# Simple two body relativity

1. Dec 27, 2013

### leehufford

Hello,

Recently I've been researching how to add velocities in a relativistic way, very interesting stuff. But every website I read uses a three body system to explain it. Object A relative to object B and object B relative to object C. I've spent a few hours looking and this is the only definition I could find.

I simply want to confirm that the following is incorrect:

Object A moves at 0.75c. Object B moves in the opposite direction as A at 0.75c. Therefore object B is moving faster than light relative to object A and vice versa.

If you could confirm that this is wrong or point me in the direction of a website that explains the phenomenon in a two body system rather than a three body system that would be great. Thanks in advance,

Lee

P.S - I am aware that "simple" and "relativity" from my subject line are oxymorons. :tongue2:

2. Dec 27, 2013

### TumblingDice

I had to read a couple of times to understand what you meant by a "three body system". The problem is where you started with, "Object A moves at 0.75c." Relative to what? There is no such thing as absolute speed, so when you say something is moving, you must also indicate what reference frame that motion is being measured 'relative' to.

Can you re-phrase with that in mind?

3. Dec 27, 2013

### leehufford

Actually, I think your response answered my question. "Object C" is an observer or a point in space. Is that right?

Thanks,

Lee

4. Dec 27, 2013

### leehufford

Is it possible for object A to move at 0.75c relative to object B and also object B moving at 0.75c relative to object A or does the question have no physical meaning? Thanks again,

Lee

5. Dec 27, 2013

### TumblingDice

If you mean object A is moving at .75c relative to object C, it sounds like you're in the right groove.

Only light is measured moving at c. Everything else will be measured moving slower. There's a formula for every possibility, but all setups must include 'relative to' for each reference frame.

6. Dec 27, 2013

### TumblingDice

There's a very valid physical meaning, perhaps not the one you were thinking of.

The 'first person' view from any reference frame is that it is at rest. So, Object A could view Object B moving at .75c relative to Object A (it's own reference frame). Likewise, Object B would view Object A as moving at .75c relative to itself.

7. Dec 27, 2013

### leehufford

Thank you so much for your responses. One last hypothetical question if I may:

Object A is moving 0.75c in one direction relative to an observer. Object B is moving at the same speed relative to the same observer but in the opposite direction of A. (The observer is in the middle). Object B shoots a beam of light at object A. Does it ever get there??

Thanks again!

-Lee

8. Dec 27, 2013

### Staff: Mentor

Yes. The flash of light travels at speed c towards A from the point where it was emitted - and according to the central observer A is moving at .75c relative to that point, so the flash overtakes A. The fact that B is heading off in the opposite direction at .75c is irrelevant to the interaction between the light and A.

B does see A moving away from him at a speed greater than .75c, but by the relativistic velocity addition rule, that speed is still less than c - so B also agrees that the flash of light, moving at speed c relative to him, eventually overtakes A.

9. Dec 28, 2013

### ghwellsjr

Einstein claimed that his theory of Special Relativity was "simple" so I don't see why you would call them oxymorons. Maybe I can show you how simple it is to address your issue.

We start with a spacetime diagram to show Object A (in blue) moving at 0.75c to the right and Object B (in red) moving at -0.75c to the left. I hope you can identify their speeds by looking at the Coordinate Time of 10 nsecs and seeing that they each have moved 7.5 feet from their starting location. The dots represent one-nanosecond increments of time on their own clocks which is dilated (stretched out) compared to the Coordinate Time by the factor gamma. I am using the speed of light to be 1 foot per nanonsecond:

Now in order for each Object to measure the speed of the other object relative to themselves, they have to measure its location at different times. They can start with the time when they are colocated and then measure the distance at some other time and dividing the distance by the time they can calculate the speed of the other Object:

The way they measure the distance is by sending a radar signal at some point in time and measuring how long it takes for the reflection to return. Then they take the midpoint in time between the emission of the radar and its reception and establish that as the time of the measurement. They also take one-half of the time interval and calculate how far light travels during that amount of time and establish that as the distance.

