"Spacetime Physics" by Taylor and Wheeler

[NoahsArk]

I've tried to get through this book a couple times in the past but didn't succeed, and am now trying it a third time. I managed to re-read and redo the problems from the first chapter recently, and now I'm on the second one where the authors get into free float frames and the force of gravity. I'm not sure why this is discussed in a book on special relativity since it seems relevant to general relativity. In any event, there is this discussion which is one of the many things confusing me so far, but this seems more related to SR:

"Regions of Spacetime

Special Relativity is limited to free float frames.

"Region of Spacetime" What is the meaning of this term? (He then refers to figure 2-5 which is two pictures showing a train dropped horizontally from 315 meters above the earth, and a train dropped vertically from the same distance above the earth. He also talks about how if you dropped one marble from each end of the train then, as the train fell horizontally, the marbles would approach each other during the fall, but if you dropped the train vertically [i.e. in a nose dive position], then the marbles would get farther and farther apart) The long narrow rail way coach in Figure 2-5 probes spacetime for a limited stretch of time and in one or another single direction in space. It can be oriented north-south or east-west or up-down. Whatever its orientation, relative acceleration of the tiny marbles released at the two ends can be measured. For all three directions- and for all intermediate directions- let it be found by calculation that the relative drift of two test particles equals half the minimum detectable amount or less. Then throughout a cube of space 20 meters on an edge and for a lapse of time of 8 seconds, (which is how long it would take the train to hit the ground of the Earth I think), 2400 million meters of light travel time being 8 seconds, test particles moving every which way depart from straight line motion by undetectable amounts. In other words, the reference frame is free-float in a local region of space-time with dimensions.

(20 meters x 20 meters x 20 meters of space) x 2400 million meters of time.

Notice that this "region of space-time is four dimensional three dimensions: of space and one of time."

I am really clueless about what he means when he says: "let it be found by calculation that the relative drift of two test particles equals half the minimum detectable amount or less"??

More interesting for me though is the calculation of the region of spacetime and it's dimensions since it seems to describe the nature of spacetime. What exactly is this 20 by 20 by 20 by 2400 million meter "region" he is referring to? I am assuming 20 meters is the length and width of the train and that it is 20 meters high as well and we need to calculate 20³ to get the volume. Why are we then multiplying the volume by 2400 million??

Thanks.

 

[Nugatory]

More interesting for me though is the calculation of the region of spacetime and it's dimensions since it seems to describe the nature of spacetime. What exactly is this 20 by 20 by 20 by 2400 million meter "region" he is referring to? I am assuming 20 meters is the length and width of the train and that it is 20 meters high as well and we need to calculate 203 to get the volume. Why are we then multiplying the volume by 2400 million??

That's a four-dimensional region of spacetime, not a three dimensional region of space, so we need to specify its extent in four dimensions: 20 meters wide, 20 meters deep, 20 meters high, and eight seconds which is 2400 million meters long along the time dimenson.

This might be easier to visualize if you consider an ordinary Minkowski diagram, the kind we use to describe problems in which we only care about one spatial direction so we can get by with a two-dimensional diagram that we can draw on a sheet of paper. The horizontal axis is distance, the vertical axis is time, and if you want to specify a region on the diagram you need to specify its horizontal width along the x-axis and its vertical width along the t axis.

 

[NoahsArk]

Thank you. If the 20³ represents the train (and I'm not sure it does?), then why isn't the distance covered by the fall of the train also represented in this?

 

[NoahsArk]

I just double checked and the train is 20 meters long, but not 20 meters wide and not 20 meters high (if so that'd be a strange looking train). So it looks like an incorrect calculation of the trains volume that the author is giving?

 

[Nugatory]

I just double checked and the train is 20 meters long, but not 20 meters wide and not 20 meters high (if so that'd be a strange looking train). So it looks like an incorrect calculation of the trains volume that the author is giving?

Were talking about a region of spacetime that is large enough to hold the train if it is oriented in any of "all possible directions" so you need 20 meters in all three spatial directions. Ask yourself what is the smallest box that can hold a meter stick in any orientation; it is one meter high, one meter deep, and one meter tall.

 

[NoahsArk]

Were talking about a region of spacetime that is large enough to hold the train if it is oriented in any of "all possible directions" so you need 20 meters in all three spatial directions.

Ok that makes sense now.

