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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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  • leafers tree style avatar for user Leonard
    At you 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)
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    • blobby green style avatar for user em.are.kish
      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)
  • leaf green style avatar for user Coco
    How would this work in circular or spiral frames of reference (since our Earth and solar syst. move in a spiral)?
    (9 votes)
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  • mr pants teal style avatar for user Udbhav Dalal
    I don't understand what Sal says at - "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)
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    • male robot hal style avatar for user Charles LaCour
      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)
  • leaf orange style avatar for user Sabith
    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)
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  • duskpin sapling style avatar for user saudj.11232000
    Why speed of light is absolute ?
    (2 votes)
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  • leafers ultimate style avatar for user jllusty
    Sal, thanks for the great video. At you 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)
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    • blobby green style avatar for user rinbar70
      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)
  • blobby green style avatar for user Sidney
    At how 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)
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  • aqualine ultimate style avatar for user Surya Shashank Sekhar H
    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)
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    • male robot hal style avatar for user Andrew M
      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)
  • leafers seed style avatar for user Hassaanghazali
    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)
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  • hopper cool style avatar for user Iron Programming
    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)
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    • male robot hal style avatar for user Charles LaCour
      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.