Are Angles Relative? - Physics Thought Experiment

[rede96]

Non-relativistically the two events can still happen with A-C at rest.

Not if B has a trajectory that is at an a 60 degree angle to the x-axis of A-C. The only way that the 2 events can happen if A-C are 'at rest' is if B's trajectory is at 90 degrees to the x axis.

You'll agree that there are a number of scenarios where the two events can happen.

1) With A-C at rest and b moving at a 90 degree angle to the x axis.

2) With B at rest and A-C moving along the y axis

3) With A-C moving along the x-axis and B moving at a 60 degree angle to A-C's x axis.

 

[Doc Al]

The only way that the 2 events can happen if A-C are 'at rest' is if B's trajectory is at 90 degrees to the x axis.

But B's trajectory is at 90 degrees to the x-axis in the A-C frame.

 

[rede96]

But B's trajectory is at 90 degrees to the x-axis in the A-C frame.

Always? Not matter what trajectory B approaches A-C?

 

[Doc Al]

Always? Not matter what trajectory B approaches A-C?

I'm talking about the original problem you started this thread with, in which B moves from A to C (as viewed from the A-C frame).

 

[rede96]

I'm talking about the original problem you started this thread with, in which B moves from A to C (as viewed from the A-C frame).

I'm talking about the original problem you started this thread with, in which B moves from A to C (as viewed from the A-C frame).

Sure, I was just trying making the point that it doesn't have to be 90 degrees and that the two events can still occur even if the angle isn't 90.

And if the two events still occur and the angle is not 90 degrees, then both A-C and B must have V>0.

Or are you saying that in every case where the A-C frame observe the two events B must be at 90 degrees? That is an honest question. :)

If that is not the case, then the issue in question is how could the A-C frame know that that the angle of B is not 90, because the A-C will always observe this.

I can think of a case where A-C ask B to accelerate after the two events have happened. If the angle was not 90, then B will be seen to move along the x axis.

Armed with this knowledge, then A-C can deduce that there observation may not be representative of the situation.

After all, there can only be one world line for B.

 

[Dale]

You'll agree that there are a number of scenarios where the two events can happen.

In absolutely any scenario where the two events happen in one frame they will happen in all frames, in particular, including the frame where A and B are at rest.

This has nothing to do with relativity and is simply the nature of coordinate systems. The angle of the path in different frames will automatically be whatever angle is required for the events to happen. This is how any coordinate transformation works.

 

[Doc Al]

Or are you saying that in every case where the A-C frame observe the two events B must be at 90 degrees? That is an honest question. :)

Well, if the first event is B leaves A and the second is B arrives at C, how can the trajectory be anything other than at 90 degrees as seen in the A-C frame.

 

[rede96]

In absolutely any scenario where the two events happen in one frame they will happen in all frames, in particular, including the frame where A and B are at rest.

Yes, I understand this. Maybe I need to try and look at my problem using coordinates. Its late right now, but I'll have a think about it tomorrow. Thanks

Well, if the first event is B leaves A and the second is B arrives at C, how can the trajectory be anything other than at 90 degrees as seen in the A-C frame.

I think the issue I am struggling to come to terms with is how can B have two trajectories. I understand how the observations in each frame are true for each frame, but they are only observations.

If throw a ball in the air it can't go up at two different angles.

Let me have a think about this for a while!

Thanks

 

[Doc Al]

If throw a ball in the air it can't go up at two different angles.

If you view it from multiple frames it can.

In your problem you are flipping back and forth between the A-C frame and that third frame that you started off with.

 

[Dale]

I think the issue I am struggling to come to terms with is how can B have two trajectories. I understand how the observations in each frame are true for each frame, but they are only observations.

I wouldn't say that B has multiple trajectories, but rather that there are multiple valid descriptions of B's trajectory.

As an analogy, suppose you and I are watching a race from opposite sides of the road. I would say they are racing to the right and you would say they are racing to the left. They only have one trajectory, but we have different (equally valid) ways to describe it. Since I know that you are facing me I know how to transform your coordinates to my coordinates so I know that your left is my right.

 

[JDoolin]

OK, so for simplicity let's consider a rocket moving at velocity v along the x-axis making an angle theta with x-axis in the x-y plane. So the worldline of a point L along the rocket is given by:
a_L(t)=\left( ct, vt + L \; cos(\theta), L \; sin(\theta), 0 \right)

I can't figure out the definition of L in this context. Is it a vector or a scalar? Is there a point on the rocket at (L cos(\theta),L sin(\theta))?

