# Are angles relative?

1. Jul 9, 2011

### rede96

2 ships, A and C are travelling on a parallel course, separated by some distance x. As their velocity is constant, I could say that they share an inertial frame of reference, or in other words are at rest wrt to each other.

Their clocks have also been synchronised and I will assume for this thought experiment no gravitational influences so their clocks tick at the same rate wrt to each other.

There is a third ship B, which passes right by A as it intersects A’s trajectory at an angle of 60°, travelling towards C’s trajectory.

As it passes ship A, it sends a signal that contains its clock time. Ship A passes that information onto ship C along with Ship A’s clock time.

It just so happens that due to B’s speed, angle of trajectory of 60°, and the distance x, that B will pass right by C as it intersects C’s trajectory. (So if you imagine a right angle triangle, ship B is travelling along the hypotenuse.)

The distance covered in this time by ship B is twice the distance ship C will travel, which is ok as ship B is travelling at twice the speed of ship C.

http://img97.imageshack.us/img97/3895/image2el.jpg [Broken]

When ship B passes right by ship C, it sends its clock time to ship C. Ship C makes a note of B’s clock time. Then it subtracts A’s clock time form its own clock time in order to work out how long it took ship B to pass between ships A and C from A and C’s frame.

It compares that result to the difference in time from Ship B’s clock times and finds that due to time dilation, ship B’s clock had registered less time.

So far so good I hope!

So here are my conundrums.

1) At no point did ships A, B or C undergo any acceleration. However time dilation still occurred.

So does that mean that acceleration is not required in order for time dilation to occur between two frames? (I was thinking of the twin paradox.)

2) As A and C are at rest wrt to each other, A nor C can say that they are ‘in motion’. So from A’s frame of reference, if ship B did not intersect ifs trajectory at 90°, then it would never meet up with ship C. (Ship A would use the tip of the ship to the stern in order to work out angle of trajectory.)

Yet ship B does meet up with ship C. So are angles relative? Does ship A have to measure ship B’s trajectory at 90°?

3) If A did see ship B pass its trajectory at 60° and then later finds out that ships B and C passed each other, does that mean that ships A and C can say that they are in absolute motion? Because if they weren’t there is no way ship B could have passed by ship C.

Last edited by a moderator: May 5, 2017
2. Jul 9, 2011

### bcrowell

Staff Emeritus
There is a sense in which angles are relative even in Galilean relativity. Suppose observer A is standing on a road and sees raindrops falling straight down. She says the angle between the raindrops' trajectories and the road is 90 degrees. Observer B is driving along the road in a convertible. B says that the angle is not 90 degrees.

3. Jul 9, 2011

### Staff: Mentor

All you need for time dilation is a relative speed, not acceleration. Note that this differs from the twin paradox as you are not directly comparing the difference in elapsed time measured by A and B's clocks directly (which would require an acceleration for the round trip). Instead you are using two clocks in the A-C frame to measure the time in that frame. B, of course, does not agree that the A-C clocks are synchronized.

In the A-C frame, B just moves from A to C. So the angle that a trajectory makes is relative to the frame measuring it.

I don't really understand this point. All that A & C can deduce is that B is moving with respect to them--relative, not absolute motion.

4. Jul 9, 2011

### rede96

Ah, I see. So in my case, B would say 60 degrees but A would 90 degrees.

Ok, thanks. Not sure why B wouldn't agree A-C clocks are synchronized, but I'll think about that one.

EDIT: After a bit of thought, I don't get this. Afterall, I could just join ships A and C together using a very long metal 'bar'. This would in effect make them one very big ship with two big rocket engines at either side. :)

So in effect the new ship would be measuring how long ship 'B' took to cross its width. No issue of synchronization.

But we agree A would have to measure the angle at 90 degrees and B and 60 degrees right?

This was only relevant if A measured B's angle at 60 degrees. So if I am correct in that A has to measure B's angle as it passes at 90 degrees, then this this point is invalid anyway.

What isn's sinking in is how two angles can be measured for the same event. If B passed over A and took a photo of A, it would see A at a 60 degree angle to B's direction of travel.

If A took a photo of B, it would see an angle of 90 degrees to A's direction.