Regarding the 8 seconds/ 2400 million meters of light travel time, I am assuming because it takes the train 8 seconds to fall from 315 meters above the Earth to hit the ground? Still not sure what the idea is of multiplying 2400 million meters to the 20³?

Thank you.

 

[Nugatory]

Still not sure what the idea is of multiplying 2400 million meters to the 20³?

We're looking at a region of space time, and its volume is 20 meters by 20 meters by 20 meters by 8 seconds.

Describing it like that is a problem for the same reason that it would be a problem calculating the volume of a box that is 8 inches by 10 inches by 1 foot - I can't just calculate the volume as 8x10x1 because my units aren't the same. Before I can calculate the volume of the box I have to convert that 1 foot length to 12 inches by using the conversion factor between inches and feet (one foot equals twelve inches) so my problem becomes a box 8 inches by 10 inches by 12 inches, and now I can calculate the volume of the box to be 960 cubic inches.

Likewise, I can't compute the volume of a region of spacetime that is 20 meters by 20 meters by 20 meters by 8 seconds until I've converted the seconds into meters. Just as the conversion factor between feet and inches is 12 in/ft, and the conversion factor between meters and centimeters is 100 cm/m, the conversion factor between seconds and meters is $c\approx 3\times{10}^8$ m/sec.

 

[NoahsArk]

It makes sense to convert into the same units. The part which doesn't make sense is why we are multiplying the fourth number (2400 million meters) together with the first 3? The first three numbers when multiplied gives us the volume, but what does multiplying the fourth number do? To me its like saying a square with width 10 and height 10 and color green has a space/color region of 10 x 10 x some number representing the color green.

With respect to the space time interval I have some what of a better idea of how there is a geometrical relationship between space and time and how we first need to convert to the same units of time and space and take the difference of their squares to find their interval. Here though it seems different since we are multiplying everything.

Also, on a 2d space time diagram, the vertical axis would show 2400 meters in this example I think? How about the horizontal axis? Shouldn't that be how far the train traveled (or fell) rather than the volume of the "box"?

 

[Nugatory]

The first three numbers when multiplied gives us the volume, but what does multiplying the fourth number do?

When we're calculating the volume of an $n$-dimension region, we need to multiply $n$ numbers, one for the extent in each dimension.
We calculate the volume of a three-dimensional region of three-dimensional space by multiplying three numbers (width, depth, height) and we calculate the volume of a two-dimensional region on the surface of a sheet of paper by multiplying two numbers, width and height (don't be confused because we have a special word "area" for two-dimensional volumes - that's just an oddity of how we use language).

Here we're calculating the volume of a four-dimensional region of spacetime so we have four numbers (width, depth, height, extent in time) to multiply, one for each of our four dimensions.

Also, on a 2d space time diagram, the vertical axis would show 2400 meters in this example I think? How about the horizontal axis? Shouldn't that be how far the train traveled (or fell) rather than the volume of the "box"?

Choose a frame in which the train is a rest (you don't have to, but it's a lot easier). In this frame, at time zero the front of the train is at the the point (x=0,t=0) and the rear of the train is at the point (x=20, t=0). Eight seconds later, the front of the train is at (x=0,t=8) and the rear of the train is at (x=20,t=8). Together, these four points define a rectangle with a base of 20 meters and a height of 8 seconds. We get the volume (which for the special case of two dimensions we call the "area") of this rectangle by multiplying the base times the height, after we've fixed the units.

 

[m4r35n357]

It makes sense to convert into the same units. The part which doesn't make sense is why we are multiplying the fourth number (2400 million meters) together with the first 3? The first three numbers when multiplied gives us the volume, but what does multiplying the fourth number do? To me its like saying a square with width 10 and height 10 and color green has a space/color region of 10 x 10 x some number representing the color green

Yes it is. Did you understand post #2?

 

[Ibix]

It makes sense to convert into the same units. The part which doesn't make sense is why we are multiplying the fourth number (2400 million meters) together with the first 3? The first three numbers when multiplied gives us the volume, but what does multiplying the fourth number do? To me its like saying a square with width 10 and height 10 and color green has a space/color region of 10 x 10 x some number representing the color green.

The point is that time is a dimension, similar but not identical to the three spatial ones. So multiplying volume by time gives you a four dimensional volume.

Multiplying time and space would make little sense in Newtonian physics (as normally constructed, anyway - I won't guess about Cartan's geometric reformulation), because time is something completely different from space. It makes perfect sense in a 4d spacetime, however, where time and space are the same kind of thing.

