Converting between frames of reference

  • #1
Hi everyone. Please be gentle with me, I am not a physicist! I am a layperson with an interest in learning more, and I’m reading a book called ‘How to Teach Relativity to Your Dog’ by Chad Orzel. It’s supposed to be physics for dummies, but it’s clearly not dumbed down enough for me because there’s something I’m really struggling with!

In the section about converting between different frames of reference, the author explains that the coordinates for measuring position are x for east-west direction, y for north-south, z for up-down and t for time. He then gives coordinates for three measurements made from two different frames of reference, Emmy’s and Nero’s:

Emmy:
(t = -1s; x = -10m; y = -10m; z = 0m)
(t = 0; x = 0m; y = -10m; z = 0m)
(t = +1s; x = +10m; y = -10m; z = 0m)

Nero:
(t = -1s; x = +10m; y = +10m; z = 0m)
(t = 0; x = 0m; y = +10m; z = 0m)
(t = +1s; x = -10m; y = +10m; z = 0m)

He goes on to say:

“If you play around with these numbers a little, you can come up with a simple recipe for converting between Nero’s measurements and Emmy’s: you simply take the east-west coordinate measured by Nero, and subtract his speed (10 m/s) multiplied by the time. A little more fiddling around will show you that getting from Emmy’s measurement to Nero’s involves just the reverse: you take the east-west coordinate measured by Emmy, and add to it Nero’s speed multiplied by the time. In this way, you can take any measurement made by Nero and convert it into a measurement that will make sense to Emmy, and vice versa.”

No matter how much I try, I can’t get from one frame of reference to the other by following these directions. I’m clearly missing something very obvious! Could anyone shed some light?
 

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  • #2
PeroK
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Hi everyone. Please be gentle with me, I am not a physicist! I am a layperson with an interest in learning more, and I’m reading a book called ‘How to Teach Relativity to Your Dog’ by Chad Orzel. It’s supposed to be physics for dummies, but it’s clearly not dumbed down enough for me because there’s something I’m really struggling with!

In the section about converting between different frames of reference, the author explains that the coordinates for measuring position are x for east-west direction, y for north-south, z for up-down and t for time. He then gives coordinates for three measurements made from two different frames of reference, Emmy’s and Nero’s:

Emmy:
(t = -1s; x = -10m; y = -10m; z = 0m)
(t = 0; x = 0m; y = -10m; z = 0m)
(t = +1s; x = +10m; y = -10m; z = 0m)

Nero:
(t = -1s; x = +10m; y = +10m; z = 0m)
(t = 0; x = 0m; y = +10m; z = 0m)
(t = +1s; x = -10m; y = +10m; z = 0m)

He goes on to say:

“If you play around with these numbers a little, you can come up with a simple recipe for converting between Nero’s measurements and Emmy’s: you simply take the east-west coordinate measured by Nero, and subtract his speed (10 m/s) multiplied by the time. A little more fiddling around will show you that getting from Emmy’s measurement to Nero’s involves just the reverse: you take the east-west coordinate measured by Emmy, and add to it Nero’s speed multiplied by the time. In this way, you can take any measurement made by Nero and convert it into a measurement that will make sense to Emmy, and vice versa.”

No matter how much I try, I can’t get from one frame of reference to the other by following these directions. I’m clearly missing something very obvious! Could anyone shed some light?
I can't imagine any dog could understand that!

Do Nero and Emmy start in the same place? In what direction is Nero moving (relative to Emmy?)?

Personally, I would draw a diagram first. Does the book have a diagram?
 
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  • #3
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  • #4
Ibix
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Woof!

I agree with @PeroK - that isn't how I'd start teaching what he's trying to teach. And I'd start with a diagram of where people and things are at t=-1 and where they are at t=1.
 
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  • #5
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There are a couple of diagrams (hopefully attached) but they involve more frames of reference. The diagrams also introduce a child riding a bike and another dog.

