Are Angles Relative? - Physics Thought Experiment

In summary, two ships, A and C, are traveling on parallel courses, with a distance x between them. They share an inertial frame of reference and their clocks are synchronized. A third ship, B, passes by A at a 60° angle and then passes by C as well. B's clock time is sent to A and then communicated to C. When B reaches C, its clock is compared to C's clock and time dilation is observed. This raises questions about the role of acceleration in time dilation, the relativity of angles, and the possibility of absolute motion for A and C. Ultimately, the angle of B's trajectory is relative to the frame measuring it and all that A and C can deduce is that
  • #1
rede96
663
16
2 ships, A and C are traveling 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°, traveling 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 traveling 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 traveling at twice the speed of ship C.

http://img97.imageshack.us/img97/3895/image2el.jpg

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.
 
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  • #2
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
rede96 said:
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.)
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.

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°?
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.

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.
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
bcrowell said:
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.

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


Doc Al said:
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.

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.


Doc Al said:
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.

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


Doc Al said:
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.

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?
 
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  • #5
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

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

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.
 
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  • #6
rede96 said:
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.
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.
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?
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
Doc Al said:
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.

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. :)

Doc Al said:
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.

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.

Doc Al said:
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.

Yes, I see that now.

Doc Al said:
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.

Yep, got that too.


Doc Al said:
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.

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
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
rede96 said:
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
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.)

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?
No, you'd need to use the relativistic velocity transformations to find the speed of B in the A-C frame.
 
  • #10
DaleSpam said:
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.

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.

Doc Al said:
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.

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


Doc Al said:
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 idea on this part. But let me do the whole thing first and see if I can figure it out.


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

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
rede96 said:
So it would keep the same orientation as its original direction once it stopped accelerating.
Sure, but what was that original direction? In space you can fire your engines while slewing sideways at a very high velocity.
 
  • #12
DaleSpam said:
Sure, but what was that original direction? In space you can fire your engines while slewing sideways at a very high velocity.

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
OK, I don't know a nice easy formula for this, but I will show you how you can work it out.
 
  • #14
DaleSpam said:
OK, I don't know a nice easy formula for this, but I will show you how you can work it out.

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
rede96 said:
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.)

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")
So are angles relative?

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

[tex]cos(\theta')=\frac{cos(\theta)-v/c}{1-\frac{v}{c} cos(\theta)}[/tex]

The above is valid for light rays but it can be extended to arbitrary trajectories.
 
  • #16
ctxyz said:
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").

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!

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

[tex]cos(\theta')=\frac{cos(\theta)-v/c}{1-\frac{v}{c} cos(\theta)}[/tex]

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

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
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:
[tex]a_L(t)=\left( ct, vt + L \; cos(\theta), L \; sin(\theta), 0 \right)[/tex]

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

Now, if we Lorentz transform that into a frame moving at velocity u along the x-axis wrt the first frame we find:
[tex]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)[/tex]

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

Substituting the solution of this back into the above gives us:
[tex]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) [/tex]

Finally, we can take the arctangent to get the angle in the primed frame:
[tex]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'[/tex]
 
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  • #18
DaleSpam said:
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:
[tex]a_L(t)=\left( ct, vt + L \; cos(\theta), L \; sin(\theta), 0 \right)[/tex]

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

Now, if we Lorentz transform that into a frame moving at velocity u along the x-axis wrt the first frame we find:
[tex]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)[/tex]

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

Substituting the solution of this back into the above gives us:
[tex]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) [/tex]

Finally, we can take the arctangent to get the angle in the primed frame:
[tex]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'[/tex]

Nice. related to the above, here is how arbitrary vectors transform
 
  • #19
DaleSpam said:
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:
[tex]a_L(t)=\left( ct, vt + L \; cos(\theta), L \; sin(\theta), 0 \right)[/tex]

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

Now, if we Lorentz transform that into a frame moving at velocity u along the x-axis wrt the first frame we find:
[tex]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)[/tex]

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

Substituting the solution of this back into the above gives us:
[tex]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) [/tex]

Finally, we can take the arctangent to get the angle in the primed frame:
[tex]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'[/tex]


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
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
ctxyz said:
[tex]cos(\theta')=\frac{cos(\theta)-v/c}{1-\frac{v}{c} cos(\theta)}[/tex]

From the above , you get:

[tex]sin(\theta')=\frac{sin(\theta)}{\gamma (1-\frac{v}{c} cos(\theta))}[/tex]

So:

[tex]tg(\theta')=\frac{tg(\theta)}{cos(\theta)-v/c}[/tex]

This is similar to what Dalespam derived for you but it might be easier for you to digest.
 
