B Understanding Spacetime Diagrams

[NoahsArk]

I got the book "An Illustrated Guide To Relativity" by Tatsu Takeuchi, and have questions on how to understand spacetime diagrams from different reference points. Before I ask, please let me know how I can draw a spacetime diagram and post it on the forum. I will want to use different colors to differentiate lines of simultaneity from different reference points. Thank you.

 

[Ibix]

You can draw it on paper and take a photo. We'd appreciate it if you used plenty of light and cropped the image to just the diagram.

You can draw one in your favourite art program. Depending on how disciplined you are this can be quite accurate, but is labour intensive.

I wrote a cheap and cheerful Minkowski diagram generator in Javascript: http://www.ibises.org.uk/Minkowski.html. It's point-and-click, but predates the ubiquity of touch screens and doesn't work particularly well with them.

I also wrote a python library for drawing Minkowski diagrams. You'd need to know the basics of python to be able to use it, but I can post it at the weekend if you like.

In any case, once you've got an image you can post it either by opening in a graphics program and copy-pasting into the editor, or by clicking on the image button (a sun over mountains in a frame) in the editor's toolbar.

 

[pervect]

Space-time diagrams aren't really any different from graphing position versus time in some chosen coordinate system. The conventions are a bit different in that time generally runs "up the page" rather than from left to right.

The theory of the space time diagram is fairly simple. To represent time on a piece of paper, we use the concept of timelines.

There's a separate timeline for every different point in space. The vertical position of a point respresent time, it's place on the timeline. The horizontal position of a point represents its position in space.

Another needed concept is the idea of the worldline. This is a line that shows the entire history of an object through time. It's a line because a point on a space-time diagram is an event, and an event happens at a particular place in space, and at a particular time. The worline plots the position of the object as a function of time, so it is a line rather than a point.

I would suggest drawing some basic diagrams in the following order:

1. The diagram representing the worldline of a stationary object
2. The diagram representing some moving object with a velocity lower than the speed of light
3. The diagram representing the worldline of a light beam
4. The diagram of a light clock "at rest". This involves putting together the diagrams for a pair of mirrors (one worldline for each mirror) and several beams of light bouncing between them.
5. The diagram of a moving light clock. Don't worry about the scale factor too much.

When you get to phase 5, you can find a simple technique (light clocks must have equal areas) in "Relativity on rotating graph papers, https://www.physicsforums.com/insights/relativity-rotated-graph-paper/

There are some other links with the same content, a peer-reviewed paper at https://aapt.scitation.org/doi/abs/10.1119/1.4943251?journalCode=ajp which may be paywalled, and a pre-print that is different from the actual published version on arxiv at https://arxiv.org/abs/1111.7254

 

[vanhees71]

The more conventional treatment of Minkowski diagrams can be found here:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
Note that it's not as trivial as claimed in the postings above. You must forget all you know about Euclidean (planar) geometry. The Minkowski diagram depicts a Minkowski plane with its specific geometry governed by the indefinite fundamental form $\mathrm{d} s^2=c^2 \mathrm{d} t^2-\mathrm{d} x^2$, and the minus-sign is crucial here!

 

[pervect]

The more conventional treatment of Minkowski diagrams can be found here:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
Note that it's not as trivial as claimed in the postings above. You must forget all you know about Euclidean (planar) geometry. The Minkowski diagram depicts a Minkowski plane with its specific geometry governed by the indefinite fundamental form $\mathrm{d} s^2=c^2 \mathrm{d} t^2-\mathrm{d} x^2$, and the minus-sign is crucial here!

Being able to draw space time diagrams, especially ones including light clocks, may not tell one everything there is to know about special relativity, but in my opinion it is an excellent first step.

The original poster is interested in drawing space-time diagrams, which I think is a great idea. The ability to draw light clocks on such diagrams will both illustrate the techniques of how to draw the diagrams, and additionally allow the diagrams to make numerical predictions about relativistic problems such as the twin paradox in a simple manner that doesn't involve any significant math, just diagrams.

I wouldn't want to discourage the original poster if he wanted to learn the math, of course.

 

[vanhees71]

I've nothing against drawing Minkowski diagrams. I do so in my FAQ article linked above, but I think it is important to emphasize that at the moment you draw a Minkowski diagram you should forget about Euclidean geometry. When I first learned special relativity in high school I got very confused with the treatment in our school book, where they did not emphasize that the Minkowski plane is not a Euclidean plane and even used Euclidean geometry to analyze the diagram! Fortunately I had a very good physics teacher, and she gave us the correct Minkowskian reasoning.

The problem of course is the construction of the axes scales. You can start with arbitrary axes scales for one inertial reference frame $\Sigma(ct,x)$, but then also the scales for axes for an arbitrary other inertial reference frame $\Sigma'(c t',x')$ is fixed. We assume that $\Sigma'$ moves with a velocity $v$ (with $|v|<c$ in $x$ direction.

