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## 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

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# Visualizing multiple Newtonian path–time diagrams

Now that we know what path–time diagrams are, let's see what happens when we bring along a friend!

## Want to join the conversation?

- I don't understand why Sal labeled the line t', the friends frame of reference, at4:00? If this is the friend's time axis, it would mean t' is longer (it's the hypotenuse of the triangle) than Sal's time t but later in the video says both people think "1 second" has gone by.(7 votes)
- Hey jp11235,

Okay, I'll try my best explaining. Yes, you are correct by saying that the t' looks longer than the t axis. However, look back at the t' against x' axis. Do you notice t' is slanted to a 45º angle, while the t and x axis are perpendicular? Well, that means that for the same position of time, you will get a smaller value of x' than x. The way I think about it is that if you kind of 'push' the slanted axis, it would make all the rhombi turn into squares (if you know what I mean)... I find it quite challenging to explain this. Hopefully this helped a little bit, even though you asked this such a long time ago...

All the best!

swishswish123(5 votes)

- I m not getting the idea the speed of photon with respect to stationary observer and moving observer is same or not?(5 votes)
- Hi Sid,

In the future videos you will learn how that the speed of light (3 x 10^8 m/s) is absolute - it is always constant no matter from what frame of reference we look at it (even if we are moving).

The Newtonium way of looking at things thought of space and time like this;

1) The passage of time is the same for all frame of references, meaning time is absolute.

2) Measured space is the same for all frames of references.

But as you will learn, this can't be.

Instead, Einstein's Theory of Special Relativity talks about how light is absolute, but space and time are not.

Now we know space and time are not absolute, they are really just different directions in this continuum spacetime. That is why people say space/the universe has 4 dimensions, the last one being time. Space = 3D and time = 1D, when you put these together you get Spacetime which is is representing that time isn't a separate absolute, and space a separate absolute, they are just different directions of spacetime.

This switch from Newtonian/Galilean thinking may take a while to sink in - just keep learning!

Hope this helps,

- Convenient Colleague(2 votes)

- Why do we use the tilted blue line(the world line of Sally) to define the system of spacetime relative to Sally, Why should we think that the coordinates along this "static path"(worldline) would be the time coordinates of Sallys spacetime system? Relative to my orange worlline photon "static path" based system of mine.(3 votes)
- Keep in mind that there is no preferred inertial frame of reference and that every inertial observer's view of the events has that observer as stationary with a world line that is vertical. Things moving relative to the observer have tilted/non-vertical world lines. Photons (or anything that moves at the speed of light) will always have a world line at a 45 degree angle.(2 votes)

- Why is that Sally's t' axis is on her movement line (the light blue), but your t axis is on the y axis (not on your orange movement line)?(3 votes)
- It looks like just a notation thing with the graph. The t values aren't different. And the orange line is not Sal's movement, but the movement of a photon emitted by Sal at time zero. So the orange line is the movement of light, not Sal. Hope this clarifies it a bit!(1 vote)

- t' axis is slanting compared to the t axis. does this mean that every second is going to be slightly longer for the moving observer as compared to the stationary observer? Can we consider this a proof of time dilation?(3 votes)
- Hi Pranav,

The reason for the axis being tilted/slanted is because the time and space happening to Sal's friend, will look different with respect to his point of view.

Basically, space and time are not absolute - they can be different depending on what frame of reference we look at them.

This is a switch from the Newtonian way of thinking (that space and time are absolute), and may take a while to sink in.

Hope this helps,

- Convenient Colleague(1 vote)

- In your videos on Sp relativity you used a system(train) of spaceships all travelling at the same speed .5c relative to Sals S system to set up the coordinate grid lines of Sally S’ system. These lines are all parallel to the tilted t’axis of Sally herself(her spaceship) at the origin of her system because they are the lines along which the time, of that particular spaceship that is at rest relative to Sally but moving at .5c relative to Sal ,evolves along. All these lines of coincidence define the time axis for all these positions that are at rest relative to each other namely all the positions of the ships and the positions in Sallys system which system moves at .5c relative to Sals system. Since the motion causes Sallys time axis to tilt relative to Sals all the time axies of all the positions in Sallys system are tilted at the same angle and are all parallel to the t’axis. They are the direction that time evolves at(in spacetime) in Sallys system.

Is there a way analogous to this system of ships that we can use to set up line of simultaneity for all time in Sallys system that is can we find a line along which all positions on it have the same time. That is can we find directions in spacetime so that all positions on the line(this axis) occurs at the same time and are simultaneous?

IN the first case the set of Sallys spaceships define all the times at each point of the Sally system along the lines of coincidence. Now I am wondering if we can find some other system of Sallys that will define at each time in Sallys system all positions which will occur at this same time (be simultaneous) and the lines a will be called simultaneity lines of Sallys system and all be parallel to the tilted x’axis of Sally who moves at .5 c relative to Sals system.

