Strange Posted July 19, 2014 Share Posted July 19, 2014 Do the laws of physics apply to something if it is not an observer? Yes. Any observer will observe light behaving the same way. But, you cannot be an observer travelling at the speed of light. So, the laws of physics apply to something (light) even if it is not an "observer". Can you just state what point you are trying to make? Are you trying to say that we can't know (or can't be sure what we know) about light because it travels at the speed of light? If so, what experimental evidence do you have to support that idea? Link to comment Share on other sites More sharing options...
Alias Moniker Posted July 19, 2014 Author Share Posted July 19, 2014 (edited) Yes. Any observer will observe light behaving the same way. But, you cannot be an observer travelling at the speed of light. So, the laws of physics apply to something (light) even if it is not an "observer". Can you just state what point you are trying to make? Are you trying to say that we can't know (or can't be sure what we know) about light because it travels at the speed of light? If so, what experimental evidence do you have to support that idea? Since you can't be an observer traveling at the speed of light, you can't know that "any" observer would observe light the same way. You can know that, all observers who are not traveling at the speed of light observe light the same way. But you can't know that "any" observer would observe light the same way. And then you'll say, it is BECAUSE we can never be an observer traveling at the speed of light, that it doesn't matter that we can't know what light would behave like if we were traveling at the speed of light. It's like asking, what would I look like if I turned myself inside out, or some other ridiculous notion. But, turning myself inside out and looking at myself is not possible and can't happen. Where as Light does travel at the speed of light, and at the speed of light, since light is no longer an observer, the laws of physics that apply to an observer are not guaranteed to apply to light. Why not? Because the laws of physics are only necessarily the same for two observers, they are not necessarily the same for an observer and a non observer. Edited July 19, 2014 by Alias Moniker Link to comment Share on other sites More sharing options...
ajb Posted July 19, 2014 Share Posted July 19, 2014 You can know that, all observers who are not traveling at the speed of light observe light the same way. But you can't know that "any" observer would observe light the same way. If you mean an inertial observer, then almost by definition any other inertial obsever must measure the relative velocity to be less than the velocity of light. Link to comment Share on other sites More sharing options...
Alias Moniker Posted July 19, 2014 Author Share Posted July 19, 2014 If you mean an inertial observer, then almost by definition any other inertial obsever must measure the relative velocity to be less than the velocity of light. I mean whatever you mean when you say that light is not an observer. Link to comment Share on other sites More sharing options...
Strange Posted July 19, 2014 Share Posted July 19, 2014 Since you can't be an observer traveling at the speed of light, you can't know that "any" observer would observe light the same way. Yes you can, because travelling at the speed of light is not a valid reference frame. What this means is that you cannot say anything about what an observer moving at the speed of light would see (because it involves dividing by zero). You can know that, all observers who are not traveling at the speed of light observe light the same way. Exactly. And that is part of the definition of observer (in relativity). Light does travel at the speed of light, and at the speed of light, since you would no longer be an observer, the laws of physics that apply to an observer not moving at the speed of light, are not guaranteed to apply to you. As I said before, you are mixing up "APPLY" and "OBSERVE". We know (as observers) what the physics associated with light is. So, again, what is your point? Link to comment Share on other sites More sharing options...
Klaynos Posted July 19, 2014 Share Posted July 19, 2014 If nothing can interact with that frame from the valid frames without following the laws we understand then how can we ever test it? If we cannot test out it isn't science of we can test it then it falls within the "laws of physics". I'm sticking with this. If you can't test it out has no impact and is not science. 1 Link to comment Share on other sites More sharing options...
ajb Posted July 19, 2014 Share Posted July 19, 2014 I mean whatever you mean when you say that light is not an observer. Okay, so it is true that things will look different in a non-inertial frame. This is not a great problem in special relativity, you can handle non-inertial frames. What are you trying to get at? Link to comment Share on other sites More sharing options...