What’s confusing me is that the maths doesn’t seem to add up. From the explanation, it seems he’s saying that if you take the east-west coordinate measured by Nero (x) and subtract his speed (10 m/s) multiplied by the time (t), you should get Emmy’s value for x in the same moment. So for t = -1, where x = +10:

10 - (10 x -1) = 20

So that gives the wrong value. I’m sure I’m doing something wrong, but I just can’t see what.
 

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PeroK
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View attachment 239039 View attachment 239040 There are a couple of diagrams (hopefully attached) but they involve more frames of reference. The diagrams also introduce a child riding a bike and another dog.

What’s confusing me is that the maths doesn’t seem to add up. From the explanation, it seems he’s saying that if you take the east-west coordinate measured by Nero (x) and subtract his speed (10 m/s) multiplied by the time (t), you should get Emmy’s value for x in the same moment. So for t = -1, where x = +10:

10 - (10 x -1) = 20

So that gives the wrong value. I’m sure I’m doing something wrong, but I just can’t see what.
Are you sure that Nero's "speed" isn't 20m/s?
 
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  • #7
The quote from the book (in my original post) gives his speed as 10 m/s. The x coordinates only change by 10m at each second interval, so surely it has to be 10 m/s?
 
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PeroK
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The quote from the book (in my original post) gives his speed as 10 m/s. The x coordinates only change by 10m at each second interval, so surely it has to be 10 m/s?
Looks like 20m/s to me.

Try with 20m/s. See if it works. You never know.
 
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  • #9
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The quote from the book (in my original post) gives his speed as 10 m/s. The x coordinates only change by 10m at each second interval, so surely it has to be 10 m/s?
Careful.... What is his speed relative to Emmy?

This example is so confusingly written that I'm almost tempted to try writing up a better one myself, but chances are that someone else already knows of one somewhere.
 
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  • #10
Ibix
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The quote from the book (in my original post) gives his speed as 10 m/s. The x coordinates only change by 10m at each second interval, so surely it has to be 10 m/s?
If you learn relativity, you'll get used to hearing the question "speed relative to what?" If we're both doing 10m/s relative to the ground but in opposite directions, what's our speed relative to each other?

Given the title of the OP's textbook I feel I ought to ask @phinds for his opinion.
 
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  • #11
The book says that Emmy is stationary and Nero is moving. So wouldn’t that make his speed relative to Emmy 10 m/s, if he is indeed moving at 10 m/s?
 
  • #12
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Given the title of the OP's textbook I feel I ought to ask @phinds for his opinion.
I was chasing Emmy past my driveway so lost track of the question.
 
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  • #13
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The book says that Emmy is stationary and Nero is moving. So wouldn’t that make his speed relative to Emmy 10 m/s, if he is indeed moving at 10 m/s?
Yes. If that's what the book says (and you haven't misread something), it's wrong. Emmy and Nero need a relative velocity of 20m/s to make the stated event coordinates consistent. That works if both are doing 10m/s in opposite directions or if one is stationary and the other doing 20m/s.
 
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  • #14
Thank you for bearing with me, but I’m still a little lost!

If the velocity is 20 m/s isn’t the calculation still wrong?

10 - (-1 x 20) = 30

Emmy’s value for x at t = -1 is -10. So it still doesn’t seem to work.
 
  • #15
PeroK
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The book says that Emmy is stationary and Nero is moving. So wouldn’t that make his speed relative to Emmy 10 m/s, if he is indeed moving at 10 m/s?
Let me tell you how I did it. First, we've got some coordinates at ##t =0##, always useful. I'll give Emmy's first:

##t=0##: ##(0, -10, 0)## and ##(0, +10, 0)##

So, where is Nero in Emmy's frame? What about ##(0, -20, 0)##? That makes sense. ##+10## for him is ##-10## for Emmy.

Now, at ##t=1##, always better for ##t## to be increasing:

##t = 1##: ##(10, -10, 0)## and ##(-10, +10, 0)##

Where is Nero now? What about ##(20, -20, 0)##? That looks right. Add on his coordinates and you get Emmy's.