Whereas the line of coincidence are all parallel to the t’axis of Sally and each line gives all the times at that one position that the line intersects the x’ axis at; the lines of simultaneity, on the other hand, are all parallel to the x’axis of Sally and each of these line gives all the positions which are simultaneous to that one time where it intersects the t’axis at.

The fact that these lines form Sallys system which moves relative to Sals at .5c means that since the t’ and x’ axis of Sally are tilted due to the motion by Q=tan^-1( v/c) results in the coordinates in Sallys system being transformed from Sals by this tilt reflected into the transformations since Y=[1/sqrt(1-v/c)^2] and x’=Y(x-Bct) and t’=Y(ct-Bx) which shows the symmetry of the concept of spacetime . B=v/c(3 votes) - These lectures are extraordinarily good, and very helpful. I would add only 2 small caveats. Sal says he's both "stationary" and "drifting". This is confusing. He should also reiterate that he is STATIONARY, at the beginning of each one of these lessons. Thanks!(2 votes)
- Drifting and stationary are relative states of motion.

As long as you are not accelerating you are always stationary with respect to yourself.

So if you are drifting through deep space far away any source of gravity you are stationary in relation to yourself.(2 votes)

- At3:25wouldn't Sal see the spaceship train at 1/4 the speed of light every second? I say this because the light would have to reflect off of the hulls that are going at 1/2 the speed of light, so wouldn't it be seen as going 1/4 the speed of light because the actual light in reference to Sal is traveling twice the distance, and so 1/2 the time the light travels is back toward Sal, doubling the denominator in reference? (I know he's doing this to represent his friend's point of view, but he intoned that he saw his friend 1/2 the speed of light away after 1 second, yet didn't mention how it would look different in his view)(2 votes)
- what will be the distance traveled by a person in a train of 100m when he moves from ?end to the front of the train if the train travels at the speed of light?(2 votes)
- Is Sal considered to be moving at the speed of light (c)? In the last video, he was stationary and turned on a flashlight. I'm confused.(2 votes)
- this video is from the person in the spaceship's perspective, and so relative to that person, Sal is moving. In the last vid, the spaceship was traveling at half the speed of light, and so sal here would be traveling away from the spaceship at the same speed(0 votes)