But for that instant, they are in the same relative space. So how can they take the same photo at the same time but get a different result?

Last edited: Jul 9, 2011
5. Jul 10, 2011

### rede96

I still can't get this thing straight in my mind!

The simple relativity part is fine. I can calculate the time dilation using various values for x and different speeds ect.

I can see how the event of the ships passing each other would look from all frames of reference and I also understand that in normal circumstances, no ship could state that 'it was moving’ while the other ship ‘stood still’ or visa-versa.

The one thing that bothers me is the observations that each ship would make do not explain the physics of what is happening.

For example, from ship A's point of view, it sees ship B passing in a straight line between it and ship C

http://img43.imageshack.us/img43/7895/shipag.jpg [Broken]

Ship B would see ships A and C passing it in a straight line (Shown as upwards in the diagram)

http://img546.imageshack.us/img546/8535/shipb.jpg [Broken]

However, the ships are at angles to each other which is not 90 degrees, which all ships would agree on.

As all ships know at some point they fired their booster rockets, which accelerated the ships in the direction they were facing, and because of the conservation of momentum, all ships know which direction they are heading.

I guess as they could measure their acceleration, they would all know their final speed too.

So in this case, armed with all this knowledge, no ship can say that it was 'at rest' while the other ship was moving relative to it.

The only way that all ships could intersect, in this model, is if all ships were moving. Thus showing there must be absolute motion.

Last edited by a moderator: May 5, 2017
6. Jul 10, 2011

### Staff: Mentor

You would still need multiple clocks--one at each end--in the big ship to measure the transit time of ship B, so synchronization issues remain the same.
Ah... I think there's been some confusion, at least on my part, of what you mean by 'angle'.

I was thinking of the angle that the trajectory of B makes in some frame. But you thinking of the angle that the ship makes (which way its nose is pointing) in some frame. Those are two different things.

There are really 3 frames of interest here. The frame of A & C (the one in which A & C are at rest); the rest frame of B; and a third frame, the one in which your first drawing was made.

In that third frame, the trajectory and orientation of ship B are both 60 degrees to the horizontal. But from A & C's frame, the trajectory of ship B is just a straight line from A to C, so it's 90 degrees to the horizontal. But that's not the angle at which A or C measures ship B to be pointing--that angle depends on the speed they all move with respect to that third frame.

7. Jul 10, 2011

### rede96

OK, thanks. I suspected that synchronization remained an issue, I just don't understand why yet. Let me think on that one for a while. :)

My fault, I guess that was my bad wording again. Sorry. :) I was looking at it from each individual frame, so the way my ship is pointing must be the way I am heading. Obviously I got my frames mixed up.

Yes, I see that now.

Yep, got that too.

This is the bit that is catching me out. I sort of understand it, but find it difficult to comprehend.

I visualise it by thinking that the reason things that move through space-time are length contracted and time dilated is because the faster the object moves, the more its 'turns' in a 4th dimensional angle. (If that makes sense!)

The effect of an object 'turning' in this 4 dimensional space time causes distances to reduce, slows down time and makes its length look shorter.

For the length contraction, I imagine it to be a similar effect of looking at something that is a distance away, measuring its length and then turning the distant object through a slight angle. Then If I measure its length again, it will be slightly shorter.

The only difference is that with the 4 dimensional effects, it works both ways. I.e. A would see the effects on B on visa-versa.

I am not sure if that is correct, but it helps me anyway!

So now the bit I try and avoid, the Math!

I really appreciate your help but could I impose a little more? Would you be able to give me an example of how A-C would see ship B's orientation please?

So looking at it from the third frame the numbers I had were
- the angle was 60 degrees
- distances covered for A and C was 10,000,000 km.
- distance covered for B was 20,000,000 km.
- distance x between A and C works out to be about 17320508.076 km

Say speeds for A-C are 0.25c and 0.5c for B

Also, if I wanted to work out the time dilation as seen from A-C's frame, would I just simply subtract speeds from one another as seen from the third frame in order to calculate B's speed relative to A-C?

8. Jul 10, 2011

### Staff: Mentor

You have to specify the orientation in some frame, then you can determine the orientation in any other frame. Remember, these rockets are coasting in space, so they can, in principle be any orientation wrt their motion. I.e. there is nothing non-physical about a rocket coasting tail-first or sideways through space.