 

[Mister T]

Consider a cube measuring 20 m by 20 m by 20 m. Study its contents for 8 seconds. Call it spacetime.

 

[NoahsArk]

Choose a frame in which the train is a rest (you don't have to, but it's a lot easier). In this frame, at time zero the front of the train is at the the point (x=0,t=0) and the rear of the train is at the point (x=20, t=0). Eight seconds later, the front of the train is at (x=0,t=8) and the rear of the train is at (x=20,t=8)

I assume this means that the observer is sitting in the front of the train? Also, I assume that for this example the train or cube that we draw out is just sitting still for 8 seconds?

The point is that time is a dimension, similar but not identical to the three spatial ones. So multiplying volume by time gives you a four dimensional volume.

Consider a cube measuring 20 m by 20 m by 20 m. Study its contents for 8 seconds. Call it spacetime.

If there are a group of apples placed next to each other in five rows and five columns, multiplying 5 by 5 gives me a square shape with 25 apples. The product of the row and columns gives me a new shape which I can understand and visualize. If the apples were stacked in 3d in five layers, each layer having 5 rows and columns of apples, I could then multiple 5x5x5 to create a cube of layered apples totaling 125 apples. Again there is a geometric relationship among the rows and columns and layers and multiplying them together gives a new shape that can be visualized.

Say I try to introduce a fourth dimension. If each row is 5 meters wide, and each column is 5 meters long, and each layer is five meters apart vertically, then the volume of the cube is 125 meters. If I looked at this cube for five seconds, or 150 million meters of light travel time, then the four dimensional volume or space-time volume is 18.75 billion meters? The math is straight forward, but what exactly is this new fourth dimensional "thing" or "spacetime region" that results? It's not something like a square which we can visualize and which results from multiplying length times width, and its not something like a cube.

If it can't be visualized, I am assuming there must be some significance of the 18.75 billion meters? Is that number used to make some other kinds of calculations?

"Spacetime" is a term that is widely used in the news, on science programs, and in many articles written for lay people. Yet, I think if you ask 1000 people from the general public to define spacetime, although I could be wrong, maybe 1 will be able to. I've even spoken to people with advanced high tech degrees who did not understand what spacetime was- it's just a term that they've heard. In the context of the spacetime interval, where space and time are related like the sides of a triangle, I thought I was beginning to understand the nature of the relationship between space and time. Say the proper time that elapsed in a flying spaceship between two events (verticle side of a right triangle) is three light meters of time. You can and add the square of that to the square of the distance that the ship traveled from the perspective of an Earth observer, say four light meters of distance (horizontal side of the right triangle), and we know that the sum of those squares = the square of the time passed on earth, and we can conclude that 5 light meters of time occurred on earth.

As a side point, the 5 light meters of time is like the hypotenuse on a right triangle. However, if I'm not mistaken, isn't this just a metaphor or visual aid we use to understand the relationship between space and time- i.e. we compare it with the relationship between sides of a triangle when in reality there is no "hypotenuse" being created out there in the above example? Seems like the same thing going on whenever we represent the relationship between two quantities on an x-y graph. A coordinate with a higher y value than x value (if both y and x represent quantities of two different things) will have a vertical line going up to the y coordinate that is "bigger" than the horizontal line going up to the x coordinate. Here we are using a spatial difference to represent a difference in quantities- two different things, one related to space and the other related to a numerical relationship. If we have 2 buildings and 4 sticks, the quantity of the sticks is double that of the buildings, and on an x-y graph the sticks will be "higher" than the buildings, but "higher" is just a metaphor for "more" and ofcourse 2 buildings are in reality higher than 4 sticks. Is any attempt to draw a space-time relationship on a graph also a metaphor?

If we compare the "triangle" example above with the "cube" example above, then I think the 18.75 billion meters in the cube example is comparable with the five meters in the triangle example. With the triangle example, though, the five meters represents something concrete, i.e. the amount of time that passed for an observer on earth. In the cube example, though, I don't see anything concrete that the 18.75 billion meters actually represents?

Thank you for the help.

 

[Nugatory]

I assume this means that the observer is sitting in the front of the train?

Not necessarily, it just means that the origin of the coordinate system is at the front of the train. We don't need any observer anywhere to make statements about where the two ends of the train are, in some frame.

Also, I assume that for this example the train or cube that we draw out is just sitting still for 8 seconds?

