# Basic doubt of special relativity

1. Jul 2, 2015

### Raman Choudhary

suppose person B was standing with person A at some point O and now person B starts to move with velocity c/2 w.r.t A on a straight track(starting from O) and both the observors(A and B) start to note time on identical clocks attached to their respective frames .Now after 1 second (according to A's clock) person B is made to stop suddenly(by applying huge amount of force that means assuming that its velocity suddenly drops to zero from c/2). therefore distance b/w A and B according to A must be = c/2(m/s) * 1s=c/2 m. but according to B's clock time just before he was made to stop as shown by his clock was "gamma*1(sec)".and therefore distance b/w(A and B) them according to B must be =c/2(m/s)*gamma(s)=c/2*gamma.

2. Jul 2, 2015

### Staff: Mentor

No, it isn't; it's $1 / \gamma$ seconds.

No, it isn't; it's $(c / 2) * (1 / \gamma)$ meters. This is just an illustration of length contraction; in B's frame A is moving, so distances in A's frame are length contracted. (To see this more clearly, imagine there is a pole, at rest relative to A, at the point where B stops, so B and the pole are co-located after B stops. Relative to A, the distance from A to the pole will be $c/2$ meters; but relative to B while he is moving, that distance will be length contracted, so it will be $(c / 2) * (1 / \gamma)$ meters.)

3. Jul 2, 2015

### Staff: Mentor

It will, however, be c/2 meters again after B has decelerated and is back at rest relative to A. Also, once B has stopped both clocks will be ticking at the same rate... however, B will have lost some time so his clock will be somewhat behind A's and B will have aged very slightly less than A.

One way of understanding what's going on is to imagine that A's clock emits a flash of light once every millisecond, and think about exactly when those flashes reach B. IUt may also help to work through the twin paradox FAQ, as this problem is really a variant of that one.

(PeterDonis knows all of this, of course - these remarks are for OP who may not).

4. Jul 3, 2015

### Raman Choudhary

thanks
but when B is asked to tell the distance b/w him and A just after he has stopped and B wishes to apply the formula d=s*t and he knows that his clock showed 1/gamma seconds then according to him the distance that A has travelled would be c/2*1/gamma meters even when he has come to halt but when A is asked the same question he would answer c/2 but isn't it a problem that both now being in a same reference frame answering differently.

5. Jul 3, 2015

### Raman Choudhary

in simple words on what basis can he find the distance b/w him and A as soon as he stops which is same as the distance found by A b/w both of them.given that B has knowledge of the velocity being c/2 and his clock.

6. Jul 3, 2015

### harrylin

Everyone is always in all reference frames! SR uses the inertial reference systems of classical physics. Hopefully you know them: they are commonly called inertial reference frames (although technically there is a difference). If you move an ideal clock relative to such a reference system then that clock will not stay in tune with the "time" of that system. That is only a problem if you don't take that in account, or if you wrongly account for it.

Call a reference system in which the two clocks were originally in rest, system S. According to S (taken as "rest system"), clock B will tick slower while it is moving, and the traveled distance is OP. On arrival at P, it will therefore indicate less time than as predicted with Newton's mechanics.

Call a reference system in which clock B is temporarily in rest, System S'. According to S' (taken as "rest system"), clock B ticks at its normal rate when the moving distance OP of S passes along, and this distance is factor γ less than as measured in S (OP is length contracted). On arrival of P, clock B will therefore indicate less time than as predicted with Newton's mechanics.

He can choose any inertial frame he likes as "rest system" (the one he finds most convenient), and then he just accounts for effects of motion accordingly. I indicated here above how that works for S and for S'.
It is similar to calculations of kinetic energy in classical physics.

Notes:
- b/w = "between"? For me it means "black and white"...
- If you want to try a calculation example, preferably use 0.6c or 0.8c for easy γ numbers.
- From this you can also work out relativity of simultaneity, if you place more clocks. S and S' cannot agree about synchronization of distant clocks.
(it may sound as if I'm complicating things, but if your brain works like mine, then this may be the start of the answer to your next question )

Last edited: Jul 3, 2015
7. Jul 3, 2015

### 1977ub

Once B is at a remote location from A, and moving, they no longer agree on when certain things happened or how much time elapsed between those events. This leads to apparent paradoxes in length, time, and velocity measurement.

8. Jul 3, 2015

### vanhees71

Let's do this example in terms of covariant objects. For each of the two particles use their proper time and four velocity $\tau_{A,B}$ and $u_{A,B}^{\mu}$. To make things simple, assume that the four velocities are constant. Then assume that at $\tau_A=\tau_B=0$ the particles are at the same place, which we choose as the origin of a Minkowski-space (inertial) reference frame. Than the world lines of the particles are straight lines given by
$$x_A^{\mu} =\tau_A u_{A}^{\mu}, \quad x_B^{\mu}=\tau_B u_{B}^{\mu}.$$
Now let's choose specific reference frames.

