Agreeing on relative separations and accelerations

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In summary, the conversation discusses the concept of agreement between two frames in terms of their relative motion, specifically in regards to the measurement of relative separations, accelerations, and velocities. It is concluded that the agreement is only valid when the particular quantity being measured is constant. The confusion stems from the relativity of simultaneity, where the concept of "when" something happens differs between the two frames. This leads to a symmetric situation where both frames will measure the same values at certain points in their respective times. The idea of agreement itself is defined as the two frames measuring the same values when their origins coincide.
  • #1
etotheipi
From the relative velocity equation we can see that two frames will both measure the same relative speed. I don't believe the same is true for relative separations and relative accelerations, but can't find anything useful online to verify this, and was wondering whether someone could point me in the right direction.

I suppose foremost, it doesn't make sense to talk about two observers agreeing on a varying quantity like relative separation, for instance, since the measured time at which the separation equals a certain value will differ between the two frames. This would be the same for relative acceleration in the case that it was not constant, and I'm guessing also for relative velocity in the case that there exists acceleration. The "agreement" in relative velocity between frames seems only to be valid since is constant for all times in both frames.

We can definitely measure the acceleration of one thing relative to another using the acceleration transformation, and the separation of one object from another with the Lorentz transformations, given that we already know what the raw values in another frame are. However, am I right in saying that the only context in which it makes sense to discuss two frames agreeing on their relative 'something' is when that particular something is a constant?

Sorry for the slightly verbose post, I promise I'll add some formulae next time...!
 
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  • #2
If they measure a different quantity (in terms of magnitude) regarding their relative motion, which will measure the higher quantity and which the lower?
 
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  • #3
PeroK said:
If they measure a different quantity (in terms of magnitude) regarding their relative motion, which will measure the higher quantity and which the lower?

This is one of the things that has confused me. For example, I was thinking about two identical spheres A and B, where B moves away at ##v##. When ##\tau## elapses for B, ##\gamma \tau## has elapsed for A. So B views A as being a distance ##v\tau## away, whilst A views B as being ##v\gamma\tau## away. Though a reciprocal argument can be applied for A, and we would end up with the opposite.

I came to the conclusion that both were "correct", and that it didn't really make sense, though I recall that in the not so dissimilar twin-non-paradox both initially observe the other ageing more slowly, so it shouldn't be an issue here either.

Although I'm inclined to say that even though we can't tell which would measure a greater value for separation, we can't say they agree either.
 
  • #4
etotheipi said:
This is one of the things that has confused me. For example, I was thinking about two identical spheres A and B, where B moves away at ##v##. When ##\tau## elapses for B, ##\gamma \tau## has elapsed for A. So B views A as being a distance ##v\tau## away, whilst A views B as being ##v\gamma\tau## away. Though a reciprocal argument can be applied for A, and we would end up with the opposite.
You've forgotten the relativity of simultaneity. "When" something happens means different things for the two objects - and this leads to symmetry between the two. Both say that "when" ##\tau## has elapsed for them, ##\tau/\gamma## has elapsed for the other.
 
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  • #5
etotheipi said:
This is one of the things that has confused me. For example, I was thinking about two identical spheres A and B, where B moves away at ##v##. When ##\tau## elapses for B, ##\gamma \tau## has elapsed for A. So B views A as being a distance ##v\tau## away, whilst A views B as being ##v\gamma\tau## away. Though a reciprocal argument can be applied for A, and we would end up with the opposite.

I came to the conclusion that both were "correct", and that it didn't really make sense, though I recall that in the not so dissimilar twin-non-paradox both initially observe the other ageing more slowly, so it shouldn't be an issue here either.

Although I'm inclined to say that even though we can't tell which would measure a greater value for separation, we can't say they agree either.

They must agree as you cannot distinguish one from the other. This follows from the assumed symmetry of space. What happens to your argument if you consider ##\tau## elapsing for A first?

The twins are different, as one stops and turns round.

PS although the explanation is given by @Ibix above, you can see that the situation must be symmetric. I.e. you can see the argument must be flawed before you find out why.
 
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  • #6
PeroK said:
They must agree as you cannot distinguish one from the other.
This depends on what simultaneity convention you apply. If your setup of that is also symmetric, you will get something symmetric. If not, you will not. The problem here is asking ”what is the separation now?” without further specifying ”now” - as usual.

Ibix said:
You've forgotten the relativity of simultaneity.
This. See my signature.
 
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  • #7
Ibix said:
You've forgotten the relativity of simultaneity. "When" something happens means different things for the two objects - and this leads to symmetry between the two. Both say that "when" ##\tau## has elapsed for them, ##\tau/\gamma## has elapsed for the other.

Oh no... I've broken Orodruin's signature rule... and I didn't even realize!

PeroK said:
They must agree as you cannot distinguish one from the other. This follows from the assumed symmetry of space. What happens to your argument if you consider ##\tau## elapsing for A first?

The twins are different, as one stops and turns round.

PS although the explanation is given by @Ibix above, you can see that the situation must be symmetric. I.e. you can see the argument must be flawed before you find out why.

Yes this makes sense, if we went backward a little bit in A's time to ##t_{A} = \tau##, A would also see B at that same distance away, so it is symmetrical.