The direction of a light signal is determined in both frames by $x=ct$ and $x'=c t'$. The time axis $x'=0$ of the new frame is determined in the old frame by $x=v t$, and through the known light cone also the spatial axis $t'=0$ is fixed.

To get the right scales you need to use the Minkowski "metric" (rather than the Euclidean metric), according to which the hyperbola $c^2 t^2-x^2=1 \text{unit}^2$ intersects the new time axis at 1 unit. For the x'-axis you have to use the hyperbola $c^2 t^2-x^2=-1 \text{unit}^2$. Of course this gives the same unit length as the one on the time axis by symmetry.

This construction can be shown to be equivalent to the construction in the "Bondi diagrams" used in the mentioned insights article

https://www.physicsforums.com/insights/relativity-rotated-graph-paper/

 

[NoahsArk]

Thank you for the helpful responses.

I will first take some pictures of diagrams from the book for the specific questions. I understood the diagrams in the book as they showed Galilean transformations and velocity addition. I even understood, for the first time, how to properly draw lines of simultaneity, which are slanted instead of horizontal when drawn for the moving frame.

I get lost though when trying to visualize Lorentz transformations or velocity addition on the digrams.

 

[Ibix]

I get lost though when trying to visualize Lorentz transformations or velocity addition on the digrams.

Look at my Javascript version, linked above. It animates the frame changes when you boost.

 

[vanhees71]

Thank you for the helpful responses.

I will first take some pictures of diagrams from the book for the specific questions. I understood the diagrams in the book as they showed Galilean transformations and velocity addition. I even understood, for the first time, how to properly draw lines of simultaneity, which are slanted instead of horizontal when drawn for the moving frame.

I get lost though when trying to visualize Lorentz transformations or velocity addition on the digrams.

Which book are you using?

 

[NoahsArk]

Ibix thank you for all the suggestions on how to post a spacetime diagram in here. Of your suggestions, I will start by using pictures since that's the easiest way for now. I tired your javascript page which is interesting, and will try using it more to get a better feel for it.

Vanhees the book is "An Illustrated Guide to Relativity" by Tatsu Takeuchi.

There are six related diagrams I am working with now, each of which deal with two frames of reference: one frame of reference is the "tree frame" where someone is standing next to a tree, and the other frame is the "car frame" where the car is driving away from the tree at .5c. On each of these diagrams, the pink colored line is the world line of the car, the grey line is the world line of the tree, and the two crimson lines are two beams of light:

The top diagram above shows points A E F G in the tree frame which are all simultaneous in the car frame, but which all occur at different times in the tree frame. I understand that from the cars point of view, this line of simultaneity ("LOS") must be slanted so that the LOS is equidistant from both beams of light.

Here's the problem: The author says "I argued earlier that line AEFG in the tree frame diagram must fall on a horizontal line on the car frame diagram, but I haven't said yet which horizontal line it is".

So I assume the author wants us to figure out where on the time axis the horizontal line needs to be drawn in the car frame? He goes on to explain the solution by saying "one is tempted to think that the line corresponds to the points $A \prime E \prime F \prime G \prime$ on the car frame diagram (second diagram from above) since the spatial separations of the tree, the car, and the two photons in $x \prime$ are the same as those in x." My only reason for thinking that the second diagram above couldn't be true is that point E, which is the location and time of the car in the tree frame, is at t = 4 in the tree frame, and $E \prime$ in the diagram below is at $t \prime = 4$. We know that t can't =$t \prime$ for even E because event E happened at a distance from the tree frame, and the car is in motion relative to the tree? I assume, though, that the author is trying to explain it another more visual way. The author continues with another pair of diagrams, the top one which is again in the tree frame and the bottom in the car frame:

He says this cannot be correct because it implies that $H \prime \prime F \prime \prime E J \prime \prime$ on the tree frame diagram correspond with $H \prime F \prime I \prime J \prime$ on the car frame diagram. This makes my head spin a bit and I'm not sure I understand his reasoning, and the only reason why I can think that it's wrong is because the event F can't be at the same time, t = 4, in both frames.

Now the author introduces a rule to use to determine where to draw the line in the car frame given where it is in the tree frame, and shows a third pair of diagrams. He calls the rule the "conservation of spacetime volume" rule:

Referring to these diagrams he says: "On the tree frame diagram (first diagram of the above pair), draw a line parallel to the car worldline that goes through A, and a line parallel to AEFG that goes through the origin O. Call the point when the two lines cross P. Note that P is at the same time as O, and at the same place as A, in the car frame. So the diamond AEOP on the tree frame diagram must correspond with a square, like the one in the bottom figure, shown on the tree frame diagram."

How can we conclude that this diamond in the tree frame must correspond with a square in the car frame? To do that, wouldn't we need to know that this distance between O and P is the same as the distance between O and E? How do we know that these distances are the same?

He then continues: "In the current case we are considering here, the area enclosed in the diamond AEOP on the tree frame diagram is $12m^2$. Just count the number of squares in the diamond."