So let's say that the blue Object sends out a radar signal at his time of 1 nsec. He gets the return signal at his time of 49 nsecs. (You'll have to count the dots to show this.) Since the time interval is 48 nsecs, he establishes the distance that light travels in 24 nsec which is 24 feet and he establishes the time of the measurement to be at the average of the two times or 25 nsec. This process is identical to Einstein's convention.

Now all he has to do is divide 24 feet by 25 nsec and get 0.96 feet per nsec as the measured speed of Object B relative to himself.

We can transform to the rest frame of the blue observer and we see that his measurements match what his rest frame shows:

Now don't you think that was simple?

#### Attached Files:

File size:
2.4 KB
Views:
118
File size:
19 KB
Views:
122
• ###### TwoBodies3.PNG
File size:
4.7 KB
Views:
116
Last edited: Dec 28, 2013
10. Dec 28, 2013

### ghwellsjr

Sure, you can have object A move at 0.75c relative to object B as shown here:

Notice that both objects measure the speed of the other one the same way as I showed in my previous post, by sending out a radar signal at their own time of 1 nanosecond and then noting that they get the return signal at their own time of 7 nanoseconds. They then establish the time of the measurement to be the average of those two times or 4 nanoseconds and the distance to be how far light travels in one-half of the difference between those two times (3 nanoseconds) or 3 feet. And so they each measure the speed of the other one relative to themselves as 3 feet divided by 4 nanoseconds or 0.75 feet per nanosecond (0.75c).

I want to emphasize that they are applying Einstein's convention and second postulate, that light propagates at c in any Inertial Rest Frame (IRF). In other words, they are not, nor can they ever, measure the speed at which the light propagates away from or toward them. It's an applied definition.

Note that the measurement is also shown more directly by object B, since we did this in object B's rest frame but we can transform to object A's rest frame (0.6c from the first) to see how the measurements are more directly shown (in other words, applying Einstein's convention):

But we can also transform to any other IRF, such as the one in which both objects are traveling at the same speed (0.451416c) in opposite directions, like you wanted in your first post:

Or we can transform to any other arbitrary IRF such as one at 0.9c from the orginal:

The two important factors (Einstein's two postulates) to note in all these IRF's is that what each object actually sees and measures according to their own clocks is the same and that light always propagates at c in each IRF. The Time Dilation of each object, determined by its speed in the IRF is a result of these two factors.

#### Attached Files:

File size:
4.3 KB
Views:
112
File size:
4.3 KB
Views:
111
File size:
4.3 KB
Views:
118
• ###### TwoBodies7.PNG
File size:
6.8 KB
Views:
115
11. Dec 28, 2013

### ghwellsjr

Ok, we'll add an observer as you described but keep in mind, he cannot observe what we observe in the diagram unless he also makes a lot of measurements and after the fact creates his own spacetime diagram which I will leave to you to figure out how he would do this. Here is a spacetime diagram (he is in black) showing what you asked for:

Note that the red object B sends a beam of light at its time of 1 nsec towards object A. The stationary observer will see this at his time of 2.645751 nsec and it will continue past him and object A will see it at its time of 7 nsec. These numbers are determined from the Relativistic Doppler factor. If you square the factor for the observer, you will get the factor for object A.

#### Attached Files:

• ###### TwoBodies8.PNG
File size:
5.9 KB
Views:
108
12. Dec 29, 2013

### phyti

Here is another case for you to graph.
U is the earth (rest) frame. A and B are launched from earth simultaneously at .6c and .8c respectively. At 5 min U time, A measures the relative speed of B. What is the result?

13. Dec 29, 2013

### Staff: Mentor

Are A and B launched in the same direction or opposite directions?

(I'm also assuming that you mean .6c and .8c as measured in U, the frame in which earth is at rest. It's a good habit to always be clear about what a speed is relative to).

