# B Special Relativity: What Time is it?

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1. Mar 4, 2017

### Dimani4

This question based on the site located here: http://galileo.phys.virginia.edu/classes/252/srelwhat.html

The question is: If the object (spaceship) moving close to the velocity of light and I'm as a static object. The time goes slower in the object moving with constant velocity (inertial frame) of light then he/she in the spaceship will age slower than me remained at the Earth. This is the situation when I look at the moving object. But if we take situation vice a versa. He/she in the moving object (spaceship) looking at my watch (me in Earth) then for her/him my watch is slowed down then I'm aging slower than her/him. "That is to say, each of them will see the other to have slower clocks, and be aging more slowly." But as I know the only the person in the "really' moving spaceship will age slower than the one stayed stationary (not moving) as explained in twin paradox.
Please explain me this contradiction. Thank you.

2. Mar 4, 2017

### Ibix

Both views are correct and consistent (assuming you mean "travelling close to the speed of light" - it's impossible to reach it). Look up "relativity of simultaneity" and/or "Lorentz transforms".

3. Mar 4, 2017

### Dimani4

Thank you for so quick response. But as I understood it's not so easy answer to that question. It seems paradox but the "real" answer to that question is the one who "really" travels will experience the less aging to his twin who remains in Earth while his twin travels with the velocity close to velocity of light of course.

4. Mar 4, 2017

### Ibix

The twin paradox is not what you were talking about. That involves a ship that stops and turns round. The resolution to that is not the same as the explanation for how two inertial observers both see the other's clock run slow. If you want to understand the twin paradox, I recommend the Insight by Orodruin on the geometric view of the twin paradox. See the Insights section of this site.

5. Mar 4, 2017

ok. thanks.

6. Mar 4, 2017

### Ibix

No problem. One of the things you find when learning relativity is that apparently quite similar phenomena can have quite different explanations.

If turns out that the elapsed time on your watch is a measure of the "distance" you travelled through spacetime. The twin paradox boils down to the fact that the twins took different routes through spacetime, and those routes had different "lengths".

The reason for the behaviour of clocks in inertial frames is essentially the same as two cars moving on non-parallel roads. They both say the other is falling behind because they both say that the other car's velocity in the forward direction is $v\cos\theta$, and they disagree on what "forward" means. Remember that "distance" through spacetime is elapsed time, and both observers see the other's clock run slow.

7. Mar 4, 2017

### Mister T

To understand what each twin actually "sees" I also recommend you go to YouTube and search for "Hewitt Twin Trip".

8. Mar 4, 2017

### robphy

The spacetime diagram is really the best way to understand what is going on.

Here are two [related] explanations of the symmetry of time-dilation
with spacetime diagrams (with likely more detail that you may want).

EXPLANATION #1:

based on my Insight https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/
and my paper "Relativity on Rotated Graph Paper" [published: http://aapt.scitation.org/doi/10.1119/1.4943251 , draft: https://arxiv.org/abs/1111.7254 ]
(Read that Insight to learn HOW I constructed these diagrams in this section.)

Note: Here, my t axis is vertical... as in my relativity books.

These are spacetime-diagrams of the light-signals in the light-clocks of Alice and Bob (where Bob has velocity (3/5)c in this frame.)
The reflection events M and N are simultaneous according to Alice [at rest in this frame], and this occurs at time (1/2)tick according to Alice.
Alice's first tick is at T.

The reflection events Y and Z are simultaneous according to Bob [in motion in this frame], and this occurs at time (1/2)tick according to Bob.
Bob's first tick is at F.

At T, draw a line parallel to MN [...this turns out to be tangent to the unit hyperbola]
and notice how that line intersects Bob's worldline OF before event F.
So, Alice says that Bob's clock takes longer to tick that her own.

At F, draw a line parallel to YZ [...this turns out to be tangent to the same unit hyperbola]
and notice how that line intersects Alice's worldline OT before event T.
So, Bob says that Alice's clock takes longer to tick that his own.

To make things easier to see, we can string up a bunch of clock-ticks for each observer... now we can count.
(from my paper)

The time-dilation factor according to Alice is OP/OQ=5/4.. she says when her 5 ticks elapse, only 4 have elasped for Bob.
The time-dilation factor according to Bob is OP'/OQ'=5/4.. he says when his 5 ticks elapse, only 4 have elasped for Alice.

EXPLANATION #2:

(It's the same argument... but using the hyperbola ["what plays the role of the unit circle in Special Relativity"] more explicitly
... and you can play the simulation (and extract numbers is you wish).)

https://www.desmos.com/calculator/ti58l2sair
Note: Here, my t axis is horizontal... like the usual position-vs-time graph.
The green dashed-line is the light-cone for the event at the origin of the diagram (call that event O).
The forward hyperbola represents one tick after that origin event for all inertial observers that meet at event O.
One observer has a solid black worldline and is at rest in this frame, and
the other has a solid red worldline who is moving with velocity (3/5)c in this frame.
Their intersections with the hyperbola identify tangent-lines (drawn as black and red dashed lines) to the hyperbola which represent lines of simultaneity for that observer.

Note the symmetry:
• The black dashed-line of simultaneity intersects the red worldline before the red worldline intersects the hyperbola...
in fact, at 4/5 of what could be called the red radius vector.
According to the black observer, it takes 5/4 longer for the red observer's clock to tick compared to his own.
• The red dashed-line of simultaneity intersects the black worldline before the black worldline intersects the hyperbola...
in fact, at 4/5 of what could be called the black radius vector.
According to the red observer, it takes 5/4 longer for the black observer's clock to tick compared to his own.
This factor of 5/4 is the time-dilation factor $\gamma$ for a relative-velocity of $v=(3/5)c$.
(In the Desmos simulation, you can find the time-dilation factor by seeing where that observer's line of simultaneity intersects the horizontal t-axis.
Click on that intersection... For the red dashed-line, it gives (0.8,0). The time-dilation factor for the red observer 1/0.8=5/4 in this frame. For black, the time-dilation factor is 1/1.0=1 since the black observer happens to be at rest in this frame.

(optional...)
You can use the sliders to modify their velocities. For general situations, you have to compare the ratios along the other worldline.
For example, [sorry for the not-so-pleasant numbers]

The relative-velocity for $v_1=4/10$ and $v_2=-3/10$ is $v_r=\frac{v_1-v_2}{1-v_1v_2}=5/8$,
which leads to a $\gamma=\frac{1}{\sqrt{1-v_r^2}}=8/\sqrt{39}\approx 1.281$.
That is equal to the ratio 1.0911/0.8517 [using the t-components] or $\frac{\sqrt{1.0911^2+.4364^2}}{\sqrt{8517^2+.3407^2}}$ [using the pythagorean theorem... since the ratio of segments on the same line is an affine invariant.]... and similarly for the other worldline.

Time-dilation goes away if you move the slider to E=0 (Galilean physics).
You get @Ibix 's argument if you move the slider to E= -1 (Euclidean geometry).

Last edited: Mar 4, 2017
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