I have re-read this multiple times and don't how we get $12m^2$ as the area of the diamond in the tree frame diagram. I can't just count the number of squares in the diamond because many of the squares in the diamond are cut up into odd fractions. Am I missing something?

"So the length of the sides of the square on the car-frame diagram must be $\sqrt 12m^2 = 2 \sqrt 3m = 3.5m.$ Therefore, points AEFG lie on the horizontal line with ## ct \prime = 3.5m, as shown on the car frame diagram."

Just looking at the diamond shape in the tree frame, each side appears to be at least 4m in length, which makes it even more unclear to me about how the author gets 3.5m as the length of the sides of the square in the car frame.

"Under the conservation of spacetime volume rule, the square on the tree frame diagram which has OF as a one of its sides will be transformed to the diamond with the same area on the car-frame diagram as shown. So under the inverse transformation F will be mapped back to where it originally was.

This conservation of spacetime area maintains the symmetry between the tree and car frames, since each is moving at the exact same speed when observed from the other frame, and ensures that the correspondence between the points on the two diagrams is one-to-one. It is a consequence of the fact that the "number of events" enclosed in AEOP must be independent of the frame of reference."

So, is the spacetime conservation rule an alternative method, in situations like this, to using the lorentz transformations? When I used the lorentz transformation in the above example to convert the time of event E in the tree frame to the time of event E in the car frame, I got the same answer of 3.5.

 

[Ibix]

I am getting the impression that this is not a great book. It seems to be taking the view that showing you a number of wrong ways to do something will help you understand the right way. I'm not convinced.

The easy way to see that the first two diagrams must be wrong is the absence of length contraction. If the tree says the car is 2m away, the car cannot agree that the tree is 2m away.

The problem with the area of the square is that $2\sqrt 3$ is about 3.46, not 3.5. The diagram doesn't have sufficient resolution to show that it's a 3.46×3.46 square, not a 3.5×3.5 square. So you cannot count the area accurately, as the text asks you to do. The author should have chosen a speed of 0.6c or 0.8c - the numbers come out rational.

Regarding lengths - you cannot (in general) just measure lengths on the diagram. It's a Euclidean representation of a plane that does not obey Euclidean geometry, and measuring lengths on it has similar pitfalls to measuring distance on a Mercator projection map. It works in special cases (parallel to the x or t axis in this case) but not in others. That's why the lengths of the sides of the representation of the parallelogram that is drawn are not the same as those of the square. If you measure $\Delta x$ and $\Delta t$ along the line (you are measuring parallel to the axes, so that works) and compute the interval, $\Delta s=\sqrt{|\Delta x^2-\Delta t^2|}$, that will be 3.46m, give or take rounding error from the author's poor choice of frame speed.

Honestly, I wouldn't faff around like this. You know the Lorentz transforms. Suppose you want to draw the line of simultaneity for the primed frame at $t'=4$ on the unprimed diagram. $t'=\gamma(t-vx/c^2)$ tells you that you want to draw $t=4/\gamma+(v/c^2)x$. Draw that line. Job done.

 

[vanhees71]

I think the book is overcomplicating the issue. With a diagram-only approach I've been always lost. For me the greatest revelation in high school was that one can do geometry without drawing diagrams but calculating things with formulas and then do the drawings to visualize the calculations.

I'm not sure yet, whether the book is really wrong. It would take me a hard time to check this carefully enough.

The construction with the areas of the "diamonds", is pretty clear algebraically. It's also the foundation of the Bondi approach and the visualization with "rotated graph paper" mentioned above. In the algebraic approach it's very simple: What's invariant is the Minkowski "length", i.e.,
$$\Delta t^2-\Delta x^2=\Delta t^{\prime 2}-\Delta x^{\prime 2}$$
for an arbitrary space-time interval. The introduction of the "rotated graph paper" is algebraically nothing else than expressing everything in light-cone coordinates $x_{\pm}=t \pm x$. Then the above invariance property simply becomes
$$\Delta x_+ \Delta x_-=\Delta x_{+}' \Delta x_-',$$
i.e., the areas of the diamonds in the two frames are the same.

So you can construct the units in the moving frame, given the units in the primed frame by making these diamonds of equal area as well as by drawing the unit hyperbolae $t^2-x^2=\pm 1$ and then getting the units on the $t'$ and $x'$ axis by the intersections of these lines with these hyperbolae.

Another way to see it with the unit hyperbola is the following, using rapidities: Written in space-time coordinates you have for e.g., the time-like unit hyperbola
$$\begin{pmatrix} t \\ x \end{pmatrix}=\begin{pmatrix} \cosh y \\ \sinh y \end{pmatrix}$$
and in light-cone coordinates
$$\begin{pmatrix} x_+ \\ x_- \end{pmatrix}=\begin{pmatrix} \exp y \\ \exp(-y)\end{pmatrix},$$
i.e., $x_+=1/x_-$, i.e., for any point on the hyperbola $x_+ x_-=1$. As I said, the algebra is much simpler than space-time diagrams ;-))).

