Question about length contraction and motion

[Dagl1]

Hi there,

I have been reading a book about time (with an equal name, by Paul Davies). In it he explains time dilation and parts of relativity. I have two questions regarding what I have read so far, one for another thread. My current question is regarding the perceived difference in distance when two observers are in relative motion to each other. A star 5 light years away from the earths point of view, might be only 3 light years away for someone traveling at 80% the speed of light towards this star, if I understand the physics right. This alone I can understand, however I cannot seem to then comprehend how this would work for motion between other observers.
Even in writing my question I notice that the ideas are very fuzzy as I keep rewording my question: If I have a map with the 'position' of different stars around the galaxy, or different galaxies in the observable universe, and we assume these bodies are moving at high speeds relative to each other, does the distance between two galaxies depend on the specific galaxy I measure these from?

I am sorry for my very fuzzy question, it is obvious I haven't wrapped my head around the length contraction yet. I think I am trying to ask what happens to the distance between different points if the motion of the observer relative to these point is what determines the distance. Does this mean that if I am in the Andromeda galaxy (or a theoretical one that is moving at very high relativistic speeds to ours), I would calculate different distances between points in the universe. 
If so then is there any 'absolute' distance between any points? Does it make sense to say that a star is x light years away, if by speeding up or slowing down, that distance is changed.

An additional question: let's say there is a galaxy A some distance from ours, and in its extension (so in the exact same direction but further away) there is a galaxy B, 'twice as far' as galaxy A is. Galaxy A is not moving relative to our Home galaxy, while galaxy B is moving at 99% the speed of light, wobbling back and forth. What are the implications for the actual distances between the galaxies. Galaxy B's distance should not change as it moves (through some unknown means) away, and then back to us. In my head, which I think is the wrong way to think about it, I would say that by measuring the distance between Home galaxy and B, I would see a much smaller distance than 2xA, while in reality they are just twice as far away from each other. I suppose the reason is that because galaxy B is in motion, we cannot say it is twice as far away from us, as we can only measure things from one frame of reference.
However my mind remains fuzzy regarding these concepts, if some people would be willing to educate me and show me the errors of my thinking, that would be greatly appreciated!

-Dagl

[md65536]

I'm assuming flat spacetime (no mass, SR only), and inertial motion unless specified.

12 hours ago, Dagl1 said:

If I have a map with the 'position' of different stars around the galaxy, or different galaxies in the observable universe, and we assume these bodies are moving at high speeds relative to each other, does the distance between two galaxies depend on the specific galaxy I measure these from?

Generally yes. Also "when are you talking about?" matters and is more complicated than a Newtonian description. The positions of things on a map are coordinates within a coordinate system, and those are different for different observers. You could have a map where the Milky Way is at a fixed location and Andromeda is moving, or one where Andromeda is fixed and the Milky Way is moving. Those correspond to the coordinate systems of 2 observers at rest in the respective galaxies.

12 hours ago, Dagl1 said:

If so then is there any 'absolute' distance between any points? Does it make sense to say that a star is x light years away, if by speeding up or slowing down, that distance is changed.

Yes, the distance to the star is different in the different frames of reference. There are invariant measures of distance, eg. the "proper length" of a 1 m stick at rest is always 1 m and everyone will agree on that, even if the stick is moving relative to some observer and is length-contracted ie. has a coordinate length less than a meter in that observer's coordinates.

12 hours ago, Dagl1 said:

I suppose the reason is that because galaxy B is in motion, we cannot say it is twice as far away from us, as we can only measure things from one frame of reference.

We can say that. We can measure the distances to both galaxies using one frame of reference (eg. the one in which we're at rest), and you can measure the motion of objects using the coordinates of that frame.

Consider the map analogy. The spatial coordinates can be represented by a grid drawn on the map. The same grid coordinates can be shown by putting a lattice of rulers throughout space. In our own frame of reference, our rulers are not moving and so they don't length-contract. An object light years away can wobble at speeds near c, but yet stay near one place in the grid of rulers.

