• I
• student34
In summary, the conversation discusses a scenario where Object A and Object B pass each other on an x-axis, causing a photon to be released and absorbed. The speaker attempts to make spacetime diagrams to represent this situation, but runs into issues with accuracy and understanding. They are advised to use calculations and not rely on intuition alone. It is also noted that the origin of the diagrams is important, and should be placed at the meeting event of A and B.
student34
Here is a simple scenario where Object A and Object B cross past each other on an x-axis (for simplicity sake, let's just assume that they somehow passed each other on one spatial dimension). As they cross past each other, a photon gets released in the positive direction, and then gets absorbed soon after at the event (in blue). Below is my attempt at making two spacetime diagrams for this situation.

In Diagram 1, we see that I tried to show approximately where the event is relative to Object A and Object B. If my diagrams are at all correct, then the event is closer to Object B in space and time.

My problem: In Diagram 2, when Object B is at rest, the event shows up much further in spacetime than what I had shown in Diagram 1. Why does everything stop making sense when I try to make a diagram for the rest frame of Object B?

If you've drawn your diagrams accurately then you will find that the ##(t,x)## coordinates of any event in one frame and its ##(t',x')## coordinates in the other satisfy ##c^2t^2-x^2=c^2t'^2-x'^2##. I don't think you have drawn the diagrams accurately, however, since the meeting event of A and B lies on the x-axis in both diagrams. That can only happen if the event is at the origin.

What calculations did you do to draw these diagrams? Specifically, what are the coordinates of the meeting point of A and B, and the coordinates of the absorption event in each frame?

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student34, malawi_glenn, PeterDonis and 1 other person
Ibix said:
If you've drawn your diagrams accurately then you will find that the ##(t,x)## coordinates of any event in one frame and its ##(t',x')## coordinates in the other satisfy ##c^2t^2-x^2=c^2t'^2-x'^2##. I don't think you have drawn the diagrams accurately, however, since the meeting event of A and B lies on the x-axis in both diagrams. That can only happen if the event is at the origin.

What calculations did you do to draw these diagrams? Specifically, what are the coordinates of the meeting point of A and B, and the coordinates of the absorption event in each frame?
I did not think it was necessary to use calculations because I just wanted to see if the diagrams made sense in principle. I will attempt using an actual calculation with help from a video from Khan Academy to make all this more clear.

In his video, he uses distance units for the y axis. So my y-axis will be t = d/c, then d =ct. To make this easier for me, I will use the numbers he uses in his example. So ct = 1 and x = 1 for the meeting event. Using Khan's calculations, this brings the values of the location of the blue event to roughly (0.58, 0.58) like in his example.

The problem seems to persist for me. I think that I am missing a major principle or factor here, but I can not find it.

student34 said:
I did not think it was necessary to use calculations because I just wanted to see if the diagrams made sense in principle.
If you haven't done the calculations, at least basic ones, you won't know if the diagrams make sense in principle. The calculations are what tell you what makes sense. At the very least, you have to look at the equations and figure out what they imply about the relationships between diagrams drawn in different frames.

student34 said:
The problem seems to persist for me. I think that I am missing a major principle or factor here, but I can not find it.
That's because you're trying to use your unaided intuition. Don't. You should look at the actual math.

student34
student34 said:
So ct = 1 and x = 1 for the meeting event.
What event is "the meeting event"? If it's the event where Object A and Object B meet, I strongly suggest that you read the first paragraph of @Ibix's post #2 very carefully.

PeterDonis said:
What event is "the meeting event"? If it's the event where Object A and Object B meet, I strongly suggest that you read the first paragraph of @Ibix's post #2 very carefully.
Yes, the meeting event is where they meet. I think a crucial point of @Ibix's post is this, "That can only happen if the event is at the origin.", but I am not sure which event is being referred to or what the origin has to be. The origin that I chose is where the meeting event is. But does the origin have some other implicit meaning in relativity?

student34 said:
The origin that I chose is where the meeting event is.
No, it isn't. If the spacetime origin were the event where A and B meet, then in Diagram A, A's worldline would be the t axis, and in Diagram B, B's worldline would be the t axis. In other words, the x coordinate of A would be 0 in Diagram 1 and the x' coordinate of B would be 0 in Diagram 2. That's not how you drew them.

