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Galilean transformation and contradictions with light

Here we'll see how classical physics predicts scenarios that disagree with what we observe in nature. Something's got to give….

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  • hopper cool style avatar for user Arnab Chowdhury
    What if I had a race with light? My friend shined a torch. Light traveled with velocity c & I ran with velocity 0.9c (both with respect to my friend). Then wouldn't the speed of light with respect to me be, (c - 0.9c) = 0.1c?
    (10 votes)
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  • primosaur ultimate style avatar for user Jacob Kalodner
    At around , Sal says that the speed is independent of the inertial frame of reference. But if Sal and Sally are traveling at different speeds, how can they observe the speed of light as being the same? Isn't Sally closer to the speed of light, so she has greater relative velocity compared to it?
    (2 votes)
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  • leaf green style avatar for user Coco
    Bear with me... I just had a dumb idea. What if we imagine this: Sally's frame of reference is perpendicular on mine (like a z axis, from a 3rd dimension). Can we imagine a situation where Sally's photon's path is a slice of an expanding sphere, such that the part of the sphere that i see, is equal to the part of the sphere in Sally's frame?
    (2 votes)
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  • primosaur ultimate style avatar for user Naresh K
    When both Sally and I shoot a photon each from our respective flashlights at exactly the same position as me at time t=0, in my understanding both the photons should move together regardless of my or Sally's relative speeds. Sally will observe the photons moving slower than I will observe them since she's moving faster relative to me in the same direction as the photons, but because she is aware of her speed, she should also be able to recognize that the speed of light has not slowed. If she were to start moving at the speed of light, she would find from that point onwards that the distance between her spaceship and the photons she's been chasing has become locked into a constant and they are both moving at the same speed. Am I missing something here?
    (3 votes)
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  • aqualine ultimate style avatar for user 鄭定宇
    So can i understand the video in one sentence like : Scientists notice that the Newtonian system doesn't work by observing our universe and find out that the velocity of light always stay the same in any frame of reference ?
    Is that right ?
    And if so, what is the experiment that can prove Galileo or Newton wrong ?
    (4 votes)
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  • aqualine seed style avatar for user Imaan
    what if i am in a black hole just before the event horizon and time would be moving at very high speeds for me according to some sources. if everything's speed is increasing what about light would it still be the same (3*10^8) in my frame of reference.
    (2 votes)
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  • blobby green style avatar for user Advait Dawane
    At , here speed of light seems to be half of the real value, and you said that it makes sense, but then you told that speed of light regardless of the frame of reference should be 3*10^8 .so i did not get it. Please help me.
    (2 votes)
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  • leaf blue style avatar for user Pavel Somesurname
    At , i'm not exactly sure what he means by "faster than the speed of light" because when you think about it, wouldn't the light travel that 1.5x10^8m with the ship and then travel the other 1.5x10^8m to be a total of 3.0x10^8m away from Sal in one second?
    (2 votes)
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    • male robot hal style avatar for user Charles LaCour
      When you measure the speed of light when traveling through a vacuum it will always travel at 3 * 10^8 m/s regardless of the velocity of the source of the light when it was emitted. So if you have a rocket going 1.5 * 10^8 m/s and they fire a laser at you they see the light as traveling at 3 * 10^8 m/s and you see it traveling at 3 * 10^8 m/s as well not 4.5 * 10^8 m/s.
      (2 votes)
  • blobby green style avatar for user xainxaphdar
    Does an observer moving with the speed of light will still observe speed of light to be 3×10^8?
    (2 votes)
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    • male robot hal style avatar for user Andrew M
      "Observer moving with speed of light" compared to what?
      The speed of light will be the same for everyone regardless of how fast they are moving compare to anyone else, or how fast they are moving compared to the source of the light.
      This was very surprising, which is why no one figured it out until Einstein did.
      But we know with great confidence that it is correct. We have over 100 years of various experimental results that would not be what they are if it weren't true.
      (2 votes)
  • blobby green style avatar for user Chandrachur27032000
    Can you tell me what are those 'OBSERVATIONS' that proved newtonian/gallelian world wrong..?
    (2 votes)
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Video transcript