## Video transcript

- [Voiceover] In the last video, we started to construct
a space-time diagram for my frame of reference and I'm just drifting through space and I'm assuming that I'm in
an intertial frame of reference which means I'm moving
at a constant velocity relative to all other frames of reference and we set up a situation where I emitted a photon right at time zero, so after one second, it would have moved 3 times 10 to the 8th meters. After two seconds, it would have moved 6 times 10 to the 8th meters, and then we had a, we added a little bit of flavor to our little scenario where right at time zero,
a friend passes me up in her spaceship and she is traveling relative to me in the positive-x direction at half the speed of light, and so we plotted her
path right at time zero. Her spaceship is right there. I could draw the spaceship. It's at the origin. Then after one, after one second, she would have traveled 1 1/2 times 10 to the 8th meters. After two seconds, she would have traveled 3 times 10 to the 8th meters, so this blue line in the last video was her path. But now I want to make
it even more interesting. Let's assume that we have
actually a whole train of spaceships, all traveling
in the positive-x direction at the same velocity as her spaceship. So this is her spaceship right over here, at time equals zero, she is exactly where I am, but let's say that 3 times
10 to the 8th meters in front of her, there's another spaceship, traveling at exactly 1 1/2 times 10 to the
8th meters per second, so they're traveling at the
same relative velocity to me, but if you think about it, from
each of their point of view, they would seem to be stationary, because the distance between them, in this x-direction is
going to stay the same. So this person, at time
zero, is going to be 3 times 10 to the 8th meters
in the positive-x direction from me. Now, if I were to wait two seconds, they are going to be, so if I were to wait two seconds, they are going to be, they are going to be 6
times 10 to the 8th meters away from me, because they're going half the speed of light, and so I could draw their path and I'm going to do it
in a slightly more muted, thinner color, so that we, what I'm essentially
going to be doing here is I'm setting up gridlines for my friend's alternate frame of reference. I'm going to put that on
top of my frame of reference and to be clear, I'm not assuming special relativity. I'm assuming a Newtonian world, classical mechanics. Just to get familiar with these ideas and to see where the Newtonian
world is going to break down but let's say that's not the only ship, let's say there was another ship, that at time equals zero is
6 times 10 to the 8th meters away from me in the positive x-direction. Well, where are they going to be? Where are they going to be after two seconds? Well, after, let's see, after one second, they're going to be 1 1/2
times 10 to the 8th meters further. Then after two seconds,
they're going to be 3 times 10 to the 8th meters further, and then after three seconds, after three seconds,
so they're going to be 1 1/2 times 10 to the 8th
meters further than that, so their path is going to look like, so let me draw this. I'm really just doing these
to show what stationary objects in my friend's frame of reference, how they actually look to me, and these should be parallel lines. These should be parallel lines, so that's a pretty good attempt, and so I'll, let me, and so let me actually just set up that alternate frame of reference, so this is going to be not only my x-axis, I'm also going to call
this my x-prime axis and I'll do it in a
second of thinking about how we can read this. So, I can color this
in with the blue color. Actually, let me do it
in that light blue color that we're going to be
using for my friend's frame of reference, so this is also, this is also going to be
our x-prime axis overlaid, and this one, let me call
this the t-prime axis, so my friend's frame of reference, I'm going to call that the
prime frame of reference, or if I called, if I called my frame of reference
the S frame of reference, I can call my friend's frame of reference the S-prime frame of reference, and so, let me draw
horizontal lines to show the passage of each second, so, that's one second, two seconds, three seconds, so on and so forth. So let's make sure we're
comfortable with reading what's going on here. So, a point, let's say this point on our diagram. Let me just see. This point on our diagram, from my frame of reference,
in the S frame of reference, this is the point time is, or space, or my distance, my distance from the origin
in the positive-x direction is going to be, I just drop a, I just go parallel to my vertical axis, so I'm just going to do that, and I get 4.5 times 10 to the 8th meters, so 4.5 times 10 to the 8th meters, and what is our, so I can write the units there if I want, and our time is at one second, one second. So this is our coordinates from, this is the coordinates
from my frame of reference, from the S frame of reference, I guess S for Sal, and so, what would be the frame? What would be those same coordinates from my friend's frame of reference? Let's say her name is Sally, or S-prime. Well, from her frame of reference, our time is still one second. We see the horizontal line there. We're just going parallel
to our horizontal axis or our x-axis so the time is still one second, but her, the distance that the, or the x-coordinate is 3
times 10 to the 8th meters, 3 times 10 to the 8th meters, and I really want you to
sit and think about this and even take out a ruler
and draw this on your own paper and feel good about this. Now, why does this make sense? Well, this, you could
view this point as where that second spaceship was after a second, and from
Sally's point of view, from the S-prime frame of reference, at time equals zero, that person, that ship was
3 times 10 to the 8th meters in front of her, and after a
second, that spaceship is still 3 times 10 to the 8th
meters in front of her. After two seconds, it's still 3 times 10 to the 8th
meters in front of her, and so from her point of view
and her frame of reference, it is stationary, so the way to read this for any point, if you want my frame of reference, you, the time is easy, you just go horizontally to either one, so on her axis, we've essentially
just taken the coordinate axis and sheared it and skewed it. This is still one second, this is two seconds,
this is three seconds, but it turned all of these kind of a grid of squares if I were to draw
a grid for my coordinates into a bunch of, into a bunch of parallelograms which we see right over here, and so for any given point in, remember we're thinking of Newtonian space and time, to read it, we go horizontally for time, in either case, but for me, you would go straight down because my time axis is vertical. You go straight down and
you read where it intersects the x-axis to figure out our position but from her frame of reference we move parallel to her t-axis, or her t-prime axis, I should say, to get what her x, I guess we could say what
her x-prime coordinates are going to be, so this
is x-prime, t-prime. This is x and this is t. Now, what's clear, at any given point, at any given point, t is going to be equal to t-prime, or we could say t-prime is equal to t, either way, but what's the relationship
between x-prime and x? Well, let's think about that for a second, so when x is 4.5 times
10 to the 8th meters, x-prime is 3 times 10 to the 8th meters, and so it looks like, it looks like x-prime
is going to be equal to x minus, well what's it going to be minus? It's going to be minus 1.5 times 10 to the 8th meters but that difference is going
to grow as time goes on, meters per second times, times the amount of
time that has passed by, and you should see that because after one second, after one second, the difference
is 1.5 times 10 to the 8th meters, after two seconds, after two seconds, so
let's look at this point right over here, in my frame of reference,
that person is sitting at 6 times 10 to the 8th meters
in the positive-x direction, but in Sally's frame of reference, in the S-prime frame of reference, they're still at 3 times
10 to the 8th meters in the positive-x direction, so we'd multiply two seconds times 1.5 times 10 to the 8th so we have 3 times 10 to the 8th meters, so we take 6 times 10 to the 8th which would be that. We subtract out 3 times 10 to the 8th and we get the coordinates in S-prime, so this might get a little bit confusing and the key here is really
try to draw this yourself and really try to plot points and think about how they
are different coordinates in the different frames of reference, or different time-space
coordinates in the different frames of reference.