I said that I was choosing to use a frame in which the train is at rest, so yes, the train is not moving in that frame so the only the t coordinates are changing as time passes.. (I could have chosen to use a frame in which the train is moving but in that frame the calculation would be way more complicated but if we do it right we'll get the same answer so I'd rather use the frame in which the calculation is easy).

The math is straight forward, but what exactly is this new fourth dimensional "thing" or "spacetime region" that results? It's not something like a square which we can visualize and which results from multiplying length times width, and its not something like a cube.

It's easy to visualize in two dimensions: We have a rectangle with a base of 20 meters and a height of 8 seconds. We know that one second is equal to $3\times{10}^8$ meters the same way that one foot is equal to 12 inches, so we rewrite the height as $20\times{3}\times{10}^8$ and calculate that the area of the rectangle is 6 billion square meters. Or we could say that one meter is equal to $1/(3\times{10}^8)$ seconds, so we have a rectangle whose base is $20/(3\times{10}^8)$ seconds and whose height is 8 seconds so its area is $160/(3\times{10}^8)$ square seconds; this is no different than saying that the area of a 12-inch by two-foot rectangle is two square feet or 288 square inches.

It's also easy enough to imagine in three dimensions. Consider the flat floor of a large room, 20 meters by 20 meters. The four corners of the room at time zero have coordinates (x=0,y=0,t=0), (x=20,y=0,t=0), (x=0,y=20,t=0), (x=20,y=20,t=0). Wait eight seconds and the four corners will be at (x=0,y=0,t=8sec), (x=20,y=0,t=8sec), (x=0,y=20,t=8sec), (x=20,y=20,t=8sec); those eight points define a three-dimensional rectangle in spacetime whose volume we calculate as $20\times\20\times\(8\times{3}\times{10}^8)$ cubic meters after we fix the seconds-to-meters problem with the units.

When we're working with four dimensions we can't visualize it because we are not equipped to visualize four-dimensional objects, but that's just a limitation of our ability to visualize stuff.

If it can't be visualized, I am assuming there must be some significance of the 18.75 billion meters? Is that number used to make some other kinds of calculations?

Well, you've just introduced a new unit when you say that you're looking at a region of space that is five apples wide. Five apple widths times five apple widths times five apple widths times eight seconds is not equal to 18.75 billion meters. We need to convert the apple widths to meters before we do the multiplication to get a result in meters⁴ (not meters! Just as the volume of a two dimensional rectangle is measured in meters² and a three dimensional volume is measured in meters³, four dimensional volumes are measured in meters⁴). That volume is the volume of the region of spacetime that your 5x5x5 array of apples will pass through during the next eight seconds.

"Spacetime" is a term that is widely used in the news, on science programs, and in many articles written for lay people. Yet, I think if you ask 1000 people from the general public to define spacetime, although I could be wrong, maybe 1 will be able to. I've even spoken to people with advanced high tech degrees who did not understand what spacetime was- it's just a term that they've heard.

That may be true, but among that fraction of the population who have takes a bachelors' degree in physics, you'll find that more like 1000 out of 1000 will know what it is.

 

[Ibix]

In the context of the spacetime interval, where space and time are related like the sides of a triangle, I thought I was beginning to understand the nature of the relationship between space and time. Say the proper time that elapsed in a flying spaceship between two events (verticle side of a right triangle) is three light meters of time. You can and add the square of that to the square of the distance that the ship traveled from the perspective of an Earth observer, say four light meters of distance (horizontal side of the right triangle), and we know that the sum of those squares = the square of the time passed on earth, and we can conclude that 5 light meters of time occurred on earth.

This isn't correct. The proper time is the hypotenuse, the coordinate time is the "vertical" side and the coordinate distance is the "horizontal" side. We're not using Euclidean geometry here, so we can't use Pythagoras' theorem. We need to use Minkowski's hyperbolic geometry version of Pythagoras' theorem, which is $\Delta s^2=\Delta t^2-\Delta x^2$ (or $\Delta s^2=\Delta x^2-\Delta t^2$ - convention on whether time or the spatial dimensions has the minus sign varies). Of course you can take the negative bits to the other side and try to disguise this as Pythagoras, but it's still the $\Delta s$ that's the frame-invariant quantity like the length of an object in Euclidean geometry.