Meanwhile, that distant wobbling object is moving relative to our lattice of rulers, and our rulers do length-contract, in its frame of reference. For simplicity consider two different inertial frames of reference F1 and F2 that the wobbling object switches between. Each of those frames has its own set of rulers making up a lattice throughout space, each at rest and not length-contracted in its own rest frame. Say I'm at 1 LY from Earth, as measured by Earth, and I'm wobbling relative to Earth. I stay near the 1 LY mark on Earth's set of rulers, but those rulers are contracted by different amounts in F1 vs F2. For example, in F1 Earth might be 0.8 LY away from me and the 1 LY mark, and only 0.6 LY away in F2's frame. The reason that the distance as measured by Earth isn't changing much, and the distance measured by me is changing drastically, is that I'm switching between different frames of reference. The distance between Earth and the 1 LY mark, which has a proper distance of 1 LY, is length-contracted by different amounts to different observers, depending on their relative speed.

[swansont]

Your map is only valid in one frame of reference

[MigL]

In case the two previous replies didn't make it abundantly clear ...
Distance and time are two ( of many ) things that are frame dependent.

And thanks for asking interesting questions 🙂 .

[studiot]

14 hours ago, Dagl1 said:

A star 5 light years away from the earths point of view, might be only 3 light years away for someone traveling at 80% the speed of light towards this star, if I understand the physics right. This alone I can understand, however I cannot seem to then comprehend how this would work for motion between other observers.

Yes you understand that physics correctly, but things are a bit weirder than you have presented since what you have said is incomplete.

 

Let us call the Earth E, the Traveller T and the Star S.

So there is a relative speed between T and S of 0.8c.

Which means that they are not in the same frame.

So we must convert the 3Ly separation to one frame or the other the T frame or the S frame, it does not matter which.

This will give us the 'proper distance' between T and S. The proper distance Lo is always longer than the measured distance between moving objects and is the same in all frames.

1 hour ago, md65536 said:

There are invariant measures of distance, eg. the "proper length" of a 1 m stick at rest is always 1 m and everyone will agree on that, even if the stick is moving relative to some observer and is length-contracted ie. has a coordinate length less than a meter in that observer's coordinates.

 

So we use the formula 

[math]L = {L_0}\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} [/math]

Substituting the given data

[math]3 = {L_0}\sqrt {1 - \frac{{{{\left( {0.8c} \right)}^2}}}{{{c^2}}}}  = {L_0}\sqrt {1 - 0.64}  = {L_0}\sqrt {0.36}  = 0.6{L_0}[/math]

[math]{L_0} = \frac{3}{{0.6}} = 5[/math] light-years

This is what I mean by your original statement was incomplete.  (no criticism intended I am being detailed in order to help)

Because you originally stated that S is 5 Ly from Earth.

Since this is the 'proper distance' it must already be corrected  for any motion between E and S, which you have omitted to say.
It is possible, though not generally the case, the S will not be also moving relative to E.

Just to add that when you start to combine the effects of two relative velocities the calculation becomes more involved. You can't simply add them up.

Note here that it does not matter whether T is travelling towards S or away from it, but T must just be passing E when the measurement is made.

 

 

 

 

 

[Dagl1]

Thanks! I will read and digest the information in detail later today!

[Dagl1]

23 hours ago, md65536 said:

Yes, the distance to the star is different in the different frames of reference. There are invariant measures of distance, eg. the "proper length" of a 1 m stick at rest is always 1 m and everyone will agree on that, even if the stick is moving relative to some observer and is length-contracted ie. has a coordinate length less than a meter in that observer's coordinates.

I generally understand your responses to my initial question, but now I wonder how we can even have any 'proper length', of course in 1 frame of reference we can agree on a specific length. But within the universe is there then anyway to say what 'proper length' is, if the distances are all different from different points of view. Is there ever a way to get down to a 'universal length', as this is what 'proper length' implies for me. But I thought that was one of the things that relavivity brought us, that there is no one main frame of reference, and thus also no one main of distance between points? 

Studiot seems to be providing me at least in part with an answer:

21 hours ago, studiot said:

So we must convert the 3Ly separation to one frame or the other the T frame or the S frame, it does not matter which.

This will give us the 'proper distance' between T and S. The proper distance Lo is always longer than the measured distance between moving objects and is the same in all frames.

