If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

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….

Want to join the conversation?

  • 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)
    Default Khan Academy avatar avatar for user
  • spunky sam blue style avatar for user abhinav.raja03
    What were the observations that led to the idea that the speed of light is constant for all inertial frames of reference?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • 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)
    Default Khan Academy avatar avatar for user
  • 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)
    Default Khan Academy avatar avatar for user
  • 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)
    Default Khan Academy avatar avatar for user
  • 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?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • duskpin sapling style avatar for user paridhi
    At , Sir drew the graph line for photon from Sally's flashlight according to her frame of reference. Wouldn't that graph be same from Sir's frame of reference? I mean to say that at t=1 s , the photon would travel , (c+0.5c)t = 1.5ct = 4.5x 10^8 (1) m, from Sir's point of view. Similarly at t=2 s, x= 9 x 10^8 m. Is the formula for position of Sally's photon from Sir's frame of reference, correct?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • mr pants green style avatar for user Manasv
    If we take a light beam and my Body then you can very well conclude that time (my Clock) Slows down. Assuming speed of light Postulate Holds.
    But if say you and I were to move in the reference Frame of a 3rd Person. Imagine that the 3rd Person is seeing you still and seeing me move away. Now for you and the 3rd Person my Clock has slowed down but from my Frame of reference your Clock and his/her has slowed down.
    That means both of our clocks must equally slow down. Now let us Imagine that the 3rd Guy shines a light beam towards both of us. Since both of us are gonna See it come at c, for you my Clock has slowed down and for me your Clock has slowed down. The two effects cancel out, so how exactly is the speed of light Constant?
    Also I had Heard that muons Live longer and are able to reach the ground from atmosphere. But if symmetry is maintained (as I've read at places) HOW does it live longer?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Charles LaCour
      This is something I have struggled with coming to grips with for a long time. This is the basis of what is know as the twin paradox. The key is that all motion through space time is relative. There is no correct or preferred passage of time so if none of the observers change their rate of motion each will see the other as moving slower in time. But to compare your clocks without having to rely on some sort of signaling between you that is also subject to this effect you have to come together and match your relative motions. To do this someone needs to break from being in an inertial reference frame and accelerate to match the others reference frame. This change of reference frame for the observer puts them into sync with the observations of the reference frame they matched.

      So if you have Alice and Bob in two space ships flying past the sun at 3/4 the speed of light. Both Alice and Bob measure the light from the sun as traveling at C and they measure their approach to the sun at 0.75 and the time on the sun as passing 1 second for every 1.5119 seconds on their clocks C but they measure the the velocity of each other at 0.96 C and the time passing in the other ship as 1 second for every 3.5714 seconds on their clock. If Bob decides that he wants to meet up with Alice and gets into a shuttle craft and fires his rockets to turn around and match Alice's frame of reference while leaving his spaceship to continue on with its original path. If he continues to take measurements of his and Alice's spaceship his reference frame changes so that his measurements of his ship shift to away from what he sees on his ship from the shuttle craft and begin to match Alice's his measurements as he gets closer to matching Alice's frame of reference even the difference in age.

      This also applies to the muons. From the muon's frame of reference there is no change in time but the distances are compressed in the direction of travel but from the surface of the earth the muon is living longer and traveling a longer distance.
      (2 votes)
  • 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.
    (1 vote)
    Default Khan Academy avatar avatar for user
  • 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.
    (1 vote)
    Default Khan Academy avatar avatar for user

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.