PeterDonis said:
No, it isn't. If the spacetime origin were the event where A and B meet, then in Diagram A, A's worldline would be the t axis, and in Diagram B, B's worldline would be the t axis. In other words, the x coordinate of A would be 0 in Diagram 1 and the x' coordinate of B would be 0 in Diagram 2. That's not how you drew them.
I could have labeled the diagram, but I did not think it would make a difference to the overall problem I have. I still do not know why it would matter where I put the t axis.

I also need to label the meeting event with the calculations as (0,0).

Here I have put in the origin (0,0) at the meeting event. I did not use the ct axis as the vertical world lines because I wanted the world lines to be visible and clear.

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student34 said:
Here I have put in the origin (0,0) at the meeting event.
That's a good start. Now go and label all the other events with their correct coordinates in both frames. (Hint: you will not be able to do that correctly with Diagram 2 without redrawing it.)

student34 said:
I did not use the ct axis as the vertical world lines because I wanted the world lines to be visible and clear.
I can understand this, but you should be aware that this is not how spacetime diagrams are standardly drawn in relativity. The ct axis in a standard diagram is the set of points that all have x coordinate zero. That is the worldline of Object A in Diagram 1. It's not some line off to the left of Object A.

student34
PeterDonis said:
That's a good start. Now go and label all the other events with their correct coordinates in both frames. (Hint: you will not be able to do that correctly with Diagram 2 without redrawing it.)
To be fair, he has done so, although he has still got the axes displaced and hasn't stated that the relative velocity of A and B is 0.5c. (Edit: and the blue event from diagram 1 is duplicated on diagram 2 for some reason - see the next two posts.)

@student34 - the only problem with your diagrams in #9 is that the event you marked as the origin is not at the origin as shown by your axes.

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student34
Ibix said:
@student34 - the only problem with your diagrams in #9 is that the event you marked as the origin is not at the origin as shown by your axes.
I moved the origin because I wanted to clearly show the vertical world lines.

Regarding my OP, do you know why there seems to be 2 different locations of the blue event for Object B? Is the event 1m and 1m away from Object B, or is it 0.58m and 0.58m. I am still confused about this.

student34 said:
Regarding my OP, do you know why there seems to be 2 different locations of the blue event for Object B?
Because you drew it twice. Why, only you can answer. I assumed you copied the event location from diagram 1 to compare its location in diagram 1 to its location in diagram 2 - it certainly does not represent any relevant event in diagram 2.
student34 said:
Is the event 1m and 1m away from Object B, or is it 0.58m and 0.58m.
In the rest frame of A, it's at (1,1). In the rest frame of B it's at (0.58,0.58).

student34 and vanhees71
student34 said:
I moved the origin because I wanted to clearly show the vertical world lines
By near-universal convention the axes cross at the origin. If you think you have a good reason to go against a convention then you can of course do so, but you absolutely must state clearly that you are doing it. Otherwise we all assume you're using the convention and point out your diagram is not correct with that convention.

I would add that you shouldn't go against convention unless you have a very good reason to do so. I would just use thicker lines for the worldlines than the axes (I usually use 3 pixels versus 1) and make sure that the axes extend further than the worldlines.

student34
student34 said:
I did not think it was necessary to use calculations because I just wanted to see if the diagrams made sense in principle. I will attempt using an actual calculation with help from a video from Khan Academy to make all this more clear.

In his video, he uses distance units for the y axis. So my y-axis will be t = d/c, then d =ct. To make this easier for me, I will use the numbers he uses in his example. So ct = 1 and x = 1 for the meeting event. Using Khan's calculations, this brings the values of the location of the blue event to roughly (0.58, 0.58) like in his example.