9. Jul 10, 2011

### Staff: Mentor

If I weren't so lazy, here's how I would do it. First, start with what you know: We have the velocity of B with respect to the third frame.

To find the velocity of B with respect to the frame A-C, I'd use the relativistic velocity transformations to move from the third frame to the frame A-C.

To find the orientation of ship B, I would imagine a laser beam tracing a path from tail to tip. Then I'd use the relativistic velocity transforms (again) to find the angle of that light beam in frame A-C. (That should work.)

No, you'd need to use the relativistic velocity transformations to find the speed of B in the A-C frame.

10. Jul 10, 2011

### rede96

Yes of course. However I did assume in my example that once the rocket accelerated in a particular direction no other forces acted upon it. So it would keep the same orientation as its original direction once it stopped accelerating.

So I guess I would use the third frame to start off with.

Ok, not too sure on this one, I'll need to find some examples.

No idea on this part. But let me do the whole thing first and see if I can figure it out.

Ok, I think I've seen some examples of this one too.

I tend to work looking at one correct example then applying that to my situation. So if anyone can save me some time by pointing me to some similar examples I would appreciate it.

Thanks guys!

11. Jul 10, 2011

### Staff: Mentor

Sure, but what was that original direction? In space you can fire your engines while slewing sideways at a very high velocity.

12. Jul 10, 2011

### rede96

Ah, you mean what was the original direction before the rockets accelerated? If that is the case, then I would have to use the third frame to 'align' all ships in the direction observed in this frame.

EDIT, so to clarify: From the third frame, ships A and C are moving horizontally, from left to right and their ships are orientated in that direction. Ship A is above Ship C by some distance x, as measured in the third frame.

Ship B is moving from A's trajectory downwards towards C's trajectory at an angle of 60 degrees as observed in the third frame and its ship is orientated in this direction.

13. Jul 10, 2011

### Staff: Mentor

OK, I don't know a nice easy formula for this, but I will show you how you can work it out.

14. Jul 10, 2011

### rede96

That'd be great. Thanks. :~) I really struggle with the math. Not that I can't do it, it has just been such a long time since I have had to do any really testing calcs!

15. Jul 10, 2011

### ctxyz

You are mixing two different contradictory things:

1. In the twin paradox acceleration is required in order to get the differences in total elapsed time (the twins must be asymetric)

2. Time dilation, on the other hand , is independent of acceleration, depends only on relative speed (google "the clock hypothesis")

Definitely. We know this since 1905, when Einstein dedicated a full paragraph to aberration, showing that , if the angle is $\theta$ in one frame then, it is $\theta'$ in another frame where:

$$cos(\theta')=\frac{cos(\theta)-v/c}{1-\frac{v}{c} cos(\theta)}$$

The above is valid for light rays but it can be extended to arbitrary trajectories.

16. Jul 10, 2011

### rede96

Yeah, for some reason I had in my head that time dilation would only become 'real' (as opposed to being observed) if someone had gone through acceleration.

I can see that I got that a bit mixed up!

As I said, I am pretty crap when it comes to the math part. I keep trying to kick start my 46 year old brain into gear, but it just wants to kick back and drink beer!

So in your formula, I take it that cos\theta' is the cosign of the anlgle seen by the third frame in my example, -v and v are the velocities, but not sure what frame they relate to. I know c is speed of light, but do not have a clue what 'frac' is. (Sorry)

17. Jul 10, 2011

### Staff: Mentor

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:
$$a_L(t)=\left( ct, vt + L \; cos(\theta), L \; sin(\theta), 0 \right)$$

We can take the arctangent to check that this is the correct expression for the worldline.
$$atan\left( a_L(t)-a_0(t) \right) = atan\left( \tan (\theta ) \right) = \theta$$

Now, if we Lorentz transform that into a frame moving at velocity u along the x axis wrt the first frame we find:
$$a'_L(t)=\left(\frac{t \left(c^2-u v\right)-L u \cos (\theta )}{c \sqrt{1-\frac{u^2}{c^2}}},\frac{L \cos (\theta )+t (v-u)}{\sqrt{1-\frac{u^2}{c^2}}},L \sin (\theta ),0\right)$$

