Converting between frames of reference

In summary, the author is trying to explain how to convert between different frames of reference, but he isn't explaining it very well. He is confusing the reader with the terminology and the maths doesn't seem to add up.
  • #1
PleaseSpeakSlowly
7
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?
 
Physics news on Phys.org
  • #2
PleaseSpeakSlowly said:
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?
 
  • Like
Likes PleaseSpeakSlowly and Ibix
  • #3
PleaseSpeakSlowly said:
No matter how much I try,...
Post what you have tried.
 
  • Like
Likes PleaseSpeakSlowly
  • #4
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.
 
  • Like
Likes PleaseSpeakSlowly
  • #5
image.jpg
image.jpg
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.
 

Attachments

  • image.jpg
    image.jpg
    15.9 KB · Views: 642
  • image.jpg
    image.jpg
    15.5 KB · Views: 648
  • #6
PleaseSpeakSlowly said:
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?
 
  • Like
Likes PleaseSpeakSlowly
  • #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?
 
  • #8
PleaseSpeakSlowly said:
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.
 
  • Like
Likes PleaseSpeakSlowly
  • #9
PleaseSpeakSlowly said:
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.
 
  • Like
Likes PleaseSpeakSlowly
  • #10
PleaseSpeakSlowly said:
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.
 
  • Like
Likes PleaseSpeakSlowly
  • #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
Ibix said:
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.
 
  • Like
Likes PleaseSpeakSlowly and Ibix
  • #13
PleaseSpeakSlowly said:
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.
 
  • Like
Likes PleaseSpeakSlowly
  • #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
PleaseSpeakSlowly said:
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.
 
  • Like
Likes PleaseSpeakSlowly
  • #16
PleaseSpeakSlowly said:
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.
 
  • Like
Likes Ibix and PleaseSpeakSlowly
  • #17
That makes a lot more sense now. Thank you! I’m going to forget the calculation - I think it’s just confusing matters!
 
  • Like
Likes PeroK
  • #18
PleaseSpeakSlowly said:
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.
 
  • Like
Likes PleaseSpeakSlowly
  • #19
Nugatory said:
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!
 

Related to Converting between frames of reference

1. What is a frame of reference?

A frame of reference is a set of coordinates used to describe the position and motion of an object. It is used to define a point of view from which measurements can be made.

2. How do you convert between frames of reference?

To convert between frames of reference, you need to use a mathematical formula or equation that takes into account the relative positions and velocities of the two frames. This can involve using vectors, matrices, or other mathematical tools.

3. Why is it important to convert between frames of reference?

Converting between frames of reference is important because different observers may have different frames of reference, and it is necessary to be able to compare and communicate measurements accurately. It also allows for a better understanding of motion and can help predict future positions and movements of objects.

4. What are some common frames of reference used in science?

Some common frames of reference used in science include the Earth's surface, the center of mass of an object, and a stationary observer. Other frames of reference may be used depending on the specific situation and the desired level of accuracy.

5. How does Einstein's theory of relativity relate to frames of reference?

Einstein's theory of relativity states that the laws of physics are the same for all observers in uniform motion, regardless of their frames of reference. This means that the conversion between frames of reference must take into account the relative motion between the two frames, rather than just the absolute positions and velocities.

Similar threads

Replies
40
Views
2K
  • Special and General Relativity
3
Replies
101
Views
3K
  • Special and General Relativity
Replies
3
Views
1K
  • Special and General Relativity
Replies
9
Views
1K
  • Special and General Relativity
Replies
16
Views
2K
  • Special and General Relativity
3
Replies
78
Views
5K
  • Special and General Relativity
Replies
4
Views
2K
Replies
13
Views
1K
  • Special and General Relativity
Replies
16
Views
2K
  • Special and General Relativity
4
Replies
123
Views
5K
Back
Top