The problem seems to persist for me. I think that I am missing a major principle or factor here, but I can not find it.View attachment 304606

I never understood the obsession with Minkowski diagrams. Compared to the usually simple calculations concerning these kinematic effects they are pretty complicated. For me the greatest obstacle is that you have to forget all your intuition about the Euclidean plane, which we are hammered in our heads since we start to learn about geometry in elementary school. The Minkowski plane is NOT a Euclidean plane. The points of equal "distances" around a point are hyperbola (lines of constant timelike or spacelike distances) and straight lines (lightlike) rather than circles. It's not clear to me what you mean by drawing the horizontal line and then write "1m" close to it. It seems not to be the right construction of the units on object B's axes!

student34 and malawi_glenn
student34 said:
do you know why there seems to be 2 different locations of the blue event for Object B?
Because you didn't actually do the math to calculate what the correct coordinates of the blue event werein Object B's frame. If you had, you would have gotten the correct answer @Ibix gave in post #14 and you would have known to only draw the blue event once in Diagram 2, at those coordinates.

student34 and vanhees71
Ibix said:
Because you drew it twice. Why, only you can answer. I assumed you copied the event location from diagram 1 to compare its location in diagram 1 to its location in diagram 2 - it certainly does not represent any relevant event in diagram 2.

In the rest frame of A, it's at (1,1). In the rest frame of B it's at (0.58,0.58).
Thanks for the answer. But I do understand why B does not get the same treatment as A. Why can't the distance be (1,1) for B like it is for A?

For example, let's say that A did not exist. Wouldn't B calculate like A did?

student34 said:
Thanks for the answer. But I do understand why B does not get the same treatment as A. Why can't the distance be (1,1) for B like it is for A?

For example, let's say that A did not exist. Wouldn't B calculate like A did?
Calculate what? You have not given a specific scenario, just an unspecified "absorbed at an event". If the event is (1,1) in A, then it is not (1,1) in B.

student34 and vanhees71
student34 said:
I do understand why B does not get the same treatment as A. Why can't the distance be (1,1) for B like it is for A?
Once more: do the math. Actually do it. Don't just wave your hands. The math will answer this question for you. But you have to do it.

student34 said:
let's say that A did not exist. Wouldn't B calculate like A did?
You could certainly set up a scenario in B's rest frame that was analogous to the scenario you have set up in A's rest frame. But it won't be the same scenario.

Once again: do the math. Actually do the calculations and see what a Lorentz transformation does to light rays.

student34 and vanhees71
student34 said:
Here I have put in the origin (0,0) at the meeting event. I did not use the ct axis as the vertical world lines because I wanted the world lines to be visible and clear.

View attachment 304608

That is wrong. In both diagrams, the red world line of object A is the ct-axis and the gray world line of object B is the ct'-axis, because the objects met at event x=x'=t=t'=0. So A stays at x=0 and B stays at x'=0. You need no extra black axes.

Diagram 1 shows the gray ct'-axis and the gray x'-axis, as labeled in the video. In diagram 2, a red x-axis is missing. The x'-axis is oriented horizontally in diagram 2 and should be gray.

You should also be aware, that in each diagram the axes of the two frames have different scales, because the "rotation" of an axis is not a normal rotation, but a hyperbolic rotation.

Source:
https://www.geogebra.org/m/dYyg5ZB8#material/NnrRvA46

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student34 and vanhees71
student34 said:
do you know why there seems to be 2 different locations of the blue event for Object B?
Becuase minkowski space is not Euclidean

student34 and vanhees71
Yes! As I said above: Forget the Minkowski diagram. Obviously you are not aware how to read or even to draw them. They are more complicated than the simple math many adviced you to do.

For simplicity put the origin of both inertial frames at the origin when A and B meet and let the wave front of the light signal be described by
$$x(t) = \lambda \begin{pmatrix} 1 \\ 1 \end{pmatrix}.$$
Now B moves with a velocity ##v=\beta c## along the ##x^1##-axis. Seting ##\gamma=1/\sqrt{1-\beta^2}## the wave front of the light signal is described in B's frame as
$$\hat{\Lambda}(v) x = \lambda \gamma \begin{pmatrix} 1-\beta \\ 1-\beta \end{pmatrix}.$$
If in ##A## the light is absorbed at ##x^1=1 \text{m}##, you have ##\lambda=1 \text{m}##. I.e. B will observe the light to be absorbed at
$$x^{\prime 1}=\gamma(1-\beta) \lambda =1\text{m} \sqrt{\frac{1-\beta}{1+\beta}}.$$