Alias Moniker Posted July 19, 2014 Author Share Posted July 19, 2014 (edited) I'm trying to keep up If nothing can interact with that frame from the valid frames without following the laws we understand then how can we ever test it? I agree with you. Okay, so it is true that things will look different in a non-inertial frame. This is not a great problem in special relativity, you can handle non-inertial frames. What are you trying to get at? Can you explain a non-inertial frame better? Does "no inertia" equate to the same thing as "no passing time"? I'm sticking with this. If you can't test it out has no impact and is not science. That's correct. But just because it isn't science doesn't mean it's not reality. Ultimately, it is testable, but to test it, you have to perform a test outside of the existence of time, because you're trying to measure a particle that exists outside of the existence of time. Edited July 19, 2014 by Alias Moniker Link to comment Share on other sites More sharing options...
ajb Posted July 19, 2014 Share Posted July 19, 2014 (edited) I'm trying to keep up Okay... I would say that what you should take away from all this is that an observer is really a choice of coordinate system used to describe the physics. In special relativity we have a nice special class of coordinates which are the inertial frames of reference. You can pass from one inertial coordinate system to another via a Lorentz transformation. However, you are free to use other so called non-inertial coordinates in special relativity, but generically things looks different. That said any meaningful physics cannot depend on your choice of coordinates. In relation to your opening post, there is no inertial frame of reference for which a photon can be considered at rest. Thus, you have to take great care in think about "the photons point of view". Edited July 19, 2014 by ajb 2 Link to comment Share on other sites More sharing options...
Alias Moniker Posted July 19, 2014 Author Share Posted July 19, 2014 (edited) Okay... I would say that what you should take away from all this is that an observer is really a choice of coordinate system used to describe the physics. In special relativity we have a nice special class of coordinates which are the inertial frames of reference. You can pass from one inertial coordinate system to another via a Lorentz transformation. However, you are free to use other so called non-inertial coordinates in special relativity, but generically things looks different. That said any meaningful physics cannot depend on your choice of coordinates. In relation to your opening post, there is no inertial frame of reference for which a photon can be considered at rest. Thus, you have to take great care in think about "the photons point of view". Well let's take great care then. But it has to be recognized at some point that the photon's point of view would not include time, so physics would not be the same for the photon as they are for us because our physics depend on an existence where there is time. It isn't accurate to measure a photon with velocity or speed because both velocity and speed require one moment to pass into a next moment and a photon is not experiencing one moment passing to the next moment. It "appears" that the photon "travels" that way if you're looking through the scope of time but it isn't actually possible for a photon to "travel" because it only ever gets one single non-existent unit of what we call "time" for its entire existence to fit into. Like the photon is created in an "extra dimensional space" where time doesn't exist, only the extra space is between two of our moments. You might think of it like cramming the DISTANCE of a thousand light years, in between 0:01 and 0:02 seconds. The photon's passing causes the disturbance in our dimension known as the electromagnetic wave, that follows the path of the photon like thunder following lightning, only the wave has to move from moment to moment where the photon passed through that space "timelessly". Edited July 19, 2014 by Alias Moniker Link to comment Share on other sites More sharing options...
Klaynos Posted July 19, 2014 Share Posted July 19, 2014 Well let's take great care then. But it has to be recognized at some point that the photon's point of view would not include time, so physics would not be the same for the photon as they are for us because our physics depend on an existence where there is time. It isn't accurate to measure a photon with velocity or speed because both velocity and speed require one moment to pass into a next moment and a photon is not experiencing one moment passing to the next moment. It "appears" that the photon "travels" that way if you're looking through the scope of time but it isn't actually possible for a photon to "travel" because it only ever gets one single non-existent unit of what we call "time" for its entire existence to fit into. The photon gets created in an extra dimension where time doesn't exist, and it's passing causes the disturbance in our dimension known as the electromagnetic wave, like thunder following lightning. Can you show that the concept of source or time translates meaningfully to a photon frame without using physics we know to not be valid for a photon frame? 1 Link to comment Share on other sites More sharing options...
Alias Moniker Posted July 19, 2014 Author Share Posted July 19, 2014 (edited) Can you show that the concept of source or time translates meaningfully to a photon frame without using physics we know to not be valid for a photon frame? I don't know what you mean? I've never taken a physics class before I'm approaching this all from a logic point of view so I appreciate your patience with my ideas and unfamiliar language. My first question is, does "translates meaningfully" basically imply "have an effect on". Can I show that time has an effect on the photon? Edited July 19, 2014 by Alias Moniker Link to comment Share on other sites More sharing options...