However, if I'm not mistaken, isn't this just a metaphor or visual aid we use to understand the relationship between space and time- i.e. we compare it with the relationship between sides of a triangle when in reality there is no "hypotenuse" being created out there in the above example?

Noting that you are identifying the wrong quantity as the hypotenuse, no, we take this as seriously as any scientific model.

On a philosophical level, one should never take any model absolutely literally. The maths of an accurate theory accurately predicts the outcomes of experiments. Any interpretation you care to put on how those results came to happen should be taken with a grain of salt, since it's not something you actually tested. Incidentally, this is why a lot of pop sci is misleading - it focuses on verbally describing the model and treats the maths as an optional extra, which is completely backwards. I know why it's done, but it's still backwards.

All of that said (with the exception of quantum theory where arguing over interpretation is a thing) most scientists will treat the models as real in day to day work. So we absolutely do take the triangle (albeit using Minkowski's hyperbolic geometry rather than Euclidean geometry) seriously. It's not just an analogy or a visualisation - it's our model of how the universe works.

Is any attempt to draw a space-time relationship on a graph also a metaphor?

Well you're stuck with Euclidean geometry on a piece of paper, so it's not perfect. But I wouldn't describe it as a metaphor. It's more like a cross-sectional drawing of something that you might see in an engineering manual. Is that a metaphor?

If we compare the "triangle" example above with the "cube" example above, then I think the 18.75 billion meters in the cube example is comparable with the five meters in the triangle example. With the triangle example, though, the five meters represents something concrete, i.e. the amount of time that passed for an observer on earth. In the cube example, though, I don't see anything concrete that the 18.75 billion meters actually represents?

It's m⁴, not meters, as Nugatory noted. And it's the volume (or hyper-volume, if giving it a different name helps) of the region of spacetime.

I don't have my copy of Taylor and Wheeler to hand at the moment so I'm not sure of the context. But it doesn't look to me from what you've quoted that they're actually expecting you to multiply those numbers. It's very common to describe a rectangle that is 10m long and 5m wide as 10m × 5m, and a cuboid 10m long, 5m wide, and 3m high as 10m × 5m × 3m. You can multiply the numbers to get the area or volume, but you aren't typically expected to do so. It looks to me like the book is simply extending this notation to 4d and describing a 4-cuboid by the length of its edges. Off the top of my head, I can't think of a reason you'd need to know its volume.

 

[Mister T]

If there are a group of apples placed next to each other in five rows and five columns, multiplying 5 by 5 gives me a square shape with 25 apples.

No, it doesn't. It simply gives you the number of apples. The square shape is something you created when you arranged the apples. The 25 apples are concrete, the square shape is a model used to describe the arrangement of the apples.

The product of the row and columns gives me a new shape which I can understand and visualize. If the apples were stacked in 3d in five layers, each layer having 5 rows and columns of apples, I could then multiple 5x5x5 to create a cube of layered apples totaling 125 apples. Again there is a geometric relationship among the rows and columns and layers and multiplying them together gives a new shape that can be visualized.

You can visualize a three dimensional cube, but what does that mean? That you can recall in your mind various concrete objects that have that same shape. By studying various concrete objects you generalize to a model, which in this case is a cube. When you study spacetime physics you build up in your mind a variety of concrete examples that you can generalize to a model.

 

[NoahsArk]

I maybe shouldn't have used the apples in my example since it made it more confusing. Point is if you take any cube with length 5m, width, 5m, and height 5m, you get a volume of 125m. That number, 125m, tells us something useful. For example if we needed to know how many objects could fit into the cube we'd want to know the cube volume. The cube's 4d "hyper volume", after observing it for 5 seconds, would be 18.75 billion meters. I don't understand though how knowing the 18.75 billion figure can be useful in the way that knowing the regular volume could be useful? Is there a parallel example to how a physicist knowing the hyper volume in meters of a region of spacetime would be useful to the physicist in the same way knowing the regular volume of a room in a house would be useful to an architect?

 

[PeroK]

I assume this means that the observer is sitting in the front of the train? Also, I assume that for this example the train or cube that we draw out is just sitting still for 8 seconds?

It seems to me that there is a fair amount of resistance to expanding your thought processes in your posts. As an alternative, you could consider all the different meanings that spacetime might have. For example:

In railway timetabling, a train at a station takes up a platform for a specified time. E.g. platform 14 for 3 minutes. That's spacetime.

Or, you could rent office space by the square metre by the week. That might be $5000 m^2 \times 16$ weeks. That's spacetime.