 

[NoahsArk]

I appreciate the responses. The book tries to explain addition velocity, time dilation, and lorentz transformations by only using diagrams. I couldn't see through the diagrams how any of those concepts are illustrated. The author, did, though, explain through a simple diagram (first one above in my previous post) how relativity of simultaneity can be seen on a graph. That alone was worth the price of the book for me, even though all the concepts that he builds on from that were confusing. To see if I can correctly explain it based on my memory, he basically explains it like this: if a car is moving at .5c in the tree frame, and two beams of light shoot out when the car is at the origin, then, according to the tree observer, the car will be half way between the right light beam and the tree, and will be much closer to the right light beam than the left. According to the car observer, though, the car will equal distance between both beams.

If you are looking at the diagram of the tree frame, in order to draw a line of simultaneity through the car's world line which is just as long to the right beam as it is to the left beam, it has to be drawn diagonally. I am assuming that based on the speed of the car, there is a formula which tells you what angle to draw that line at?

The author, as well as other posters here, have argued that understanding relativity of simultaneity is key to understanding how time dilation works. I am trying to understand that connection, but so far I am not sure the reason.

It appears that my understanding of time dilation and why it occurs was fundamentally wrong for a long time, and I am trying to get the right understanding. Typically time dilation is explained using the example of a light clock. It's easy to see that for someone in the rocket the light beam is going a shorter distance than someone observing it from Earth (since in the rocket it's only bouncing up and down whereas for someone on Earth it's going both up and across), and that therefore time should be passing more slowly in the rocket. What if, though, we constructed another light clock: say a laser gun which is positioned in the middle of the rocket and shooting beams of light toward the tail. Each time a beam hit the tail, a new beam would fire. One tick on this clock would be the time it takes for the light to go from the gun to the tail. From someone's perspective on earth, each tick is faster than it is for the person in the rocket since for the person on Earth the beam is traveling a shorter distance due to the fact that the tail is catching up to the beam. From this why not conclude that the person on Earth will measure time moving faster on the rocket? Somewhere I've gone very wrong I just don't know where?

 

[Pencilvester]

Each time a beam hit the tail, a new beam would fire.

This is where relativity of simultaneity comes into play. The two events “beam hitting the tail” and “firing a new beam” might be simultaneous in the rocket frame, but what makes you think those two events will be simultaneous in the Earth frame?

 

[Ibix]

I am assuming that based on the speed of the car, there is a formula which tells you what angle to draw that line at?

Yes - the Lorentz transform, as I said in #11. The point is that the line of simultaneity is defined as the line of equal $t'$. So if you set $t'=T'$, where $T'$ is some constant (I used 4 in #11), then the line of simultaneity on your unprimed frame's diagram is the $(x,t)$ values that Lorentz transform to $T'$. Thus $T'=\gamma(t-vx/c^2)$, which you can rearrange into $t=(v/c^2)x+T'/\gamma$.

From this why not conclude that the person on Earth will measure time moving faster on the rocket? Somewhere I've gone very wrong I just don't know where?

As @Pencilvester comments, you've forgotten the relativity of simultaneity. I sketched a Minkowski diagram below - please excuse the lack of straight lines.

This is drawn in the frame of the rocket. The black line up the middle is the laser and the blue lines are the front and rear of the rocket. The red lines are the laser pulses, arriving at the front and rear at the same times as each other in this frame, and at the same time in this frame as the next pulse is emitted. But I've also added a purple dashed line, which is the line of simultaneity of the Earth frame. Note that the three events are not simultaneous in the Earth frame - which is what is wrong with your analysis. The reason for using a round-trip style clock in the standard approach is that the next emission event happens in the same place as the pulse return, so there is no ambiguity about simultaneity.

You can go a huge way towards avoiding this type of error by reviewing anything you write and searching for any word like "simultaneously" or "at the same time" and adding which frame regards the events as simultaneous.

 

[NoahsArk]

This is where relativity of simultaneity comes into play. The two events “beam hitting the tail” and “firing a new beam” might be simultaneous in the rocket frame, but what makes you think those two events will be simultaneous in the Earth frame?

Yes, I didn't take that into account!

Ibix thank you for the diagram. Doesn't this show, though, that the typical way of explaining time dilation, i.e., with a light clock, is not the real explanation for time dilation? The whole reason why the light clock is supposed to show time dilation is that the beam of light in the Earth frame is traveling a longer path than the beam in the rocket, and c is the same speed in both frames. In the laser gun example, though, the beam is traveling a shorter distance than it is in the rocket frame. So, is it true that we can't say that time dilation occurs due to the longer distance of the light traveling in the Earth frame? As you pointed out, the longer time has to do with ROS.

Also, when you said that it might not be a good book to learn from, do you mean that in general it's difficult to learn SR concepts purely through diagrams, or did you just mean that the specific way the book is describing things is difficult? Can we, for example, illustrate time dilation with spacetime diagrams? If so, I would appreciate seeing how. I still don't quite understand the relationship between ROS and time dilation.