Simply find the coordinates of the two events in any frame and then use the Lorentz transform to get them in any other frame.

That's essentially what I did in the demo. Calculated the coordinates of the two events in one frame (which is a simple task) and then Lorentz Transformed those events to another frame, (which is also a simple task if you've got a computer to handle it for you.)

In hindsight, another thing I could have put in the demo would have been the lorentz contracted frame of the body. An X design on a passing space-ship would be mashed in the direction of its motion. So Lorentz Contraction is another effect on angles

And then there are the things that would have been hard to put in the demo, since it is 2Dimensional, such as effects like Terrell Rotation, and that effect that makes stars bunch up in front of you (I forgot what that effect is called.)

Edit: Stellar aberration is what its called. And now that I look at it, I think that the demo handles stellar aberration just fine; the gist of it can be shown in 2 Dimensional space.

 

[rede96]

I made this demo about 10 years ago. It fires off a pulse at a ninety degree angle and at equal angles to the front and to the back. You have to click the right arrow buton a few times to read all the instructions.

http://www.wiu.edu/users/jdd109/stuff/relativity/gardner.swf

In any case, yes, the angles are different.

(Actually I'm not sure if this even addresses a similar question to what you're asking.)

Thanks for that, I found quite useful. Cheers.

I wouldn't say that B has multiple trajectories, but rather that there are multiple valid descriptions of B's trajectory.

Yes, this was exactly the point I was trying to make, or one of them. (Albeit very badly!)

So this would apply to my original question, are angles relative. The answer as I understand it is that there is only one angle, but multiple descriptions of it. And as I understand now, we use coordinate transformation to change whatever angle is seen in another reference frame into our reference frame.

If you view it from multiple frames it can.

In your problem you are flipping back and forth between the A-C frame that third frame that you started off with.

OK, let me see if I can clear this up.

This post got me thinking about whether or not we could establish an ‘absolute’ frame. I suspect that this should have been a new post, but let me explain my thought process.

If there is only one trajectory or angle etc, but multiple descriptions of it, could the same be the same for a 'series of events' observed by different frames? Is there just one absolute frame that we can coordinate transform all other frames into?

I know that this is probably a non-starter, but I wanted to explore it a bit further anyway.

So that got me thinking, if there was an absolute frame, there must be absolute velocity.

Relativity says that no frame can say ‘I am in motion’. Whilst I don’t wish to challenge relativity, I couldn’t help but ask:

is this because the laws of nature say no frame can have absolute motion or is that we just don’t always have enough information to establish it?

Hence I built on my original thought experiment and believed that I found a way in which frame A-C could prove that it must be in motion.


So to build on that, imagine 3 ships, A, C and D, all at rest wrt each other and all sat on the y-axis at points y0, y-1 and y-2. (Forgive me if I don’t get the notation correct.)

Obviously, because we are describing events from the A-C-D frame v=0, x=0 and z = 0.

So along comes ship B, passing A and creating event 1 at y0, x0 and z0 (I assumed that B would have to be at z1 so it doesn’t collide with A)

A watches B as it continues its journey and passes C creating event 2 at y-1, x0 and z0 and concludes that B‘s trajectory is at 90 degrees to the x axis. A notices that B’s ship is not orientated in the y direction, but doesn’t think much about it.

So for A, he would expect that as B’s trajectory is at 90 degrees to the x axis, B would soon pass D creating event 3 at y-2, x0, z0

He gets a bit fed up of waiting and contacts B to accelerate. B accelerates but the guy watching from A sees B move away from the y axis, along the x-axis and doesn’t pass D. In fact he passes well in front of D along the x axis.

How can that be?

A assumes that B must have changed trajectory as he knows that A, C and D have not moved as there has been no acceleration experienced. He fires a laser beam down the y-axis and sees that A, C and D are still aligned.

So he contacts B and asks why he changed trajectory. B says he accelerated in the direction he was heading, but didn’t change trajectory and can prove he didn’t. As B has been following a laser beam that comes from a space station further back in B’s trajectory.

So I guess for A, this creates a bit of paradox. As for his frame B has changed trajectory but obviously this can’t be. (No pun intended!)

So A thinks about it for a while and works out that the only way that B could pass A and C but not D, whilst B remained in the same trajectory, was if his frame moved out of the way or B’s trajectory was not at 90 degrees. He knows his frame did not accelerate, so therefore B must not have been at 90 degrees. And therefore, the only way events 1 and 2 could have happened is if both B and A-C-D had a velocity > 0.