Although I still don't entirely get how this is a 'proper distance', does the proper distance represent the amount of protons I could put in a row to reach such a point? But even those can get length contracted? How do we know that the proper distance is not actually still measured from a moving frame, or that there is even a 'stiller frame'?

21 hours ago, studiot said:

his is what I mean by your original statement was incomplete.  (no criticism intended I am being detailed in order to help)

No criticims perceived , I love the detailed responses and am happy you guys are sticking by me even as I am floundering through incomplete comprehension!

I then also have another question as to how the expansion of the universe comes into play with this, how does length contraction work for galaxies far away and which are accelerating away from us, already at speeds faster than the speed of light. When we say nothing can go faster than the speed of light, I am aware that due to how this is calculated the 'actual speed' of two things at relative speeds is never faster than the speed of light (I hope I am correct in my understanding here?), so I suppose we use that calculation for length contraction?
What I am getting at is that in my mind, given that galaxy A and galaxy B are already at such distances that the expansion of the universe leads to faster movement away from each other than light will be able to travel (so the galaxies that are on the edge of the observable universe and will then 'move out' of that bubble). How does lenght contraction not lead us to say that these are actually very close (because relatively we are moving at enormous speeds)? I really seem to have some trouble with this whole concept;p

 

[studiot]

2 hours ago, Dagl1 said:

Although I still don't entirely get how this is a 'proper distance', does the proper distance represent the amount of protons I could put in a row to reach such a point? But even those can get length contracted? How do we know that the proper distance is not actually still measured from a moving frame, or that there is even a 'stiller frame'?

Let us do this a little bit at a time.

Then it might 'click' for you.

 

23 hours ago, studiot said:

1) Let us call the Earth E, the Traveller T and the Star S.

2) So there is a relative speed between T and S of 0.8c.

3) Which means that they are not in the same frame.

Did you understand lines 1, 2, and 3 above ?

In particular did you understand how you get from line 2 to line 3  ?

[Dagl1]

Yes this seems to be clear, as each has its own relative motion, they are in differnt frames.

If the T and S are not moving at the same speed relative to each other, they by definiton can't be in the same frame right?

[Bufofrog]

15 hours ago, Dagl1 said:

Although I still don't entirely get how this is a 'proper distance', does the proper distance represent the amount of protons I could put in a row to reach such a point? But even those can get length contracted? How do we know that the proper distance is not actually still measured from a moving frame, or that there is even a 'stiller frame'?

Proper distance or proper length is the distance you measure in your own frame.  In your frame your meter stick is not moving hence it is the proper length. Any other frame will see your meter stick as length contracted.  

[swansont]

4 hours ago, Dagl1 said:

Yes this seems to be clear, as each has its own relative motion, they are in differnt frames.

If the T and S are not moving at the same speed relative to each other, they by definiton can't be in the same frame right?

How do they not move at the same speed relative to each other?   

If they are moving relative to each other, at all, they are not in the same frame.

[studiot]

4 hours ago, Dagl1 said:

Yes this seems to be clear, as each has its own relative motion, they are in differnt frames.

If the T and S are not moving at the same speed relative to each other, they by definiton can't be in the same frame right?

Oh dear and Oh good, at least you understand something.

Every object has a frame in which it is 'at rest'.
We call this its own frame.

So

S is at rest in the S frame
T is at rest in the T frame
E is at rest in the E frame

and so on.

S sees other objects moving 'relative to itself' in the S frame
T sees objects moving relative to itself in the T frame
E sees objects moving relative to itself in the E frame

and so on.

If the velocity of such movements  constant the frame is an inertial frame

If the velocity is not constant then there is acceleration (or deceleration) and the frame is not an inertial frame.

 

Now taking just two objects, T and S, 

I said "The relative velocity between them" and you said "not moving at the same speed relative to each other"

The point of relativity is that whatever T sees in the T frame, S sees in the S frame, but in reverse.

So if T sees S approaching at velocity v (in this case 0.8c), then S sees T appoaching at -v (or -0.8c)
(Note that we are saying T sees S approaching, it does not see itself as travelling it sees itself as at rest)

Which is what I said, but not what you said. It is important to get this straight, before moving on to proper distance and proper length.

v is the velocity you enter into the formulae for relativity and is called the relative velocity between T and S.

(Note that observers in other frames will observe a different value for v)

 

If you look back at my maths you will see two things.