 

[Pencilvester]

is it true that we can't say that time dilation occurs due to the longer distance of the light traveling in the Earth frame?

The universe works the way that it works, and it doesn’t matter too much which mathematical route you choose to describe certain phenomena as long as it accurately describes what is actually happening.

You could look at time dilation as a result of the path light takes in a light clock looking different between different frames, or you could look at it as a special case of the Lorentz transformation, which itself can be thought of as resulting from the principle of relativity and the constancy of the speed of light in inertial frames. They’re just different routes to get to the same answer.

 

[vanhees71]

Also, when you said that it might not be a good book to learn from, do you mean that in general it's difficult to learn SR concepts purely through diagrams, or did you just mean that the specific way the book is describing things is difficult? Can we, for example, illustrate time dilation with spacetime diagrams? If so, I would appreciate seeing how. I still don't quite understand the relationship between ROS and time dilation.

I personally find it very difficult to learn SR from spacetime diagrams, but that's just an individual thing. It may well be that you can understand the math of SR better from spacetime diagrams than from algebraic formulae. Just try it out. A more algebraic approach you can find in my SRT FAQ (not finished, but the kinematical part is complete):

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
I also included Minkowski diagrams (as the most used type of spacetime diagrams in SR). You find the depiction of time dilation and length contraction in Fig. 1.2. Note also the construction of the unit lengths on the different coordinate axes via the hyperbolae in Fig. 1.1. I think the latter issue is the most difficult thing with Minkwoski diagrams: You have to force yourself not to think about length in the Euclidean way we are all used to from elementary school on!

 

[NoahsArk]

When I first learned special relativity in high school I got very confused with the treatment in our school book, where they did not emphasize that the Minkowski plane is not a Euclidean plane and even used Euclidean geometry to analyze the diagram!

When I think of Euclid I remember intersecting lines, supplementary angles, and things like that. What does it mean in this context, that a Minkowski plane is not Euclidean? What Euclidean rules, for example, doesn't it follow.

 

[Pencilvester]

When I think of Euclid I remember intersecting lines, supplementary angles, and things like that. What does it mean in this context, that a Minkowski plane is not Euclidean? What Euclidean rules, for example, doesn't it follow.

The primary difference between the Euclidean plane and the Minkowski plane is that the line element in the MP is not positive definite. In the EP (with two dimensions and using cartesian coordinates), the line element can be written as $$ds^2 = dx^2 + dy^2$$However, in the MP (with cartesian coordinates again, except instead of $x$ and $y$ you typically use $t$ and $x$) it is usually written as $$ds^2 = -dt^2 + dx^2$$This implies that pairs of nearby points can have positive, negative, or even zero “distance” (in the Minkowski sense of the word) between them. In relativity texts, instead of “distance” you’ll often see “spacetime interval”.

If you have a curve on the MP on which every pair of points has a negative spacetime interval (given the line element I mentioned earlier), then it’s called a timelike curve—the worldlines of all massive particles are timelike. Conversely, if every pair of points has a positive spacetime interval, then the curve is “spacelike”—you can think of that as a curve literally drawn in space at a single instant of time (I won’t go into all the caveats that go with that last bit). Finally, if the spacetime interval is everywhere zero on the curve, then it is “lightlike”, which, as the name implies, applies to curves traced out in spacetime by light.

 

[Ibix]

What Euclidean rules, for example, doesn't it follow.

See Ben Crowell's GR text, free to download from his website at http://www.lightandmatter.com/genrel/. The first chapter covers it, and there's a quick reference on p411.

As Pencilvester's answer shows, we don't generally use an axiomatic approach but a metric based approach. The difference between Minkowski and Euclidean geometry is that the sign on one of the metric elements is flipped with respect to the others.

 

[vanhees71]

When I think of Euclid I remember intersecting lines, supplementary angles, and things like that. What does it mean in this context, that a Minkowski plane is not Euclidean? What Euclidean rules, for example, doesn't it follow.

The notion of "distance" and "angle" is different in Minkowski space is different. The most important thing to keep in mind on this context is that the unit intervals on the axes of a Minkowski diagram have not the same Euclidean length when drawn on paper, but you have to gauge the unit intervals on the axes of different inertial reference frames relative to each other correctly by using hyperbolae in the Minkowski plane. The analogous construction in the Euclidean plane uses of course circles, and at least for me it's hard to remember this always and don't get confused when reading off space-time intervals in different frames of reference.

 

[Herman Trivilino]

What Euclidean rules, for example, doesn't it follow.

The hypotenuse is not the longest side of a right triangle.

 

[NoahsArk]

Thank you for the responses. I will have to think about this more to be able to try and better visualize what is going on.

ds2=−dt2+dx2

The way I've seen the space time interval written is $t \prime^2 = t^2 - x^2$ Are these equivalents?