- [Voiceover] Things are starting to get interesting. In the first video, we set up a space-time diagram from my frame of reference and started to plot things past in that space-time diagram. And then we thought about our friend, Sally, who right at time equals zero is at x equals zero, but she's passing me up at a relative velocity of half the speed of light in the positive x direction. So after one second, she has gone 1 1/2 times 10 to the 8th meters in the positive x direction. After two seconds, she's gone three times 10 to the 8th. And then we assume that she was part of a train of spaceships. So three times 10 to the 8th meters in front of her was another spaceship and it's moving with the same relative velocity relative to me, or in my frame of reference. So at time equals zero is three times 10 to the 8th meters in front of me. But then at time equals one, if we take this point in space-time, in my frame of reference, that ship is now, it now has, so t, let me write this. If we look at my space-time coordinates, t is equal to one second and x is equal to 4.5 times 10 to the 8th meters. That's if you look at my space-time coordinates. But what about Sally's space-time coordinates? Well, to figure out Sally's space-time coordinates, we go parallel to the x prime axis and see where we intersect the t prime axis, but that's still t is equal to one second, this is t is equal to two seconds, this is t is equal to three seconds. So t, or I should say t prime is equal to one second, t prime is two seconds, t prime is three seconds. So t prime is equal to one second still. And so in general, we can say that t prime is going to be equal to t, but what is x prime going to be equal to? So x prime, well, to figure out x prime, you go horizontal to the t prime axis and see where you intersect the x prime axis, and you see that it's three times 10 to the 8th meters. And hopefully, this makes intuitive sense. If it doesn't, pause the video and really think about it because in her frame of reference, that spaceship looks stationary because it's moving with the exact same relative velocity to me. It's going to continue to stay three times 10 to the 8th meters in front of her, which is exactly what we see. So that's why its x prime coordinates stay three times 10 to the 8th meters. From my point of view, it's getting further and further away from me at the relative velocity, at 1.5 times 10 to the 8th meters per second. So how do we translate between our, between our x coordinates? Let me do this in a different color. Between our x coordinates and our x prime coordinates? Well, you see for these examples, we see x prime is going to be less than x, and that should also make sense because especially for this case, it's stationary from the s prime point of view, but its x is continuously increasing as time passes on from my frame of reference, from Sally's frame of reference, from the S frame of reference. So if we start with x, we should subtract something, and the difference between the two, the discrepancy between the two is going to be the relative velocity times time. And so for this particular example, we saw that three times 10 to the 8th meters was equal to 4.5 times 10 to the 8th, minus the relative velocity, 1.5 times 10 to the 8th meters per second times time, which was one second. And I didn't write the units but if you write the units, it all works out and you get exactly this. So hopefully that's starting to get you comfortable with having these two coordinate planes or two space-time diagrams over on top of each other. And the reason why the blue one is distorted is because it's on top, they're moving with a relative velocity relative to what I'm considering to be a stationary frame of reference, which is mine. Obviously, there's no such thing as an absolute stationary frame of reference, and we'll talk more about that in the future. But what I now want to focus on is that photon that I emitted at time equal zero. Because we saw it moves with the speed of light in my frame of reference. After one second, it has moved, its x coordinate is three times 10 to the 8th meters. After two seconds, after two seconds, the photon is at six times 10 to the 8th meters. But let's see what that photon looks like from the s prime frame of reference, from Sally's frame of reference. Well, from Sally's frame of reference, let's think about that photon after two seconds. So the photon is right over there. So t prime is equal to two seconds, two seconds, but what is x prime? What is x prime going to be equal to? Well, x prime, we go parallel to the t prime axis, is three times 10 to the 8th meters. Three times 10 to the 8th meters. So in her frame of reference, it took that photon of light two seconds to go three times 10 to the 8th meters, or it looks like the velocity of that photon is one and a half times 10 to the 8th meters per second in the positive x direction. And this should hopefully makes sense from a Newtonian point of view, or a Galilean point of view. These are called Galilean transformations because if I'm in a car and there's another car and you see this on the highway all the time, if I'm in a car going 60 miles per hour, there's another car going 65 miles per hour, from my point of view, it looks like it's only moving forward at five miles per hour. So that photon will look slower to Sally. Similarly, if we assume this Newtonian, this Galilean world, if she had a flashlight, if she had a flashlight right over here and right at time equals zero she turned it on, and that first photon we were to plot it on her frame of reference, well, it should go the speed of light in her frame of reference. So it starts here at the origin. And then after one second, in the s prime, in the s prime coordinates, it should have gone three times 10 to the 8th meters. After two seconds, it should've gone six times 10 to the 8th meters. And so it's path on her space-time diagram should look like that. That's her photon, that first photon that was emitted from it. So you might be noticing something interesting. That photon from my point of view is going faster than the speed of light. After one second, its x coordinate is 4.5 times 10 to the 8th meters. It's going 4.5 times 10 to the 8th meters per second. It's going faster than the speed of light. It's going faster than my photon, and that might make intuitive sense except it's not what we actually observe in nature. And anytime we try to make a prediction that's not what's observed in nature, it means that our understanding of the universe is not complete because it turns out that regardless of which inertial reference frame we are in, the speed of light, regardless of the speed or the relative velocity of the source of that light, is always going three times 10 to the 8th meters per second. So we know from observations of the universe that Sally, when she looked at my photon, she wouldn't see it going half the speed of light, she would see it going three times 10 to the 8th meters per second. And we know from observations of the universe that Sally's photon, I would not observe it as moving at 4.5 times 10 to the 8th meters per second, that it would actually still be moving at three times 10 to the 8th meters per second. So something has got to give. This is breaking down our classical, our Newtonian, our Galilean views of the world. It's very exciting. We need to think of some other way to conceptualize things, some other way to visualize these space-time diagrams for the different frames of reference.