This expression uses the time coordinate of the unprimed frame, so we have to solve the first coordinate for t in terms of t':
$$c t'=\frac{t \left(c^2-u v\right)-L u \cos (\theta )}{c \sqrt{1-\frac{u^2}{c^2}}}$$

Substituting the solution of this back into the above gives us:
$$a'_L(t') = \left(c t',\frac{c^2 \left(L \sqrt{1-\frac{u^2}{c^2}} \cos (\theta )+t' (v-u)\right)}{c^2-u v},L \sin (\theta ),0\right)$$

Finally, we can take the arctangent to get the angle in the primed frame:
$$atan\left( a'_L(t')-a'_0(t') \right) = atan\left( \frac{\tan (\theta ) \left(c^2-u v\right)}{c^2 \sqrt{1-\frac{u^2}{c^2}}} \right) = \theta'$$

Last edited: Jul 10, 2011
18. Jul 10, 2011

### ctxyz

Nice. related to the above, here is how arbitrary vectors transform

19. Jul 10, 2011

### rede96

Wow, I think I am in a bit over my head! lol

I've tried to convert Tex to word so I can see what is going on, but even then I don't fully understand the formulas.

I really appreciate your help, but I can't follow this just yet. Even just calculating the world line has got me stumped as I don't fully understand the notation.

a_L (t)=(ct,vt+L cos(θ),L sin(θ),0)

Obviously I know the simple stuff like cos, sin, however I don't understand the notation or tex. For example, what are the commas for? (i.e. ,0) What is L? Do I work out t from v for a given distance?

Sorry, please forgive my ignorance but I think I need to back to school! (And I haven't even got to ctxyz's Vector transformations yet :)

20. Jul 10, 2011

### Staff: Mentor

Sorry about that. I don't know a non-mathematical way to derive the angle in a different frame.

Regarding the notation, this is four-vector notation: (ct,x,y,z). So the bit before the first comma is the time coordinate of the four-vector, and the remaining commas separate out the x, y, and z coordinates.

http://en.wikipedia.org/wiki/Four-vector

21. Jul 10, 2011

### ctxyz

From the above , you get:

$$sin(\theta')=\frac{sin(\theta)}{\gamma (1-\frac{v}{c} cos(\theta))}$$

So:

$$tg(\theta')=\frac{tg(\theta)}{cos(\theta)-v/c}$$

This is similar to what Dalespam derived for you but it might be easier for you to digest.

22. Jul 10, 2011

### Staff: Mentor

Is there something wrong with your browser? Mine displays the latex as it should look, without needing to paste it into word.

23. Jul 10, 2011

### Staff: Mentor

So if you have had a chance to go through the Wikipedia article on four-vectors then let's look at this again. Remember that the expression is describing the worldline of a point along the rocket, basically the above expression is equal to the coordinates in four-vector notation: (ct,x,y,z). So that is nothing more than shorthand for the following four equations:
ct = ct
x = vt + L cos(θ)
y = L sin(θ)
z = 0

Do you see how that defines the x,y,z coordinates of the point L on the rocket at any time t?

24. Jul 10, 2011

### rede96

Yeah, ie8 is really screwed. I have had a lot of problems since I upgraded. I can see the formula fine using Firefox.

No need to appolgise, its me that is just a bit dumb! lol. Seriously I really need to do some sort of refresher in order to progress my interest in SR/GR. I can't thank you enough for helping me out here.

Yes, that makes sense. However I think it will sink in more once I've managed to go through an example.

I take it that when we describe the event, we are doing it from the rocket's frame of reference first. Then transforming for other frames?

Would it be just as valid to describe the event from a different frame and then transform for the rocket's frame?

25. Jul 10, 2011

### Staff: Mentor

Actually, in the above I am starting first from a frame of reference where the rocket is moving at velocity v along the x axis. That is only the rocket's frame of reference for the special case v=0. (The unprimed frame is the rocket's frame for v=0)

Certainly. To transform to the rocket's frame simply use u=v. (The primed frame is the rocket's frame for u=v)