# Relative velocity of third object

1. Apr 27, 2008

### newTonn

With reference to the sketch attached,
if A and B is having a relative velocity of 0.9c,and
B and C having a relative velocity 0.9c
What will be the relative velocity between A and C?
(all in x direction)

#### Attached Files:

• ###### rel.pdf
File size:
7.1 KB
Views:
61
2. Apr 27, 2008

### pam

Just use the relativistic formula for addition of velocities. If you don't know it, look at a book.

3. Apr 27, 2008

### newTonn

Ok i looked at the book and got an answer of 0.9944c(approx).
How can i justify this with the relative displacement after one second?

4. Apr 27, 2008

### Staff: Mentor

What do you mean? According to A, C moves 0.9944c*1s closer in one second.

5. Apr 27, 2008

### newTonn

Sorry i din't mean it.Initally let all of them have same x co-ordinates =0; According to A,what will be the positions of B & C in after 1 second in A's frame?

Last edited: Apr 27, 2008
6. Apr 27, 2008

### HallsofIvy

Staff Emeritus
Since you now know the velocity of each of B and C relative to A just multiply by 1 second to find how far they have moved, relative to A in one second.

7. Apr 27, 2008

### Staff: Mentor

According to A, B has a speed of -.9c. So B will be at x = -.9c*1sec (~ -2.7 10^8 m) after one second. Similarly, C will be at position x = -.9944c*1sec (~ -3 10^8 m).

8. Apr 28, 2008

### newTonn

Is there any length contraction to be considered or not?

9. Apr 28, 2008

### Staff: Mentor

Each observer will view the "metersticks" used by the others to be contracted. But you don't need to explicitly use that fact to figure out the coordinates of each at a given time, since you are not changing reference frames.

10. May 4, 2008

### newTonn

Ok what about if B and C are long rods with length 'c' in their frames?

11. May 4, 2008

### Staff: Mentor

length contraction

If B and C carry long rods (held parallel to the direction of motion) that have length $L_0$ in their own frames, then the length of those rods as measured by A will be:

$$L = L_0/\gamma$$

Where:
$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$

And v is the relative velocity of B or C with respect to A (discussed earlier).

12. May 4, 2008

### newTonn

This shortening will be from one end or it would be symmetrical from centre of Rod?

13. May 5, 2008

### Staff: Mentor

I don't understand what you mean. All segments of the rod will be uniformly "shortened".

14. May 5, 2008

### Antenna Guy

I think he's asking if the rod will also shrink in dimensions not parallel to the direction of travel.

Regards,

Bill

15. May 8, 2008

### newTonn

Ok.i will come to the point.Consider A,B and C are at rest to each other.B & C are each carrying a rod of length 'L'.This rods are made of 'n' number of molecules in x direction.Now let us say the spacing from centre to centre of molecules are 's'.Let us mark it on C's Rod.
Now B,with the rod goes for a trip in x direction.He came back to cross his initial position with a relative velocity of 0.9c(in x-direction).The rod now is contracted as per A.Does the spacing between the molecules of the rod in B's hand change,with respect to the spacing marked on C's rod(in A's frame).?
If so,can we say the space itself is contracted ?

16. May 8, 2008

### Staff: Mentor

I don't quite understand what you're asking, especially in that last sentence. (Not sure what you mean by "spacing marked on C's rod in A's frame".)

In any case, I'll rephrase what I already said. You can imagine each rod as having markings every meter (or other interval of your choice). According to frame A, the markings on B and C are closer together by the factor given in post #11. (Similar statements can be made about measurements according to any of the three frames.)

17. May 8, 2008

### newTonn

Actualy i mean the centre to centre distance of molecules(or atoms of the rod)
I will try to explain my question with another example.A,B,C and D are four persons.Initialy they are all at rest with respect to each other.
B,Cand D are in a line in x direction.
C is in the middle of B and D
C is holding a long bar whose end are just touching B and D.
Now this B,C and D started a Journey in x direction with same velocity(they are at rest with respect to each other).
They all came back in the x direction to cross A with 0.9c velocity with respect to the stationary observer A.
Do A see the ends of the rod still touching B and D or not?

18. May 8, 2008

### Staff: Mentor

It all depends on how B, C, and D accelerate to reach their final speeds. If they do it in such a way as to maintain a constant proper length between themselves, then their distances will be uniformly Lorentz contracted as seen by A--and everyone will agree that the ends of the rod still touch B and D. (Note that this is equivalent to having A accelerate to speed 0.9c, while the others remain at rest in their original inertial frame.)

Whether the ends of the rod still touch B and D (or not) will be a fact that all observers will agree upon.

19. May 11, 2008

### newTonn

In fact the space itself is contracted isn't it?.
If only the rod was contracted,the ends would not have touch B and D.
The contraction of space leads to an ambiguity.Because the formula for velocity(derived from time dilation) doesn't take into consideration of this reduced distance.

Last edited: May 11, 2008
20. May 11, 2008

### Staff: Mentor

Yes, all distances are contracted.
What ambiguity in what formula? Velocity = Distance/Time works just fine as long as you stick to a single frame.