(a) Rest frame of particle A
------------------------------------

Then after some time $t$, corresponding to the proper times $\tau_A$ and $\tau_B$ of the particles an observer at rest in this frame reads off the spatial coordinates of these particles in his reference frame and calls the modulus of the difference its length. The important point is that this observer does this for both particles at the same time wrt. his reference frame. Since it's the rest frame of particle A, this condition reads in covariant form
$$t=u_A \cdot x_A=u_A \cdot x_B \; Rightarrow \; t=\tau_A=\gamma \tau_B,$$
because in this frame we have
$$u_A=(1,0,0,0), \quad u_B=\gamma(1,v,0,0),$$
where $v$ is the constant velocity of particle B in the rest frame of A.

The distance between the particles $L_A(t)$ is then given by
$$L_A^2=-(x_A-x_B)^2|_{\tau_A=t, \quad \tau_B=t/\gamma} \; \Rightarrow \; L_A=|\gamma v \tau_B|=|v t|.$$

(b) Rest frame of particle B
------------------------------------

In the very same way the distance for the analogous situation in the rest frame of particle B you get the same result, as it should be
$$L_B=|v t|,$$
where $t$ is now the time measured in the rest frame of particle B.

(c) General frame of reference
-----------------------------------------

The most interesting case is of course for an arbitrary reference frame of an observer C. Then the equal-time condition reads
$$t=u_C \cdot x_A=u_C \cdot x_B$$ or $$t=u_C \cdot u_A \tau_A=u_C \cdot u_B \tau_B$$
and one has
$$L_C^2=-(x_A-x_B)^2|_{\tau_A=t/(u_c \cdot u_A), \quad \tau_B=t/(u_C \cdot u_B)}$$
which gives after some calculation
$$L_C=\left (2 \frac{u_A \cdot u_B}{(u_C \cdot u_A) (u_C \cdot u_B)}- \frac{1}{(u_C \cdot u_A)^2}-\frac{1}{(u_C \cdot u_A)^2} \right)^{1/2} t.$$
This is pretty complicated, and the reason is, how to define an instantaneous distance in a given reference frame. By convention, you measure the location of both particles at a given instant of coordinate time and take the Euclidean distance between these vectors. So in general you get a different result in any frame of reference.

The only exception are the two reference frames where one of the particles is at rest and the other is moving. From the fundamental symmetry assumptions for inertial observers' reference frames it already follows that the velocity of B in the rest frame of A is opposite equal of the velocity of A in the restframe of B, and thus the distances between the particles as measured after a given coordinate time $t$ in each frame is the same.

That's also the reason, why one defines the proper relative speed between the particles as the speed of one of the particles in the rest frame of the other, because that's a unique definition for any pair of particles.

Last edited: Jul 3, 2015
9. Jul 3, 2015

### Staff: Mentor

If B wants to apply that formula after he has stopped, he has to use quantities that are valid in the frame in which he is at rest after he has stopped. Those quantities give a distance of $c/2$ meters. The fact that only $1 / \gamma$ seconds have elapsed on his own clock is irrelevant; his clock does not correctly show elapsed time in the frame in which he is at rest after he has stopped, because of the period during which he was not at rest in that frame.

10. Jul 3, 2015

### Raman Choudhary

b/w=between
Suppose at the instant he was made to come to rest he also paused his watch and now asked to tell the distance b/w him and A ,now he would answer c/2*1/γ as he knows that 1/γ was the time A took to travel and c/2 was A's velocity. therefore he would calculate the distance to be c/2*1/γ meteres. but the same question to A would give c/2 meters as an answer...so what's going on???

11. Jul 3, 2015

### Staff: Mentor

B made his calculation based on measurements made in his frame as he was moving with respect to A. He certainly agrees that in A's rest frame the distance traveled is c/2. Distances depend on reference frame.

12. Jul 3, 2015

### Staff: Mentor

If he measures the distance between A and his stopping point before he stops moving, he will find it to be $c/(2\gamma)$. After he stops moving, he will measure that distance to be $c/2$. The time lost his clock is lost either way.

13. Jul 3, 2015

### harrylin

I and others already explained in different ways how everything is consistent. However, there is another way to interpret your question: what is going on here is a very interesting option for far distance travel. It may be possible for a human to travel all the way to a star that is 100 light years away, and still manage to arrive there before dying.

14. Jul 3, 2015

### PWiz

A digression to the topic, but it is generally agreed upon that a rocket with constant proper acceleration can take you farther (actually to the edge of the observable universe within the passenger's lifetime for a proper acceleration which equals $g$). That hyperbolic sine gives a good disparity between coordinate time on Earth and proper time