14. Dec 29, 2013

### ghwellsjr

Do you mean for A and B to be launched in opposite directions like in the earlier examples or in the same direction?

And when you say "At 5 min U time", do you mean the time when A initiates his measurement or the time that he establishes his measurement or do you really care?

It will take me till later today to get to this, but I hope I have described the process well enough that you or anyone else could do it in the meantime.

And there are other ways for the measurement to be made, are you interested in any of those?

15. Dec 30, 2013

### ghwellsjr

Now that I have had a chance to work on this, I see why you wanted me to do this particular example. It's choice!! Is this one related to the 5-12-13 right-triangle relationship?

OK, here's the diagram you requested:

Since A (in blue) sends his radar signal at his time of 4 minutes and receives the reflected return signal at his time of 9 minutes, he considers his measurement to have been made at his time of 6.5 minutes (the average of the two times) and the distance to be how far light travels in one-half the difference between the two times which is 2.5 light-minutes. The resulting speed is 2.5/6.5 which is the same as 5/13 or 0.3846c.

To confirm this, I will transform to the rest frame of A:

I have extended the worldline of B so that we can see that it goes through the coordinates of x=5 and t=13.

We can also use the relativistic addition formula:

(0.8-0.6)/(1-0.8*0.6) = 0.2/(1-0.48) = 0.2/0.52 = 0.1/0.26 = 0.05/0.13 = 5/13

Do you know any more magic examples?

#### Attached Files:

File size:
4.8 KB
Views:
108
• ###### TwoBodies10.PNG
File size:
3.9 KB
Views:
100
Last edited: Dec 30, 2013
16. Dec 30, 2013

### ghwellsjr

Just for the fun of it, I decided to try it with A going in the opposite direction as B and it too produces an integer relationship.

But to make the diagram more manageable, I had A send his radar signal at his time of 1 (we can multiply everything by 4 to see the effect at the earth's frame at 5 minutes):

A sends his signal at 1 minute and gets the return at 36 minutes so he establishes B's speed as 35/37 = 0.954954c (I left off the two divisions by 2 since they cancel out). And here is the diagram transformed to A's rest frame but with B's worldline extended:

It's easy to see that B passes through x=35 at t=37 and we can do the exact calculation as:

(0.6+0.8)/(1+0.6*0.8) = 1.4/1.48 = 0.7/0.74 = 3.5/3.7 = 35/37

#### Attached Files:

File size:
24 KB
Views:
93
• ###### TwoBodies12.PNG
File size:
7.5 KB
Views:
100
17. Dec 30, 2013

### phyti

You got it!
No it's not about Pythagorean triples.
What's interesting about this case is using the SR convention, A overestimates the speed of B as measured by U. A's measurement should not change based on the presence of U. Also without communicating with earth (U), A cannot make the RAF calculation since he doesn't know his speed relative to light, and he only has two time recordings in addition to the origin, to work with. Only U knows the speed of A and B, and makes the RAF calculation as a correction to the A measurement.
This is true for any pair of observers moving relative to U. There is a reason for this difference in speed measurement by the 'designated rest frame' and the moving observers. I'll offer it as a problem to anyone interested.

Last edited: Dec 30, 2013
18. Dec 31, 2013

### phyti

I don't try for integer solutions, just used some more familiar values that are easy to calculate.

An analysis of the general case with two observers moving at different speeds a and b results in an expression for the separating/closing velocity w as
w = (b-a)/(1-ab).

Using example #15: w= (.80-.60)/(1-.48) = (.39).

Using op example of .75 in opposite directions from the center, and remembering to change the sign for neg x, w= (.75+.75)/(1+.56) = (.96).

Using example #16: w= (.80+.60)/(1+.48) = (.95).

Also note you don't need to use dilated times since the calculation is free from td effects.

Ignore the last part of my post 17, it isn't important.