So, Nero's velocity (in Emmy's frame) is ##(20, 0, 0)##.

Hmm, so isn't the rule really that:

Emmy's coordinates for an event = Nero's position in Emmy's frame + Nero's coordinates for the event?

Isn't that quite simple?

And, Nero's position is: Nero's starting position + (Nero's velocity x time).

That's the way I'd think about it anyway.
 
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  • #16
PeroK
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Thank you for bearing with me, but I’m still a little lost!

If the velocity is 20 m/s isn’t the calculation still wrong?

10 - (-1 x 20) = 30

Emmy’s value for x at t = -1 is -10. So it still doesn’t seem to work.
It sees to me that Professor Orzel, whatever other merits his book may have, has successfully made a simple matter seem very complicated.

This is pointless fiddling about with numbers when coordinates represent the physical position of things on a diagram at a given time.
 
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  • #17
That makes a lot more sense now. Thank you! I’m going to forget the calculation - I think it’s just confusing matters!
 
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  • #18
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If the velocity is 20 m/s isn’t the calculation still wrong?
10 - (-1 x 20) = 30
Emmy’s value for x at t = -1 is -10. So it still doesn’t seem to work.
Again, careful.... Emmy's speed relative to Nero is the negative of Nero's speed relative to Emmy. That is, if Emmy is moving to the right (the positive x direction) at 20 m/sec relative to Nero her velocity is +20 m/s; and then Nero's velocity relative to Emmy is -20 m/sec because he is moving to the left (the negative x direction).

There's an easier way to get started with this stuff (which, by the way, goes under the name "Galilean transformations" - google will find many references, and yes, Galileo is the person who worked this all out for the first time).
Alice is sitting at the side of a road. At time zero Roger passes her driving to the right at 20 m/sec; Leonard in other lane passes her at the same time but driving left at 10 m/sec. Roger's velocity relative to Alice is 20 m/sec, Leonard's velocity relative to Alice is -10 m/sec, Leonard's velocity relative to Roger is -30 m/sec, Alice's velocity relative to Leonard is 10 m/sec, Roger's velocity relative to Leonard is 30 m/sec, Alice's velocity relative to Roger is -20 m/sec.

Try using these velocities and you'll find that the transformations work between the three frames "Alice at rest, Leonard moving left, Roget moving right", "Roger at rest, Alice moving to left, Leonard moving even faster to left", and "Leonard at rest, Alice moving right, Roger moving even faster right". Once you have that under your belt, you can return to Orzel's example and it will make more sense.
 
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  • #19
Again, careful.... Emmy's speed relative to Nero is the negative of Nero's speed relative to Emmy. That is, if Emmy is moving to the right (the positive x direction) at 20 m/sec relative to Nero her velocity is +20 m/s; and then Nero's velocity relative to Emmy is -20 m/sec because he is moving to the left (the negative x direction).

There's an easier way to get started with this stuff (which, by the way, goes under the name "Galilean transformations" - google will find many references, and yes, Galileo is the person who worked this all out for the first time).
Alice is sitting at the side of a road. At time zero Roger passes her driving to the right at 20 m/sec; Leonard in other lane passes her at the same time but driving left at 10 m/sec. Roger's velocity relative to Alice is 20 m/sec, Leonard's velocity relative to Alice is -10 m/sec, Leonard's velocity relative to Roger is -30 m/sec, Alice's velocity relative to Leonard is 10 m/sec, Roger's velocity relative to Leonard is 30 m/sec, Alice's velocity relative to Roger is -20 m/sec.

Try using these velocities and you'll find that the transformations work between the three frames "Alice at rest, Leonard moving left, Roget moving right", "Roger at rest, Alice moving to left, Leonard moving even faster to left", and "Leonard at rest, Alice moving right, Roger moving even faster right". Once you have that under your belt, you can return to Orzel's example and it will make more sense.

Working on it now - thank you!
 

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