The most important thing to keep in mind on this context is that the unit intervals on the axes of a Minkowski diagram have not the same Euclidean length when drawn on paper, but you have to gauge the unit intervals on the axes of different inertial reference frames relative to each other correctly by using hyperbolae in the Minkowski plane.

This is something that is hard for me to picture. I was thinking, we can draw the diagram from the tree frame and have each t unit and each x unit be the same length (e.g. one box like in the picture from the book I posted). The tree frame's lines of simultaneity will be horizontal which is intuitive. The car frame's lines of simultaneity will be slanted. The distance between each line of simultaneity in the car frame will also be different, I think, in the car frame than in the tree frame, and so will the world lines. What is the formula for determining the distance between lines of simulaneity and worldlines in the primed (car) frame? Also could we draw a space time diagram with numbered lines of simultaneity in the tree frame, and, in the same diagram, do the same for the car frame?

The hypotenuse is not the longest side of a right triangle.

In the space time interval equation, $t \prime^2 = t^2 - x^2$, would the $t \prime^2$ term be the hypotenuse? It's hard for me to picture this on a diagram.

 

[NoahsArk]

I also wanted to say something more general about space time diagrams: I am a more visual learner, which is why I am trying to take this approach. From my understanding, a space time diagram, can, in one graph, show all the major concepts of SR, why time travel to the past is not possible, how the twin paradox works, the nature of causality, and many other concepts which are fascinating for me. Correct me if I'm wrong, but it can show these things if one is properly interpreting the space time diagram. It is still tricky for me to interpret a diagram from multiple frames of reference due to the fact that not only the time access gets tilted, but also the space axes, so I feel I'll need some time drilling it in and getting used to how to read them.

 

[Herman Trivilino]

It's hard for me to picture this on a diagram.

Suppose a rocket ship is moving in a straight line, just above a long straight section of roadway at a speed of $0.6 c$ (relative to the road). A tiny explosion on the ship generates a spark that leaves a mark on the roadway, and then 4 seconds later (as measured by a clock on board the ship) another tiny explosion on the ship generates a spark that leaves another mark on the roadway. A person stands still on top of each mark, They measure the distance between the marks to be 3 light-seconds. Each person has a clock with them (they have synchronized their clocks with each other) and determine that the time that elapsed between explosions is 5 seconds.

$(4)^2=(5)^2-(3)^2$

 

[Herman Trivilino]

It is still tricky for me to interpret a diagram from multiple frames of reference due to the fact that not only the time access gets tilted, but also the space axes, so I feel I'll need some time drilling it in and getting used to how to read them.

You need to focus on events. Each of the tiny explosions in my previous post is an event. Focus on the time that elapses between events as well as the distance between those events.

 

[Pencilvester]

The way I've seen the space time interval written is $t′^2=t^2−x^2$ Are these equivalents?

If you are looking at a spacetime diagram that plots the worldline of an inertial observer moving with respect to the given frame with velocity $v$ such that her time coordinate is called $t’$ and the origin of her frame coincides with the origin of the frame of the diagram we’re looking at (basically the quintessential beginners’ spacetime diagram, but I wanted to explicitly state all of the conditions that constrain your equation), then that equation will only hold on the moving observer’s worldline where $x=vt$ and therefore could just as easily be written as $t’^2=t^2(1-v^2)$ or $t’=\frac{t}{\gamma}$, which is just the time dilation formula.

The equation $ds^2=dt^2 -dx^2$ is much more general—it applies to any pair of points on the spacetime diagram (using cartesian coordinates on flat spacetime). You might notice that I flipped the signs on the dt and dx. This is a matter of convention. The important thing is, as @Ibix said, that one of the components of the metric have a different sign than the rest (I say “rest” instead of “other” because the full metric in SR is four-dimensional).

 

[robphy]

In the space time interval equation, $t \prime^2 = t^2 - x^2$, would the $t \prime^2$ term be the hypotenuse? It's hard for me to picture this on a diagram.

Here are two 3-4-5 triangles in Minkowski spacetime.
Can you see them?

This is how I use graph paper to draw spacetime diagrams,
as @pervect mentioned in an early post in this thread.

These diamond edges are traced out by the light rays of a light clock carried by inertial observers.
All of my clock diamonds have the same area... and this follows from the Lorentz transformations,
which preserve the light-cone and (since its determinant equals one) preserves the areas of regions in the plane. So, you can see the tickmarks are marked off according to this "invariant unit area" ... and you can calculate by counting.

And, as @vanhees71 said, back here in #6 and here in #12, I am secretly (or not secretly) using light-cone coordinates, which is actually the simplest coordinate system because it's the eigenbasis of the Lorentz transformation: the eigenvectors are along the light-cone and the eigenvalues are the Doppler factors (the "k" in Bondi's "k- calculus). In this diagram, k=2 defining the stretch along the forward lightcone is the Doppler factor corresponding to v=(3/5)c.The timelike diagonal follows the worldline of the inertial observer.
The other diagonal is spacelike and is Minkowski-perpendicular to that observer.
This is an instant of time for that observer.
"Her spaceline is Minkowski-perpendicular to her worldline ["timeline"]."