1) The calculations come to the same figure whether you use v or -v since it is squared.

If v has a value other than zero T and S are moving relative to each other but calculations in either frame will give the same result ie S measures the same distance to T as T measures to S.

2) If v = 0 then T and S are not moving relative to each other, and the measured distance between T and S is a maximum.

This distance between them is not changing with time as the movement occurs.

This maximum distance is called the proper distance.
I will state that the proper distance has the same numerical value measured by all observers, not just T and S.

Since T and S are not moving then they are in the same frame.
That is we have reduced or transformed the motion of T to the S frame or S to the T frame.

The proper distance is the distance measured between any two points in the same frame, ie at rest with rspect to each other.

 

 

 

 

[Dagl1]

I was not very careful in that last response;

Two observers that are (in their frame) both at rest, are in the same frame. If from the perspective of one of them, the other is moving, then they are not in the same frames. This is at least how I understood it, but I see now that my response is not saying that. What I meant with "If the T and S are not moving at the same speed relative to each other, they by definiton can't be in the same frame right?" is that if they aren't at rest (both moving along line A-B at the same speed so that the distance between them does not change), they must be not in the same frame. But I see now that what I said was pretty vague/incorrect, my bad.

Okay just to reiterate what you said (to make sure that I am following you). Each observer has its own frame, and one observes movement relative to their frame.
I thought I understood inertial frames, but even if you are accelerating, you can just take your own frame as inertial right, and thus say that whatever you are comparing your frame too is decelerating (or vice versa)?

If I am in my own frame, and I observe someone moving accelerating towards me, am I not still in my own inertial frame though? 

On 9/3/2021 at 2:24 PM, studiot said:

So if T sees S approaching at velocity v (in this case 0.8c), then S sees T appoaching at -v (or -0.8c)
(Note that we are saying T sees S approaching, it does not see itself as travelling it sees itself as at rest)

Which is what I said, but not what you said. It is important to get this straight, before moving on to proper distance and proper length.

This I cannot understand, how can one approach at a minus velocity, they both see each other approaching at 0.8c? Maybe this is the big hiccup but... if I run towards you at 5 km per hour, then you are approaching me at 5km per hour, from your perspective you, I would be approaching you at 5km per hour. Ignoring that in this example I am obviously the one running, but let's say we are in space and neither of us is doing anything, we are just in motion. Why would one of us think the other person is going backwards (this is what negative velocity implies, or at least how I interpret it, but that is probably wrong then). 
You call it the relative velocity, but that relative velocity must remain a positive number right, otherwise one would think someone is receding while the other thinks they are approaching?

On 9/3/2021 at 2:24 PM, studiot said:

1) The calculations come to the same figure whether you use v or -v since it is squared.

If v has a value other than zero T and S are moving relative to each other but calculations in either frame will give the same result ie S measures the same distance to T as T measures to S.

I suppose as you said, I should first understand the previous points, but this makes sense a little. If I move towards a star at 80% the speed of light, and I measure the distance to be 6LY, then if I moved very slowly towards it it would have been 10LY? And measured from the star, I too am 6 LY away when moving at at 0.8c, right?

On 9/3/2021 at 2:24 PM, studiot said:

2) If v = 0 then T and S are not moving relative to each other, and the measured distance between T and S is a maximum.

This distance between them is not changing with time as the movement occurs.

This maximum distance is called the proper distance.
I will state that the proper distance has the same numerical value measured by all observers, not just T and S.

Since T and S are not moving then they are in the same frame.
That is we have reduced or transformed the motion of T to the S frame or S to the T frame.

The proper distance is the distance measured between any two points in the same frame, ie at rest with rspect to each other.

Okay! That is pretty understandable. How does gravity (acceleration) then come into play with distances (although maybe that is a question I can ask in another thread afterwards, after making sure the last points were understood).

Thanks everyone for their answers and continued responses!

[studiot]

24 minutes ago, Dagl1 said:

On 9/3/2021 at 1:24 PM, studiot said:

So if T sees S approaching at velocity v (in this case 0.8c), then S sees T appoaching at -v (or -0.8c)
(Note that we are saying T sees S approaching, it does not see itself as travelling it sees itself as at rest)

Which is what I said, but not what you said. It is important to get this straight, before moving on to proper distance and proper length.

