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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
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Angle of x' axis in Minkowski spacetime
It looks like there might be some relationship between the relative angles of the axes in the two references frames we've considered. Let's take a closer look.
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- At6:20you show that the for faster speeds the ct' axis converges with the photon. I was wondering, do they actually meet at full light speed? And if so, since the path of the photon would cover the entire axis, does that mean a photon is really everywhere at once?(17 votes)
- the photon is still at a 45 degree angle between ct' and ct". imagine the ct'-x' axis as having been rotated off of the ct-x axis so that the ct' axis is pointing into the screen and the x' axis is pointing up and out of the screen. the ct"-x" is just rotated a little more. the angle between the ct'-x' axis and between the ct"-x" axis is 90 degrees to each other. the 2d drawing makes the 3d situation look like the axes and photons converge, but they don't. the x'/x" axis just swings out in front of the photon path line(0 votes)
- How would this work in circular or spiral frames of reference (since our Earth and solar syst. move in a spiral)?(9 votes)
- A bit late to the party for my response, but...
As I see it (with my limited understanding), the circular movement wouldn't be inertial, as there is some apparent relative acceleration - so it would be a bit more complicated.(4 votes)
- I don't understand what Sal says at5:50- "In one light meter the thing travels a meter" How does this prove that light travels at light speed in Sally's frame of reference?(4 votes)
- From experimental results we have been able to show that regardless the relative velocity of two inertial frames of reference the speed of light is always measured as the same. So since Sally is moving at a constant speed she is in a inertial frame of reference and will measure the speed of light to be the same as Sal does.(5 votes)
- I understand that if Sally travels at 1.5 * 10^8 m/s in our frame of reference and shoots a beam of light it appears to travel at 3 * 10^8 m/s regardless of the frame of reference, and I understand that this is the case only for light.
But what would happen in our frame of reference if Sally shoots a ball so that in her frame of reference it travels at 2 * 10^8 m/s.
In her frame of reference the ball obeys the laws of physics; nothing travels faster than light. What about our frame of reference? I know that objects' mass increase as they approach the speed of light which should decrease speed to conserve momentum. In Sally's frame of reference it isn't travelling too close to the speed of light so 2 * 10^8 m/s could be achieved, right? But in our frame of reference it should travel beyond the speed of light. How is that possible?(3 votes)- The addition of velocity does not work the way you are used to, if the speeds are close to the speed of light.
The way that you add the ball's speed to sally's speed with relativity is
(1.5 + 2) / (1 + (1.5*2)/3^2) = 2.62 * 10^8 m/s(4 votes)
- Why speed of light is absolute ?(2 votes)
- No one can really answer why that's the case, it's just the way it is in our universe.(6 votes)
- Sal, thanks for the great video. At3:50you remark that the triangle formed by the spacetime path of the photon from the x' axis to the ct' axis is an isosceles triangle because the corresponding lengths of the frame in uniform motion are equal. Why are we allowed to assume that those lengths along the axes of the frame in uniform motion are preserved?(3 votes)
- We are not allowed to assume they are preserved, and in fact they are not preserved. However, they are equal because of the following argument: the angle between the two yellow lines is 90 degrees. By the converse of Thales’ Theorem, you can draw a circle centered at the origin that will pass through both of the pink triangle’s base vertices. This shows that they are equal.(3 votes)
- At5:50how does the graph show that in one light meter that the thing travels one meter? Also what is the thing? Is it the photon emitted by Sally?(3 votes)
- how do we know it travels one light meter? we know because this light ray is parallel to the other light ray which originated from -3*10^8 m. so if they are parallel they end up with the same values of the space time continuum(2 votes)
- I understand from the video that the speed of light is absolute but not relative. Does this mean that the motion of light is absolute? And if it is, then all motion in the universe could also be made absolute by comparing it to frame of reference of a photon! So is motion really relative or can it be viewed as absolute?(2 votes)
- The speed of light being absolute means that all observers will measure the same speed, regardless of what frame of reference they are in. There are an infinite number of photons moving in an infinte number of directions so I don't know what you have in mind to make measurements of the universe absolute by using a reference frame of a photon.(4 votes)
- Shouldn't the x' and t' axis have a 90 degree angle between each other? The axis are supposed to be perpendicular to each other, no?(2 votes)
- It is from Sally's references frame. But space and time change at super fast speeds.(3 votes)
- What is the reasoning behind that the angle of x' increases as the object increases in speed?
I basically understand, but a quick answer for this would be nice. :)(2 votes)- When you trace a path of an object moving at a constant rate in a Minkowski spacetime diagram you have a line. The angle of this line depends on the velocity of the object's velocity. At a velocity of 0 this line is a it is vertical and is parallel with the Y axis, as the velocity increases this angle away from vertical increases. This line is essentially the Y' axis of the objects frame of reference.
For the X' axis you have need to translate coordinates of one frame of reference to the other. Since the metric of Minkowski spacetime is Δs^2 -Δt^2 + Δx^2 + Δy^2 + Δz^2 and the 2 dimensional version is Δs^2 -Δt^2 + Δx^2. You will notice that the t component is negative. With the more familiar Pythagorean metric as the Y' axis rotates the X' rotates in the same direction but with the negative on the time element for Minkowski spacetime causes the X' rotate in the opposite direction.