Or, a motion picture is a kind of spacetime. The film itself has a physical length, a running time per showing, a running time at a given cinema and a time when the film itself takes place: it might run for two hours, play at Cinema X for six weeks and take place in Paris in the summer of 1968. That's all spacetime.

 

[kith]

I just checked Taylor & Wheeler and they talk about the spacetime volume in the context of local inertial systems.

I would put it this way (beware of errors, my knowledge of relativity is quite rusty): The significance of the calculated spacetime volume is that if you perform measurements to a certain accuracy within the given spacetime volume, the rest frame of the falling train can be approximated as an inertial system. If part of your measurement happens outside the given spacetime volume (because you consider particles with a bigger spatial separation or the duration of your experiment exceeds the time of 8 seconds), you cannot treat the rest frame of the falling train as an inertial system. This is because in this case, you get measureable effects on your results from the presence of Earth.

 

[NoahsArk]

The significance of the calculated spacetime volume is that if you perform measurements to a certain accuracy within the given spacetime volume, the rest frame of the falling train can be approximated as an inertial system.

So, using the example from the initial post, the spacetime volume is 20x20x20x2,400,000,000 = 19,200,000,000,000 meters. That means that the particles can't be separated more than that without having the effects of Earth's gravity on them? The example the book uses is complicated for me, but this starts to show at least the significance of the spacetime volume. So, is the 19,200,000,000,000 meters some larger cube in space that we form, and we say that all measurements have to happen inside that region?

Or, you could rent office space by the square metre by the week. That might be 5000m2×165000m2×165000 m^2 \times 16 weeks. That's spacetime.

It's a subtle subject, and it's not been easy for me to even phrase the question right. If someone had no idea what "circumference" meant, but knew what a radius of a circle was and knew the value of pi, then you could still teach that person how to correctly calculate every time what the circumference of a circle was. They'd get the right numerical value of the circumference every time, if they know the diameter, by multiplying by pi. Even though they'd know the value of the circumference in meters and could calculate it for any given circle, they still wouldn't know what a circumference is. I think spacetime is the circumference here and the volume is like the diameter and the time is like pi. Someone who knows what volume and time are can correctly calculate the spacetime region correctly each time, but still not know what spacetime is.

 

[Ibix]

20x20x20x2,400,000,000 = 19,200,000,000,000 meters

No! It's m⁴ not m! Measuring a volume in metres is like measuring a speed in kilograms.

 

[NoahsArk]

I guess the simplest way to ask my question is why is the measurement for spacetime an important number?

Thank you for all the help and sorry if some of my questions have been repetetive.

 

[Ibix]

I don't think the volume is particularly important. Having an idea of the extent (in both space and time) of a region that you can treat as flat is important. So twenty-by-twenty-by-twenty-by-two-thousand-four-hundred-million is of interest because you can ignore the more complex maths of GR if your experiment fits in that space and doesn't run longer than that time. The actual product isn't particularly interesting.

It's like I sell you a desk. It's 1m by 2m by 1.2m. You need to know the three extents of the desk to see if it'll fit in the space you have in your house. You can also calculate that the volume is 2.4m³, but so what? No one cares in this case.

In relativity you need duration as well as width, depth, height.

 

[Mister T]

I maybe shouldn't have used the apples in my example since it made it more confusing. Point is if you take any cube with length 5m, width, 5m, and height 5m, you get a volume of 125m. That number, 125m, tells us something useful.

The volume of 125 m³ is indeed useful. You know many of the uses.

Likewise, if you studied that space for, say, 10 m of light travel time, you'd have 1250 m⁴ of spacetime. That tells me something useful, and one example of that usefulness is the discussion of the freely falling train car in Taylor and Wheeler.

I don't know if your other question ever got answered. If you were able to measure the separation distance to the nearest, say, 0.2 mm. You can't do any better than that because that is the limit of precision of your measuring device. If you calculate that the separation distance changes by 0.1 mm then you can declare that any changes in the separation distance are negligible.

 

[kith]

So, is the 19,200,000,000,000 meters some larger cube in space that we form, and we say that all measurements have to happen inside that region?

No, it is the volume of a fourdimensional hypercube with units $m \times m \times m \times m = m^4$. Just like a cube isn't some "larger square in a [twodimensional] plane", a fourdimensional hypercube isn't "some larger cube in [threedimensional] space". In addition to that measurements need to happen inside a certain region of space, they also mustn't take longer than a certain time.