So A can conclude that its frame is moving! He can’t say with what velocity he is moving, only with what velocity he is moving relative to B

However, is there was an absolute frame or ‘zero’ frame that we could use to calculate absolute velocity?

I haven’t thought this through fully (obviously!) but if we assume that in order for the two events to happen both A-C-D and B had velocity >0 and that B’s velocity had to be double that of A-C-D’s velocity, then the faster these two frames were travelling, even though the velocities remained double, the more time dilation they would experience between the two frames.

So if A-C-D measured the time dilation between the two events, then they might be able to work out at what velocities would give this amount of time dilation. Thus would know absolute velocity.

Now that is a very long winded way to think about I’m sure! But as far as I can tell, it is still valid.

Now I have to say that I do not present this as an alternative theory, god no. Just a way of interpreting relativity differently as I understand it.

 

[Dale]

The answer as I understand it is that there is only one angle, but multiple descriptions of it.

The angle IS the description of the trajectory. If left is an angle of 180º then right is an angle of 360º.

He can’t say with what velocity he is moving, only with what velocity he is moving relative to B

Yes. A clearly knows that he is moving relative to B by the simple fact that he knows that B is moving relative to him. The rest of the thought experiment is extraneous.

However, is there was an absolute frame or ‘zero’ frame that we could use to calculate absolute velocity?

No.

 

[rede96]

The angle IS the description of the trajectory.

Ok. Thanks.

 

[JDoolin]

However, is there was an absolute frame or ‘zero’ frame that we could use to calculate absolute velocity?

No.

...unless you say we are moving at around 600 km/s relative to the CMBR "rest frame." This implies that there is some universal rest-frame.

http://www.astro.ubc.ca/people/scott/faq_basic.html]There[/PLAIN] clearly is a frame where the CMB is at rest, and so this is, in some sense, the rest frame of the Universe. But for doing any physics experiment, any other frame is as good as this one. So the only difference is that in the CMB rest frame you measure no velocity with respect to the CMB photons, but that does not imply any fundamental difference in the laws of physics.

There is an untested alternative hypothesis out there originally developed by A.E. Milne, and generally ignored. It assumes there is no "rest frame" and that the galaxies really are flying apart at the speeds they appear to be flying apart. I believe, after a couple of tweaks, (namely acknowledging that the early universe must have been a period of intense acceleration) the Milne model should be seriously revisited. And the data from astronomy should be analyzed under the context of Milne's Model where changing scale factors are due to changes in the non-local observer's reference frame, instead of being analyzed under the context of changing local scale-factors in time (i.e. stretching space).

But I still have homework to do on this, and maybe I will eventually be able to make it clear, what I'm trying to say. Figure 2, page 24 of Weinberg's "The First Three Minutes" is one of my favorite examples, though, where the relativity of simultaneity is very pointedly ignored, and leads him to an incorrect conclusion.

 

[Dale]

...unless you say we are moving at around 600 km/s relative to the CMBR "rest frame." This implies that there is some universal rest-frame.

That still wouldn't be an absolute velocity. That would only be velocity relative to the local CMBR. That is no more absolute than your velocity relative to Mt. Everest or some other specific nearby physical object.

An absolute velocity does not simply mean a velocity relative to some physical feature, but a velocity where the laws of physics take on a different form. Such a velocity simply does not exist to the best of modern experimental precision.

PS Hijacking a thread is considered rude. If you wish to discuss the CMBR please find an appropriate thread or make your own. This thread is about discussing how trajectory angles transform under the Lorentz transform, not about the CMBR.

 

[rede96]

This thread is about discussing how trajectory angles transform under the Lorentz transform, not about the CMBR.

Good point, however I might have had something to do with that as I think my post #43 may have been taking this thread off-topic.

I found the info posted here very useful, thanks to all. However I would suggest that we leave this topic here for now and I'll make another post regarding absolute motion.

Thanks!

 

[JDoolin]

Good point, however I might have had something to do with that as I think my post #43 may have been taking this thread off-topic.

I found the info posted here very useful, thanks to all. However I would suggest that we leave this topic here for now and I'll make another post regarding absolute motion.

Thanks!

Could you make a diagram of what you were describing? I don't think you were off topic. I just couldn't quite piece it all together, from the word description.