(Sorry not a short answer)(2 votes)
Video transcript
- [Voiceover] We have been
doing some interesting things in the last few videos. We let go of our Newtonian assumptions that the passage of time is the same in all initial frame of reference that time is absolute that one second in my frame of reference is the same as one second passing in
your frame of reference. We even let go of the idea
that space is absolute that one meter in a certain direction in my frame of reference
is going to be the same as one meter in your frame of reference if we're are moving at relative velocities with respect to each other. And what that allows us to
do, it allowed us to reconcile what is actually observed in the universe. And that's the idea that
speed of light is an absolute. That regardless of what inertial
frame of reference we're in regardless of the speed
of the source of the light that we will always measure
light travelling at three times or roughly three times 10 to
the eighth meters per second. And when we let go of our assumptions about absolute time and absolute space and we did assume absolute
velocity for light. It gave us a very interesting diagram because it took our x
prime axis from being on top of the x axis and
put it an angle relative to the x-axis. Now, one question that you might have is well what is this angle,
the way I've drawn it, it looks like the angle
between x prime and x is the same as the angle
between t prime and t. Or as we did in the last videos,
we turned our units in time where the units along
this axis, the ct-axis, instead of saying seconds, we're now going to measure it in meters, and once again watch the last video if you're having an issue with that. But it looks like this angle
and this angle is the same. What I want to do in this video
is feel good about the fact that they really are the same. So, one of the convenient
things when a couple videos ago, we went through this thought experiment of in my friend's frame of reference, her admitting a photon of light bouncing off of the spaceship that's three times 10 to the eighth meters in front of her, and
then coming back to her. As we assumed that as
we mark off the meters, and as we mark off the seconds that the distance from the origin for three times 10 to the
eighth meters is the same as the distance from the
origin is one second. And now it makes even more
sense 'cause now we call this three times 10 to the eighth meters and this times three times 10
to the eighth meters as well and so the speed... When we depict the path
of light of a photon in either of these space time diagrams, it's always going to be at a either a positive 45 degree angle
or a negative 45 degree angle depending on the direction
that light is traveling in. In fact, that was what
we used to establish that this x prime axis is going to be, it's not going to sit
on top of the x-axis, it's going to be at an angle. But I don't actually
think about that angle. So, if we say that this angle
right over here is alpha and if we were to continue
this 45-degree line that we had to remember,
any path of a photon is gonna be at 45 degree
angle, so let me continue that. So, this is going to look like that and then we could go back and continue it right like that. Well, we know that this
going to be 45 degrees. 45 degrees right over there
this is going to be... Whoops, this is going to be
45 degrees right over there and we know, we that this
green triangle that I'm... It's not green, it's not different enough, let me get a better color. This??, no that's still green. We know that this, this purplish triangle that I'm setting up right here. This is an isosceles triangle. How do I know that this
is an isosceles triangle? Well, this hash mark right
over here, on the ct prime axis that is three times 10
to the eighth meters and this hash mark on
the x prime axis is also three times 10 to the eighth meters. And, so, this side is equal to that side and so we know that the base
angles of an isosceles triangle are going to congruent. So, this angle is going to
congruent to that angle. Well, if those two angles are congruent, then this angle is going
to congruent to that angle. Because they are supplementary
to those two base angles. And so notice, if we look at... If we look at this triangle... Let me do this in a new color... If we look at this
triangle right over here and compare it to this
triangle right over here notice they both have a 45 degree angle, they both have this blue
angle which is congruent so the third angle has to be congruent. They always have to add up to 180 degrees if you have two angles of
two different triangles of the same, then the third
angle has to be the same and so, if this is alpha, I shall do... Let's see, I've already used... I'll do four arcs there,
if that angle is alpha then this angle is going
to be alpha as well. And, so, this is really interesting, it's a beautiful kind of symmetry, and it really comes out of the fact that the speed of light
should always be measured at three times 10 to the
eighth meters per second in any inertial frame of reference. Notice, if we go back
to my frame of reference that photon that I emit
from my, my flashlight it will look like this. It will look like this. It's path in this Minkowski space time from my point of view will look like that and notice, from my friend's point of view the one that's travelling
on a ship for every amount let's say, let's pick a
certain point in space time right over there. Well, her x time coordinate is
going to be right over there and her ct prime coordinate is
going to be right over there so that makes sense. In one, in one I guess
we could say light meter that thing travels a meter. So, it is still travelling
at the speed of light from her point of view. Even though, she is moving
at half the speed of light relative to me. And you can think about what would happen if she was moving even faster. Someone moving even faster. Their ct prime axis
would be yet at an even, at an even more severe angle so it might look something like that. And then, what would their
x prime angle look like? Well then, their x prime
angle is going to symmetric around that line that
shows the path of light, so it would look like this. So, this is someone,
let's say who's moving even faster relative to me,
let's call that ct prime prime and this would be their x prime prime. As you can imagine, you get close as that second frame of reference gets closer and closer to speed of light from my frame of reference,
their coordinate axis are just going to get
more and more squanched up around, around this 45 degree
angle, so that's really neat. Now, another thing I wanted to think about and now, I won't draw it in this video just 'cause I think I've
given you enough to digest is there's a symmetry here. If my friend is moving at
half the speed of light moving at half the speed of light in the positive extraction
according to me, then from her point of view, I'm moving at half the speed of light in the negative x prime direction,
so you should think about well what would if I were to, if we were to look at this right over here what would my frame of reference look like if I were to project it on top of that.