If you can't wrap your head around the spacetime volume, just ignore it. Most authors don't talk about it at all. Spacetime intervals and Minkowski diagrams are much more important and if you have a good understanding of them you have a good picture of what spacetime is supposed to be in special relativity.

Taylor & Wheeler calculate it because they want to talk about systems which are only locally inertial systems. Their definition of what an inertial system is is therefor a bit more general than the one I learned a long time ago. They admit also systems where Newton's first law holds only approximately. In order to talk about the limits of such approximations you need a spatial and a temporal border. That's the important point to understand in this section, not what the volume of a spacetime region "actually" is.

 

[Mister T]

I'm not sure why this is discussed in a book on special relativity since it seems relevant to general relativity.

The authors are showing the reader the limits of validity of special relativity. Evidently they think it's important to understand the limits of validity of a theory just as you are beginning to study that theory. That seems like a perfectly reasonable thing to do, and in fact to omit it is a serious shortcoming.

 

[NoahsArk]

These responses were all helpful.

The volume of 125 m3 is indeed useful. You know many of the uses.

Likewise, if you studied that space for, say, 10 m of light travel time, you'd have 1250 m4 of spacetime. That tells me something useful, and one example of that usefulness is the discussion of the freely falling train car in Taylor and Wheeler.

The example they used was a bit complicated for me since it's talking about the effects of gravity, etc. If there is a simpler or different example showing the usefulness of knowing a spacetime region I'd appreciate it.

If you can't wrap your head around the spacetime volume, just ignore it. Most authors don't talk about it at all. Spacetime intervals and Minkowski diagrams are much more important and if you have a good understanding of them you have a good picture of what spacetime is supposed to be in special relativity.

That's good to know. Chapter 2 was difficult for me and I am going to skip over the problems for now and maybe go back to them. I feel like having taken a course in classical mechanics is a prerequisite to understanding how to do those problems well.

 

[Mister T]

The example they used was a bit complicated for me since it's talking about the effects of gravity, etc.

What is it about the example that you find complicated? Do you understand why gravity causes the distance between the test particles to change? The point the authors are making is that if those effects are small enough we can ignore them.

Note that we do this in every branch of physics.

If there is a simpler or different example showing the usefulness of knowing a spacetime region I'd appreciate it.

Why? Understanding the limits of validity of a theory is part of understanding that theory. But a better way to do that is to study the theory itself, and as you do so remind yourself, from time to time, of those limits. As you understand the theory better you will understand the limits better.

 

[NoahsArk]

What is it about the example that you find complicated? Do you understand why gravity causes the distance between the test particles to change?

I understand that if the train is dropped in a nose dive position, then the increase in distance between two particles dropped from the front and back end of the train will be due to stronger gravitational effects on the front end particle since it's closer to earth. Whether or not that increase in distance is linear or not I don't know but I assume it doesn't matter for this example. What I don't understand is how these gravitational effects only occur (or don't occur) within the given spacetime region, and what the boundaries of that region are. The spacetime region is astronomical (bigger than the train of course). I assume the book is not saying you can separate those particles by 20x20x20x2400 million meters or less and the experiment will still work? To be honest I'm not even sure what the experiment is? Is it to test if there is any gravitational force between two particles that is present by seeing whether or not the distance between the two particles increases after they are dropped?

I am also confused about why this experiment can't be done for more than 8 seconds unless ofcourse the reason is that the train will have crashed into Earth in 8 seconds.

 

[Mister T]

What I don't understand is how these gravitational effects only occur (or don't occur) within the given spacetime region,

Well, they do occur.

and what the boundaries of that region are.

But you've already told us the boundaries: 20 by 20 by 20 by 2400 million meters.

The spacetime region is astronomical (bigger than the train of course).

A spacetime region cannot be said to be bigger or smaller than a train. A spacetime region is a four-dimensional quantity. The volume of a train is a three-dimensional quantity.

To be honest I'm not even sure what the experiment is? Is it to test if there is any gravitational force between two particles that is present by seeing whether or not the distance between the two particles increases after they are dropped?

Not the gravitational interaction between the two objects. It's the gravitational interaction each particle has with the planet below. If the two particles behave the same, the gravitational interaction they each have with the planet is the same. That's the condition necessary to be able to ignore spacetime curvature.