  • #22
rede96 said:
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.
Is there something wrong with your browser? Mine displays the latex as it should look, without needing to paste it into word.
 
  • #23
rede96 said:
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?
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
DaleSpam said:
Is there something wrong with your browser? Mine displays the latex as it should look, without needing to paste it into word.

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

DaleSpam said:
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

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.

DaleSpam said:
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?

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
rede96 said:
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?
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)

rede96 said:
Would it be just as valid to describe the event from a different frame and then transform for the rocket's frame?
Certainly. To transform to the rocket's frame simply use u=v. (The primed frame is the rocket's frame for u=v)
 
  • #26
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.)
 
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  • #27
First of all, thanks for the help guys. However, for now I have had to concede that I need a bit more time (and help!) to get my head around the math.

It isn’t the math per se that is the problem, it is my understanding of how to populate the formula with the right values that I haven’t grasped yet!

Anyway, I still have one thing about my thought experiment that is bugging me. That is if I understand it right, in SR it is not valid to say that ‘I am moving’. We have to say that I am moving wrt another FoR.

However doesn’t my thought experiment show that frame A-C has to have a velocity of v>0 in order for their frame to experience the two events of B passing A and B passing C? Even looking at it from the A-C FoR?

We know that for the two events to happen either the A-C frame or the B frame must have V>0. I understand that it is just as valid for either frame to say that the other is moving wrt their frame.

However, assuming that A and C are together on the Y axis, and B’s path is at an angle <> to 90 degrees to the x axis, then the two events can only happen if both the A-C frame and the B frame have a velocity > 0.

I know the obvious question is how can the A-C frame show that the path of B (I hesitate to use world line as I don’t fully understand it yet.) is not along the Y axis. Well if B accelerates in the direction the ship is facing, then it should continue along the Y axis.

If it doesn’t, then A-C will observe B moving along the X axis for example. So assume this is what they observe, then the A-C frame can say that in order for the events to have happened, then both frames must have a velocity >0

Because of the conservation of momentum, the A-C frame can say that they their velocity will always be > 0

Does that make sense?
 
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  • #28
rede96 said:
However doesn’t my thought experiment show that frame A-C has to have a velocity of v>0 in order for their frame to experience the two events of B passing A and B passing C? Even looking at it from the A-C FoR?
No. Simply find the coordinates of the two events in any frame and then use the Lorentz transform to get them in any other frame.
 
  • #29
DaleSpam said:
No. Simply find the coordinates of the two events in any frame and then use the Lorentz transform to get them in any other frame.

I assume that the Lorentz transform only has a major effect on what we observe at relativistic speeds.

So assume that the velocity and distances in question are small enough to make the Lorentz transform negligible.

Just to help my understanding, if we look at this problem from a 'non relativistic' point of view, we agree that the both A-C and B must have a velocity > 0 yes in order for the two events to happen right?
 
  • #30
rede96 said:
I assume that the Lorentz transform only has a major effect on what we observe at relativistic speeds.

So assume that the velocity and distances in question are small enough to make the Lorentz transform negligible.
At non-relativistic speeds the Lorentz transform is approximately equal to the Galilean transform, but it is never negligible.

rede96 said:
Just to help my understanding, if we look at this problem from a 'non relativistic' point of view, we agree that the both A-C and B must have a velocity > 0 yes in order for the two events to happen right?
Non-relativistically the two events can still happen with A-C at rest.
 
  • #31
DaleSpam said:
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.
 
  • #32
rede96 said:
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.
 
  • #33
Doc Al said:
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?
 
  • #34
rede96 said:
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).
 
  • #35
Doc Al said:
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).

Doc Al said:
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.
 

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