This I cannot understand, how can one approach at a minus velocity, they both see each other approaching at 0.8c? Maybe this is the big hiccup but... if I run towards you at 5 km per hour, then you are approaching me at 5km per hour, from your perspective you, I would be approaching you at 5km per hour. Ignoring that in this example I am obviously the one running, but let's say we are in space and neither of us is doing anything, we are just in motion. Why would one of us think the other person is going backwards (this is what negative velocity implies, or at least how I interpret it, but that is probably wrong then). 
You call it the relative velocity, but that relative velocity must remain a positive number right, otherwise one would think someone is receding while the other thinks they are approaching?

Since you are listed as still online, let's try and deal withthis one quickly.

 

Let us say that T is to the west of S and travelling towards S.  That is T is travelling east.

Then travel to the east is positive and travel to the west is negative.

So the velocity of T is positive.

 

However T may also himself as standing still and S to be travelling towards himself as he stands still.

Then S must be travelling west in this view.

So the velocity of S, as seen by T is then negative.

 

Now for yourself work out the same situation with regard to the point of view of S.

 

[swansont]

49 minutes ago, Dagl1 said:

Two observers that are (in their frame) both at rest, are in the same frame.

They must be at rest with respect to each other to be in the same frame

[Dagl1]

3 hours ago, swansont said:

They must be at rest with respect to each other to be in the same frame

Is that not what this means? If I am not moving, and I see you not moving, then we are both in the same frame? How could it otherwise be? If A sees B not moving, and B sees A not moving, assuming both take themselves to be at rest, is there a possibility for them to not be at rest? This is what 'two observes that are in their frame both at rest' means or can it be taken differently? Am I still missing something or is this just more semantics? 

4 hours ago, studiot said:

Since you are listed as still online, let's try and deal withthis one quickly.

Sorry Studiot! I come and go, didn't notice your message earlier;

I can definitely go with your explanation of East and West and using negative there, maybe it was a bit pedantic of me, but approaching has no directionality (as far as I understand the term). 
Anyway, for S who is in the East, S will see T is moving East, or S is the one moving Westwards to a stationary T?

[zapatos]

6 minutes ago, Dagl1 said:

Is that not what this means? If I am not moving, and I see you not moving, then we are both in the same frame?

True. But you said "two observers that are both at rest". They may not be at rest with respect to a third observer. 

No one is simply "at rest". They are only "at rest with respect to something else".

[swansont]

21 minutes ago, Dagl1 said:

Is that not what this means? If I am not moving, and I see you not moving, then we are both in the same frame? How could it otherwise be? If A sees B not moving, and B sees A not moving, assuming both take themselves to be at rest, is there a possibility for them to not be at rest? This is what 'two observes that are in their frame both at rest' means or can it be taken differently? Am I still missing something or is this just more semantics? 

 All inertial observers are at rest in their own frame, so saying two observers are at rest does not fully specify that they are at rest with respect to each other. We can be moving with respect to each other and each claim to be at rest. So as zapatos notes, rest (and motion) always has to be specified with respect to something.

[Dagl1]

As long as we understand each other now;p I feel my sentence meant that, but I see it can mean multiple things, sorry for being vague. The semantics discussion are not as important for this thread.

Edit: changed sentence

Edited September 4, 2021 by Dagl1

[studiot]

1 hour ago, Dagl1 said:

I can definitely go with your explanation of East and West and using negative there, maybe it was a bit pedantic of me, but approaching has no directionality (as far as I understand the term). 
Anyway, for S who is in the East, S will see T is moving East, or S is the one moving Westwards to a stationary T?

Of course it has directionality.

 

First we establish a coordinate sytem based at C.
This is the C frame and C is not moving in this frame.

A, and B are moving and both are approaching C at the same speed.

But their velocities are different, because they are going in different directions. The exact opposite in this example.

The negative is how we show which side of C the approach is made.

 

Never be afraid to draw yourself a simple sketch.

 

Have we got this straight now?

I would like to move on to the next item in my calculations and I can then use this sketch to show what I am doing with the formulae so you can do less translation and concetrate on the physics/maths.