I am also confused about why this experiment can't be done for more than 8 seconds unless of course the reason is that the train will have crashed into Earth in 8 seconds.

You get around the crashing concern by imagining a tunnel bored through the center of the planet large enough for the train to pass through freely. The experiment can be conducted for any length of time. The 8 seconds is a more or less randomly chosen number.

 

[whatif]

I am also confused about why this experiment can't be done for more than 8 seconds unless ofcourse the reason is that the train will have crashed into Earth in 8 seconds.

This is the way I think about it. I think it is less abstract and I hope it is correct.

For all three directions- and for all intermediate directions- let it be found by calculation that the relative drift of two test particles equals half the minimum detectable amount or less.

The required accuracy is that the measuring instruments are not able to detect the effects of gravity. Taylor and Wheeler are saying that at the particular location (relative to the Earth's surface) and for the required accuracy, the experiment must be performed within the spatial dimensions and within the time dimension, without explaining how they know and without reference to the dimensions (however, that is not clear from the way they express it).

Then throughout a cube of space 20 meters on an edge and for a lapse of time of 8 seconds, (which is how long it would take the train to hit the ground of the Earth I think), 2400 million meters of light travel time being 8 seconds, test particles moving every which way depart from straight line motion by undetectable amounts.

What is meant is that for those spatial dimensions the experiment must be done within 8 seconds to achieve the required accuracy, without explaining how they know. Any longer than 8 seconds and gravitational effects would be detectable, meaning the region of spacetime could not be treated as a free-float frame (this is clear). It so happens that the train will crash into the Earth just after 8 seconds but that is not a consideration as to whether the region of spacetime can be treated as a free float frame.

I am really clueless about what he means when he says: "let it be found by calculation that the relative drift of two test particles equals half the minimum detectable amount or less"??

Taylor and Wheeler are just telling the reader that region of spacetime meets the requirements of accuracy but they have expressed themselves clumsily.
They are not explaining how they arrived at that knowledge. They are only trying to explain that the dimensions of regions of spacetime being treated as free float frames are limited by the required accuracy (based on the measuring instruments not being able to detect the effects of gravity).

More interesting for me though is the calculation of the region of spacetime and it's dimensions since it seems to describe the nature of spacetime. What exactly is this 20 by 20 by 20 by 2400 million meter "region" he is referring to?

I do not think Taylor and Wheeler intended for the result of the product to be calculated. It was just stating the dimensions. Taylor and Wheeler do not make any use of the calculated product. As per Mister T:

Consider a cube measuring 20 m by 20 m by 20 m. Study its contents for 8 seconds.

8 seconds being the time that region of space can be treated as a free float frame to meet the required accuracy because Taylor and Wheeler told us so (effectively telling the reader that the effects of gravity will be undetectable by the instruments used in that region of space time).

Multiplying time and space would make little sense in Newtonian physics (as normally constructed, anyway - I won't guess about Cartan's geometric reformulation), because time is something completely different from space. It makes perfect sense in a 4d spacetime, however, where time and space are the same kind of thing.

These kinds of statements apply to the mathematics (geometric units being used). They are abstract statements. It allows, for example, for time to be measured in meters by use of the speed of light constant, c. However, a meter of distance is not the same as a meter of time. I have no problem with the mathematics, because it simplifies the analysis. However, these kinds of statements and analogies used to explain them are misleading, in my opinion. A volume of spacetime, being a calculation based on 3 dimensions of space and one dimension of time is a mathematical number with units. However, one needs to remain aware that the same volume can represent two different things, space and time, in different proportions (to say space and time are the same kind of thing is misleading ,in my opinion).

 

[Ibix]

However, one needs to remain aware that the same volume can represent two different things, space and time, in different proportions

If you are going to say this, you should also say that one needs to remain aware that the exact same region in spacetime (not the volume number but the geometric figure, the 20x20x20x(2.4x10⁹) hypercuboid in this case) is itself different amounts of space and time, because the foliation of spacetime into space and time is arbitrary and can be done in different ways.

 

[PeterDonis]

You get around the crashing concern by imagining a tunnel bored through the center of the planet large enough for the train to pass through freely. The experiment can be conducted for any length of time.

No, it can't. The train can indeed execute simple harmonic motion passing through the tunnel, but tidal gravity will affect the marbles inside it. The point of the 8 seconds is that, given the limits of detectability of tidal gravity that the textbook is assuming, that's how long you can let the train fall freely without detecting tidal gravity effects.