Main content

## Physics library

### Course: Physics library > Unit 16

Lesson 2: Minkowski spacetime- Starting to set up a Newtonian path–time diagram
- Visualizing multiple Newtonian path–time diagrams
- Galilean transformation and contradictions with light
- Introduction to special relativity and Minkowski spacetime diagrams
- Measuring time in meters in Minkowski spacetime
- Angle of x' axis in Minkowski spacetime

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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?

- 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)
- No, light would move at c for you. That's the whole point of the postulate "Light speed is constant for any frame of reference".(30 votes)

- At around7:15, 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)
- What you are saying makes common sense but that is not how velocities that are a significant fraction of the speed of light work. Any observer that is not accelerating will always measure light in a vacuum traveling at the same speed.(6 votes)

- 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)- For one thing, time dilation can be proved with radar. I studied that on some web pages a few years ago, and I thought I understood it for about a day, then lost it. But maybe Sal can "bring me back".(1 vote)

- 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?(1 vote)
- can you elaborate , the thought seems interesting(5 votes)

- 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)
- No all observers will always measure the speed of light in a vacuum as the same value. All constant velocity motion is relative, you can't tell if you are moving or other things are.(2 votes)

- What were the observations that led to the idea that the speed of light is constant for all inertial frames of reference?(2 votes)
- At6:26, 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)
- 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)- 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)

- 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)
- Speed of light is always the same in all frames of reference. That is the primary boundary condition for special and general relativity.(2 votes)

- At5:11, 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)
- He's telling you what the Galilean reasoning would suggest and then he's showing that it's wrong when you try to apply it to light, which is why the video has the title it has.(1 vote)

## 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.