For the timelike line OP, this spacelike diagonals of her diamonds are parallel to the tangent to the hyperbola at P.
This is how Minkowski defined "perpendicular" (but he used "normal", according to the translation).
The tangent line [to the hyperbola] is Minkowski-perpendicular to the radius.
The Minkowski-right-angle is at P... so the other side OQ is the hypotenuse.
See OP' and OQ' for the other 3-4-5 triangle.There are many things that can be read off of this diagram, with practice.

My Insights might be good places to start.

https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/
https://www.physicsforums.com/insights/relativity-rotated-graph-paper/
postscript:
Diagrams alone might be difficult. Equations alone might be difficult
That's why some equations (hopefully simple[r] ones) and some diagrams are helpful,
and easy-to-see physical interpretations are also helpful.

Does anyone do Euclidean geometry by whipping out the rotation equations?

 

[NoahsArk]

Suppose a rocket ship is moving in a straight line, just above a long straight section of roadway at a speed of 0.6c (relative to the road). A tiny explosion on the ship generates a spark that leaves a mark on the roadway, and then 4 seconds later (as measured by a clock on board the ship) another tiny explosion on the ship generates a spark that leaves another mark on the roadway. A person stands still on top of each mark, They measure the distance between the marks to be 3 light-seconds. Each person has a clock with them (they have synchronized their clocks with each other) and determine that the time that elapsed between explosions is 5 seconds.

I drew a picture of that situation here.

I am drawing this from the "tree frame" where a tree is standing at the origin 0,0 along the road as the rocket flies by and emits the first spark at 0,0. The world line of the tree is green, and the world line of the rocket is red. The yellow lines are two beams of light. Just to get some practice with different concepts using this same example, I tried to draw a line of simultaneity for the rocket at the point of the second explosion (the explosions are marked with asterisks). That looks like this:

This shows that from the rocket's perspective, the tree frame's clock read 3.2 seconds during the time when the second explosion occurred. The orange line is the line of simultaneity of the rocket, and it goes through 3.2 on the tree frame on the left tree. I drew another tree in this frame on the right side as well which shows that at the second explosion, a clock on the right tree would read about 6.8 (I should have drawn the second tree at x= 5 in the tree frame to make it easier, but this should do for now). Is this so far correct?

I also wanted to use this example to see how far apart the lines of simultaneity should be in the rocket frame. In other words, how to show time dilation in the diagram:

This diagram shows that each line of simultaneity should be .8 meters apart. Since v = .6, gamma = 1.25. So, when t = 1, $t \prime$ should be .8, when t = 2, $t \prime$ should be 1.6, etc. Is this correct as well?

Regarding the hypotenuse not being the longest side of the triangle, if you take a literal measurement of this diagram with a rule and measure the length of the red line segment between the two sparks, the length will come out to be 5.83, which would be the longest side of the triangle. So, I must be missing something? To show the line which represents the time between sparks from the rocket point of view, I assume we would start at the origin an draw a diagonal line going up and to the left crossing through each of the orange lines.

In another thread, Nugatory said that it was relativity of simultaneity that is really causing time dilation. I wanted to be able to see that through a diagram, which I think I can see here. When the rocket is at t = 1 in the tree frame, in order to find out what it is in the rocket frame we have to draw the line downward and to the left which therefore means that $t \prime$ will cross the tree frames time access at a lesser time. If t = 1, $t \prime = .8$, if t = 2, $t \prime = 1.6$, if t = 3, $t \prime = 2.4$, if t = 4, $t \prime = 3.2$. I think though a flaw in the way I worked this out might be that just because the rocket frame's clocks are reading a certain time, that doesn't mean they cross the tree frame's time access at that particular time?

I also am still not sure how length contraction can be shown on a spacetime diagram.

The equation ds2=dt2−dx2 is much more general—it applies to any pair of points on the spacetime diagram

I sort of see. But aren't the equations for the Lorentz transformations the more general form of the equation $t \prime ^2 = t^2-x^2$ (which is for situations where $x \prime = 0$)? I assume you are not saying that the equation $ds^2 = dt^2 - dx^2$ is another way of writing one of the Lorentz transformations?

Here are two 3-4-5 triangles in Minkowski spacetime.

I tried to read through the diagram, but, in general, because of my level, I could not understand what is going on in it. There is a lot of detail, and I wasn't sure which lines represent the world lines of which frames, whose frame of reference the diagram was drawn in, etc. Also, all the lines are diagonal and there is no set of vertical lines and horizontal lines for the time and space axis of the main frame. Things need to be greatly simplified for me:)

 

[Pencilvester]

I assume you are not saying that the equation $ds^2=dt^2−dx^2$ is another way of writing one of the Lorentz transformations?