I think the source of my confusion is actually on how the agreement itself is defined. That is, if the origins coincide, when the time in each frame reaches 5 seconds respectively they will both measure the same values - in that sense, they agree.
 
  • #8
Orodruin said:
The problem here is asking ”what is the separation now?” without further specifying ”now” - as usual.

Ah, right I get it now. I can't say I don't feel slightly stupid for not getting that earlier but it's all a learning process I guess!
 
  • #9
etotheipi said:
I think the source of my confusion is actually on how the agreement itself is defined. That is, if the origins coincide, when the time in each frame reaches 5 seconds respectively they will both measure the same values - in that sense, they agree.

The starting point is that each measures the other to be moving with a constant speed ##v##. Constant motion between two reference frames can always be reduced to one dimension. (Using spatial homogeneity to move the origin.) From that you get the symmetry in terms of relative speed.

Once each has decided on coordinates, the motion of the other must be of the form ##\vec{x}(t) = \vec{x_0} + \vec{v}t##.

Where ##|\vec{v}| = |\vec{v}'|##.

The only asymmetry is the arbitrary choice of origin.
 
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  • #10
PeroK said:
The starting point is that each measures the other to be moving with a constant speed ##v##. Constant motion between two reference frames can always be reduced to one dimension. (Using spatial homogeneity to move the origin.) From that you get the symmetry in terms of relative speed.

Once each has decided on coordinates, the motion of the other must be of the form ##\vec{x}(t) = \vec{x_0} + \vec{v}t##.

Where ##|\vec{v}| = |\vec{v}'|##.

The only asymmetry is the arbitrary choice of origin.

Thank you, this makes it clear!
 
  • #11
I thought I'd post this just for sake of closure; just a little spacetime diagram that helped me to understand!

IMG_0533.jpeg
 
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  • #12
Ibix said:
Both say that "when" ##\tau## has elapsed for them, ##\tau/\gamma## has elapsed for the other.

Huhh?! I was assuming the usual convention where ##\tau## is a proper time. Thus ##\gamma \tau## would be the time that each of them say has elapsed in the other frame. And likewise, each will say that he has a traveled a distance of ##v \tau## while the other has traveled a distance of ##v \tau/\gamma##.

In other words, each will observe that the other has experienced dilated time and contracted length.

Edit: See Post #3.
 
  • #13
Mister T said:
Huhh?! I was assuming the usual convention where ##\tau## is a proper time. Thus ##\gamma \tau## would be the time that each of them say has elapsed in the other frame.
Due to time dilation, according to me your clock must read nearer zero. Since ##\gamma\geq 1##, if my clock reads ##\tau## yours must read ##\tau/\gamma## using my simultaneity convention. I was using notation slightly differently from #3 (##\tau## in my post is the proper time of whichever object regards itself as at rest, while I think @etotheipi was using it for the proper time of object B) which probably wasn't the best thing I could have done. I don't think I'm wrong, though.
Mister T said:
And likewise, each will say that he has a traveled a distance of ##v \tau## while the other has traveled a distance of ##v \tau/\gamma##.
I would expect each object to say it has traveled a distance of zero, while the other traveled ##v\tau## (using my definition of ##\tau##) or ##v\gamma\tau## (using etotheipi's). That is, two objects regarding themself as "at rest" and the other as moving at speed ##\pm v##. You seem to be interpreting the setup somewhat differently, but I'm not sure quite what you have in mind.
 
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  • #14
Ibix said:
if my clock reads τ yours must read τ/γ using my simultaneity convention. I was using notation slightly differently from #3 (ττ\tau in my post is the proper time of whichever object regards itself as at rest, while I think @etotheipi was using it for the proper time of object B) which probably wasn't the best thing I could have done. I don't think I'm wrong, though.

I think my confusion must have been somewhere in there. I can't remember, though, what I was thinking then! I think I got the frame-dependent simultaneity conventions switched.
 
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1. What is the concept of agreeing on relative separations and accelerations?

The concept of agreeing on relative separations and accelerations is a fundamental principle in physics that states that the laws of motion and the concept of relative motion are the same for all observers, regardless of their relative velocities.

2. Why is it important to agree on relative separations and accelerations?

It is important to agree on relative separations and accelerations because it allows for consistent and accurate measurements and predictions in physics. Without this agreement, different observers would have different measurements and interpretations of motion, making it difficult to understand and study the physical world.

3. How do scientists agree on relative separations and accelerations?

Scientists agree on relative separations and accelerations by using a reference frame, which is a specific point of view or perspective from which measurements are made. By choosing a common reference frame, scientists can compare and analyze data from different observers and ensure consistency in their results.

4. Can relative separations and accelerations change depending on the reference frame?

Yes, relative separations and accelerations can change depending on the reference frame. This is because the laws of motion are relative, meaning they depend on the observer's point of view. However, the overall principles and equations remain the same, regardless of the reference frame used.

5. What are some real-life applications of agreeing on relative separations and accelerations?

Agreeing on relative separations and accelerations is essential in various fields, including physics, engineering, and astronomy. It allows for the accurate prediction and analysis of motion in everyday objects, vehicles, and celestial bodies. It also plays a crucial role in the development of technologies such as GPS systems and spacecraft navigation.

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