Twin Paradox: Is Time Speeding Up?

[Xori]

A common concept used in the explanation of the twin paradox is that upon acceleration, one twin sees the other twin's time speed up. Is this correct?

Assuming it is, how would they observe a light clock next to that other twin? The light beam can't bounce faster, right? But (assuming a vertical orientation of a light clock) the clock can't get shorter either, since contraction is only allowed in the direction of the velocity, correct?

 

[genneth]

See http://www.math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

 

[JesseM]

When you talk about "seeing" the other twin speed up, are you talking about what the traveling twin would see using her eyes, so that the Doppler effect must be taken into account (see fig. 2 here), or are you talking about the speed of the other twin's clock in a non-inertial coordinate system where the traveling twin is at rest throughout the trip? (see fig. 3 from the same link as above) If the first, keep in mind that what you see is different than what happens in your frame, an object's speed in your frame will be different than how fast it appears to move visually (because if it's moving relative to you, then light signals from different points in its journey have a different distance to travel to reach you). If the second, the speed of light is only c in inertial coordinate systems, it can be different than c in non-inertial coordinate systems.

 

[Xori]

When you talk about "seeing" the other twin speed up, are you talking about what the traveling twin would see using her eyes, so that the Doppler effect must be taken into account (see fig. 2 here)

Yes, this.

I understand that there's a doppler shift, but does that change the fact that light always has to appear to move at speed c?

In that example, light pulses are used to transmit clock speed. What if the twins could see each other's light beams? One twin would see the other twin's light beam move faster than c, would they not?

 

[JesseM]

Yes, this.

I understand that there's a doppler shift, but does that change the fact that light always has to appear to move at speed c?

No, it has to travel at c in your rest frame, which is different. For instance, suppose there is a rod whose near end is 1 light year away, and whose far end is 2 light years away, and there are clocks at either end of the rod which are synchronized in my frame. If we have two machines which are designed to set off fireworks displays when they detect a laser beam passing them, then if someone fires a laser which passes the first machine when the clock there reads Jan. 1, 2010, then naturally the second machine will detect the laser when its clock reads Jan. 1, 2011, since the rod is 1 light-year long. But since the light from the first machine's fireworks display also travels at the speed of light alongside the laser, naturally that light is passing the second machine at the same time that it's detecting the laser and setting off its own fireworks display, so the light from both displays reaches me at the same time on Jan. 1, 2012. If I pay attention to the time on each machine's clock when I look through the telescope, I'll conclude that both fireworks displays happened a year apart in my frame (as measured locally by synchronized clocks at rest in my frame), so the laser must have been moving at c, but I see both at exactly the same time.

In that example, light pulses are used to transmit clock speed. What if the twins could see each other's light beams? One twin would see the other twin's light beam move faster than c, would they not?

No, "speed" in relativity is always understood in terms of the time events happen in your frame, not when you see them (would you say that since I saw the two fireworks displays simultaneously in the example above, I should conclude the laser was traveling infinitely fast?)

 

[Xori]

No, "speed" in relativity is always understood in terms of the time events happen in your frame, not when you see them (would you say that since I saw the two fireworks displays simultaneously in the example above, I should conclude the laser was traveling infinitely fast?)

I'm having some trouble relating your rod/fireworks analogy.

You say that the light beams wouldn't appear faster, but what would they look like?

I'm not sure if you're understaing exactly where I'm confused, or you might be, idk...

Let's say we have Twin A with Clock A next him. The clock is a light clock with beams bouncing up and down at speed c. Likewise for Twin B, Clock B.

Now when Twin A "sees" Twin B's clock speed up, does the light beam not appear to be moving faster than c?

You said no, so what does it appear like?

 

[JesseM]

You say that the light beams wouldn't appear faster

No I didn't, I said that you're not even supposed to define "speed" in terms of when you see things, only in terms of when they happen in your frame. For example, if a moving object passes marker A first and marker B a little later, and the distance between the markers is d, the speed would be d/t, where t is not the difference in time between when you saw each event, but the difference in time between when you calculate the object actually passed the markers in your frame (you can find the times either by having synchronized clocks at A and B, or by subtracting the delay between when the event happened and when you saw it--for example, if I see an event 5 light-years away in 2010, I can calculate it actually happened in 2005 in my frame).

Let's say we have Twin A with Clock A next him. The clock is a light clock with beams bouncing up and down at speed c. Likewise for Twin B, Clock B.

Now when Twin A "sees" Twin B's clock speed up, does the light beam not appear to be moving faster than c?

How exactly are you calculating the speed? If it's based on taking the distance between the mirrors of the light clock and dividing by the difference in time between when A sees the light hitting each mirror, this is simply not a meaningful way to calculate "speed" in relativity (as in my fireworks example, if the two mirrors were arranged parallel to the line between twin A and twin B and the light were traveling towards A and the mirrors were at rest with respect to A, A would actually see the light hitting both mirrors simultaneously). If it's instead based on dividing by the difference in time between when A calculates the light actually hit each mirror in his frame, then the beam will be found to be moving at c.

 

[Xori]

No I didn't, I said that you're not even supposed to define "speed" in terms of when you see things, only in terms of when they happen in your frame. For example, if a moving object passes marker A first and marker B a little later, and the distance between the markers is d, the speed would be d/t, where t is not the difference in time between when you saw each event, but the difference in time between when you calculate the object actually passed the markers in your frame (you can find the times either by having synchronized clocks at A and B, or by subtracting the delay between when the event happened and when you saw it--for example, if I see an event 5 light-years away in 2010, I can calculate it actually happened in 2005 in my frame). How exactly are you calculating the speed? If it's based on taking the distance between the mirrors of the light clock and dividing by the difference in time between when A sees the light hitting each mirror, this is simply not a meaningful way to calculate "speed" in relativity (as in my fireworks example, if the two mirrors were arranged parallel to the line between twin A and twin B and the light were traveling towards A and the mirrors were at rest with respect to A, A would actually see the light hitting both mirrors simultaneously). If it's instead based on dividing by the difference in time between when A calculates the light actually hit each mirror in his frame, then the beam will be found to be moving at c.

Ah ok I see how the previous analogy relates now.

It makes sense now. Thank you!