Klaynos Posted July 19, 2014 Share Posted July 19, 2014 We can measure time and space in one frame and translate those measurements into another using maths. For inertial frames this is done using the Lorentz transformations. Your statements require there to be transformations into a photon frame for any comment on time it space to be valid. Link to comment Share on other sites More sharing options...
Alias Moniker Posted July 19, 2014 Author Share Posted July 19, 2014 (edited) The physics of a photon = The physics of an observer(-time). Observer's frame = Space, Space, Space, TimePhoton's frame = Space, Space, SpaceWhy do you think it's not possible for you to translate one to the other?The observer's frame would, by definition, include variables for time.The Photon's frame would, by definition, have no place settings to insert those variables for time.It's like trying to run Direct X 11 software on a Direct X 10 GPU. There's more information in the software than the hardware knows what to do with. Edited July 19, 2014 by Alias Moniker Link to comment Share on other sites More sharing options...
Klaynos Posted July 19, 2014 Share Posted July 19, 2014 The physics of a photon = The physics of an observer(-time). Observer's frame = Space, Space, Space, Time Photon's frame = Space, Space, Space Why do you think it's not possible for you to translate one to the other? The observer's frame would, by definition, include variables for time. The Photon's frame would, by definition, have no place settings to insert those variables for time. It's like trying to run Direct X 11 software on a Direct X 10 GPU. There's more information in the software than the hardware knows what to do with. You have made several assertions here with no proof. You need to show (which in physics requires mathematics) how the first three statements you've made are true. Just stating something as true does not make it so. Link to comment Share on other sites More sharing options...
Strange Posted July 19, 2014 Share Posted July 19, 2014 Well let's take great care then. But it has to be recognized at some point that the photon's point of view would not include time, so physics would not be the same for the photon as they are for us because our physics depend on an existence where there is time. It isn't accurate to measure a photon with velocity or speed because both velocity and speed require one moment to pass into a next moment .... It isn't accurate (or meaninful) from the photon's point of view. But that is not what we do. We measure the speed of light in our frame of reference. As photons very obviously do travel, I'm not quite sure what the point of your post is. Link to comment Share on other sites More sharing options...
Alias Moniker Posted July 19, 2014 Author Share Posted July 19, 2014 (edited) It isn't accurate (or meaninful) from the photon's point of view. But that is not what we do. We measure the speed of light in our frame of reference. As photons very obviously do travel, I'm not quite sure what the point of your post is. That doesn't agree with what Neil deGrasse Tyson implies in the video in my first post. In the video that I linked to on the very first comment, Neil deGrasse Tyson explains the "timelessness" of photons, so you should watch that. Edited July 19, 2014 by Alias Moniker Link to comment Share on other sites More sharing options...
Strange Posted July 19, 2014 Share Posted July 19, 2014 The Photon's frame would, by definition, have no place settings to insert those variables for time. Right. And we all agree that you cannot define or measure these things from the photon's frame (because there is no such thing as the "photon's frame"; it is undefined). But that is irrelevant because there is no reason to try and do this. I'm approaching this all from a logic point of view You are being consistently illogical by simply repeating the same statements, even after things have been explained to you. This is a science forum. Are you interested in discussing (and perhaps learning) science or are you just here to air your misunderstandings? The physics of a photon = The physics of an observer(-time). Observer's frame = Space, Space, Space, Time Photon's frame = Space, Space, Space. Actually, if this extrapolation to a photon's frame were valid (it isn't) then it would actually be: Photon's frame = Space, Space. Which raises another flaw in your argument. You are saying that time disappears. But so does space (in the direction of travel). You can take the limit as these approach zero and show that photons still move at the speed of light. So your application of "logic" (which appears to mean "what make sense to me") fails. Link to comment Share on other sites More sharing options...
Alias Moniker Posted July 19, 2014 Author Share Posted July 19, 2014 (edited) Right. And we all agree that you cannot define or measure these things from the photon's frame (because there is no such thing as the "photon's frame"; it is undefined). But that is irrelevant because there is no reason to try and do this. You are being consistently illogical by simply repeating the same statements, even after things have been explained to you. This is a science forum. Are you interested in discussing (and perhaps learning) science or are you just here to air your misunderstandings? Actually, if this extrapolation to a photon's frame were valid (it isn't) then it would actually be: Photon's frame = Space, Space. Which raises another flaw in your argument. You are saying that time disappears. But so does space (in the direction of travel). You can take the limit as these approach zero and show that photons still move at the speed of light. So your application of "logic" (which appears to mean "what make sense to me") fails. Space doesn't disappear in the direction of travel because if there is no time then there is no direction of travel. Or do you believe that travel can happen without change over time? Just because you're horrible at understanding a point of view that isn't your own doesn't mean my logic is flawed. Edited July 19, 2014 by Alias Moniker Link to comment Share on other sites More sharing options...
