Question

To whom does the elapsed time for a process seem to be longer, an observer moving relative to the process or an observer moving with the process?

Answer 1

I understand that the elapsed time will be longer for the observer moving relative to it, but why?

Answer 2

'Cause of light! Making the speed of light constant in separate reference frames has tricky effects. If we look at two reference frames, the stationary (the observer moving relative to the process) and the moving frame (the observer moving with the process), we can describe the relationship between their positions with a linear transformation.



\[ \textrm{To find those constants, we can consider 3 special cases of motion.}
\\ \
\\ \

\textrm{1) An object fixed at the origin of frame S' would move with speed v relative to x:}
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\\
\ \ \ \ \ \ \ \ \ \ \ \ x' = 0 \ \ \textrm{for all time, and} \ \ \ x = vt
\\ \
\\
\ \ \ \ \ \ \textrm{Inserting that into x' = Ax + Bt gives}
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\\ \ \ \ \ \ \ \ \ \ \ \ \ x' = 0 = Avt + Bt
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\\ \ \ \ \ \ \ \ \ \ \ \ \ B = -Av
\\ \
\\ \
\]

Answer 3

\[
\\ \textrm{2) Object fixed at origin in S would have velocity -v according to S'}
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\\ \ \ \ \ \ \ \ \ \ \ \ \ x' = -vt' \ \ \ and \ \ \ x=0
\\ \
\\ \
\\ \ \ \ \ \ \ \ \ \textrm{Inserting them into} \ \ \ x ' = Ax + Bt \ \ \ ;\ \ \ t' = Cx + Dt
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\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x' = \cancel{Ax}^0 + Bt = -vt' \ \ \ \ ; \ \ \ \ t' = \cancel{Cx}^0 + Dt
\\ \
\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Bt = -vt' \ \ \ \ ; \ \ \ \ t' = Dt
\\ \
\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Bt = -vDt
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\\ \ \ \ \ \ \ \ \ \ \ \ \ \ D = \frac{-B}{v} = A
\]

Answer 4

But how would you explain it in terms of words (without the equations)? Like is it something to do with time dilation, and if so, how does that work?

Answer 5

\[
\\ \textrm{3) Einstein's 2nd postulate - speed of light's the same in all frames}
\\ \
\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x' = ct' \ \ \ \ \ \ \ ; \ \ \ \ \ \ \ x = ct
\\ \
\\ \ \ \ \ \ \ \ \ \ \ \ \ x' = Act + Bt = ct' \ \ \ \ ; \ \ \ \ t' = Cct + Dt
\\ \
\\ \\ \ \ \ \ \ \ \ \ \ \ \ Act + Bt = c(Cct + Dt)
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\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Ac + B = Cc^2 + Dc
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\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cancel{Ac} -Av = Cc^2 + \cancel{Ac}
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\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C = -A\frac{v}{c^2}
\\ \
\\ \ \ \ \ \ \ \ \ \textrm{Now all of our transformation constants are in terms of A}
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\\ \ \ \ \ \ \ \ \ \ \ \ A = A \ \ ; \ \ B = -vA \ \ ; \ \ C = -A\frac{v}{c^2} \ \ ; \ \ D =A
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\\ \ \ \ \ \ \ \ \ \ \ \ x' = A(x - vt) \ \ \ \ ; \ \ \ \ t' = A(\frac{v}{c^2}x+t)
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\\ \
\\ \textrm{A bit of algebra and we can solve for x and t}
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\\ x = \frac{1}{A(1-\frac{v^2}{c^2})}(x'+vt') \ \ \ \ ; \ \ \ \ t = \frac{1}{A(1-\frac{v^2}{c^2})} \left(\frac{v}{c^2}x' + t' \right)
\]
almost there1

Answer 6

This derivation is specifically used in time dilation problems, yeah. Lorentz transformations. In terms of words, since speed of light is constant in both frames, the closer you get to the speed of light (the faster you go), the more you change how your frame "feels" time in relation to a 'stationary' observer. Because time is woven into the way things move, when you go really quickly things 'get shorter' in your perspective. In the stationary observer's perspective, you "get longer."

