Timeline for Why do we say that light travels at a speed?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 16, 2014 at 6:14 | comment | added | user12262 | Derek Roberts: "So what do you want me to say, at the speed of light, time "nullifies"" -- No: duration "nullifies", if you like. "Explain how your equation proves that you've observed a photon" -- I've been referring to a sender and a receiver exchanging a signal between each other. (If any arguments are to be made then about which notions should be considered and admitted for anybody as self-evident.) | |
| Jul 16, 2014 at 6:08 | comment | added | Derek Roberts | Take that conversation to that question and stop arguing out of spite or whatever. We're going to assume this website's understanding and Einstein's understanding of time dilation not yours. | |
| Jul 16, 2014 at 6:07 | comment | added | user12262 | Derek Roberts: "When traveling at the speed of light (v=c), left under the radical you would have 0." -- Right. "This answer would be undefined or infinity if you will (let's go with infinity)." -- Nonsense. Square root of zero is: zero. Consequently: Any duration attributable to the signal (e.g. "photon") having been exchanged, in comparison to any (non-zero) duration of sender or receiver is: zero. ""[...] you could infer that time is 'frozen' to an object traveling at the speed of light."" -- That's misusing the words "time"; and (related) "object". | |
| Jul 16, 2014 at 6:01 | comment | added | Derek Roberts | So what do you want me to say, at the speed of light, time "nullifies", does that make you happy? Are you gonna move beyond the single point you've made three times already and address the question? Explain how your equation proves that you've observed a photon, start there. | |
| Jul 16, 2014 at 5:59 | comment | added | user12262 | Derek Roberts: "Einstein's theory of time dilation concludes that at the speed of light, time stops." -- That's an unbearably sloppy account of evaluating $$\lim_{ \beta \rightarrow 1 } \left[ \sqrt{ 1 - \beta^2 } \right],$$ and of the related convention to set the magnitude of "light-like intervals" to $0$, calling them "null intervals". | |
| Jul 16, 2014 at 5:47 | comment | added | Derek Roberts | User12262 do you understand the concept that I'm presenting? | |
| Jul 16, 2014 at 5:44 | comment | added | Derek Roberts | Neil deGrasse Tyson: "That whole phenomena is called time dilation, and if you take it to the limit... the equations tell us that, at the speed of light, time stops..." It's a lecture he did so that must be why it's sloppy. | |
| Jul 16, 2014 at 5:34 | history | edited | user12262 | CC BY-SA 3.0 |
(v3.141592: addendum referring to "travelling".)
|
| Jul 16, 2014 at 5:29 | comment | added | Derek Roberts | Yes. The photon is an instantaneous particle. The photon does not exist for a measure of time. Einstein's theory of time dilation concludes that at the speed of light, time stops. Time does not pass for you. So in the same instant that a photon is emitted, it is also absorbed at its final destination (from any POV other than the photon). But the photon isn't keeping track of its existence with "time" either, it would measure how "old" it is by the distance it has traveled. | |
| Jul 16, 2014 at 5:20 | comment | added | user12262 | Derek Roberts: "So, you set up your experiment [...] In that same single moment when you create the photon that you are attempting to measure, the photon that you have created also ceases to exist." -- No. The "moment" ("indication") of the sender stating a signal is not the same as the "moment" ("indication") of the detector observing that the signal had been stated by the sender (a.k.a. the detector "receiving the signal"); at least if sender and detector are distinct participants, or even separated from each other. | |
| Jul 16, 2014 at 5:17 | comment | added | Derek Roberts | You measured the disturbance that the photon created by traveling through time. | |
| Jul 16, 2014 at 5:14 | comment | added | user12262 | Derek Roberts: "How do you account [...]" -- As far as I understand your comment (which, since I first read it, has been removed) the only "accounting" is pretty much summed up in the derivation of the factor $\sqrt{1 - \beta^2}$, which requires (or presumes) mutually equal determination of the value $\beta^2$. The determination of $\beta^2 = 1$ is only unilateral, since signal fronts don't exchange signals between each other, therefore signal fronts cannot be said "being at rest to each other" (or constituting an "inertial frame"), etc. | |
| Jul 16, 2014 at 5:09 | comment | added | Derek Roberts | The speed of the signal front is not the same as the speed of the photon. The speed of the wake is not the same as the speed of the boat. So, you set up your experiment to measure photons. You have all the equipment ready to gather data. You press "start". In that same single moment when you create the photon that you are attempting to measure, the photon that you have created also ceases to exist. Then the experiment begins, and all the data is gathered. But the photon is gone already. So what you're measuring is not light, it's the aftermath of light. | |
| Jul 16, 2014 at 5:02 | comment | added | Derek Roberts | Neil deGrasse Tyson: "As you travel faster, or if you find yourself in the vicinity of a higher source of gravity, time ticks more slowly for you then for those who are not". | |
| Jul 15, 2014 at 22:26 | history | answered | user12262 | CC BY-SA 3.0 |