student34 said:
Thanks for the answer. But I do understand why B does not get the same treatment as A. Why can't the distance be (1,1) for B like it is for A?
If something is fixed to the ground in front of me and I turn around, why can't it still be in front of me? In that case I've rotated my coordinates; in your example they are boosted. But the principle is the same - the x direction is not the x' direction and the t direction is not the t' direction, so why would you expect coordinates specified in terms of those directions to be the same?

vanhees71 and student34
Sagittarius A-Star said:
That is wrong. In both diagrams, the red world line of object A is the ct-axis and the gray world line of object B is the ct'-axis, because the objects met at event x=x'=t=t'=0. So A stays at x=0 and B stays at x'=0. You need no extra black axes.
I did not want to use the y-axis for each object's ct axis because I wanted their respective world lines clear to me. I didn't think it mattered. But you are right, it makes things more confusing in different ways.

Sagittarius A-Star said:
Diagram 1 shows the gray ct'-axis and the gray x'-axis, as labeled in the video. In diagram 2, a red x-axis is missing. The x'-axis is oriented horizontally in diagram 2 and should be gray.

You should also be aware, that in each diagram the axes of the two frames have different scales, because the "rotation" of an axis is not a normal rotation, but a hyperbolic rotation.

Source:
https://www.geogebra.org/m/dYyg5ZB8#material/NnrRvA46
I think I know what I did wrong now. I think that I changed the rest frame to B while also keeping the rest frame of A. This is probably why I am getting two events for B. Do you agree?

student34 said:
I did not want to use the y-axis for each object's ct axis because I wanted their respective world lines clear to me.
The "y" axis can anyway not be each object's ct (ct') axis because in each diagram only the ct (ct') axis of the object at rest is vertically. The ct' (ct) axis of the moving object has an angle, for example ##\arctan(-0.5) \approx -26.57°## for v/c=-0.5. So there is no overlapping of axes.

student34 said:
I think I know what I did wrong now. I think that I changed the rest frame to B while also keeping the rest frame of A. This is probably why I am getting two events for B. Do you agree?
Yes.

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student34
Sagittarius A-Star said:
The y-axis can anyway not be each object's ct (ct') axis because in each diagram only the ct (ct') axis of the object at rest is vertically. The ct' (ct) axis of the moving object has an angle, for example arctan⁡(−0.5)≈−26.57° for v/c=-0.5. So there is no overlapping of axes.
He means that he did not want the time axes to overlap with the world lines of the object at rest in each respective frame.

vanhees71 and Sagittarius A-Star

## 1. What is a spacetime diagram?

A spacetime diagram is a graphical representation of the relationship between space and time in the theory of relativity. It shows how objects move through both space and time, and how these two dimensions are interconnected.

## 2. How do I read a spacetime diagram?

In a spacetime diagram, the horizontal axis represents space and the vertical axis represents time. The diagonal lines on the diagram, called world lines, show the path of an object through both space and time. The steeper the line, the faster the object is moving through space. The closer the lines are together, the slower the object is moving through time.

## 3. What is the purpose of using a spacetime diagram?

The purpose of a spacetime diagram is to help visualize and understand the concepts of space and time in the theory of relativity. It allows scientists to see how objects move through both dimensions and how they are affected by factors such as gravity and velocity.

## 4. How do I use a spacetime diagram to solve a relativity problem?

To solve a relativity problem using a spacetime diagram, you must first understand the problem and identify the relevant variables such as velocity, distance, and time. Then, plot these variables on the diagram and use the relationships between space and time to analyze the problem and find the solution.

## 5. Can a spacetime diagram be used for any type of relativity problem?

Yes, a spacetime diagram can be used for any type of relativity problem as long as the problem involves the concepts of space and time and their relationship. It is a useful tool for visualizing and understanding the complex concepts of relativity and can be applied to various scenarios and problems.

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