Strange Posted July 19, 2014 Share Posted July 19, 2014 Space doesn't disappear in the direction of travel because if there is no time then there is on direction of travel. Or do you believe that travel can happen without change over time? Er, then time doesn't disappear either. You can't accept time dilation without length contraction. Unless you are rejecting the theory of special relativity in its entirety and replacing it with your own theory. In your theory, please show how you derive time dilation as an effect due to motion, while not length contraction does not occur. The math for this in SR is very simple. So it should be straightforward for you to present your alternative here. Then you can show us how well it matches experimental results... Link to comment Share on other sites More sharing options...
Alias Moniker Posted July 19, 2014 Author Share Posted July 19, 2014 (edited) Er, then time doesn't disappear either. You can't accept time dilation without length contraction. Unless you are rejecting the theory of special relativity in its entirety and replacing it with your own theory. In your theory, please show how you derive time dilation as an effect due to motion, while not length contraction does not occur. The math for this in SR is very simple. So it should be straightforward for you to present your alternative here. Then you can show us how well it matches experimental results... Ok check this out... Time dilation and length contraction Both are laws of observer physics So neither are required to apply to a non observer like light. Light does not need to follow the physics of an observer because it is not an observer. In the instance of length contraction: "Length contraction is only in the direction parallel to the direction in which the observed body is travelling." Again, the photon would need to be traveling, which would require the passage of time. Since the photon is not traveling during its existence, the photon would not experience length contraction. In the instance of time dilation: The photon would not experience time dilation, it would be created already under the full effect of time dilation. The Photon is created at C, so it's time is already 0. Edited July 19, 2014 by Alias Moniker Link to comment Share on other sites More sharing options...
Strange Posted July 19, 2014 Share Posted July 19, 2014 Light does not need to follow the physics of an observer because it is not an observer. Sigh. What is the point of this endless trolling. We observe the physics of light. The physics applies to light. Light does not cannot observe anything. Light does not observe anything. If you have a theory or some evidence that contradicts current physics, then feel free to present it. Simply repeatedly saying "hey guys, I don't any math or science but here my nonsensical ideas" isn't very productive. Link to comment Share on other sites More sharing options...
Alias Moniker Posted July 19, 2014 Author Share Posted July 19, 2014 (edited) You totally believe that you observe the physics of light, yes. I don't know why you don't understand. Physics under the influence of time, not the same as, physics not under the influence of time. Edited July 19, 2014 by Alias Moniker Link to comment Share on other sites More sharing options...
Klaynos Posted July 19, 2014 Share Posted July 19, 2014 You have made several assertions here with no proof. You need to show (which in physics requires mathematics) how the first three statements you've made are true. Just stating something as true does not make it so. Quoting this as it seems to have been missed. Link to comment Share on other sites More sharing options...
Alias Moniker Posted July 19, 2014 Author Share Posted July 19, 2014 (edited) The physics of a photon = The physics of an observer(-time). Observer's frame = Space, Space, Space, Time Photon's frame = Space, Space, Space The physics of a photon = The physics of an observer(-time) Observer physics: Speed = Distance / Time Photon physics: Speed = Distance... there is no time for a photon so the photon's distance can't be divided by time and it can't have a speed. If you want "proof" that there is no time for a photon you'll have to ask Neil deGrasse Tyson, I'm sure he knows the math that led him to the conclusion that photons are "timeless" as he states in the video in the opening post. "If you hit the speed of light, which we don't know how to do yet, but if you hit the speed of light, then time stops... time does not exist." So then the following two statements are conclusions from the idea that time does not exist to a photon, but does exist to an observer. Observer physics = space, space, space, time, gravity But for a photon, time doesn't exist so, Photon physics = space, space, space, gravity (or whatever, but not including time) How can physics that require time, explain a reality where time does not exist? Edited July 19, 2014 by Alias Moniker Link to comment Share on other sites More sharing options...
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