Time dilation and length contraction. Hand in hand!

Answer 7

Well, my statement is wrong. You always "feel" time the same - it's just everything else around you changes.

GGAAHH UPDATES

Answer 8

How come when you move quicker, the lengths of things get shorter in your perspective though? I thought that it would mean velocity is greater, and thus result in a greater length.

(And ughh, I'm experiencing them too)

Answer 9

It's all about the maths unfortunately, and the really precise way the language is set up! Plus I'm bad at explaining; but I'll try! ^_^

When you're moving really fast, you don't actually perceive yourself to be moving - everything else is perceived to be moving around you. So when you see something coming towards you very quickly , weird things happen, because of something called relative simultaneity. An instant in one frame is "split" into two instances in another, kinda.

Gawd, I'm terrible at explaining this... sorry...

Answer 10

That's just talking in circles and giving you no useful information :P

Answer 11

Okay, I think I get the gist of it. I'll just read over what you wrote and let it sink in, haha. Thanks for your time and also for all your help!

Answer 12

There's one more quick "paradox" that deals well with contraction if you want it.

Answer 13

Is it the twin paradox? I understand that one, just not so much the length contraction. :P

Answer 14

No, it's called the pole-barn "paradox." Spoiler, it's not really a paradox. It deals with measuring different events in different frames, which is the crux of what you're looking for. And I've just copied this from a wikibook about special relativity. It seems decent enough, and this part at least is correct ^_^ (link at bottom of page)

"Firstly, how can a succession of events be observed as simultaneous events by another observer?"

"Secondly, if a rod is simultaneously between two points in one frame how can it be observed as being successively between those points in another frame? For instance, if a pole enters a building at high speed how can one observer find it is fully within the building and another find that the two ends of the rod are opposed to the two ends of the building at successive times? What happens if the rod hits the end of the building? The second question is known as the "pole-barn paradox" or "ladder paradox."

"The pole-barn paradox states the following: suppose a superhero running at 0.75c and carrying a horizontal pole 15 m long towards a barn 10m long, with front and rear doors. When the runner and the pole are inside the barn, a ground observer closes and then opens both doors (by remote control) so that the runner and pole are momentarily captured inside the barn and then proceed to exit the barn from the backdoor.

One may be surprised to see a 15-m pole fit inside a 10-m barn. But the pole is in motion with respect to the ground observer, who measures the pole to be contracted to a length of 9.9 m (check using equations).

The “paradox” arises when we consider the runner’s point of view. The runner sees the barn contracted to 6.6 m. Because the pole is in the rest frame of the runner, the runner measures it to have its proper length of 15 m. Now, how can our superhero make it safely through the barn?

The resolution of the “paradox” lies in the relativity of simultaneity. The closing of the two doors is measured to be simultaneous by the ground observer. However, since the doors are at different positions, the runner says that they do not close simultaneously.The rear door closes and then opens first, allowing the leading edge of the pole to exit.The front door of the barn does not close until the trailing edge of the pole passes by.If the rear door is kept closed and made out of some impenetrable material then in the frame of the runner a shock wave will travel at the speed of light from the rear door that compresses the rod so that it fits within the barn. This shock wave will appear like an instantaneous explosion in the frame of the barn and a progressive wave in the frame of the runner."

http://en.wikibooks.org/wiki/Special_relativity

Answer 15

Here's the outcome of the derivation too - what's up there isn't worth much without it

\[ x' = \gamma_v(x-vt) \ \ \ \ \ \ \ \ ; \ \ \ \ \ \ \ \ t' = \gamma_v \left(-\frac{v}{c^2}x+t \right)
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\\ x = \gamma_v(x' + vt') \ \ \ \ \ \ \ \ ; \ \ \ \ \ \ \ \ t = \gamma_v \left(\frac{v}{c^2}x' + t' \right)
\\ \
\\ \
\\ \ \gamma_v \equiv \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} ≥ 1 \]

v is the velocity of the frame. As v approaches c, gamma starts getting really, really large, so has a greater effect on the terms.

Answer 16

happy physics ^^

Answer 17

I also lied above. Things never appear longer, there are only contractions.