Relativity~Speed of Light~Twins Paradox

Factual Questions
1

A few questions:

  1. Two ships move in opposing directions, Each traveling 1/2 the speed of light. Wouldn’t they each be traveling at the speed of light in reference to each other?

  2. Since everything is assumed to be truly relative in the Twins Paradox, why can’t it be assumed that the Earth moving away from the Twin in the rocket then back towards it?

  3. I’ve seen mention of the Doppler effect of light in the Twins Paradox, how can this be if Light is traveling at a constant?

2
  1. Velocity is relative, acceleration is not. This is because acceleration is caused by a force, and nearly all [1] parties in the Universe can agree which object has the forces acting on it. “Turning around” involves decelerating (negative acceleration) to a standstill, then acceleration to the new velocity. These require forces, and they are acting on the rocket, not the Earth.
    [1] Not including trouble mongers.

-k-
Karen Lingel, Physicist

3

Velocities in relativity are not additive, as in Newtonian mechanics. If two ships set out from the earth, moving in opposite directions, each with a speed of 0.5 c with respect to the earth, the relative speed between the 2 ships is 5/9 c.

v(rel)=v-V/{[1-(v*V)]/c^2]}

v and V are of opposite sign because they are in opposite directions.

Work is the curse of the drinking classes. (Oscar Wilde)

4

Miscalculation. Sorry. The relative velocity is 4/5 c.

5

No. Velocities don’t add that way. At the kind of speeds we encounter in everyday life, the errors introduced by ordinary adding of velocities are insignificant; but at c/2 it really matters.

Consider ship 1 moving at velocity c/2 relative to some point and ship 2 moving at velocity -c/2 relative to the same point. What is the velocity of ship 2 relative to ship 1? We have to “add” a velocity to all velocities so the velocity of ship 1 comes out to be zero. The formula for “adding” velocities is:

(V1 + V2)/(1 + V1V2/(cc))

(which is almost V1 + V2 when V1 and V2 are much smaller than c).

If V1 is the velocity of ship 1, c/2, what V2 do we add to come out with zero? Looking at the formula, we can immediatly see that only -c/2 will do this. This should not be surprising. Now, we calculate the velocity of ship 2 relative to ship 1 by substituting -c/2 (the original velocity of ship 2) for V1 and -c/2 (the velocity we are “adding”) for V2 and we get … 4c/5. Near to c, but not quite c … as always.

Of course, you can. But you have to do it right.

karen Two’s answer is the most common one, but it can raise philosophical questions (Mach’s principle). It’s not necessary to invoke acceleration and forces, though; you can do the same thing with three people. One person stays on Earth, one leaves Earth, and the third rockets in from out of nowhere passing both people (and interchanging messages at the instant that they pass) at different times. But this formulation contains the key to the question.

There are three natural reference frames in the problem; one for the person who stays on Earth, one for the person who leaves earth, and one for the person who rockets in out of nowhere. (In the original formulation, the third reference frame is the second person returning to Earth). You can calculate what happens in any reference frame, or you can switch frames midstream … but you must use the relativistic transformations when switching frames. You implicitly switched frames (from the second frame to the third frame) when you considered the “paradox” from the point of view of the person who left Earth, but you did not apply the relativistic transformation between frames; you just added velocities. Not kosher.

See Twin Paradox.

The speed of light in vacuum is indeed constant, but the rate at which clocks run is not constant, and the length of the rulers we use to measure distance is not constant. Is it surprising, then, that frequency shift (the Doppler effect) is observed when you are moving relative to a source of light?

Of course, there’s a special version of the Doppler shift equation, that must be used when the velocities involved are significant fractions of c; see Relativistic Doppler Shift.

jrf

6

What if the ships are travelling at 99% of light speed? Now they’re separating almost twice light speed!

Notice I said “separating.” You measure each ship moving near light speed relative to you. What object do you measure moving faster than light relative to you? None! You measure a separation increasing at a certain rate. You infer from this that ship A should measure the ship B flying away from it faster than light (and vice versa). This is an incorrect inference. The formulas given above tell you what speed ship A will actually measure.

To summarize:

A. You are not in the frame of reference of either ship, so your attempted measurement of their speed relative to each other can be greater than light speed. Special relativity only applies within a frame of reference.

B. You are really measuring a separation distance, and a separation is not a physical object, so it can increase faster than light.

More info at The Relativity FAQ.

…this is another Moebius sig…b!s sn!qaoW jay+oue s! s!y+…
(adaptation of a WallyM7Sig™ a la quadell)

For business reasons, I must preserve the outward signs of sanity. - Mark Twain

7
  1. The doppler effect is a change in frequency (with light as with sound). It does not imply a change in velocity.

Virtually yours,

DrMatrix
If I’ve told you once, I’ve told you 0.99999999… times.

8

To tag on:

As has been said, if two ships (A and B) leave you heading in opposite directions at 90% the speed of light, you perceive that their distance from each other is increasing faster than the speed of light. However, from your perspective, each one is still traveling less than the speed of light from you (this is your ‘frame of reference’ which the others talk about).

Now, you must think, “Well, ship A sees me receding at .9c (90% the speed of light), and since ship B is leaving me at .9c in the other direction, ship A must be seeing ship B recede at 1.8c (180% the speed of light, which is ftl).”

Nope. When ship a measures the speed you are receding, it will see you going away at .9c. However, when ship A measures how fast B is moving away from itself, ship A they will not get 1.8c. The will measure a fraction of c (I’m not going to do the math, let’s say .96c, or 96% the speed of light). From its ‘frame of reference’ neither you nor ship B is going ftl.

Why the paradox? The closer to the speed of light you get, the more your ‘measuring device’ will ‘distort’ to prevent you from measuring an excess of the speed of light. Any ‘ruler’ ship A will use to measure the speed of recession of shop B will distort to give a speed less than c.

It’s a universal conspiracy.

Peace.

9

Thank you. You should also see twin paradox using general relativity, as it treats the case where the “stay at home” twin is the one that feels the acceleration.

rocks