[Dagl1]

6 minutes ago, studiot said:

The negative is how we show which side of C the approach is made.

Never be afraid to draw yourself a simple sketch.

 

Have we got this straight now?

I would like to move on to the next item in my calculations and I can then use this sketch to show what I am doing with the formulae so you can do less translation and concetrate on the physics/maths.

The problem wasn't the sketch, I see now that the difference between speed/velocity was something I wasn't really taking into account. Yes we got it straight;p 

[studiot]

37 minutes ago, Dagl1 said:

The problem wasn't the sketch, I see now that the difference between speed/velocity was something I wasn't really taking into account. Yes we got it straight;p 

 

OK so in the sketch, A and B are moving in respect of the C frame.

So if we want to find (calculate) quantities concerning A and B with repect to the C frame we have to correct for this motion.

This is where my formula come in.

[math]L = {L_0}\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} [/math]

 

Now L₀ is a constant, as is c and v is the relative velocity as already discussed.

L is the distance that is measured in the C frame from C to A (or B).

This can be seen to depend upon v.

Further the formula allows us to calculate the contstant Lo by putting in v = 0.

It can also be seen from the formula that L < L_o for any non zero value of of v.

If v = 0 then L = L_o and the distance between C and A does not change, justifying my earlier claim that it is a constant.

It is the distance between C and A that would be measured if they were in the same frame, which can only happen if v=0 (swansont)

Since L_o is a constant and is evaluated by all observers as the same constant, it is a very useful value to calculate as I did in my first post.

Because I can then use this value in the formula to calculate the distance C observes to A at any given velocity v.

This I also did in my first post.

 

Because it is so useful it is given the name proper distance.

The proper distance is the distance between two objects corrected for the motion between them and is the same for all observers.

 

I prefer to look at proper distance first then proper length, which is more confusing, because books talk about rulers and you then have to be careful to speciy which frame the ruler is laid out in.

We can do that next time

 

[MigL]

It may help the explanations if Dagl1 was to express his opinions/misconceptions of what a frame ( inertial or otherwise ) is.
I get the impression he thinks it is something other than the accepted definition.

[Dagl1]

Hmmm, I realise my words are pretty fuzzy when I describe it, but here it goes: A frame is a particular set of I suppose coordinates that are based upon a specific observer. I would say that a grid centered around a specific observer is a frame of reference. Inertial would then imply that it isn't accelerating (but that part I still don't really get, as you can always set yourself to be non-moving and non-accelerating by just thinking of your surrounding as decelerating). 

That is a pretty ambigious definition; I never really thought of the definition as described in words; If I am on Earth, then since the Earth is moving with me (ignoring its rotation and stuff, because I suppose that might give some additional problems) the Earth and I are in the same frame of reference. A rocket flying away from me, or a person on a train (very fast moving) is in a different frame of reference.

Is this correct, or am I missing it;/

Regarding the proper distance, so far I can understand (I think); with no movement relative to two points, we find the proper distance, which is also the largest (with all movement reducing the distance?). If I moving with the speed of light, that then also means there is no distance right? Constant * sqrt(1-(1/1)) = Constant *sqrt(0) = 0 right?
And any speed in between 0 and c will be between the proper distance and 0, relatively? 

I see now that this also answers my question regarding how we know this distance is the actual distance, as this function doesn't really allow for anything bigger than Constant*1, and therefore that is the proper distance (hope I am correct here).

Still very interested in what you can then tell me about the proper length! 

also thank you for the continued responses! 

[pzkpfw]

6 minutes ago, Dagl1 said:

... Inertial would then imply that it isn't accelerating (but that part I still don't really get, as you can always set yourself to be non-moving and non-accelerating by just thinking of your surrounding as decelerating).  ...

 

 

If you are in an accelerating rocket ship, you will feel yourself pressed into your seat. It's hard to see how the rest of the Universe decelerating would cause you and your seat to press together.

 

Acceleration is an absolute, in that if you are holding a coffee cup, you might spill some coffee.

Sitting at my desk my coffee does not spill, whether I consider myself at rest and the spaceship flying past Earth (at constant speed) to be moving; or if the alien in the ship considers itself as at rest and Earth is flying past.

An Earthquake, which moves me around (accelerating) will make me spill my coffee.