duration, which means the light must have traveled 186,000 miles in that one second, BUT, the fact remains, the light is 93,000 miles in front of me at the end of that one second duration. Illusion for me, not a reality. All I observe is that the light is 93,000 miles in front of me one second after I turned it on. If I disregard the 93,000 miles I traveled in the calculations, that is my mistake. I should have been smart enough to realize the light actually traveled 186,000 miles in that one second, because it started its journey 93,000 miles behind me and is 93,000 miles in front of me one second later. Merged post follows: Consecutive posts mergedCould you stick to my scenario so I can see where I go wrong? Thanks. You need to understand the equations to see how it works. This is not easy, or intuitive. Merged post follows: Consecutive posts mergedTime and distance are not measured the same by people in different rest frames. Link to comment Share on other sites More sharing options...
Duration Posted March 21, 2009 Share Posted March 21, 2009 (edited) No, you don't in your rest frame you are not moving, so you observe the light to have travelled 186,000 miles. You already told me I am traveling at .5c, correct? Am I supposed to disregard that information when I perform the calculations? How could I have made an observation of the light traveling 186,000 mi/sec when it is clearly only 93,000 miles in front of me after one second? If light travels at a velocity of 186,000 mi/sec, and it is 186,000 miles away from me after a one second duration, that means the light accelerated at a rate of 372,000 mi/sec^2. Isn't that just a tad greater than Einsteins E=MC^2, as according to that, light ACCELERATES at the rate of 186,000 mi/sec^2, which means it can only be 93,000 miles away after one second. Edited March 21, 2009 by Duration Link to comment Share on other sites More sharing options...
Sisyphus Posted March 21, 2009 Share Posted March 21, 2009 You already told me I am traveling at .5c, correct? Am I supposed to disregard that information when I perform the calculations? How could I have made an observation of the light traveling 186,000 mi/sec when it is clearly only 93,000 miles in front of me after one second? If light travels at a velocity of 186,000 mi/sec, and it is 186,000 miles away from me after a one second duration, that means the light accelerated at a rate of 372,000 mi/sec^2. Isn't that just a tad greater than Einsteins E=MC^2, as according to that, light ACCELERATES at the rate of 186,000 mi/sec^2, which means it can only be 93,000 miles away after one second. In your own frame of reference, you are not moving, and the light is moving at C away from you. After one second it is 186,000 miles away from you. In the Earth's frame of reference, you are moving at 0.5C and the light is moving at C towards the earth and 0.5C away from you, and after one second it is 93,000 miles away from you. This is possible because time and distance are are different depending on what frame of reference you're in. Also, light does not accelerate. It only ever moves at exactly C. The photon, which has no mass, comes into existence already traveling. Link to comment Share on other sites More sharing options...
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