Correct (I’m not saying that). It is an expression of the line element, or the metric, for the Minkowski plane, i.e. what makes the Minkowski plane fundamentally different from the Euclidean plane. The quantity $ds^2$ is invariant—for any two points on the plane, regardless of the frame of reference or the choice of coordinates, $ds^2$ will be the same. The analogous version of this in Euclidean space is the distance between two points (squared). It doesn’t matter how you rotate the space or move it around or assign coordinates to the points—the distance between the points stays the same.

 

[robphy]

Here are two 3-4-5 triangles in Minkowski spacetime.

I tried to read through the diagram, but, in general, because of my level, I could not understand what is going on in it. There is a lot of detail, and I wasn't sure which lines represent the world lines of which frames, whose frame of reference the diagram was drawn in, etc. Also, all the lines are diagonal and there is no set of vertical lines and horizontal lines for the time and space axis of the main frame. Things need to be greatly simplified for me:)

The diagrams below are based on the standard spacetime diagram that was drawn by @Mister T above .

I have reproduced my diagram with usual [unrotated] graph paper
along with the rotated graph paper version.

Note that the observer's frames of reference are determined by the joining opposite corners of the diamonds (that is, the diagonals of the diamonds). So, it's drawn in the frame where the red-observer (along the vertical OP) is at rest. The space-direction for red is along the horizontal PQ... just like the ordinary spacetime diagrams like what you are drawing. To get the grid for red, just join the corresponding corners parallel to the red diamond diagonals.)

(The rotated grid lines represent light-cones, which in my opinion is at the heart of special relativity...
and it's this grid that allows calculations to be done easier, when the method is followed... because it's easier to count little diamonds in the grid to compute the area.
One advantage of the rotated graph paper is that
it can display many reference frames on the same diagram.
If you want the grid for blue, just join the corresponding corners parallel to blue diamond diagonals. )

 

[Herman Trivilino]

Regarding the hypotenuse not being the longest side of the triangle, if you take a literal measurement of this diagram with a rule and measure the length of the red line segment between the two sparks, the length will come out to be 5.83, which would be the longest side of the triangle. So, I must be missing something?

Yes, you are missing the fact that the length of the hypotenuse can't be measured with a ruler on a spacetime diagram. In this case (see Post #26) the length is 4 seconds.

You are writing $t'^2=t^2-x^2$, so in this case we have $t'^2=5^2-3^2$, so $t'= 4$.

A more general expression is $t'^2-x'^2=t^2-x^2$, but in this case $x'=0$ because both events happened at the same place in the rocket frame. When two events happen at the same place, the time that elapses between them is called the proper time $\tau$. Therefore $\tau^2=t^2-x^2$. It's a relativistic invariant.

 

[Pencilvester]

I tried to draw a line of simultaneity for the rocket at the point of the second explosion (the explosions are marked with asterisks). That looks like this:

This shows that from the rocket's perspective, the tree frame's clock read 3.2 seconds during the time when the second explosion occurred.

Just a helpful tip: in spacetime diagrams where light traces out lines at 45 degree angles (which is the case here as well as in most spacetime diagrams you’ll see), the worldline of a physical object and that object’s lines of simultaneity will always be symmetrical with respect to light rays—i.e. light rays bisect the angles made between a worldline and it’s line of simultaneity. I see in your otherwise well drawn diagram that the rocket’s worldline goes over 2 units for every 3 units up, which means the lines of simultaneity should go over 3 units for every 2 units up, but it looks like yours is going over 4. The rocket should say the tree’s clock read 3.0 at the time of the second explosion.

 

[Pencilvester]

the rocket’s worldline goes over 2 units for every 3 units up

Never mind, I was looking at too small a portion of your lines. It does look like on average your slopes are 5/3 and 3/5.

 

[NoahsArk]

I see in your otherwise well drawn diagram that the rocket’s worldline goes over 2 units for every 3 units up, which means the lines of simultaneity should go over 3 units for every 2 units up

This is a very helpful tip! So, to get the slope of the line of simultaneity given the X and Y coordinates of the worldline, we just flip the numbers. E.g. if the car is moving at .1 c, then for every movement of .1 to the right, the y movement is 1 (.1, 1). The line of simultaneity, then, would have coordinates of (1, .1) and the slope is .1?

The rotated grid lines represent light-cones, which in my opinion is at the heart of special relativity

It was through watching a video about light cones that I first started to understand the rule about why two events can never occur in reverse order, no matter which frame of reference is observing them, when the first event had caused the second event to occur:

 

[Herman Trivilino]

It was through watching a video about light cones that I first started to understand the rule about why two events can never occur in reverse order, no matter which frame of reference is observing them, when the first event had caused the second event to occur:

If the first event caused the second event then their separation is timelike and what you say about the order is true. If the events have a spacelike separation then the order of the events varies with different reference frames. In particular, there will be a frame where the events are simultaneous.

 

[Pencilvester]

the slope is .1?

Yes, $\frac{dt}{dx}=0.1$ for the lines of simultaneity of an object moving with $\frac{dx}{dt}=0.1$.